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https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/20%3A_Electrochemistry/20.6%3A_Corrosion%3A_Unwanted_Voltaic_Cells |
is a galvanic process by which metals deteriorate through oxidation—usually but not always to their oxides. For example, when exposed to air, iron rusts, silver tarnishes, and copper and brass acquire a bluish-green surface called a patina. Of the various metals subject to corrosion, iron is by far the most important commercially. An estimated $100 billion per year is spent in the United States alone to replace iron-containing objects destroyed by corrosion. Consequently, the development of methods for protecting metal surfaces from corrosion constitutes a very active area of industrial research. In this section, we describe some of the chemical and electrochemical processes responsible for corrosion. We also examine the chemical basis for some common methods for preventing corrosion and treating corroded metals. Corrosion is a process. Under ambient conditions, the oxidation of most metals is thermodynamically spontaneous, with the notable exception of gold and platinum. Hence it is actually somewhat surprising that any metals are useful at all in Earth’s moist, oxygen-rich atmosphere. Some metals, however, are resistant to corrosion for kinetic reasons. For example, aluminum in soft-drink cans and airplanes is protected by a thin coating of metal oxide that forms on the surface of the metal and acts as an impenetrable barrier that prevents further destruction. Aluminum cans also have a thin plastic layer to prevent reaction of the oxide with acid in the soft drink. Chromium, magnesium, and nickel also form protective oxide films. Stainless steels are remarkably resistant to corrosion because they usually contain a significant proportion of chromium, nickel, or both. In contrast to these metals, when iron corrodes, it forms a red-brown hydrated metal oxide (\(\ce{Fe2O3 \cdot xH2O}\)), commonly known as rust, that does not provide a tight protective film (Figure \(\Page {1}\)). Instead, the rust continually flakes off to expose a fresh metal surface vulnerable to reaction with oxygen and water. Because both oxygen and water are required for rust to form, an iron nail immersed in deoxygenated water will not rust—even over a period of several weeks. Similarly, a nail immersed in an organic solvent such as kerosene or mineral oil will not rust because of the absence of water even if the solvent is saturated with oxygen. In the corrosion process, iron metal acts as the anode in a galvanic cell and is oxidized to Fe ; oxygen is reduced to water at the cathode. The relevant reactions are as follows: The \(\ce{Fe^{2+}}\) ions produced in the initial reaction are then oxidized by atmospheric oxygen to produce the insoluble hydrated oxide containing \(\ce{Fe^{3+}}\), as represented in the following equation: \[\ce{4Fe^{2+}(aq) + O2(g) + (2 + 4x)H2O \rightarrow 2Fe2O3 \cdot xH2O + 4H^{+}(aq)} \label{Eq4} \] The sign and magnitude of \(E^o_{cell}\) for the corrosion process (Equation \(\ref{Eq3}\)) indicate that there is a strong driving force for the oxidation of iron by O under standard conditions (1 M H ). Under neutral conditions, the driving force is somewhat less but still appreciable (E = 1.25 V at pH 7.0). Normally, the reaction of atmospheric CO with water to form H and HCO provides a low enough pH to enhance the reaction rate, as does acid rain. Automobile manufacturers spend a great deal of time and money developing paints that adhere tightly to the car’s metal surface to prevent oxygenated water, acid, and salt from coming into contact with the underlying metal. Unfortunately, even the best paint is subject to scratching or denting, and the electrochemical nature of the corrosion process means that two scratches relatively remote from each other can operate together as anode and cathode, leading to sudden mechanical failure (Figure \(\Page {2}\)). One of the most common techniques used to prevent the corrosion of iron is applying a protective coating of another metal that is more difficult to oxidize. Faucets and some external parts of automobiles, for example, are often coated with a thin layer of chromium using an electrolytic process. With the increased use of polymeric materials in cars, however, the use of chrome-plated steel has diminished in recent years. Similarly, the “tin cans” that hold soups and other foods are actually consist of steel container that is coated with a thin layer of tin. While neither chromium nor tin metals are intrinsically resistant to corrosion, they both form protective oxide coatings that hinder access of oxygen and water to the underlying steel (iron alloy). As with a protective paint, scratching a protective metal coating will allow corrosion to occur. In this case, however, the presence of the second metal can actually increase the rate of corrosion. The values of the standard electrode potentials for \(\ce{Sn^{2+}}\) (E° = −0.14 V) and Fe (E° = −0.45 V) in show that \(\ce{Fe}\) is more easily oxidized than \(\ce{Sn}\). As a result, the more corrosion-resistant metal (in this case, tin) accelerates the corrosion of iron by acting as the cathode and providing a large surface area for the reduction of oxygen (Figure \(\Page {3}\)). This process is seen in some older homes where copper and iron pipes have been directly connected to each other. The less easily oxidized copper acts as the cathode, causing iron to dissolve rapidly near the connection and occasionally resulting in a catastrophic plumbing failure. One way to avoid these problems is to use a more easily oxidized metal to protect iron from corrosion. In this approach, called cathodic protection, a more reactive metal such as \(\ce{Zn}\) (E° = −0.76 V for \(\ce{Zn^{2+} + 2e^{−} -> Zn}\)) becomes the anode, and iron becomes the cathode. This prevents oxidation of the iron and protects the iron object from corrosion. The reactions that occur under these conditions are as follows: \[ \underbrace{O_{2(g)} + 4e^− + 4H^+_{(aq)} \rightarrow 2H_2O_{(l)} }_{\text{reduction at cathode}}\label{Eq5} \] \[ \underbrace{Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^−}_{\text{oxidation at anode}} \label{Eq6} \] \[ \underbrace{ 2Zn_{(s)} + O_{2(g)} + 4H^+_{(aq)} \rightarrow 2Zn^{2+}_{(aq)} + 2H_2O_{(l)} }_{\text{overall}}\label{Eq7} \] The more reactive metal reacts with oxygen and will eventually dissolve, “sacrificing” itself to protect the iron object. Cathodic protection is the principle underlying galvanized steel, which is steel protected by a thin layer of zinc. Galvanized steel is used in objects ranging from nails to garbage cans. In a similar strategy, using magnesium, for example, are used to protect underground tanks or pipes (Figure \(\Page {4}\)). Replacing the sacrificial electrodes is more cost-effective than replacing the iron objects they are protecting. Suppose an old wooden sailboat, held together with iron screws, has a bronze propeller (recall that bronze is an alloy of copper containing about 7%–10% tin). identity of metals corrosion reaction, \(E^o°_{cell}\), and preventive measures Over time, the iron screws will dissolve, and the boat will fall apart. Suppose the water pipes leading into your house are made of lead, while the rest of the plumbing in your house is iron. To eliminate the possibility of lead poisoning, you call a plumber to replace the lead pipes. He quotes you a very low price if he can use up his existing supply of copper pipe to do the job. Not unless you plan to sell the house very soon because the \(\ce{Cu/Fe}\) pipe joints will lead to rapid corrosion. Any existing \(\ce{Pb/Fe}\) joints should be examined carefully for corrosion of the iron pipes due to the \(\ce{Pb–Fe}\) junction; the less active \(\ce{Pb}\) will have served as the cathode for the reduction of \(\ce{O2}\), promoting oxidation of the more active \(\ce{Fe}\) nearby. Corrosion is a galvanic process that can be prevented using cathodic protection. The deterioration of metals through oxidation is a galvanic process called corrosion. Protective coatings consist of a second metal that is more difficult to oxidize than the metal being protected. Alternatively, a more easily oxidized metal can be applied to a metal surface, thus providing cathodic protection of the surface. A thin layer of zinc protects galvanized steel. Sacrificial electrodes can also be attached to an object to protect it. | 8,357 | 3,291 |
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At the macroscopic level, a complete physical description of a sample of a gas requires four quantities: (expressed in kelvins), (expressed in liters), (expressed in moles), and (in atmospheres). As we explain in this section and , these variables are independent. If we know the values of any of these quantities, we can calculate the fourth and thereby obtain a full physical description of the gas. Temperature, volume, and amount have been discussed in previous chapters. We now discuss pressure and its units of measurement. Any object, whether it is your computer, a person, or a sample of gas, exerts a force on any surface with which it comes in contact. The air in a balloon, for example, exerts a force against the interior surface of the balloon, and a liquid injected into a mold exerts a force against the interior surface of the mold, just as a chair exerts a force against the floor because of its mass and the effects of gravity. If the air in a balloon is heated, the increased kinetic energy of the gas eventually causes the balloon to burst because of the increased pressure( ) of the gas, the force ( ) per unit area ( ) of surface:
\( P=\dfrac{F}{A} \tag{10.2.1}\) Pressure is dependent on the force exerted the size of the area to which the force is applied. We know from that applying the same force to a smaller area produces a higher pressure. When we use a hose to wash a car, for example, we can increase the pressure of the water by reducing the size of the opening of the hose with a thumb. The units of pressure are derived from the units used to measure force and area. In the English system, the units of force are pounds and the units of area are square inches, so we often see pressure expressed in pounds per square inch (lb/in , or psi), particularly among engineers. For scientific measurements, however, the SI units for force are preferred. The SI unit for pressure, derived from the SI units for force (newtons) and area (square meters), is the newton per square meter (N/m ), which is called the pascal (Pa) , after the French mathematician Blaise Pascal (1623–1662):
\( 1\;Pa=1\;N/m^{2} \tag{10.2.1}\) To convert from pounds per square inch to pascals, multiply psi by 6894.757 [1 Pa = 1 psi (6894.757)]. In addition to his talents in mathematics (he invented modern probability theory), Pascal did research in physics and was an author and a religious philosopher as well. His accomplishments include invention of the first syringe and the first digital calculator and development of the principle of hydraulic pressure transmission now used in brake systems and hydraulic lifts. Assuming a paperback book has a mass of 2.00 kg, a length of 27.0 cm, a width of 21.0 cm, and a thickness of 4.5 cm, what pressure does it exert on a surface if it is mass and dimensions of object pressure Calculate the force exerted by the book and then compute the area that is in contact with a surface. Substitute these two values into to find the pressure exerted on the surface in each orientation. The force exerted by the book does depend on its orientation. Recall from that the force exerted by an object is = , where is its mass and is its acceleration. In Earth’s gravitational field, the acceleration is due to gravity (9.8067 m/s at Earth’s surface). In SI units, the force exerted by the book is therefore The pressure exerted by the book in this position is thus Thus the exerted by the book varies by a factor of about six depending on its orientation, although the exerted by the book does not vary. Exercise What pressure does a 60.0 kg student exert on the floor Just as we exert pressure on a surface because of gravity, so does our atmosphere. We live at the bottom of an ocean of gases that becomes progressively less dense with increasing altitude. Approximately 99% of the mass of the atmosphere lies within 30 km of Earth’s surface, and half of it is within the first 5.5 km ( ). Every point on Earth’s surface experiences a net pressure called . The pressure exerted by the atmosphere is considerable: a 1.0 m column, measured from sea level to the top of the atmosphere, has a mass of about 10,000 kg, which gives a pressure of about 100 kPa:
\( Pressure = \dfrac{\left ( 1.0\times 10^{4} \; kg \right )\left ( 9,807 \cancel{m}/s^{2} \right )}{1.0 \; m^{\cancel{2}}} = 0.98 \times 10^{5} \; Pa =98 \; kPa \tag{10.2.3}\) In English units, this is about 14 lb/in. , but we are so accustomed to living under this pressure that we never notice it. Instead, what we notice are in the pressure, such as when our ears pop in fast elevators in skyscrapers or in airplanes during rapid changes in altitude. We make use of atmospheric pressure in many ways. We can use a drinking straw because sucking on it removes air and thereby reduces the pressure inside the straw. The atmospheric pressure pushing down on the liquid in the glass then forces the liquid up the straw. Atmospheric pressure can be measured using a barometer , a device invented in 1643 by one of Galileo’s students, Evangelista Torricelli (1608–1647). A barometer may be constructed from a long glass tube that is closed at one end. It is filled with mercury and placed upside down in a dish of mercury without allowing any air to enter the tube. Some of the mercury will run out of the tube, but a relatively tall column remains inside ( ). Why doesn’t all the mercury run out? Gravity is certainly exerting a downward force on the mercury in the tube, but it is opposed by the pressure of the atmosphere pushing down on the surface of the mercury in the dish, which has the net effect of pushing the mercury up into the tube. Because there is no air above the mercury inside the tube in a properly filled barometer (it contains a ), there is no pressure pushing down on the column. Thus the mercury runs out of the tube until the pressure exerted by the mercury column itself exactly balances the pressure of the atmosphere. Under normal weather conditions at sea level, the two forces are balanced when the top of the mercury column is approximately 760 mm above the level of the mercury in the dish, as shown in This value varies with meteorological conditions and altitude. In Denver, Colorado, for example, at an elevation of about 1 mile, or 1609 m (5280 ft), the height of the mercury column is 630 mm rather than 760 mm. Mercury barometers have been used to measure atmospheric pressure for so long that they have their own unit for pressure: the millimeter of mercury (mmHg) , often called the torr , after Torricelli. Standard atmospheric pressure is the atmospheric pressure required to support a column of mercury exactly 760 mm tall; this pressure is also referred to as 1 atmosphere (atm) . These units are also related to the pascal:
\( 1 \; atm = 760 \; mmHg = 760 \; torr = 1.01325 \times × 10^{5} \;Pa = 101.325 \; kPa \tag{10.2.4}\) Thus a pressure of 1 atm equals 760 mmHg exactly and is approximately equal to 100 kPa. While mercury barometers were the workhorses for pressure measurement into the last quarter of the 20th century, they have been replaced by electronic gauges. One motivation was safety. Mercury and mercury vapor are heavy metal poisons. The oldest and simplest replacement for mercury barometers are aneroid gauges. The simplest version of an aneroid gauge (Figure 6.2.3) has a thin metal diaphram which expands and contracts. The diaphragm is connected to a dial by a mechanical linkage. Aneroid gauges can either be absolute or differential. Differential aneroid gauges compare the pressure in the gauge housing to that being measured. A common use for aneroid gauges was airplane altimeters. Most modern pressure sensors are based on strain gauge which convert pressure into a strain on a semiconductor element. The strain is converted into an electrical signal by the piezoelectric effect as a change in resistance monitored by a resistance bridge. Alternatively pressure is related to the change in frequency of a quartz crystal oscillator. Modern balances are also based on this later principle. A very accurate type of pressure sensor uses a metal diaphragm as one part of a capacitor. As the pressure changes the diaphragm moves altering the capacitance. Handbooks are available from sensor manufacturers with details including and . Below atmospheric pressures another type of gauge is used based on measuring temperature change of a hot wire as a function of pressure. This is best suited to pressures below atmospheric. Older types called thermocouple or Pirani gauges measure only up to a few Torr. Modern variations called convection gauges can measure up to atmospheric pressure A summary of the various types of pressure gauge in use today can be found in a technical note at the Pressure gauges specify whether the measurement is absolute (relative to zero pressure) or gauge (relative to the standard atmosphere). One must be careful about which kind a gauge is when buying or using one.. One of the authors visited Rocky Mountain National Park several years ago. After departing from an airport at sea level in the eastern United States, he arrived in Denver (altitude 5280 ft), rented a car, and drove to the top of the highway outside Estes Park (elevation 14,000 ft). He noticed that even slight exertion was very difficult at this altitude, where the atmospheric pressure is only 454 mmHg. Convert this pressure to pressure in millimeters of mercury pressure in atmospheres and kilopascals Use the conversion factors in to convert from millimeters of mercury to atmospheres and kilopascals. From , we have 1 atm = 760 mmHg = 101.325 kPa. The pressure at 14,000 ft in atm is thus \( P=\left ( 454 \; \cancel{mmHg} \right )\left ( \dfrac{1 \; atm}{760 \; \cancel{mmHg}} \right )=0.597 \; atm\) The pressure in kPa is given by \( P=\left ( 0.597 \; \cancel{atm} \right )\left ( \dfrac{101.325 \; kPa}{1 \; \cancel{atm}} \right )=80.5 \; kPa\) Exercise Mt. Everest, at 29,028 ft above sea level, is the world’s tallest mountain. The normal atmospheric pressure at this altitude is about 0.308 atm. Convert this pressure to a. 234 mmHg; b. 31.2 kPa Barometers measure atmospheric pressure, but manometers measure the pressures of samples of gases contained in an apparatus. The key feature of a manometer is a U-shaped tube containing mercury (or occasionally another nonvolatile liquid). A closed-end manometer is shown schematically in part (a) in . When the bulb contains no gas (i.e., when its interior is a near vacuum), the heights of the two columns of mercury are the same because the space above the mercury on the left is a near vacuum (it contains only traces of mercury vapor). If a gas is released into the bulb on the right, it will exert a pressure on the mercury in the right column, and the two columns of mercury will no longer be the same height. The between the heights of the two columns is equal to the pressure of the gas. If the tube is open to the atmosphere instead of closed, as in the open-end manometer shown in part (b) in , then the two columns of mercury have the same height only if the gas in the bulb has a pressure equal to the atmospheric pressure. If the gas in the bulb has a pressure, the mercury in the open tube will be forced up by the gas pushing down on the mercury in the other arm of the U-shaped tube. The pressure of the gas in the bulb is therefore the sum of the atmospheric pressure (measured with a barometer) and the difference in the heights of the two columns. If the gas in the bulb has a pressure than that of the atmosphere, then the height of the mercury will be greater in the arm attached to the bulb. In this case, the pressure of the gas in the bulb is the atmospheric pressure minus the difference in the heights of the two columns. Suppose you want to construct a closed-end manometer to measure gas pressures in the range 0.000–0.200 atm. Because of the toxicity of mercury, you decide to use water rather than mercury. How tall a column of water do you need? (At 25°C, the density of water is 0.9970 g/cm ; the density of mercury is 13.53 g/cm .) pressure range and densities of water and mercury column height Calculate the height of a column of mercury corresponding to 0.200 atm in millimeters of mercury. This is the height needed for a mercury-filled column. From the given densities, use a proportion to compute the height needed for a water-filled column. In millimeters of mercury, a gas pressure of 0.200 atm is \( P=\left ( 0.200 \; \cancel{atm} \right )\left ( \dfrac{760 \; mmHg}{1 \; \cancel{atm}} \right )=152 \; mmHg\) Using a mercury manometer, you would need a mercury column at least 152 mm high. Because water is less dense than mercury, you need a column of water to achieve the same pressure as a given column of mercury. The height needed for a water-filled column corresponding to a pressure of 0.200 atm is proportional to the ratio of the density of mercury (d )(d ) \( \begin{matrix}
\left ( height_{H_{2}O} \right )\left ( d_{H_{2}O} \right )=\left ( height_{Hg} \right )\left ( d_{Hg} \right )
\\
\\height_{H_{2}O}=\left ( height_{Hg} \right )\dfrac{d_{Hg}}{d_{H_{2}O}}
\\
\\= \left ( 152 \; mmHg\right )\left ( \dfrac{13.53 \; \cancel{g/cm^{2}} }{0.9970 \; \cancel{g/cm^{2}}} \right )
\\
\\=2.06\times 10^{3} \; mm\; H_{2}O \end{matrix}\) This answer makes sense: it takes a taller column of a less dense liquid to achieve the same pressure. Exercise Suppose you want to design a barometer to measure atmospheric pressure in an environment that is always hotter than 30°C. To avoid using mercury, you decide to use gallium, which melts at 29.76°C; the density of liquid gallium at 25°C is 6.114 g/cm . How tall a column of gallium do you need if = 1.00 atm? 1.68 m The answer to Example 4 also tells us the maximum depth of a farmer’s well if a simple suction pump will be used to get the water out. If a column of water 2.06 m high corresponds to 0.200 atm, then 1.00 atm corresponds to a column height of \( \begin{matrix}
\dfrac{h}{2.06 \; m} = \dfrac{1.00 \; \cancel{atm}}{0.200 \; \cancel{atm}}
\\
\\h= 10.3 \; m
\end{matrix}\) A suction pump is just a more sophisticated version of a straw: it creates a vacuum above a liquid and relies on atmospheric pressure to force the liquid up a tube. If 1 atm pressure corresponds to a 10.3 m (33.8 ft) column of water, then it is physically impossible for atmospheric pressure to raise the water in a well higher than this. Until electric pumps were invented to push water mechanically from greater depths, this factor greatly limited where people could live because obtaining water from wells deeper than about 33 ft was difficult. Four quantities must be known for a complete physical description of a sample of a gas: , , , and . is force per unit area of surface; the SI unit for pressure is the , defined as 1 newton per square meter (N/m ). The pressure exerted by an object is proportional to the force it exerts and inversely proportional to the area on which the force is exerted. The pressure exerted by Earth’s atmosphere, called , is about 101 kPa or 14.7 lb/in. at sea level. Atmospheric pressure can be measured with a , a closed, inverted tube filled with mercury. The height of the mercury column is proportional to atmospheric pressure, which is often reported in units of , also called . , the pressure required to support a column of mercury 760 mm tall, is yet another unit of pressure: 1 . A is an apparatus used to measure the pressure of a sample of a gas. : What four quantities must be known to completely describe a sample of a gas? What units are commonly used for each quantity? If the applied force is constant, how does the pressure exerted by an object change as the area on which the force is exerted decreases? In the real world, how does this relationship apply to the ease of driving a small nail versus a large nail? As the force on a fixed area increases, does the pressure increase or decrease? With this in mind, would you expect a heavy person to need smaller or larger snowshoes than a lighter person? Explain. What do we mean by ? Is the atmospheric pressure at the summit of Mt. Rainier greater than or less than the pressure in Miami, Florida? Why? Which has the highest atmospheric pressure—a cave in the Himalayas, a mine in South Africa, or a beach house in Florida? Which has the lowest? Mars has an average atmospheric pressure of 0.007 atm. Would it be easier or harder to drink liquid from a straw on Mars than on Earth? Explain your answer. Is the pressure exerted by a 1.0 kg mass on a 2.0 m area greater than or less than the pressure exerted by a 1.0 kg mass on a 1.0 m area? What is the difference, if any, between the pressure of the atmosphere exerted on a 1.0 m piston and a 2.0 m piston? If you used water in a barometer instead of mercury, what would be the major difference in the instrument? Because pressure is defined as the force per unit area ( = ), increasing the force on a given area increases the pressure. A heavy person requires larger snowshoes than a lighter person. Spreading the force exerted on the heavier person by gravity (that is, their weight) over a larger area decreases the pressure exerted per unit of area, such as a square inch, and makes them less likely to sink into the snow. Calculate the pressure in atmospheres and kilopascals exerted by a fish tank that is 2.0 ft long, 1.0 ft wide, and 2.5 ft high and contains 25.0 gal of water in a room that is at 20°C; the tank itself weighs 15 lb (d = 1.00 g/cm at 20°C). If the tank were 1 ft long, 1 ft wide, and 5 ft high, would it exert the same pressure? Explain your answer. Calculate the pressure in pascals and in atmospheres exerted by a carton of milk that weighs 1.5 kg and has a base of 7.0 cm × 7.0 cm. If the carton were lying on its side (height = 25 cm), would it exert more or less pressure? Explain your reasoning. If atmospheric pressure at sea level is 1.0 × 10 Pa, what is the mass of air in kilograms above a 1.0 cm area of your skin as you lie on the beach? If atmospheric pressure is 8.2 × 10 Pa on a mountaintop, what is the mass of air in kilograms above a 4.0 cm patch of skin? Complete the following table: The SI unit of pressure is the pascal, which is equal to 1 N/m . Show how the product of the mass of an object and the acceleration due to gravity result in a force that, when exerted on a given area, leads to a pressure in the correct SI units. What mass in kilograms applied to a 1.0 cm area is required to produce a pressure of If you constructed a manometer to measure gas pressures over the range 0.60–1.40 atm using the liquids given in the following table, how tall a column would you need for each liquid? The density of mercury is 13.5 g/cm . Based on your results, explain why mercury is still used in barometers, despite its toxicity. 5.4 kPa or 5.3 × 10 atm; 11 kPa, 1.1 × 10 atm; the same force acting on a smaller area results in a greater pressure. | 19,084 | 3,292 |
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Diborane reacts readily with alkynes, but the formation of substituted alkene products leaves open the possibility of a second addition reaction. A clever technique for avoiding this event takes advantage of the fact that alkynes do not generally suffer from steric hindrance near the triple-bond (the configuration of this functional group is linear). Consequently, large or bulky electrophilic reagents add easily to the triple-bond, but the resulting alkene is necessarily more crowded or sterically hindered and resists further additions. The bulky hydroboration reagent needed for this strategy is prepared by reaction of diborane with 2-methyl-2-butene, a highly branched alkene. Because of the alkyl branching, only two alkenes add to a BH moiety (steric hindrance again), leaving one B-H covalent bond available for reaction with an alkyne, as shown below. The resulting dialkyl borane is called disiamylborane, a contraction of di-secondary-isoamylborane (amyl is an old name for pentyl). An important application of disiamylborane is its addition reaction to terminal alkynes. As with alkenes, the B-H reagent group adds in an apparently anti-Markovnikov manner, due to the fact that the boron is the electrophile, not the hydrogen. Further addition to the resulting boron-substituted alkene does not occur, and the usual oxidative removal of boron by alkaline hydrogen peroxide gives an enol which rapidly rearranges to the aldehyde tautomer. Thus, by the proper choice of reagents, terminal alkynes may be converted either to methyl ketones (mercuric ion catalyzed hydration) or aldehydes (hydroboration followed by oxidation). Hydroboration of internal alkynes is not a particularly useful procedure because a mixture of products will often be obtained, unless the triple-bond is symmetrically substituted. Mercuric ion catalyzed hydration gives similar results. Reactions of alkynes with oxidizing agents such as potassium permanganate and ozone usually result in cleavage of the triple-bond to give carboxylic acid products. A general equation for this kind of transformation follows. The symbol is often used in a general way to denote an oxidation. RC≡CR' + [O] RCO H + R'CO H | 2,210 | 3,293 |
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In our use of the ideal gas law thus far, we have focused entirely on the properties of pure gases with only a single chemical species. But what happens when two or more gases are mixed? In this section, we describe how to determine the contribution of each gas present to the total pressure of the mixture. The ideal gas law that all gases behave identically and that their behavior is independent of attractive and repulsive forces. If volume and temperature are held constant, the ideal gas equation can be rearranged to show that the pressure of a sample of gas is directly proportional to the number of moles of gas present:
\[P=n \bigg(\dfrac{RT}{V}\bigg) = n \times \rm const. \tag{10.5.1}\] Nothing in the equation depends on the of the gas—only the amount. With this assumption, let’s suppose we have a mixture of two ideal gases that are present in equal amounts. What is the total pressure of the mixture? Because the pressure depends on only the total number of particles of gas present, the total pressure of the mixture will simply be twice the pressure of either component. More generally, the total pressure exerted by a mixture of gases at a given temperature and volume is the sum of the pressures exerted by each gas alone. Furthermore, if we know the volume, the temperature, and the number of moles of each gas in a mixture, then we can calculate the pressure exerted by each gas individually, which is its partial pressure , the pressure the gas would exert if it were the only one present (at the same temperature and volume). To summarize, . This law was first discovered by John Dalton, the father of the atomic theory of matter. It is now known as Dalton’s law of partial pressures . We can write it mathematically as
where is the total pressure and the other terms are the partial pressures of the individual gases ( ). For a mixture of two ideal gases, A and B, we can write an expression for the total pressure:
More generally, for a mixture of components, the total pressure is given by
. restates in a more general form and makes it explicitly clear that, at constant temperature and volume, the pressure exerted by a gas depends on the total number of moles of gas present, whether the gas is a single chemical species or a mixture of dozens or even hundreds of gaseous species. For to be valid, the identity of the particles present cannot have an effect. Thus an ideal gas must be one whose properties are not affected by either the size of the particles or their intermolecular interactions because both will vary from one gas to another. The calculation of total and partial pressures for mixtures of gases is illustrated in Example 11. For reasons that we will examine in , deep-sea divers must use special gas mixtures in their tanks, rather than compressed air, to avoid serious problems, most notably a condition called “the bends.” At depths of about 350 ft, divers are subject to a pressure of approximately 10 atm. A typical gas cylinder used for such depths contains 51.2 g of O and 326.4 g of He and has a volume of 10.0 L. What is the partial pressure of each gas at 20.00°C, and what is the total pressure in the cylinder at this temperature? masses of components, total volume, and temperature partial pressures and total pressure Calculate the number of moles of He and O present. Use the ideal gas law to calculate the partial pressure of each gas. Then add together the partial pressures to obtain the total pressure of the gaseous mixture. The number of moles of He is \[n_{\rm He}=\rm\dfrac{326.4\;g}{4.003\;g/mol}=81.54\;mol \notag \] The number of moles of O is \[n_{\rm O_2}=\rm \dfrac{51.2\;g}{32.00\;g/mol}=1.60\;mol \notag \] We can now use the ideal gas law to calculate the partial pressure of each: \[P_{\rm He}=\dfrac{n_{\rm He}RT}{V}=\rm\dfrac{81.54\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=196.2\;atm \notag \] \[P_{\rm O_2}=\dfrac{n_{\rm O_2}RT}{V}=\rm\dfrac{1.60\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=3.85\;atm \notag \] The total pressure is the sum of the two partial pressures: \[P_{\rm tot}=P_{\rm He}+P_{\rm O_2}=\rm(196.2+3.85)\;atm=200.1\;atm \notag \] Exercise A cylinder of compressed natural gas has a volume of 20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate the partial pressure of each gas at 22.0°C and the total pressure in the cylinder. P(CH ) = 137 atm; P(C H ) = 13.4 atm; P = 151 atm. The composition of a gas mixture can be described by the mole fractions of the gases present. The mole fraction ( ) of any component of a mixture is the ratio of the number of moles of that component to the total number of moles of all the species present in the mixture ( ):
\[x_A=\dfrac{\text{moles of A}}{\text{total moles}}= \dfrac{n_A}{n_{tot}} =\dfrac{n_A}{n_A+n_B+\cdots}\tag{10.5.6}\] The mole fraction is a dimensionless quantity between 0 and 1. If = 1.0, then the sample is pure A, not a mixture. If = 0, then no A is present in the mixture. The sum of the mole fractions of all the components present must equal 1. To see how mole fractions can help us understand the properties of gas mixtures, let’s evaluate the ratio of the pressure of a gas A to the total pressure of a gas mixture that contains A. We can use the ideal gas law to describe the pressures of both gas A and the mixture: = / and = / . The ratio of the two is thus
\[\dfrac{P_A}{P_{tot}}=\dfrac{n_ART/V}{n_{tot}RT/V} = \dfrac{n_A}{n_{tot}}=x_A \tag{10.5.7}\] Rearranging this equation gives
\[P_A = x_AP_{tot} \tag{10.5.8}\] That is, the partial pressure of any gas in a mixture is the total pressure multiplied by the mole fraction of that gas. This conclusion is a direct result of the ideal gas law, which assumes that all gas particles behave ideally. Consequently, the pressure of a gas in a mixture depends on only the percentage of particles in the mixture that are of that type, not their specific physical or chemical properties. Recall from ( ) that by volume, Earth’s atmosphere is about 78% N , 21% O , and 0.9% Ar, with trace amounts of gases such as CO , H O, and others. This means that 78% of the particles present in the atmosphere are N ; hence the mole fraction of N is 78%/100% = 0.78. Similarly, the mole fractions of O and Ar are 0.21 and 0.009, respectively. Using .4, we therefore know that the partial pressure of N is 0.78 atm (assuming an atmospheric pressure of exactly 760 mmHg) and, similarly, the partial pressures of O and Ar are 0.21 and 0.009 atm, respectively. We have just calculated the partial pressures of the major gases in the air we inhale. Experiments that measure the composition of the air we yield different results, however. The following table gives the measured pressures of the major gases in both inhaled and exhaled air. Calculate the mole fractions of the gases in exhaled air. pressures of gases in inhaled and exhaled air mole fractions of gases in exhaled air Calculate the mole fraction of each gas using . The mole fraction of any gas A is given by \[x_A=\dfrac{P_A}{P_{tot}} \notag \] where is the partial pressure of A and is the total pressure. In this case, \[x_{\rm CO_2}=\rm\dfrac{48\;mmHg}{767\;mmHg}=0.063 \notag \] The following table gives the values of and for exhaled air. Exercise We saw in Example 10 that Venus is an inhospitable place, with a surface temperature of 560°C and a surface pressure of 90 atm. The atmosphere consists of about 96% CO and 3% N , with trace amounts of other gases, including water, sulfur dioxide, and sulfuric acid. Calculate the partial pressures of CO and N . \(P_{\rm CO_2}=\rm86\; atm\), \(P_{\rm N_2}=\rm2.7\;atm\) The pressure exerted by each gas in a gas mixture (its ) is independent of the pressure exerted by all other gases present. Consequently, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the components ( ). The amount of gas present in a mixture may be described by its partial pressure or its mole fraction. The of any component of a mixture is the ratio of the number of moles of that substance to the total number of moles of all substances present. In a mixture of gases, the partial pressure of each gas is the product of the total pressure and the mole fraction of that gas. : \( X_{A}= \dfrac{moles A}{total moles} = \dfrac{n_{A}}{n_{t}} \) .8: = Dalton’s law of partial pressures makes one key assumption about the nature of the intermolecular interactions in a mixture of gases. What is it? What is the relationship between the partial pressure of a gas and its mole fraction in a mixture? What is the partial pressure of each gas if the following amounts of substances are placed in a 25.0 L container at 25°C? What is the total pressure of each mixture? What is the partial pressure of each gas in the following 3.0 L mixtures at 37°C as well as the total pressure? In a mixture of helium, oxygen, and methane in a 2.00 L container, the partial pressures of He and O are 13.6 kPa and 29.2 kPa, respectively, and the total pressure inside the container is 95.4 kPa. What is the partial pressure of methane? If the methane is ignited to initiate its combustion with oxygen and the system is then cooled to the original temperature of 30°C, what is the final pressure inside the container (in kilopascals)? A 2.00 L flask originally contains 1.00 g of ethane (C H ) and 32.0 g of oxygen at 21°C. During ignition, the ethane reacts completely with oxygen to produce CO and water vapor, and the temperature of the flask increases to 200°C. Determine the total pressure and the partial pressure of each gas before and after the reaction. If a 20.0 L cylinder at 19°C is charged with 5.0 g each of sulfur dioxide and oxygen, what is the partial pressure of each gas? The sulfur dioxide is ignited in the oxygen to produce sulfur trioxide gas, and the mixture is allowed to cool to 19°C at constant pressure. What is the final volume of the cylinder? What is the partial pressure of each gas in the piston? The highest point on the continent of Europe is Mt. Elbrus in Russia, with an elevation of 18,476 ft. The highest point on the continent of South America is Mt. Aconcagua in Argentina, with an elevation of 22,841 ft. 52.6 kPa, 66.2 kPa Thumbnail from | 10,360 | 3,294 |
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There are numerous natural indicators present in plants. The dye in red cabbage, the purple color of grapes, even the color of some flowers are some examples. What is the cause for some fruits to change color when they ripen? \(\ce{[H+]}\) of the juice changes. The changes in pH or \(\ce{[H+]}\) cause the dye to change color if their conjugate acid-base pairs have different colors. There may be other reasons too. Do colors indicate how good or bad they taste? Choose the true statement: d. Color change is a requirement for indicators. Do all indicators change color at pH 7 (y/n)? No! Phenolphthalein changes color at pH ~9. Bromothymol blue has a p value of 7.1. At pH 7, its color changes from yellow to blue. Some indicators change color at pH other than 7. | 784 | 3,295 |
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Imagine that you need to do a lab experiment where hydrogen gas is generated. In order to calculate the yield of gas, you have to know the pressure inside the tube where the gas is collected. But how can you get a barometer in there? Very simple—you don't. All you need is the atmospheric pressure in the room. As the gas pushes out the water, it is pushing against the atmosphere, so the pressure inside is equal to the pressure outside. Gases that are produced in laboratory experiments are often collected by a technique called (see figure below). A bottle is filled with water and placed upside-down in a pan of water. The reaction flask is fitted with rubber tubing which is then fed under the bottle of water. As the gas is produced in the reaction flask, it exits through the rubber tubing and displaces the water in the bottle. When the bottle is full of the gas, it can be sealed with a lid. Because the gas is collected over water, it is not pure, but is mixed with vapor from the evaporation of the water. Dalton's law can be used to calculate the amount of the desired gas by subtracting the contribution of the water vapor: \[\begin{array}{ll} P_\text{Total} = P_g + P_{H_2O} & P_g \: \text{is the pressure of the desired gas} \\ P_g = P_{Total} - P_{H_2O} & \end{array}\nonumber \] In order to solve a problem, it is necessary to know the vapor pressure of water at the temperature of the reaction (see table below). The sample problem illustrates the use of Dalton's law when a gas is collected over water. A certain experiment generates \(2.58 \: \text{L}\) of hydrogen gas, which is collected over water. The temperature is \(20^\text{o} \text{C}\) and the atmospheric pressure is \(98.60 \: \text{kPa}\). Find the volume that the dry hydrogen would occupy at . The atmospheric pressure is converted from \(\text{kPa}\) to \(\text{mm} \: \ce{Hg}\) in order to match units with the table. The sum of the pressures of the hydrogen and the water vapor is equal to the atmospheric pressure. The pressure of the hydrogen is found by subtraction. Then, the volume of the gas at STP can be calculated by using the combined gas law. \[P_{H_2} = P_\text{Total} - P_{H_2O} = 739.7 \: \text{mm} \: \ce{Hg} - 17.54 \: \text{mm} \: \ce{Hg} = 722.2 \: \text{mm} \: \ce{Hg}\nonumber \] Now the combined gas law is used, solving for \(V_2\), the volume of hydrogen at STP. \[V_2 = \frac{P_1 \times V_1 \times T_2}{P_2 \times T_1} = \frac{722.2 \: \text{mm} \: \ce{Hg} \times 2.58 \: \text{L} \times 273 \: \text{K}}{760 \: \text{mm} \: \ce{Hg} \times 293 \: \text{K}} = 2.28 \: \text{L} \: \ce{H_2}\nonumber \] If the hydrogen gas were to be collected at STP and without the presence of the water vapor, its volume would be \(2.28 \: \text{L}\). This is less than the actual collected volume because some of that is water vapor. The conversion using STP is useful for stoichiometry purposes. | 2,916 | 3,297 |
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Ring compounds containing nitrogen, oxygen, sulfur, or other elements as ring atoms generally are known as compounds, and the ring atoms other than carbon are the atoms. Over the years, the more common heterocyclic compounds have acquired a hodge-podge of trivial names, such as ethylene oxide, tetrahydrofuran, and dioxane. Systematic naming of ring compounds is necessary but, unfortunately, several competing systems have been developed. We prefer the simplest procedure, which is to name the simple heterocycles as , , and -derivatives of cycloalkanes. However, this procedure has not been accepted (or adopted) universally and we are obliged to deal with the usages in existing literature. Having lived through at least two cycles of drastic changes in these matters, the authors hope that the simple procedure will prevail in the long run, but the long run is still ahead. We summarize here the rules of the so-called Hantzsch-Widman nomenclature system for heterocycles, which currently is the fashionable procedure, although relegated to second-class status by a recent, very practical approach to organic nomenclature.\(^1\) 4. and proceeds around the ring so as to give substituents (or other hetero atoms) the lowest numbered positions. When two or more hetero atoms are present, oxygen takes precedence over sulfur and sulfur over nitrogen for the number one position. Examples follow to illustrate both the heterocycloalkane and the Hantzsch-Widman systems. Trivial names also are included. Although Hantzsch-Widman system works satisfactorily (if you can remember the rules) for monocyclic compounds, it is cumbersome for polycyclic compounds. In the case of oxiranes it is simplest for purposes to name them as oxides of the cycloalkenes or derivatives of the corresponding cycloalkanes. The oxabicycloalkane names seem preferable for indexing purposes, particularly because the word “oxide” is used in many other connections. Oxacyclopropane (oxirane), the simplest cyclic ether, is an outstanding exception to the generalization that most ethers are resistant to cleavage. Like cyclopropane, the three-membered ring is highly strained and readily opens under mild conditions. Indeed, the importance of oxacyclopropane as an industrial chemical lies in its readiness to form other important compounds. The major products derived from it are shown in Figure 15-5. The lesser known four-membered cyclic ether, oxacyclobutane (oxetane), \(\ce{(CH_2)_3O}\), also is cleaved readily, but less so than oxacyclopropane. Oxacyclopentane (oxolane, tetrahydrofuran) is a relatively unreactive water-miscible compound with desirable properties as an organic solvent. It often is used in place of diethyl ether in Grignard reactions and reductions with lithium aluminum hydride. Three-membered cyclic ethers are important as reactive intermediates in organic synthesis. Like the cyclopropanes, the vicinal\(^2\) disubstituted compounds have cis and trans isomers: The most important method of preparation involves oxidation, or "epoxidation", of an alkene with a peroxycarboxylic acid, \(\ce{RCO_3H}\). This reaction achieves suprafacial addition of oxygen across the double bond, and is a type of electrophilic addition to alkenes: Oxacyclopropanes also can be prepared from vicinal chloro- or bromoalcohols and a base. This is an \(S_\text{N}2\) reaction and, if the stereochemistry is correct, proceeds quite rapidly, even if a strained ring is formed: Unlike most ethers, oxacyclopropanes react readily with nucleophilic reagents. These reactions are no different from the nucleophilic displacements previously encountered in Chapter 8, except that the leaving group, which is the oxygen of the oxide ring, remains a part of the original molecule. The stereochemistry is consistent with an \(S_\text{N}2\) mechanism because inversion of configuration at the site of attack occurs. Thus cyclopentene oxide yields products with the trans configuration: Acidic conditions also can be used for the cleavage of oxacyclopropane rings. An oxonium ion is formed first, which subsequently is attacked by the nucleophile in an \(S_\text{N}2\) displacement or forms a carbocation in an \(S_\text{N}1\) reaction. Evidence for the \(S_\text{N}2\) mechanism, which produces inversion, comes not only from the stereochemistry but also from the fact that the rate is dependent on the concentration of the nucleophile. An example is ring opening with hydrogen bromide: The same kind of mechanism can operate in the formation of 1,2-diols by acid-catalyzed ring-opening with water as the nucleophile: Some acid-catalyzed solvolysis reactions of oxacyclopropanes appear to proceed by \(S_\text{N}1\) mechanisms involving carbocation intermediates. Evidence for the \(S_\text{N}1\) mechanism is available from the reactions of unsymmetrically substituted oxacyclopropanes. For example, we would expect the conjugate acid of 2,2-dimethyloxacyclopropane to be attacked by methanol at the primary carbon by an \(S_\text{N}2\) reaction and at the tertiary carbon by an \(S_\text{N}1\) reaction: Because both products actually are obtained, we can conclude that both the \(S_\text{N}1\) and \(S_\text{N}2\) mechanisms occur. The \(S_\text{N}1\) product, the tertiary ether, is the major product. We have emphasized the contrasts in properties between the ionic compounds such as sodium chloride, and the nonpolar organic compounds, such as the alkanes and arenes. There are great difficulties in dissolving the extreme types of these substances in a mutually compatible medium for carrying on chemical reactions, as, for example, in \(S_\text{N}\) reactions of organic halides with alkali-metal salts ( and 8-7F). The essence of the problem is that electrostatic forces in ionic crystals of inorganic salts are strong, and nonpolar solvents simply do not have the solvating power for ions to make dissolution of the crystals a favorable process. However, it has long been known that polyethers, such as the "glymes" ( ), are able to assist in the dissolution of ionic compounds through their ability to solvate metal cations by providing multiple complexing sites: In 1967, C. J. Pedersen reported the synthesis of a series of cyclic polyethers, which he called " ", that have shown great promise for brining together what traditionally have been regarded as wholly incompatible substances - even achieving measurable solubilities of salts such as \(\ce{NaCl}\), \(\ce{KOH}\), and \(\ce{KMnO_4}\) in benzene. The crown ethers can be regarded as cyclic "glymes" and are available by \(S_\text{N}2\)-type cyclization reactions: The crown ethers and many modifications of them (especially with nitrogen replacing one or more of the oxygens) function by coordinating with metal cations and converting them into less polar entities that are more stable in solution, even in a nonpolar solvent, than they are in the crystal. Many of the crown ethers have considerable specificity with regard to the metal with which they complex. Ring size as well as the number and kind of hetero atoms are very important in this connection. 18-Crown-6 is especially effective for potassium: An important application for the crown ethers in synthetic work is for solubilization of salts such as \(\ce{KCN}\) in nonpolar solvents for use in \(S_\text{N}2\) displacements. If the solvent has a low anion-solvating capability, then the reactivity of the anion is enhanced greatly. Consequently many displacement reactions that proceed slowly at elevated temperatures will proceed at useful rates at room temperatures, because the energy of "desolvating" the anion before it undergoes \(S_\text{N}2\) displacement is low (Section 8-7F). For example, potassium fluoride becomes a potent nucleophilic reagent in nonpolar solvents when complexed with 18-crown-6: The grouping \(\ce{C-O-C-O-C}\) is characteristic of an acetal or a ketal (see ), but it also can be regarded as an ether with two ether links to one carbon. Compared to other ethers (except for the oxacyclopropanes), substances with the \(\ce{C-O-C-O-C}\) group are very active toward acidic reagents, as pointed out in connection with their formation from alcohols ( ) and their sue as protecting groups for the \(\ce{OH}\) function ( ). \(^1\)J. H. Fletcher, O. C. Dermer, and R. B. Fox (Editors), "Nomenclature of Organic Compounds, Principles and Practice." , American Chemical Society, Washington, D.C., 1974. \(^2\) means substituted on adjacent carbons. and (1977) | 8,528 | 3,298 |
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Thus far, we have used two-dimensional Lewis structures to represent molecules. However, molecular structure is actually three-dimensional, and it is important to be able to describe molecular bonds in terms of their distances, angles, and relative arrangements in space (Figure \(\Page {1}\)). A is the angle between any two bonds that include a common atom, usually measured in degrees. A (or bond length) is the distance between the nuclei of two bonded atoms along the straight line joining the nuclei. Bond distances are measured in Ångstroms (1 Å = 10 m) or picometers (1 pm = 10 m, 100 pm = 1 Å). enables us to predict the molecular structure, including approximate bond angles around a central atom, of a molecule from an examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement that minimizes repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom form either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far from each other as possible. VSEPR theory predicts the arrangement of electron pairs around each central atom and, usually, the correct arrangement of atoms in a molecule. We should understand, however, that the theory only considers electron-pair repulsions. Other interactions, such as nuclear-nuclear repulsions and nuclear-electron attractions, are also involved in the final arrangement that atoms adopt in a particular molecular structure. As a simple example of VSEPR theory, let us predict the structure of a gaseous BeF molecule. The Lewis structure of BeF (Figure \(\Page {2}\)) shows only two electron pairs around the central beryllium atom. With two bonds and no lone pairs of electrons on the central atom, the bonds are as far apart as possible, and the electrostatic repulsion between these regions of high electron density is reduced to a minimum when they are on opposite sides of the central atom. The bond angle is 180° (Figure \(\Page {2}\)). Figure \(\Page {3}\) illustrates this and other electron-pair geometries that minimize the repulsions among regions of high electron density (bonds and/or lone pairs). Two regions of electron density around a central atom in a molecule form a geometry; three regions form a geometry; four regions form a geometry; five regions form a geometry; and six regions form an geometry. It is important to note that electron-pair geometry around a central atom is the same thing as its molecular structure. The electron-pair geometries shown in Figure \(\Page {3}\) describe all regions where electrons are located, bonds as well as lone pairs. Molecular structure describes the location of the , not the electrons. We differentiate between these two situations by naming the geometry that includes electron pairs the . The structure that includes only the placement of the atoms in the molecule is called the . The electron-pair geometries will be the same as the molecular structures when there are no lone electron pairs around the central atom, but they will be different when there are lone pairs present on the central atom. For example, the methane molecule, CH , which is the major component of natural gas, has four bonding pairs of electrons around the central carbon atom; the electron-pair geometry is tetrahedral, as is the molecular structure (Figure \(\Page {4}\)). On the other hand, the ammonia molecule, NH , also has four electron pairs associated with the nitrogen atom, and thus has a tetrahedral electron-pair geometry. One of these regions, however, is a lone pair, which is not included in the molecular structure, and this lone pair influences the shape of the molecule (Figure \(\Page {5}\)). Small distortions from the ideal angles in Figure \(\Page {5}\) can result from differences in repulsion between various regions of electron density. VSEPR theory predicts these distortions by establishing an order of repulsions and an order of the amount of space occupied by different kinds of electron pairs. The order of electron-pair repulsions from greatest to least repulsion is: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair This order of repulsions determines the amount of space occupied by different regions of electrons. A lone pair of electrons occupies a larger region of space than the electrons in a triple bond; in turn, electrons in a triple bond occupy more space than those in a double bond, and so on. The order of sizes from largest to smallest is: lone pair > triple bond > double bond > single bond Consider formaldehyde, H CO, which is used as a preservative for biological and anatomical specimens. This molecule has regions of high electron density that consist of two single bonds and one double bond. The basic geometry is trigonal planar with 120° bond angles, but we see that the double bond causes slightly larger angles (121°), and the angle between the single bonds is slightly smaller (118°). In the ammonia molecule, the three hydrogen atoms attached to the central nitrogen are not arranged in a flat, trigonal planar molecular structure, but rather in a three-dimensional trigonal pyramid (Figure \(\Page {6}\)) with the nitrogen atom at the apex and the three hydrogen atoms forming the base. The ideal bond angles in a trigonal pyramid are based on the tetrahedral electron pair geometry. Again, there are slight deviations from the ideal because lone pairs occupy larger regions of space than do bonding electrons. The H–N–H bond angles in NH are slightly smaller than the 109.5° angle in a regular tetrahedron (Figure \(\Page {6}\)) because the lone pair-bonding pair repulsion is greater than the bonding pair-bonding pair repulsion. The ideal molecular structures are predicted based on the electron-pair geometries for various combinations of lone pairs and bonding pairs. According to VSEPR theory, the terminal atom locations (Xs in Figure \(\Page {7}\)) are equivalent within the linear, trigonal planar, and tetrahedral electron-pair geometries (the first three rows of the table). It does not matter which X is replaced with a lone pair because the molecules can be rotated to convert positions. For trigonal bipyramidal electron-pair geometries, however, there are two distinct X positions (Figure \(\Page {7}\)a): an (if we hold a model of a trigonal bipyramid by the two axial positions, we have an axis around which we can rotate the model) and an (three positions form an equator around the middle of the molecule). The axial position is surrounded by bond angles of 90°, whereas the equatorial position has more space available because of the 120° bond angles. In a trigonal bipyramidal electron-pair geometry, lone pairs always occupy equatorial positions because these more spacious positions can more easily accommodate the larger lone pairs. Theoretically, we can come up with three possible arrangements for the three bonds and two lone pairs for the ClF molecule (Figure \(\Page {7}\)). The stable structure is the one that puts the lone pairs in equatorial locations, giving a T-shaped molecular structure. When a central atom has two lone electron pairs and four bonding regions, we have an octahedral electron-pair geometry. The two lone pairs are on opposite sides of the octahedron (180° apart), giving a square planar molecular structure that minimizes lone pair-lone pair repulsions. The following procedure uses VSEPR theory to determine the electron pair geometries and the molecular structures: The following examples illustrate the use of VSEPR theory to predict the molecular structure of molecules or ions that have no lone pairs of electrons. In this case, the molecular structure is identical to the electron pair geometry. Predict the electron-pair geometry and molecular structure for each of the following: (a) We write the Lewis structure of CO as: This shows us two regions of high electron density around the carbon atom—each double bond counts as one region, and there are no lone pairs on the carbon atom. Using VSEPR theory, we predict that the two regions of electron density arrange themselves on opposite sides of the central atom with a bond angle of 180°. The electron-pair geometry and molecular structure are identical, and CO molecules are linear. (b) We write the Lewis structure of BCl as: Thus we see that BCl contains three bonds, and there are no lone pairs of electrons on boron. The arrangement of three regions of high electron density gives a trigonal planar electron-pair geometry. The B–Cl bonds lie in a plane with 120° angles between them. BCl also has a trigonal planar molecular structure. The electron-pair geometry and molecular structure of BCl are both trigonal planar. Note that the VSEPR geometry indicates the correct bond angles (120°), unlike the Lewis structure shown above. Carbonate, \(\ce{CO3^2-}\), is a common polyatomic ion found in various materials from eggshells to antacids. What are the electron-pair geometry and molecular structure of this polyatomic ion? The electron-pair geometry is trigonal planar and the molecular structure is trigonal planar. Due to resonance, all three C–O bonds are identical. Whether they are single, double, or an average of the two, each bond counts as one region of electron density. Two of the top 50 chemicals produced in the United States, ammonium nitrate and ammonium sulfate, both used as fertilizers, contain the ammonium ion. Predict the electron-pair geometry and molecular structure of the \(\ce{NH4+}\) cation. We write the Lewis structure of \(\ce{NH4+}\) as: Identify a molecule with trigonal bipyramidal molecular structure. Any molecule with five electron pairs around the central atoms including no lone pairs will be trigonal bipyramidal. \(\ce{PF5}\) is a common example The next several examples illustrate the effect of lone pairs of electrons on molecular structure. Predict the electron-pair geometry and molecular structure of a water molecule. The Lewis structure of H O indicates that there are four regions of high electron density around the oxygen atom: two lone pairs and two chemical bonds: The hydronium ion, H O , forms when acids are dissolved in water. Predict the electron-pair geometry and molecular structure of this cation. electron pair geometry: tetrahedral; molecular structure: trigonal pyramidal SF , is extremely valuable for the preparation of fluorine-containing compounds used as herbicides (i.e., SF is used as a fluorinating agent). Predict the electron-pair geometry and molecular structure of a SF molecule. The Lewis structure of SF indicates five regions of electron density around the sulfur atom: one lone pair and four bonding pairs: Predict the electron pair geometry and molecular structure for molecules of XeF . The electron-pair geometry is trigonal bipyramidal. The molecular structure is linear. Of all the noble gases, xenon is the most reactive, frequently reacting with elements such as oxygen and fluorine. Predict the electron-pair geometry and molecular structure of the XeF molecule. The Lewis structure of XeF indicates six regions of high electron density around the xenon atom: two lone pairs and four bonds: In a certain molecule, the central atom has three lone pairs and two bonds. What will the electron pair geometry and molecular structure be? electron pair geometry: trigonal bipyramidal; molecular structure: linear When a molecule or polyatomic ion has only one central atom, the molecular structure completely describes the shape of the molecule. Larger molecules do not have a single central atom, but are connected by a chain of interior atoms that each possess a “local” geometry. The way these local structures are oriented with respect to each other also influences the molecular shape, but such considerations are largely beyond the scope of this introductory discussion. For our purposes, we will only focus on determining the local structures. The Lewis structure for the simplest amino acid, glycine, H NCH CO H, is shown here. Predict the local geometry for the nitrogen atom, the two carbon atoms, and the oxygen atom with a hydrogen atom attached: Consider each central atom independently. The electron-pair geometries: The local structures: Another amino acid is alanine, which has the Lewis structure shown here. Predict the electron-pair geometry and local structure of the nitrogen atom, the three carbon atoms, and the oxygen atom with hydrogen attached: electron-pair geometries: nitrogen––tetrahedral; carbon ( H)—tetrahedral; carbon ( H )—tetrahedral; carbon ( O )—trigonal planar; oxygen ( H)—tetrahedral; local structures: nitrogen—trigonal pyramidal; carbon ( H)—tetrahedral; carbon ( H )—tetrahedral; carbon ( O )—trigonal planar; oxygen ( H)—bent (109°) Using this molecular shape simulator allows us to control whether bond angles and/or lone pairs are displayed by checking or unchecking the boxes under “Options” on the right. We can also use the “Name” checkboxes at bottom-left to display or hide the electron pair geometry (called “electron geometry” in the simulator) and/or molecular structure (called “molecular shape” in the simulator). Build the molecule in the simulator based on the following Lewis structure: on each bond type or lone pair at right to add that group to the central atom. Once you have the complete molecule, rotate it to examine the predicted molecular structure. What molecular structure is this? The molecular structure is linear. Build a more complex molecule in the simulator. Identify the electron-group geometry, molecular structure, and bond angles. Then try to find a chemical formula that would match the structure you have drawn. Answers will vary. For example, an atom with four single bonds, a double bond, and a lone pair has an octahedral electron-group geometry and a square pyramidal molecular structure. XeOF is a molecule that adopts this structure. As discussed previously, polar covalent bonds connect two atoms with differing electronegativities, leaving one atom with a partial positive charge (δ+) and the other atom with a partial negative charge (δ–), as the electrons are pulled toward the more electronegative atom. This separation of charge gives rise to a bond dipole moment. The magnitude of a bond dipole moment is represented by the Greek letter mu (µ) and is given by \[μ=Qr \label{7.6.X} \] where This bond moment can be represented as a vector, a quantity having both direction and magnitude (Figure \(\Page {12}\)). Dipole vectors are shown as arrows pointing along the bond from the less electronegative atom toward the more electronegative atom. A small plus sign is drawn on the less electronegative end to indicate the partially positive end of the bond. The length of the arrow is proportional to the magnitude of the electronegativity difference between the two atoms. A whole molecule may also have a separation of charge, depending on its molecular structure and the polarity of each of its bonds. If such a charge separation exists, the molecule is said to be a (or dipole); otherwise the molecule is said to be nonpolar. The measures the extent of net charge separation in the molecule as a whole. We determine the dipole moment by adding the bond moments in three-dimensional space, taking into account the molecular structure. For diatomic molecules, there is only one bond, so its bond dipole moment determines the molecular polarity. Homonuclear diatomic molecules such as Br and N have no difference in electronegativity, so their dipole moment is zero. For heteronuclear molecules such as CO, there is a small dipole moment. For , there is a larger dipole moment because there is a larger difference in electronegativity. When a molecule contains more than one bond, the geometry must be taken into account. If the bonds in a molecule are arranged such that their bond moments cancel (vector sum equals zero), then the molecule is nonpolar. This is the situation in CO (Figure \(\Page {13A}\)). Each of the bonds is polar, but the molecule as a whole is nonpolar. From the Lewis structure, and using VSEPR theory, we determine that the CO molecule is linear with polar C=O bonds on opposite sides of the carbon atom. The bond moments cancel because they are pointed in opposite directions. In the case of the water molecule (Figure \(\Page {13B}\)), the Lewis structure again shows that there are two bonds to a central atom, and the electronegativity difference again shows that each of these bonds has a nonzero bond moment. In this case, however, the molecular structure is bent because of the lone pairs on O, and the two bond moments do not cancel. Therefore, water does have a net dipole moment and is a polar molecule (dipole). The molecule has a structure similar to CO , but a sulfur atom has replaced one of the oxygen atoms. To determine if this molecule is polar, we draw the molecular structure. VSEPR theory predicts a linear molecule: The C–O bond is considerably polar. Although C and S have very similar electronegativity values, S is slightly more electronegative than C, and so the C-S bond is just slightly polar. Because oxygen is more electronegative than sulfur, the oxygen end of the molecule is the negative end. Chloromethane, CH Cl, is another example of a polar molecule. Although the polar C–Cl and C–H bonds are arranged in a tetrahedral geometry, the C–Cl bonds have a larger bond moment than the C–H bond, and the bond moments do not completely cancel each other. All of the dipoles have a upward component in the orientation shown, since carbon is more electronegative than hydrogen and less electronegative than chlorine: When we examine the highly symmetrical molecules BF (trigonal planar), CH (tetrahedral), PF (trigonal bipyramidal), and SF (octahedral), in which all the polar bonds are identical, the molecules are nonpolar. The bonds in these molecules are arranged such that their dipoles cancel. However, just because a molecule contains identical bonds does not mean that the dipoles will always cancel. Many molecules that have identical bonds and lone pairs on the central atoms have bond dipoles that do not cancel. Examples include H S and NH . A hydrogen atom is at the positive end and a nitrogen or sulfur atom is at the negative end of the polar bonds in these molecules: To summarize, to be polar, a molecule must: Polar molecules tend to align when placed in an electric field with the positive end of the molecule oriented toward the negative plate and the negative end toward the positive plate (Figure \(\Page {14}\)). We can use an electrically charged object to attract polar molecules, but nonpolar molecules are not attracted. Also, polar solvents are better at dissolving polar substances, and nonpolar solvents are better at dissolving nonpolar substances. Open the and select the “Three Atoms” tab at the top. This should display a molecule with three electronegativity adjustors. You can display or hide the bond moments, molecular dipoles, and partial charges at the right. Turning on the Electric Field will show whether the molecule moves when exposed to a field, similar to Figure \(\Page {14}\). Use the electronegativity controls to determine how the molecular dipole will look for the starting bent molecule if: Determine the partial charges that will give the largest possible bond dipoles. The largest bond moments will occur with the largest partial charges. The two solutions above represent how unevenly the electrons are shared in the bond. The bond moments will be maximized when the electronegativity difference is greatest. The controls for A and C should be set to one extreme, and B should be set to the opposite extreme. Although the magnitude of the bond moment will not change based on whether B is the most electronegative or the least, the direction of the bond moment will. VSEPR theory predicts the three-dimensional arrangement of atoms in a molecule. It states that valence electrons will assume an electron-pair geometry that minimizes repulsions between areas of high electron density (bonds and/or lone pairs). Molecular structure, which refers only to the placement of atoms in a molecule and not the electrons, is equivalent to electron-pair geometry only when there are no lone electron pairs around the central atom. A dipole moment measures a separation of charge. For one bond, the bond dipole moment is determined by the difference in electronegativity between the two atoms. For a molecule, the overall dipole moment is determined by both the individual bond moments and how these dipoles are arranged in the molecular structure. Polar molecules (those with an appreciable dipole moment) interact with electric fields, whereas nonpolar molecules do not. | 21,091 | 3,299 |
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A free-moving 30 cc glass syringe is plugged with a luer lok stopper with a trapped volume of 20 cc at atmospheric pressure. The plunger is 7/8" in diameter (0.88") so the area is πr = 0.60 in . Push down on the scale with the plunger until V = 15 cc and measure weight (force), ~3 lb New pressure = 3 lb/0.60 in external + 14.7 atmospheric = 19.6 lb/in . Push down on the scale witht the plunger until V = 10 cc and measure weight (force), ~9 lb New pressure = 9 lb/0.60 in external + 14.7 atmospheric = 29.4 lb/in . For the three cases, PV = ~294, so P V = k = P V Connect a vacuum pump to the top of a 100 cm tube dipped in mercury in a vial in a sidearm flask, turn on pump, measure height of Hg corresponding to 5 mi of air in atmosphere. Put a syringe, marshmallow, small bag of chips or pretzels, etc. in a bell jar and turn on vacuum. | 858 | 3,300 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Standard_Reduction_Potential |
The standard reduction potential is in a category known as the standard cell potentials or standard electrode potentials. The standard cell potential is the potential difference between the cathode and anode. For more information view Cell Potentials. The standard potentials are all measured at 298 K, 1 atm, and with 1 M solutions. As stated above, the standard reduction potential is the likelihood that a species will be reduced. It is written in the form of a reduction half reaction. An example can be seen below where "A" is a generic element and C is the charge. \[ A^{C+} + C \,e^- \rightarrow A\] For example, copper's Standard Reduction Potential of \(E^o =+0.340 \;V)\) is for this reaction: \[ Cu^{2+} + 2 \,e^- \rightarrow Cu\] The standard oxidation potential is much like the standard reduction potential. It is the tendency for a species to be at standard conditions. It is also written in the form of a half reaction, and an example is shown below. \[ A(s) \rightarrow A^{c+} + C\,e^-\] Copper's Standard Oxidation Potential \[ Cu (s) \rightarrow Cu^{2+}+ 2e^- \] \[ E_0^o (SOP) = -0.34\, V\] The standard oxidation potential and the standard reduction potential are opposite in sign to each other for the same chemical species. \[ E_0^o (SRP) = -E_0^o (SOP)\] Standard reduction or oxidation potentials can be determined using a SHE (standard hydrogen electrode). Universally, hydrogen has been recognized as having reduction and oxidation potentials of zero. Therefore, when the standard reduction and oxidation potential of chemical species are measured, it is actually the difference in the potential from hydrogen. By using a galvanic cell in which one side is a SHE, and the other side is half cell of the unknown chemical species, the potential difference from hydrogen can be determined using a voltmeter. Standard reduction and oxidation potentials can both be determined in this fashion. When the standard reduction potential is determined, the unknown chemical species is being reduced while hydrogen is being oxidized, and when the standard oxidation potential is determined, the unknown chemical species is being oxidized while hydrogen is being reduced. The following diagrams show how a standard reduction potential is determined. Standard reduction potentials are used to determine the standard cell potential. The standard reduction cell potential and the standard oxidation cell potential can be combined to determine the overall Cell Potentials of a galvanic cell. The equations that relate these three potentials are shown below: \[ E^o_{cell} = E^o_{reduction} \text{ of reaction at cathode} + E^o_{oxidation} \text{ of reaction at anode}\] or alternatively \[ E^o_{cell} = E^o_{reduction} \text{ of reaction at cathode} - E^o_{reduction} \text{ of reaction at anode}\] When solving for the standard cell potential, the species oxidized and the species reduced must be identified. This can be done using an activity series. The table shown below is simply a table of standard reduction potentials in decreasing order. The species at the top have a greater likelihood of being reduced while the ones at the bottom have a greater likelihood of being oxidized. Therefore, when a species at the top is coupled with a species at the bottom, the one at the top will become reduced while the one at the bottom will become oxidized. Below is a table of standard reduction potentials. | 3,433 | 3,306 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/01%3A_AcidBase_Reactions |
As we will see, organic reactions can be classified using a small set of reaction types—the largest and most all-encompassing of which are those involving acid–base reactions. Understanding acid–base reactions, therefore, provides a broadly useful conceptual framework within which to consider a wide range of organic reactions. Although it is likely that you have already been introduced to acid–base reactions (especially if you used the CLUE general chemistry curriculum ), we are going to review this class of reactions in order to emphasize their general features. Our goal is that you learn how to recognize their role in a range of reaction mechanisms; understanding how and why acid–base reactions occur will give you to a set of tools to understand phenomena as diverse as why most drugs are usually administered as in their salt form (a conjugate acid or base), why biological systems are buffered to specific pH levels (and why different pH levels are found in different cellular and organismic compartments), and why molecular oxygen (O ) transport systems require a metal ion complex (within the proteins involved, e.g. myoglobin, hemoglobin, cytochromes). As we will see, acid–base reactions are by far the most common types of reactions in biological systems.
| 1,289 | 3,309 |
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" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact Until recently, benzidine was used in forensic medicine to detect the presence of human blood: when mixed with human blood, benzidine turns a characteristic blue color. Because benzidine has recently been identified as a carcinogen, other indicators have replaced it. Draw the complete Lewis dot structure for benzidine. Would you expect this compound to behave as a Lewis acid or a Lewis base? There are three possible ways to connect carbon, nitrogen, and oxygen to form a monoanion: CNO , CON , and OCN . One is the cyanate ion, a common and stable species; one is the fulminate ion, salts of which are used as explosive detonators; and one is so unstable that it has never been isolated. Use Lewis electron structures and the concept of formal charge to determine which isomer is cyanate, which is the fulminate, and which is the least stable. The colorless gas N O is a deadly poison that has been used as an oxidizing agent in rocket fuel. The compound has a single N–N bond, with a formal charge of +1 on each nitrogen atom. Draw resonance structures for this molecule. Naphthalene is an organic compound that is commonly used in veterinary medicine as the active ingredient in dusting powders; it is also used internally as an intestinal antiseptic. From its chemical structure and the Δ of CO and H O, estimate the molar enthalpy of combustion and the enthalpy of formation of naphthalene. ♦ Compare the combustion of hydrazine (N H ), which produces nitrogen and water, with the combustion of methanol. Use the chemical structures to estimate which has a higher heat of combustion. Given equal volumes of hydrazine ( = 1.004 g/mL) and methanol ( = 0.791 g/mL), which is the better fuel (i.e., which provides more energy per unit volume during combustion)? Can you think of a reason why hydrazine is not used in internal combustion engines? ♦ Race car drivers frequently prefer methanol to isooctane as a fuel. Is this justified based on enthalpies of combustion? If you had a choice between 10 gal of methanol ( = 0.791 g/mL) and the same volume of isooctane ( = 0.688 g/mL), which fuel would you prefer? ♦ An atmospheric reservoir species is a molecule that is rather unreactive, but it contains elements that can be converted to reactive forms. For example, chlorine nitrate (ClONO ) is a reservoir species for both chlorine and nitrogen dioxide. In fact, most of the chlorine in the atmosphere is usually bound up in chlorine nitrate as a result of the reaction of ClO with NO . Draw Lewis electron structures for each species in this reaction. What difficulty is associated with the structure of the reactants? How does this affect the reactivity of the compounds? Chlorine nitrate can react in a surface reaction with water to form HClO and nitric acid. ♦ Aniline is an oily liquid used to prepare organic dyes, varnishes, and black shoe polishes. The –NH bound to the ring contains a lone pair of electrons that can participate in resonance in the following way: Draw a second Lewis structure for aniline that takes this interaction into account. This molecule is likely to serve as a Lewis base because of the lone pair of electrons on each nitrogen atom. The balanced chemical reaction is: Hydrazine: Methanol (CH OH): Hydrazine is both extremely toxic and potentially explosive. Both reactants have one unpaired electron, which makes them more reactive than might otherwise be expected. ClONO + H O → HClO + HONO | 3,585 | 3,310 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Catalytic_Hydrogenation_of_Alkenes_II |
The carbon-carbon double bond in alkenes react with hydrogen in the presence of a metal catalyst. This is called hydrogenation. It includes the manufacture of margarine from animal or vegetable fats and oils. Ethene reacts with hydrogen in the presence of a finely divided nickel catalyst at a temperature of about 150°C. Ethane is produced. This is a fairly pointless reaction because ethene is a far more useful compound than ethane! However, what is true of the reaction of the carbon-carbon double bond in ethene is equally true of it in much more complicated cases. Some margarine is made by hydrogenating carbon-carbon double bonds in animal or vegetable fats and oils. You can recognize the presence of this in foods because the ingredients list will include words showing that it contains "hydrogenated vegetable oils" or "hydrogenated fats". The impression is sometimes given that all margarine is made by hydrogenation - that's simply not true. These are similar molecules, differing in their melting points. If the compound is a solid at room temperature, you usually call it a fat. If it is a liquid, it is often described as an oil. Their melting points are largely determined by the presence of carbon-carbon double bonds in the molecule. The higher the number of carbon-carbon double bonds, the lower the melting point. If there aren't any carbon-carbon double bonds, the substance is said to be saturated. A typical saturated fat might have the structure: Molecules of this sort are usually solid at room temperature. If there is only one carbon-carbon double bond in each of the hydrocarbon chains, it is called a mono-unsaturated fat (or mono-unsaturated oil, because it is likely to be a liquid at room temperature.) A typical mono-unsaturated oil might be: If there are two or more carbon-carbon double bonds in each chain, then it is said to be polyunsaturated. For example: For simplicity, in all these diagrams, all three hydrocarbon chains in each molecule are the same. That doesn't have to be the case - you can have a mixture of types of chain in the same molecule. Vegetable oils often contain high proportions of polyunsaturated and mono-unsaturated fats (oils), and as a result are liquids at room temperature. That makes them messy to spread on your bread or toast, and inconvenient for some baking purposes. You can "harden" (raise the melting point of) the oil by hydrogenating it in the presence of a nickel catalyst. Conditions (like the precise temperature, or the length of time the hydrogen is passed through the oil) are carefully controlled so that some, but not necessarily all, of the carbon-carbon double bonds are hydrogenated. This produces a "partially hydrogenated oil" or "partially hydrogenated fat". You need to hydrogenate enough of the bonds to give the final texture you want. However, there are possible health benefits in eating mono-unsaturated or polyunsaturated fats or oils rather than saturated ones - so you wouldn't want to remove all the carbon-carbon double bonds. The flow diagram below shows the complete hydrogenation of a typical mono-unsaturated oil. There are some probable health risks from eating hydrogenated fats or oils. Consumers are becoming more aware of this, and manufacturers are increasingly finding alternative ways of converting oils into spreadable solids. One of the problems arises from the hydrogenation process. The double bonds in unsaturated fats and oils tend to have the groups around them arranged in the "cis" form. The relatively high temperatures used in the hydrogenation process tend to flip some of the carbon-carbon double bonds into the "trans" form. If these particular bonds aren't hydrogenated during the process, they will still be present in the final margarine in molecules of trans fats. The consumption of trans fats has been shown to increase cholesterol levels (particularly of the more harmful LDL form) - leading to an increased risk of heart disease. Any process which tends to increase the amount of trans fat in the diet is best avoided. Read food labels, and avoid any food which contains (or is cooked in) hydrogenated oil or hydrogenated fat. Jim Clark ( ) | 4,188 | 3,314 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Carbohydrates/Carbohydrates_Fundamentals/Carbohydrate_Isomers |
Glyceraldehyde, the simplest carbohydrate, exhibits properties of a chiral or optical isomer compound. This molecule forms the basis for the designation of the isomers of all of the carbohydrates. Glyceraldehyde can exist in two isomeric forms that are mirror images of each other which are shown below. The absolute configuration is defined by the molecule on the far left as the D-glyceraldehyde. With the aldehyde group in the "up" direction, the the -OH group must project to the right side of the molecule for the D isomer. Chemists have used this configuration of D-glyceraldehyde to determine the optical isomer families of the rest of the carbohydrates. All naturally occurring monosaccharides belong to the D optical family. It is remarkable that the chemistry and enzymes of all living things can tell the difference between the geometry of one optical isomer over the other. Monosaccharides are assigned to the D-family according to the projection of the -OH group to the on the chiral carbon that is the farthest from the carbonyl (aldehyde) group. This is on carbon # 5 if the carbonyl carbon is # 1. Note: For whatever reason, the ball and stick model does not completely match the projections of the -OH groups on carbons # 2 and 4. It is in the way that the flat Fischer model has been defined. How many chiral carbons can you find? List them. If necessary Review Chiral Compounds to find the definitions.Then check the answer from the drop down menu. Examine the structures of glucose and galactose carefully. Which -OH group determines that they both are the D isomer? Then check the answer from the drop down menu. Isomers have different arrangements of atoms. Which carbon bonding to -OH and -H is different in glucose vs. galactose? This single difference makes glucose and galactose isomers. Then check the answer from the drop down menu. | 1,882 | 3,315 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_OpenStax/08%3A_Advanced_Theories_of_Covalent_Bonding_(Exercises) |
1. : Both types of bonds result from overlap of atomic orbitals on adjacent atoms and contain a maximum of two electrons. : σ bonds are stronger and result from end-to-end overlap and all single bonds are σ bonds; π bonds between the same two atoms are weaker because they result from side-by-side overlap, and multiple bonds contain one or more π bonds (in addition to a σ bond). 2 When H and Cl are separate (the x axis) the energy is at a particular value. As they approach, it decreases to a minimum at 127 pm (the bond distance), and then it increases sharply as you get closer. \[−184.7 kJ/mol=(ΔH∘BDE(H–H)+ΔH∘BDE(Cl–Cl))−(2ΔH∘BDE(H–Cl))\] \[H–H is 436 kJ/mol and Cl–Cl is 243\] \[–184.7 kJ/mol = (436 + 243) – 2x = 679 – 2x\] \[2x = 863.7 kJ/mol\] \[x = 432\; kJ/mol\] This is very close to the value from part (a). 3. The specific average bond distance is the distance with the lowest energy. At distances less than the bond distance, the positive charges on the two nuclei repel each other, and the overall energy increases. 4. The single bond present in each molecule results from overlap of the relevant orbitals: F 2 orbitals in F , the H 1 and F 2 orbitals in HF, and the Cl 3 orbital and Br 4 orbital in ClBr. 5. Bonding: One σ bond and one π bond. The orbitals are filled and do not overlap. The orbitals overlap along the axis to form a σ bond and side by side to form the π bond. 6. \(\ce{H–C≡N}\) has two σ (H–C and C–N) and two π (making the CN triple bond). 7. No, two of the orbitals (one on each N) will be oriented end-to-end and will form a σ bond. 8. (a) 2 σ 2 π; (b) 1 σ 2 π; Why is the concept of hybridization required in valence bond theory? Hybridization is introduced to explain the geometry of bonding orbitals in valance bond theory. Give the shape that describes each hybrid orbital set: (a) (b) (c) (d) Explain why a carbon atom cannot form five bonds using hybrid orbitals. There are no orbitals in the valence shell of carbon. What is the hybridization of the central atom in each of the following? (a) BeH (b) SF (c) \(\ce{PO4^3-}\) (d) PCl A molecule with the formula AB could have one of four different shapes. Give the shape and the hybridization of the central A atom for each. trigonal planar, ; trigonal pyramidal (one lone pair on A) ; T-shaped (two lone pairs on A , or (three lone pairs on A) Methionine, CH SCH CH CH(NH )CO H, is an amino acid found in proteins. Draw a Lewis structure of this compound. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur? Sulfuric acid is manufactured by a series of reactions represented by the following equations: \(\ce{S8}(s)+\ce{8O2}(g)⟶\ce{8SO2}(g)\) \(\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g)\) \(\ce{SO3}(g)+\ce{H2O}(l)⟶\ce{H2SO4}(l)\) Draw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following: (a) circular S molecule (b) SO molecule (c) SO molecule (d) H SO molecule (the hydrogen atoms are bonded to oxygen atoms) (a) Each S has a bent (109°) geometry, (b) Bent (120°), (c) Trigonal planar, (d) Tetrahedral, Two important industrial chemicals, ethene, C H , and propene, C H , are produced by the steam (or thermal) cracking process: For each of the four carbon compounds, do the following: (a) Draw a Lewis structure. (b) Predict the geometry about the carbon atom. (c) Determine the hybridization of each type of carbon atom. For many years after they were discovered, it was believed that the noble gases could not form compounds. Now we know that belief to be incorrect. A mixture of xenon and fluorine gases, confined in a quartz bulb and placed on a windowsill, is found to slowly produce a white solid. Analysis of the compound indicates that it contains 77.55% Xe and 22.45% F by mass. (a) What is the formula of the compound? (b) Write a Lewis structure for the compound. (c) Predict the shape of the molecules of the compound. (d) What hybridization is consistent with the shape you predicted? (a) XeF (b) (c) linear (d) Consider nitrous acid, HNO (HONO). (a) Write a Lewis structure. (b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the HNO molecule? (c) What is the hybridization on the internal oxygen and nitrogen atoms in HNO ? Strike-anywhere matches contain a layer of KClO and a layer of P S . The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. KClO contains the \(\ce{ClO3-}\) ion. P S is an unusual molecule with the skeletal structure. (a) Write Lewis structures for P S and the \(\ce{ClO3-}\) ion. (b) Describe the geometry about the P atoms, the S atom, and the Cl atom in these species. (c) Assign a hybridization to the P atoms, the S atom, and the Cl atom in these species. (d) Determine the oxidation states and formal charge of the atoms in P S and the \(\ce{ClO3-}\) ion. (a) (b) P atoms, trigonal pyramidal; S atoms, bent, with two lone pairs; Cl atoms, trigonal pyramidal; (c) Hybridization about P, S, and Cl is, in all cases, ; (d) Oxidation states P +1, S \(−1\dfrac{1}{3}\), Cl +5, O –2. Formal charges: P 0; S 0; Cl +2: O –1 Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; you need to determine how many bonds connect each pair of atoms.) Write Lewis structures for NF and PF . On the basis of hybrid orbitals, explain the fact that NF , PF , and PF are stable molecules, but NF does not exist. Phosphorus and nitrogen can form hybrids to form three bonds and hold one lone pair in PF and NF , respectively. However, nitrogen has no valence orbitals, so it cannot form a set of hybrid orbitals to bind five fluorine atoms in NF . Phosphorus has orbitals and can bind five fluorine atoms with hybrid orbitals in PF . In addition to NF , two other fluoro derivatives of nitrogen are known: N F and N F . What shapes do you predict for these two molecules? What is the hybridization for the nitrogen in each molecule? The bond energy of a C–C single bond averages 347 kJ mol ; that of a C≡C triple bond averages 839 kJ mol . Explain why the triple bond is not three times as strong as a single bond. A triple bond consists of one σ bond and two π bonds. A σ bond is stronger than a π bond due to greater overlap. For the carbonate ion, \(\ce{CO3^2-}\), draw all of the resonance structures. Identify which orbitals overlap to create each bond. A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, H CCN. It is present in paint strippers. (a) Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule. (b) Identify the hybrid orbitals used by the carbon atoms in the molecule to form σ bonds. (c) Describe the atomic orbitals that form the π bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom. (a) (b) The terminal carbon atom uses hybrid orbitals, while the central carbon atom is hybridized. (c) Each of the two π bonds is formed by overlap of a 2 orbital on carbon and a nitrogen 2 orbital. For the molecule allene, \(\mathrm{H_2C=C=CH_2}\), give the hybridization of each carbon atom. Will the hydrogen atoms be in the same plane or perpendicular planes? Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds: (a) ClNO (N is the central atom) (b) CS (c) Cl CO (C is the central atom) (d) Cl SO (S is the central atom) (e) SO F (S is the central atom) (f) XeO F (Xe is the central atom) (g) \(\ce{ClOF2+}\) (Cl is the central atom) (a) ; (b) ; (c) ; (d) ; (e) ; (f) ; (g) Describe the molecular geometry and hybridization of the N, P, or S atoms in each of the following compounds. (a) H PO , phosphoric acid, used in cola soft drinks (b) NH NO , ammonium nitrate, a fertilizer and explosive (c) S Cl , disulfur dichloride, used in vulcanizing rubber (d) K [O POPO ], potassium pyrophosphate, an ingredient in some toothpastes For each of the following molecules, indicate the hybridization requested and whether or not the electrons will be delocalized: (a) ozone (O ) central O hybridization (b) carbon dioxide (CO ) central C hybridization (c) nitrogen dioxide (NO ) central N hybridization (d) phosphate ion (\(\ce{PO4^3-}\)) central P hybridization (a) , delocalized; (b) , localized; (c) , delocalized; (d) , delocalized For each of the following structures, determine the hybridization requested and whether the electrons will be delocalized: (a) Hybridization of each carbon (b) Hybridization of sulfur (c) All atoms Draw the orbital diagram for carbon in CO showing how many carbon atom electrons are in each orbital. Each of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom. Sketch the distribution of electron density in the bonding and antibonding molecular orbitals formed from two orbitals and from two orbitals. How are the following similar, and how do they differ? (a) σ molecular orbitals and π molecular orbitals (b) for an atomic orbital and for a molecular orbital (c) bonding orbitals and antibonding orbitals (a) Similarities: Both are bonding orbitals that can contain a maximum of two electrons. Differences: σ orbitals are end-to-end combinations of atomic orbitals, whereas π orbitals are formed by side-by-side overlap of orbitals. (b) Similarities: Both are quantum-mechanical constructs that represent the probability of finding the electron about the atom or the molecule. Differences: for an atomic orbital describes the behavior of only one electron at a time based on the atom. For a molecule, represents a mathematical combination of atomic orbitals. (c) Similarities: Both are orbitals that can contain two electrons. Differences: Bonding orbitals result in holding two or more atoms together. Antibonding orbitals have the effect of destabilizing any bonding that has occurred. If molecular orbitals are created by combining five atomic orbitals from atom A and five atomic orbitals from atom B combine, how many molecular orbitals will result? Can a molecule with an odd number of electrons ever be diamagnetic? Explain why or why not. An odd number of electrons can never be paired, regardless of the arrangement of the molecular orbitals. It will always be paramagnetic. Can a molecule with an even number of electrons ever be paramagnetic? Explain why or why not. Why are bonding molecular orbitals lower in energy than the parent atomic orbitals? Bonding orbitals have electron density in close proximity to more than one nucleus. The interaction between the bonding positively charged nuclei and negatively charged electrons stabilizes the system. Calculate the bond order for an ion with this configuration: Explain why an electron in the bonding molecular orbital in the H molecule has a lower energy than an electron in the 1 atomic orbital of either of the separated hydrogen atoms. The pairing of the two bonding electrons lowers the energy of the system relative to the energy of the nonbonded electrons. Predict the valence electron molecular orbital configurations for the following, and state whether they will be stable or unstable ions. (a) \(\ce{Na2^2+}\) (b) \(\ce{Mg2^2+}\) (c) \(\ce{Al2^2+}\) (d) \(\ce{Si2^2+}\) (e) \(\ce{P2^2+}\) (f) \(\ce{S2^2+}\) (g) \(\ce{F2^2+}\) (h) \(\ce{Ar2^2+}\) Determine the bond order of each member of the following groups, and determine which member of each group is predicted by the molecular orbital model to have the strongest bond. (a) H , \(\ce{H2+}\), \(\ce{H2-}\) (b) O , \(\ce{O2^2+}\), \(\ce{O2^2-}\) (c) Li , \(\ce{Be2+}\), Be (d) F , \(\ce{F2+}\), \(\ce{F2-}\) (e) N , \(\ce{N2+}\), \(\ce{N2-}\) (a) H bond order = 1, \(\ce{H2+}\) bond order = 0.5, \(\ce{H2-}\) bond order = 0.5, strongest bond is H ; (b) O bond order = 2, \(\ce{O2^2+}\) bond order = 3; \(\ce{O2^2-}\) bond order = 1, strongest bond is \(\ce{O2^2+}\); (c) Li bond order = 1, \(\ce{Be2+}\) bond order = 0.5, Be bond order = 0, strongest bond is \(\ce{Li2}\);(d) F bond order = 1, \(\ce{F2+}\) bond order = 1.5, \(\ce{F2-}\) bond order = 0.5, strongest bond is \(\ce{F2+}\); (e) N bond order = 3, \(\ce{N2+}\) bond order = 2.5, \(\ce{N2-}\) bond order = 2.5, strongest bond is N For the first ionization energy for an N molecule, what molecular orbital is the electron removed from? Compare the atomic and molecular orbital diagrams to identify the member of each of the following pairs that has the highest first ionization energy (the most tightly bound electron) in the gas phase: (a) H and H (b) N and N (c) O and O (d) C and C (e) B and B (a) H ; (b) N ; (c) O; (d) C ; (e) B Which of the period 2 homonuclear diatomic molecules are predicted to be paramagnetic? A friend tells you that the 2 orbital for fluorine starts off at a much lower energy than the 2 orbital for lithium, so the resulting σ molecular orbital in F is more stable than in Li . Do you agree? Yes, fluorine is a smaller atom than Li, so atoms in the 2 orbital are closer to the nucleus and more stable. True or false: Boron contains 2 2 valence electrons, so only one orbital is needed to form molecular orbitals. What charge would be needed on F to generate an ion with a bond order of 2? 2+ Predict whether the MO diagram for S would show s-p mixing or not. Explain why \(\ce{N2^2+}\) is diamagnetic, while \(\ce{O2^4+}\), which has the same number of valence electrons, is paramagnetic. N has s-p mixing, so the π orbitals are the last filled in \(\ce{N2^2+}\). O does not have s-p mixing, so the σ orbital fills before the π orbitals. Using the MO diagrams, predict the bond order for the stronger bond in each pair: (a) B or \(\ce{B2+}\) (b) F or \(\ce{F2+}\) (c) O or \(\ce{O2^2+}\) (d) \(\ce{C2+}\) or \(\ce{C2-}\) | 13,973 | 3,316 |
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The rate of a chemical reaction (or the ) can be defined by the time needed for a change in concentration to occur. But there is a problem in that this allows for the definition to be made based on concentration changes for either the reactants or the products. Plus, due to stoichiometric concerns, the rates at which the concentrations are generally different! Toward this end, the following convention is used. For a general reaction \[a A + b B \rightarrow c C + d D \nonumber \] the reaction rate can be defined by any of the ratios \[\text{rate} = - \dfrac{1}{a} \dfrac{\Delta [A]}{dt} = - \dfrac{1}{b} \dfrac{\Delta[B]}{dt} = + \dfrac{1}{c} \dfrac{\Delta [C]}{dt} = + \dfrac{1}{d} \dfrac{ \Delta [D]}{dt} \nonumber \] Or for infinitesimal time intervals \[\text{rate} = - \dfrac{1}{a} \dfrac{d[A]}{dt} = - \dfrac{1}{b} \dfrac{d[C]}{dt} = + \dfrac{1}{c} \dfrac{d[C]}{dt} = + \dfrac{1}{d} \dfrac{d[D]}{dt} \nonumber \] Under a certain set of conditions, the rate of the reaction \[N_2 + 3 H_2 \rightarrow 2 NH_3 \nonumber \] the reaction rate is \(6.0 \times 10^{-4}\, M/s\). Calculate the time-rate of change for the concentrations of N , H , and NH . Due to the stoichiometry of the reaction, \[\text{rate} = - \dfrac{d[N_2]}{dt} = - \dfrac{1}{3} \dfrac{d[H_2]}{dt} = + \dfrac{1}{2} \dfrac{d[NH_3]}{dt} \nonumber \] so \[\dfrac{d[N_2]}{dt} = -6.0 \times 10^{-4} \,M/s \nonumber \] \[\dfrac{d[H_2]}{dt} = -2.0 \times 10^{-4} \,M/s \nonumber \] \[\dfrac{d[NH_3]}{dt} = 3.0 \times 10^{-4} \,M/s \nonumber \] : The time derivatives for the reactants are negative because the reactant concentrations are decreasing, and those of products are positive since the concentrations of products increase as the reaction progresses. | 1,740 | 3,317 |
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In precipitation gravimetry an insoluble compound forms when we add a precipitating reagent, or , to a solution that contains our analyte. In most cases the precipitate is the product of a simple metathesis reaction between the analyte and the precipitant; however, any reaction that generates a precipitate potentially can serve as a gravimetric method. Most precipitation gravimetric methods were developed in the nineteenth century, or earlier, often for the analysis of ores. in Chapter 1, for example, illustrates a precipitation gravimetric method for the analysis of nickel in ores. All precipitation gravimetric analyses share two important attributes. First, the precipitate must be of low solubility, of high purity, and of known composition if its mass is to reflect accurately the analyte’s mass. Second, it must be easy to separate the precipitate from the reaction mixture. To provide an accurate result, a precipitate’s solubility must be minimal. The accuracy of a total analysis technique typically is better than ±0.1%, which means the precipitate must account for at least 99.9% of the analyte. Extending this requirement to 99.99% ensures the precipitate’s solubility will not limit the accuracy of a gravimetric analysis. A total analysis technique is one in which the analytical signal—mass in this case—is proportional to the absolute amount of analyte in the sample. See for a discussion of the difference between total analysis techniques and concentration techniques. We can minimize solubility losses by controlling the conditions under which the precipitate forms. This, in turn, requires that we account for every equilibrium reaction that might affect the precipitate’s solubility. For example, we can determine Ag gravimetrically by adding NaCl as a precipitant, forming a precipitate of AgCl. \[\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}(s) \label{8.1}\] If this is the only reaction we consider, then we predict that the precipitate’s solubility, , is given by the following equation. \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]=\frac{K_{\mathrm{sp}}}{\left[\mathrm{Cl}^{-}\right]} \label{8.2}\] Equation \ref{8.2} suggests that we can minimize solubility losses by adding a large excess of Cl . In fact, as shown in Figure 8.2.1
, adding a large excess of Cl increases the precipitate’s solubility. To understand why the solubility of AgCl is more complicated than the relationship suggested by Equation \ref{8.2}, we must recall that Ag also forms a series of soluble silver-chloro metal–ligand complexes. \[\operatorname{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}(a q) \quad \log K_{1}=3.70 \label{8.3}\] \[\operatorname{AgCl}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}_{2}(a q) \quad \log K_{2}=1.92 \label{8.4}\] \[\mathrm{AgCl}_{2}^{-}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}_{3}^{2-}(a q) \quad \log K_{3}=0.78 \label{8.5}\] Note the difference between reaction \ref{8.3}, in which we form AgCl( ) as a product, and reaction \ref{8.1}, in which we form AgCl( ) as a product. The formation of AgCl( ) from AgCl( ) \[\operatorname{AgCl}(s)\rightleftharpoons\operatorname{AgCl}(a q) \nonumber\] is called AgCl’s intrinsic solubility. The actual solubility of AgCl is the sum of the equilibrium concentrations for all soluble forms of Ag . \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]+[\mathrm{AgCl}(a q)]+\left[\mathrm{AgCl}_{2}^-\right]+\left[\mathrm{AgCl}_{3}^{2-}\right] \label{8.6}\] By substituting into Equation \ref{8.6} the equilibrium constant expressions for reaction \ref{8.1} and reactions \ref{8.3}–\ref{8.5}, we can define the solubility of AgCl as \[S_\text{AgCl} = \frac {K_\text{sp}} {[\text{Cl}^-]} + K_1K_\text{sp} + K_1K_2K_\text{sp}[\text{Cl}^-]+K_1K_2K_3K_\text{sp}[\text{Cl}^-]^2 \label{8.7}\] Equation \ref{8.7} explains the solubility curve for AgCl shown in . As we add NaCl to a solution of Ag , the solubility of AgCl initially decreases because of reaction \ref{8.1}. Under these conditions, the final three terms in Equation \ref{8.7} are small and Equation \ref{8.2} is sufficient to describe AgCl’s solubility. For higher concentrations of Cl , reaction \ref{8.4} and reaction \ref{8.5} increase the solubility of AgCl. Clearly the equilibrium concentration of chloride is important if we wish to determine the concentration of silver by precipitating AgCl. In particular, we must avoid a large excess of chloride. The predominate silver-chloro complexes for different values of pCl are shown by the ladder diagram along the -axis in . Note that the increase in solubility begins when the higher-order soluble complexes of \(\text{AgCl}_2^-\) and \(\text{AgCl}_3^{2-}\) are the predominate species. Another important parameter that may affect a precipitate’s solubility is pH. For example, a hydroxide precipitate, such as Fe(OH) , is more soluble at lower pH levels where the concentration of OH is small. Because fluoride is a weak base, the solubility of calcium fluoride, \(S_{\text{CaF}_2}\), also is pH-dependent. We can derive an equation for \(S_{\text{CaF}_2}\) by considering the following equilibrium reactions \[\mathrm{CaF}_{2}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{F}^{-}(a q) \quad K_{\mathfrak{sp}}=3.9 \times 10^{-11} \label{8.8}\] \[\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l )\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q) \quad K_{\mathrm{a}}=6.8 \times 10^{-4} \label{8.9}\] and the following equation for the solubility of CaF . \[S_{\mathrm{Ca} \mathrm{F}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\frac{1}{2}\left\{\left[\mathrm{F}^{-}\right]+[\mathrm{HF}]\right\} \label{8.10}\] Be sure that Equation \ref{8.10} makes sense to you. Reaction \ref{8.8} tells us that the dissolution of CaF produces one mole of Ca for every two moles of F , which explains the term of 1/2 in Equation \ref{8.10}. Because F is a weak base, we must account for both chemical forms in solution, which explains why we include HF. Substituting the equilibrium constant expressions for reaction \ref{8.8} and reaction \ref{8.9} into Equation \ref{8.10} allows us to define the solubility of CaF in terms of the equilibrium concentration of H O . \[S_{\mathrm{CaF}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\left\{\frac{K_{\mathrm{p}}}{4}\left(1+\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{K_{\mathrm{a}}}\right)^{2}\right\}^{1 / 3} \label{8.11}\] Figure 8.2.2
shows how pH affects the solubility of CaF . Depending on the solution’s pH, the predominate form of fluoride is either HF or F . When the pH is greater than 4.17, the predominate species is F and the solubility of CaF is independent of pH because only reaction \ref{8.8} occurs to an appreciable extent. At more acidic pH levels, the solubility of CaF increases because of the contribution of reaction \ref{8.9}. You can use a ladder diagram to predict the conditions that will minimize a precipitate’s solubility. Draw a ladder diagram for oxalic acid, H C2O , and use it to predict the range of pH values that will minimize the solubility of CaC O . Relevant equilibrium constants are in the appendices. The solubility reaction for CaC O is \[\mathrm{CaC}_{2} \mathrm{O}_{4}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \nonumber\] To minimize solubility, the pH must be sufficiently basic that oxalate, \(\text{C}_2\text{O}_4^{2-}\), does not react to form \(\text{HC}_2\text{O}_4^{-}\) or H C O . The ladder diagram for oxalic acid, including approximate buffer ranges, is shown below. Maintaining a pH greater than 5.3 ensures that \(\text{C}_2\text{O}_4^{2-}\) is the only important form of oxalic acid in solution, minimizing the solubility of CaC O . When solubility is a concern, it may be possible to decrease solubility by using a non-aqueous solvent. A precipitate’s solubility generally is greater in an aqueous solution because of water’s ability to stabilize ions through solvation. The poorer solvating ability of a non-aqueous solvent, even those that are polar, leads to a smaller solubility product. For example, the of PbSO is \(2 \times 10^{-8}\) in H O and \(2.6 \times 10^{-12}\) in a 50:50 mixture of H O and ethanol. In addition to having a low solubility, a precipitate must be free from impurities. Because precipitation usually occurs in a solution that is rich in dissolved solids, the initial precipitate often is impure. To avoid a determinate error, we must remove these impurities before we determine the precipitate’s mass. The greatest source of impurities are chemical and physical interactions that take place at the precipitate’s surface. A precipitate generally is crystalline—even if only on a microscopic scale—with a well-defined lattice of cations and anions. Those cations and anions at the precipitate’s surface carry, respectively, a positive or a negative charge because they have incomplete coordination spheres. In a precipitate of AgCl, for example, each silver ion in the precipitate’s interior is bound to six chloride ions. A silver ion at the surface, however, is bound to no more than five chloride ions and carries a partial positive charge (Figure 8.2.3
). The presence of these partial charges makes the precipitate’s surface an active site for the chemical and physical interactions that produce impurities. One common impurity is an , in which a potential interferent, whose size and charge is similar to a lattice ion, can substitute into the lattice structure if the interferent precipitates with the same crystal structure (Figure 8.2.4
a). The probability of forming an inclusion is greatest when the interfering ion’s concentration is substantially greater than the lattice ion’s concentration. An inclusion does not decrease the amount of analyte that precipitates, provided that the precipitant is present in sufficient excess. Thus, the precipitate’s mass always is larger than expected. An inclusion is difficult to remove since it is chemically part of the precipitate’s lattice. The only way to remove an inclusion is through in which we isolate the precipitate from its supernatant solution, dissolve the precipitate by heating in a small portion of a suitable solvent, and then reform the precipitate by allowing the solution to cool. Because the interferent’s concentration after dissolving the precipitate is less than that in the original solution, the amount of included material decreases upon reprecipitation. We can repeat the process of reprecipitation until the inclusion’s mass is insignificant. The loss of analyte during reprecipitation, however, is a potential source of determinate error. Suppose that 10% of an interferent forms an inclusion during each precipitation. When we initially form the precipitate, 10% of the original interferent is present as an inclusion. After the first reprecipitation, 10% of the included interferent remains, which is 1% of the original interferent. A second reprecipitation decreases the interferent to 0.1% of the original amount. An forms when an interfering ions is trapped within the growing precipitate. Unlike an inclusion, which is randomly dispersed within the precipitate, an occlusion is localized, either along flaws within the precipitate’s lattice structure or within aggregates of individual precipitate particles (Figure 8.2.4
b). An occlusion usually increases a precipitate’s mass; however, the precipitate’s mass is smaller if the occlusion includes the analyte in a lower molecular weight form than that of the precipitate. We can minimize an occlusion by maintaining the precipitate in equilibrium with its supernatant solution for an extended time, a process called digestion. During a , the dynamic nature of the solubility–precipitation equilibria, in which the precipitate dissolves and reforms, ensures that the occlusion eventually is reexposed to the supernatant solution. Because the rates of dissolution and reprecipitation are slow, there is less opportunity for forming new occlusions. After precipitation is complete the surface continues to attract ions from solution (Figure 8.2.4
c). These comprise a third type of impurity. We can minimize surface adsorption by decreasing the precipitate’s available surface area. One benefit of digestion is that it increases a precipitate’s average particle size. Because the probability that a particle will dissolve completely is inversely proportional to its size, during digestion larger particles increase in size at the expense of smaller particles. One consequence of forming a smaller number of larger particles is an overall decrease in the precipitate’s surface area. We also can remove surface adsorbates by washing the precipitate, although we cannot ignore the potential loss of analyte. Inclusions, occlusions, and surface adsorbates are examples of —otherwise soluble species that form along with the precipitate that contains the analyte. Another type of impurity is an interferent that forms an independent precipitate under the conditions of the analysis. For example, the precipitation of nickel dimethylglyoxime requires a slightly basic pH. Under these conditions any Fe in the sample will precipitate as Fe(OH) . In addition, because most precipitants rarely are selective toward a single analyte, there is a risk that the precipitant will react with both the analyte and an interferent. In addition to forming a precipitate with Ni , dimethylglyoxime also forms precipitates with Pd and Pt . These cations are potential interferents in an analysis for nickel. We can minimize the formation of additional precipitates by controlling solution conditions. If an interferent forms a precipitate that is less soluble than the analyte’s precipitate, we can precipitate the interferent and remove it by filtration, leaving the analyte behind in solution. Alternatively, we can mask the analyte or the interferent to prevent its precipitation. Both of the approaches outline above are illustrated in Fresenius’ analytical method for the determination of Ni in ores that contain Pb , Cu , and Fe (see in Chapter 1). Dissolving the ore in the presence of H SO selectively precipitates Pb as PbSO . Treating the resulting supernatant with H S precipitates Cu as CuS. After removing the CuS by filtration, ammonia is added to precipitate Fe as Fe(OH) . Nickel, which forms a soluble amine complex, remains in solution. Masking was introduced in . Size matters when it comes to forming a precipitate. Larger particles are easier to filter and, as noted earlier, a smaller surface area means there is less opportunity for surface adsorbates to form. By controlling the reaction conditions we can significantly increase a precipitate’s average particle size. The formation of a precipitate consists of two distinct events: nucleation, the initial formation of smaller, stable particles of the precipitate, and particle growth. Larger particles form when the rate of particle growth exceeds the rate of nucleation. Understanding the conditions that favor particle growth is important when we design a gravimetric method of analysis. We define a solute’s , , as \[R S S=\frac{Q-S}{S} \label{8.12}\] where is the solute’s actual concentration and is the solute’s concentration at equilibrium [Von Weimarn, P. P. . , , 217–242]. The numerator of Equation \ref{8.12}, – , is a measure of the solute’s supersaturation. A solution with a large, positive value of has a high rate of nucleation and produces a precipitate with many small particles. When the is small, precipitation is more likely to occur by particle growth than by nucleation. A supersaturated solution is one that contains more dissolved solute than that predicted by equilibrium chemistry. A supersaturated solution is inherently unstable and precipitates solute to reach its equilibrium position. How quickly precipitation occurs depends, in part, on the value of . Equation \ref{8.12} suggests that we can minimize if we decrease the solute’s concentration, , or if we increase the precipitate’s solubility, . A precipitate’s solubility usually increases at higher temperatures and adjusting pH may affect a precipitate’s solubility if it contains an acidic or a basic ion. Temperature and pH, therefore, are useful ways to increase the value of . Forming the precipitate in a dilute solution of analyte or adding the precipitant slowly and with vigorous stirring are ways to decrease the value of There are practical limits to minimizing . Some precipitates, such as Fe(OH) and PbS, are so insoluble that is very small and a large is unavoidable. Such solutes inevitably form small particles. In addition, conditions that favor a small may lead to a relatively stable supersaturated solution that requires a long time to precipitate fully. For example, almost a month is required to form a visible precipitate of BaSO under conditions in which the initial is 5 [Bassett, J.; Denney, R. C.; Jeffery, G. H. Mendham. J. , Longman: London, 4th Ed., 1981, p. 408]. A visible precipitate takes longer to form when is small both because there is a slow rate of nucleation and because there is a steady decrease in as the precipitate forms. One solution to the latter problem is to generate the precipitant as the product of a slow chemical reaction, which effectively maintains a constant . Because the precipitate forms under conditions of low , initial nucleation produces a small number of particles. As additional precipitant forms, particle growth supersedes nucleation, which results in larger particles of precipitate. This process is called a [Gordon, L.; Salutsky, M. L.; Willard, H. H. , Wiley: NY, 1959]. Two general methods are used for homogeneous precipitation. If the precipitate’s solubility is pH-dependent, then we can mix the analyte and the precipitant under conditions where precipitation does not occur, and then increase or decrease the pH by chemically generating OH or H O . For example, the hydrolysis of urea, CO(NH ) , is a source of OH because of the following two reactions. \[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons2 \mathrm{NH}_{3}(a q)+\mathrm{CO}_{2}(g) \nonumber\] \[\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons\mathrm{OH}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q) \nonumber\] Because the hydrolysis of urea is temperature-dependent—the rate is negligible at room temperature—we can use temperature to control the rate of hydrolysis and the rate of precipitate formation. Precipitates of CaC O , for example, have been produced by this method. After dissolving a sample that contains Ca , the solution is made acidic with HCl before adding a solution of 5% w/v (NH ) C O . Because the solution is acidic, a precipitate of CaC O does not form. The solution is heated to approximately 50 C and urea is added. After several minutes, a precipitate of CaC O begins to form, with precipitation reaching completion in about 30 min. In the second method of homogeneous precipitation, the precipitant is generated by a chemical reaction. For example, Pb is precipitated homogeneously as PbCrO by using bromate, \(\text{BrO}_3^-\), to oxidize Cr to \(\text{CrO}_4^{2-}\). \[6 \mathrm{BrO}_{3}^{-}(a q)+10 \mathrm{Cr}^{3+}(a q)+22 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons 3 \mathrm{Br}_{2}(a q)+10 \mathrm{CrO}_{4}^{2-}(a q)+44 \mathrm{H}^{+}(a q) \nonumber\] Figure 8.2.5
shows the result of preparing PbCrO by direct addition of K CrO (Beaker A) and by homogenous precipitation (Beaker B). Both beakers contain the same amount of PbCrO . Because the direct addition of K CrO leads to rapid precipitation and the formation of smaller particles, the precipitate remains less settled than the precipitate prepared homogeneously. Note, as well, the difference in the color of the two precipitates. The effect of particle size on color is well-known to geologists, who use a streak test to help identify minerals. The color of a bulk mineral and its color when powdered often are different. Rubbing a mineral across an unglazed porcelain plate leaves behind a small streak of the powdered mineral. Bulk samples of hematite, Fe O , are black in color, but its streak is a familiar rust-red. Crocite, the mineral PbCrO , is red-orange in color; its streak is orange-yellow. A homogeneous precipitation produces large particles of precipitate that are relatively free from impurities. These advantages, however, are offset by the increased time needed to produce the precipitate and by a tendency for the precipitate to deposit as a thin film on the container’s walls. The latter problem is particularly severe for hydroxide precipitates generated using urea. An additional method for increasing particle size deserves mention. When a precipitate’s particles are electrically neutral they tend to coagulate into larger particles that are easier to filter. Surface adsorption of excess lattice ions, however, provides the precipitate’s particles with a net positive or a net negative surface charge. Electrostatic repulsion between particles of similar charge prevents them from coagulating into larger particles. Let’s use the precipitation of AgCl from a solution of AgNO using NaCl as a precipitant to illustrate this effect. Early in the precipitation, when NaCl is the limiting reagent, excess Ag ions chemically adsorb to the AgCl particles, forming a positively charged primary adsorption layer (Figure 8.2.6
a). The solution in contact with this layer contains more inert anions, \(\text{NO}_3^-\) in this case, than inert cations, Na , giving a secondary adsorption layer with a negative charge that balances the primary adsorption layer’s positive charge. The solution outside the secondary adsorption layer remains electrically neutral. cannot occur if the secondary adsorption layer is too thick because the individual particles of AgCl are unable to approach each other closely enough. We can induce coagulation in three ways: by decreasing the number of chemically adsorbed Ag ions, by increasing the concentration of inert ions, or by heating the solution. As we add additional NaCl, precipitating more of the excess Ag , the number of chemically adsorbed silver ions decreases and coagulation occurs (Figure 8.2.6
b). Adding too much NaCl, however, creates a primary adsorption layer of excess Cl with a loss of coagulation. The coagulation and decoagulation of AgCl as we add NaCl to a solution of AgNO can serve as an endpoint for a titration. See for additional details. A second way to induce coagulation is to add an inert electrolyte, which increases the concentration of ions in the secondary adsorption layer (Figure 8.2.6
c). With more ions available, the thickness of the secondary absorption layer decreases. Particles of precipitate may now approach each other more closely, which allows the precipitate to coagulate. The amount of electrolyte needed to cause spontaneous coagulation is called the critical coagulation concentration. Heating the solution and the precipitate provides a third way to induce coagulation. As the temperature increases, the number of ions in the primary adsorption layer decreases, which lowers the precipitate’s surface charge. In addition, heating increases the particles’ kinetic energy, allowing them to overcome the electrostatic repulsion that prevents coagulation at lower temperatures. After precipitating and digesting a precipitate, we separate it from solution by filtering. The most common filtration method uses filter paper, which is classified according to its speed, its size, and its ash content on ignition. Speed, or how quickly the supernatant passes through the filter paper, is a function of the paper’s pore size. A larger pore size allows the supernatant to pass more quickly through the filter paper, but does not retain small particles of precipitate. Filter paper is rated as fast (retains particles larger than 20–25 μm), medium–fast (retains particles larger than 16 μm), medium (retains particles larger than 8 μm), and slow (retains particles larger than 2–3 μm). The proper choice of filtering speed is important. If the filtering speed is too fast, we may fail to retain some of the precipitate, which causes a negative determinate error. On the other hand, the precipitate may clog the pores if we use a filter paper that is too slow. A filter paper’s size is just its diameter. Filter paper comes in many sizes, including 4.25 cm, 7.0 cm, 11.0 cm, 12.5 cm, 15.0 cm, and 27.0 cm. Choose a size that fits comfortably into your funnel. For a typical 65-mm long-stem funnel, 11.0 cm and 12.5 cm filter paper are good choices. Because filter paper is hygroscopic, it is not easy to dry it to a constant weight. When accuracy is important, the filter paper is removed before we determine the precipitate’s mass. After transferring the precipitate and filter paper to a covered crucible, we heat the crucible to a temperature that coverts the paper to CO and H O , a process called . Igniting a poor quality filter paper leaves behind a residue of inorganic ash. For quantitative work, use a low-ash filter paper. This grade of filter paper is pretreated with a mixture of HCl and HF to remove inorganic materials. Quantitative filter paper typically has an ash content of less than 0.010% w/w. Gravity filtration is accomplished by folding the filter paper into a cone and placing it in a long-stem funnel (Figure 8.2.7
). To form a tight seal between the filter cone and the funnel, we dampen the paper with water or supernatant and press the paper to the wall of the funnel. When prepared properly, the funnel’s stem fills with the supernatant, increasing the rate of filtration. The precipitate is transferred to the filter in several steps. The first step is to decant the majority of the through the filter paper without transferring the precipitate (Figure 8.2.8
). This prevents the filter paper from clogging at the beginning of the filtration process. The precipitate is rinsed while it remains in its beaker, with the rinsings decanted through the filter paper. Finally, the precipitate is transferred onto the filter paper using a stream of rinse solution. Any precipitate that clings to the walls of the beaker is transferred using a rubber policeman (a flexible rubber spatula attached to the end of a glass stirring rod). An alternative method for filtering a precipitate is to use a filtering crucible. The most common option is a fritted-glass crucible that contains a porous glass disk filter. Fritted-glass crucibles are classified by their porosity: coarse (retaining particles larger than 40–60 μm), medium (retaining particles greater than 10–15 μm), and fine (retaining particles greater than 4–5.5 μm). Another type of filtering crucible is the Gooch crucible, which is a porcelain crucible with a perforated bottom. A glass fiber mat is placed in the crucible to retain the precipitate. For both types of crucibles, the pre- cipitate is transferred in the same manner described earlier for filter paper. Instead of using gravity, the supernatant is drawn through the crucible with the assistance of suction from a vacuum aspirator or pump (Figure 8.2.9
). Because the supernatant is rich with dissolved inert ions, we must remove residual traces of supernatant without incurring loss of analyte due to solubility. In many cases this simply involves the use of cold solvents or rinse solutions that contain organic solvents such as ethanol. The pH of the rinse solution is critical if the precipitate contains an acidic or a basic ion. When coagulation plays an important role in determining particle size, adding a volatile inert electrolyte to the rinse solution prevents the precipitate from reverting into smaller particles that might pass through the filter. This process of reverting to smaller particles is called . The volatile electrolyte is removed when drying the precipitate. In general, we can minimize the loss of analyte if we use several small portions of rinse solution instead of a single large volume. Testing the used rinse solution for the presence of an impurity is another way to guard against over-rinsing the precipitate. For example, if Cl is a residual ion in the supernatant, we can test for its presence using AgNO . After we collect a small portion of the rinse solution, we add a few drops of AgNO and look for the presence or absence of a precipitate of AgCl. If a precipitate forms, then we know Cl is present and continue to rinse the precipitate. Additional rinsing is not needed if the AgNO does not produce a precipitate. After separating the precipitate from its supernatant solution, we dry the precipitate to remove residual traces of rinse solution and to remove any volatile impurities. The temperature and method of drying depend on the method of filtration and the precipitate’s desired chemical form. Placing the precipitate in a laboratory oven and heating to a temperature of 110 C is sufficient to remove water and other easily volatilized impurities. Higher temperatures require a muffle furnace, a Bunsen burner, or a Meker burner, and are necessary if we need to decompose the precipitate before its weight is determined. Because filter paper absorbs moisture, we must remove it before we weigh the precipitate. This is accomplished by folding the filter paper over the precipitate and transferring both the filter paper and the precipitate to a porcelain or platinum crucible. Gentle heating first dries and then chars the filter paper. Once the paper begins to char, we slowly increase the temperature until there is no trace of the filter paper and any remaining carbon is oxidized to CO . Fritted-glass crucibles can not withstand high temperatures and are dried in an oven at a temperature below 200 C. The glass fiber mats used in Gooch crucibles can be heated to a maximum temperature of approximately 500 C. For a quantitative application, the final precipitate must have a well-defined composition. A precipitate that contains volatile ions or substantial amounts of hydrated water, usually is dried at a temperature that completely removes these volatile species. For example, one standard gravimetric method for the determination of magnesium involves its precipitation as MgNH PO •6H O. Unfortunately, this precipitate is difficult to dry at lower temperatures without losing an inconsistent amount of hydrated water and ammonia. Instead, the precipitate is dried at a temperature greater than 1000 C where it decomposes to magnesium pyrophosphate, Mg P O . An additional problem is encountered if the isolated solid is nonstoichiometric. For example, precipitating Mn as Mn(OH) and heating frequently produces a nonstoichiometric manganese oxide, MnO , where varies between one and two. In this case the nonstoichiometric product is the result of forming a mixture of oxides with different oxidation state of manganese. Other nonstoichiometric compounds form as a result of lattice defects in the crystal structure [Ward, R., ed., , American Chemical Society: Washington, D. C., 1963]. The best way to appreciate the theoretical and practical details discussed in this section is to carefully examine a typical precipitation gravimetric method. Although each method is unique, the determination of Mg in water and wastewater by precipitating MgNH PO • 6H O and isolating Mg P O provides an instructive example of a typical procedure. The description here is based on Method 3500-Mg D in , 19th Ed., American Public Health Asso- ciation: Washington, D. C., 1995. With the publication of the 20th Edition in 1998, this method is no longer listed as an approved method. Magnesium is precipitated as MgNH PO •6H O using (NH ) HPO as the precipitant. The precipitate’s solubility in a neutral solution is relatively high (0.0065 g/100 mL in pure water at 10 C), but it is much less soluble in the presence of dilute ammonia (0.0003 g/100 mL in 0.6 M NH ). Because the precipitant is not selective, a preliminary separation of Mg from potential interferents is necessary. Calcium, which is the most significant interferent, is removed by precipitating it as CaC O . The presence of excess ammonium salts from the precipitant, or from the addition of too much ammonia, leads to the formation of Mg(NH ) (PO ) , which forms Mg(PO ) after drying. The precipitate is isolated by gravity filtration, using a rinse solution of dilute ammonia. After filtering, the precipitate is converted to Mg P O and weighed. Transfer a sample that contains no more than 60 mg of Mg into a 600-mL beaker. Add 2–3 drops of methyl red indicator, and, if necessary, adjust the volume to 150 mL. Acidify the solution with 6 M HCl and add 10 mL of 30% w/v (NH ) HPO . After cooling and with constant stirring, add concentrated NH dropwise until the methyl red indicator turns yellow (pH > 6.3). After stirring for 5 min, add 5 mL of concentrated NH and continue to stir for an additional 10 min. Allow the resulting solution and precipitate to stand overnight. Isolate the precipitate by filtering through filter paper, rinsing with 5% v/v NH . Dissolve the precipitate in 50 mL of 10% v/v HCl and precipitate a second time following the same procedure. After filtering, carefully remove the filter paper by charring. Heat the precipitate at 500 C until the residue is white, and then bring the precipitate to constant weight at 1100 C. 1. Why does the procedure call for a sample that contains no more than 60 mg of Mg ? A 60-mg portion of Mg generates approximately 600 mg of MgNH PO •6H O, which is a substantial amount of precipitate. A larger quantity of precipitate is difficult to filter and difficult to rinse free of impurities. 2. Why is the solution acidified with HCl before we add the precipitant? The HCl ensures that MgNH PO • 6H O does not precipitate immediately upon adding the precipitant. Because \(\text{PO}_4^{3-}\) is a weak base, the precipitate is soluble in a strongly acidic solution. If we add the precipitant under neutral or basic conditions (that is, a high ), then the resulting precipitate will consist of smaller, less pure particles. Increasing the pH by adding base allows the precipitate to form under more favorable (that is, a low ) conditions. 3. Why is the acid–base indicator methyl red added to the solution? The indicator changes color at a pH of approximately 6.3, which indicates that there is sufficient NH to neutralize the HCl added at the beginning of the procedure. The amount of NH is crucial to this procedure. If we add insufficient NH , then the solution is too acidic, which increases the precipitate’s solubility and leads to a negative determinate error. If we add too much NH , the precipitate may contain traces of Mg(NH ) (PO ) , which, on drying, forms Mg(PO ) instead of Mg P O . This increases the mass of the ignited precipitate, and gives a positive determinate error. After adding enough NH to neutralize the HCl, we add an additional 5 mL of NH to complete the quantitative precipitation of MgNH PO • 6H O. 4. Explain why forming Mg(PO ) instead of Mg P O increases the precipitate’s mass. Each mole of Mg P O contains two moles of magnesium and each mole of Mg(PO ) contains only one mole of magnesium. A conservation of mass, therefore, requires that two moles of Mg(PO ) form in place of each mole of Mg P O . One mole of Mg P O weighs 222.6 g. Two moles of Mg(PO ) weigh 364.5 g. Any replacement of Mg P O with Mg(PO ) must increase the precipitate’s mass. 5. What additional steps, beyond those discussed in questions 2 and 3, help improve the precipitate’s purity? Two additional steps in the procedure help to form a precipitate that is free of impurities: digestion and reprecipitation. 6. Why is the precipitate rinsed with a solution of 5% v/v NH ? This is done for the same reason that the precipitation is carried out in an ammonical solution; using dilute ammonia minimizes solubility losses when we rinse the precipitate. Although no longer a common analytical technique, precipitation gravimetry still provides a reliable approach for assessing the accuracy of other methods of analysis, or for verifying the composition of standard reference materials. In this section we review the general application of precipitation gravimetry to the analysis of inorganic and organic compounds. Table 8.2.1
provides a summary of precipitation gravimetric methods for inorganic cations and anions. Several methods for the homogeneous generation of precipitants are shown in Table 8.2.2
. The majority of inorganic precipitants show poor selectivity for the analyte. Many organic precipitants, however, are selective for one or two inorganic ions. Table 8.2.3
lists examples of several common organic precipitants. Ba \(\text{SO}_4^{2-}\) Precipitation gravimetry continues to be listed as a standard method for the determination of \(\text{SO}_4^{2-}\) in water and wastewater analysis [Method 4500-SO42– C and Method 4500-SO42– D as published in , 20th Ed., American Public Health Association: Wash- ington, D. C., 1998]. Precipitation is carried out using BaCl in an acidic solution (adjusted with HCl to a pH of 4.5–5.0) to prevent the precipitation of BaCO or Ba (PO ) , and at a temperature near the solution’s boiling point. The precipitate is digested at 80–90 C for at least two hours. Ashless filter paper pulp is added to the precipitate to aid in its filtration. After filtering, the precipitate is ignited to constant weight at 800 C. Alternatively, the precipitate is filtered through a fine porosity fritted glass crucible (without adding filter paper pulp), and dried to constant weight at 105 C. This procedure is subject to a variety of errors, including occlusions of Ba(NO ) , BaCl , and alkali sulfates. Other standard methods for the determination of sulfate in water and wastewater include ion chromatography (see ), capillary ion electrophoresis (see ), turbidimetry (see ), and flow injection analysis (see ). Several organic functional groups or heteroatoms can be determined using precipitation gravimetric methods. Table 8.2.4
provides a summary of several representative examples. Note that the determination of alkoxy functional groups is an indirect analysis in which the functional group reacts with and excess of HI and the unreacted I determined by precipitating as AgCl. The stoichiometry of a precipitation reaction provides a mathematical relationship between the analyte and the precipitate. Because a precipitation gravimetric method may involve additional chemical reactions to bring the analyte into a different chemical form, knowing the stoichiometry of the precipitation reaction is not always sufficient. Even if you do not have a complete set of balanced chemical reactions, you can use a conservation of mass to deduce the mathematical relationship between the analyte and the precipitate. The following example demonstrates this approach for the direct analysis of a single analyte. To determine the amount of magnetite, Fe O , in an impure ore, a 1.5419-g sample is dissolved in concentrated HCl, resulting in a mixture of Fe and Fe . After adding HNO to oxidize Fe to Fe and diluting with water, Fe is precipitated as Fe(OH) using NH . Filtering, rinsing, and igniting the precipitate provides 0.8525 g of pure Fe O . Calculate the %w/w Fe O in the sample. A conservation of mass requires that the precipitate of Fe O contain all iron originally in the sample of ore. We know there are 2 moles of Fe per mole of Fe O (FW = 159.69 g/mol) and 3 moles of Fe per mole of Fe O (FW = 231.54 g/mol); thus \[0.8525 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} \times \frac{2 \ \mathrm{mol} \ \mathrm{Fe}}{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}} \times \frac{231.54 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} \nonumber\] The % w/w Fe O in the sample, therefore, is \[\frac{0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{1.5419 \ \mathrm{g} \ \text { sample }} \times 100=53.44 \% \nonumber\] A 0.7336-g sample of an alloy that contains copper and zinc is dissolved in 8 M HCl and diluted to 100 mL in a volumetric flask. In one analysis, the zinc in a 25.00-mL portion of the solution is precipitated as ZnNH PO , and isolated as Zn P O , yielding 0.1163 g. The copper in a separate 25.00-mL portion of the solution is treated to precipitate CuSCN, yielding 0.2383 g. Calculate the %w/w Zn and the %w/w Cu in the sample. A conservation of mass requires that all zinc in the alloy is found in the final product, Zn P O . We know there are 2 moles of Zn per mole of Zn P O ; thus \[0.1163 \ \mathrm{g} \ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7} \times \frac{2 \ \mathrm{mol} \ \mathrm{Zn}}{304.70 \ \mathrm{g}\ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}} \times \frac{65.38 \ \mathrm{g} \ \mathrm{Zn}}{\mathrm{mol} \ \mathrm{Zn}}=0.04991 \ \mathrm{g} \ \mathrm{Zn}\nonumber\] This is the mass of Zn in 25% of the sample (a 25.00 mL portion of the 100.0 mL total volume). The %w/w Zn, therefore, is \[\frac{0.04991 \ \mathrm{g} \ \mathrm{Zn} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=27.21 \% \ \mathrm{w} / \mathrm{w} \mathrm{Zn} \nonumber\] For copper, we find that \[\begin{array}{c}{0.2383 \ \mathrm{g} \ \mathrm{CuSCN} \times \frac{1 \ \mathrm{mol} \ \mathrm{Zn}}{121.63 \ \mathrm{g} \ \mathrm{CuSCN}} \times \frac{63.55 \ \mathrm{g} \ \mathrm{Cu}}{\mathrm{mol} \ \mathrm{Cu}}=0.1245 \ \mathrm{g} \ \mathrm{Cu}} \\ {\frac{0.1245 \ \mathrm{g} \ \mathrm{Cu} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=67.88 \% \ \mathrm{w} / \mathrm{w} \mathrm{Cu}}\end{array} \nonumber\] In Practice Exercise 8.2.2
the sample contains two analytes. Because we can precipitate each analyte selectively, finding their respective concentrations is a straightforward stoichiometric calculation. But what if we cannot separately precipitate the two analytes? To find the concentrations of both analytes, we still need to generate two precipitates, at least one of which must contain both analytes. Although this complicates the calculations, we can still use a conservation of mass to solve the problem. A 0.611-g sample of an alloy that contains Al and Mg is dissolved and treated to prevent interferences by the alloy’s other constituents. Aluminum and magnesium are precipitated using 8-hydroxyquinoline, which yields a mixed precipitate of Al(C H NO) and Mg(C H NO) that weighs 7.815 g. Igniting the precipitate converts it to a mixture of Al O and MgO that weighs 1.002 g. Calculate the %w/w Al and %w/w Mg in the alloy. The masses of the solids provide us with the following two equations. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{g} \ \mathrm{MgO}=1.002 \ \mathrm{g} \nonumber\] With two equations and four unknowns, we need two additional equations to solve the problem. A conservation of mass requires that all the aluminum in Al(C H NO) also is in Al O ; thus \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.43 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{101.96 \ \mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Al}_{2} \mathrm{O}_{3}} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \nonumber\] Using the same approach, a conservation of mass for magnesium gives \[\mathrm{g} \ \mathrm{MgO}=\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{40.304 \ \mathrm{g} \ \mathrm{MgO}}{\mathrm{mol} \ \mathrm{MgO}} \nonumber\] \[\mathrm{g} \ \mathrm{MgO}=0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \nonumber\] Substituting the equations for g MgO and g Al O into the equation for the combined weights of MgO and Al O leaves us with two equations and two unknowns. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ 0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=1.002 \ \mathrm{g} \nonumber\] Multiplying the first equation by 0.11096 and subtracting the second equation gives \[-0.01797 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=-0.1348 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.504 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}=7.815 \ \mathrm{g}-7.504 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}, \mathrm{H}_{6} \mathrm{NO}\right)_{2}=0.311 \ \mathrm{g} \nonumber\] Now we can finish the problem using the approach from . A conservation of mass requires that all the aluminum and magnesium in the original sample of Dow metal is in the precipitates of Al(C H NO) and the Mg(C H NO) . For aluminum, we find that \[0.311 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.45 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{26.982 \ \mathrm{g} \ \mathrm{Al}}{\mathrm{mol} \ \mathrm{Al}}=0.01826 \ \mathrm{g} \ \mathrm{Al} \nonumber\] \[\frac{0.01826 \ \mathrm{g} \ \mathrm{Al}}{0.611 \ \mathrm{g} \text { sample }} \times 100=2.99 \% \mathrm{w} / \mathrm{w} \mathrm{Al} \nonumber\] and for magnesium we have \[7.504 \ \text{g Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{24.305 \ \mathrm{g} \ \mathrm{Mg}}{\mathrm{mol} \ \mathrm{MgO}}=0.5834 \ \mathrm{g} \ \mathrm{Mg} \nonumber\] \[\frac{0.5834 \ \mathrm{g} \ \mathrm{Mg}}{0.611 \ \mathrm{g} \text { sample }} \times 100=95.5 \% \mathrm{w} / \mathrm{w} \mathrm{Mg} \nonumber\] A sample of a silicate rock that weighs 0.8143 g is brought into solution and treated to yield a 0.2692-g mixture of NaCl and KCl. The mixture of chloride salts is dissolved in a mixture of ethanol and water, and treated with HClO , precipitating 0.3314 g of KClO . What is the %w/w Na O in the silicate rock? The masses of the solids provide us with the following equations \[\mathrm{g} \ \mathrm{NaCl}+\mathrm{g} \ \mathrm{KCl}=0.2692 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{KClO}_{4} = 0.3314 \ \mathrm{g} \nonumber\] With two equations are three unknowns—g NaCl, g KCl, and g KClO —we need one additional equation to solve the problem. A conservation of mass requires that all the potassium originally in the KCl ends up in the KClO ; thus \[\text{g KClO}_4 = \text{g KCl} \times \frac{1 \text{ mol Cl}}{74.55 \text{ g KCl}} \times \frac {138.55 \text{ g KClO}_4}{\text{mol Cl}} = 1.8585 \times \text{ g KCl} \nonumber\] Given the mass of KClO , we use the third equation to solve for the mass of KCl in the mixture of chloride salts \[\text{ g KCl} = \frac{\text{g KClO}_4}{1.8585} = \frac{0.3314 \text{ g}}{1.8585} = 0.1783 \text{ g KCl} \nonumber\] The mass of NaCl in the mixture of chloride salts, therefore, is \[\text{ g NaCl} = 0.2692 \text{ g} - \text{g KCl} = 0.2692 \text{ g} - 0.1783 \text{ g KCl} = 0.0909 \text{ g NaCl} \nonumber\] Finally, to report the %w/w Na O in the sample, we use a conservation of mass on sodium to determine the mass of Na O \[0.0909 \text{ g NaCl} \times \frac{1 \text{ mol Na}}{58.44 \text{ g NaCl}} \times \frac{61.98 \text{ g Na}_2\text{O}}{2 \text{ mol Na}} = 0.0482 \text{ g Na}_2\text{O} \nonumber\] giving the %w/w Na O as \[\frac{0.0482 \text{ g Na}_2\text{O}}{0.8143 \text{ g sample}} \times 100 = 5.92\% \text{ w/w Na}_2\text{O} \nonumber\] The previous problems are examples of direct methods of analysis because the precipitate contains the analyte. In an indirect analysis the precipitate forms as a result of a reaction with the analyte, but the analyte is not part of the precipitate. As shown by the following example, despite the additional complexity, we still can use conservation principles to organize our calculations. An impure sample of Na PO that weighs 0.1392 g is dissolved in 25 mL of water. A second solution that contains 50 mL of 3% w/v HgCl , 20 mL of 10% w/v sodium acetate, and 5 mL of glacial acetic acid is prepared. Adding the solution that contains the sample to the second solution oxidizes \(\text{PO}_3^{3-}\) to \(\text{PO}_4^{3-}\) and precipitates Hg Cl . After digesting, filtering, and rinsing the precipitate, 0.4320 g of Hg Cl is obtained. Report the purity of the original sample as % w/w Na PO . This is an example of an indirect analysis because the precipitate, Hg Cl , does not contain the analyte, Na PO . Although the stoichiometry of the reaction between Na PO and HgCl is given earlier in the chapter, let’s see how we can solve the problem using conservation principles. ( ) The reaction between Na PO and HgCl is an oxidation-reduction reaction in which phosphorous increases its oxidation state from +3 in Na PO to +5 in Na PO and in which mercury decreases its oxidation state from +2 in HgCl to +1 in Hg Cl . A redox reaction must obey a conservation of electrons because all the electrons released by the reducing agent, Na PO , must be accepted by the oxidizing agent, HgCl . Knowing this, we write the following stoichiometric conversion factors: \[\frac{2 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}} \text { and } \frac{1 \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \nonumber\] Now we are ready to solve the problem. First, we use a conservation of mass for mercury to convert the precipitate’s mass to the moles of HgCl . \[0.4320 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2} \times \frac{2 \ \mathrm{mol} \ \mathrm{Hg}}{472.09 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{HgCl}_{2}}{\mathrm{mol} \ \mathrm{Hg}}=1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \nonumber\] Next, we use the conservation of electrons to find the mass of Na PO . \[1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \times \frac{1 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{2 \ \mathrm{mol} \ e^{-}} \times \frac{147.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}=0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] Finally, we calculate the %w/w Na PO in the sample. \[\frac{0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{0.1392 \ \mathrm{g} \text { sample }} \times 100=97.26 \% \mathrm{w} / \mathrm{w} \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] As you become comfortable using conservation principles, you will see ways to further simplify problems. For example, a conservation of electrons requires that the electrons released by Na PO end up in the product, Hg Cl , yielding the following stoichiometric conversion factor: \[\frac{2 \ \operatorname{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \nonumber\] This conversion factor provides a direct link between the mass of Hg Cl and the mass of Na PO . One approach for determining phosphate, \(\text{PO}_4^{3-}\), is to precipitate it as ammonium phosphomolybdate, (NH ) PO •12MoO . After we isolate the precipitate by filtration, we dissolve it in acid and precipitate and weigh the molybdate as PbMoO . Suppose we know that our sample is at least 12.5% Na PO and that we need to recover a minimum of 0.600 g of PbMoO ? What is the minimum amount of sample that we need for each analysis? To find the mass of (NH ) PO •12MoO that will produce 0.600 g of PbMoO , we first use a conservation of mass for molybdenum; thus \[0.600 \ \mathrm{g} \ \mathrm{PbMoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mo}}{351.2 \ \mathrm{g} \ \mathrm{PbMoO}_{3}} \times \frac{1876.59 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}}{12 \ \mathrm{mol} \ \mathrm{Mo}}= 0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \nonumber\] Next, to convert this mass of (NH ) PO •12MoO to a mass of Na PO , we use a conservation of mass on \(\text{PO}_4^{3-}\). \[0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{PO}_{4}^{3-}}{1876.59 \ \mathrm{g \ }\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}} \times \frac{163.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{\mathrm{mol} \ \mathrm{PO}_{4}^{3-}}=0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \nonumber\] Finally, we convert this mass of Na PO to the corresponding mass of sample. \[0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \times \frac{100 \ \mathrm{g} \text { sample }}{12.5 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}=0.187 \ \mathrm{g} \text { sample } \nonumber\] A sample of 0.187 g is sufficient to guarantee that we recover a minimum of 0.600 g PbMoO . If a sample contains more than 12.5% Na PO , then a 0.187-g sample will produce more than 0.600 g of PbMoO . A precipitation reaction is a useful method for identifying inorganic and organic analytes. Because a qualitative analysis does not require quantitative measurements, the analytical signal is simply the observation that a precipitate forms. Although qualitative applications of precipitation gravimetry have been replaced by spectroscopic methods of analysis, they continue to find application in spot testing for the presence of specific analytes [Jungreis, E. ; 2nd Ed., Wiley: New York, 1997]. Any of the precipitants listed in , , and can be used for a qualitative analysis. The scale of operation for precipitation gravimetry is limited by the sensitivity of the balance and the availability of sample. To achieve an accuracy of ±0.1% using an analytical balance with a sensitivity of ±0.1 mg, we must isolate at least 100 mg of precipitate. As a consequence, precipitation gravimetry usually is limited to major or minor analytes, in macro or meso samples. The analysis of a trace level analyte or a micro sample requires a microanalytical balance. For a macro sample that contains a major analyte, a relative error of 0.1– 0.2% is achieved routinely. The principal limitations are solubility losses, impurities in the precipitate, and the loss of precipitate during handling. When it is difficult to obtain a precipitate that is free from impurities, it often is possible to determine an empirical relationship between the precipitate’s mass and the mass of the analyte by an appropriate calibration. The relative precision of precipitation gravimetry depends on the sample’s size and the precipitate’s mass. For a smaller amount of sample or precipitate, a relative precision of 1–2 ppt is obtained routinely. When working with larger amounts of sample or precipitate, the relative precision extends to several ppm. Few quantitative techniques can achieve this level of precision. For any precipitation gravimetric method we can write the following general equation to relate the signal (grams of precipitate) to the absolute amount of analyte in the sample \[\text { g precipitate }=k \times \mathrm{g} \text { analyte } \label{8.13}\] where , the method’s sensitivity, is determined by the stoichiometry between the precipitate and the analyte. Equation \ref{8.13} assumes we used a suitable blank to correct the signal for any contributions of the reagent to the precipitate’s mass. Consider, for example, the determination of Fe as Fe O . Using a conservation of mass for iron, the precipitate’s mass is \[\mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{\text{AW Fe}} \times \frac{\text{FW Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}} \nonumber\] and the value of is \[k=\frac{1}{2} \times \frac{\mathrm{FW} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{\mathrm{AW} \ \mathrm{Fe}} \label{8.14}\] As we can see from Equation \ref{8.14}, there are two ways to improve a method’s sensitivity. The most obvious way to improve sensitivity is to increase the ratio of the precipitate’s molar mass to that of the analyte. In other words, it helps to form a precipitate with the largest possible formula weight. A less obvious way to improve a method’s sensitivity is indicated by the term of 1/2 in Equation \ref{8.14}, which accounts for the stoichiometry between the analyte and precipitate. We can also improve sensitivity by forming a precipitate that contains fewer units of the analyte. Suppose you wish to determine the amount of iron in a sample. Which of the following compounds—FeO, Fe O , or Fe O —provides the greatest sensitivity? To determine which form has the greatest sensitivity, we use a conservation of mass for iron to find the relationship between the precipitate’s mass and the mass of iron. \[\begin{aligned} \mathrm{g} \ \mathrm{FeO} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{71.84 \ \mathrm{g} \ \mathrm{FeO}}{\mathrm{mol} \ \mathrm{Fe}}=1.286 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}}=1.430 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{231.53 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=1.382 \times \mathrm{g} \ \mathrm{Fe} \end{aligned} \nonumber\] Of the three choices, the greatest sensitivity is obtained with Fe O because it provides the largest value for . Due to the chemical nature of the precipitation process, precipitants usually are not selective for a single analyte. For example, silver is not a selective precipitant for chloride because it also forms precipitates with bromide and with iodide. Interferents often are a serious problem and must be considered if accurate results are to be obtained. Precipitation gravimetry is time intensive and rarely practical if you have a large number of samples to analyze; however, because much of the time invested in precipitation gravimetry does not require an analyst’s immediate supervision, it is a practical alternative when working with only a few samples. Equipment needs are few—beakers, filtering devices, ovens or burners, and balances—inexpensive, routinely available in most laboratories, and easy to maintain. | 59,802 | 3,318 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.13%3A_Titrations |
A is a volumetric technique in which a solution of one reactant (the titrant) is added to a solution of a second reactant (the "analyte") until the equivalence point is reached. The equivalence point is the point at which titrant has been added in exactly the right quantity to react stoichiometrically with the analyten (when moles of titrant = moles of analyte). If either the titrant or analyte is colored, the equivalence point is evident from the disappearance of color as the reactants are consumed. Otherwise, an may be added which has an "endpoint" (changes color) at the equivalence point, or the equivalence point may be determined from a titration curve. The amount of added titrant is determined from its concentration and volume: mol mol and the amount of titrant can be used in the usual stoichiometric calculation to determine the amount of analyte. The titration process can be observed in the video below. A measured volume of the solution to be titrated, in this case, colorless aqueous acetic acid, CH COOH( ) is placed in a beaker. The colorless sodium hydroxide NaOH( ), which is the , is added carefully by means of a buret. The volume of titrant added can then be determined by reading the level of liquid in the buret before and after titration. This reading can usually be estimated to the nearest hundredth of a milliliter, so precise additions of titrant can be made rapidly. As the first few milliliters of titrant flow into the flask, some indicator briefly changes to pink, but returns to colorless rapidly. This is due to a large excess of acetic acid. The limiting reagent NaOH is entirely consumed. The added indicator changes to pink when the titration is complete, indicating that all of the aqueous acetic acid has been consumed by NaOH( ). The reaction which occurs is \[ \text{C} \text{H}_{3} \text{COOH} (aq) + \text{ NaOH} (aq) \rightarrow \text{ Na}^{+} (aq) + \text{CH}_{3} \text{COO}^{-} (aq) + \text{H}_{2} \text{O} (l) \label{2} \] Eventually, all the acetic acid is consumed. Addition of even a fraction of a drop of titrant produces a lasting pink color due to unreacted NaOH in the flask. The color change that occurs at the of the indicator signals that all the acetic acid has been consumed, so we have reached the of the titration. If slightly more NaOH solution were added, there would be an excess and the color of the solution in the flask would get much darker. The endpoint appears suddenly, and care must be taken not to overshoot the endpoint. After the titration has reached the endpoint, a final volume is read from the buret. Using the initial and final reading, the volume added can be determined quite precisely: The object of a titration is always to add just the amount of titrant needed to consume the amount of substance being titrated. In the NaOH—CH COOH reaction Eq. \(\ref{2}\), the equivalence point occurs when an equal molar amount of NaOH has been added from the graduated cylinder for every mole of CH COOH originally in the titration flask. That is, at the equivalence point the ratio of the amount of NaOH, added to the amount of CH COOH consumed must equal the stoichiometric ratio \[\dfrac{n_{\text{NaOH}}\text{(added from graduated cylinder)}}{n_{\text{CH}_{\text{3}}{\text{COOH}}}\text{(initially in flask)}}=\text{S}( \dfrac{\text{NaOH}}{\text{CH}_{\text{3}}\text{COOH}} ) \nonumber \] \[=\dfrac{\text{1 mol NaOH}}{\text{1 mol CH}_{\text{3}}\text{COOH}} \nonumber \] What volume of 0.05386 KMnO would be needed to reach the endpoint when titrating 25.00 ml of 0.1272 H O , given S(KMnO /H O ) = 2/5 At the equivalence point, the stoichiometric ratio will apply, and we can use it to calculate the amount of KMnO which must be added: \[n_{\text{KMnO}_{\text{4}}}\text{(added)}=n_{\text{H}_{\text{2}}\text{O}_{\text{2}}}\text{(in flask)}\times \text{S}\left( \dfrac{\text{KMnO}_{\text{4}}}{\text{H}_{\text{2}}\text{O}_{\text{2}}} \right) \nonumber \] The amount of H O is obtained from the volume and concentration: \[n_{\text{H}_{\text{2}}\text{O}_{\text{2}}}\text{(in flask)}=25.00\text{ cm}^{\text{3}}\times \text{0}\text{.1272 }\dfrac{\text{mmol}}{\text{cm}^{\text{3}}}=\text{3}\text{.180 mmol H}_{\text{2}}\text{O}_{\text{2}} \nonumber \] Then \[n_{\text{KMnO}_{\text{4}}}\text{(added)}=\text{3}\text{.180 mmol H}_{\text{2}}\text{O}_{\text{2}}\times \dfrac{\text{2 mol KMnO}_{\text{4}}}{\text{5 mol H}_{\text{2}}\text{O}_{\text{2}}}\times \dfrac{\text{10}^{\text{-3}}}{\text{10}^{\text{-3}}} \nonumber \] \[=\text{3}\text{.180 mmol H}_{\text{2}}\text{O}_{\text{2}}\times \dfrac{\text{2 mmol KMnO}_{\text{4}}}{\text{5 mmol H}_{\text{2}}\text{O}_{\text{2}}} \nonumber \] To obtain we use the concentration as a conversion factor: \[V_{\text{KMnO}_{\text{4}}\text{(}aq\text{)}}=\text{1}\text{.272 mmol KMnO}_{\text{4}}\times \dfrac{\text{1 cm}^{\text{3}}}{\text{5}\text{.386}\times \text{10}^{\text{-2}}\text{ mmol KMnO}_{\text{4}}} \nonumber \] Note that overtitrating [adding more than 23.62 cm of KMnO ( ) would involve an excess (more than 1.272 mmol) of KMnO . Titration is often used to determine the concentration of a solution. In many cases it is not a simple matter to obtain a pure substance, weigh it accurately, and dissolve it in a volumetric flask as was done in . NaOH, for example, combines rapidly with H O and CO from the air, and so even a freshly prepared sample of solid NaOH will not be pure. Its weight would change continuously as CO ( ) and H O( ) were absorbed. Hydrogen chloride (HCl) is a gas at ordinary temperatures and pressures, making it very difficult to handle or weigh. Aqueous solutions of both of these substances must be ; that is, their concentrations must be determined by titration. A sample of pure potassium hydrogen phthalate (KHC H O ) weighing 0.3421 g is dissolved in distilled water. Titration of the sample requires 27.03 ml NaOH( ). The titration reaction is \[ \text{NaOH} (aq) + \text{KHC}_{8} \text{H}_{4} \text{O}_{4} (aq) \rightarrow \text{NaKC}_{8} \text{H}_{4} \text{O}_{4} (aq) + \text{H}_{2} \text{O} \nonumber \] What is the concentration of NaOH( ) ? To calculate concentration, we need to know the amount of NaOH and the volume of solution in which it is dissolved. The former quantity could be obtained via a stoichiometric ratio from the amount of KHC H O , and that amount can be obtained from the mass \[m_{\text{KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}\text{ }\xrightarrow{M_{\text{KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}}\text{ }n_{\text{KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}\text{ }\xrightarrow{S\text{(NaOH/KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}\text{)}}\text{ }n_{\text{NaOH}} \nonumber \] \[n_{\text{NaOH}}=\text{3}\text{.180 g}\times \dfrac{\text{1 mol KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}}{\text{204}\text{.22 g}}\times \dfrac{\text{1 mol NaOH}}{\text{1 mol KHC}_{\text{8}}\text{H}_{\text{4}}\text{O}_{\text{4}}} \nonumber \] \[=\text{1}\text{.674 }\times 10^{\text{-3}}\text{ mol NaOH}=\text{1}\text{.675 mmol NaOH} \nonumber \] The concentration is \[c_{\text{NaOH}}=\dfrac{n_{\text{NaOH}}}{V}=\dfrac{\text{1}\text{.675 mmol NaOH}}{\text{27}\text{.03 cm}^{\text{3}}}=\text{0}\text{.06197 mmol cm}^{\text{-3}} \nonumber \] or 0.06197 . By far the most common use of titrations is in determining unknowns, that is, in determining the concentration or amount of substance in a sample about which we initially knew nothing. The next example involves an unknown that many persons encounter every day. Vitamin C tablets contain ascorbic acid (C H O ) and a starch “filler” which holds them together. To determine how much vitamin C is present, a tablet can be dissolved in water andwith sodium hydroxide solution, NaOH( ). The equation is \[ \text{C}_{6} \text{H}_{8} \text{O}_{6} (aq) + \text{NaOH} (aq) \rightarrow \text{ Na C}_{6} \text{H}_{7} \text{O}_{6} (aq) + \text{H}_{2} \text{O} (l) \nonumber \] If titration of a dissolved vitamin C tablet requires 16.85 cm³ of 0.1038 NaOH, how accurate is the claim on the label of the bottle that each tablet contains 300 mg of vitamin C? The known volume and concentration allow us to calculate the amount of NaOH( ) which reacted with all the vitamin C. Using the stoichiometric ratio \[\text{S}\left( \dfrac{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}{\text{NaOH}} \right)=\dfrac{\text{1 mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}{\text{1 mmol NaOH}} \nonumber \] we can obtain the amount of C H O . The molar mass converts that amount to a mass which can be compared with the label. Schematically \[ \begin{align} & V_{\text{NaOH}}\rightarrow{c_{\text{NaOH}}}n_{\text{NaOH}}\rightarrow{\text{S(C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}\text{/NaOH)}}n_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}\rightarrow{M_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}}\text{m}_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}} \\ & \text{m}_{\text{C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}=\text{16}\text{.85 cm}^{\text{3}}\times \dfrac{\text{0}\text{.1038 mmol NaOH}}{\text{1 cm}^{\text{3}}}\times \dfrac{\text{1 mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}{\text{1 mmol NaOH}}\times \dfrac{\text{176}\text{.1 mg }}{\text{mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}} \\ & = 308.0 \text{ mg} \end{align} \nonumber \] Note that the molar mass of C H O \[\dfrac{\text{176}\text{.1 g }}{\text{1 mol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}=\dfrac{\text{176}\text{.1 g }}{\text{1 mol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}\times \dfrac{\text{10}^{\text{-3}}}{\text{10}^{\text{-3}}} \nonumber \] \[=\dfrac{\text{176}\text{.1 g}\times \text{10}^{\text{-3}}\text{ }}{\text{10}^{\text{-3}}\text{ mol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}}=\dfrac{\text{176}\text{.1 mg }}{\text{1 mmol C}_{\text{6}}\text{H}_{\text{8}}\text{O}_{\text{6}}} \nonumber \] can be expressed in milligrams per millimole as well as in grams per mole. The 308.0 mg obtained in this example is in reasonably close agreement with the manufacturer’s claim of 300 mg. The tablets are stamped out by machines, not weighed individually, and so some variation is expected. | 10,297 | 3,319 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/05%3A_Thermochemistry/5.S%3A_Thermochemistry_(Summary) |
\[w= F \times d \nonumber \] : energy of motion where is the kinetic energy, is the mass, and is the velocity. : energy that an object possesses as a result of its composition or its position with respect to another object : aka ; a statement of our experience that energy is conserved in any process. We can express the first law in many ways. One of the more useful expressions is that the change in internal energy , of a system in any process is equal to the heat, , added to the system, plus the work, , done the system by its surroundings: . : total energy of the system (see above) changes – E > E = system has gained energy E < E = system has lost energy Initial state refers to reactants, final state to products. ?E = q + w where ∆E is internal energy, q is heat, and w is work Ex. A system absorbs 50 J of heat and does 10 J of work on its surroundings, = 50 J and = 10 J. Thus, ∆E = 50 J + (10 J) = 40 J. : a property of a system that is determined by the state or condition of the system and not by how it got to that state; its value is fixed when temperature, pressure, composition, and physical form are specified. The internal energy of a system is a state function. However, the work done by a system in a given process is not a state function!!! Heat flows between system and surroundings until . : a process in which a system absorbs heat from its surroundings : a process in which a system releases heat to its surroundings : represented by ; deals with the amount of heat absorbed or released during a chemical reaction under constant pressure ?H = q (The subscript P on q is a reminder that we are considering a special case where pressures is constant) = positive ; system has gained heat from surroundings (endothermic) = negative ; system has lost heat to the surroundings (exothermic) = (products) – (reactants) : balanced chemical equations that show the associated enthalpy change : measurement of heat flow : apparatus that measures heat flow : amount of energy required to raise the temperature of a given object by 1°C. The greater the capacity of a body, the more heat it requires to raise its temperature. : when referring to pure substances; heat capacity of 1 mol of substance : heat capacity of 1 g of a substance (as opposed to a mole) Specific heat= quantity of heat transferred / (grams of substance) X (temperature change) = q / m X ∆T q = (specific heat) X (grams of substance) X Calorimeters are simple instruments to control pressure. Because the calorimeter prevents the gain or loss of heat from its surroundings, the heat released by the reaction, , equals that gained by the solution, . Thus, = = (specific heat) X (grams of solution) X : device for measuring the heat evolved in the combustion of a substance under constantvolume conditions. Most used for combustion reactions. = X ∆ , where is the heat capacity of the calorimeter Because enthalpy is a state function, the enthalpy change, ∆ , associated with any chemical process depends only on the , and on the nature of the and the . CH + 2O ? CO + 2H O ∆ = 802 kJ (Add) 2H O ? 2H O ? = 88 kJ __________________________________________________________ CH + 2O + 2H O → CO + 2H O + 2H O Net equation: CH + 2O ? CO + 2H O To obtain the net equation, the sum of the reactants of the two equations is placed on one side of the arrow, and the sum of the products on the other side. Because 2H O occurs on both sides of the equation, it can be cancelled. : states that The overall enthalpy change for the process is independent of the number of steps or the particular nature of the path by which the reaction is carried out. : aka ; enthalpy change that accompanies the formation of a substance from the most stable forms of its component elements. Labeled ∆ , where the subscript indicates that the substance has been formed from its elements. : for a substance is the form most stable at the particular temperature of interest and at standard atmospheric pressure. : ∆H° . chang in enthalpy that accompanies the formation of 1 mol of that substance from its elements, with all substances in their standard states. ∆H° = ∑ n∆H° (products) ∑ m∆H° (reactants) ∑ = the sum of m and n = stoichiometric coefficients of the chemical reaction C H + 5O → 3CO + 4H O ∆H° = [3∆H° (CO ) + 4∆H° (H O)] – [∆H° (C H ) + 5∆H° (O )] ∆H° = [(3 mol CO ) (393.5 kJ/mol) + (4 mol H O) (285.5 kJ/mol)] = (2324 kJ) – (103.85 kJ) =2220 kJ : energy released when 1g of a material is combusted | 4,565 | 3,320 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.12%3A_Multi-Step_Problems_with_Changes_of_State |
If you have a cube of ice, which process will take more energy—the melting of that ice cube or the conversion of the water to steam? The short answer is that more energy is needed to convert the water to steam. The long answer is really a series of questions: How do you get from one point to the other? What is the temperature of the ice? What is the mass of that ice cube? A long process is involved to take the material from the starting point to the end point. Heating curves show the phase changes that a substance undergoes as heat is continuously absorbed. The specific heat of a substance allows us to calculate the heat absorbed or released as the temperature of the substance changes. It is possible to combine that type of problem with a change of state to solve a problem involving multiple steps. The figure above shows ice at \(-30^\text{o} \text{C}\) being converted in a five-step process to gaseous water (steam) at \(140^\text{o} \text{C}\). It is now possible to calculate the heat absorbed during that entire process. The process and the required calculations are summarized below. 1. Ice is heated from \(-30^\text{o} \text{C}\) to \(0^\text{o} \text{C}\). The heat absorbed is calculated by using the specific heat of ice and the equation \(\Delta H = c_p \times m \times \Delta T\). 2. Ice is melted at \(0^\text{o} \text{C}\). The heat absorbed is calculated by multiplying the moles of ice by the molar heat of fusion. 3. Water at \(0^\text{o} \text{C}\) is heated to \(100^\text{o} \text{C}\). The heat absorbed is calculated by using the specific heat of water and the equation \(\Delta H = c_p \times m \times \Delta T\). 4. Water is vaporized to steam at \(100^\text{o} \text{C}\). The heat absorbed is calculated by multiplying the moles of water by the molar heat of vaporization. 5. Steam is heated from \(100^\text{o} \text{C}\) to \(140^\text{o} \text{C}\). The heat absorbed is calculated by using the specific heat of steam and the equation \(\Delta H = c_p \times m \times \Delta T\). Calculate the total amount of heat absorbed (in \(\text{kJ}\)) when \(2.00 \: \text{mol}\) of ice at \(-30^\text{o} \text{C}\) is converted to steam at \(140.0^\text{o} \text{C}\). The required specific heats can be found in the table in . Follow the steps previously described. Note that the mass of the water is needed for the calculations that involve the specific heat, while the moles of water is needed for the calculations that involve changes of state. All heat quantities must be in kilojoules so that they can be added together to get a total for the five-step process. \[\Delta H_\text{total} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 + \Delta H_5 = 113.4 \: \text{kJ}\nonumber \] The total heat absorbed as the ice at \(-30^\text{o} \text{C}\) is heated to steam at \(140^\text{o} \text{C}\) is \(133.4 \: \text{kJ}\). The largest absorption of heat comes during the vaporization of the liquid water. | 2,966 | 3,321 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/09%3A_Mixtures/9.09%3A_Chapter_9_Problems |
An underlined problem number or problem-part letter indicates that the numerical answer appears in Appendix I.
9.1 For a binary solution, find expressions for the mole fractions \(x_{\mathrm{B}}\) and \(x_{\mathrm{A}}\) as functions of the solute molality \(m_{\mathrm{B}}\). 9.2 Consider a binary mixture of two liquids, \(\mathrm{A}\) and \(\mathrm{B}\). The molar volume of mixing, \(\Delta V(\mathrm{mix}) / n\), is given by Eq. 9.2.19. (a) Find a formula for calculating the value of \(\Delta V(\mathrm{mix}) / n\) of a binary mixture from values of \(x_{\mathrm{A}}, x_{\mathrm{B}}, M_{\mathrm{A}}, M_{\mathrm{B}}, \rho, \rho_{\mathrm{A}}^{*}\), and \(\rho_{\mathrm{B}}^{*} .\) Table \(9.7\) Molar volumes of mixing of binary mixtures of 1-hexanol (A) and 1 -octene \((\mathrm{B})\) at \(25^{\circ} \mathrm{C} .^{a}\) \begin{tabular}{lccc}
\hline\(x_{\mathrm{B}}\) & {\([\Delta V(\mathrm{mix}) / n] / \mathrm{cm}^{3} \mathrm{~mol}^{-1}\)} & \(x_{\mathrm{B}}\) & {\([\Delta V(\mathrm{mix}) / n] / \mathrm{cm}^{3} \mathrm{~mol}^{-1}\)} \\
\hline 0 & 0 & \(0.555\) & \(0.005\) \\
\(0.049\) & \(-0.027\) & \(0.597\) & \(0.011\) \\
\(0.097\) & \(-0.050\) & \(0.702\) & \(0.029\) \\
\(0.146\) & \(-0.063\) & \(0.716\) & \(0.035\) \\
\(0.199\) & \(-0.077\) & \(0.751\) & \(0.048\) \\
\(0.235\) & \(-0.073\) & \(0.803\) & \(0.056\) \\
\(0.284\) & \(-0.074\) & \(0.846\) & \(0.058\) \\
\(0.343\) & \(-0.065\) & \(0.897\) & \(0.057\) \\
\(0.388\) & \(-0.053\) & \(0.944\) & \(0.049\) \\
\(0.448\) & \(-0.032\) & 1 & 0 \\
\(0.491\) & \(-0.016\) & & \\
\hline
\end{tabular}
\({ }^{a}\) Ref. [170].
(b) The molar volumes of mixing for liquid binary mixtures of 1-hexanol (A) and 1-octene (B) at \(25^{\circ} \mathrm{C}\) have been calculated from their measured densities. The data are in Table 9.7. The molar volumes of the pure constituents are \(V_{\mathrm{A}}^{*}=125.31 \mathrm{~cm}^{3} \mathrm{~mol}^{-1}\) and \(V_{\mathrm{B}}^{*}=\) \(157.85 \mathrm{~cm}^{3} \mathrm{~mol}^{-1}\). Use the method of intercepts to estimate the partial molar volumes of both constituents in an equimolar mixture \(\left(x_{\mathrm{A}}=x_{\mathrm{B}}=0.5\right)\), and the partial molar volume \(V_{\mathrm{B}}^{\infty}\) of B at infinite dilution. 9.3 Extend the derivation of Prob. 8.1, concerning a liquid droplet of radius \(r\) suspended in a gas, to the case in which the liquid and gas are both mixtures. Show that the equilibrium conditions are \(T^{\mathrm{g}}=T^{\mathrm{l}}, \mu_{i}^{\mathrm{g}}=\mu_{i}^{1}\) (for each species \(i\) that can equilibrate between the two phases), and \(p^{1}=p^{g}+2 \gamma / r\), where \(\gamma\) is the surface tension. (As in Prob. 8.1, the last relation is the Laplace equation.) 9.4 Consider a gaseous mixture of \(4.0000 \times 10^{-2} \mathrm{~mol}\) of \(\mathrm{N}_{2}\) (A) and \(4.0000 \times 10^{-2} \mathrm{~mol}\) of \(\mathrm{CO}_{2}\) (B) in a volume of \(1.0000 \times 10^{-3} \mathrm{~m}^{3}\) at a temperature of \(298.15 \mathrm{~K}\). The second virial coefficients at this temperature have the values \({ }^{14}\)
\[
\begin{aligned}
B_{\mathrm{AA}} &=-4.8 \times 10^{-6} \mathrm{~m}^{3} \mathrm{~mol}^{-1} \\
B_{\mathrm{BB}} &=-124.5 \times 10^{-6} \mathrm{~m}^{3} \mathrm{~mol}^{-1} \\
B_{\mathrm{AB}} &=-47.5 \times 10^{-6} \mathrm{~m}^{3} \mathrm{~mol}^{-1}
\end{aligned}
\]
Compare the pressure of the real gas mixture with that predicted by the ideal gas equation. See Eqs. 9.3.20 and 9.3.23.
\({ }^{14}\) Refs. [3], [49], and [50]. 9.5 At \(25^{\circ} \mathrm{C}\) and 1 bar, the Henry's law constants of nitrogen and oxygen dissolved in water are \(k_{\mathrm{H}, \mathrm{N}_{2}}=8.64 \times 10^{4}\) bar and \(k_{\mathrm{H}, \mathrm{O}_{2}}=4.41 \times 10^{4}\) bar. \({ }^{15}\) The vapor pressure of water at this temperature and pressure is \(p_{\mathrm{H}_{2} \mathrm{O}}=0.032\) bar. Assume that dry air contains only \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) at mole fractions \(y_{\mathrm{N}_{2}}=0.788\) and \(y_{\mathrm{O}_{2}}=0.212\). Consider liquid-gas systems formed by equilibrating liquid water and air at \(25^{\circ} \mathrm{C}\) and \(1.000 \mathrm{bar}\), and assume that the gas phase behaves as an ideal gas mixture. Hint: The sum of the partial pressures of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) must be \((1.000-0.032)\) bar \(=0.968\) bar. If the volume of one of the phases is much larger than that of the other, then almost all of the \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) will be in the predominant phase and the ratio of their amounts in this phase must be practically the same as in dry air. Determine the mole fractions of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) in both phases in the following limiting cases: (a) A large volume of air is equilibrated with just enough water to leave a small drop of liquid. (b) A large volume of water is equilibrated with just enough air to leave a small bubble of gas. 9.6 Derive the expression for \(\gamma_{m, \mathrm{~B}}\) given in Table 9.4, starting with Eq. 9.5.18. 9.7 Consider a nonideal binary gas mixture with the simple equation of state \(V=n R T / p+n B\) (Eq. \(9.3 .21\) ). (a) The rule of Lewis and Randall states that the value of the mixed second virial coefficient \(B_{\mathrm{AB}}\) is the average of \(B_{\mathrm{AA}}\) and \(B_{\mathrm{BB}}\). Show that when this rule holds, the fugacity coefficient of \(\mathrm{A}\) in a binary gas mixture of any composition is given by \(\ln \phi_{\mathrm{A}}=B_{\mathrm{AA}} p / R T\). By comparing this expression with Eq. \(7.8 .18\) for a pure gas, express the fugacity of \(\mathrm{A}\) in the mixture as a function of the fugacity of pure \(A\) at the same temperature and pressure as the mixture. (b) The rule of Lewis and Randall is not accurately obeyed when constituents A and B are chemically dissimilar. For example, at \(298.15 \mathrm{~K}\), the second virial coefficients of \(\mathrm{H}_{2} \mathrm{O}\) (A) and \(\mathrm{N}_{2}\) (B) are \(B_{\mathrm{AA}}=-1158 \mathrm{~cm}^{3} \mathrm{~mol}^{-1}\) and \(B_{\mathrm{BB}}=-5 \mathrm{~cm}^{3} \mathrm{~mol}^{-1}\), respectively, whereas the mixed second virial coefficient is \(B_{\mathrm{AB}}=-40 \mathrm{~cm}^{3} \mathrm{~mol}^{-1}\). When liquid water is equilibrated with nitrogen at \(298.15 \mathrm{~K}\) and 1 bar, the partial pressure of \(\mathrm{H}_{2} \mathrm{O}\) in the gas phase is \(p_{\mathrm{A}}=0.03185\) bar. Use the given values of \(B_{\mathrm{AA}}, B_{\mathrm{BB}}\), and \(B_{\mathrm{AB}}\) to calculate the fugacity of the gaseous \(\mathrm{H}_{2} \mathrm{O}\) in this binary mixture. Compare this fugacity with the fugacity calculated with the value of \(B_{\mathrm{AB}}\) predicted by the rule of Lewis and Randall. Table \(9.8\) Activity coefficient of benzene (A) in mixtures of benzene and 1 -octanol at \(20^{\circ} \mathrm{C}\). The reference state
is the pure liquid. \begin{tabular}{lccc}
\hline\(x_{\mathrm{A}}\) & \(\gamma_{\mathrm{A}}\) & \(x_{\mathrm{A}}\) & \(\gamma_{\mathrm{A}}\) \\
\hline 0 & \(2.0^{a}\) & \(0.7631\) & \(1.183\) \\
\(0.1334\) & \(1.915\) & \(0.8474\) & \(1.101\) \\
\(0.2381\) & \(1.809\) & \(0.9174\) & \(1.046\) \\
\(0.4131\) & \(1.594\) & \(0.9782\) & \(1.005\) \\
\(0.5805\) & \(1.370\) & & \\
\hline \multicolumn{3}{l}{\(a_{\text {extrapolated }}\)}
\end{tabular}
\({ }^{15}\) Ref. [184]. 9.8 Benzene and 1-octanol are two liquids that mix in all proportions. Benzene has a measurable vapor pressure, whereas 1-octanol is practically nonvolatile. The data in Table \(9.8\) on the preceding page were obtained by Platford \({ }^{16}\) using the isopiestic vapor pressure method. (a) Use numerical integration to evaluate the integral on the right side of Eq. \(9.6 .10\) at each of the values of \(x_{\mathrm{A}}\) listed in the table, and thus find \(\gamma_{\mathrm{B}}\) at these compositions. (b) Draw two curves on the same graph showing the effective mole fractions \(\gamma_{\mathrm{A}} x_{\mathrm{A}}\) and \(\gamma_{\mathrm{B}} x_{\mathrm{B}}\) as functions of \(x_{\mathrm{A}}\). Are the deviations from ideal-mixture behavior positive or negative? Table \(9.9\) Liquid and gas compositions in the two-phase system of methanol (A) and benzene (B) at \(45^{\circ} \mathrm{C}^{a}\) \begin{tabular}{llllll}
\hline\(x_{\mathrm{A}}\) & \(y_{\mathrm{A}}\) & \(p / \mathrm{kPa}\) & \(x_{\mathrm{A}}\) & \(y_{\mathrm{A}}\) & \(p / \mathrm{kPa}\) \\
\hline 0 & 0 & \(29.894\) & \(0.4201\) & \(0.5590\) & \(60.015\) \\
\(0.0207\) & \(0.2794\) & \(40.962\) & \(0.5420\) & \(0.5783\) & \(60.416\) \\
\(0.0314\) & \(0.3391\) & \(44.231\) & \(0.6164\) & \(0.5908\) & \(60.416\) \\
\(0.0431\) & \(0.3794\) & \(46.832\) & \(0.7259\) & \(0.6216\) & \(59.868\) \\
\(0.0613\) & \(0.4306\) & \(50.488\) & \(0.8171\) & \(0.6681\) & \(58.321\) \\
\(0.0854\) & \(0.4642\) & \(53.224\) & \(0.9033\) & \(0.7525\) & \(54.692\) \\
\(0.1811\) & \(0.5171\) & \(57.454\) & \(0.9497\) & \(0.8368\) & \(51.009\) \\
\(0.3217\) & \(0.5450\) & \(59.402\) & 1 & 1 & \(44.608\) \\
\hline
\end{tabular} \({ }^{a}\) Ref. [169]. 9.9 Table \(9.9\) lists measured values of gas-phase composition and total pressure for the binary two-phase methanol-benzene system at constant temperature and varied liquid-phase composition. \(x_{\mathrm{A}}\) is the mole fraction of methanol in the liquid mixture, and \(y_{\mathrm{A}}\) is the mole fraction of methanol in the equilibrated gas phase. (a) For each of the 16 different liquid-phase compositions, tabulate the partial pressures of \(\mathrm{A}\) and \(\mathrm{B}\) in the equilibrated gas phase. (b) Plot \(p_{\mathrm{A}}\) and \(p_{\mathrm{B}}\) versus \(x_{\mathrm{A}}\) on the same graph. Notice that the behavior of the mixture is far from that of an ideal mixture. Are the deviations from Raoult's law positive or negative? (c) Tabulate and plot the activity coefficient \(\gamma_{\mathrm{B}}\) of the benzene as a function of \(x_{\mathrm{A}}\) using a pure-liquid reference state. Assume that the fugacity \(f_{\mathrm{B}}\) is equal to \(p_{\mathrm{B}}\), and ignore the effects of variable pressure. (d) Estimate the Henry's law constant \(k_{\mathrm{H}, \mathrm{A}}\) of methanol in the benzene environment at \(45^{\circ} \mathrm{C}\) by the graphical method suggested in Fig. 9.7(b). Again assume that \(f_{\mathrm{A}}\) and \(p_{\mathrm{A}}\) are equal, and ignore the effects of variable pressure. 9.10 Consider a dilute binary nonelectrolyte solution in which the dependence of the chemical potential of solute B on composition is given by
\[
\mu_{\mathrm{B}}=\mu_{m, \mathrm{~B}}^{\mathrm{ref}}+R T \ln \frac{m_{\mathrm{B}}}{m^{\circ}}+k_{m} m_{\mathrm{B}}
\]
where \(\mu_{m, \mathrm{~B}}^{\mathrm{ref}}\) and \(k_{m}\) are constants at a given \(T\) and \(p\). (The derivation of this equation is sketched in Sec. 9.5.4.) Use the Gibbs-Duhem equation in the form \(\mathrm{d} \mu_{\mathrm{A}}=-\left(n_{\mathrm{B}} / n_{\mathrm{A}}\right) \mathrm{d} \mu_{\mathrm{B}}\) to obtain an expression for \(\mu_{\mathrm{A}}-\mu_{\mathrm{A}}^{*}\) as a function of \(m_{\mathrm{B}}\) in this solution.
\({ }^{16}\) Ref. [145]. 9.11 By means of the isopiestic vapor pressure technique, the osmotic coefficients of aqueous solutions of urea at \(25^{\circ} \mathrm{C}\) have been measured at molalities up to the saturation limit of about \(20 \mathrm{~mol} \mathrm{~kg}^{-1} .{ }^{17}\) The experimental values are closely approximated by the function
\[
\phi_{m}=1.00-\frac{0.050 m_{\mathrm{B}} / m^{\circ}}{1.00+0.179 m_{\mathrm{B}} / m^{\circ}}
\]
where \(m^{\circ}\) is \(1 \mathrm{~mol} \mathrm{~kg}^{-1}\). Calculate values of the solvent and solute activity coefficients \(\gamma_{\mathrm{A}}\) and \(\gamma_{m, \mathrm{~B}}\) at various molalities in the range \(0-20 \mathrm{~mol} \mathrm{~kg}^{-1}\), and plot them versus \(m_{\mathrm{B}} / m^{\circ}\). Use enough points to be able to see the shapes of the curves. What are the limiting slopes of these curves as \(m_{\mathrm{B}}\) approaches zero? 9.12 Use Eq. \(9.2 .49\) to derive an expression for the rate at which the logarithm of the activity coefficient of component \(i\) of a liquid mixture changes with pressure at constant temperature and composition: \(\left(\partial \ln \gamma_{i} / \partial p\right)_{T,\left\{n_{i}\right\}}=\) ? 9.13 Assume that at sea level the atmosphere has a pressure of \(1.00\) bar and a composition given by \(y_{\mathrm{N}_{2}}=0.788\) and \(y_{\mathrm{O}_{2}}=0.212\). Find the partial pressures and mole fractions of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), and the total pressure, at an altitude of \(10.0 \mathrm{~km}\), making the (drastic) approximation that the atmosphere is an ideal gas mixture in an equilibrium state at \(0^{\circ} \mathrm{C}\). For \(g\) use the value of the standard acceleration of free fall listed in Appendix B. 9.14 Consider a tall column of a dilute binary liquid solution at equilibrium in a gravitational field. (a) Derive an expression for \(\ln \left[c_{\mathrm{B}}(h) / c_{\mathrm{B}}(0)\right]\), where \(c_{\mathrm{B}}(h)\) and \(c_{\mathrm{B}}(0)\) are the solute concentrations at elevations \(h\) and 0 . Your expression should be a function of \(h, M_{\mathrm{B}}, T, \rho\), and the partial specific volume of the solute at infinite dilution, \(v_{\mathrm{B}}^{\infty}\). For the dependence of pressure on elevation, you may use the hydrostatic formula \(\mathrm{d} p=-\rho g \mathrm{~d} h\) (Eq. \(8.1 .14\) on page 200) and assume the solution density \(\rho\) is the same at all elevations. Hint: use the derivation leading to Eq. \(9.8 .22\) as a guide. (b) Suppose you have a tall vessel containing a dilute solution of a macromolecule solute of molar mass \(M_{\mathrm{B}}=10.0 \mathrm{~kg} \mathrm{~mol}^{-1}\) and partial specific volume \(v_{\mathrm{B}}^{\infty}=0.78 \mathrm{~cm}^{3} \mathrm{~g}^{-1}\). The solution density is \(\rho=1.00 \mathrm{~g} \mathrm{~cm}^{-3}\) and the temperature is \(T=300 \mathrm{~K}\). Find the height \(h\) from the bottom of the vessel at which, in the equilibrium state, the concentration \(c_{\mathrm{B}}\) has decreased to 99 percent of the concentration at the bottom. 9.15 FhuA is a protein found in the outer membrane of the Escherichia coli bacterium. From the known amino acid sequence, its molar mass is calculated to be \(78.804 \mathrm{~kg} \mathrm{~mol}^{-1}\). In aqueous solution, molecules of the detergent dodecyl maltoside bind to a FhuA molecule to form an aggregate that behaves as a single solute species. Figure \(9.13\) on the next page shows data collected in a sedimentation equilibrium experiment with a dilute solution of the aggregate. \({ }^{18}\) In the graph, \(A\) is the absorbance measured at a wavelength of \(280 \mathrm{~nm}\) (a property that is a linear function of the aggregate concentration) and \(r\) is the radial distance from the axis of rotation of the centrifuge rotor. The experimental points fall very close to the straight line shown in the graph. The sedimentation conditions were \(\omega=838 \mathrm{~s}^{-1}\) and \(T=293 \mathrm{~K}\). The authors used the values \(v_{\mathrm{B}}^{\infty}=0.776 \mathrm{~cm}^{3} \mathrm{~g}^{-1}\) and \(\rho=1.004 \mathrm{~g} \mathrm{~cm}^{-3}\). (a) The values of \(r\) at which the absorbance was measured range from \(6.95 \mathrm{~cm}\) to \(7.20 \mathrm{~cm}\). Find the difference of pressure in the solution between these two positions. (b) Find the molar mass of the aggregate solute species, and use it to estimate the mass binding ratio (the mass of bound detergent divided by the mass of protein).
\({ }^{17}\) Ref. [160]. \({ }^{18}\) Ref. [18]. | 15,661 | 3,322 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Properties_of_Alkynes/Properties_and_Bonding_in_the_Alkynes |
The characteristic of the triple bond helps to explain the properties and bonding in the alkynes. Hybridization due to triple bonds allows the uniqueness of alkyne structure. This triple bond contributes to the nonpolar bonding strength, linear, and the acidity of alkynes. Physical Properties include nonpolar due to slight solubility in polar solvents and insoluble in water. This solubility in water and polar solvents is a characteristic feature to alkenes as well. Alkynes dissolve in organic solvents. Compared to alkanes and alkenes, alkynes have a slightly higher boiling point. Ethane has a boiling point of -88.6 ?C, while Ethene is -103.7 ?C and Ethyne has a higher boiling point of -84.0 ?C. The acidity of terminal alkynes compared to alkenes and alkanes are stronger. Compared with Ethane which has a pKa of 62 (least acidic) and Ethene of a pKa of 45, Ethyne has a pKa of 26. With alkynes having the sp hybridization, this makes it the most acidic hydrocarbon. Therefore terminal alkynes must be deprotonated by stronger bases. The importance of the s orbital being attracted to the nucleus contributes to the electronegativity Alkynes are involved in a high release of energy because of repulsion of electrons. The content of energy involved in the alkyne molecule contributes to this high amount of energy. The pi-bonds however, do not encompass a great amount of energy even though the concentration is small within the molecule. The combustion of Ethyne is a major contributor from CO , water, and the ethyne molecule ? H = -311 kcal/mol To help understand the relative stabilities of alkyne isomers, heats of hydrogenation must be used. Hydrogenation of the least energy, results in the release of the internal alkyne. With the result of the production of butane, the stability of internal versus terminal alkynes has significant relative stability due to hyperconjugation. | 1,911 | 3,323 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Concept_Development_Studies_in_Chemistry_(Hutchinson)/08_Molecular_Structure_and_Physical_Properties |
We begin with our knowledge of the structure and properties of atoms. We know that atoms have a nuclear structure, meaning that all of the positive charge and virtually all of the mass of the atom are concentrated in a nucleus which is a very small fraction of the volume of the atom. In addition, we know that many of the properties of atoms can be understood by a model in which the electrons in the atom are arranged in "shells" about the nucleus, with each shell farther from the nucleus than the previous. The electrons in outer shells are more weakly attached to the atom than the electrons in the inner shells, and only a limited number of electrons can fit in each shell. Within each shell are subshells, each of which can also hold a limited number of electrons. The electrons in different subshells have different energies and different locations for motion about the nucleus. We also assume a knowledge of the for chemical bonding based on valence shell electron pair sharing and the octet rule. A covalent chemical bond is formed when the two bonded atoms share a pair of valence shell electrons between them. In general, atoms of Groups IV through VII bond so as to complete an octet of valence shell electrons. We finally assume the for understanding and predicting molecular geometries. The pairs of valence shell electrons are arranged in bonding and non-bonding domains, and these domains are separated in space to minimize electron-electron repulsions. This electron domain arrangement determines the molecular geometry. We should expect that the properties of molecules, and correspondingly the substances which they comprise, should depend on the details of the structure and bonding in these molecules. Now that we have developed an understanding of the relationship between molecular structure and chemical bonding, we analyze physical properties of the molecules and compounds of these molecules to relate to this bonding and structure. Simple examples of physical properties which can be related to molecular properties are the melting and boiling temperatures. These vary dramatically from substance to substance, even for substances which appear similar in molecular formulae, with some melting temperatures in the hundreds or thousands of degrees Celsius and others well below \(0^\text{o} \text{C}\). We seek to understand these variations by analyzing molecular structures. To develop this understanding, we will have to apply more details of our understanding of atomic structure and electronic configurations. In our covalent bonding model, we have assumed that atoms "share" electrons to form a bond. However, our knowledge of the properties of atoms reveals that different atoms attract electrons with different strengths, resulting in very strong variations in ionization energies, atomic radii, and electron affinities. We seek to incorporate this information into our understanding of chemical bonding. We begin by analyzing compounds formed from elements from and (e.g. sodium and magnesium). These compounds are not currently part of our Lewis structure model. For example, Sodium, with a single valence electron, is unlikely to gain seven additional electrons to complete an octet. Indeed, the common valence of the alkali metals in Group 1 is 1, not 7, and the common valence of the alkaline earth metals is 2, not 6. Thus, our current model of bonding does not apply to elements in these groups. To develop an understanding of bonding in these compounds, we focus on the halides of these elements. In Table 8.1, we compare physical properties of the chlorides of elements in Groups I and II to the chlorides of the elements of Groups IV, V, and VI, and we see enormous differences. All of the alkali halides and alkaline earth halides are solids at room temperature and have melting points in the hundreds of degrees centigrade. The melting point of \(\ce{NaCl}\) is \(808^\text{o} \text{C}\), for example. By contrast, the melting points of the non-metal halides from Periods 2 and 3, such as \(\ce{CCl_4}\), \(\ce{PCl_3}\), and \(\ce{SCl_2}\), are below \(0^\text{o} \text{C}\), so that these materials are liquids at room temperature. Furthermore, all of these compounds have low boiling points, typically in the range of \(50^\text{o} \text{C}\) to \(80^\text{o} \text{C}\). Second, the non-metal halide liquids are electrical insulators, that is, they do not conduct an electrical current. By contrast, when we melt an alkali halide or alkaline earth halide, the resulting liquid is an excellent electrical conductor. This indicates that these molten compounds consist of ions, whereas the non-metal halides do not. We must conclude that the bonding of atoms in alkali halides and alkaline earth halides differs significantly from bonding in non-metal halides. We need to extend our valence shell electron model to account for this bonding, and in particular, we must account for the presence of ions in the molten metal halides. Consider the prototypical example of \(\ce{NaCl}\). We have already deduced that \(\ce{Cl}\) atoms react so as to form a complete octet of valence shell electrons. Such an octet could be achieved by covalently sharing the single valence shell electron from a sodium atom. However, such a covalent sharing is clearly inconsistent with the presence of ions in molten sodium chloride. Furthermore, this type of bond would predict that \(\ce{NaCl}\) should have similar properties to other covalent chloride compounds, most of which are liquids at room temperature. By contrast, we might imagine that the chlorine atom completes its octet by taking the valence shell electron from a sodium atom, without covalent sharing. This would account for the presence of \(\ce{Na^+}\) and \(\ce{Cl^-}\) ions in molten sodium chloride. In the absence of a covalent sharing of an electron pair, though, what accounts for the stability of sodium chloride as a compound? It is relatively obvious that a negatively charged chloride ion will be attracted electrostatically to a positively charged sodium ion. We must also add to this model, however, the fact that individual molecules of \(\ce{NaCl}\) are not generally observed at temperatures less than \(1465^\text{o} \text{C}\), the boiling point of sodium chloride. Note that, if solid sodium chloride consists of individual sodium ions in proximity to individual chloride ions, then each positive ion is not simply attracted to a single specific negative ion but rather to all of the negative ions in its near vicinity. Hence, solid sodium chloride cannot be viewed as individual \(\ce{NaCl}\) molecules, but must be viewed rather as a lattice of positive sodium ions interacting with negative chloride ions. This type of "ionic" bonding, which derives from the electrostatic attraction of interlocking lattices of positive and negative ions, accounts for the very high melting and boiling points of the alkali halides. We can now draw modified Lewis structures to account for ionic bonding, but these are very different from our previous drawings. Sodium chloride can be represented as shown in Figure 8.1. This indicates explicitly that the bonding is due to positive-negative ion attraction, and not due to sharing of an electron pair. The only sense in which the \(\ce{Na^+}\) ion has obeyed an octet rule is perhaps that, in having emptied its valence shell of electrons, the remaining outer shell of electrons in the ion has the same octet as does a neon atom. We must keep in mind, however, that the positive sodium ion is attracted to many negative chloride ions, and not just the single chloride ion depicted in the Lewis structure. Our Lewis model of bonding, as currently developed, incorporates two extreme views of the distribution of electrons in a bond. In a covalent bond, we have assumed up to this point that the electron pair is shared perfectly. In complete contrast, in ionic bonding we have assumed that the electrons are not shared at all. Rather, one of the atoms is assumed to entirely extract one or more electrons from the other. We might expect that a more accurate description of the reality of chemical bonds falls in general somewhere between these two extremes. To observe this intermediate behavior, we can examine molecular dipole moments. An electric dipole is a spatial separation of positive and negative charges. In the simples case, a positive charge \(Q\) and a negative charge \(-Q\) separated by a distance \(R\) produce a measurable , \(\mu\) equal to \(Q \times R\). An electric field can interact with an electric dipole and can even orient the dipole in the direction of the field. We might initially expect that molecules do not in general have dipole moments. Each atom entering into a chemical bond is electrically neutral, with equal numbers of positive and negative charges. Consequently, a molecule formed from neutral atoms must also be electrically neutral. Although electron pairs are shared between bonded nuclei, this does not affect the total number of negative charges. We might from these simple statements predict that molecules would be unaffected by electric or magnetic fields, each molecule behaving as a single uncharged particle. This prediction is incorrect, however. To illustrate, a stream of water can be deflected by an electrically charged object near the stream, indicating that individual water molecules exhibit a dipole moment. A water molecule is rather more complicated than a simple separation of a positive and negative charges, however. Recall though that a water molecule has equal total numbers of positive and negative charges, consisting of three positively charged nuclei surrounded by ten electrons. Nevertheless, measurements reveal that water has a dipole moment of \(6.17 \times 10^{-30} \: \text{C} \cdot \text{m} =\) 1.85 debye. (The debye is a unit used to measure dipole moments: 1 debye \(= 3.33 \times 10^{-30} \: \text{C} \cdot \text{m}\).) Water is not unique: the molecules of most substances have dipole moments. A sampling of molecules and their dipole moments is given in Table 8.2. Focusing again on the water molecule, how can we account for the existence of a dipole moment in a neutral molecule? The existence of the dipole moment reveals that a water molecule must have an internal separation of positive partial charge \(\delta\) and negative partial charge \(-\delta\). Thus, it must be true that the electrons in the covalent bond between the hydrogen and oxygen are not shared. Rather, the shared electrons must spend more time in the vicinity of one nucleus than the other. The molecule thus has one region where, on average, there is a net surplus of negative charge and one region where, on average, there is a compensating surplus of positive charge, thus producing a molecular dipole. Additional observations reveal that the oxygen "end" of the molecule holds the partial negative charge. Hence, the covalently shared electrons spend more time near the oxygen atom than near the hydrogen atoms. We conclude that oxygen atoms have a greater ability to attract the shared electrons in the bond than do hydrogen atoms. We should not be surprised by the fact that individual atoms of different elements have differing abilities to attract electrons to themselves. We have previously seen that different atoms have greatly varying ionization energies, representing great variation in the extent to which atoms cling to their electrons. We have also seen great variation in the electron affinities of atoms, representing variation in the extent to which atoms attract an added electron. We now define the of an atom as the ability of the atom to attract electrons in a chemical bond. This is different than either ionization energy or electron affinity, because electronegativity is the attraction of electrons , whereas ionization energy and electron affinity refer to removal and attachment of electrons in free atoms. However, we can expect electronegativity to be correlated with electron affinity and ionization energy. In particular, the electronegativity of an atom arises from a combination of properties of the atom, including the size of the atom, the charge on the nucleus, the number of electrons about the nuclei, and the number of electrons in the valence shell. Because electronegativity is an abstractly defined property, it cannot be directly measured. In fact, there are many definitions of electronegativity, resulting in many different scales of electronegativities. However, relative electronegativities can be observed indirectly by measuring molecular dipole moments: in general, the greater the dipole moment, the greater the separation of charges must be, and therefore, the less equal the sharing of the bonding electrons must be. With this in mind, we refer back to the dipoles given in Table 8.2. There are several important trends in these data. Note that each hydrogen halide (\(\ce{HF}\), \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\)) has a significant dipole moment. Moreover, the dipole moments increase as we move the periodic table in the halogen group. We can conclude that fluorine atoms have a greater electronegativity than do chlorine atoms, etc. Note also that \(\ce{HF}\) has a greater dipole moment than \(\ce{H_2O}\), which is in turn greater than that of \(\ce{NH_3}\). We can conclude that electronegativity increases as we move the periodic table from left to right in a single period. These trends hold generally in comparisons of the electronegativities of the individual elements. One set of relative electronegativities of atoms in the first three rows of the periodic table is given in Table 8.3. We might reasonably expect from our analysis to observe a dipole moment in any molecule formed from atoms with different electronegativities. Although this must be the case for a diatomic molecule, this is not necessarily true for a polyatomic molecule, i.e. one with more than two atoms. For example, carbon is more electronegative than hydrogen. However, the simplest molecule formed from carbon and hydrogen (e.g. \(\ce{CH_4}\)) does possess a dipole moment, as we see Table 8.2. Similarly, oxygen is significantly more electronegative than carbon, yet \(\ce{CO_2}\) is a non-polar molecule. An analysis of molecular dipole moments in polyatomic molecules requires us to apply our understanding of molecular geometry. Note that each \(\ce{C-O}\) bond is expected to be polar, due to the unequal sharing of the electron pairs between the carbon and the oxygen. Thus, the carbon atom should have a slight positive charge and the oxygen atom a slight negative charge in each \(\ce{C-O}\) bond. However, since each oxygen atom should have the same net negative charge, neither end of the molecule would display a greater affinity for an electric field. Moreover, because \(\ce{CO_2}\) is linear, the dipole in one \(\ce{C-O}\) bond is exactly offset by the dipole in the opposite direction due to the other \(\ce{C-O}\) bond. As measured by an electric field from a distance, the \(\ce{CO_2}\) molecule does not appear to have separated positive and negative charges and therefore does not display polarity. Thus, in predicting molecular dipoles we must take into account both differences in electronegativity, which affect bond polarity, and overall molecular geometry, which can produce cancelation of bond polarities. Using this same argument, we can rationalize the zero molecular dipole moments observed for other molecules, such as methane, ethene, and acetylene. In each of these molecules, the individual \(\ce{C-H}\) bonds are polar. However, the symmetry of the molecule produces a cancelation of these bond dipoles overall, and none of these molecules have a molecular dipole moment. As an example of how a molecular property like the dipole moment can affect the macroscopic property of a substance, we can examine the boiling points of various compounds. The boiling point of a compound is determined by the strength of the forces between molecules of the compound: the stronger the force, the more energy is required to separate the molecules, the higher the temperature required to provide this energy. Therefore, molecules with strong intermolecular forces have high boiling points. We begin by comparing molecules which are similar in size, such as the hydrides \(\ce{SiH_4}\), \(\ce{PH_3}\), and \(\ce{SH_2}\) from the third period. The boiling points at standard pressure for these molecules are, respectively, \(-11.8^\text{o} \text{C}\), \(-87.7^\text{o} \text{C}\), and \(-60.7^\text{o} \text{C}\). All three compounds are thus gases at room temperature and well below. These molecules have very similar masses and have exactly the same number of electrons. However, the dipole moments of these molecules are very different. The dipole moment of \(\ce{SiH_4}\) is \(0.0 \: \text{D}\), the dipole moment of \(\ce{PH_3}\) is \(0.58 \: \text{D}\), and the dipole moment of \(\ce{SH_2}\) is \(0.97 \: \text{D}\). Note that, for these similar molecules, the higher the dipole moment, the higher the boiling point. Thus, molecules with larger dipole moments generally have stronger intermolecular forces than similar molecules with smaller dipole moments. This is because the positive end of the dipole in one molecule can interact electrostatically with the negative end of the dipole in another molecule, and vice versa. We note, however, that one cannot generally predict from dipole moment information only the relative boiling points of compounds of very dissimilar molecules. ; Chemistry) | 17,620 | 3,324 |
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In all the examples discussed thus far, the reactants were assumed to be present in stoichiometric quantities. Consequently, none of the reactants was left over at the end of the reaction. This is often desirable, as in the case of a space shuttle, where excess oxygen or hydrogen was not only extra freight to be hauled into orbit but also an explosion hazard. More often, however, reactants are present in mole ratios that are not the same as the ratio of the coefficients in the balanced chemical equation. As a result, one or more of them will not be used up completely but will be left over when the reaction is completed. In this situation, the amount of product that can be obtained is limited by the amount of only one of the reactants. The reactant that restricts the amount of product obtained is called the limiting reactant. The reactant that remains after a reaction has gone to completion is in excess. Consider a nonchemical example. Assume you have invited some friends for dinner and want to bake brownies for dessert. You find two boxes of brownie mix in your pantry and see that each package requires two eggs. The balanced equation for brownie preparation is thus \[ 1 \,\text{box mix} + 2 \,\text{eggs} \rightarrow 1 \, \text{batch brownies} \label{4.4.1}\] If you have a dozen eggs, which ingredient will determine the number of batches of brownies that you can prepare? Because each box of brownie mix requires two eggs and you have two boxes, you need four eggs. Twelve eggs is eight more eggs than you need. Although the ratio of eggs to boxes in is 2:1, the ratio in your possession is 6:1. Hence the eggs are the ingredient (reactant) present in excess, and the brownie mix is the limiting reactant. Even if you had a refrigerator full of eggs, you could make only two batches of brownies. Now consider a chemical example of a limiting reactant: the production of pure titanium. This metal is fairly light (45% lighter than steel and only 60% heavier than aluminum) and has great mechanical strength (as strong as steel and twice as strong as aluminum). Because it is also highly resistant to corrosion and can withstand extreme temperatures, titanium has many applications in the aerospace industry. Titanium is also used in medical implants and portable computer housings because it is light and resistant to corrosion. Although titanium is the ninth most common element in Earth’s crust, it is relatively difficult to extract from its ores. In the first step of the extraction process, titanium-containing oxide minerals react with solid carbon and chlorine gas to form titanium tetrachloride (TiCl ) and carbon dioxide. Titanium tetrachloride is then converted to metallic titanium by reaction with magnesium metal at high temperature: \[ TiCl_4 (g) + 2 \, Mg (l) \rightarrow Ti (s) + 2 \, MgCl_2 (l) \label{4.4.2}\] Because titanium ores, carbon, and chlorine are all rather inexpensive, the high price of titanium (about $100 per kilogram) is largely due to the high cost of magnesium metal. Under these circumstances, magnesium metal is the limiting reactant in the production of metallic titanium. With 1.00 kg of titanium tetrachloride and 200 g of magnesium metal, how much titanium metal can be produced according to the equation above? Solving this type of problem requires that you carry out the following steps: 1. To determine the number of moles of reactants present, calculate or look up their molar masses: 189.679 g/mol for titanium tetrachloride and 24.305 g/mol for magnesium. The number of moles of each is calculated as follows: \[ moles \, TiCl_4 = {mass \, TiCl_4 \over molar \, mass \, TiCl_4} \] \[ = 1000 \, g \, TiCl_4 \times {1 \, mol \, TiCl_4 \over 189.679 \, g \, TiCl_4} = 5.272 \, mol \, TiCl_4 \] \[ moles \, Mg = {mass \, Mg \over molar \, mass \, Mg}\] \[ = 200 \, g \, Mg \times {1 \, mol \, Mg \over 24.305 \, g \, Mg } = 8.23 \, mol \, Mg \] 2. There are more moles of magnesium than of titanium tetrachloride, but the ratio is only the following: \[ {mol \, Mg \over mol \, TiCl_4} = {8.23 \, mol \over 5.272 \, mol } = 1.56 \] Because the ratio of the coefficients in the balanced chemical equation is, \[{ 2 \, mol \, Mg \over 1 \, mol \, TiCl_4} = 2 \] there is not have enough magnesium to react with all the titanium tetrachloride. If this point is not clear from the mole ratio, calculate the number of moles of one reactant that is required for complete reaction of the other reactant. For example, there are 8.23 mol of Mg, so (8.23 ÷ 2) = 4.12 mol of TiCl are required for complete reaction. Because there are 5.272 mol of TiCl , titanium tetrachloride is present in excess. Conversely, 5.272 mol of TiCl requires 2 × 5.272 = 10.54 mol of Mg, but there are only 8.23 mol. Therefore, magnesium is the limiting reactant. 3. Because magnesium is the limiting reactant, the number of moles of magnesium determines the number of moles of titanium that can be formed: \[ moles \, Ti = 8.23 \, mol \, Mg = {1 \, mol \, Ti \over 2 \, mol \, Mg} = 4.12 \, mol \, Ti \] Thus only 4.12 mol of Ti can be formed. 4. To calculate the mass of titanium metal that can obtain, multiply the number of moles of titanium by the molar mass of titanium (47.867 g/mol): \[ moles \, Ti = mass \, Ti \times molar \, mass \, Ti = 4.12 \, mol \, Ti \times {47.867 \, g \, Ti \over 1 \, mol \, Ti} = 197 \, g \, Ti \] Here is a simple and reliable way to identify the limiting reactant in any problem of this sort: Density is the mass per unit volume of a substance. If we are given the density of a substance, we can use it in stoichiometric calculations involving liquid reactants and/or products, as Example \(\Page {1}\) demonstrates. Determining the Limiting Reactant and Theoretical Yield for a Reaction: Ethyl acetate (CH CO C H ) is the solvent in many fingernail polish removers and is used to decaffeinate coffee beans and tea leaves. It is prepared by reacting ethanol (C H OH) with acetic acid (CH CO H); the other product is water. A small amount of sulfuric acid is used to accelerate the reaction, but the sulfuric acid is not consumed and does not appear in the balanced chemical equation. Given 10.0 mL each of acetic acid and ethanol, how many grams of ethyl acetate can be prepared from this reaction? The densities of acetic acid and ethanol are 1.0492 g/mL and 0.7893 g/mL, respectively. : reactants, products, and volumes and densities of reactants : mass of product : : Always begin by writing the balanced chemical equation for the reaction: \[ C_2H_5OH (l) + CH_3CO_2H (aq) \rightarrow CH_3CO_2C_2H_5 (aq) + H_2O (l) \] We need to calculate the number of moles of ethanol and acetic acid that are present in 10.0 mL of each. Recall from that the density of a substance is the mass divided by the volume: \[ density = {mass \over volume } \] Rearranging this expression gives mass = (density)(volume). We can replace mass by the product of the density and the volume to calculate the number of moles of each substance in 10.0 mL (remember, 1 mL = 1 cm ): \[ moles \, C_2H_5OH = { mass \, C_2H_5OH \over molar \, mass \, C_2H_5OH } \] \[ = {volume \, C_2H_5OH \times density \, C_2H_5OH \over molar \, mass \, C_2H_5OH}\] \[ = 10.0 \, ml \, C_2H_5OH \times {0.7893 \, g \, C_2H_5OH \over 1 \, ml \, C_2H_5OH} \times {1 \, mole \, C_2H_5OH \over 46.07 \, g\, C_2H_5OH}\] \[ = 0.171 \, mol \, C_2H_5OH \] \[moles \, CH_3CO_2H = {mass \, CH_3CO_2H \over molar \, mass \, CH_3CO_2H} \] \[= {volume \, CH_3CO_2H \times density \, CH_3CO_2H \over molar \, mass \, CH_3CO_2H} \] \[= 10.0 \, ml \, CH_3CO_2H \times {1.0492 \, g \, CH_3CO_2H \over 1 \, ml \, CH_3CO_2H} \times {1 \, mol \, CH_3CO_2H \over 60.05 \, g \, CH_3CO_2H } \] \[= 0.175 \, mol \, CH_3CO_2H \] The number of moles of acetic acid exceeds the number of moles of ethanol. Because the reactants both have coefficients of 1 in the balanced chemical equation, the mole ratio is 1:1. We have 0.171 mol of ethanol and 0.175 mol of acetic acid, so ethanol is the limiting reactant and acetic acid is in excess. The coefficient in the balanced chemical equation for the product (ethyl acetate) is also 1, so the mole ratio of ethanol and ethyl acetate is also 1:1. This means that given 0.171 mol of ethanol, the amount of ethyl acetate produced must also be 0.171 mol: \[ moles \, ethyl \, acetate = molethanol \times {1 \, mol \, ethyl \, acetate \over 1 \, mol \, ethanol } \] \[ = 0.171 \, mol \, C_2H_5OH \times {1 \, mol \, CH_3CO_2C_2H_5 \over 1 \, mol \, C_2H_5OH} \] \[ = 0.171 \, mol \, CH_3CO_2C_2H_5 \] The final step is to determine the mass of ethyl acetate that can be formed, which we do by multiplying the number of moles by the molar mass: \[mass \, of \, ethyl \, acetate = moleethyl \, acetate \times molar \, mass \, ethyl \, acetate\] \[ = 0.171 \, mol \, CH_3CO_2C_2H_5 \times {88.11 \, g \, CH_3CO_2C_2H_5 \over 1 \, mol \, CH_3CO_2C_2H_5}\] \[ = 15.1 \, g \, CH_3CO_2C_2H_5 \] Thus 15.1 g of ethyl acetate can be prepared in this reaction. If necessary, you could use the density of ethyl acetate (0.9003 g/cm3) to determine the volume of ethyl acetate that could be produced: \[ volume \, of \, ethyl \, acetate = 15.1 \, g \, CH_3CO_2C_2H_5 \times { 1 \, ml \, CH_3CO_2C_2H_5 \over 0.9003 \, g\, CH_3CO_2C_2H_5} \] \[ = 16.8 \, ml \, CH_3CO_2C_2H_5 \] Under appropriate conditions, the reaction of elemental phosphorus and elemental sulfur produces the compound P S . How much P S can be prepared starting with 10.0 g of P and 30.0 g of S ? : 35.9 g The concept of limiting reactants applies to reactions carried out in solution as well as to reactions involving pure substances. If all the reactants but one are present in excess, then the amount of the limiting reactant may be calculated as illustrated in Example \(\Page {2}\). Because the consumption of alcoholic beverages adversely affects the performance of tasks that require skill and judgment, in most countries it is illegal to drive while under the influence of alcohol. In almost all US states, a blood alcohol level of 0.08% by volume is considered legally drunk. Higher levels cause acute intoxication (0.20%), unconsciousness (about 0.30%), and even death (about 0.50%). The Breathalyzer is a portable device that measures the ethanol concentration in a person’s breath, which is directly proportional to the blood alcohol level. The reaction used in the Breathalyzer is the oxidation of ethanol by the dichromate ion: \[ 3CH_3 CH_2 OH(aq) + \underset{yellow-orange}{2Cr_2 O_7^{2 -}}(aq) + 16H ^+ (aq) \underset{H_2 SO_4 (aq)}{\xrightarrow{\hspace{10px} Ag ^+\hspace{10px}} } 3CH_3 CO_2 H(aq) + \underset{green}{4Cr^{3+}} (aq) + 11H_2 O(l) \] When a measured volume (52.5 mL) of a suspect’s breath is bubbled through a solution of excess potassium dichromate in dilute sulfuric acid, the ethanol is rapidly absorbed and oxidized to acetic acid by the dichromate ions. In the process, the chromium atoms in some of the Cr O ions are reduced from Cr to Cr . In the presence of Ag ions that act as a catalyst, the reaction is complete in less than a minute. Because the Cr O ion (the reactant) is yellow-orange and the Cr ion (the product) forms a green solution, the amount of ethanol in the person’s breath (the limiting reactant) can be determined quite accurately by comparing the color of the final solution with the colors of standard solutions prepared with known amounts of ethanol. When a measured volume of a suspect’s breath is bubbled through the solution, the ethanol is oxidized to acetic acid, and the solution changes color from yellow-orange to green. The intensity of the green color indicates the amount of ethanol in the sample. A typical Breathalyzer ampul contains 3.0 mL of a 0.25 mg/mL solution of K Cr O in 50% H SO as well as a fixed concentration of AgNO (typically 0.25 mg/mL is used for this purpose). How many grams of ethanol must be present in 52.5 mL of a person’s breath to convert all the Cr to Cr ? volume and concentration of one reactant mass of other reactant needed for complete reaction In any stoichiometry problem, the first step is always to calculate the number of moles of each reactant present. In this case, we are given the mass of K Cr O in 1 mL of solution, which can be used to calculate the number of moles of K Cr O contained in 1 mL: \( \dfrac{moles\: K_2 Cr_2 O_7} {1\: mL} = \dfrac{(0 .25\: \cancel{mg}\: K_2 Cr_2 O_7 )} {mL} \left( \dfrac{1\: \cancel{g}} {1000\: \cancel{mg}} \right) \left( \dfrac{1\: mol} {294 .18\: \cancel{g}\: K_2 Cr_2 O_7} \right) = 8.5 \times 10 ^{-7}\: moles \) Because 1 mol of K Cr O produces 1 mol of Cr O when it dissolves, each milliliter of solution contains 8.5 × 10 mol of Cr O . The total number of moles of Cr O in a 3.0 mL Breathalyzer ampul is thus \( moles\: Cr_2 O_7^{2-} = \left( \dfrac{8 .5 \times 10^{-7}\: mol} {1\: \cancel{mL}} \right) ( 3 .0\: \cancel{mL} ) = 2 .6 \times 10^{-6}\: mol\: Cr_2 O_7^{2–} \) The balanced chemical equation tells us that 3 mol of C H OH is needed to consume 2 mol of Cr O ion, so the total number of moles of C H OH required for complete reaction is \( moles\: of\: C_2 H_5 OH = ( 2.6 \times 10 ^{-6}\: \cancel{mol\: Cr_2 O_7 ^{2-}} ) \left( \dfrac{3\: mol\: C_2 H_5 OH} {2\: \cancel{mol\: Cr _2 O _7 ^{2 -}}} \right) = 3 .9 \times 10 ^{-6}\: mol\: C _2 H _5 OH \) As indicated in the strategy, this number can be converted to the mass of C H OH using its molar mass: \( mass\: C _2 H _5 OH = ( 3 .9 \times 10 ^{-6}\: \cancel{mol\: C _2 H _5 OH} ) \left( \dfrac{46 .07\: g} {\cancel{mol\: C _2 H _5 OH}} \right) = 1 .8 \times 10 ^{-4}\: g\: C _2 H _5 OH \) Thus 1.8 × 10 g or 0.18 mg of C H OH must be present. Experimentally, it is found that this value corresponds to a blood alcohol level of 0.7%, which is usually fatal. The compound -nitrophenol (molar mass = 139 g/mol) reacts with sodium hydroxide in aqueous solution to generate a yellow anion via the reaction Because the amount of -nitrophenol is easily estimated from the intensity of the yellow color that results when excess NaOH is added, reactions that produce -nitrophenol are commonly used to measure the activity of enzymes, the catalysts in biological systems. What volume of 0.105 M NaOH must be added to 50.0 mL of a solution containing 7.20 × 10 g of -nitrophenol to ensure that formation of the yellow anion is complete? 4.93 × 10 L or 49.3 μL In Examples 4.4.1 and 4.4.2, the identities of the limiting reactants are apparent: [Au(CN) ] , LaCl , ethanol, and -nitrophenol. When the limiting reactant is not apparent, it can be determined by comparing the molar amounts of the reactants with their coefficients in the balanced chemical equation. The only difference is that the volumes and concentrations of solutions of reactants, rather than the masses of reactants, are used to calculate the number of moles of reactants, as illustrated in Example \(\Page {3}\). When aqueous solutions of silver nitrate and potassium dichromate are mixed, an exchange reaction occurs, and silver dichromate is obtained as a red solid. The overall chemical equation for the reaction is as follows: \(2AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2KNO_3(aq) \) What mass of Ag Cr O is formed when 500 mL of 0.17 M K Cr O are mixed with 250 mL of 0.57 M AgNO ? balanced chemical equation and volume and concentration of each reactant mass of product The balanced chemical equation tells us that 2 mol of AgNO (aq) reacts with 1 mol of K Cr O (aq) to form 1 mol of Ag Cr O (s) (Figure \(\Page {2}\)). The first step is to calculate the number of moles of each reactant in the specified volumes: \[ moles\: K_2 Cr_2 O_7 = 500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .17\: mol\: K_2 Cr_2 O_7} {1\: \cancel{L}} \right) = 0 .085\: mol\: K_2 Cr_2 O_7 \] \[ moles\: AgNO_3 = 250\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .57\: mol\: AgNO_3} {1\: \cancel{L}} \right) = 0 .14\: mol\: AgNO_3 \] Now determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient: \[K_2 Cr_2 O_7: \: \dfrac{0 .085\: mol} {1\: mol} = 0 .085 \] \[ AgNO_3: \: \dfrac{0 .14\: mol} {2\: mol} = 0 .070 \] Because 0.070 < 0.085, we know that AgNO is the limiting reactant. Each mole of Ag Cr O formed requires 2 mol of the limiting reactant (AgNO ), so we can obtain only 0.14/2 = 0.070 mol of Ag Cr O . Finally, convert the number of moles of Ag Cr O to the corresponding mass: \[ mass\: of\: Ag_2 Cr_2 O_7 = 0 .070\: \cancel{mol} \left( \dfrac{431 .72\: g} {1 \: \cancel{mol}} \right) = 30\: g \: Ag_2 Cr_2 O_7 \] The Ag and Cr O ions form a red precipitate of solid Ag Cr O , while the K and NO ions remain in solution. (Water molecules are omitted from molecular views of the solutions for clarity.) Aqueous solutions of sodium bicarbonate and sulfuric acid react to produce carbon dioxide according to the following equation: \(2NaHCO_3(aq) + H_2SO_4(aq) \rightarrow 2CO_2(g) + Na_2SO_4(aq) + 2H_2O(l)\) If 13.0 mL of 3.0 M H SO are added to 732 mL of 0.112 M NaHCO , what mass of CO is produced? 3.4 g Limiting Reactant Problems Using Molarities: The stoichiometry of a reaction describes the relative amounts of reactants and products in a balanced chemical equation. A stoichiometric quantity of a reactant is the amount necessary to react completely with the other reactant(s). If a quantity of a reactant remains unconsumed after complete reaction has occurred, it is in excess. The reactant that is consumed first and limits the amount of product(s) that can be obtained is the limiting reactant. To identify the limiting reactant, calculate the number of moles of each reactant present and compare this ratio to the mole ratio of the reactants in the balanced chemical equation. The maximum amount of product(s) that can be obtained in a reaction from a given amount of reactant(s) is the theoretical yield of the reaction. The actual yield is the amount of product(s) actually obtained in the reaction; it cannot exceed the theoretical yield. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield, expressed as a percentage. ( ) | 18,291 | 3,325 |
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Chemical formulas such as \(\ce{HClO4}\) can be divided into empirical formula, molecular formula, and structural formula. Chemical symbols of elements in the chemical formula represent the elements present, and subscript numbers represent mole proportions of the proceeding elements. Note that no subscript number means a subscript of 1. From a chemical point of view, an element contained in the substance is a fundamental question, and we represent the elemental composition by a chemical formula, such as \(\ce{H2O}\) for water. This formula implies that the water molecules consist of 2 hydrogen and 1 oxygen atoms. The formula \(\ce{H2O}\) is also the molecular formula of water. For non-molecular substances such as table salt, we represent the composition with an empirical formula. Sodium chloride is represented by \(\ce{NaCl}\), meaning that sodium and chlorine ratio in sodium chloride is 1 to 1. Again, the subscript 1 is omitted. Since table salt is an ionic compound, the formula implies that numbers of \(\ce{Na+}\) ions and \(\ce{Cl-}\) ions are the same in the solid. The subscript numbers in an empirical formula should have no common divisor. A structural formula reflects the bonding of atoms in a molecule or ion. For example, ethanol can be represented by \(\ce{CH3CH2OH}\). This is a simple way of representing a more elaborated structure shown on your left. Molecular structures are often beautiful, but the representation is an artwork. For example, a 3-dimensional structure of cyclohexane is shown on the right. This is a chair form, and another structure has a boat form. You will learn more about it in organic chemistry. The molecular formula of benzene is \(\ce{C6H6}\), and its empirical formula is \(\ce{CH}\). You may refer to a substance by its name, and recognize it by its properties. Properties are related to the structure and the composition of the molecules. Knowing the chemical formula is a giant step towards understanding a substance. The formula weight is the sum of all the atomic weights in a formula. The evaluation of formula weight is illustrated in this example. What is the formula weight of sufuric acid \(\ce{H2SO4}\)? The formula also indicates a mass as the sum of masses calculated this way \(\mathrm{2\times1.008 + 32.0 + 4\times16.0 = 98.0}\) where 1.008, 32.0 and 16.0 are the atomic weights of \(\ce{H}\), \(\ce{S}\), and \(\ce{O}\) respectively. If the formula is a molecular formula, the mass associated with it is called molecular mass or molecular weight. As an exercise, work out the following problem. What is the molecular weight of caffeine, \(\ce{C8H10N4O2}\)? The diagram shown here is a model of the caffeine molecule. With the aid of a table of atomic weights, a formula indirectly represents the formula weight. If the formula is a molecular formula, it indirectly represents the molecular weight. For simplicity, we may call these weights molar masses, which can be formula weights or molecular weights. A chemical formula not only represents what a substance is made of, it provides a great deal of information about the substance. Do you know that chemical formulas are used all over the world, regardless of the language? Chinese, Russian, Japanese, African, and South Americans use the same notations we do. Thus, \(\ce{H2S}\) is recognized as a smelly gas all over the world. Chemical formula is an international or universal language. A chemical formula not only gives the formula weight, it accurately represents the percentages of elements in a compound. On the other hand, if you know the percentage of a compound, you may figure out its formula. Percentage based on weights is called weight percentage, and percentage based on the numbers of atoms or moles is called mole percentage. What are the weight and mole percentages of \(\ce{S}\) in sufuric acid? From example 1, we know that there are 32.0 g of \(\ce{S}\) in 98.0 g of sulfuric acid. Thus the weight percentage is \(\mathrm{Weight\: percentage = \dfrac{32}{98} = 32.7\%}\) From the formula, there is one \(\ce{S}\) atom among 7 atoms in \(\ce{H2SO4}\) \(\mathrm{Mole\: percentage = \dfrac{1}{7} = 14.3\%}\) You have learned what weight and mole percentages are and how to evaluate them in this example. As an exercise, work out the following problem: What are the weight and mole percentages of \(\ce{C}\), \(\ce{H}\), \(\ce{N}\), and \(\ce{O}\) for caffeine, \(\ce{C8H10N4O2}\)? How would you find the chemical formula of a substance? If you know the substance, its formula and other information is usually listed in a handbook. Handbooks such as the contain information on millions of substances. If you are a researcher and you made a new compound that no one has ever made before, then you need to determine its empirical or molecular formula. For an organic compound, you burn it completely to convert all carbon (\(\ce{C}\)) to \(\ce{CO2}\), and all hydrogen (\(\ce{H}\)) to \(\ce{H2O}\). \(\mathrm{C_xH_{2y} \xrightarrow{(burned\: in\: O_2)} x\, CO_2 + y\, H_2O}\) Thus, from the weight of \(\ce{CO2}\) and \(\ce{H2O}\) produced by burning a definite amount of the substance, you can figure out the percent of \(\ce{C}\) and \(\ce{H}\) in the compound. Nitrogen is determined by converting it to \(\ce{NH3}\). The amount of \(\ce{NH3}\) can be determined by titration, and the percentage can also be determined. Percentage of \(\ce{O}\) is usually obtained by subtracting all percentages of \(\ce{C}\), \(\ce{H}\), and \(\ce{N}\), if the compound does not contain any other element. A compound contains 92.3 weight percent of carbon and 7.7 weight percent of \(\ce{H}\). What is the empirical formula? Assume that you have 100 g of the compound, then you have 92.3 g of carbon and 7.7 g of hydrogen. Thus the mole ratio of \(\ce{C}\) to \(\ce{H}\) should be \(\mathrm{\dfrac{92.3}{12}:\dfrac{7.7}{1.008} = 7.7 : 7.7 = 1 : 1}\) Thus, the empirical formula is \(\ce{CH}\). You have learned how to determine a chemical formula if the percentages of various elements present in the compound are known in this example. To test your skill, you may be asked to work out the empirical formula of any compound. Try this problem: Aspartic acid contains 36.09% \(\ce{C}\), 5.30% \(\ce{H}\), 10.52% \(\ce{N}\), and 48.08 \(\ce{O}\) by weight. What is the empirical formula for aspartic acid?
Aspartic acid is one of the non-essential amino acids, usually present in young plants. It is obtained by hydrolysis of asparagine, which is abundant in asparagus A compound with an empirical formula of \(\ce{CH}\) has a molecular weight of 78 g/mol. What is the molecular formula? The formula weight of \(\ce{CH}\) is 13.0. Since 78/13 = 6, the molecular formula is \(\ce{C6H6}\), the formula for benzene. This example illustrates the difference between empirical and molecular formula, for which the molecular weight must be known. When 1.00 g of benzene is burned, how much \(\ce{CO2}\) and \(\ce{H2O}\) should be produced? \(\mathrm{1\:g\:CH \times \dfrac{1\:mol\:C}{13\:g\:CH}\times\dfrac{1\:mol\:CO_2}{1\:mol\:C}\times\dfrac{44.0\:g\:CO_2}{1\:mol\:CO_2} = 3.38\:g\:CO_2}\) Use the same method to calculate the amount of \(\ce{H2O}\) produced (Ans. 0.692 g). When 1.00 g of a compound containing only carbon and hydrogen is burned completely, 3.14 g of \(\ce{CO2}\) and 1.29 g of \(\ce{H2O}\) is produced. What is the empirical formula? Amounts of carbon and \(\ce{H}\) in 3.14 g of carbon dioxide, \(\mathrm{3.14\: g\: CO_2\times \dfrac{12\: g\: C}{44\: g\: CO_2}\times\dfrac{1\: mol\: C}{12\: g\: C}= 0.0714\: mol\: C}\) \(\mathrm{1.29\: g\: H_2O\times \dfrac{2\: g\: H}{18\: g\: H_2O}\times\dfrac{1\: mol\: H}{1\: g\: H}= 0.143\: mol\: H}\) Thus, mole ratio of \(\ce{C}\) : \(\ce{H}\) is 0.0714 : 0.143 = 1 : 2. Therefore, the empirical formula is \(\ce{CH2}\) The molecular formula for ethylene is \(\ce{C2H4}\) and cyclohexane is \(\ce{C6H12}\). What are their molecular weights? Chloroform is a common solvent used in chemical labs. It has a molecular formula of \(\ce{CHCl3}\). What is the weight percentage of chlorine (\(\ce{Cl}\))? (Atomic weight, \(\ce{Cl}\), 35.453; \(\ce{H}\), 1.00794; \(\ce{C}\), 12.0110) :
You should understand the reason for using this formula to calculate it: \(\mathrm{\dfrac{3\times 35.453}{3\times 35.453 + 12.011 + 1.00794}= 89.094\%\: (weight\: percentage)}\) What is the weight percentage of \(\ce{C}\) in \(\ce{CCl4}\)? These examples illustrate some fundamental types of problems related to the understanding of the chemical formula. Their solutions are related to the skills outlined at the start of this page. What is the molecular weight of sugar \(\ce{C12H22O11}\)? ~12*12+22+11*16 Skill - Know where to find atomic weights and evaluate molecular weights for:
caffeine, \(\ce{C8H10N4O2}\) (194.2);
calcium aspirin, \(\ce{Ca(OOCC6H4OCOCH3)2}\) (398.4);
ascorbic acid (vitamin C), \(\ce{C6H8O6}\);
quartz (silicon), \(\ce{SiO2}\);
ruby, nearly all \(\ce{Al2O3}\). 12*12/(12*12+22+11*16) and (12/45) Skill - Evaluate weight and mole percentages of any element in any substance. \(\ce{C9H8O}\) Skill - Find a chemical formula when weight percentages are known. 2.64 kg Skill - Calculate the amount of water produced in this case. 111 g Skill - Calculate the amount (kg) of \(\ce{C}\) in a block of dry ice weighing 88 kg. \(\ce{CH}\) Skill - Describe elemental analysis for the determination of chemical formula. 0.0017 mol Skill - Determination of chemical formula from elemental analysis. Two \(\ce{H}\) atoms in an \(\ce{H2O}\) molecule. Skill: Explain the meaning of chemical formula. How many hydrogen atoms are there in an ethanol molecule, \(\ce{C2H5OH}\)? 27.3% Skill: The weight percentage and mole percentage can be calculated if you know the chemical formula.
If you reduce 44.0 grams of \(\ce{CO2}\) to carbon, how many grams of \(\ce{C}\) will you get? 50% by mole Skill: Differentiate mole percentage and weight percentage. The empirical formula for benzene is \(\ce{CH}\). Its molecular weight was determined to be 78. What is the molecular formula? \(\ce{C6H6}\) Skill: Differentiate empirical and molecular formulas. | 10,147 | 3,327 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.09%3A_Standard_Pressure |
You will often find enthalpy changes indicated by Δ and called . The superscript is added to indicate that the enthalpy change has occurred at the . Unless very high pressures are involved, Δ changes very little with a change in pressure, and so we have ignored superscripts up to this point. However, two other properties of matter called the entropy, symbol , and the free energy, symbol , are . These are quite sensitive to pressure, and the inclusion of this superscript is important. For reasons of consistency therefore we will indicate standard enthalpy changes as Δ from now on. | 606 | 3,328 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.09%3A_Calculating_the_Molar_Mass_of_a_Gas |
Helium has long been used in balloons and blimps. Since it is much less dense than air, it will float above the ground. Small balloons filled with helium are often affordable and available at stores, but large ones are much more expensive (and require a lot more helium). A chemical reaction, which produces a gas, is performed. The produced gas is then collected and its mass and volume are determined. The molar mass and volume are determined. The molar mass of the unknown gas can be found using the ideal gas law, provided the temperature and pressure of the gas are also known. A certain reaction occurs, producing an oxide of nitrogen as a gas. The gas has a mass of \(1.211 \: \text{g}\) and occupies a volume of \(677 \: \text{mL}\). The temperature in the laboratory is \(23^\text{o} \text{C}\) and the air pressure is \(0.987 \: \text{atm}\). Calculate the molar mass of the gas and deduce its formula. Assume the gas is ideal. First the ideal gas law will be used to solve for the moles of unknown gas \(\left( n \right)\). Then the mass of the gas divided by the moles will give the molar mass. \[n = \frac{PV}{RT} = \frac{0.987 \: \text{atm} \times 0.677 \: \text{L}}{0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol} \times 296 \: \text{K}} = 0.0275 \: \text{mol}\nonumber \] Now divide \(\text{g}\) by \(\text{mol}\) to get the molar mass. \[\text{molar mass} = \frac{1.211 \: \text{g}}{0.0275 \: \text{mol}} = 44.0 \: \text{g/mol}\nonumber \] Since \(\ce{N}\) has a molar mass of \(14 \: \text{g/mol}\) and \(\ce{O}\) has a molar mass of \(16 \: \text{g/mol}\), the formula \(\ce{N_2O}\) would produce the correct molar mass. The \(R\) value that corresponds to a pressure in \(\text{atm}\) was chosen for this problem. The calculated molar mass gives a reasonable formula for dinitrogen monoxide. The ideal gas law can be used to find the density of a gas at conditions that are not standard. For example, we will determine the density of ammonia gas \(\left( \ce{NH_3} \right)\) at \(0.913 \: \text{atm}\) and \(20^\text{o} \text{C}\), assuming the ammonia is ideal. First, the molar mass of ammonia is calculated to be \(17.04 \: \text{g/mol}\). Next, assume exactly \(1 \: \text{mol}\) of ammonia \(\left( n = 1 \right)\) and calculate the volume that such an amount would occupy at the given temperature and pressure. \[V = \frac{nRT}{P} = \frac{1.00 \: \text{mol} \times 0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol} \times 293 \: \text{K}}{0.913 \: \text{atm}} = 26.3 \: \text{L}\nonumber \] Now the density can be calculated by dividing the mass of one mole of ammonia by the volume above. \[\text{Density} = \frac{17.04 \: \text{g}}{26.3 \: \text{L}} = 0.648 \: \text{g/L}\nonumber \] As a point of comparison, this density is slightly less than the density of ammonia at , which is equal to \(\frac{\left( 170.4 \: \text{g/mol} \right)}{\left( 22.4 \: \text{L/mol} \right)} = 0.761 \: \text{g/L}\). It makes sense that the density should be lower compared to that at STP since both the increase in temperature (from \(0^\text{o} \text{C}\) to \(20^\text{o} \text{C}\)) and the decrease in pressure (from \(1 \: \text{atm}\) to \(0.913 \: \text{atm}\)) would cause the \(\ce{NH_3}\) molecules to spread out a bit further from one another. | 3,301 | 3,329 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/06%3A_Electron_Transfer/6.01%3A_Biological_Redox_Components |
Three types of oxidation-reduction (redox) centers are found in biology: protein side chains, small molecules, and redox cofactors. The first class is frequently overlooked by mechanistic enzymologists. The sulfhydryl group of cysteine is easily oxidized to produce a dimer, known as cystine: \[2R-SH \xrightarrow[-2H^{+}]{-2e^{-}} R-S-S-R \tag{6.1}\] This type of interconversion is known to occur in several redox proteins, including xanthine oxidase, mercuric ion reductase, and thioredoxin. Other enzyme systems display spectral evidence pointing to the presence of a protein-based radical in at least one intermediate. EPR spectroscopy provides a powerful tool in studying such systems; the observation of a g = 2.0 signal that cannot be attributed to impurities or an organic redox cofactor is generally taken to be evidence for a protein-based radical. Radicals localized on tyrosine (e.g., in photosystem II and the B2 subunit of ribonucleotide reductase ) and tryptophan (e.g., in yeast cytochrome c peroxidase ) have been unambiguously identified using EPR techniques together with protein samples containing isotopically labeled amino acids (e.g., perdeuterated Tyr) or single amino-acid mutations (e.g., Trp → Phe). A variety of small molecules, both organic and inorganic, can function as redox reagents in biological systems. Of these, only the nicotinamide and quinone coenzymes are found throughout the biosphere. Nicotinamide adenine dinucleotide (NAD) and nicotinamide adenine dinucleotide phosphate (NADP) participate in a wide variety of biological redox reactions. The 4-position of the pyridine ring is the reactive portion of both molecules (Figure 6.1). Both typically function as 2-electron redox reagents. In contrast, quinones may function as either 1- or 2-electron carriers: \[Q \xrightleftharpoons{e^{-},H^{+}} QH \cdotp \xrightleftharpoons{e^{-},H^{+}} QH_{2} \tag{6.2}\] Free-radical semiquinone (QH•) intermediates have been detected by EPR spectroscopy in some electron transfers. Coenzyme Q, also called ubiquinone because it occurs in virtually all cells, contains a long isoprenoid tail that enables it to diffuse through membranes rapidly. This quinone derivative, which occurs in both free and protein-bound forms, is called ubiquinol when reduced (Figure 6.2). Other types of quinones are less frequently found in cells. Metalloproteins containing a single type of redox cofactor can be divided into two general classes: electron carriers and proteins involved in the transport or activation of small molecules. Adman has identified some of the factors that seem to be characteristic of electron-transfer proteins (these proteins are sometimes called "electron transferases"): (a) possession of a suitable cofactor to act as an electron sink; (b) placement of the cofactor close enough to the protein surface to allow electrons to move in and out; (c) existence of a hydrophobic shell adjacent to, but not always entirely surrounding, the cofactor; (d) small structural changes accompanying electron transfer; and (e) an architecture that permits slight expansion or contraction in preferred directions upon electron transfer. Proteins that function as electron transferases typically place their prosthetic groups in a hydrophobic environment and may provide hydrogen bonds (in addition to ligands) to assist in stabilizing both the oxidized and the reduced forms of the cofactor. Metal-ligand bonds remain intact upon electron transfer to minimize inner-sphere reorganization (discussed in Section III). Many of the complex multisite metalloenzymes (e.g., cytochrome c oxidase, xanthine oxidase, the nitrogenase FeMo protein) contain redox centers that function as intramolecular electron transferases, shuttling electrons to/from other metal centers that bind exogenous ligands during enzymatic turnover. There are four classes of electron transferases, each of which contains many members that exhibit important structural differences: flavodoxins, blue copper proteins, iron-sulfur proteins, and cytochromes. The flavodoxins are atypical in that they contain an organic redox cofactor, flavin mononucleotide (FMN; see Figure 6.3). These proteins have molecular weights in the 8-13 kDa range, and are found in many species of bacteria and algae. The FMN cofactor is found at one end of the protein, near the molecular surface, but only the dimethylbenzene portion of FMN is significantly exposed to the solvent (Figure 6.4). FMN can act as either a 1- or a 2-electron redox center. In solution, the semiquinone form of free FMN is unstable, and disproportionates to the quinone (oxidized) and hydroquinone (reduced) forms. Hence, free FMN functions in effect as a 2-electron reagent. FMN in flavodoxins, on the other hand, can function as a single-electron carrier. This is easily discerned by comparing reduction potentials for free and protein-bound FMN (Table 6.1). Clearly, the protein medium is responsible for this drastic alteration in oxidation-state stability. From an NMR study of the flavodoxin quinone/semiquinone and semiquinone/hydroquinone electron self-exchange rates, it was concluded that the latter is approximately 300 times faster than the former, in keeping with the view that the physiologically relevant redox couple is semiquinone/hydroquinone. The blue copper proteins are characterized by intense S(Cys) → Cu chargetransfer absorption near 600 nm, an axial EPR spectrum displaying an unusually small hyperfine coupling constant, and a relatively high reduction potential. With few exceptions (e.g., photosynthetic organisms), their precise roles in bacterial and plant physiology remain obscure. X-ray structures of several blue copper proteins indicate that the geometry of the copper site is approximately trigonal planar, as illustrated by the azurin structure (Figure 6.5). In all these proteins, three ligands (one Cys, two His) bind tightly to the copper in a trigonal arrangement. Differences in interactions between the copper center and the axially disposed ligands may significantly contribute to variations in reduction potential that are observed for the blue copper electron transferases. For example, E°' = 276 mV for azurin, whereas that of plastocyanin is 360 mV. In azurin, the Cu-S(Met) bond is 0.2 Å longer than in poplar plastocyanin, and there is a carbonyl oxygen 3.1 Å from the copper center, compared with 3.8 Å in plastocyanin. These differences in bond lengths are expected to stabilize Cu in azurin to a greater extent than in plastocyanin, and result in a lower E°' value for azurin. The iron-sulfur proteins play important roles as electron carriers in virtually all living organisms, and participate in plant photosynthesis, nitrogen fixation, steroid metabolism, and oxidative phosphorylation, as well as many other processes (Chapter 7). The optical spectra of all iron-sulfur proteins are very broad and almost featureless, due to numerous overlapping charge-transfer transitions that impart red-brown-black colors to these proteins. On the other hand, the EPR spectra of iron-sulfur clusters are quite distinctive, and they are of great value in the study of the redox chemistry of these proteins. The simplest iron-sulfur proteins, known as rubredoxins, are primarily found in anaerobic bacteria, where their function is unknown. Rubredoxins are small proteins (6 kDa) and contain iron ligated to four Cys sulfurs in a distorted tetrahedral arrangement. The E°' value for the Fe / couple in water is 770 mV; that of rubredoxin is -57 mV. The reduction potentials of iron-sulfur proteins are typically quite negative, indicating a stabilization of the oxidized form of the redox couple as a result of negatively charged sulfur ligands. The [2Fe-2S] ferredoxins (10-20 kDa) are found in plant chloroplasts and mammalian tissue. The structure of ferredoxin confirmed earlier suggestions, based on EPR and Mössbauer studies, that the iron atoms are present in a spin-coupled [2Fe-2S] cluster structure. One-electron reduction (E°' ~ -420 mV) of the protein results in a mixed-valence dimer (Equation 6.3): \[[Fe_{2}S_{2}(SR)_{4}]^{2-} \xrightleftharpoons[-e^{-}]{+e^{-}} [Fe_{2}S_{2}(SR)_{4}]^{3-} \tag{6.3}\] \[Fd_{ox} \qquad \qquad \qquad Fd_{red}\] \[2Fe(III) \qquad \qquad Fe(II) + Fe(III)\] The additional electron in Fd is associated with only one of the iron sites, resulting in a so-called trapped-valence structure. The [Fe S (SR) ] cluster oxidation state, containing two ferrous ions, can be produced when strong reductants are used. Four-iron clusters [4Fe-4S] are found in many strains of bacteria. In most of these bacterial iron-sulfur proteins, also termed ferredoxins, two such clusters are present in the protein. These proteins have reduction potentials in the -400 mV range and are rather small (6-10 kDa). Each of the clusters contains four iron centers and four sulfides at alternate comers of a distorted cube. Each iron is coordinated to three sulfides and one cysteine thiolate. The irons are strongly exchange-coupled, and the [4Fe-4S] cluster in bacterial ferredoxins is paramagnetic when reduced by one electron. The so-called "high-potential ironsulfur proteins" (HiPIPs) are found in photosynthetic bacteria, and exhibit anomalously high (~350 mV) reduction potentials. The HiPIP (10 kDa) structure demonstrates that HiPIPs are distinct from the [4Fe-4S] ferredoxins, and that the reduced HiPIP cluster structure is significantly distorted, as is also observed for the structure of the oxidized ferredoxin. In addition, oxidized HiPIP is paramagnetic, whereas the reduced protein is EPR-silent. This bewildering set of experimental observations can be rationalized in terms of a "three-state" hypothesis (i.e., [4Fe-4S(SR) ] clusters exist in three physiological oxidation states). This hypothesis nicely explains the differences in magnetic behavior and redox properties observed for these iron-sulfur proteins (Equation 6.4): \[[4Fe-4S(SR)_{4}]^{-} \xrightleftharpoons[-e^{-}]{+e^{-}} [4Fe-4S(SR)_{4}]^{2-} \xrightleftharpoons[-e^{-}]{+e^{-}} [4Fe-4S(SR)_{4}]^{3-} \tag{6.4}\] \[HiPIP_{ox} \qquad \qquad \qquad HiPIP_{red} \qquad \qquad \qquad Ferredoxin_{red}\] \[Ferredoxing_{ox}\] The bacterial ferredoxins and HiPIPs all possess tetracubane clusters containing thiolate ligands, yet the former utilize the -2/-3 cluster redox couple, whereas the latter utilize the -1/-2 cluster redox couple. The protein environment thus exerts a powerful influence over the cluster reduction potentials. This observation applies to classes of electron transferases—the factors that are critical determinants of cofactor reduction potentials are poorly understood at present but are thought to include the low dielectric constants of protein interiors (~4 for proteins vs. ~78 for H O), electrostatic effects due to nearby charged amino-acid residues, hydrogen bonding, and geometric constraints imposed by the protein. As a class, the cytochromes are the most thoroughly characterized of the electron transferases. By definition, a cytochrome contains one or more heme cofactors. These proteins were among the first to be identified in cellular extracts because of their distinctive optical properties, particularly an intense absorption in the 410-430 nm region (called the Soret band). Cytochromes are typically classified on the basis of heme type. Figure 6.6 displays the three most commonly encountered types of heme: heme a possesses a long phytyl "tail" and is found in cytochrome c oxidase; heme b is found in b-type cytochromes and globins; heme c is covalently bound to c-type cytochromes via two thioether linkages. Cytochrome nomenclature presents a real challenge! Some cytochromes are designated according to the historical order of discovery, e.g., cytochrome c in bacterial photosynthesis. Others are designated according to the \(\lambda_{max}\)of the \(\alpha\) band in the absorption spectrum of the reduced protein (e.g., cytochrome c ). Cytochromes c are widespread in nature. Ambler divided these electron carriers into three classes on structural grounds. The Class I cytochromes c contain axial His and Met ligands, with the heme located near the N-terrninus of the protein. These proteins are globular, as indicated by the ribbon drawing of tuna cytochrome c (Figure 6.7). X-ray structures of Class I cytochromes c from a variety of eukaryotes and prokaryotes clearly show an evolutionarily conserved "cytochrome fold," with the edge of the heme solvent-exposed. The reduction potentials of these cytochromes are quite positive (200 to 320 mV). Mammalian cytochrome c, because of its distinctive role in the mitochondrial electron-transfer chain, will be discussed later. Class II cytochromes c (E°' ~ -100 mV) are found in photosynthetic bacteria, where they serve an unknown function. Unlike their Class I cousins, these c-type cytochromes are high-spin: the iron is five-coordinate, with an axial His ligand. These proteins, generally referred to as cytochromes c' , are four-\(\alpha\)-helix bundles (Figure 6.8). The vacant axial coordination site is buried in the protein interior. Finally, Class III cytochromes c, also called cytochromes c , contain four hemes, each ligated by two axial histidines. These proteins are found in a restricted class of sulfate-reducing bacteria and may be associated with the cytoplasmic membrane. The low molecular weights of cytochromes c (~14. 7 kDa) require that the four hemes be much more exposed to the solvent than the hemes of other cytochromes (see Figure 6.9), which may be in part responsible for their unusually negative (-200 to -350 mV) reduction potentials. These proteins possess many aromatic residues and short heme-heme distances, two properties that could be responsible for their anomalously large solid-state electrical conductivity. | 13,892 | 3,330 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/03%3A_Chemical_Compounds/3.3%3A_Composition_of_Chemical_Compounds |
\[2\times(1.0079\;amu) + 1 \times (15.9994 \;amu) = 18.01528 \;amu\] If a substance exists as discrete molecules (as with atoms that are together) then the the , and the the . For example, carbon, hydrogen and oxygen can chemically bond to form a molecule of the sugar with the chemical and molecular formula of C H O . The formula weight and the molecular weight of glucose is thus: Ionic substances are not chemically bonded and do not exist as discrete molecules. However, they do associate in discrete ratios of ions. Thus, we can describe their formula weights, but not their . Table salt (NaCl), for example, has a formula weight of: \[23.0\; amu + 35.5 \;amu = 58.5 \;amu\] In some types of analyses of it is important to know the of each type of element in a compound. states that a chemical compound always contains the same proportion of elements by mass; that is, the percent composition—the percentage of each element present in a pure substance—is (although there are exceptions to this law). Take for example methane (\(CH_4\)) with a Formula and molecular weight: \[1\times (12.011 \;amu) + 4 \times (1.008) = 16.043 \;amu\] the relative (mass) percentages of carbon and hydrogen are \[\%C = \dfrac{1 \times (12.011\; amu)}{16.043 amu} = 0.749 = 74.9\%\] \[\%H = \dfrac{4 \times (1.008 \;amu)}{16.043\; amu} = 0.251 = 25.1\%\] A more complex example is sucrose (table sugar) is 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. This means that 100.00 g of sucrose always contains 42.11 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen. First the molecular formula of (C H O ) is used to calculate the mass percentage of the component elements; the mass percentage can then be used to determine an . According to its molecular formula, each molecule of sucrose contains 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. A mole of sucrose molecules therefore contains 12 mol of carbon atoms, 22 mol of hydrogen atoms, and 11 mol of oxygen atoms. This information can be used to calculate the mass of each element in 1 mol of sucrose, which gives the molar mass of sucrose. These masses can then be used to calculate the percent composition of sucrose. To three decimal places, the calculations are the following: \[ \text {mass of C/mol of sucrose} = 12 \, mol \, C \times {12.011 \, g \, C \over 1 \, mol \, C} = 144.132 \, g \, C \label{3.3.1a}\] \[ \text {mass of H/mol of sucrose} = 22 \, mol \, H \times {1.008 \, g \, H \over 1 \, mol \, H} = 22.176 \, g \, C \label{3.3.1b}\] \[ \text {mass of O/mol of sucrose} = 11 \, mol \, O \times {15.999 \, g \, O \over 1 \, mol \, O} = 175.989 \, g \, O \label{3.3.1c}\] Thus 1 mol of sucrose has a mass of 342.297 g; note that more than half of the mass (175.989 g) is oxygen, and almost half of the mass (144.132 g) is carbon. The mass percentage of each element in sucrose is the mass of the element present in 1 mol of sucrose divided by the molar mass of sucrose, multiplied by 100 to give a percentage. The result is shown to two decimal places: \[ \text {mass % C in Sucrose} = {\text {mass of C/mol sucrose} \over \text {molar mass of sucrose} } \times100 = {144.132 \, g \, C \over 342.297 \, g/mol } \times 100 = 42.12 \% \] \[ \text {mass % H in Sucrose} = {\text {mass of H/mol sucrose} \over \text {molar mass of sucrose} } \times100 = {22.176 \, g \, H \over 342.297 \, g/mol } \times 100 = 6.48 \% \] \[ \text {mass % O in Sucrose} = {\text {mass of O/mol sucrose} \over \text {molar mass of sucrose} } \times100 = {175.989 \, g \, O \over 342.297 \, g/mol } \times 100 = 51.41 \% \] This can be checked by verifying that the sum of the percentages of all the elements in the compound is 100%: \[ 42.12\% + 6.48\% + 51.41\% = 100.01\%\] If the sum is not 100%, an error has been made in calculations. (Rounding to the correct number of decimal places can, however, cause the total to be slightly different from 100%.) Thus 100.00 g of sucrose contains 42.12 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen; to two decimal places, the percent composition of sucrose is indeed 42.12% carbon, 6.48% hydrogen, and 51.41% oxygen. It is also possible to calculate mass percentages using atomic masses and molecular masses, with atomic mass units. Because the answer is a ratio, expressed as a percentage, the units of mass cancel whether they are grams (using molar masses) or atomic mass units (using atomic and molecular masses). Aspartame is the artificial sweetener sold as NutraSweet and Equal. Its molecular formula is C H N O . : molecular formula and mass of sample : mass percentage of all elements and mass of one element in sample : : a. We calculate the mass of each element in 1 mol of aspartame and the molar mass of aspartame, here to three decimal places: \[ 14 \,C (14 \, mol \, C)(12.011 \, g/mol \, C) = 168.154 \, g\] \[ 18 \,H (18 \, mol \, H)(1.008 \, g/mol \, H) = 18.114 \, g\] \[ 2 \,N (2 \, mol \, N)(14.007 \, g/mol \, N) = 28.014 \, g\] \[ +5 \,O (5 \, mol \, O)(15.999 \, g/mol \, O) = 79.995 \, g\] \[C_{14}H_{18}N_2O_5 \text {molar mass of aspartame} = 294.277 \, g/mol \] Thus more than half the mass of 1 mol of aspartame (294.277 g) is carbon (168.154 g). To calculate the mass percentage of each element, we divide the mass of each element in the compound by the molar mass of aspartame and then multiply by 100 to obtain percentages, here reported to two decimal places: \[ mass \% \, C = {168.154 \, g \, C \over 294.277 \, g \, aspartame } \times 100 = 57.14 \% C\] \[ mass \% \, H = {18.114 \, g \, H \over 294.277 \, g \, aspartame } \times 100 = 6.16 \% H\] \[ mass \% \, N = {28.014 \, g \, N \over 294.277 \, g \, aspartame } \times 100 = 9.52 \% \] \[ mass \% \, O = {79.995 \, g \, O \over 294.277 \, g \, aspartame } \times 100 = 27.18 \% \] As a check, we can add the percentages together: \[ 57.14\% + 6.16\% + 9.52\% + 27.18\% = 100.00\% \] If you obtain a total that differs from 100% by more than about ±1%, there must be an error somewhere in the calculation. b. The mass of carbon in 1.00 g of aspartame is calculated as follows: \[ \text {mass of C} = 1.00 \, g \, aspartame \times {57.14 \, g \, C \over 100 \, g \, aspartame } = 0.571 \, g \, C \] Calculate the mass percentage of each element in aluminum oxide (Al O ). Then calculate the mass of aluminum in a 3.62 g sample of pure aluminum oxide. : 52.93% aluminum; 47.08% oxygen; 1.92 g Al Percent Composition: Just as the empirical formula of a substance can be used to determine its percent composition, the percent composition of a sample can be used to determine its empirical formula, which can then be used to determine its molecular formula. Such a procedure was actually used to determine the empirical and molecular formulas of the first antibiotic to be discovered: penicillin. Antibiotics are chemical compounds that selectively kill microorganisms, many of which cause diseases. Although antibiotics are often taken for granted today, penicillin was discovered only about 80 years ago. The subsequent development of a wide array of other antibiotics for treating many common diseases has contributed greatly to the substantial increase in life expectancy over the past 50 years. The discovery of penicillin is a historical detective story in which the use of mass percentages to determine empirical formulas played a key role. In 1928, Alexander Fleming, a young microbiologist at the University of London, was working with a common bacterium that causes boils and other infections such as blood poisoning. For laboratory study, bacteria are commonly grown on the surface of a nutrient-containing gel in small, flat culture dishes. One day Fleming noticed that one of his cultures was contaminated by a bluish-green mold similar to the mold found on spoiled bread or fruit. Such accidents are rather common, and most laboratory workers would have simply thrown the cultures away. Fleming noticed, however, that the bacteria were growing everywhere on the gel except near the contaminating mold (part (a) in ), and he hypothesized that the mold must be producing a substance that either killed the bacteria or prevented their growth. To test this hypothesis, he grew the mold in a liquid and then filtered the liquid and added it to various bacteria cultures. The liquid killed not only the bacteria Fleming had originally been studying but also a wide range of other disease-causing bacteria. Because the mold was a member of the Penicillium family (named for their pencil-shaped branches under the microscope) (part (b) in ), Fleming called the active ingredient in the broth penicillin. Although Fleming was unable to isolate penicillin in pure form, the medical importance of his discovery stimulated researchers in other laboratories. Finally, in 1940, two chemists at Oxford University, Howard Florey (1898–1968) and Ernst Chain (1906–1979), were able to isolate an active product, which they called penicillin G. Within three years, penicillin G was in widespread use for treating pneumonia, gangrene, gonorrhea, and other diseases, and its use greatly increased the survival rate of wounded soldiers in World War II. As a result of their work, Fleming, Florey, and Chain shared the Nobel Prize in Medicine in 1945. As soon as they had succeeded in isolating pure penicillin G, Florey and Chain subjected the compound to a procedure called combustion analysis (described later in this section) to determine what elements were present and in what quantities. The results of such analyses are usually reported as mass percentages. They discovered that a typical sample of penicillin G contains 53.9% carbon, 4.8% hydrogen, 7.9% nitrogen, 9.0% sulfur, and 6.5% sodium by mass. The sum of these numbers is only 82.1%, rather than 100.0%, which implies that there must be one or more additional elements. A reasonable candidate is oxygen, which is a common component of compounds that contain carbon and hydrogen; do not assume that the “missing” mass is always due to oxygen. It could be any other element. For technical reasons, however, it is difficult to analyze for oxygen directly. Assuming that all the missing mass is due to oxygen, then penicillin G contains (100.0% − 82.1%) = 17.9% oxygen. From these mass percentages, the empirical formula and eventually the molecular formula of the compound can be determined. To determine the empirical formula from the mass percentages of the elements in a compound such as penicillin G, the mass percentages must be converted to relative numbers of atoms. For convenience, assume a 100.0 g sample of the compound, even though the sizes of samples used for analyses are generally much smaller, usually in milligrams. This assumption simplifies the arithmetic because a 53.9% mass percentage of carbon corresponds to 53.9 g of carbon in a 100.0 g sample of penicillin G; likewise, 4.8% hydrogen corresponds to 4.8 g of hydrogen in 100.0 g of penicillin G; and so forth for the other elements. Each mass is then divided by the molar mass of the element to determine how many moles of each element are present in the 100.0 g sample: \[ { mass \, (g) \over molar \,\, mass \,\, (g/mol)} = (g) \left ({mol \over g } \right ) = mol \label{3.3.2a}\] \[ 53.9 \, g \, C \left ({1 \, mol \, C \over 12.011 \, g \, C} \right ) = 4.49 \, mol \, C \label{3.3.2b}\] \[ 4.8 \, g \, H \left ({1 \, mol \, H \over 1.008 g \, H} \right ) = 4.8 \, mol \, H \label{3.3.2c}\] \[ 7.9 \, g \, N \left ({1 \, mol \, N \over 14.007 \, g \, N} \right ) = 0.56 \, mol \, N \label{3.3.2d}\] \[ 9 \, g \, S \left ({1 \, mol \, S \over 32.065 \, g \, S} \right ) = 0.28 \, mol \, S \label{3.3.2e}\] \[ 6.5 \, g \, Na \left ({1 \, mol \, Na \over 22.990 \, g \, Na} \right ) = 0.28 \, mol \, Na \label{3.3.2f}\] Thus 100.0 g of penicillin G contains 4.49 mol of carbon, 4.8 mol of hydrogen, 0.56 mol of nitrogen, 0.28 mol of sulfur, 0.28 mol of sodium, and 1.12 mol of oxygen (assuming that all the missing mass was oxygen). The number of significant figures in the numbers of moles of elements varies between two and three because some of the analytical data were reported to only two significant figures. These results give the ratios of the moles of the various elements in the sample (4.49 mol of carbon to 4.8 mol of hydrogen to 0.56 mol of nitrogen, and so forth), but they are not the whole-number ratios needed for the empirical formula—the empirical formula expresses the relative numbers of atoms in the smallest whole numbers possible. To obtain whole numbers, divide the numbers of moles of all the elements in the sample by the number of moles of the element present in the lowest relative amount, which in this example is sulfur or sodium. The results will be the subscripts of the elements in the empirical formula. To two significant figures, the results are as follows: \[C: {4.49 \over 0.28} = 16 \, \, \, \, \, H: {4.8 \over 0.28} = 17 \, \, \, \, \, N: {0.56 \over 0.28} = 2.0 \label{3.3.3a}\] \[S: {0.28 \over 0.28 } = 1.0 \, \, \, \, \, Na: {0.28 \over 0.28 } = 1.0 \, \, \, \, \, O: {1.12 \over 0.28} = 4.0 \label{3.3.3b}\] The empirical formula of penicillin G is therefore C H N NaO S. Other experiments have shown that penicillin G is actually an ionic compound that contains Na cations and [C H N O S] anions in a 1:1 ratio. The complex structure of penicillin G ( ) was not determined until 1948. In some cases, one or more of the subscripts in a formula calculated using this procedure may not be integers. Does this mean that the compound of interest contains a nonintegral number of atoms? No; rounding errors in the calculations as well as experimental errors in the data can result in nonintegral ratios. When this happens, judgment must be exercised in interpreting the results, as illustrated in Example 6. In particular, ratios of 1.50, 1.33, or 1.25 suggest that you should multiply all subscripts in the formula by 2, 3, or 4, respectively. Only if the ratio is within 5% of an integral value should one consider rounding to the nearest integer. Calculate the empirical formula of the ionic compound calcium phosphate, a major component of fertilizer and a polishing agent in toothpastes. Elemental analysis indicates that it contains 38.77% calcium, 19.97% phosphorus, and 41.27% oxygen. : percent composition : empirical formula : : A 100 g sample of calcium phosphate contains 38.77 g of calcium, 19.97 g of phosphorus, and 41.27 g of oxygen. Dividing the mass of each element in the 100 g sample by its molar mass gives the number of moles of each element in the sample: \[ \text {moles Ca} = 38.77 \, g \, Ca \times {1 \, mol \, Ca \over 40.078 \, g \, Ca} = 0.9674 \, mol \, Ca\] \[ \text {moles P} = 19.97 \, g \, P \times {1 \, mol \, P \over 30.9738 \, g \, P} = 0.6447 \, mol \, Ca\] \[ \text {moles O} = 41.27 \, g \, O \times {1 \, mol \, O \over 15.9994 \, g \, O} = 2.5800 \, mol \, O\] To obtain the relative numbers of atoms of each element in the compound, divide the number of moles of each element in the 100-g sample by the number of moles of the element in the smallest amount, in this case phosphorus: \[P: {0.6447 \, mol\, P \over 0.6447 \, mol\, P} = 1.000 \, \, \, \, Ca: {0.9674 \over 0.6447} = 1.501\, \, \, \, O: {2.5800\over 0.6447}= 4.002\] We could write the empirical formula of calcium phosphate as Ca P O , but the empirical formula should show the ratios of the elements as small whole numbers. To convert the result to integral form, multiply all the subscripts by 2 to get Ca P O . The deviation from integral atomic ratios is small and can be attributed to minor experimental errors; therefore, the empirical formula is Ca P O . The calcium ion (Ca ) is a cation, so to maintain electrical neutrality, phosphorus and oxygen must form a polyatomic anion. We know from that phosphorus and oxygen form the phosphate ion (PO ; see ). Because there are two phosphorus atoms in the empirical formula, two phosphate ions must be present. So we write the formula of calcium phosphate as Ca (PO ) . Calculate the empirical formula of ammonium nitrate, an ionic compound that contains 35.00% nitrogen, 5.04% hydrogen, and 59.96% oxygen by mass. Although ammonium nitrate is widely used as a fertilizer, it can be dangerously explosive. For example, it was a major component of the explosive used in the 1995 Oklahoma City bombing. The Alfred P. Murrah Federal Building destroyed in the Oklahoma City bombing via chemical explosives (rapid chemical reactions that generate massive quantities of gases). N H O is NH NO , written as NH NO Determining Empirical and Molecular Formulas from % Composition: The empirical formula gives only the relative numbers of atoms in a substance in the smallest possible ratio. For a covalent substance, chemists are usually more interested in the molecular formula, which gives the actual number of atoms of each kind present per molecule. Without additional information, however, it is impossible to know whether the formula of penicillin G, for example, is C H N NaO S or an integral multiple, such as C H N Na O S , C H N Na O S , or (C H N NaO S) , where n is an integer. (The actual structure of penicillin G is shown in ). Consider glucose, the sugar that circulates in our blood to provide fuel for the body and brain. Results from combustion analysis of glucose report that glucose contains 39.68% carbon and 6.58% hydrogen. Because combustion occurs in the presence of oxygen, it is impossible to directly determine the percentage of oxygen in a compound by using combustion analysis; other more complex methods are necessary. Assuming that the remaining percentage is due to oxygen, then glucose would contain 53.79% oxygen. A 100.0 g sample of glucose would therefore contain 39.68 g of carbon, 6.58 g of hydrogen, and 53.79 g of oxygen. To calculate the number of moles of each element in the 100.0 g sample, divide the mass of each element by its molar mass: \[ moles \, C = 39.68 \, g \, C \times {1 \, mol \, C \over 12.011 \, g \, C } = 3.304 \, mol \, C \label{3.3.4a}\] \[ moles \, H = 6.58 \, g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H } = 6.53 \, mol \, H \label{3.3.4b}\] \[ moles \, O = 53.79 \, g \, O \times {1 \, mol \, O \over 15.9994 \, g \, O } = 3.362 \, mol \, O \label{3.3.4c} \] Once again, the subscripts of the elements in the empirical formula are found by dividing the number of moles of each element by the number of moles of the element present in the smallest amount: \[ C: {3.304 \over 3.304} = 1.000 \, \, \, \, H: {6.53 \over 3.304} = 1.98 \, \, \, \, O: {3.362 \over 3.304} = 1.018 \] The oxygen:carbon ratio is 1.018, or approximately 1, and the hydrogen:carbon ratio is approximately 2. The empirical formula of glucose is therefore CH O, but what is its molecular formula? Many known compounds have the empirical formula CH O, including formaldehyde, which is used to preserve biological specimens and has properties that are very different from the sugar circulating in the blood. At this point, it cannot be known whether glucose is CH O, C H O , or any other (CH O) . However, the experimentally determined molar mass of glucose (180 g/mol) can be used to resolve this dilemma. First, calculate the formula mass, the molar mass of the formula unit, which is the sum of the atomic masses of the elements in the empirical formula multiplied by their respective subscripts. For glucose, \[ \text {formula mass of} CH_2O = \left [ 1 \, mol C \left ( {12.011 \, g \over 1 \, mol \, C} \right ) \right ] + \left [ 2 \, mol \, H \left ({1.0079 \, g \over 1 \, mol \, H }\right )\right ] + \left [ 1 \, mole \, O \left ( {15.5994 \, mol \, O \over 1 \, mol \, O} \right ) \right ] = 30.026 g \label{3.3.5}\] This is much smaller than the observed molar mass of 180 g/mol. Second, determine the number of formula units per mole. For glucose, calculate the number of (CH O) units—that is, the n in (CH O) —by dividing the molar mass of glucose by the formula mass of CH O: \[n={180 \, g \over 30.026\, g/CH_2O} = 5.99 \approx 6 CH_2O \, \text {formula units} \label{3.3.6}\] Each glucose contains six CH O formula units, which gives a molecular formula for glucose of (CH O) , which is more commonly written as C H O . The molecular structures of formaldehyde and glucose, both of which have the empirical formula CH O, are shown in Calculate the molecular formula of caffeine, a compound found in coffee, tea, and cola drinks that has a marked stimulatory effect on mammals. The chemical analysis of caffeine shows that it contains 49.18% carbon, 5.39% hydrogen, 28.65% nitrogen, and 16.68% oxygen by mass, and its experimentally determined molar mass is 196 g/mol. : percent composition and molar mass : molecular formula : : A We begin by dividing the mass of each element in 100.0 g of caffeine (49.18 g of carbon, 5.39 g of hydrogen, 28.65 g of nitrogen, 16.68 g of oxygen) by its molar mass. This gives the number of moles of each element in 100 g of caffeine. \[ moles \, C = 49.18 \, g \, C \times {1 \, mol \, C \over 12.011 \, g \, C} = 4.095 \, mol \, C \] \[ moles \, H = 5.39 \, g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H} = 5.35 \, mol \, H \] \[ moles \, N = 28.65 \, g \, N \times {1 \, mol \, N \over 14.0067 \, g \, N} = 2.045 \, mol \, N \] \[ moles \, O = 16.68 \, g \, O \times {1 \, mol \, O \over 15.9994 \, g \, O} = 1.043 \, mol \, O \] To obtain the relative numbers of atoms of each element present, divide the number of moles of each element by the number of moles of the element present in the least amount: \[O: {1.043 \over 1.043} = 1.000 \, \, \, \, C: {4.095 \over 1.043} = 3.926 \, \, \, \, H: {5.35 \over 1.043} = 5.13 \, \, \, \, N: {2.045 \over 1.043} = 1.960 \] These results are fairly typical of actual experimental data. None of the atomic ratios is exactly integral but all are within 5% of integral values. Just as in Example 6, it is reasonable to assume that such small deviations from integral values are due to minor experimental errors, so round to the nearest integer. The empirical formula of caffeine is thus C H N O. The molecular formula of caffeine could be C H N O, but it could also be any integral multiple of this. To determine the actual molecular formula, we must divide the experimentally determined molar mass by the formula mass. The formula mass is calculated as follows: \[ 4C \, \, \, ( 4 \, atoms \, C) (12.011 \, g/ atom \, C) = 48.044 \, g \] \[ 5H \, \, \, ( 5 \, atoms \, H ) (1.0079 \, g/ atom \, H) = 5.0395 \, g \] \[ 2N \, \, \, (2 \, atoms \, N) (14.0067 \, g/ atom \, N) = 28.0134 \, g \] \[ +1O \, \, \, (1 \, atom \, O) (15.9994 \, g/ atom \, O) = 15.9994 \, g \] \[ C_4H_5N_2O \, \, \, \, \text {formula mass of caffeine} = 97.096 \, g \] Dividing the measured molar mass of caffeine (196 g/mol) by the calculated formula mass gives C There are two C H N O formula units in caffeine, so the molecular formula must be (C H N O) = C H N O . The structure of caffeine is as follows: Calculate the molecular formula of Freon-114, which has 13.85% carbon, 41.89% chlorine, and 44.06% fluorine. The experimentally measured molar mass of this compound is 171 g/mol. Like Freon-11, Freon-114 is a commonly used refrigerant that has been implicated in the destruction of the ozone layer. : \[C_2Cl_2F_4\] One of the most common ways to determine the elemental composition of an unknown hydrocarbon is an analytical procedure called combustion analysis. A small, carefully weighed sample of an unknown compound that may contain carbon, hydrogen, nitrogen, and/or sulfur is burned in an oxygen atmosphere,Other elements, such as metals, can be determined by other methods. and the quantities of the resulting gaseous products (CO , H O, N , and SO , respectively) are determined by one of several possible methods. One procedure used in combustion analysis is outlined schematically in and a typical combustion analysis is illustrated in Example \(\Page {4}\). Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO and 11.30 mg of H O. Determine the empirical formula of naphthalene. : mass of sample and mass of combustion products : empirical formula : : Upon combustion, 1 mol of CO is produced for each mole of carbon atoms in the original sample. Similarly, 1 mol of H O is produced for every 2 mol of hydrogen atoms present in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO and H O, and their molar masses. Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams: \[ mass \, of \, C = 69.00 \, mg \, CO_2 \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, CO_2 \over 44.010 \, g \, CO_2} \times {1 \, mol C \over 1 \, mol \, CO_2 } \times {12.011 \,g \over 1 \, mol \, C} \] \[ = 1.883 \times 10^{-2} \, g \, C \] \[ mass \, of \, H = 11.30 \, mg \, H_2O \times {1 \, g \over 1000 \, mg } \times {1 \, mol \, H_2O \over 18.015 \, g \, H_2O} \times {2 \, mol H \over 1 \, mol \, H_2O } \times {1.0079 \,g \over 1 \, mol \, H} \] \[ = 1.264 \times 10^{-3} \, g \, H \] To obtain the relative numbers of atoms of both elements present, we need to calculate the number of moles of each and divide by the number of moles of the element present in the smallest amount: \[ moles \, C = 1.883 \times 10^{-2} \,g \, C \times {1 \, mol \, C \over 12.011 \, g \, C} = 1.568 \times 10^{-3} \, mol C \] \[ moles \, H = 1.264 \times 10^{-3} \,g \, H \times {1 \, mol \, H \over 1.0079 \, g \, H} = 1.254 \times 10^{-3} \, mol H \] Dividing each number by the number of moles of the element present in the smaller amount gives \[H: {1.254\times 10^{−3} \over 1.254 \times 10^{−3}} = 1.000 \, \, \, C: {1.568 \times 10^{−3} \over 1.254 \times 10^{−3}}= 1.250\] Thus naphthalene contains a 1.25:1 ratio of moles of carbon to moles of hydrogen: C H . Because the ratios of the elements in the empirical formula must be expressed as small whole numbers, multiply both subscripts by 4, which gives C H as the empirical formula of naphthalene. In fact, the molecular formula of naphthalene is C H , which is consistent with our results. : The empirical formula of a substance can be calculated from the experimentally determined percent composition, the percentage of each element present in a pure substance by mass. In many cases, these percentages can be determined by combustion analysis. If the molar mass of the compound is known, the molecular formula can be determined from the empirical formula. ( ) | 26,731 | 3,331 |
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In , we saw that there are some salts which dissolve in water to only a very limited extent. For example, if BaSO crystals are shaken with water, so little dissolves that it is impossible to see that anything has happened, as you will see in the video below. Nevertheless, the few Ba ( ) and SO ( ) ions that do go into solution increase the conductivity of the water, allowing us to measure their concentration. The video below shows the creation of Barium Sulfate in a precipitation reaction between barium chloride and sodium sulfate. Notice the white precipitate that forms, which is barium sulfate. We find that at 25°C \[[\text{Ba}^{2+}]=\text{0.97} \times \text{10}^{-5}\text{ mol L}^{-1} = [\text{SO}_{4}^{2-}]\label{1} \] that we would describe the solubility of BaSO as 0.97 × 10 mol L at this temperature. The solid salt and its ions are in dynamic equilibrium, and so we can write the equation \[\text{BaSO}_{4} ({s}) \rightleftharpoons \text{Ba}^{2+} ({aq}) + \text{SO}_{4}^{2-} ({aq})\label{2} \] As in other dynamic equilibria we have discussed, a particular Ba ion will sometimes find itself part of a crystal and at other times find itself hydrated and in solution. Since the concentration of BaSO has a constant value, it can be incorporated into for Equation \(\ref{2}\). This gives a special equilibrium constant called the : \[K_{sp}= K_{c}[\text{BaSO}_{4}] = [\text{Ba}^{2+},\text{SO}_{4}^{2-}]\label{3} \] For BaSO , is easily calculated from the solubility by substituting Equation \(\ref{1}\) into \(\ref{3}\): \[\begin{align}K_{sp} & = (\text{0.97} \times \text{10}^{-5}\text{ mol L}^{-1}) (\text{0.97} \times \text{10}^{-5}\text{ mol L}^{-1})\\ &= \text{0.94} \times \text{10}^{-10}\text{ mol}^{2} \text{ L}^{-2}\end{align} \nonumber \] In the general case of an ionic compound whose formula is \(A_xB_y\), the equilibrium can be written \[\text{A}_{x}\text{B}_{y} ({s}) \rightleftharpoons {x}\text{A}^{m+} ({aq}) + {y}\text{A}^{n+} ({aq}) \nonumber \] The solubility product is then \[\text{K}_{sp} = [\text{A}^{m+}]^{x}[\text{B}^{n+}]^{y} \nonumber \] Solubility products for some of the more common sparingly soluble compounds are given in the table below. 1. Taken from Patnaik, Pradyot, Dean’s Analytical Chemistry Handbook, 2nd ed., New York: McGraw-Hill, 2004, Table 4.2 (published on the Web by Knovel, ). 2. Taken from Meites, L. ed., Handbook of Analytical Chemistry, 1st ed., New York: McGraw-Hill, 1963. 3. Because [H O] does not appear in equilibrium constants for equilibria in aqueous solution in general, it does not appear in the expressions for hydrated solids. No metal sulfides are listed in this table because sulfide ion is such a strong base that the usual solubility product equilibrium equation does not apply. See Myers, R. J. , Vol. 63, 1986; pp. 687-690. When crystals of PbCl are shaken with water at 25°C, it is found that 1.62 × 10 mol PbCl dissolves per cubic decimeter of solution. Find the value of at this temperature. We first write out the equation for the equilibrium: \[\text{PbCl}_{2}({s})\rightleftharpoons \text{Pb}^{2+}({aq}) = \text{2Cl}^{-}({aq}) \nonumber \] so that \[\text{K}_{sp}\text{PbCl}_{2} = [\text{Pb}^{2+},\text{Cl}^{-}]^{2} \nonumber \] Since 1.62 × 10 mol PbCl dissolves per cubic decimeter, we have \[[\text{Pb}^{2+}]=\text{1.62} \times \text{10}^{-2} \text{mol L}^{-1} \nonumber \] while \[[\text{Cl}^{-}]=\text{2} \times \text{1.62} \times \text{10}^{-2} \text{mol L}^{-1} \nonumber \] since 2 mol Cl ions are produced for each mol PbCl which dissolves. Thus \[\begin{align}{K}_{sp}= (\text{1.62}\times \text{10}^{-2}\text{mol L}^{-1})(\text{2 } \times \text{ 1.62 } \times \text{ 10}^{-2} \text{mol L}^{-1})\text{ }^{2}\\\text{ } = \text{1.70 } \times \text{ 10}^{-5} \text{ mol}^{3} \text{L} ^{-3}\end{align} \nonumber \] The solubility product of silver chromate, Ag CrO , is 1.0 × 10 mol L . Find the solubility of this salt. Again we start by writing the equation \[\text{Ag}_{2}\text{CrO } ({s}) \rightleftharpoons \text {2Ag}^{2+} ({aq}) + \text{CrO}_{4}^{2-} ({aq}) \nonumber \] from which \[{K}_{sp}(\text{Ag}_{2}\text{CrO}_{4})= [\text{Ag}^{+}]^{2} [\text{CrO}_{4}^{2-}]= \text{1.0} \times \text{10}^{-12} \text{mol}^{3} \text{L}^{-3} \nonumber \] Let the solubility be mol L . Then \[[\text{CrO}_{4}^{2-}]= {x } \text{ mol } \text{L}^{-1} \nonumber \] and \[[\text{Ag}^{+}] = 2x \text{mol L}^{-1} \nonumber \] Thus \[\begin{align}{K}_{sp}= (\text{2}{x} \text{ mol } \text{L}^{-1})^{2} {x} \text{ mol L}^{-1}\\\text{ }= (\text{2}{x})^{2} { x}\text{ mol}^{3} \text{ L}^{-3} = \text{1.0} \times \text{10}^{-12}\text{ mol}^{3} \text{ L}^{-3} \, \end{align} \nonumber \] or \[4x^{3} = \text{1.0 x 10}^{-12} \nonumber \] and \[x^{\text{3}}=\frac{\text{1.0}}{\text{4}}\text{ }\times \text{ 10}^{12}=\text{2.5 }\times \text{ 10}^{-13}=\text{250 }\times \text{ 10}^{-15} \nonumber \] so that \[x=\sqrt[\text{3}]{\text{250}}\text{ }\times \text{ }\sqrt[\text{3}]{\text{10}^{-\text{15}}}=\text{6}\text{.30 }\times \text{ 10}^{-\text{5}} \nonumber \] Thus the solubility is 6.30 × 10 mol L . | 5,155 | 3,333 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Enzymes/The_Effect_of_Changing_Conditions_on_Enzyme_Catalysis |
This page looks at the effect of changing substrate concentration, temperature and pH on reactions involving enzymes. It follows on from a page describing in simple terms how enzymes function as catalysts. Remember that in biology or biochemistry, the reactant in an enzyme reaction is known as the "substrate". What follows is a very brief and simple look at a very complicated topic. Anything beyond this is the stuff of biochemistry degree courses! If you have done any work on rates of reaction (especially if you have done orders of reaction), you will have come across cases where the rate of reaction is proportional to the concentration of a reactant, or perhaps to the square of its concentration. You would discover this by changing the concentration of one of the reactants, keeping everything else constant, and measuring the initial rate of the reaction. If you measure the rate after the reaction has been going for a while, the concentration of the reactant(s) will have changed and that just complicates things. That's why initial rates are so useful - you know exactly how much you have of everything. If you plotted a graph of initial reaction rate against the concentration of a reactant, then there are various possibilities depending on the relationship between the concentration and the rate. This is called a zero order reaction. This is called a first order reaction. In this case, you get a curve. If the rate was proportional to the square of the concentration, that's a second order reaction. With reactions controlled by enzymes, you get a completely different type of graph. The graph for enzyme controlled reactions looks like this: Two minor things to notice before we discuss it . . . So why is the graph the shape it is? For very, very low substrate concentrations, the graph is almost a straight line - like the second chemistry rate graph above. In other words, at very, very low concentrations, the rate is proportional to the substrate concentration. But as concentration increases, increasing the concentration more has less and less effect - and eventually the rate reaches a maximum. Increasing the concentration any more makes no difference to the rate of the reaction. If you know about orders of reaction, the reaction has now become zero order with respect to the substrate. The reason for this is actually fairly obvious if you think about it. After a certain concentration of substrate is reached, every enzyme molecule present in the mixture is working as fast as it can. If you increase the substrate concentration any more, there aren't any enzyme molecules free to help the extra substrate molecules to react. Is this unique to enzyme-controlled reactions? No! It can happen in some ordinary chemistry cases as well, usually involving a solid catalyst working with gases. At very high gas pressures (in other words, very high concentrations of gas molecules), the surface of the catalyst can be completely full of gas molecules. If you increase the amount of gas any more, there isn't any available surface for it to stick to and react. The maximum rate for a particular enzyme reaction is known as V . (That's V for velocity - a bit confusing for chemists where V is almost always used for volume!) This is easily measured by drawing a line on the graph: This is sometimes reported as a "turnover number", measured as the number of molecules of substrate processed by a single enzyme molecule per second, or per minute. is known as the Michaelis constant or the Michaelis-Menten constant (for reasons which needn't concern us), and is a useful measure of the efficiency of an enzyme. K is the concentration of the substrate in mol dm which produces a reaction rate of half V . So it is found like this . . . A low value of K means that the reaction is going quickly even at low substrate concentrations. A higher value means the enzyme isn't as effective. Remember that for molecules to react, they have to collide with an energy equal to or greater than the activation energy for the reaction. Heating a reaction makes the molecules move faster and so collide more often. More collisions in a given time means a faster reaction. But far more importantly, increasing the temperature has a very big effect on the number of collisions with enough energy to react. Quite a small temperature rise can produce a large increase in rate. As a reasonable approximation for many (although not all) reactions close to room temperature, a 10°C increase in temperature will double the rate of a chemical reaction. This is only an approximation - it may take 9°C or 11°C or whatever, but it gives you an idea of what to expect. For low temperatures up to about 40°C, enzyme-controlled reactions behave much as you would expect any other chemical reaction to behave. But above about 40°C (the exact temperature varies from enzyme to enzyme), there is a dramatic fall in reaction rate. A typical graph of rate against temperature might look like this: The temperature at which the rate is fastest is called the optimum temperature for that enzyme. Different enzymes have different optimum temperatures. Some enzymes, for example, in organisms known as thermophiles or extremophiles are capable of working at temperatures like 80°C or even higher. The optimum temperature for a particular enzyme varies depending on how long it is exposed to the higher temperatures. Enzymes can withstand temperatures higher than their normal optimum if they are only exposed to the higher temperatures for a very short time. At lower temperatures, the shape of the graph is exactly what you would expect for any reaction. All that needs explaining is why the rate falls above the optimum temperature. At extreme pH's, something more drastic can happen. Remember that the tertiary structure of the protein is in part held together by ionic bonds just like those we've looked at between the enzyme and its substrate. At very high or very low pH's, these bonds within the enzyme can be disrupted, and it can lose its shape. If it loses its shape, the active site will probably be lost completely. This is essentially the same as denaturing the protein by heating it too much. | 6,211 | 3,335 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/11%3A_Solutions_and_Colloids/11.4%3A_Colligative_Properties?sectionId=11 |
The properties of a solution are different from those of either the pure solute(s) or solvent. Many solution properties are dependent upon the chemical identity of the solute. Compared to pure water, a solution of hydrogen chloride is more acidic, a solution of ammonia is more basic, a solution of sodium chloride is more dense, and a solution of sucrose is more viscous. There are a few solution properties, however, that depend upon the total concentration of solute species, regardless of their identities. These include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. This small set of properties is of central importance to many natural phenomena and technological applications, as will be described in this module. Several units commonly used to express the concentrations of solution components were introduced in an earlier chapter of this text, each providing certain benefits for use in different applications. For example, molarity (M) is a convenient unit for use in stoichiometric calculations, since it is defined in terms of the molar amounts of solute species: \[M=\dfrac{\text{mol solute}}{\text{L solution}} \label{11.5.1} \] Because solution volumes vary with temperature, molar concentrations will likewise vary. When expressed as molarity, the concentration of a solution with identical numbers of solute and solvent species will be different at different temperatures, due to the contraction/expansion of the solution. More appropriate for calculations involving many colligative properties are mole-based concentration units whose values are not dependent on temperature. Two such units are (introduced in the previous chapter on gases) and . The mole fraction, \(\chi\), of a component is the ratio of its molar amount to the total number of moles of all solution components: \[\chi_\ce{A}=\dfrac{\text{mol A}}{\text{total mol of all components}} \label{11.5.2} \] Molality is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms: \[m=\dfrac{\text{mol solute}}{\text{kg solvent}} \label{11.5.3} \] Since these units are computed using only masses and molar amounts, they do not vary with temperature and, thus, are better suited for applications requiring temperature-independent concentrations, including several colligative properties, as will be described in this chapter module. The antifreeze in most automobile radiators is a mixture of equal volumes of ethylene glycol and water, with minor amounts of other additives that prevent corrosion. What are the (a) mole fraction and (b) molality of ethylene glycol, C H (OH) , in a solution prepared from \(\mathrm{2.22 \times 10^3 \;g}\) of ethylene glycol and \(\mathrm{2.00 \times 10^3\; g}\) of water (approximately 2 L of glycol and 2 L of water)? (a) The mole fraction of ethylene glycol may be computed by first deriving molar amounts of both solution components and then substituting these amounts into the unit definition. \(\mathrm{mol\:H_2O=2000\:g×\dfrac{1\:mol\:H_2O}{18.02\:g\:H_2O}=111\:mol\:H_2O}\) \(\chi_\mathrm{ethylene\:glycol}=\mathrm{\dfrac{35.8\:mol\:C_2H_4(OH)_2}{(35.8+111)\:mol\: total}=0.245}\) Notice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles). (b) To find molality, we need to know the moles of the solute and the mass of the solvent (in kg). First, use the given mass of ethylene glycol and its molar mass to find the moles of solute: \[\mathrm{2220\:g\:C_2H_4(OH)_2\left(\dfrac{mol\:C_2H_2(OH)_2}{62.07\:g}\right)=35.8\:mol\:C_2H_4(OH)_2} \nonumber \] Then, convert the mass of the water from grams to kilograms: \[\mathrm{2000\: g\:H_2O\left(\dfrac{1\:kg}{1000\:g}\right)=2\: kg\:H_2O} \nonumber \] Finally, calculate molarity per its definition: \[\begin{align*}
\ce{molality}&=\mathrm{\dfrac{mol\: solute}{kg\: solvent}}\\
\ce{molality}&=\mathrm{\dfrac{35.8\:mol\:C_2H_4(OH)_2}{2\:kg\:H_2O}}\\
\ce{molality}&=17.9\:m
\end{align*} \nonumber \] What are the mole fraction and molality of a solution that contains 0.850 g of ammonia, NH , dissolved in 125 g of water? 7.14 × 10 ; 0.399 Calculate the mole fraction of solute and solvent in a 3.0 solution of sodium chloride. Converting from one concentration unit to another is accomplished by first comparing the two unit definitions. In this case, both units have the same numerator (moles of solute) but different denominators. The provided molal concentration may be written as: \[\mathrm{\dfrac{3.0\;mol\; NaCl}{1.0\; kg\; H_2O}} \nonumber \] The numerator for this solution’s mole fraction is, therefore, 3.0 mol NaCl. The denominator may be computed by deriving the molar amount of water corresponding to 1.0 kg \[\mathrm{1.0\:kg\:H_2O\left(\dfrac{1000\:g}{1\:kg}\right)\left(\dfrac{mol\:H_2O}{18.02\:g}\right)=55\:mol\:H_2O} \nonumber \] and then substituting these molar amounts into the definition for mole fraction. \[\begin{align*}
X_\mathrm{H_2O}&=\mathrm{\dfrac{mol\:H_2O}{mol\: NaCl + mol\:H_2O}}\\
X_\mathrm{H_2O}&=\mathrm{\dfrac{55\:mol\:H_2O}{3.0\:mol\: NaCl+55\:mol\:H_2O}}\\
X_\mathrm{H_2O}&=0.95\\
X_\mathrm{NaCl}&=\mathrm{\dfrac{mol\: NaCl}{mol\: NaCl+mol\:H_2O}}\\
X_\mathrm{NaCl}&=\mathrm{\dfrac{3.0\:mol\:NaCl}{3.0\:mol\: NaCl+55\:mol\:H_2O}}\\
X_\mathrm{NaCl}&=0.052
\end{align*} \nonumber \] The mole fraction of iodine, \(\ce{I_2}\), dissolved in dichloromethane, \(\ce{CH_2Cl_2}\), is 0.115. What is the molal concentration, , of iodine in this solution? 1.50 As described in the chapter on liquids and solids, the equilibrium vapor pressure of a liquid is the pressure exerted by its gaseous phase when vaporization and condensation are occurring at equal rates: \[ \text{liquid} \rightleftharpoons \text{gas} \label{11.5.4} \] Dissolving a nonvolatile substance in a volatile liquid results in a lowering of the liquid’s vapor pressure. This phenomenon can be rationalized by considering the effect of added solute molecules on the liquid's vaporization and condensation processes. To vaporize, solvent molecules must be present at the surface of the solution. The presence of solute decreases the surface area available to solvent molecules and thereby reduces the rate of solvent vaporization. Since the rate of condensation is unaffected by the presence of solute, the net result is that the vaporization-condensation equilibrium is achieved with fewer solvent molecules in the vapor phase (i.e., at a lower vapor pressure) (Figure \(\Page {1}\)). While this kinetic interpretation is useful, it does not account for several important aspects of the colligative nature of vapor pressure lowering. A more rigorous explanation involves the property of , a topic of discussion in a later text chapter on thermodynamics. For purposes of understanding the lowering of a liquid's vapor pressure, it is adequate to note that the greater entropy of a solution in comparison to its separate solvent and solute serves to effectively stabilize the solvent molecules and hinder their vaporization. A lower vapor pressure results, and a correspondingly higher boiling point as described in the next section of this module. The relationship between the vapor pressures of solution components and the concentrations of those components is described by : . \[P_\ce{A}=X_\ce{A}P^\circ_\ce{A} \label{11.5.5} \] where is the partial pressure exerted by component A in the solution, \(P^\circ_\ce{A}\) is the vapor pressure of pure A, and is the mole fraction of A in the solution. (Mole fraction is a concentration unit introduced in the chapter on gases.) Recalling that the total pressure of a gaseous mixture is equal to the sum of partial pressures for all its components (Dalton’s law of partial pressures), the total vapor pressure exerted by a solution containing components is \[P_\ce{solution}=\sum_{i}P_i=\sum_{i}X_iP^\circ_i \label{11.5.6} \] A nonvolatile substance is one whose vapor pressure is negligible ( ° ≈ 0), and so the vapor pressure above a solution containing only nonvolatile solutes is due only to the solvent: \[P_\ce{solution}=X_\ce{solvent}P^\circ_\ce{solvent} \label{11.5.7} \] Compute the vapor pressure of an ideal solution containing 92.1 g of glycerin, C H (OH) , and 184.4 g of ethanol, C H OH, at 40 °C. The vapor pressure of pure ethanol is 0.178 atm at 40 °C. Glycerin is essentially nonvolatile at this temperature. Since the solvent is the only volatile component of this solution, its vapor pressure may be computed per Raoult’s law as: \(P_\ce{solution}=X_\ce{solvent}P^\circ_\ce{solvent}\) First, calculate the molar amounts of each solution component using the provided mass data. \(\mathrm{92.1\cancel{g\:C_3H_5(OH)_3}×\dfrac{1\:mol\:C_3H_5(OH)_3}{92.094\cancel{g\:C_3H_5(OH)_3}}=1.00\:mol\:C_3H_5(OH)_3}\) Next, calculate the mole fraction of the solvent (ethanol) and use Raoult’s law to compute the solution’s vapor pressure. \(X_\mathrm{C_2H_5OH}=\mathrm{\dfrac{4.000\:mol}{(1.00\:mol+4.000\:mol)}=0.800}\) \(P_\ce{solv}=X_\ce{solv}P^\circ_\ce{solv}=\mathrm{0.800×0.178\:atm=0.142\:atm}\) A solution contains 5.00 g of urea, CO(NH ) (a nonvolatile solute) and 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution? 23.4 torr As described in the chapter on liquids and solids, the of a liquid is the temperature at which its vapor pressure is equal to ambient atmospheric pressure. Since the vapor pressure of a solution is lowered due to the presence of nonvolatile solutes, it stands to reason that the solution’s boiling point will subsequently be increased. Compared to pure solvent, a solution, therefore, will require a higher temperature to achieve any given vapor pressure, including one equivalent to that of the surrounding atmosphere. The increase in boiling point observed when nonvolatile solute is dissolved in a solvent, \(ΔT_b\), is called boiling point elevation and is directly proportional to the molal concentration of solute species: \[ΔT_b=K_bm \label{11.5.8} \] where Boiling point elevation constants are characteristic properties that depend on the identity of the solvent. Values of for several solvents are listed in Table \(\Page {1}\). The extent to which the vapor pressure of a solvent is lowered and the boiling point is elevated depends on the total number of solute particles present in a given amount of solvent, not on the mass or size or chemical identities of the particles. A 1 aqueous solution of sucrose (342 g/mol) and a 1 aqueous solution of ethylene glycol (62 g/mol) will exhibit the same boiling point because each solution has one mole of solute particles (molecules) per kilogram of solvent. What is the boiling point of a 0.33 solution of a nonvolatile solute in benzene? Use the equation relating boiling point elevation to solute molality to solve this problem in two steps. What is the boiling point of the antifreeze described in Example \(\Page {4}\)? 109.2 °C Find the boiling point of a solution of 92.1 g of iodine, \(\ce{I2}\), in 800.0 g of chloroform, \(\ce{CHCl3}\), assuming that the iodine is nonvolatile and that the solution is ideal. We can solve this problem using four steps. Result: 0.363 mol Result: 0.454 Result: 1.65 °C Result: 62.91 °C Check each result as a self-assessment. What is the boiling point of a solution of 1.0 g of glycerin, \(\ce{C3H5(OH)3}\), in 47.8 g of water? Assume an ideal solution. 100.12 °C Distillation is a technique for separating the components of mixtures that is widely applied in both in the laboratory and in industrial settings. It is used to refine petroleum, to isolate fermentation products, and to purify water. This separation technique involves the controlled heating of a sample mixture to selectively vaporize, condense, and collect one or more components of interest. A typical apparatus for laboratory-scale distillations is shown in Figure \(\Page {2}\). Oil refineries use large-scale to separate the components of crude oil. The crude oil is heated to high temperatures at the base of a tall , vaporizing many of the components that rise within the column. As vaporized components reach adequately cool zones during their ascent, they condense and are collected. The collected liquids are simpler mixtures of hydrocarbons and other petroleum compounds that are of appropriate composition for various applications (e.g., diesel fuel, kerosene, gasoline), as depicted in Figure \(\Page {3}\). Solutions freeze at lower temperatures than pure liquids. This phenomenon is exploited in “de-icing” schemes that use salt (Figure \(\Page {4}\)), calcium chloride, or urea to melt ice on roads and sidewalks, and in the use of ethylene glycol as an “antifreeze” in automobile radiators. Seawater freezes at a lower temperature than fresh water, and so the Arctic and Antarctic oceans remain unfrozen even at temperatures below 0 °C (as do the body fluids of fish and other cold-blooded sea animals that live in these oceans). The decrease in freezing point of a dilute solution compared to that of the pure solvent, Δ , is called the and is directly proportional to the molal concentration of the solute \[ΔT_\ce{f}=K_\ce{f}m \label{11.5.9} \] where Just as for boiling point elevation constants, these are characteristic properties whose values depend on the chemical identity of the solvent. Values of for several solvents are listed in Table \(\Page {1}\). What is the freezing point of the 0.33 solution of a nonvolatile nonelectrolyte solute in benzene described in Example \(\Page {4}\)? Use the equation relating freezing point depression to solute molality to solve this problem in two steps. What is the freezing point of a 1.85 solution of a nonvolatile nonelectrolyte solute in nitrobenzene? −9.3 °C Sodium chloride and its group 2 analogs calcium and magnesium chloride are often used to de-ice roadways and sidewalks, due to the fact that a solution of any one of these salts will have a freezing point lower than 0 °C, the freezing point of pure water. The group 2 metal salts are frequently mixed with the cheaper and more readily available sodium chloride (“rock salt”) for use on roads, since they tend to be somewhat less corrosive than the NaCl, and they provide a larger depression of the freezing point, since they dissociate to yield three particles per formula unit, rather than two particles like the sodium chloride. Because these ionic compounds tend to hasten the corrosion of metal, they would not be a wise choice to use in antifreeze for the radiator in your car or to de-ice a plane prior to takeoff. For these applications, covalent compounds, such as ethylene or propylene glycol, are often used. The glycols used in radiator fluid not only lower the freezing point of the liquid, but they elevate the boiling point, making the fluid useful in both winter and summer. Heated glycols are often sprayed onto the surface of airplanes prior to takeoff in inclement weather in the winter to remove ice that has already formed and prevent the formation of more ice, which would be particularly dangerous if formed on the control surfaces of the aircraft ( \(\Page {1}\)). The colligative effects on vapor pressure, boiling point, and freezing point described in the previous section are conveniently summarized by comparing the phase diagrams for a pure liquid and a solution derived from that liquid. Phase diagrams for water and an aqueous solution are shown in Figure \(\Page {5}\). The liquid-vapor curve for the solution is located the corresponding curve for the solvent, depicting the vapor pressure , Δ , that results from the dissolution of nonvolatile solute. Consequently, at any given pressure, the solution’s boiling point is observed at a higher temperature than that for the pure solvent, reflecting the boiling point elevation, Δ , associated with the presence of nonvolatile solute. The solid-liquid curve for the solution is displaced left of that for the pure solvent, representing the freezing point depression, Δ , that accompanies solution formation. Finally, notice that the solid-gas curves for the solvent and its solution are identical. This is the case for many solutions comprising liquid solvents and nonvolatile solutes. Just as for vaporization, when a solution of this sort is frozen, it is actually just the molecules that undergo the liquid-to-solid transition, forming pure solid solvent that excludes solute species. The solid and gaseous phases, therefore, are composed solvent only, and so transitions between these phases are not subject to colligative effects. A number of natural and synthetic materials exhibit , meaning that only molecules or ions of a certain size, shape, polarity, charge, and so forth, are capable of passing through (permeating) the material. Biological cell membranes provide elegant examples of selective permeation in nature, while dialysis tubing used to remove metabolic wastes from blood is a more simplistic technological example. Regardless of how they may be fabricated, these materials are generally referred to as . Consider the apparatus illustrated in Figure \(\Page {6}\), in which samples of pure solvent and a solution are separated by a membrane that only solvent molecules may permeate. Solvent molecules will diffuse across the membrane in both directions. Since the concentration of is greater in the pure solvent than the solution, these molecules will diffuse from the solvent side of the membrane to the solution side at a faster rate than they will in the reverse direction. The result is a net transfer of solvent molecules from the pure solvent to the solution. Diffusion-driven transfer of solvent molecules through a semipermeable membrane is a process known as . When osmosis is carried out in an apparatus like that shown in Figure \(\Page {6}\), the volume of the solution increases as it becomes diluted by accumulation of solvent. This causes the level of the solution to rise, increasing its hydrostatic pressure (due to the weight of the column of solution in the tube) and resulting in a faster transfer of solvent molecules back to the pure solvent side. When the pressure reaches a value that yields a reverse solvent transfer rate equal to the osmosis rate, bulk transfer of solvent ceases. This pressure is called the e (\(\Pi\)) of the solution. The osmotic pressure of a dilute solution is related to its solute molarity, , and absolute temperature, , according to the equation \[Π=MRT \label{11.5.10} \] where \(R\) is the universal gas constant. What is the osmotic pressure (atm) of a 0.30 solution of glucose in water that is used for intravenous infusion at body temperature, 37 °C? We can find the osmotic pressure, \( , using Equation \ref{11.5.10}, where is on the Kelvin scale (310 K) and the value of is expressed in appropriate units (0.08206 L atm/mol K). \[\begin{align*}
Π&=MRT\\
&=\mathrm{0.03\:mol/L×0.08206\: L\: atm/mol\: K×310\: K}\\
&=\mathrm{7.6\:atm}
\end{align*} \nonumber \] What is the osmotic pressure (atm) a solution with a volume of 0.750 L that contains 5.0 g of methanol, CH OH, in water at 37 °C? 5.3 atm If a solution is placed in an apparatus like the one shown in Figure \(\Page {7}\), applying pressure greater than the osmotic pressure of the solution reverses the osmosis and pushes solvent molecules from the solution into the pure solvent. This technique of reverse osmosis is used for large-scale desalination of seawater and on smaller scales to produce high-purity tap water for drinking. Examples of osmosis are evident in many biological systems because cells are surrounded by semipermeable membranes. Carrots and celery that have become limp because they have lost water can be made crisp again by placing them in water. Water moves into the carrot or celery cells by osmosis. A cucumber placed in a concentrated salt solution loses water by osmosis and absorbs some salt to become a pickle. Osmosis can also affect animal cells. Solute concentrations are particularly important when solutions are injected into the body. Solutes in body cell fluids and blood serum give these solutions an osmotic pressure of approximately 7.7 atm. Solutions injected into the body must have the same osmotic pressure as blood serum; that is, they should be with blood serum. If a less concentrated solution, a solution, is injected in sufficient quantity to dilute the blood serum, water from the diluted serum passes into the blood cells by osmosis, causing the cells to expand and rupture. This process is called . When a more concentrated solution, a solution, is injected, the cells lose water to the more concentrated solution, shrivel, and possibly die in a process called (Figure 11.5.8). Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements. A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at 2.32 °C. What is the molar mass of this compound? We can solve this problem using the following steps. \(ΔT_\ce{f}=K \(m=\dfrac{ΔT_\ce{f}}{K_\ce{f}}=\dfrac{3.2\:°\ce C}{5.12\:°\ce C m^{−1}}=0.63\:m\) \(\ dfrac mol A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. What is the molar mass of this compound? 1.8 × 10 g/mol A 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at 22 °C. What is the molar mass of hemoglobin? Here is one set of steps that can be used to solve the problem: \[\Pi=MRT \nonumber \] \(M=\dfrac{Π}{RT}=\mathrm{\dfrac{7.8×10^{−3}\:atm}{(0.08206\:L\: atm/mol\: K)(295\:K)}=3.2×10^{−4}\:M}\) \(\mathrm{molar\: mass=\dfrac{10.0\:g}{1.6×10^{−4}\:mol}=6.2×10^4\:g/mol}\) What is the molar mass of a protein if a solution of 0.02 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 °C? 2.7 × 10 g/mol As noted previously in this module, the colligative properties of a solution depend only on the number, not on the kind, of solute species dissolved. For example, 1 mole of any nonelectrolyte dissolved in 1 kilogram of solvent produces the same lowering of the freezing point as does 1 mole of any other nonelectrolyte. However, 1 mole of sodium chloride (an electrolyte) forms of ions when dissolved in solution. Each individual ion produces the same effect on the freezing point as a single molecule does. The concentration of ions in seawater is approximately the same as that in a solution containing 4.2 g of NaCl dissolved in 125 g of water. Assume that each of the ions in the NaCl solution has the same effect on the freezing point of water as a nonelectrolyte molecule, and determine the freezing temperature the solution (which is approximately equal to the freezing temperature of seawater). We can solve this problem using the following series of steps. Check each result as a self-assessment. Assume that each of the ions in calcium chloride, CaCl , has the same effect on the freezing point of water as a nonelectrolyte molecule. Calculate the freezing point of a solution of 0.724 g of CaCl in 175 g of water. −0.208 °C Assuming complete dissociation, a 1.0 aqueous solution of NaCl contains 2.0 mole of ions (1.0 mol Na and 1.0 mol Cl ) per each kilogram of water, and its freezing point depression is expected to be \[ΔT_\ce{f}=\mathrm{2.0\:mol\: ions/kg\: water×1.86\:°C\: kg\: water/mol\: ion=3.7\:°C.} \label{11.5.11} \] When this solution is actually prepared and its freezing point depression measured, however, a value of 3.4 °C is obtained. Similar discrepancies are observed for other ionic compounds, and the differences between the measured and expected colligative property values typically become more significant as solute concentrations increase. These observations suggest that the ions of sodium chloride (and other strong electrolytes) are not completely dissociated in solution. To account for this and avoid the errors accompanying the assumption of total dissociation, an experimentally measured parameter named in honor of Nobel Prize-winning German chemist Jacobus Henricus van’t Hoff is used. The is defined as the ratio of solute particles in solution to the number of formula units dissolved: \[i=\dfrac{\textrm{moles of particles in solution}}{\textrm{moles of formula units dissolved}} \label{11.5.12} \] Values for measured van’t Hoff factors for several solutes, along with predicted values assuming complete dissociation, are shown in Table \(\Page {2}\). In 1923, the chemists Peter and Erich proposed a theory to explain the apparent incomplete ionization of strong electrolytes. They suggested that although interionic attraction in an aqueous solution is very greatly reduced by solvation of the ions and the insulating action of the polar solvent, it is not completely nullified. The residual attractions prevent the ions from behaving as totally independent particles (Figure \(\Page {9}\)). In some cases, a positive and negative ion may actually touch, giving a solvated unit called an ion pair. Thus, the , or the effective concentration, of any particular kind of ion is less than that indicated by the actual concentration. Ions become more and more widely separated the more dilute the solution, and the residual interionic attractions become less and less. Thus, in extremely dilute solutions, the effective concentrations of the ions (their activities) are essentially equal to the actual concentrations. Note that the van’t Hoff factors for the electrolytes in Table \(\Page {2}\) are for 0.05 solutions, at which concentration the value of for NaCl is 1.9, as opposed to an ideal value of 2. Properties of a solution that depend only on the concentration of solute particles are called colligative properties. They include changes in the vapor pressure, boiling point, and freezing point of the solvent in the solution. The magnitudes of these properties depend only on the total concentration of solute particles in solution, not on the type of particles. The total concentration of solute particles in a solution also determines its osmotic pressure. This is the pressure that must be applied to the solution to prevent diffusion of molecules of pure solvent through a semipermeable membrane into the solution. Ionic compounds may not completely dissociate in solution due to activity effects, in which case observed colligative effects may be less than predicted. | 26,714 | 3,339 |
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Water, a natural occurring and abundant substance that exists in solid, liquid, and gas forms on the planet Earth, has attracted the attention of artists, engineers, poets, writers, philosophers, environmentalists, scientists, and politicians. Every aspect of life involves water as food, as a medium in which to live, or as the essential ingredient of life. The food-science aspects of water range from agriculture, aquaculture, biology, biochemistry, cookery, microbiology, nutrition, photosynthesis, power generation, to zoology. Even in the narrow sense of food technology, water is intimately involved in the production, washing, preparation, manufacture, cooling, drying, and hydration of food. Water is eaten, absorbed, transported, and utilized by cells. Facts and data about water are abundant and diverse. This article can only selectively present some fundamental characteristics of water molecules and their collective properties for readers when they ponder food science at the molecular level. The physics and chemistry of water is the backbone of engineering and sciences. The basic data for the properties of pure water, which are found in the CRC Handbook of Chemistry and Physics (1), are useful for food scientists. However, water is a universal solvent, and natural waters contain dissolved substances present in the environment. All solutes in the dilute solutions modify the water properties. Lang’s Handbook of Chemistry (2) gives solubilities of various gases and salts in water. Water usage in the food processing industry is briefly described in the Nalco Water Handbook (3). For water supplies and treatments, the Civil Engineering Handbook (4) provides practical guides. The Handbook of Drinking Water Quality (5) sets guidelines for waters used in food services and technologies. Wastewater from the food industry needs treatment, and the technology is usually dealt with in industrial chemistry (6). Most fresh food contains large amounts of water. Modifying the water content of foodstuffs to extend storage life and enhance quality is an important and widely used process (7). A very broad view and deep insight on water can be found in “Water – A Matrix of Life” (8). Research leading to our present-day understanding of water has been reviewed in the series “Water – A Comprehensive Treatise” (9). The interaction of water with proteins (10, 11) is a topic in life science and food science. Water is the elixir of life and H O is a biomolecule. Water is an essential component of food (12). Philosophical conjectures abound as to how Earth evolved to provide the mantle, crust, atmosphere, hydrosphere, and life. Debates continue, but some scientists believe that primitive forms of life began to form in water (13). Complicated life forms developed, and their numbers grew. Evolution produced anaerobic, aerobic and photosynthetic organisms. The existence of abundant life forms enabled parasites to appear and utilize plants and other organisms. From water all life began (14). Homo sapiens are integral parts of the environment, and constant exchange of water unites our internal space with the external. The proper amount of water is also the key for sustaining and maintaining a healthy life. Water transports nutrients and metabolic products throughout the body to balance cell contents and requirements. Water maintains biological activities of proteins, nucleotides, and carbohydrates, and participates in hydrolyses, condensations, and chemical reactions that are vital for life (15). On average, an adult consumes 2 to 3 L of water: 1-2 L as fluid, 1 L ingested with food, and 0.3 L from metabolism. Water is excreted via the kidney, skin, lung, and anus (16). The amount of water passing through us in our lifetimes is staggering. Aside from minute amounts of minerals, food consists of plant and animal parts. Water is required for cultivating, processing, manufacturing, washing, cooking and digesting food. During or after eating, a drink, which consists of mostly water, is a must to hydrate or digest the food. Furthermore, water is required in the metabolic process. Cells and living organisms require, contain and maintain a balance of water. An imbalance of water due to freezing, dehydration, exercise, overheating, etc. leads to the death of cells and eventually the whole body. Dehydration kills far more quickly than starvation. In the human body, water provides a medium for the transportation, digestion and metabolism of food in addition to many other physiological functions such as body temperature regulation (17). Two-thirds of the body mass is water, and in most soft tissues, the contents can be as high as 99% (16). Water molecules interact with biomolecules intimately (9); it is part of us. Functions of water and biomolecules collectively manifest life. Water is also required for running households, making industrial goods, and generating electric power. Water has shaped the landscape of Earth for trillions of years, and it covers 70% of the Earth’s surface. Yet, for food production and technology it is a precious commodity. Problems with water supply can lead to disaster (5). Few brave souls accept the challenge to stay in areas with little rainfall. Yet, rainfall can be a blessing or a curse depending on the timing and amount. Praying for timely and bounty rainfall used to be performed by emperors and politicians, but water for food challenges scientists and engineers today. Plato hypothesized four : water, fire, earth, and air. His doctrine suggested that a combination and permutation of various amounts of these four produced all the materials of the world. Scholars followed this doctrine for 2000 years, until it could not explain experimental results. The search of fundamental substances led to the discovery of hydrogen, oxygen, nitrogen, etc., as chemical elements. Water is made up of hydrogen (H) and oxygen (O). Chemists use H O as the universal symbol for water. The molecular formula, H O, implies that a water molecule consists of two H atoms and one O atom. However, many people are confused with its other chemical names such as hydrogen oxide, dihydrogen oxide, dihydrogen monoxide, etc. The discoveries of electrons, radioactivity, protons, and neutrons implied the existence of isotopes. Natural isotopes for all elements have been identified. Three isotopes of hydrogen are protium ( H), deuterium (D, D or H), and radioactive tritium (T, T or H), and the three stable oxygen isotopes are O, O, and O. The masses and abundances of these isotopes are given in Table 1. For radioactive isotopes, the half-lives are given. 1.007825 2.00141018 3.0160493 15.9949146 16.9991315 17.9991604 99.985% 0.015% 12.33 years 99.762% 0.038% 0.200%
molar mass (amu) and relative abundance (%, ppm or trace) 18.010564 99.78% 20.014810 0.20% 19.014781 0.03% 19.00415 0.0149% 19.997737 0.022 ppm 20.018789 trace Random combination of these isotopes gives rise to the various isotopic water molecules, the most abundant one being H O (99.78%, its mass is 18.010564 atomic mass units (amu)). Water molecules with molecular masses about 19 and 20 are present at some fractions of a percent. Although HD O (0.0149%) is much more abundant than D O (heavy water, 0.022 part per million), D O can be concentrated and extracted from water. In the extraction process, HDO molecules are converted to D O due to isotopic exchange. Rather pure heavy water (D O) is produced on an industrial scale especially for its application in nuclear technology, which provides energy for the food industry. A typical mass spectrum for water shows only mass-over-charge ratio of 18 and 17 respectively for H O and OH ions in the gas phase. Other species are too weak for detection, partly due to condensation of water in mass spectrometers. The isotopic composition of water depends on its source and age. Its study is linked to other sciences (18). For the isotopic analysis of hydrogen in water, the hydrogen is reduced to a hydrogen gas and then the mass spectrum of the gas is analyzed. For isotopes of oxygen, usually the oxygen in H O is allowed to exchange with CO , and then the isotopes of the CO are analyzed. These analyses are performed on archeological food remains and unusual food samples in order to learn their origin, age, and history. Chemical bonding is a force that binds atoms into a molecule. Thus, chemists use H-O-H or HOH to represent the bonding in water. Furthermore, spectroscopic studies revealed the H–O–H bond angle to be 104.5 and the H–O bond length to be 96 picometers (pm) for gas H O molecules (19). For solid and liquid, the values depend on the temperature and states of water. The bond length and bond angles are fundamental properties of a molecule. However, due to the vibration and rotational motions of the molecule, the measured values are average or equilibrium bond lengths and angles. The mean van der Waals diameter of water has been reported as nearly identical with that of isoelectronic neon (282 pm). Some imaginary models of the water molecule are shown in Fig. 1 An isolated water molecule is hardly static. It constantly undergoes a vibration motion that can be a combination of any or all of the three principle modes: symmetric stretching, asymmetric stretching, and bending (or deformation). These vibration modes are indicated in Fig. 2. Absorption of light (photons) excites water molecules to higher energy levels. Absorption of photons in the infrared (IR) region excites the vibration motion. Photons exciting the symmetric stretching, bending, and asymmetric stretching to the next higher energy levels have wave numbers 3656, 1594, and 3756 cm respectively, for H O (20). These values and those for other water molecules involving only O are given in Table 2 The spectrum of water depends on temperature and density of the gaseous H O. A typical IR spectrum for the excitation of only the fundamental vibration modes consists of three peaks around 1594, 3656 and 3756 cm . Additional peaks due to excitation to mixed modes appear at higher wave numbers. Rotating the H O molecule around the line bisecting the HOH angle by 180° (360°/2) results in the same figure. Thus, the molecules have a 2-fold rotation axis. There are two mirror planes of symmetry as well. The 2-fold rotation and mirror planes give the water molecules the symmetry point group Rutherford alpha scattering experiment in 1909 showed that almost all atomic mass is in a very small atomic nucleus. In a neutral atom the number of protons in the nucleus is the same as the number of electrons around the nucleus. A proton and an electron have the same amount, but different kind of charge. Electrons occupy nearly all of the atomic volume, because the radius of an atom is 100,000 times that of the nucleus. Electrons, in quantum mechanical view, are waves confined in atoms, and they exist in several energy states called orbitals. Electrons in atoms and molecules do not have fixed locations or orbits. Electron states in an element are called electronic configurations, and their designation for H and O are 1 , and 1 2 2 , respectively. The superscripts indicate the number of electrons in the orbitals 1 , 2 or 2 . The electronic configuration for the inert helium (He) is 1 , and 1 is a stable core of electrons. Bonding or valence electrons are 1 and 2 2 for H and O respectively. The valence bond approach blends one 2 and three 2 orbitals into four bonding orbitals, two of which accommodate two electron pairs. The other two orbitals have only one electron each, and they accommodate electrons of the H atoms bonded to O, thus forming the two H–O bonds. An electron pair around each H atom and four electron pairs around the O atom contribute stable electronic configurations for H and O, respectively. The Lewis dot-structure, Fig. 3, represents this simple view. The two bonding and two lone pairs are asymmetrically distributed with major portions pointing to the vertices of a slightly distorted tetrahedron in 3-dimensional space. The two lone pairs mark slightly negative sites and the two H atoms are slightly positive. This charge distribution around a water molecule is very important in terms of its microscopic and macroscopic, chemical and physical properties described later. Of course, the study of water continues and so does the evolution of bonding theories. Moreover, the distribution of electrons in a single water molecule is different from those of dimers, clusters, and bulk water. The asymmetric distribution of H atoms and electrons around the O atom results in positive and negative sites in the water molecule. Thus, water consists of polar molecules. The is a measure of polarity and a useful concept. A pair of opposite charge, , separated by a distance, , has a dipole moment of = with the direction pointing towards the positive charge as shown in Fig. 4. The dipole moment of individual water molecules is 6.187 10 C m (or 1.855 D) (21). This quantity is the vector resultant of two dipole moments due to the O–H bonds. The bond angle H–O–H of water is 104.5 . Thus, the dipole moment of an O–H bond is 5.053 10 C m. The bond length between H and O is 0.10 nm, and the partial charge at the O and the H is therefore = 5.053 10 C, 32 % of the charge of an electron (1.6022 10 C). Of course, the dipole moment may also be considered as separation of the electron and positive charge by a distance 0.031 nm. It should be pointed out that the dipole moments of liquid and solid water appear to be higher due to the influence of neighboring molecules. For the liquid and solid, macroscopic properties need be considered. Attraction among water molecules is more than polar-polar in nature. The O atoms are small and very electronegative. As a result, the positive H atoms (protons) are very attractive to the negative O atoms of neighboring molecules. This O- - -H – O strong attraction is called a , a concept popularized by L. Pauling (22). Furthermore, hydrogen atoms bonded to atoms of N and F, neighboring elements of O in the periodic table, are positive, and they form hydrogen bonds with atoms of N, O, or F. The strength of hydrogen bonds depends on the X–H - - - Y (X or Y are N, O, or F atoms) distances and angles; the shorter the distances, the stronger are the hydrogen bonds. When two isolated water molecules approach each other, a dimer is formed due to hydrogen bonding. The dimer may have one or two hydrogen bonds. Dimers exist in gaseous and liquid water. When more water molecules are in close proximity, they form trimmers, tetramers and clusters. Hydrogen bonds are not static, they exchange protons and partners constantly. Hydrogen bonding is a prominent feature in the structures of various solid phases of water usually called ice as we shall see later. Water molecules not only form hydrogen bonds among themselves, they form hydrogen bonds with any molecule that contains N–H, O–H and F–H bonds. Foodstuffs such as starch, cellulose, sugars, proteins, DNA, and alkaloids contain N–H and O–H groups, and these are both H-donors and H-acceptors of hydrogen bonds of the type N- - H–O, O- -H–N, N- -H–N, etc. A dimer depicting the hydrogen bond and the van der Waal sphere of two molecules is shown in Fig. 6 (23). Carbohydrates (starch, cellulose and sugars) contain H–C–O–H groups. The O–H groups are similar to those of water molecules, and they are H-acceptors and H-donors for hydrogen bonds. Proteins contain O–H, R–NH or R >NH groups, and the O–H and N–H groups are both H-donors and H-acceptors for the formation of hydrogen bonds. Thus, water molecules have intimate interactions with carbohydrates and proteins. Collectively, water molecules exist as gas, liquid, or solid depending on the temperature and pressure. These phases of water exhibit collective or macroscopic properties such as phase transitions, crystal structures, liquid structures, vapor pressures, and volume-pressure relationships of vapor. In addition, energies or enthalpies for melting, vaporization and heating are also important for applications in food technology. Thermodynamic constants for phase transitions given in Table 3 are those of pure water. Natural waters, of course, contain dissolved air, carbon dioxide, organic substances, microorganisms, and minerals. Water in food or used during food processing usually contains various organic and inorganic substances. These solutes modify the properties of water and caution should be taken to ensure proper values are applied in food technology. The triple point and boiling points of water are defined as 273.16 and 373.16 K (kelvin) in the SI unit of temperature, respectively. Thus, the temperature differences can be in units of K or oC. Water has many unusual properties due to its ability to form hydrogen bonds and its large dipole moment. As a result, the melting, boiling and critical points for water are very high compared to substances of similar molar masses. In general, the higher the molar masses, the higher are the melting and boiling points of the material. Associated with these properties are its very large heat of melting, heat capacity, heat of vaporization and heat of sublimation. Moreover, its surface tension and viscosity are also very large. Thermodynamic energies, and volume changes for phase transitions of H O are summarized in Table 3. These data are mostly taken from the Encyclopedia of Chemical Technology, Vol. 25 (1991) (24). Hydrogen bonding is prominent in the crystal structures of various solid phases of H O. The triple point of water is at 273.16 K and 4.58 torr (611 Pa). The melting point at 1.00 atm (760 torr or 101.325 kPa) is used to define the Kelvin scale as 273.15 K. When water freezes at these temperatures and atmospheric pressure or lower, the solids are hexagonal ice crystals usually designated as I . Properties of I are given in Table 4. Snowflakes have many shapes because their growth habit depends on temperature and vapor pressure, but they all exhibit hexagonal symmetry, due to the hexagonal structure of ice (25). However, from a geometric point of view, the same bonding may also be arranged to have cubic symmetry. The existence of cubic ice has been confirmed. When water vapor deposits onto a very cold, 130 – 150 K, surface or when small droplets are cold under low pressure at high altitude, the ice has a cubic symmetry usually designated as I . At still higher pressures, different crystal forms designated as ice II, III, IV, … etc., up to 13 phases of cubic, hexagonal, tetragonal, monoclinic, and orthorhombic symmetries have been identified (26). The polymorphism of solid water is very complicated. Some of these ice forms are made under very high pressures, and water crystallizes into solid at temperatures above the normal melting or even boiling temperatures. Ice VII is formed above 10 G Pa (gigapascal) at 700 K (26). The basic relationships between nearest neighboring water molecules are the same in both I and I . All O atoms are bonded to four other O atoms by hydrogen bonds, which extend from an oxygen atom towards the vertices of a tetrahedron. A sketch of the crystal structure of hexagonal I is shown in Fig. 7 (27). In I , hydrogen positions are somewhat random due to thermal motion, disorder and exchanges. For example, the hydrogen may shift between locations to form H O and OH ions dynamically throughout the structure. In this structure, bond angles or hydrogen-bond angles around oxygen atoms are those of the idealized tetrahedral arrangement of 109.5 rather than 104.5 observed for isolated molecules. Formation of the hydrogen bond in ice lengthens the O–H bond distance, 100 pm compared to 96 pm in a single water molecule. The diagram illustrates a crystal structure that is completely hydrogen bonded, except for the molecules at the surface. Each O atom of hexagonal ice I is surrounded by four almost linear O–H - - O hydrogen bonds of length 275 pm, in a tetrahedral fashion. Each C atom of cubic diamond is also surrounded by four C–C covalent bonds of length 154 pm. Thus, the tetrahedral coordination can either be cubic or hexagonal, from a geometrical viewpoint. Indeed, the uncommon cubic ice and hexagonal diamond have been observed, giving a close relationship in terms of spatial arrangement of atoms between ice and diamond (26). Strong hydrogen bonds make ice hard, but brittle. The structure is related to its physical properties, which vary with temperature. The pressure of H O vapor in equilibrium with ice is called the which decreases as the temperature decreases. At the triple point or 0° C, the pressure is 611.15 Pa. When ice is slightly overheated to 0.01° C, the pressure increases to 611.657 Pa. However, at this temperature, the vapor pressure of liquid water is lower. The vapor pressures of ice between 0° C and – 40° C are listed in Table 5 at 1°C interval. Various models can be used to estimate the vapor pressure at other temperatures. One method uses the Clausius-Clapeyron differential equation
d
––– = –
d
where is the pressure, is the temperature (K), is the latent heat or of phase transition, and is the difference in volume of the phases. The enthalpy of sublimation for ice depends on the temperature. At the freezing point, the enthalpy of sublimation for ice is 51 (51.06 in Table 3) kJ mol , estimated from the vapor pressure at 0 and –1 C. The enthalpy of sublimation is required to overcome hydrogen bonding, dipole, and intermolecular attractions. The energy required in freeze-drying processes varies, depending on temperature and other conditions. Water in solutions and in food freezes below 0 C. The number of hydrogen bonds is twice the number of water molecules, when surface water molecules are ignored. The energy required to separate water molecules from the solid is the enthalpy of sublimation (55.71 J mol ). Half of this value, 26 kJ mol , is the energy to separate the H- -O linkages, and it translates into 0.26 eV, per H- -O bond. These values are close to those obtained by other means (25, 26, 28, 29, 30). Several factors contribute to this linkage, and the hydrogen-bond energy is less than 0.26 eV. The macroscopic physical properties of this common but eccentric fluid at 298 K (25 C) are given in Table 6. Water has an unusually high melting and boiling points for a substance of molar mass of only 18 daltons. Strong hydrogen bonds and high polarity accounts for this. The heat of formation is the energy released when a mole of hydrogen and half a mole of oxygen at 298 K and 1.00 atm react to give one mole of water at 298 K. This value differs from that for ice in Table 4 due to both temperature and phase differences. As temperature increases, the average kinetic energy of molecules increases, and this affects water’s physical properties. For example, surface tension of water decreases, whereas the thermal conductance increases as the temperature increases. Heat capacity at constant pressure ( p), vapor pressure, viscosity, thermal conductance, dielectric constant, and surface tension in the temperature range 273–373 K (0–100 C) are given in Table 7. conductance constant tension ----- More detailed data can be found in the CRC Handbook of Chemistry and Physics (1) Liquid water has the largest heat capacity per unit mass of all substances. Large quantities of energy are absorbed or release when its temperatures changes. The large heat capacity makes water an excellent reservoir and transporter of energy. A large body of water moderates climate. The heat capacity of water varies between 4.1 to 4.2 J g K (74 to 76 J mol K ) even at temperature above 100 C and high pressure. The enthalpy of vaporization for water is also very large (55.71 kJ mol at 298 K). Thus, energy consumption is high for food processing when water is involved. Water and aqueous solutions containing only low molar-mass solutes are typical Newtonian fluids for which the is proportional to . Viscosity is the ratio of shear stress to shear strain rate. On the other hand, viscosity of solutions containing high molar-mass substances depends on shear strain rate. For pure water, the viscosity decreases from 1.793 to 0.282 mPa s (millipascal seconds; identical to centipoise (cp)) as temperature increases from 0 to 100 C. Thus, the flow rate through pipes increases as water or solution temperature increases. The dielectric constant of water is very large, and this enables water to separate ions of electrolytes, because it reduces the electrostatic attraction between positive and negative ions. Many salts dissolve in water. When an electric field is applied to water, its dipole molecules orient themselves to decrease the field strength. Thus, its dielectric constant is very large. The dielectric constant decreases as temperature increases, because the percentage of molecules involved in hydrogen bonding and the degree of order decrease (28, 29). The measured dielectric constant also depends on the frequency of the applied electric field used in the measurement, but the variation is small when the frequency of the electric field is less than 100 MHz. The dielectric behavior of water allows water vapor pressure to be sensed by capacitance changes when moisture is absorbed by a substance that lies between the plates of a capacitor. These sensors have been developed for water activity measurement (31). The light absorption coefficients are high in the infrared and ultraviolet regions, but very low in the visible region. Thus, water is transparent to human vision. The variation of vapor pressure as a function of temperature is the bases for defining of food. Liquid water exists between the triple-point and the critical-point temperatures (0 – 373.98 C) at pressures above the vapor pressures in this range. As with ice, the vapor pressure of liquid water increases as the temperature increases. Vapor pressures of water (in kPa instead of Pa for ice in Table 5) between the triple and critical points, at 10 C interval, are given in Table 8. When the vapor pressure is 1.00 atm (101.32 kPa) the temperature is the boiling point (100 C). At slightly below 221 C, the vapor pressure is 2.00 atm. The critical pressure at the critical temperature, 373.98 C, is 217.67 atm (22,055 kPa). Above this temperature, water cannot be liquefied, and the phase is called . The partial pressure of H O in the air at any temperature is the . When the air is saturated with water vapor, the is 100%. The unsaturated vapor pressure divided by the vapor pressure of water as given in Table 8 at the temperature of the air is the . The temperature at which the vapor pressure in the air becomes saturated is the , at which dew begins to form. However when the dew point is below 273 K or 0 C, ice crystals (frost) begin to form. Thus, the relative humidity can be measured by finding the dew point. Dividing the vapor pressure at the dew point by the vapor pressure of water at the temperature of the air gives the relative humidity. The transformations between solid, liquid, and gaseous water play important roles in hydrology and in the transformation of the environment on Earth. Phase transitions of water combined with the energy from the sun make the weather. Density is a collective property, and it varies with temperature, isotopic composition, purity, etc. The International Union of Pure and Applied Chemistry (IUPAC) has adopted the density of pure water from the ocean as the density standard. The isotopic composition of ordinary water is constant, and the density of pure water between 0 and 39 C extracted from (32) is given in Table 9. The density of cage-like ice I , due to 100% of its molecules involved in hydrogen-bonded is only 9% lower than that of water. This indicates that water has a high percentage of molecules involved in the transient and dynamic hydrogen bonding. The percentage of hydrogen-bonded water molecules in water decreases as temperature increases, causing water density to increase. As temperature increases the thermal expansion causes its density to decreases. The two effects cause water density to increase from 0 to 3.98 C, reaching its maximum of 1.0000 g mL and then decrease as temperature increases. Incidentally at 8 C, the density of water is about the same as that at 0 C. At 25 C, the density decreases 0.3 % with respect to its maximum density, whereas at 100 C, it decreases by 4 %. Dense water sinks, and convection takes place when temperature fluctuates at the surface of lakes and ponds, bringing dissolved air and nutrients to various depths of waters for the organism living in them. On the other hand, the pattern of density dependence on temperature of water makes temperatures at the bottoms of lakes and oceans vary little if the water is undisturbed. When water freezes, ice begins to form at the surface, leaving the water at some depth undisturbed. Water at the bottom remains at 4 C preserving various creatures living in water. When hydrogen-bonded to tissues and cells or in food, water has a unique order and structure, and the vapor pressure and density differ from those of pure water. Yet, the collective behavior of water molecules sheds some light regarding their properties in food, cells, tissues, and solutions. Water is a chemical as is any substance, despite the confusion and distrust of the public regarding the term “chemical”. Thus, water has lots of interesting chemical properties. It interacts intimately with components of food particularly as a solvent, due to its dipole moment and its tendency to form hydrogen bond. These interactions affect the chemical properties of nutrients, including their tendency to undergo oxidation or reduction, to act as acids or bases, and to ionize. Water is dubbed a universal solvent, because it dissolves many substances due to strong interactions between water molecules and those of other substances. Entropy is another driving force for a liquid to dissolve or mix with other substances. Mixing increases disorder or entropy. Because of its large dielectric constant, high dipole moment and ability to donate and accept protons for hydrogen bonding, water is an excellent solvent for polar substances and electrolytes, which consist of ions. Molecules strongly interact with or love water molecules are , due to hydrogen bonding, polar-ionic or polar-polar attractions. Nonpolar molecules that do not mix with water are or , because they tend to dissolve in oil. Large molecules such as proteins and fatty acids that have hydrophilic and hydrophobic portions are or Water molecules strongly intermingle with hydrophilic portions by means of dipole-dipole interaction or hydrogen bonding. The lack of strong interactions between water molecules and lipophilic molecules or the nonpolar portions of amphipathic molecules is called the , a term coined by Charles Tanford (33). Instead of a direct interaction with such solutes, water molecules tend to form hydrogen-bonded cages around small nonpolar molecules when the latter are dispersed into water. Hydrogen-bonded water molecules form cages, called or , For example, the clathrate of methane forms stable crystals at low temperatures (34). Nonpolar chains in proteins prefer to stay together as they avoid contact with water molecules. Hydrophilic and hydrophobic effects play important roles for the stability and state of large molecules such as enzymes, proteins, and lipids. Hydrophobic portions of these molecules stay together forming pockets in globular proteins. Hydrophilic and hydrophobic effects cause nonpolar portions of phospholipids, proteins, and cholesterol to assemble into bilayers or biological membranes (34). Due to its high dielectric constant, water reduces the attractions among positive and negative ions of electrolytes and dissolves them. The polar water molecules coordinate around ions forming hydrated ions such as Na(H O) , Ca(H O) , Al(H O) etc. Six to eight water molecules form the first sphere of hydration around these ions. Fig. 8 is a sketch of the interactions of water molecules with ions. The water molecules point their negative ends of their dipoles towards positive ions, and their positive ends towards negative ions. Molecules in the hydration sphere constantly and dynamically exchange with those around them. The number of hydrated-water molecules and their lifetimes have been studied by various methods. These studies reveal that the hydration sphere is one-layer deep, and the life times of these hydrated-water molecules are in the order of picoseconds (10 s). The larger negative ions also interact with the polar water molecules, not as strong as those of cations. The presence of ions in the solution changes the ordering of molecules even if they are not in the first hydration sphere (9). The hydration of ions releases energy, but breaking up ions from a solid requires energy. The amounts of energy depends on the substance, and for this reason, some are more soluble than others. Natural waters in the ocean, streams, rivers and lakes are in contact with minerals and salts. The concentrations of various ions depend on the solubility of salts (35) and the contact time. Drinking water includes all waters used in growth, processing, and manufacturing of food. J. De Zuane divides ions in natural water into four types in The Handbook on Drinking Water Quality (5). Type A includes arsenic, barium, cadmium, chromium, copper, fluoride, mercury, nitrate, nitrite, and selenium ions. They are highly toxic, yet abundant. Type B includes aluminum, nickel, sodium, cyanide, silver, zinc, molybdenum, and sulfate ions. Their concentrations are also high, but they are not very toxic. Type C consists of calcium, carbonate, chloride, iron, lithium, magnesium, manganese, oxygen, phosphate, potassium, silica, bromine, chlorine, and iodine and ozone. They are usually present at reasonable levels. Type D ions are present usually at low levels: antimony, beryllium, cobalt, tin, thorium, vanadium and thallium. Most metals are usually present in water as cations, with a few as anions. However, some chemical analyses may not distinguish their state in water. The most common anions are chloride, sulfate, carbonate, bicarbonate, phosphate, bromide, iodide, etc. Toxicity is a concern for ions in water, but some of these ions are essential for humans. Pure water has a very low electric conductivity, but ions in solutions move in an electric field making electrolyte solutions highly conductive. The conductivity is related to total dissolved solids (TDS), salts of carbonate, bicarbonate, chloride, sulfate, and nitrate. Sodium, potassium, calcium and magnesium ions are often present in natural waters because their soluble salts are common minerals in the environment. The solubilities of clay (alumina), silicates, and most common minerals in the Earth crust, are low. Waters containing plenty of dissolved CO (H CO ) are acidic and they dissolve CaCO and MgCO . Waters with dissolved Ca , Mg , HCO and CO are called as the hardness can be removed by boiling, which reduces the solubility of CO . When CO is driven off, the solution becomes less acidic due to the following equilibria (the double arrows, , indicate reversible reactions): H (aq) + HCO (aq) H CO (aq) H O + CO (g) HCO (aq) H (aq) + CO (aq)
Reducing the acidity increases the concentration of CO and solids CaCO and MgCO precipitate:
Ca (aq) + CO (aq) CaCO (s) Mg (aq) + CO (aq) MgCO (s) Water containing less than 50 mg L of these substances is considered soft; 50-150 mg L moderately hard; 150 - 300 mg L hard; and more than 300 mg L very hard. For the , we determine the amount of dissolved Ca and Mg first; and then add an equal number of moles of lime, Ca(OH) , to remove them by these reactions: Mg + Ca(OH) (s) Mg(OH) (s) + Ca Ca + 2 HCO + Ca(OH) (s) 2 CaCO (s) + 2 H O contain sulfate (SO ) ions with Ca and Mg . Calcium ions, Ca , of the sulfate solution can be removed by adding sodium carbonate: Ca + Na CO CaCO (s) + 2 Na Hard waters cause scales or deposits to build up in boilers and pipes, and they are usually softened by ion exchange with resins or zeolites. In these processes, the calcium and magnesium ions are taken up by the zeolite or resin that releases sodium or hydrogen ions into the water. Reverse osmosis has also being used to soften hard water. However, water softening replaces desirable calcium and other ions by sodium ions. Thus, soft waters are not suitable drinking waters. Incidentally, bakers use hard water because the calcium ions strengthen the gluten proteins in dough mixing. Some calcium salts are added to dough to enhance bread quality. Waters containing dissolved substances are aqueous solutions; their physical properties differ from those of pure water. For example, at the same temperature, the H O vapor pressures of solutions are lower than that of pure water, resulting in boiling point elevation (higher), freezing point depression (lower) and osmotic pressure. Several ways can be used to express concentrations: part per million (ppm), percent, moles per liter, mole fraction, etc. The mole fraction of water is the fraction of water molecules among all molecules and ions in the system. The vapor pressure of an ideal solution, , is the vapor pressure of water (at a given temperature), , modified by the mole fraction . = ( < 1)
If the solute has a significant vapor pressure, is also modified by its mole fraction, = For non-ideal solutions, in which water and solute strongly interact, the formulas require modifications. A practical method is to use an defined by:
/ = In any case, the vapor pressures of solutions containing nonvolatile electrolytes are lower than those of pure water at their corresponding temperature. Phase transitions take place when the vapor pressures of the two phases are the same. Because solutions’ vapor pressures are lower, their melting points are lower but their boiling points are higher. The difference in temperature, , is proportional to the concentrations of all solutes, (molality),
where is either the molar boiling point elevation constant, , or the molar freezing point depression constant . For water, = 1.86 K L kg , and = 0.52 K L kg . Due to ionization of electrolytes, positive and negative ions should be treated as separate species and all species should be included in . The tendency of water molecules from a dilute solution to diffuse into a more concentrated solution, through semipermeable membranes, has a measurable quantity called , , which is proportional to the concentration (mol per kg of water) of all dissolved species, in mol kg and temperature in K, = –
where is the gas constant 8.3145 J mol K . Water molecules diffuse from pure water ( = 0) into the solution, and the osmotic pressure is therefore given as a negative value here. Theoretically, a solution with = 1.0 mol kg of water, = – 2477 J kg or
– 2.477 kJ kg at 298 K. Note = ; ( being the number of ions produced by the solute) in the van’t Hoff equation, which is often used in other literature. Solutions having the identical osmotic pressure are isotonic. Applying more pressure to a solution to compensate for the osmotic pressure causes water molecules to diffuse through membranes, generating pure or fresh waters. This process is called , and it has been used to soften water or desalinate seawater, converting it to fresh water. The lowering of vapor pressure and the osmotic pressure of solutions play important roles in hydration and dehydration of food and in living cells. Solutions containing proper concentrations of nutrients and electrolyte have been used to medically treat dehydrated patients. J.R. Cade and his coworkers applied these principles to formulate drinks for athletes; he and his coworkers were credited as the inventor of the sports drink Gaterade (36). The concept of a balanced solution for hydration became a great business decades after its invention. Acidity and alkalinity are also important characteristics of water due to its dynamic self-ionization equilibrium,
H O (l) H + OH , = [H ,OH ] = 1 10 at 298 K and 1 atm
where [H ] and [OH ] represent the molar concentrations of H (or H O ) and OH ions, respectively, and is called the , (see tables in ref. 1 and 36). Values of under various conditions have been evaluated theoretically (37, 38). Solutions in which [H ] = [OH ] are said to be neutral. At 298 K, for a neutral solution,
pH = – log [H+] = pOH = – log [OH ] = 7 (at 298 K) The H ions or protons dynamically exchange with protons in other water molecules. The self-ionization and equilibrium are present in all aqueous solutions, including acid and base solutions, as well as in pure water. Water is both an acid and a base. Strong acids such as HClO , HClO , HCl, HNO , and H SO ionize completely in their solutions to give H (H O ) ions and anions: ClO , ClO , Cl , NO , and HSO , respectively. Strong bases such as NaOH, KOH, and Ca(OH) also ionize completely giving OH ions and Na , K , and Ca ions respectively. In an acidic solution, [H ] is greater than [OH ]. In a 1.0 mol L HCl solution, [H ] = 1.0 mol L , pH = 0. Weak acids such as formic acid, HCOOH, acetic acid (CH COOH), ascorbic acid (C H O ), oxalic acid (H C O ), carbonic acid (H CO ), benzoic acid (C H COOH), malic acid (C H O ), lactic acid H CCH(OH)COOH, and phosphoric acid (H PO ) also ionize in their aqueous solutions, but not completely. The ionization of acetic acid, is represented by the equilibrium,
CH COOH (aq) H (aq) + CH COO (aq), = 1.75 x 10 at 298 K
where K is the acid dissociation constant. The solubility of CO in water increases with its (CO partial) pressure, according to Henry’s law, and the chemical equilibria for the dissolution is, H O + CO (g) H CO (aq) Of course, H CO dynamically exchanges H , and H O with other water molecules, and this weak diprotic acid ionizes in two stages with their acid constants, and . H O + CO (aq) H (aq) + HCO (aq), = 4.30 x 10 (at 298 K) HCO (aq) H (aq) + CO (aq), = 5.61 x 10
Constants and increase with temperature. At 298 K, the pH of a solution containing 0.1 mol L H CO is 3.7, acidophilic organisms may grow, but most pathogenic organisms are neutrophiles and they cease growing. Soft drinks contain other acids – citric, malic, phosphoric, ascorbic acids etc; they lower the pH further. Ammonia and many nitrogen-containing compounds are weak bases. The ionization equilibrium of NH in water and the base dissociation constant are,
NH + H O NH (aq) + OH = 1.70 x 10 at 298 K.
Other weak bases react with H O similarly. The ionization or dissociation constants of inorganic and organic acids and bases are extensive, and they have been tabulated in various books (39, 40, 41). Amino acids and proteins contain acidic and basic groups. At some specific pH called the , they carry no charge, but exist as zwitterions. For example, the isoelectric point for glycine is pH = 6.00 and it exists as the zwitterion H C(NH )COO . Oxidation of hydrogen by oxygen not only produces water, but also releases energy. At the standard conditions, the electrochemical half reaction equations are: H = 2 H + 2 e E = 0.000 V (defined) O + 4 H + 4 e = 2 H O E = 1.229 V The cell reaction and the cell potential at the standard condition for it are: 2 H + O = 2 H O E = 1.229 V Proper setups for harvesting this energy are the goals of hydrogen-fuel-cell technology. The cell potential E for non-standard conditions depends on pH and temperature. Its value is related to the energy released in the reaction. A plot of E versus pH yields a Pourbaix diagram, which is useful to evaluate the stability of various species in water. Water can be a reducing or oxidizing reagent, because it offers protons or electrons. Applying a voltage to pass electrons through a chemical cell decomposes water by electrolysis. Waters containing dissolved oxygen cause additional reactions, for example: 2 H O + 2 e = H + 2 OH E = -0.828 V O + 2 H + 2 e = H O E = 0.682 V O + H O + 2 e = HO + OH E = 0.076 V O + 2 H O + 2 e = 4 OH E = 0.401 V At the proper conditions, a suitable chemical reaction driven by the potential takes place. Oxidation-reduction reactions involving water usually are due to proton or electron transfer. These oxidation-reduction reactions occur for the growth, production, manufacture, digestion, and metabolism of food. Water participates in oxidation-reduction reactions in many steps of photosynthesis, resulting in the fixation of CO into biomolecules, releasing oxygen atoms of water as O . Engineering a new generation of plants with greater photosynthetic capacity facing lack of waters challenges geneticists (42) and botanists. We now understand photosynthesis to great details, from the studies by many scientists. Photosynthetic reactions are related to food production, but they are so complex that we can only mention them (43). The oxidation-reduction reactions of water cause corrosion on metal surfaces. Not only deterioration of facilities is very costly for the food industry, corrosion of pipes results in having toxic metal ions Cu and Pb in drinking water. The concern of lead ions in drinking water led the Environmental Protection Agency to ban the use of high-lead solders for water pipes. These reactions are electrochemical processes. Galvanic effects, high acidity, high flow rate, high water temperature, and the presence of suspended solids accelerate corrosion, as do lack of Ca and Mg ions in purified waters. The formation of scales protects the metal surface. However, balancing the clogging against surface protection of pipes is a complicated problem, requiring scientific testing and engineering techniques for a satisfactory solution. Enzymes are mostly large protein molecules, and they are selective and specific catalysts responsible for most of reactions in biological bodies. Folding of the long protein provides specific 3-dimensional selective pockets for their substrates. The pockets not only fix the substrates in position, they also weaken certain bonds to facilitate specific reactions. This is the mechanism by which enzymes select their substrates and facilitate their specific reactions. Hydrogen bond strength is stronger in nonaqueous media than in aqueous solutions as the charge densities on the donor and acceptor atoms increase (44). Hydrogen bonds between the enzyme and its substrate can be stronger than those in an aqueous environment, thus speeding up the reaction rate even further. The hydrolysis of peptide linkage is the reaction of a protein with water, R-C(=O)-NH-R’ + H O R-C(=O)OH + H N-R’ This type of reaction can be catalyzed by acids, bases, and enzymes. Water is a nutrient and a component in food groups: grains, meat, dairy, fruits and vegetables. Furthermore, major nutrients such as carbohydrates, proteins, water-soluble vitamins, and minerals are hydrophilic. Even parts of fat or lipid molecules are hydrophilic, but the alkyl chains of fats and proteins experience the hydrophobic effect in an aqueous environment. (45). Foodstuffs interact with water by means of polar, hydrogen-bonding and hydrophobic interactions. The results of these interactions change the chemical potential (properties) of water. Foodstuffs dissolve in or absorb water. Thus, water within food may be divided into bound water, affected water, and free water in the order of their interaction strength. The bound water molecules are similar to those in the first hydration sphere of ions, and those close to the first sphere are affected water molecules. Further away from the interface are free water molecules. The structure and properties of the first two types change. Interaction of water with dietary fiber is an example (46). Thus, properties of water in food are different from those of pure water. Water molecules in both liquid and vapor phases can participate in hydration reactions. At equilibrium in a system with two or more phases, their vapor pressure or chemical potential, , must be equal. The chemical potential, , of a solution or water-containing foodstuff must be equal at a given temperature , and = + ln ( ),
where is the gas constant (8.3145 J mol K ), and is the vapor pressure of the solution or of water in foodstuff, and is the vapor pressure of pure water at the same temperature. The ratio is called the , which is related to the water chemical potential of water in solutions or in the foodstuff. For ideal solutions and for most moist foods, is less than unity, < 1.0 (31). Both water activity and relative humidity are fractions of the vapor pressure of pure water. Methods for their measurements are the same. We have mentioned the measurement by changes in capacitance earlier. Water contents have a sigmoidal relationship with water activities, = 1.0 for infinitely dilute solutions, > 0.7 for dilute solutions and moist foods and < 0.6 for dry foods. Of course, the precise relationship depends on the material in question. In general, if the water vapor of the atmosphere surrounding the food is greater than the activity of the food, water is absorbed. Otherwise, dehydration takes place. The water activity reflects the combined effects of water-solute, water-surface, capillary, hydrophilic and hydrophobic interactions. The water activity of a foodstuff is a vital parameter, because it affects its texture, taste, safety, shelf life, and appearance. Furthermore, controlling water activity, rather than water content is important. When < 0.9, most molds are inhibited. Growth of yeasts and bacteria also depends on . Microorganisms cease growing if < 0.6. Similar to water activity in food, is a term used in plant, soil, and crop sciences. Water potential, represented by (psi) or w, is a measure of the free energy of water in a system: soil, material, seeds, plants, roots, leaves, or an organism. Water potential is the difference between the chemical potential of pure water and water in the system at the same temperature. Pure water has the highest free energy: = 0 for pure water by convention, and < 0 for solutions. Water diffuses from high potential to low potential. Physiological processes decrease as the water potential decreases. In general, water potential, w ,is a combined effect of osmotic ( s), matrix (interface and water binding m), turgor ( t) pressures, and gravity ( g). w = s + m + t + g Osmotic pressure, s, is always present due to solutes in the fluids. The metric pressure, m, is related to bound-, affected- and free-waters in the system. The outwardly directed pressure extended by the swelling protoplast against the wall is called turgor pressure, t. Usually, this term is insignificant until the cell is full, and at such point, t increases rapidly and stops when w = t. Otherwise, the cell ruptures. The mechanical rigidity of succulent plant parts, the opening of stomata and the blossom are usually the results of turgor pressure. In systems such as tall plants and soil science, pressure due to the gravitational pull of water, g, is also included in the water potential. For example, the water potential of potato tissues can be measured by incubating them in a serious of solutions of known osmotic pressures. The potato will neither lose nor gain water if the osmotic pressure of the solution equals the water potential of potato tissues. The osmotic pressure ( = – ) may be evaluated from a known concentration , using the equation given earlier. Instead of energy units, water potential is often expressed in units of pressure (megapascal, MPa), which is derived by dividing the energy by the molar volume (0.018 L mol for H O) of water (47). There are many other methods for water potential measurements depending on the system: soil, leaf, stem, organism, etc. The soil water potential is related to the water available for the plants growing on the soil. Water potential of a plant or leaf indicates its health or state with respect to water. Thus, water potential is a better indicator for plant, agriculture, irrigation, and environmental managements than water content. Water moves through plants because
= 0 > > > > Thus, the concept of water potential and water activity are very useful in growth, manufacture, handling, storage, and management of food. The closer we look, the more we see. Living organisms on Earth are so complicated that their classification and phylogeny are still being studied and revised. New relationships are proposed to modify the five kingdoms proposed by Robert Whittaker in 1969. Nevertheless, most of the earliest unicellular living organisms in the Monera, and Protista kingdoms are still living in water. Both the numbers of species and individuals are staggering. For example, photosynthesis by algae in oceans consumes more CO than that by all plants on land. Algae were probably present on Earth before other organisms. Many phyla (divisions) of fungi, plantae, and animalia kingdoms also make water their homes. Both numbers and species of organisms living in water are probably more than those on land. The subject on living organisms in water is fascinating, but we can only mention some fundamentals about their relationships to water here. Certainly, every aspect of living organisms in water is related to food, because Homo sapiens is part of the food chain, if not at the top of it. All life requires energy or food. Some living organisms receive their energy from the sun whereas others get their energy from chemical reactions in the aquatic media. Chemical reactions are vital during their lives. For example, some bacteria derive energy by catalyzing the oxidation of iron sulfide, FeS , to produce iron ions Fe(H O) and elemental sulfur. Water is the oxidant, which in turn reduces oxygen (48). Chemical reactions provide energy for bacteria to sustain their lives and to reproduce. Factors affecting life in water are minerals, solubility of the mineral, electrochemical potentials of the material, acidity (pH), sunlight, dissolved oxygen level, presence of ions, chemical equilibria, etc. Properties of water influence life in general, and in the aquatic system in particular. As the population grows, aquaculture probably will be seen as a more efficient way of supplying protein for the ever-increasing population. Regarding drinking water, we are concerned with aquatic organisms invisible to the naked eye. Pathogenic organisms present in drinking water cause intestinal infections, dysentery, hepatitis, typhoid fever, cholera, and other illnesses. Pathogens are usually present in waters contaminated with human and animal wastes that enter the water system via discharge, run offs, flood, and accidents at sewage treatment facilities. Insects, rodents, and animals can also bring bacteria to the water system (49, 50). Testing for all pathogenic organisms is impossible, but some organisms have common living conditions to some pathogenic bacteria. Thus, water testing can use these harmless bacteria as indicators for drinking water safety. About 70% of the Earth surface is covered with water, but only about 2% is covered by fresh water. Ocean waters are salty, and only the small percentages that is fresh water resources (lakes, rivers, and underground). Fresh water is needed for drinking, food, farming, washing, and manufacturing. When salty water freezes, the ice so formed contains very little salt, if any. Thus, nearly all ice, including the massive ice at the polar cap, is fresh water. In fact, the ice cap in the Antarctic contains a lot of fresh-water ice, but that cannot be considered a water resource. Hydrologists, environmentalists, and scientists, engineers, sociologists, economists, and politicians are all concerned with problems associated with water resources. Solutions to these problems require experts and social consensus. Waters at temperatures between the normal boiling and critical points (0 to 373.98 C) are called , whereas the phase above the critical point is . In the 17 century, Denis Papin (a physicist) generated high-pressure steam using a closed boiler, and thereafter pressure canners were used to preserve food. Pressure cookers were popular during the 20 century. Analytical chemists have used subcritical waters to extract chemicals from solids for analysis since 1994 (51). Water vapor pressures up to its critical point are given in Table 8, but data on polarity, dielectric constant, surface tension, density and viscosity above 100 C are scarce. In general, these properties decrease as the temperature increases. In fact, some drop dramatically for supercritical water. On the other hand, some of them increase with pressure. Thus, properties of sub- and super-critical waters can be manipulated by adjusting temperature and pressure to attain desirable properties. As the polarity and dielectric constant decrease, water becomes an excellent solvent for non-polar substances such as those for flavor and fragrance. However, foodstuffs may degrade at high temperatures. Applications of sub- and super-critical water are relatively recent events, but applications of supercritical CO (critical temperature 32 C) for chemical analyses started in the 1980s, and investigations of supercritical water followed. However, research and development have been intensified in recent years (52). Scientists and engineers explore the usage of supercritical water for waste treatment, polymer degradation, pharmaceutical manufacturing, chromatographic analysis, nuclear reactor cooling, etc. Significant advances have also been made in material processing, ranging from fine particle manufacture to the creation of porous materials. Water has been called a green solvent compared to the polluting organic solvents. Sub- and super-critical waters have been explored as replacement of organic solvents in many applications including the food industry (53). However, supercritical water is very reactive, and it is corrosive for stainless steels that are inert to ordinary water. Yet, the application of sub- and super-critical waters is a wide-open field. Water, ice, and vapor are collections of H O molecules, whose characteristics determine the properties of all phases of water. Together and in concert water molecules shape the landscape, nurture lives, fascinate poets, and captivate scientists. Human efforts in understanding water have accumulated a wealth of science applicable in almost all disciplines, while some people take it for granted. Water molecules are everywhere, including outer space. They not only intertwine with our history and lives, they are parts of us. How blessed we are to be able to associate and correlate the phenomena we see or experience to the science of water. An article has a beginning and an end, but in the science of water, no one has the last word, as research and exploration, including its presence in outer space, on water continue (54). Writing this article induced my fascination on this subject, and for this reason, I am grateful to Professors Wai-kit Nip, Lewis Brubacher, and Peter F. Bernath for their helpful discussions and encouragement. | 59,580 | 3,340 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Key_Elements_of_Green_Chemistry_(Lucia)/02%3A_Life-Cycle_Analysis/2.01%3A_Introduction_to_Life_Cycles |
Green chemistry, in addition to being a science, it is also a philosophy and nearly a religion. Attendance at American Chemical Society Green Chemistry & Engineering Conferences will instill such an ideal into any attendant because of the universal appeal and possibilities in this novel approach to radicalizing the business of doing science and engineering. Life Cycle Assessment (LCA) is a comprehensive life cycle approach that quantifies ecological and human health impacts of a product or system over its complete life cycle. It uses credible scientific methods to model steady-state, global environmental and human health impacts. It also helps decision-makers understand the scale of many environmental and human health impacts of competing products, services, policies or actions. Triple Bottom Line accounting enables enterprises to support sustainability analysis in their operations, products and services. LCA contributes to the Triple Bottom Line reporting by quantifying the ecological and human health performance of competing products and services (Figure 2-1). Adding the social and economic performance reporting of a product or service to the LCA results of the product or service is one way to deliver Triple Bottom Line reporting.
Eco-labels that require LCA: Type 1) Third-party certified multi-criteria environmental labeling Type 3) Independently verified label with preset quantified indices Life Cycle Assessment (LCA) enables the creation of Type 1 and Type 3 eco-labels. These eco-labels can be powerful tools in obtaining larger shares of a specific market sector. Life cycle assessment (LCA) is a standardized programmatic tool to determine the environmental impacts of products or services. It can be described by a four-part framework as outlined by the 14044 ISO standard. This integrated framework was inspired by earlier forms of life-cycle thought from life cycle financial analysis. Examining a product from origination to use and disposal provides more holistic analysis to identify where environmental impacts originate and guide efforts in reducing impacts. The ISO standards ( ) provides guidance on structure framework, reuse requirements of data, study assumptions, and methods. By using more standard LCA methodologies, studies are more comparable and of greater scientific rigor. A standardized method allows LCA practitioners to manage complex datasets, provide comparisons among products, and allow benchmarking. In the absence of a standardized method, the results of LCA studies are even more variable depending on study assumptions and methods. The ISO standards help reduce the influence of the practitioner influence on study results. | 2,702 | 3,341 |
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" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact Application Problems Problems marked with a ♦ involve multiple concepts. The analytical concentration of lactic acid in blood is generally less than 1.2 × 10−3 M, corresponding to the sum of [lactate] and [lactic acid]. During strenuous exercise, however, oxygen in the muscle tissue is depleted, and overproduction of lactic acid occurs. This leads to a condition known as lactic acidosis, which is characterized by elevated blood lactic acid levels (approximately 5.0 × 10−3 M). The pKa of lactic acid is 3.86. What is the actual lactic acid concentration under normal physiological conditions? What is the actual lactic acid concentration during lactic acidosis? When the internal temperature of a human reaches 105°F, immediate steps must be taken to prevent the person from having convulsions. At this temperature, Kw is approximately 2.94 × 10−14. Calculate the pKw and the pH and pOH of a neutral solution at 105°F. Is the pH greater than or less than that calculated in Exercise 1 for a neutral solution at a normal body temperature of 98.6°F? ♦ The compound diphenhydramine (DPH) is the active ingredient in a number of over-the-counter antihistamine medications used to treat the runny nose and watery eyes associated with hay fever and other allergies. DPH is a derivative of trimethylamine (one methyl group is replaced by a more complex organic “arm” containing two phenyl rings): The compound is sold as the water-soluble hydrochloride salt (DPH+Cl−). A tablet of diphenhydramine hydrochloride contains 25.0 mg of the active ingredient. Calculate the pH of the solution if two tablets are dissolved in 100 mL of water. The pKb of diphenhydramine is 5.47, and the formula mass of diphenhydramine hydrochloride is 291.81 amu. ♦ Epinephrine, a secondary amine, is used to counter allergic reactions as well as to bring patients out of anesthesia and cardiac arrest. The pKb of epinephrine is 4.31. What is the percent ionization in a 0.280 M solution? What is the percent ionization after enough solid epinephrine hydrochloride is added to make the final epinephrineH+ concentration 0.982 M? What is the final pH of the solution? ♦ Fluoroacetic acid is a poison that has been used by ranchers in the western United States. The ranchers place the poison in the carcasses of dead animals to kill coyotes that feed on them; unfortunately, however, eagles and hawks are also killed in the process. How many milliliters of 0.0953 M Ca(OH)2 are needed to completely neutralize 50.0 mL of 0.262 M fluoroacetic acid solution (pKa = 2.59)? What is the initial pH of the solution? What is the pH of the solution at the equivalence point? Accidental ingestion of aspirin (acetylsalicylic acid) is probably the most common cause of childhood poisoning. Initially, salicylates stimulate the portion of the brain that controls breathing, resulting in hyperventilation (excessively intense breathing that lowers the PCO2 in the lungs). Subsequently, a potentially serious rebound effect occurs, as the salicylates are converted to a weak acid, salicylic acid, in the body. Starting with the normal values of PCO2 = 40 mmHg, pH = 7.40, and [HCO3−] = 24 nM, show what happens during the initial phase of respiratory stimulation and the subsequent phase of acid production. Why is the rebound effect dangerous? Emphysema is a disease that reduces the efficiency of breathing. As a result, less CO2 is exchanged with the atmosphere. What effect will this have on blood pH, PCO2, and [HCO3−]? 68.7 mL; 1.60; 7.85 | 3,648 | 3,343 |
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The cycloalkanes with one ring have the general formula \(C_nH_{2n}\), and are named by adding the prefix cyclo- to the name of the corresponding continuous-chain alkane having the same number of carbon atoms as the ring. Substituents are assigned numbers consistent with their position in such a way as to give the lowest numbers possible for the substituent positions: The substituent groups derived from cycloalkanes by removing one hydrogen are named by replacing the ending - of the hydrocarbon with - to give cycloalkyl. Thus cyclohex becomes cyclohex , cyclopentane becomes cyclopentyl, and so on. Remember, the numbering of the cycloalkyl substituent starts at the position of attachment, and larger rings take precedence over smaller rings: When a cycloalkane has an alkyl substituent, the compound could be called either an alkylcycloalkane or a cycloalkylalkane. The alkylcycloalkane name is the proper one in naming alkyl substituted cycloalkanes. and (1977) | 992 | 3,344 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Conjugation/Overlap_of_Adjacent_p_Orbitals-Electron_Delocalization |
Have you ever wondered why color exists? If you know it is because of molecules, do you know what makes the different colors appear? Most of it can be explained by the presence of the bonds, and how many there are. When photons hit the atoms and excite the electrons, the bonds that the electrons are in affect the frequency of light that is reflected when the electrons go back to th ground state and is perceived by our eyes. are formed from the overlap of two adjacent parallel orbitals (Fig.1). bonds are important in the . It is vital in synthesising many products like the polymers we use in modern day society. These systems have incredibly unique thermodynamic and photochemical properties. In this section we discover the special properties that three adjacent orbitals (a double bond and a p orbital) can have on the reactivity of a carbon center. We call a carbon next to a double bonded carbon an allylic carbon. The electrons that are shared between all the atomic centers are commonly said to be . A Bond is the interaction between the two orbitals of the bonded carbons. Lets take a look at the energy required to homolytically cleave (bond enthalpies) a fig.2 shows the decreasing energy required to a cleave a bond as we add methyl groups (for stabilization) As you can see, the more stable the , the easier it is to form. The tertiary carbocation takes less energy to cleave than the primary carbocation. This is due to which stabilizes the free electron by the slight interaction with nearby bond orbitals in the methyl groups. Now we look at a propene molecule Figure 3. propene radical formation Even though the carbocation is a primary one, the energy required to cleave the bond is than that of a seemingly stable tertiary structure. Also, the p of propene is about 40, compared to propane, which is about 50. This means the propene is much more willing to form the propenyl anion, about ten orders of magnitude more. Let's take a look at why: One way to explain the stability is in terms of , the allylic effect can be shown in all carbon center forms: Cation, Radical, and Anion in figure 4. The double bond's ability to shift in between different carbons gives it extra resonance forms, which lends extra stability. It can also be clearly shown through orbitals in figure 5. The stability of the carbocation of propene is due to a conjugated electron system. A "double bond" doesn't really exist. Instead, it is a group of 3 adjacent, overlapping, non-hybridized orbitals we call a . You can clearly see the interactions between all three of the orbitals from the three carbons resulting in a really stable cation. It all comes down to where the location of the electron-deficient carbon is. Molecular orbital descriptions can explain allylic stability in yet another way using 2-propenyl. Fig.6 If we just take the molecular orbital and not any of the s, we get three of them. is bonding with no nodes, is nonbonding (In other words, the same energy as a regular -orbital) with a node, and is antibonding with 2 nodes (none of the orbitals are interacting). The first two electrons will go into the molecular orbital, regardless of whether it is a cation, radical, or anion. If it is a radical or anion, the next electron goes into the molecular orbital. The last anion electron goes into the nonbonding orbital also. So no matter what kind of carbon center exists, no electron will ever go into the antibonding orbital. The Bonding orbitals are the lowest energy orbitals and are favorable, which is why they are filled first. Even though the nonbonding orbitals can be filled, the overall energy of the system is still lower and more stable due to the filled bonding molecular orbitals. This figure also shows that is the only molecular orbital where the electrion differs, and it is also where a single node passes through the middle. Because of this, the charges of the molecule are mainly on the two terminal carbons and not the middle carbon. This description can also illustrate the stability of allylic carbon centers in figure 7. The bonding orbital is lower in energy than the nonbonding orbital. Since every carbon center shown has two electrons in the lower energy, bonding orbitals, the energy of each system is lowered overall (and thus more stable), regardless of cation, radical, or anion. An electronic spectra can indicate the relative amounts of delocalization in a π electron system. The more conjugated π bonds there are, the longer the wavelength is reflected. Ethene has a maximum wavelength of 171 nm compared to 1,4-Pentadiene, which has a maximum wavelength of 178 nm and 1,3-Butadiene which has a maximum wavelength of 217 nm. The higher wavelength of the 1,3-Butadiene compared to only 178 nm for 1,4-Pentadiene shows the unique difference between a conjugated an non conjugated system. | 4,905 | 3,345 |
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The maximum solubility of a solute can be determined using the same methods we have used to describe colligative properties. The chemical potential of the solute in a liquid solution can be expressed \[ \mu_{B} (solution) = \mu_B^o (liquid) + RT \ln \chi_B \nonumber \] If this chemical potential is lower than that of a pure solid solute, the solute will dissolve into the liquid solvent (in order to achieve a lower chemical potential!) So the point of saturation is reached when the chemical potential of the solute in the solution is equal to that of the pure solid solute. \[ \mu_B^o (solid) = \mu_B^o (liquid) + RT \ln \chi_B \nonumber \] Since the mole fraction at saturation is of interest, we can solve for \(\ln(\chi_B)\). \[\ln \chi_B = \dfrac{\mu_B^o (solid) = \mu_B^o (liquid)}{RT} \nonumber \] The difference in the chemical potentials is the molar Gibbs function for the phase change of fusion. So this can be rewritten \[\ln \chi_B = \dfrac{-\Delta G_{fus}^o}{RT} \nonumber \] It would be convenient if the solubility could be expressed in terms of the enthalpy of fusion for the solute rather than the Gibbs function change. Fortunately, the Gibbs-Helmholtz equation gives us a means of making this change. Noting that \[ \left( \dfrac{\partial \left( \dfrac{\Delta G}{T} \right)}{\partial T} \right)_p = \dfrac{\Delta H}{T^2} \nonumber \] Differentiation of the above expression for \(\ln(\chi_B)\) with respect to \(T\) at constant \(p\) yields \[ \left( \dfrac{\partial \ln \chi_B}{\partial T} \right)_p = \dfrac{1}{R} \dfrac{\Delta H_{fus}}{T^2} \nonumber \] Separating the variables puts this into an integrable form that can be used to see how solubility will vary with temperature: \[ \int_0^{\ln \chi_B} d \ln \chi_B = \dfrac{1}{R} \int_{T_f}^{T} \dfrac{\Delta H_{fus} dT}{T^2} \nonumber \] So if the enthalpy of fusion is constant over the temperature range of \(T_f\) to the temperature of interest, \[ \ln \chi_B = \dfrac{\Delta H_{fus}}{R} \left( \dfrac{1}{T_f} - \dfrac{1}{T} \right) \nonumber \] And \(\chi_B\) will give the mole fraction of the solute in a saturated solution at the temperature \(T\). The value depends on both the enthalpy of fusion, and the normal melting point of the solute. | 2,240 | 3,347 |
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The carbon-carbon double bond in alkenes such as ethene react with concentrated sulfuric acid. It includes the conversion of the product into an alcohol. Alkenes react with concentrated sulfuric acid in the cold to produce alkyl hydrogensulfates. Ethene reacts to give ethyl hydrogensulfate. The structure of the product molecule is sometimes written as CH CH HSO , but the version in the equation is better because it shows how all the atoms are linked up. You may also find it written as CH CH OSO H. Confused by all this? Don't be! All you need to do is to learn the structure of sulfuric acid. A hydrogen from the sulfuric acid joins on to one of the carbon atoms, and the rest joins on to the other one. Make sure that you can see how the structure of the sulfuric acid relates to the various ways of writing the formula for the product. This is typical of the reaction with unsymmetrical alkenes. An unsymmetrical alkene has different groups at either end of the carbon-carbon double bond. If sulfuric acid adds to an unsymmetrical alkene like propene, there are two possible ways it could add. You could end up with one of two products depending on which carbon atom the hydrogen attaches itself to. However, in practice, there is only one major product. This is in line with Markovnikov's Rule which says: When a compound HX is added to an unsymmetrical alkene, the hydrogen becomes attached to the carbon with the most hydrogens attached to it already. In this case, the hydrogen becomes attached to the CH group, because the CH group has more hydrogens than the CH group. Notice that only the hydrogens directly attached to the carbon atoms at either end of the double bond count. The ones in the CH group are totally irrelevant. Ethene is passed into concentrated sulfuric acid to make ethyl hydrogensulphate (as above). The product is diluted with water and then distilled. The water reacts with the ethyl hydrogensulphate to produce ethanol which distils off. More complicated alkyl hydrogensulphates react with water in exactly the same way. For example: Notice that the position of the -OH group is determined by where the HSO group was attached. You get propan-2-ol rather than propan-1-ol because of the way the sulfuric acid originally added across the double bond in propene. These reactions were originally used as a way of manufacturing alcohols from alkenes in the petrochemical industry. These days, alcohols like ethanol or propan-2-ol tend to be manufactured by direct hydration of the alkene because it is cheaper and easier. Jim Clark ( ) | 2,581 | 3,348 |
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Make sure you thoroughly understand the following essential ideas which are presented below. The concept of chemical bonding lies at the very core of Chemistry; it is what enables about one hundred elements to form the more than fifty million known chemical substances that make up our physical world. Before we get into the theory of chemical bonding, we need to define what we are talking about: Exactly what is a chemical bond? And what observable properties can we use to distinguish one kind of bond from another? This is the first of ten lessons that will help familiarize you with the fundamental concepts of this very broad subject. You probably learned some time ago that chemical bonds are what hold atoms together to form the more complicated aggregates that we know as molecules and extended solids. Chemists talk about bonds all the time, and draw pictures of them as lines joining atom symbols. Teachers often identify them as the little sticks that connect the spheres that represent atoms in a plastic molecular model. So it's not surprising that we sometimes tend to think of chemical bonds as “things”. But no one has ever a chemical bond, and there is no reason to believe that they really even exist as physical objects. "SOMETIMES IT SEEMS to me that a bond between two atoms has become so real, so tangible, so friendly, that I can almost see it. Then I awake with a little shock, for a chemical bond is not a real thing. It does not exist. No one has ever seen one. No one ever can. It is a figment of our own imagination." (1910-1974) was an English theoretical chemist who played a central role in the development of quantum theories of chemical bonding. So although the "chemical bond" is no more than a convenient fiction, , which leads to the near-infinity of substances (31 million in mid-2007), lies at the very core of chemistry. The forces that hold bonded atoms together are basically just the same kinds of electrostatic attractions that bind the electrons of an atom to its positively-charged nucleus; Chemical bonding occurs when one or more electrons are simultaneously attracted to (or more) nuclei. This is the most important fact about chemical bonding that you should know, but it is not of itself a workable of bonding because it does not describe the conditions under which bonding occurs, nor does it make useful predictions about the properties of the bonded atoms. Even at the end of the 19th century, when compounds and their formulas had long been in use, some prominent chemists doubted that molecules (or atoms) were any more than a convenient model. Molecules suddenly became real in 1905, when Albert Einstein showed that Brownian motion, the irregular microscopic movements of tiny pollen grains floating in water, could be directly attributed to collisions with molecule-sized particles. Most people think of molecules as the particles that result when atoms become joined together in some way. This conveys the general picture, but a somewhat better definition that we will use in these lessons is: A more restrictive definition distinguishes between a "true" molecule that exists as an independent particle, and an that can only be represented by its simplest formula. Methane, CH , is an example of the former, while sodium chloride, which does not contain any discrete NaCl units, is the most widely-known extended solid. But because we want to look at chemical bonding in the most general way, we will avoid making this distinction here except in a few special cases. In order to emphasize this "aggregate of atoms" definition, we will often use terms such as "chemical species" and "structures" in place of "molecules" in this lesson. The definition written above is an one; that is, it depends on our ability to observe and measure the molecule's properties. Clearly, this means that the molecule must retain its identity for a period of time long enough to carry out the measurements. For most of the molecules that chemistry deals with, this presents no difficulty. But it does happen that some structures that we can write formulas for, such as He , have such brief lives that no significant properties have been observed. So to some extent, what we consider to be a molecule depends on the technology we use to observe them, and this will necessarily change with time. And what are those properties that characterize a particular kind of molecule and distinguish it from others? Just as real estate is judged by "location, location, location", the identity of a chemical species is defined by its . In its most fundamental sense, the structure of a molecule is specified by the identity of its constituent atoms and the sequence in which they are joined together, that is, by the . This, in turn, defines the — the spatial relationship between the bonded atoms. The importance of bonding connectivity is nicely illustrated by the structures of the two compounds and , both of which have the C H O. The reveal the very different connectivities of these two molecules whose physical and chemistry properties are quite different: The precise definition of bonding energy is described in another lesson and is not important here. For the moment you only need to know that in any stable structure, the potential energy of its atoms is lower than that of the individual isolated atoms. Thus the formation of methane from its gaseous atoms (a reaction that cannot be observed under ordinary conditions but for which the energetics are known from indirect evidence) \[ \ce{ 4 H(g) + C(g) → CH4}\] is accompanied by the release of heat, and is thus an process. The quantity of heat released is related to the stability of the molecule. The smaller the amount of energy released, the more easily can the molecule absorb thermal energy from the environment, driving the above reaction in reverse and leading to the molecule's decomposition. A highly stable molecule such as methane must be subjected to temperatures of more than 1000°C for significant decomposition to occur. But the noble-gas molecule KrF is so weakly bound that it decomposes even at 0°C, and the structure He has never been observed. If a particular arrangement of atoms is too unstable to reveal its properties at any achievable temperature, then it does not qualify to be called a molecule. There are many molecules that are energetically stable enough to meet the above criterion, but are so that their lifetimes are too brief to make their observation possible. The molecule CH , , is a good example: it can be formed by electrical discharge in gaseous CH , but it is so reactive that it combines with almost any molecule it strikes (even another CH )within a few collisions. It was not until the development of spectroscopic methods (in which a molecule is characterized by the wavelengths of light that it absorbs or emits) that methyl was recognized as a stable albeit shamelessly promiscuous molecule that is an important intermediate in many chemical processes ranging from flames to atmospheric chemistry. Chemical species are traditionally represented by such as the one for ascorbic acid (vitamin C) which we show here. The lines, of course, represent the "chemical bonds" of the molecule. More importantly, the structural formula of a molecule defines its , as was illustrated in the comparison of ethanol and dimethyl ether shown above. One limitation of such formulas is that they are drawn on a two-dimensional paper or screen, whereas most molecules have a three-dimensional shape. The wedge-shaped lines in the structural formula are one way of indicating which bonds extend above or below the viewing plane. You will probably be spared having to learn this convention until you get into second-year Chemistry. Three-dimensional models (either real plastic ones or images that incorporate perspective and shading) reveal much more about a molecule's structure. The ball-and-stick and space-filling renditions are widely employed, but each has its limitations, as seen in the following examples: But what would a molecule "really" look like if you could view it through a magical microscope of some kind? A possible answer would be this computer-generated view of nicotine. At first you might think it looks more like a piece of abstract sculpture than a molecule, but it does reveal the shape of the negative charge-cloud that envelops the collection of atom cores and nuclei hidden within. This can be very important for understanding how the molecule interacts with the similar charge-clouds that clothe solvent and bioreceptor molecules. In 2009, IBM scientists in Switzerland succeeded in imaging a real molecule, using a technique known as atomic force microscopy in which an atoms-thin metallic probe is drawn ever-so-slightly above the surface of an immobilized pentacene molecule cooled to nearly absolute zero. In order to improve the image quality, a molecule of carbon monoxide was placed on the end of the probe. The image produced by the AFM probe is shown at the very bottom. What is actually being imaged is the surface of the electron clouds of the molecule, which consists of five fused hexagonal rings of carbon atoms with hydrogens on its periphery. The tiny bumps that correspond to these hydrogen atom attest to the remarkable resolution of this experiment. The purpose of rendering a molecular structure in a particular way is not to achieve "realism" (whatever that might be), but rather to convey useful information of some kind. Modern computer rendering software takes its basic data from various kinds of standard structural databases which are compiled either from experimental X-ray scattering data, or are calculated from theory. As was mentioned above, it is often desirable to show the "molecular surface"— the veil of negative charge that originates in the valence electrons of the atoms but which tends to be spread over the entire molecule to a distance that can significantly affect with neighboring molecules. It is often helpful to superimpose images representing the atoms within the molecule, scaled to their average covalent radii, and to draw the "bonding lines" expressing their connectivity. Knowing the properties of molecular surfaces is vitally important to understanding any process that depends on one molecule remaining in physical contact with another. Catalysis is one example, but one of the main interests at the present time is , in which a relatively small molecule binds to or "docks" with a receptor site on a much larger one, often a protein. Sophisticated molecular modeling software such as was used to produce these images is now a major tool in many areas of research biology. Visualizing such as carbohydrates and proteins that may contain tens of thousands of atoms presents obvious problems. The usual technique is to simplify the major parts of the molecule, representing major kinds of extended structural units by shapes such as ribbons or tubes which are twisted or bent to approximate their conformations. These are then gathered to reveal the geometrical relations of the various units within the overall structure. Individual atoms, if shown at all, are restricted to those of special interest. Study of the surface properties of large molecules is crucial for understanding how proteins, carbohydrates, and DNA interact with smaller molecules, especially those involved in transport of ions and small molecule across cell membranes, immune-system behavior, and signal transduction processes such as the "turning on" of genes. When we talk about the properties of a particular chemical bond, we are really discussing the relationship between two adjacent atoms that are part of the molecule. molecules are of course the easiest to study, and the information we derive from them helps us interpret various kinds of experiments we carry out on more complicated molecules. It is important to bear in mind that the exact properties of a specific kind of bond will be determined in part by the nature of the other bonds in the molecule; thus the energy and length of the C–H bond will be somewhat dependent on what other atoms are connected to the carbon atom. Similarly, the C-H bond length can vary by as much a 4 percent between different molecules. For this reason, the values listed in tables of bond energy and bond length are usually averages taken over a variety of compounds that contain a specific atom pair.. In some cases, such as C—O and C—C, the variations can be much greater, approaching 20 percent. In these cases, the values fall into groups which we interpret as representative of single- and bonds: double, and triple. The energy of a system of two atoms depends on the distance between them. At large distances the energy is zero, meaning “no interaction”. At distances of several atomic diameters attractive forces dominate, whereas at very close approaches the force is repulsive, causing the energy to rise. The attractive and repulsive effects are balanced at the minimum point in the curve. Plots that illustrate this relationship are known as , and they are quite useful in defining certain properties of a chemical bond. The internuclear distance at which the potential energy minimum occurs defines the . This is more correctly known as the bond length, because thermal motion causes the two atoms to vibrate about this distance. In general, the stronger the bond, the smaller will be the bond length. Attractive forces operate between all atoms, but unless the potential energy minimum is at least of the order of RT, the two atoms will not be able to withstand the disruptive influence of thermal energy long enough to result in an identifiable molecule. Thus we can say that a chemical bond exists between the two atoms in \(\ce{H2}\). The weak attraction between argon atoms does not allow \(\ce{Ar2}\) to exist as a molecule, but it does give rise to the van Der Waals force that holds argon atoms together in its liquid and solid forms. Potential energy and kinetic energy Quantum theory tells us that an electron in an atom possesses kinetic energy K as well as potential energy P, so the total energy E is always the sum of the two: E = P + K. The relation between them is surprisingly simple: K = –0.5 P. This means that when a chemical bond forms (an exothermic process with ΔE < 0), the decrease in potential energy is accompanied by an increase in the kinetic energy (embodied in the momentum of the bonding electrons), but the magnitude of the latter change is only half as much, so the change in potential energy always dominates. The bond energy –ΔE has half the magnitude of the fall in potential energy. The is the amount of work that must be done to pull two atoms completely apart; in other words, it is the same as the depth of the “well” in the potential energy curve shown above. This is almost, but not quite the same as the actually required to break the chemical bond; the difference is the very small This relationship will be clarified below in the section on bond vibrational frequencies. Bond energies are usually determined indirectly from thermodynamic data, but there are two main experimental ways of measuring them directly: involves separating the two atoms by an electrical discharge or some other means, and then measuring the heat given off when they recombine. Thus the energy of the C—C single bond can be estimated from the heat of the recombination reaction between methyl radicals, yielding ethane: \[CH_3 + CH_3 → H_3C–CH_3\] Although this method is simple in principle, it is not easy to carry out experimentally. The highly reactive components must be prepared in high purity and in a stream of moving gas. is based on the principle that absorption of light whose wavelength corresponds to the bond energy will often lead to the breaking of the bond and dissociation of the molecule. For some bonds, this light falls into the green and blue regions of the spectrum, but for most bonds ultraviolet light is required. The experiment is carried out by observing the absorption of light by the substance being studied as the wavelength is decreased. When the wavelength is sufficiently small to break the bond, a characteristic change in the absorption pattern is observed. Spectroscopy is quite easily carried out and can yield highly precise results, but this method is only applicable to a relatively small number of simple molecules. The major problem is that the light must first be by the molecule, and relatively few molecules happen to absorb light of a wavelength that corresponds to a bond energy. \[CH_3 + H → CH_4\] \[CH_2 + H → CH_3\] One can often get a very good idea of how much heat will be absorbed or given off in a reaction by simply finding the difference in the total bond energies contained in the reactants and products. The strength of an individual bond such as O–H depends to some extent on its environment in a molecule (that is, in this example, on what other atom is connected to the oxygen atom), but tables of "average" energies of the various common bond types are widely available and can provide useful estimates of the quantity of heat absorbed or released in many chemical reaction. Average bond energies are the averages of bond dissociation energies (see Table T3 for more complete list). For example the average bond energy of O-H in H O is 464 kJ/mol. This is due to the fact that the H-OH bond requires 498.7 kJ/mol to dissociate, while the O-H bond needs 428 kJ/mol. \[\dfrac{498.7\; kJ/mol + 428\; kJ/mol}{2}=464\; kJ/mol\] Consider the reaction of chlorine with methane to produce dichloromethane and hydrogen chloride: \[\ce{CH4(g) + 2 Cl2(g) → CH2Cl2(g) + 2 HCl(g)}\] In this reaction, two C–Cl bonds and two H–Cl bonds are broken, and two new C–Cl and H–Cl bonds are formed. The net change associated with the reaction is 2(C–H) + 2(Cl–Cl) – 2(C–Cl) – 2(H–Cl) = (830 + 486 –660 – 864) kJ which comes to –208 kJ per mole of methane; this agrees quite well with the observed heat of reaction, which is –202 kJ/mol. The length of a chemical bond the distance between the centers of the two bonded atoms (the .) Bond lengths have traditionally been expressed in Ångstrom units, but picometers are now preferred (1Å = 10 cm = 100 pm.) Bond lengths are typically in the range 1-2 Å or 100-200 pm. Even though the bond is vibrating, equilibrium bond lengths can be determined experimentally to within ±1 pm. The most common method of measuring bond lengths in solids is by analysis of the diffraction or scattering of X-rays when they pass through the regularly-spaced atoms in the crystal. For gaseous molecules, neutron- or electron-diffraction can also be used. The complete structure of a molecule requires a specification of the coordinates of each of its atoms in three-dimensional space. This data can then be used by computer programs to construct of the molecule as discussed above. One such visualization of the water molecule, with bond distances and the HOH bond angle superimposed on a space-filling model, is shown here. (It is taken from an excellent reference source on water). The colors show the results of calculations that depict the way in which electron charge is distributed around the three nuclei. When an atom is displaced from its equilibrium position in a molecule, it is subject to a restoring force which increases with the displacement. A spring follows the same law (Hooke’s law); a chemical bond is therefore formally similar to a spring that has weights (atoms) attached to its two ends. A mechanical system of this kind possesses a natural vibrational frequency which depends on the masses of the weights and the stiffness of the spring. These vibrations are initiated by the thermal energy of the surroundings; chemically-bonded atoms are never at rest at temperatures above absolute zero. On the atomic scale in which all motions are , a vibrating system can possess a series of vibrational frequencies, or . These are depicted by the horizontal lines in the potential energy curve shown here. Notice that the very bottom of the curve does not correspond to an allowed state because at this point the positions of the atoms are precisely specified, which would violate the uncertainty principle. The lowest-allowed, or vibrational state is the one denoted by 0, and it is normally the only state that is significantly populated in most molecules at room temperature. In order to jump to a higher state, the molecule must absorb a photon whose energy is equal to the distance between the two states. For ordinary chemical bonds, the energy differences between these natural vibrational frequencies correspond to those of . Each wavelength of infrared light that excites the vibrational motion of a particular bond will be absorbed by the molecule. In general, the stronger the bond and the lighter the atoms it connects, the higher will be its natural stretching frequency and the shorter the wavelength of light absorbed by it. Studies on a wide variety of molecules have made it possible to determine the wavelengths absorbed by each kind of bond. By plotting the degree of absorption as a function of wavelength, one obtains the of the molecule which allows one to "see" what kinds of bonds are present. The low points in the plot below indicate the frequencies of infrared light that are absorbed by ethanol (ethyl alcohol), CH CH OH. Notice how stretching frequencies involving hydrogen are higher, reflecting the smaller mass of that atom. Only the most prominent absorption bands are noted here. Actual infrared spectra are complicated by the presence of more complex motions (stretches involving more than two atoms, wagging, etc.), and absorption to higher quantum states (overtones), so infrared spectra can become quite complex. This is not necessarily a disadvantage, however, because such spectra can serve as a "fingerprint" that is unique to a particular molecule and can be helpful in identifying it. Largely for this reason, infrared spectrometers are standard equipment in most chemistry laboratories. Now that you know something about bond stretching vibrations, you can impress your friends by telling them The aspect of bond stretching and bending frequencies that impacts our lives most directly is the way that some of the gases of the atmosphere absorb infrared light and thus affect the heat balance of the Earth. Owing to their symmetrical shapes, the principal atmospheric components N and O do not absorb infrared light, but the minor components water vapor and carbon dioxide are strong absorbers, especially in the long-wavelength region of the infrared. Absorption of infrared light by a gas causes its temperature to rise, so any source of infrared light will tend to warm the atmosphere; this phenomenon is known as the . The incoming radiation from the Sun (which contains relatively little long-wave infrared light) passes freely through the atmosphere and is absorbed by the Earth's surface, warming it up and causing it to re-emit some of this energy as long-wavelength infrared. Most of the latter is absorbed by the H O and CO , the major greenhouse gas is in the unpolluted atmosphere, effectively trapping the radiation as heat. Thus the atmosphere is heated by the Earth, rather than by direct sunlight. Without the “ ” in the atmosphere, the Earth's heat would be radiated away into space, and our planet would be too cold for life. In order to maintain a constant average temperature, the quantity of radiation (sunlight) absorbed by the surface must be exactly balanced by the quantity of long-wavelength infrared emitted by the surface and atmosphere and radiated back into space. Atmospheric gases that absorb this infrared light (depicted in red on the right part of this diagram) partially block this emission and become warmer, raising the Earth's temperature. This diagram is from the U. of Oregon Web page referenced below. Since the beginning of the Industrial Revolution in the 19th century, huge quantities of additional greenhouse gases have been accumulating in the atmosphere. Carbon dioxide from fossil fuel combustion has been the principal source, but intensive agriculture also contributes significant quantities of methane (CH ) and nitrous oxide (N O) which are also efficient far-infrared absorbers. The measurable increase in these gases is believed by many to be responsible for the increase in the average temperature of the Earth that has been noted over the past 50 years— a trend that could initiate widespread flooding and other disasters if it continues. | 24,666 | 3,350 |
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This page looks at the facts about the nitration of benzene and methylbenzene. The mechanisms for these reactions are covered elsewhere on the site, and you will find links to these. Nitration happens when one (or more) of the hydrogen atoms on the benzene ring is replaced by a nitro group, NO . Benzene is treated with a mixture of concentrated nitric acid and concentrated sulfuric acid at a temperature not exceeding 50°C. The mixture is held at this temperature for about half an hour. Yellow oily nitrobenzene is formed. You could write this in a more condensed form as: \[ C_6H_6 + HNO_3 \rightarrow C_6H_5NO_2 + H_2O \] The concentrated sulfuric acid is acting as a catalyst and so is not written into the equations. At higher temperatures there is a greater chance of getting more than one nitro group substituted onto the ring. You will get a certain amount of 1,3-dinitrobenzene formed even at 50°C. Some of the nitrobenzene formed reacts with the nitrating mixture of concentrated acids. Notice that the new nitro group goes into the 3 position on the ring. Nitro groups "direct" new groups into the 3 and 5 positions. It is also possible to get a third nitro group attached to the ring (in the 5 position). However, nitro groups make the ring much less reactive than the original benzene ring. Two nitro groups on the ring make its reactions so slow that virtually no trinitrobenzene is produced under these conditions. Methylbenzene reacts rather faster than benzene - in nitration, the reaction is about 25 times faster. That means that you would use a lower temperature to prevent more than one nitro group being substituted - in this case, 30°C rather than 50°C. Apart from that, the reaction is just the same - using the same nitrating mixture of concentrated sulphuric and nitric acids. You get a mixture of mainly two isomers formed: 2-nitromethylbenzene and 4-nitromethylbenzene. Only about 5% of the product is 3-nitromethylbenzene. Methyl groups are said to be 2,4-directing. For 2-nitromethylbenzene: and 4-nitromethylbenzene: Just as with benzene, you will get a certain amount of dinitro compound formed under the conditions of the reaction, but virtually no trinitro product because the reactivity of the ring decreases for every nitro group added (bythe way, trinitromethylbenzene used to be called trinitrotoluene or TNT). The reactivity of a benzene ring is governed by the electron density around the ring. Methyl groups tend to "push" electrons towards the ring - increasing the density, and so making the ring more attractive to attacking reagents. This is actually a simplification. In order to understand the rate effect properly you have to think about the stability of the intermediate ions formed during the reactions, because this affects the activation energy of the reactions. Jim Clark ( ) | 2,843 | 3,351 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO17._Oxidation |
You may recall that conversion of an aldehyde or ketone to an alcohol is referred to as a reduction. The hydride from an NADH molecule or a BH anion acts as a nucleophile, adding H to the carbonyl carbon. A proton source can then protonate the oxygen of the resulting alkoxide ion, forming an alcohol. In this reduction, two electrons and two protons are donated to the carbonyl compound to produce an alcohol. The opposite process, the loss of two protons and two electrons from an alcohol to form a ketone or aldehyde, is an oxidation. In biological pathways, oxidation is often the microscopic reverse of reduction. That means that the products of a reduction, NAD and an alcohol, could react together under the right circumstances to form NADH and a carbonyl. The reduction of NAD by a hydride donor is possible because, although the NAD loses the aromaticity of its nicotinamide ring upon becoming NADH, it also loses its positive charge. Charge stabilization is frequently an energetic problem for molecules. Notice that this same argument has been used to look at biological oxidation and reduction in both directions. That's because each side of the equation has some energetic advantage. The NAD is aromatic. The NADH is neutral. This is a well-balanced system energetically, and the balance of the reaction can be tipped in either direction. The direction of the reaction is influenced by the surroundings. In biology, NADH and NAD are just cofactors in a reaction. A reduction would take place in an enzyme that specifically carries out reductions, and an oxidation would occur in an enzyme specific for the oxidation. The enzyme is just a large protein that holds the substrate (such as the alcohol) and cofactor (such as NAD ) together in close proximity. Acidic and basic sites are provided by nearby amino acid residues, and other amino acid side chains may push the reactants into optimum position for one reaction or another. In the laboratory, enzymes and cofactors can sometimes be added to reaction flasks in order to oxidize or reduce substrates. Sometimes these reactions are not convenient, however. In addition, there are a number of other ways to carry out oxidations and reductions. For example, addition of a hydride could be accomplished via addition of sodium borohydride. That would result in reduction of a carbonyl to an alcohol. By analogy to the NADH / NAD approach in nature, the easiest way to oxidize an alcohol to a carbonyl would be to remove a hydride and a proton. Nature uses enzymes to bring reactants together for transformations like this. In the laboratory, metals are often used to bring two reactants together. This is sometimes true in enzymes, too: a metal at the enzyme active site may tether two molecules together, or even activate one so that it is ready to react. Metals can "hold onto" reactants because molecules with lone pairs will often coordinate to metals; a lone pair is shared with the electrophilic metal. In a process known as an Oppenauer oxidation, a Lewis acidic metal such as aluminum is used for this tethering role. Aluminum tris(isopropoxide), Al(OCH(CH ) ) carries out the oxidation of an alcohol when dissolved in acetone or propanone, (CH ) C=O. In this reaction, the acetone is a sacrificial oxidant. When both the alcohol and a molecule of acetone are coordinated to the aluminum, a hydride is transferred from the alcohol carbon to the carbonyl carbon of the coordinated acetone. The alcohol is converted to an aldehyde or ketone and the acetone is converted to isopropanol. This general type of reaction, hydride transfer reduction, has been adapted by , of Nagoya University in Japan, to produce a single enantiomer of a chiral alcohol product. Noyori's work on this reaction, and others, led to him being awarded the Nobel Prize in Chemistry in 2001. Show the two enantiomers that could be produced from reduction of acetophenone, CH (CO)C H . Provide a mechanism with arrows for the Oppenauer oxidation of benzyl alcohol, C H CH OH. Explain why acetone is used as the solvent in an Oppenauer oxidation. Meerwein-Ponndorf-Verley reduction of a ketone is carried out with aluminum tris(isopropoxide) in isopropanol as solvent. Provide a mechanism for the reduction of acetophenone, CH (CO)C H , via this reaction. Explain why isopropanol is used as the solvent in a Meerwein-Pondorf-Verley reduction. A second general method for alcohol oxidation employs a "redox-active" transition metal to accept a pair of electrons from from an alcohol during the oxidation. Because oxidation of an alcohol formally involves the loss of two electrons and two protons, a proton acceptor is also involved in this oxidation. There are many redox-active metals, but one of the most commonly used is Cr(VI). When Cr(VI) accepts a pair of electrons, it becomes Cr(IV). In order to look at how chromium oxidation works, we'll use chromium oxide, CrO , as an oxidant and water as a solvent. Note that water could also act as a proton acceptor or proton shuttle, moving protons from one place to another as needed. To carry out an oxidation, a number of events need to happen. In reality, CrO isn't used that often as an oxidant. It tends to catch fire when mixed with organic compounds. Instead, a variety of other chromium compounds are used. In determining an oxidation state, we imagine giving both electrons in a bond to the more electronegative atom and looking at the resulting charges on the ions that result. Assuming all of the oxygens in chromium oxide can be thought of as dianions, confirm that the chromium can be thought of as a Cr cation (in other words, in oxidation state Cr(VI)). By the reasoning used in the previous question, determine the oxidation state of the transition metal in the following compounds. Note that in some cases, there is an anion and cation in the compound. a) KMnO b) NaIO c) Ag O d) OsO d) (CH CH CH ) N RuO , | 5,940 | 3,352 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Roadmap_Problems_in_Natural_Product_synthesis |
Topics required for successful completion are listed under each link. Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic substitution, Alkene addition, Pericyclics Carbonyl addition, Elimination, Alkene addition, Pericyclics Carbonyl Addition, Aldol, Carboxyl Substitution, Oxidation, Nucleophilic Substitution, Elimination, Radicals Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic Substitution & Elimination Carbonyl addition, Carboxylic substitution, Nucleophilic Substitution, Bioassay study Nucleophilic Alkene Carbonyl addition, Carboxylic substitution, Nucleophilic substitution & elimination, Oxidation, Alkene oxidation, Pericyclics Carbonyl addition, Carboxylic substitution, Nucleophilic Substitution Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic substitution, Alkene addition, Radicals Carbonyl addition, Conjugate addition, Enolates, Carboxylic substitution, Nucleophilic substitution, Alkene addition, Rearrangements, Radicals Nucleophilic Contains a bioassay exercise. Aldol Claisen A-lpha-alkylation Nucleophilic Carbonyl addition, Phosphorus ylides, Carboxylic substitution, Oxidation, Nucleophilic substitution, Cuprate addition, Alkene addition Nucleophilic Alkene Carbonyl addition, Conjugate addition, Carboxylic substitution, Oxidation, Nucleophilic Substitution & Elimination Carbonyl addition, Aldols, Carboxylic substitution, Oxidation, Nucleophilic Substitution & Elimination, Pericyclics Carbonyl addition, Enolates, Conjugate addition, Carboxylic substitution, Oxidation, Nucleophilic substitution & elimination, Radicals Aldols Nucleophilic Pericyclics Nucleophilic Nucleophilic Nucleophilic Alkene Contains a bioassay exercise. Nucleophilic Alkene Pericyclics Nucleophilic Alkene Carbonyl addition, Enolates, Conjugate addition, Carboxylic substitution, Nucleophilic substitution & elimination, Alkene oxidation Exercise: catalytic cycles in transition metal organometallics. Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic substitution & elimination, Alkene oxidation, Pericyclics, Radicals Carbonyl addition, Carboxylic substitution, Aldol Carbonyl addition, Carboxylic substitution, Oxidation, Nucleophilic substitution, Alkene addition, Alkene oxidation, Pericyclics, Rearrangements Carbonyl addition, Carboxylic Substitution, Sulfur ylides, Oxidation, Nucleophilic substitution, Alkene oxidation, Radicals, Organo-transition metal reactions Nucleophilic Alkene , | 2,488 | 3,353 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/28%3A_Photochemistry/28.04%3A_Chemiluminescence |
The most common means of generating electronically excited states of molecules is by the absorption of electromagnetic radiation. But excited states are accessible by other routes. Indeed, as shown in , the excited singlet state of molecular oxygen can be produced by chemical reactions (Equations 28-9 and 28-10). Many other reactions are known that generate products in electronically excited states, and this is especially evident when the electronically excited products go to the ground state by the emission of visible light. This behavior is known as and is transduction of chemical energy \(\left( \Delta H \right)\) into radiant energy \(\left( h \nu \right)\). Chemiluminescence is possible only when the \(\Delta H\) of the reaction is sufficiently large to allow for production of at least one of the products in an electronically excited state \(\left( ^* \right)\). Chemiluminescence amounts to \(\Delta H \rightarrow ^* \rightarrow h \nu\), which is opposite to most photochemistry which involves \(h \nu \rightarrow ^* \rightarrow \Delta H\). A beautiful example of a chemiluminescent reaction is the thermal dissociation of the cyclic peroxide, \(8\), into two molecules of 2-propanone: We should not be surprised at the high exothermicity of Reaction 28-11. The peroxide is of high energy (thermochemically unstable) because it combines the strain-energy characteristics of small rings with the weakness of \(\ce{O-O}\) bonds, whereas the product is a stable substance with a strong carbonyl bond. Chemiluminescence in many reactions is hard to detect because the efficiency of light emission is low. Thus, even though the excited state may be formed in high yield, it may be quenched by other species more efficiently than it loses energy by emission. This fact can be used to advantage by adding a substance that quenches the excited state efficiently and, after energy transfer, gives bright fluorescence or phosphorescence: \[\Delta H \rightarrow ^* \underset{\text{energy transfer}}{\overset{\text{dye}}{\longrightarrow}} \text{dye}^* \overset{\text{luminescence}}{\longrightarrow} \text{dye} + h \nu\] Chemiluminescence can be greatly amplified by this process and it forms the basis of spectacular demonstrations of "cold light". An example is the hydrolysis of ethanedioic (oxalic) esters with hydrogen peroxide in the presence of a fluorescent substance (Equation 28-12). The reaction is believed to pass through the highly unstable dioxacyclobutanedione, which dissociates into two moles of carbon dioxide with such exothermicity that electronic excitation occurs, as evident from the intense light produced in the presence of fluorescent dyes: \(\tag{28-12}\) This reaction has been developed into a commercial product, marketed under the trade name "Coolite", which can be used as an emergency light source by simply shaking a tube to bring the reactants in contact with one another. Of major interest is the identity of the excited state (singlet or triplet) produced by chemiluminescent reactions. Little is known about excited states produced chemically except in a few cases, as in Reaction 28-11. Here the chemiluminescence dissociation gives a ratio of triplet 2-propanone to excited singlet 2-propanone of 100:1. This is a surprising result because it means that spectroscopic selection rules of electron-spin conservation are not followed in this chemiexcitation. The reaction has generated a triplet state from a singlet state. How can this be? Some idea of what is involved can be obtained from Figure 28-7, in which we see that breaking of the two sigma \(\ce{C-C}\) and \(\ce{O-O}\) bonds gives directly one molecule of ground-state ketone (all spins paired) and one molecule of triplet ketone. In this process, the electrons associated with the orbitals on one of the oxygen atoms appear to interact in such a way as to interchange electrons between orbitals with a spin inversion. This is called . The emission of visible light by living organisms is a mysterious and fascinating phenomenon. The magical glow of the firefly and of certain plants and marine animals is a familiar sight and one that has stimulated man's curiosity and imagination for centuries. Despite intense interest in bioluminescence, it is only recently that substantial progress has been made in our understanding of how it occurs. One of the earliest studies of bioluminescence was made by the French scientist R. Dubois toward the end of the nineteenth century. He demonstrated that bioluminescent organisms emitted light as a consequence of chemical change. He succeeded in isolating the active chemical from fireflies (luciferin) and the activating enzyme (luciferase, named by Dubois from the Latin , meaning light bearer). Luciferin and the enzyme in the presence of oxygen were found to reproduce the natural bioluminescence: \[\text{luciferin} + \ce{O_2} \overset{\text{luciferase}}{\longrightarrow} \text{oxyluciferin} + h \nu\] Further progress required elucidation of the structures of luciferin and its oxidation product. It turned out that there are several luciferins, depending on the organism. Firefly luciferin has the benzothiazol structure, \(9\); the luciferins from the marine crustacean and the sea pansy have structures \(10\) and \(11\), respectively. Their oxidation products \(12\), \(13\), and \(14\) also are shown: Although the luciferins \(9\)-\(11\) may not seem closely related, each appears to react with oxygen (at the direction of the appropriate enzyme) to give cyclic peroxylacetone intermediates \(12\)-\(14\). Luminescence is the consequence of the energetically favorable dissociation of the dioxacyclobutanone ring system to carbon dioxide and a carbonyl component. This mechanism is suggested by experiments with the peroxy acid, \(15\), which with \(\ce{N}\),\(\ce{N}\)-dicyclohexylcarbodiimide gives a very reactive compound presumed to be the peroxylactone, \(16\). This substance liberates \(\ce{CO_2}\) rapidly at room temperature with luminescence: \(^4\)Production of two molecules of excited 2-propanone per molecule of \(8\) is not possible under the same conditions because this would correspond to a reaction with \(\Delta H^0\) of at least \(156 \: \text{kcal}\) above formation of two moles of ground-state ketone. and (1977) | 6,326 | 3,354 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Lattice_Defects/Schottky_Defects |
Lattice structures are not perfect; in fact most of the time they experience defects. Lattice structures (or crystals) are prone to defects especially when their temperature is greater than 0 K [1]. One of these defects is known as the Schottky defect, which occurs when oppositely charged ions vacant their sites [1]. Like the human body, lattice structures (most commonly known as crystals) are far from perfection. Our body works hard to keep things proportional but occasionally our right foot is bigger than our left; similarly, crystals may try to arrange it's ions under a strict layout, but occasionally an ion slips to another spot or simply goes missing. Realistically speaking, it should be expected that crystals will depart itself from order (not surprising considering defects occurs at temperature greater than 0 K). There are many ways a crystal can depart itself from order (thus experiences defects); these defects can be grouped in different categories such as Point Defects, Line Defects, Planar Defects, or Volume or Bulk Defects [2]. We will focus on Point Defects, specifically the defect that occurs in ionic crystal structures (i.e. NaCl) called the Schottky Defect. Lattice structures (or crystals) undergoing point defects experience one of two types: By the simplest definition, the Schottky defect is defined by type one, while type two defects are known as the Frenkel defect. The Schottky defect is often visually demonstrated using the following layout of anions and cations: In addition, this layout is applicable only for ionic crystal compounds of the formula MX--layout for ionic crystals with formula MX and M X will be discussed later--where M is metal and X is nonmetal. Notice the figure has exactly one cation and one anion vacating their sites; that is what defines a (one) Schottky Defect for a crystal of MX formula--for every cation that vacant its site, the same number of anion will follow suit; essentially the vacant sites come in pairs. This also means the crystal will neither be too positive or too negative because the crystal will always be in equilibrium in respect to the number of anions and cations. It is possible to approximate the number of Schottky defects (n ) in a MX ionic crystal compound by using the equation: \[N= \exp^{-\dfrac{\Delta H}{2RT}} \label{3}\] where N can be calculated by: \[N = \dfrac{\text{density of the ionic crystal compound} \times N_A}{\text{molar mass of the ionic crystal compound}} \label{4}\] From Equation \(\ref{3}\), it is also possible to calculate the fraction of vacant sites by using the equation: \[\dfrac{n_s}{N} = \exp^{-\dfrac{\Delta H}{2RT}} \label{5}\] As mentioned earlier, a Schottky defect will always result a crystal structure in equilibrium--where no crystal is going to be too positive or too negative; thus in the case of: | 2,860 | 3,355 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.17%3A_Polyatomic_Ions |
Our discussion of was confined to monatomic ions. However, more complex ions, containing several atoms to one another, but having a positive or negative charge, occur quite frequently in chemistry. The charge arises because the total number of valence electrons from the atoms cannot produce a stable structure. With one or more electrons added or removed, a stable structure results. Well-known examples of such are the sulfate ion (SO ), the hydroxide ion (OH ), the hydronium ion (H O ), and the ammonium ion (NH ). The atoms in these ions are joined together by covalent electron-pair bonds, and we can draw Lewis structures for the ions just as we can for molecules. The only difference is that the number of electrons in the ion does not exactly balance the sum of the nuclear charges. Either there are too many electrons, in which case we have an anion, or too few, in which case we have a cation. Consider, for example, the hydroxide ion (OH ) for which the Lewis structure is A neutral molecule containing one O and one H atom would contain only seven electrons, six from O and one from H. The hydroxide ion, though, contains an of electrons, than the neutral molecule. The hydroxide ion must thus carry a single negative charge. In order to draw the Lewis structure for a given ion, we must first determine how many valence electrons are involved. Suppose the structure of H O is required. The total number of electrons is obtained by adding the valence electrons for each atom, 6 + 1 + 1 + 1 = 9 electrons. We must now subtract 1 electron since the species under consideration is not H O but H O . The total number of electrons is thus 9 – 1 = 8. Since this is an octet of electrons, we can place them all around the O atom. The final structure then follows very easily: In more complicated cases it is often useful to calculate the number of before drawing a Lewis structure. This is particularly true when the ion in question is an (i.e., a central atom is surrounded by several O atoms). A well-known oxyanion is the carbonate ion, which has the formula CO . (Note that the central atom C is written first, as was done earlier for molecules.) The total number of valence electrons available in CO is \( 4 \text{(for C)} + 3 \times 6 \text{(for O)} + 2 \text{(for the –2 charge)} = 24 \) We must distribute these electrons over 4 atoms, giving each an octet, a requirement of 4 × 8 = 32 electrons. This means that 32 – 24 = 8 electrons need to he counted twice for octet purposes; i.e., 8 electrons are shared. The a ion thus contains electron-pair bonds. Presumably the C atom is double-bonded to one of the O’s and singly bonded to the other two: In this diagram the 4C electrons have been represented by dots, the 18 O electrons by ×’s, and the 2 extra electrons by colored dots, for purposes of easy reference. Real electrons do not carry labels like this; they are all the same. There is a serious objection to the Lewis structure just drawn. How do the electrons know which oxygen atom to single out and form a double bond with, since there is otherwise nothing to differentiate the oxygens? The answer is that they do not. To explain the bonding in the CO ion and some other molecules requires an extension of the Lewis theory. We pursue this matter further when we discuss . Now we end with an example. Draw a Lewis structure for the sulfite ion, SO . The safest method here is to count electrons. The total number of valence electrons available is Note that each of the S—O bonds is coordinate covalent. | 3,566 | 3,356 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Pharmaceuticals/Drugs_Acting_Upon_the_Central_Nervous_System |
The central nervous system directs the functions of all tissues of the body. The peripheral nervous system receives thousands of sensory inputs and transmits them to the brain via the spinal cord. The brain processes this incoming information and discards 99% as unimportant. After sensory information has been evaluated, selected areas of the central nervous system initiate nerve impulses to organs or tissue to make an appropriate response. Chemical influences are capable of producing a myriad of effects on the activity and function of the central nervous system. Since our knowledge of different regions of brain function and the neurotransmitters in the brain is limited, the explanations for the mechanisms of drug action may be vague. The known neurotransmitters are: acetylcholine which is involved with memory and learning; which is involved with mania-depression and emotions; and which is involved with biological rhythms, sleep, emotion, and pain. Stimulants are drugs that exert their action through excitation of the central nervous system. Psychic stimulants include caffeine, cocaine, and various amphetamines. These drugs are used to enhance mental alertness and reduce drowsiness and fatigue. However, increasing the dosage of caffeine above 200 mg (about 2 cups of coffee) does not increase mental performance but may increase nervousness, irritability, tremors, and headache. Heavy coffee drinkers become psychically dependent upon caffeine. If caffeine is withheld, a person may experience mild withdrawal symptoms characterized by irritability, nervousness, and headache. and the chemically related xanthines, theophylline and theobromine, decrease in the order given in their stimulatory action. They may be included in some over-the-counter drugs. The action of caffeine is to block adenosine receptors as an antagonist. As caffeine has a similar structure to the adenosine group. This means that caffeine will fit adenosine receptors as well as adenosine itself. It inhibits the release of neurotransmitters from presynaptic sites but works in concert with norepinephrine or angiotensin to augment their actions. Antagonism of adenosine receptors by caffeine would appear to promote neurotransmitter release, thus explaining the stimulatory effects of caffeine. The stimulation caused by amphetamines is caused by excessive release of norepinephrine from storage sites in the peripheral nervous system. It is not known whether the same action occurs in the central nervous system. Two other theories for their action are that they are degraded slower than norepinephrine or that they could act on serotonin receptor sites. Therapeutic doses of amphetamine elevate mood, reduce feelings of fatigue and hunger, facilitate powers of concentration, and increase the desire and capacity to carry out work. They induce exhilarating feelings of power, strength, energy, self-assertion, focus and enhanced motivation. The need to sleep or eat is diminished. Levoamphetamine (Benzedrine), dextroamphetamine (Dexedrine), and methamphetamine (Methedrine) are collectively referred to as amphetamines. Benzedrine is a mixture of both the dextro and levoamphetamine isomers. The dextro isomer is several times more potent than the levo isomer. The misuse and abuse of amphetamines is a significant problem which may include the house wife taking diet pills, athletes desiring an improved performance, the truck driver driving non-stop coast to-coast, or a student cramming all night for an exam. | 3,533 | 3,357 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.1%3A_Getting_Started%3A_Some_Terminology |
Because energy takes many forms, only some of which can be seen or felt, it is defined by its effect on matter. For example, microwave ovens produce energy to cook food, but we cannot see that energy. In contrast, we can see the energy produced by a light bulb when we switch on a lamp. In this section, we describe the forms of energy and discuss the relationship between energy, heat, and work. The forms of energy include thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy (Figure \(\Page {1}\)). Thermal energy results from atomic and molecular motion; the faster the motion, the greater the thermal energy. The temperature of an object is a measure of its thermal energy content. Radiant energy is the energy carried by light, microwaves, and radio waves. Objects left in bright sunshine or exposed to microwaves become warm because much of the radiant energy they absorb is converted to thermal energy. Electrical energy results from the flow of electrically charged particles. When the ground and a cloud develop a separation of charge, for example, the resulting flow of electrons from one to the other produces lightning, a natural form of electrical energy. Nuclear energy is stored in the nucleus of an atom, and chemical energy is stored within a chemical compound because of a particular arrangement of atoms. Electrical energy, nuclear energy, and chemical energy are different forms of potential energy (PE), which is energy stored in an object because of the relative positions or orientations of its components. A brick lying on the windowsill of a 10th-floor office has a great deal of potential energy, but until its position changes by falling, the energy is contained. In contrast, kinetic energy (KE) is energy due to the motion of an object. When the brick falls, its potential energy is transformed to kinetic energy, which is then transferred to the object on the ground that it strikes. The electrostatic attraction between oppositely charged particles is a form of potential energy, which is converted to kinetic energy when the charged particles move toward each other. Energy can be converted from one form to another (Figure \(\Page {2}\)) or, as we saw with the brick, transferred from one object to another. For example, when you climb a ladder to a high diving board, your body uses chemical energy produced by the combustion of organic molecules. As you climb, the chemical energy is converted to to overcome the force of gravity. When you stand on the end of the diving board, your potential energy is greater than it was before you climbed the ladder: the greater the distance from the water, the greater the potential energy. When you then dive into the water, your potential energy is converted to kinetic energy as you fall, and when you hit the surface, some of that energy is transferred to the water, causing it to splash into the air. Chemical energy can also be converted to radiant energy; one common example is the light emitted by fireflies, which is produced from a chemical reaction. Although energy can be converted from one form to another, . This is known as the : The kinetic energy of an object is related to its mass \(m\) and velocity \(v\): For example, the kinetic energy of a 1360 kg (approximately 3000 lb) automobile traveling at a velocity of 26.8 m/s (approximately 60 mi/h) is \[KE=\dfrac{1}{2}(1360 kg)(26.8 ms)^2= 4.88 \times 10^5 g \cdot m^2 \tag{7.1.5}\] Because all forms of energy can be interconverted, energy in any form can be expressed using the same units as kinetic energy. The SI unit of energy, the joule (J), is defined as 1 kilogram·meter /second (kg·m /s ). Because a joule is such a small quantity of energy, chemists usually express energy in kilojoules (1 kJ = 10 J). For example, the kinetic energy of the 1360 kg car traveling at 26.8 m/s is 4.88 × 10 J or 4.88 × 10 kJ. It is important to remember that , whether thermal, radiant, chemical, or any other form. Because heat and work result in changes in energy, their units must also be the same. To demonstrate, let’s calculate the potential energy of the same 1360 kg automobile if it were parked on the top level of a parking garage 36.6 m (120 ft) high. Its potential energy is equivalent to the amount of work required to raise the vehicle from street level to the top level of the parking garage, which is = . According to Equation 7.1.2, the force ( ) exerted by gravity on any object is equal to its mass ( , in this case, 1360 kg) times the acceleration ( ) due to gravity ( , 9.81 m/s at Earth’s surface). The distance ( ) is the height ( ) above street level (in this case, 36.6 m). Thus the potential energy of the car is as follows: \[ PE= F\;d = m\,a\;d = m\,g\,h \tag{7.1.6a}\] \[PE=(1360, Kg)\left(\dfrac{9.81\, m}{s^2}\right)(36.6\;m) = 4.88 \times 10^5\; \frac{Kg \cdot m}{s^2} \tag{7.1.6b}\] \[=4.88 \times 10^5 J = 488\; kJ \tag{7.1.6c}\] The units of potential energy are the same as the units of kinetic energy. Notice that in this case the potential energy of the stationary automobile at the top of a 36.6 m high parking garage is the same as its kinetic energy at 60 mi/h. If the vehicle fell from the roof of the parking garage, its potential energy would be converted to kinetic energy, and it is reasonable to infer that the vehicle would be traveling at 60 mi/h just before it hit the ground, neglecting air resistance. After the car hit the ground, its potential and kinetic energy would both be zero. The units of energy are the same for all forms of energy. Energy can also be expressed in the non-SI units of calories (cal), where 1 cal was originally defined as the amount of energy needed to raise the temperature of exactly 1 g of water from 14.5°C to 15.5°C. The name is derived from the Latin , meaning “heat.” Although energy may be expressed as either calories or joules, calories were defined in terms of heat, whereas joules were defined in terms of motion. Because calories and joules are both units of energy, however, the calorie is now defined in terms of the joule: \[1 \;cal = 4.184 \;J \;\text{exactly} \tag{7.1.7a}\] \[1 \;J = 0.2390\; cal \tag{7.1.7b}\] In this text, we will use the SI units—joules (J) and kilojoules (kJ)—exclusively, except when we deal with nutritional information. mass and velocity or height kinetic and potential energy Use Equation 7.1.4 to calculate the kinetic energy and Equation 7.1.6 to calculate the potential energy, as appropriate. The kinetic energy of the baseball is therefore \[
KE= 1492 \;\cancel{g} \left(\dfrac{1\; kg}{1000 \;\cancel{g}} \right) \left(\dfrac{44.7 \;m}{s} \right)^2= 1.49 \times 10^2 \dfrac{kg⋅m^2}{s^2}= 1.49 \times10^2\; J\] is a branch of chemistry that qualitatively and quantitatively describes the energy changes that occur during chemical reactions. is the capacity to do work. is the amount of energy required to move an object a given distance when opposed by a force. is due to the random motions of atoms, molecules, or ions in a substance. The of an object is a measure of the amount of thermal energy it contains. is the transfer of thermal energy from a hotter object to a cooler one. Energy can take many forms; most are different varieties of , energy caused by the relative position or orientation of an object. is the energy an object possesses due to its motion. Energy can be converted from one form to another, but the states that energy can be neither created nor destroyed. The most common units of energy are the , defined as 1 (kg·m )/s , and the , defined as the amount of energy needed to raise the temperature of 1 g of water by 1°C (1 cal = 4.184 J). | 7,716 | 3,358 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Hydration_of_Alkynes_and_Tautomerism |
As with alkenes, the addition of water to alkynes requires a strong acid, usually sulfuric acid, and is facilitated by mercuric sulfate. However, unlike the additions to double bonds which give alcohol products, addition of water to alkynes gives ketone products ( except for acetylene which yields acetaldehyde ). The explanation for this deviation lies in , illustrated by the following equation. The initial product from the addition of water to an alkyne is an enol (a compound having a hydroxyl substituent attached to a double-bond), and this immediately rearranges to the more stable keto tautomer. Tautomers are defined as rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom (colored red here) and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers ( acetone, for example, is 99.999% keto tautomer ). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples. The three examples shown below illustrate these reactions for different substitutions of the triple-bond. The tautomerization step is indicated by a red arrow. For terminal alkynes the addition of water follows the Markovnikov rule, as in the second example below, and the final product ia a methyl ketone ( except for acetylene, shown in the first example ). For internal alkynes ( the triple-bond is within a longer chain ) the addition of water is not regioselective. If the triple-bond is not symmetrically located ( i.e. if R & R' in the third equation are not the same ) two isomeric ketones will be formed. Two factors have an important influence on the enol-keto tautomerizations described here. The first is the potential energy difference between the tautomeric isomers. . The second factor is the activation energy for the interconversion of one tautomer to the other. . Since the potential energy or stability of a compound is in large part a function of its covalent bond energies, we can estimate the relative energy of keto and enol tautomers by considering the bonds that are changed in the rearrangement. From the following diagram, we see that only three significant changes occur, and the standard bond energies for those changes are given to the right of the equation. The keto tautomer has a 17.5 kcal/mole advantage in bond energy, so its predominance at equilibrium is expected. The rapidity with which enol-keto tautomerization occurs suggests that the activation energy for this process is low. We have noted that the rearrangement is acid & base catalyzed, and very careful experiments have shown that interconversion of tautomers is much slower if such catalysts are absent. A striking example of the influence of activation energy on such transformations may be seen in the following hypothetical rearrangement. Here we have substituted a methyl group (colored maroon) for the proton of a conventional tautomerism, and the methyl shifts from oxygen to carbon just as the proton does in going from an enol to a ketone. H C=CH-O- -CH -CH=O The potential energy change for this rearrangement is even more advantageous than for enol-keto tautomerism, being estimated at over 25 kcal/mole from bond energy changes. Despite this thermodynamic driving force, the enol ether described above is completely stable to base treatment, and undergoes rapid acid-catalyzed hydrolysis with loss of methanol, rather than rearrangement. The controlling difference in this case must be a prohibitively high activation energy for the described rearrangement, combined with lower energy alternative reaction paths. | 3,884 | 3,359 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.02%3A_Concentration |
To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as or are used to describe solutions that have a little or a lot of dissolved solute, respectively, but these are relative terms whose meanings depend on various factors. There is usually a limit to how much solute will dissolve in a given amount of solvent. This limit is called the of the solute. Some solutes have a very small solubility, while other solutes are soluble in all proportions. Table \(\Page {1}\) lists the solubilities of various solutes in water. Solubilities vary with temperature, so Table \(\Page {1}\) includes the temperature at which the solubility was determined. If a solution contains so much solute that its solubility limit is reached, the solution is said to be , and its concentration is known from information contained in Table \(\Page {1}\). If a solution contains less solute than the solubility limit, it is . Under special circumstances, more solute can be dissolved even after the normal solubility limit is reached; such solutions are called and are not stable. If the solute is solid, excess solute can easily recrystallize. If the solute is a gas, it can bubble out of solution uncontrollably, like what happens when you shake a soda can and then immediately open it. Recrystallization of excess solute from a supersaturated solution usually gives off energy as heat. Commercial heat packs containing supersaturated sodium acetate (NaC H O ) take advantage of this phenomenon. You can probably find them at your local drugstore. Watered-down sodium acetate trihydrate. Needle crystal is truly wonderful structures Most solutions we encounter are unsaturated, so knowing the solubility of the solute does not accurately express the amount of solute in these solutions. There are several common ways of specifying the concentration of a solution. There are several ways of expressing the concentration of a solution by using a percentage. The (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100: \[\mathrm{\% \:m/m = \dfrac{mass\: of\: solute}{mass\: of\: solution}\times100\%} \nonumber \] If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. Each mass must be expressed in the same units to determine the proper concentration. A saline solution with a mass of 355 g has 36.5 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solution? We can substitute the quantities given in the equation for mass/mass percent: \(\mathrm{\%\: m/m=\dfrac{36.5\: g}{355\: g}\times100\%=10.3\%}\) A dextrose (also called D-glucose, C H O ) solution with a mass of 2.00 × 10 g has 15.8 g of dextrose dissolved in it. What is the mass/mass percent concentration of the solution? 7.90% For gases and liquids, volumes are relatively easy to measure, so the concentration of a liquid or a gas solution can be expressed as a (% v/v): the volume of a solute divided by the volume of a solution times 100: \[\mathrm{\%\: v/v = \dfrac{volume\: of\: solute}{volume\: of\: solution}\times100\%} \nonumber \] Again, the units of the solute and the solution must be the same. A hybrid concentration unit, (% m/v), is commonly used for intravenous (IV) fluids (Figure \(\Page {1}\)). It is defined as the mass in grams of a solute, divided by volume in milliliters of solution times 100: \[\mathrm{\%\: m/v = \dfrac{mass\: of\: solute\: (g)}{volume\: of\: solution\: (mL)}\times100\%} \nonumber \] The can be used to produce a between the and the . As such, concentrations can be useful in a variety of stoichiometry problems as discussed in Chapter 6. In many cases, it is best to use the original definition of the concentration unit; it is that definition that provides the conversion factor. As an example, if the given concentration is , this means that there are 5 mL of alcohol dissolved in every 100 mL solution. The two possible conversion factors are written as follows: or Use the first conversion factor to convert from a given amount of solution to amount of solute. The second conversion factor is used to convert from a given amount of solute to amount of solution. Given any two quantities in any percent composition, the third quantity can be calculated, as the following example illustrates. A sample of 45.0% v/v solution of ethanol (C H OH) in water has a volume of 115 mL. What volume of ethanol solute does the sample contain? A percentage concentration is simply the number of parts of solute per 100 parts of solution. Thus, the percent concentration of 45.0% v/v implies the following: \(\mathrm{45.0\%\: v/v \rightarrow \dfrac{45\: mL\: C_2H_5OH}{100\: mL\: solution}}\) That is, there are 45 mL of C H OH for every 100 mL of solution. We can use this fraction as a to determine the amount of C H OH in 115 mL of solution: \(\mathrm{115\: mL\: solution\times\dfrac{45\: mL\: C_2H_5OH}{100\: mL\: solution}=51.8\: mL\: C_2H_5OH}\) What volume of a 12.75% m/v solution of glucose (C H O ) in water is needed to obtain 50.0 g of C H O ? \(\mathrm{50.0\: g\: C_6H_12O_6\times\dfrac{100\: mL\: solution}{12.75\: g\: C_6H_12O_6}=392\: mL\: solution}\) A normal saline IV solution contains 9.0 g of NaCl in every liter of solution. What is the mass/volume percent of normal saline? We can use the definition of mass/volume percent, but first we have to express the volume in milliliter units: Because this is an exact relationship, it does not affect the significant figures of our result. \(\mathrm{\%\: m/v = \dfrac{9.0\: g\: NaCl}{1,000\: mL\: solution}\times100\%=0.90\%\: m/v}\) The chlorine bleach that you might find in your laundry room is typically composed of 27.0 g of sodium hypochlorite (NaOCl), dissolved to make 500.0 mL of solution. What is the mass/volume percent of the bleach? \(\mathrm{\%\: m/v = \dfrac{27.0\: g\: NaOCl}{500.0\: mL\: solution}\times100\%=5.40\%\: m/v}\) In addition to percentage units, the units for expressing the concentration of extremely dilute solutions are parts per million (ppm) and parts per billion (ppb). Both of these units are mass based and are defined as follows: \[\mathrm{ppm=\dfrac{mass\: of\: solute}{mass\: of\: solution}\times1,000,000} \nonumber \] \[\mathrm{ppb=\dfrac{mass\: of\: solute}{mass\: of\: solution}\times1,000,000,000} \nonumber \] Similar to parts per million and parts per billion, related units include parts per thousand (ppth) and parts per trillion (ppt). Concentrations of in the body—elements that are present in extremely low concentrations but are nonetheless necessary for life—are commonly expressed in parts per million or parts per billion. Concentrations of poisons and pollutants are also described in these units. For example, cobalt is present in the body at a concentration of 21 ppb, while the State of Oregon’s Department of Agriculture limits the concentration of arsenic in fertilizers to 9 ppm. In aqueous solutions, 1 ppm is essentially equal to 1 mg/L, and 1 ppb is equivalent to 1 µg/L. If the concentration of cobalt in a human body is 21 ppb, what mass in grams of Co is present in a body having a mass of 70.0 kg? A concentration of 21 ppb means “21 g of solute per 1,000,000,000 g of solution.” Written as a this concentration of Co is as follows: \(\mathrm{21\: ppb\: Co \rightarrow \dfrac{21\: g\: Co}{1,000,000,000\: g\: solution}}\) We can use this as a conversion factor, but first we must convert 70.0 kg to gram units: \(\mathrm{70.0\: kg\times\dfrac{1,000\: g}{1\: kg}=7.00\times10^4\: g}\) Now we determine the amount of Co: \(\mathrm{7.00\times10^4\: g\: solution\times\dfrac{21\: g\: Co}{1,000,000,000\: g\: solution}=0.0015\: g\: Co}\) This is only 1.5 mg. An 85 kg body contains 0.012 g of Ni. What is the concentration of Ni in parts per million? 0.14 ppm Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Such concentration units are useful for discussing chemical reactions in which a solute is a product or a reactant. Molar mass can then be used as a conversion factor to convert amounts in moles to amounts in grams. Molarity is defined as the number of moles of a solute dissolved per liter of solution: \[\mathrm{molarity=\dfrac{number\: of\: moles\: of\: solute}{number\: of\: liters\: of\: solution}} \nonumber \] Molarity is abbreviated M (often referred to as “molar”), and the units are often abbreviated as mol/L. It is important to remember that “mol” in this expression refers to moles of solute and that “L” refers to liters of solution. For example, if you have 1.5 mol of NaCl dissolved in 0.500 L of solution, its molarity is therefore \[\mathrm{\dfrac{1.5\: mol\: NaCl}{0.500\: L\: solution}=3.0\: M\: NaCl} \nonumber \] which is read as “three point oh sodium chloride.” Sometimes (aq) is added when the solvent is water, as in “3.0 M NaCl(aq).” Before a molarity concentration can be calculated, the amount of the , and the must be expressed in . If the quantity of the solute is given in mass units, you must convert mass units to mole units before using the definition of molarity to calculate concentration. For example, what is the molar concentration of a solution of 22.4 g of HCl dissolved in 1.56 L? Step 1: convert the to using the \[22.4\cancel{gHCl}\times \dfrac{1\: mol\: HCl}{36.46\cancel{gHCl}}=0.614\, mol\; HCl \nonumber \] Step 2: use the to determine the concentration: \[M \: =\: \dfrac{0.614\: mol\: HCl}{1.56L\: solution}=0.394\, M HCl \nonumber \] What is the molarity of an aqueous solution of 25.0 g of NaOH in 750 mL? Before we substitute these quantities into the definition of molarity, we must convert them to the proper units. The mass of NaOH must be converted to moles of NaOH. The molar mass of NaOH is 40.00 g/mol: \(\mathrm{25.0\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}=0.625\: mol\: NaOH}\) Next, we convert the volume units from milliliters to liters: \(\mathrm{750\: mL\times\dfrac{1\: L}{1,000\: mL}=0.750\: L}\) Now that the quantities are expressed in the proper units, we can substitute them into the definition of molarity: \(\mathrm{M=\dfrac{0.625\: mol\: NaOH}{0.750\: L}=0.833\: M\: NaOH}\) If a 350 mL cup of coffee contains 0.150 g of caffeine (C H N O ), what is the molarity of this caffeine solution? 0.00221 M The definition of molarity can also be used to calculate a needed volume of solution, given its concentration and the number of moles desired, or the number of moles of solute (and subsequently, the mass of the solute), given its concentration and volume. As in the percent concentration, molarity can also be expressed as a . Molarity is defined as moles solute per liter solution. There is an understood 1 in the denominator of the conversion factor. For example, a means that there are three moles of sucrose dissolved in every . Mathematically, this is stated as follows: Dividing both sides of this expression by either side, we generate two possible conversion factors: or The first conversion factor can be used to convert from , and the . For example, suppose we are asked how many moles of sucrose are present in 0.108 L of a 3.0 M sucrose solution. The given volume (0.108 L) is multiplied by the first conversion factor to cancel the L units, and find that 0.32 moles of sucrose are present. \[0.108\cancel{L\, solution}\times \dfrac{3.0\, mol\, sucrose}{\cancel{1L\, solution}}=0.32\, mol\, sucrose \nonumber \] How many liters of 3.0 M sucrose solution are needed to obtain 4.88 mol of sucrose? In such a conversion, we multiply the given (4.88 moles sucrose) with the second conversion factor. This cancels the moles units and converts it to liters of solution. \[4.88\cancel{mol\, sucrose}\times \dfrac{1\, L\, solution}{\cancel{3.0\, mol\, sucrose}}=1.63\, L\, solution \nonumber \] 1. To solve for the volume, multiply the "given" (0.450 mol of dimethylamine) with the molarity conversion factor (0.0753 M). Use the proper conversion factor to cancel the unit "mol" and get the unit volume (L) of solution: \(\mathrm{0.450\: mol\: dimethylamine\times\dfrac{1\: L\: solution}{0.0753\: mol\: dimethylamine}=5.98\: L\: solution}\) 2. The strategy in solving this problem is to convert the given volume (5.00 L) using the 6.00 M (conversion factor) to solve for moles of ethylene glycol, which can then be converted to grams. Step 1: Convert the given volume (5.00 L) to moles ethylene glycol. \(\mathrm{5.00\: L\: solution\times\dfrac{6.00\: mol\: C_2H_6O_2}{1\: L\: solution}=30.0\: mol\: C_2H_6O_2}\) Step 2: Convert 30.0 mols C H O to grams C H O . Molar mass of C H O = 62.08 g/mol \(\mathrm{30.0\: mol\: C_2H_6O_2\times\dfrac{62.08\: g\: C_2H_6O_2}{1\: mol\: C_2H_6O_2}=1,860\: g\: C_2H_6O_2}\) The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows: \(\mathrm{5.00\: L\: solution\times\dfrac{6.00\: mol\: C_2H_6O_2}{1\: L\: solution}\times\dfrac{62.08\: g\: C_2H_6O_2}{1\: mol\: C_2H_6O_2}=1,860\: g\: C_2H_6O_2}\) The final answer is rounded off to 3 significant figures. Thus, there are 1,860 g of C H O in the specified amount of engine coolant. Note: Dimethylamine has a “fishy” odor. In fact, organic compounds called cause the odor of decaying fish. This is an alternative method in case you don't want to use the conversion factor for molarity. In both parts, we will use the definition of molarity to solve for the desired quantity. To solve for the volume of solution, we multiply both sides by volume of solution and divide both sides by the molarity value to isolate the volume of solution on one side of the equation: \(\mathrm{volume\:of\:solution = \dfrac{0.450\:mol\:(CH_3)_2NH}{0.0753\:M}=5.98\:L}\) Note that because the definition of molarity is mol/L, the division of mol by M yields L, a unit of volume. \(\mathrm{6.00\: M=\dfrac{moles\: of\: solute}{5.00\: L}}\) To solve for the number of moles of solute, we multiply both sides by the volume: Note that because the definition of molarity is mol/L, the product M × L gives mol, a unit of amount. Now, using the molar mass of C H O , we convert mol to g: \(\mathrm{30.0\: mol\times\dfrac{62.08\: g}{mol}=1,860\: g}\) Thus, there are 1,860 g of C H O in the specified amount of engine coolant. a. 9.84 L b. 32.4 g Of all the ways of expressing concentration, molarity is the one most commonly used in stoichiometry problems because it is directly related to the mole unit. Consider the following chemical equation: Suppose we want to know how many liters of aqueous HCl solution will react with a given mass of NaOH. A typical approach to answering this question is as follows: In itself, each step is a straightforward conversion. It is the combination of the steps that is a powerful quantitative tool for problem solving. How many milliliters of a 2.75 M HCl solution are needed to react with 185 g of NaOH? The balanced chemical equation for this reaction is as follows: We will follow the flowchart to answer this question. First, we convert the mass of NaOH to moles of NaOH using its molar mass, 40.00 g/mol: \(\mathrm{185\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}=4.63\: mol\: NaOH}\) Using the balanced chemical equation, we see that there is a one-to-one ratio of moles of HCl to moles of NaOH. We use this to determine the number of moles of HCl needed to react with the given amount of NaOH: \(\mathrm{4.63\: mol\: NaOH\times\dfrac{1\: mol\: HCl}{1\: mol\: NaOH}=4.63\: mol\: HCl}\) Finally, we use the definition of molarity to determine the volume of 2.75 M HCl needed: \(\mathrm{2.75\: M\: HCl=\dfrac{4.63\: mol\: HCl}{volume\: of\: HCl\: solution}}\) \(\mathrm{volume\: of\: HCl=\dfrac{4.63\: mol\: HCl}{2.75\: M\: HCl}=1.68\: L\times\dfrac{1,000\: mL}{1\: L}=1,680\: mL}\) We need 1,680 mL of 2.75 M HCl to react with the NaOH. The same multi-step problem can also be worked out in a single line, rather than as separate steps, as follows: \(\mathrm{185\: g\: NaOH\times\dfrac{1\: mol\: NaOH}{40.00\: g\: NaOH}\times\dfrac{1\: mol\: HCl}{1\: mol\: NaOH}\times\dfrac{1\: L\: HCl\: solution}{2.75\: mol\: HCl}\times\dfrac{1000\: mL\: HCl\: solution}{1\: L\: HCl\: solution}=1,680\: mL\: HCl\: solution}\) Our final answer (rounded off to three significant figures) is 1,680 mL HCl solution. How many milliliters of a 1.04 M H SO solution are needed to react with 98.5 g of Ca(OH) ? The balanced chemical equation for the reaction is as follows: \[H_2SO_{4(aq)} + Ca(OH)_{2(s)} \rightarrow 2H_2O_{(ℓ)} + CaSO_{4(aq)} \nonumber \] 1,280 mL The general steps for performing stoichiometry problems such as this are shown in Figure \(\Page {3}\). You may want to consult this figure when working with solutions in chemical reactions. The double arrows in Figure \(\Page {3}\) indicate that you can start at either end of the chart and, after a series of simple conversions, determine the quantity at the other end. Many of the fluids found in our bodies are solutions. The solutes range from simple ionic compounds to complex proteins. Table \(\Page {2}\) lists the typical concentrations of some of these solutes. Why is it that we can drink 1 qt of water when we are thirsty and not be harmed, but if we ingest 0.5 g of arsenic, we might die? There is an old saying: . This means that what may be dangerous in some amounts may not be dangerous in other amounts. Take arsenic, for example. Some studies show that arsenic deprivation limits the growth of animals such as chickens, goats, and pigs, suggesting that arsenic is actually an essential trace element in the diet. Humans are constantly exposed to tiny amounts of arsenic from the environment, so studies of completely arsenic-free humans are not available; if arsenic is an essential trace mineral in human diets, it is probably required on the order of 50 ppb or less. A toxic dose of arsenic corresponds to about 7,000 ppb and higher, which is over 140 times the trace amount that may be required by the body. Thus, arsenic is not poisonous in and of itself. Rather, it is the amount that is dangerous: the dose makes the poison. Similarly, as much as water is needed to keep us alive, too much of it is also risky to our health. Drinking too much water too fast can lead to a condition called water intoxication, which may be fatal. The danger in water intoxication is not that water itself becomes toxic. It is that the ingestion of too much water too fast dilutes sodium ions, potassium ions, and other salts in the bloodstream to concentrations that are not high enough to support brain, muscle, and heart functions. Military personnel, endurance athletes, and even desert hikers are susceptible to water intoxication if they drink water but do not replenish the salts lost in sweat. As this example shows, even the right substances in the wrong amounts can be dangerous! Concentrations of ionic solutes are occasionally expressed in units called equivalents (Eq). One equivalent equals 1 mol of positive or negative charge. Thus, 1 mol/L of Na (aq) is also 1 Eq/L because sodium has a 1+ charge. A 1 mol/L solution of Ca (aq) ions has a concentration of 2 Eq/L because calcium has a 2+ charge. Dilute solutions may be expressed in milliequivalents (mEq)—for example, human blood plasma has a total concentration of about 150 mEq/L. (For more information about the ions present in blood plasma, see Chapter 3, Section 3.3.) When additional water is added to an aqueous solution, the concentration of that solution decreases. This is because the number of , while the . We can set up an equality between the moles of the solute before the dilution (1) and the moles of the solute after the dilution (2). \[\text{mol}_1 = \text{mol}_2 \nonumber \] Since the moles of solute in a solution is equal to the molarity multiplied by the liters, we can set those equal. \[M_1 \times L_1 = M_2 \times L_2 \nonumber \] Finally, because the two sides of the equation are set equal to one another, the volume can be in any units we choose, as long as that unit is the same on both sides. Our equation for calculating the molarity of a diluted solution becomes: \[M_1 \times V_1 = M_2 \times V_2 \nonumber \] Suppose that you have \(100. \: \text{mL}\) of a \(2.0 \: \text{M}\) solution of \(\ce{HCl}\). You dilute the solution by adding enough water to make the solution volume \(500. \: \text{mL}\). The new molarity can easily be calculated by using the above equation and solving for \(M_2\). \[M_2 = \frac{M_1 \times V_1}{V_2} = \frac{2.0 \: \text{M} \times 100. \: \text{mL}}{500. \: \text{mL}} = 0.40 \: \text{M} \: \ce{HCl} \nonumber \] The solution has been diluted by one-fifth since the new volume is five times as great as the original volume. Consequently, the molarity is one-fifth of its original value. Any units of concentration and volume can be used, as long as both concentrations and both volumes have the same unit. For example, if we are using M (molarity), then we can express the equation as follows: Molarity Molarity If we are using percent, the dilution equation is as follows: % % A 125 mL sample of 0.900 M NaCl is diluted to 1,125 mL. What is the final concentration of the diluted solution? Because the volume units are the same, and we are looking for the molarity of the final solution, we can use (concentration × volume) = (concentration × volume) : We solve by isolating the unknown concentration by itself on one side of the equation. Dividing by 1,125 mL gives \(\mathrm{concentration = \dfrac{0.900\: M\times125\: mL}{1,125\: mL}=0.100\: M}\) as the final concentration. a. A nurse uses a syringe to inject 5.00 mL of 0.550 M heparin solution (heparin is an anticoagulant drug) into a 250 mL IV bag, for a final volume of 255 mL. What is the concentration of the resulting heparin solution? b. A 0.885 M solution of KBr whose initial volume is 76.5 mL has more water added until its concentration is 0.500 M. What is the new volume of the solution? a. 0.0108 M b. 135.4 mL In a hospital emergency room, a physician orders an intravenous (IV) delivery of 100 mL of 0.5% KCl for a patient suffering from hypokalemia (low potassium levels). Does an aide run to a supply cabinet and take out an IV bag containing this concentration of KCl? Not likely. It is more probable that the aide must make the proper solution from an IV bag of sterile solution and a more concentrated, sterile solution, called a , of KCl. The aide is expected to use a syringe to draw up some stock solution and inject it into the waiting IV bag and dilute it to the proper concentration. Thus the aide must perform a dilution calculation. If the stock solution is 10.0% KCl and the final volume and concentration need to be 100 mL and 0.50%, respectively, then it is an easy calculation to determine how much stock solution to use: Of course, the addition of the stock solution affects the total volume of the diluted solution, but the final concentration is likely close enough even for medical purposes. Medical and pharmaceutical personnel are constantly dealing with dosages that require concentration measurements and dilutions. It is an important responsibility: calculating the dose can be useless, harmful, or even fatal! | 23,296 | 3,360 |
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In precipitation gravimetry an insoluble compound forms when we add a precipitating reagent, or , to a solution that contains our analyte. In most cases the precipitate is the product of a simple metathesis reaction between the analyte and the precipitant; however, any reaction that generates a precipitate potentially can serve as a gravimetric method. Most precipitation gravimetric methods were developed in the nineteenth century, or earlier, often for the analysis of ores. in Chapter 1, for example, illustrates a precipitation gravimetric method for the analysis of nickel in ores. All precipitation gravimetric analyses share two important attributes. First, the precipitate must be of low solubility, of high purity, and of known composition if its mass is to reflect accurately the analyte’s mass. Second, it must be easy to separate the precipitate from the reaction mixture. To provide an accurate result, a precipitate’s solubility must be minimal. The accuracy of a total analysis technique typically is better than ±0.1%, which means the precipitate must account for at least 99.9% of the analyte. Extending this requirement to 99.99% ensures the precipitate’s solubility will not limit the accuracy of a gravimetric analysis. A total analysis technique is one in which the analytical signal—mass in this case—is proportional to the absolute amount of analyte in the sample. See for a discussion of the difference between total analysis techniques and concentration techniques. We can minimize solubility losses by controlling the conditions under which the precipitate forms. This, in turn, requires that we account for every equilibrium reaction that might affect the precipitate’s solubility. For example, we can determine Ag gravimetrically by adding NaCl as a precipitant, forming a precipitate of AgCl. \[\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}(s) \label{8.1}\] If this is the only reaction we consider, then we predict that the precipitate’s solubility, , is given by the following equation. \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]=\frac{K_{\mathrm{sp}}}{\left[\mathrm{Cl}^{-}\right]} \label{8.2}\] Equation \ref{8.2} suggests that we can minimize solubility losses by adding a large excess of Cl . In fact, as shown in Figure 8.2.1
, adding a large excess of Cl increases the precipitate’s solubility. To understand why the solubility of AgCl is more complicated than the relationship suggested by Equation \ref{8.2}, we must recall that Ag also forms a series of soluble silver-chloro metal–ligand complexes. \[\operatorname{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}(a q) \quad \log K_{1}=3.70 \label{8.3}\] \[\operatorname{AgCl}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\operatorname{AgCl}_{2}(a q) \quad \log K_{2}=1.92 \label{8.4}\] \[\mathrm{AgCl}_{2}^{-}(a q)+\mathrm{Cl}^{-}(a q)\rightleftharpoons\mathrm{AgCl}_{3}^{2-}(a q) \quad \log K_{3}=0.78 \label{8.5}\] Note the difference between reaction \ref{8.3}, in which we form AgCl( ) as a product, and reaction \ref{8.1}, in which we form AgCl( ) as a product. The formation of AgCl( ) from AgCl( ) \[\operatorname{AgCl}(s)\rightleftharpoons\operatorname{AgCl}(a q) \nonumber\] is called AgCl’s intrinsic solubility. The actual solubility of AgCl is the sum of the equilibrium concentrations for all soluble forms of Ag . \[S_{\mathrm{AgCl}}=\left[\mathrm{Ag}^{+}\right]+[\mathrm{AgCl}(a q)]+\left[\mathrm{AgCl}_{2}^-\right]+\left[\mathrm{AgCl}_{3}^{2-}\right] \label{8.6}\] By substituting into Equation \ref{8.6} the equilibrium constant expressions for reaction \ref{8.1} and reactions \ref{8.3}–\ref{8.5}, we can define the solubility of AgCl as \[S_\text{AgCl} = \frac {K_\text{sp}} {[\text{Cl}^-]} + K_1K_\text{sp} + K_1K_2K_\text{sp}[\text{Cl}^-]+K_1K_2K_3K_\text{sp}[\text{Cl}^-]^2 \label{8.7}\] Equation \ref{8.7} explains the solubility curve for AgCl shown in . As we add NaCl to a solution of Ag , the solubility of AgCl initially decreases because of reaction \ref{8.1}. Under these conditions, the final three terms in Equation \ref{8.7} are small and Equation \ref{8.2} is sufficient to describe AgCl’s solubility. For higher concentrations of Cl , reaction \ref{8.4} and reaction \ref{8.5} increase the solubility of AgCl. Clearly the equilibrium concentration of chloride is important if we wish to determine the concentration of silver by precipitating AgCl. In particular, we must avoid a large excess of chloride. The predominate silver-chloro complexes for different values of pCl are shown by the ladder diagram along the -axis in . Note that the increase in solubility begins when the higher-order soluble complexes of \(\text{AgCl}_2^-\) and \(\text{AgCl}_3^{2-}\) are the predominate species. Another important parameter that may affect a precipitate’s solubility is pH. For example, a hydroxide precipitate, such as Fe(OH) , is more soluble at lower pH levels where the concentration of OH is small. Because fluoride is a weak base, the solubility of calcium fluoride, \(S_{\text{CaF}_2}\), also is pH-dependent. We can derive an equation for \(S_{\text{CaF}_2}\) by considering the following equilibrium reactions \[\mathrm{CaF}_{2}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{F}^{-}(a q) \quad K_{\mathfrak{sp}}=3.9 \times 10^{-11} \label{8.8}\] \[\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l )\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q) \quad K_{\mathrm{a}}=6.8 \times 10^{-4} \label{8.9}\] and the following equation for the solubility of CaF . \[S_{\mathrm{Ca} \mathrm{F}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\frac{1}{2}\left\{\left[\mathrm{F}^{-}\right]+[\mathrm{HF}]\right\} \label{8.10}\] Be sure that Equation \ref{8.10} makes sense to you. Reaction \ref{8.8} tells us that the dissolution of CaF produces one mole of Ca for every two moles of F , which explains the term of 1/2 in Equation \ref{8.10}. Because F is a weak base, we must account for both chemical forms in solution, which explains why we include HF. Substituting the equilibrium constant expressions for reaction \ref{8.8} and reaction \ref{8.9} into Equation \ref{8.10} allows us to define the solubility of CaF in terms of the equilibrium concentration of H O . \[S_{\mathrm{CaF}_{2}}=\left[\mathrm{Ca}^{2+}\right]=\left\{\frac{K_{\mathrm{p}}}{4}\left(1+\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{K_{\mathrm{a}}}\right)^{2}\right\}^{1 / 3} \label{8.11}\] Figure 8.2.2
shows how pH affects the solubility of CaF . Depending on the solution’s pH, the predominate form of fluoride is either HF or F . When the pH is greater than 4.17, the predominate species is F and the solubility of CaF is independent of pH because only reaction \ref{8.8} occurs to an appreciable extent. At more acidic pH levels, the solubility of CaF increases because of the contribution of reaction \ref{8.9}. You can use a ladder diagram to predict the conditions that will minimize a precipitate’s solubility. Draw a ladder diagram for oxalic acid, H C2O , and use it to predict the range of pH values that will minimize the solubility of CaC O . Relevant equilibrium constants are in the appendices. The solubility reaction for CaC O is \[\mathrm{CaC}_{2} \mathrm{O}_{4}(s)\rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \nonumber\] To minimize solubility, the pH must be sufficiently basic that oxalate, \(\text{C}_2\text{O}_4^{2-}\), does not react to form \(\text{HC}_2\text{O}_4^{-}\) or H C O . The ladder diagram for oxalic acid, including approximate buffer ranges, is shown below. Maintaining a pH greater than 5.3 ensures that \(\text{C}_2\text{O}_4^{2-}\) is the only important form of oxalic acid in solution, minimizing the solubility of CaC O . When solubility is a concern, it may be possible to decrease solubility by using a non-aqueous solvent. A precipitate’s solubility generally is greater in an aqueous solution because of water’s ability to stabilize ions through solvation. The poorer solvating ability of a non-aqueous solvent, even those that are polar, leads to a smaller solubility product. For example, the of PbSO is \(2 \times 10^{-8}\) in H O and \(2.6 \times 10^{-12}\) in a 50:50 mixture of H O and ethanol. In addition to having a low solubility, a precipitate must be free from impurities. Because precipitation usually occurs in a solution that is rich in dissolved solids, the initial precipitate often is impure. To avoid a determinate error, we must remove these impurities before we determine the precipitate’s mass. The greatest source of impurities are chemical and physical interactions that take place at the precipitate’s surface. A precipitate generally is crystalline—even if only on a microscopic scale—with a well-defined lattice of cations and anions. Those cations and anions at the precipitate’s surface carry, respectively, a positive or a negative charge because they have incomplete coordination spheres. In a precipitate of AgCl, for example, each silver ion in the precipitate’s interior is bound to six chloride ions. A silver ion at the surface, however, is bound to no more than five chloride ions and carries a partial positive charge (Figure 8.2.3
). The presence of these partial charges makes the precipitate’s surface an active site for the chemical and physical interactions that produce impurities. One common impurity is an , in which a potential interferent, whose size and charge is similar to a lattice ion, can substitute into the lattice structure if the interferent precipitates with the same crystal structure (Figure 8.2.4
a). The probability of forming an inclusion is greatest when the interfering ion’s concentration is substantially greater than the lattice ion’s concentration. An inclusion does not decrease the amount of analyte that precipitates, provided that the precipitant is present in sufficient excess. Thus, the precipitate’s mass always is larger than expected. An inclusion is difficult to remove since it is chemically part of the precipitate’s lattice. The only way to remove an inclusion is through in which we isolate the precipitate from its supernatant solution, dissolve the precipitate by heating in a small portion of a suitable solvent, and then reform the precipitate by allowing the solution to cool. Because the interferent’s concentration after dissolving the precipitate is less than that in the original solution, the amount of included material decreases upon reprecipitation. We can repeat the process of reprecipitation until the inclusion’s mass is insignificant. The loss of analyte during reprecipitation, however, is a potential source of determinate error. Suppose that 10% of an interferent forms an inclusion during each precipitation. When we initially form the precipitate, 10% of the original interferent is present as an inclusion. After the first reprecipitation, 10% of the included interferent remains, which is 1% of the original interferent. A second reprecipitation decreases the interferent to 0.1% of the original amount. An forms when an interfering ions is trapped within the growing precipitate. Unlike an inclusion, which is randomly dispersed within the precipitate, an occlusion is localized, either along flaws within the precipitate’s lattice structure or within aggregates of individual precipitate particles (Figure 8.2.4
b). An occlusion usually increases a precipitate’s mass; however, the precipitate’s mass is smaller if the occlusion includes the analyte in a lower molecular weight form than that of the precipitate. We can minimize an occlusion by maintaining the precipitate in equilibrium with its supernatant solution for an extended time, a process called digestion. During a , the dynamic nature of the solubility–precipitation equilibria, in which the precipitate dissolves and reforms, ensures that the occlusion eventually is reexposed to the supernatant solution. Because the rates of dissolution and reprecipitation are slow, there is less opportunity for forming new occlusions. After precipitation is complete the surface continues to attract ions from solution (Figure 8.2.4
c). These comprise a third type of impurity. We can minimize surface adsorption by decreasing the precipitate’s available surface area. One benefit of digestion is that it increases a precipitate’s average particle size. Because the probability that a particle will dissolve completely is inversely proportional to its size, during digestion larger particles increase in size at the expense of smaller particles. One consequence of forming a smaller number of larger particles is an overall decrease in the precipitate’s surface area. We also can remove surface adsorbates by washing the precipitate, although we cannot ignore the potential loss of analyte. Inclusions, occlusions, and surface adsorbates are examples of —otherwise soluble species that form along with the precipitate that contains the analyte. Another type of impurity is an interferent that forms an independent precipitate under the conditions of the analysis. For example, the precipitation of nickel dimethylglyoxime requires a slightly basic pH. Under these conditions any Fe in the sample will precipitate as Fe(OH) . In addition, because most precipitants rarely are selective toward a single analyte, there is a risk that the precipitant will react with both the analyte and an interferent. In addition to forming a precipitate with Ni , dimethylglyoxime also forms precipitates with Pd and Pt . These cations are potential interferents in an analysis for nickel. We can minimize the formation of additional precipitates by controlling solution conditions. If an interferent forms a precipitate that is less soluble than the analyte’s precipitate, we can precipitate the interferent and remove it by filtration, leaving the analyte behind in solution. Alternatively, we can mask the analyte or the interferent to prevent its precipitation. Both of the approaches outline above are illustrated in Fresenius’ analytical method for the determination of Ni in ores that contain Pb , Cu , and Fe (see in Chapter 1). Dissolving the ore in the presence of H SO selectively precipitates Pb as PbSO . Treating the resulting supernatant with H S precipitates Cu as CuS. After removing the CuS by filtration, ammonia is added to precipitate Fe as Fe(OH) . Nickel, which forms a soluble amine complex, remains in solution. Masking was introduced in . Size matters when it comes to forming a precipitate. Larger particles are easier to filter and, as noted earlier, a smaller surface area means there is less opportunity for surface adsorbates to form. By controlling the reaction conditions we can significantly increase a precipitate’s average particle size. The formation of a precipitate consists of two distinct events: nucleation, the initial formation of smaller, stable particles of the precipitate, and particle growth. Larger particles form when the rate of particle growth exceeds the rate of nucleation. Understanding the conditions that favor particle growth is important when we design a gravimetric method of analysis. We define a solute’s , , as \[R S S=\frac{Q-S}{S} \label{8.12}\] where is the solute’s actual concentration and is the solute’s concentration at equilibrium [Von Weimarn, P. P. . , , 217–242]. The numerator of Equation \ref{8.12}, – , is a measure of the solute’s supersaturation. A solution with a large, positive value of has a high rate of nucleation and produces a precipitate with many small particles. When the is small, precipitation is more likely to occur by particle growth than by nucleation. A supersaturated solution is one that contains more dissolved solute than that predicted by equilibrium chemistry. A supersaturated solution is inherently unstable and precipitates solute to reach its equilibrium position. How quickly precipitation occurs depends, in part, on the value of . Equation \ref{8.12} suggests that we can minimize if we decrease the solute’s concentration, , or if we increase the precipitate’s solubility, . A precipitate’s solubility usually increases at higher temperatures and adjusting pH may affect a precipitate’s solubility if it contains an acidic or a basic ion. Temperature and pH, therefore, are useful ways to increase the value of . Forming the precipitate in a dilute solution of analyte or adding the precipitant slowly and with vigorous stirring are ways to decrease the value of There are practical limits to minimizing . Some precipitates, such as Fe(OH) and PbS, are so insoluble that is very small and a large is unavoidable. Such solutes inevitably form small particles. In addition, conditions that favor a small may lead to a relatively stable supersaturated solution that requires a long time to precipitate fully. For example, almost a month is required to form a visible precipitate of BaSO under conditions in which the initial is 5 [Bassett, J.; Denney, R. C.; Jeffery, G. H. Mendham. J. , Longman: London, 4th Ed., 1981, p. 408]. A visible precipitate takes longer to form when is small both because there is a slow rate of nucleation and because there is a steady decrease in as the precipitate forms. One solution to the latter problem is to generate the precipitant as the product of a slow chemical reaction, which effectively maintains a constant . Because the precipitate forms under conditions of low , initial nucleation produces a small number of particles. As additional precipitant forms, particle growth supersedes nucleation, which results in larger particles of precipitate. This process is called a [Gordon, L.; Salutsky, M. L.; Willard, H. H. , Wiley: NY, 1959]. Two general methods are used for homogeneous precipitation. If the precipitate’s solubility is pH-dependent, then we can mix the analyte and the precipitant under conditions where precipitation does not occur, and then increase or decrease the pH by chemically generating OH or H O . For example, the hydrolysis of urea, CO(NH ) , is a source of OH because of the following two reactions. \[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons2 \mathrm{NH}_{3}(a q)+\mathrm{CO}_{2}(g) \nonumber\] \[\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}( l)\rightleftharpoons\mathrm{OH}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q) \nonumber\] Because the hydrolysis of urea is temperature-dependent—the rate is negligible at room temperature—we can use temperature to control the rate of hydrolysis and the rate of precipitate formation. Precipitates of CaC O , for example, have been produced by this method. After dissolving a sample that contains Ca , the solution is made acidic with HCl before adding a solution of 5% w/v (NH ) C O . Because the solution is acidic, a precipitate of CaC O does not form. The solution is heated to approximately 50 C and urea is added. After several minutes, a precipitate of CaC O begins to form, with precipitation reaching completion in about 30 min. In the second method of homogeneous precipitation, the precipitant is generated by a chemical reaction. For example, Pb is precipitated homogeneously as PbCrO by using bromate, \(\text{BrO}_3^-\), to oxidize Cr to \(\text{CrO}_4^{2-}\). \[6 \mathrm{BrO}_{3}^{-}(a q)+10 \mathrm{Cr}^{3+}(a q)+22 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons 3 \mathrm{Br}_{2}(a q)+10 \mathrm{CrO}_{4}^{2-}(a q)+44 \mathrm{H}^{+}(a q) \nonumber\] Figure 8.2.5
shows the result of preparing PbCrO by direct addition of K CrO (Beaker A) and by homogenous precipitation (Beaker B). Both beakers contain the same amount of PbCrO . Because the direct addition of K CrO leads to rapid precipitation and the formation of smaller particles, the precipitate remains less settled than the precipitate prepared homogeneously. Note, as well, the difference in the color of the two precipitates. The effect of particle size on color is well-known to geologists, who use a streak test to help identify minerals. The color of a bulk mineral and its color when powdered often are different. Rubbing a mineral across an unglazed porcelain plate leaves behind a small streak of the powdered mineral. Bulk samples of hematite, Fe O , are black in color, but its streak is a familiar rust-red. Crocite, the mineral PbCrO , is red-orange in color; its streak is orange-yellow. A homogeneous precipitation produces large particles of precipitate that are relatively free from impurities. These advantages, however, are offset by the increased time needed to produce the precipitate and by a tendency for the precipitate to deposit as a thin film on the container’s walls. The latter problem is particularly severe for hydroxide precipitates generated using urea. An additional method for increasing particle size deserves mention. When a precipitate’s particles are electrically neutral they tend to coagulate into larger particles that are easier to filter. Surface adsorption of excess lattice ions, however, provides the precipitate’s particles with a net positive or a net negative surface charge. Electrostatic repulsion between particles of similar charge prevents them from coagulating into larger particles. Let’s use the precipitation of AgCl from a solution of AgNO using NaCl as a precipitant to illustrate this effect. Early in the precipitation, when NaCl is the limiting reagent, excess Ag ions chemically adsorb to the AgCl particles, forming a positively charged primary adsorption layer (Figure 8.2.6
a). The solution in contact with this layer contains more inert anions, \(\text{NO}_3^-\) in this case, than inert cations, Na , giving a secondary adsorption layer with a negative charge that balances the primary adsorption layer’s positive charge. The solution outside the secondary adsorption layer remains electrically neutral. cannot occur if the secondary adsorption layer is too thick because the individual particles of AgCl are unable to approach each other closely enough. We can induce coagulation in three ways: by decreasing the number of chemically adsorbed Ag ions, by increasing the concentration of inert ions, or by heating the solution. As we add additional NaCl, precipitating more of the excess Ag , the number of chemically adsorbed silver ions decreases and coagulation occurs (Figure 8.2.6
b). Adding too much NaCl, however, creates a primary adsorption layer of excess Cl with a loss of coagulation. The coagulation and decoagulation of AgCl as we add NaCl to a solution of AgNO can serve as an endpoint for a titration. See for additional details. A second way to induce coagulation is to add an inert electrolyte, which increases the concentration of ions in the secondary adsorption layer (Figure 8.2.6
c). With more ions available, the thickness of the secondary absorption layer decreases. Particles of precipitate may now approach each other more closely, which allows the precipitate to coagulate. The amount of electrolyte needed to cause spontaneous coagulation is called the critical coagulation concentration. Heating the solution and the precipitate provides a third way to induce coagulation. As the temperature increases, the number of ions in the primary adsorption layer decreases, which lowers the precipitate’s surface charge. In addition, heating increases the particles’ kinetic energy, allowing them to overcome the electrostatic repulsion that prevents coagulation at lower temperatures. After precipitating and digesting a precipitate, we separate it from solution by filtering. The most common filtration method uses filter paper, which is classified according to its speed, its size, and its ash content on ignition. Speed, or how quickly the supernatant passes through the filter paper, is a function of the paper’s pore size. A larger pore size allows the supernatant to pass more quickly through the filter paper, but does not retain small particles of precipitate. Filter paper is rated as fast (retains particles larger than 20–25 μm), medium–fast (retains particles larger than 16 μm), medium (retains particles larger than 8 μm), and slow (retains particles larger than 2–3 μm). The proper choice of filtering speed is important. If the filtering speed is too fast, we may fail to retain some of the precipitate, which causes a negative determinate error. On the other hand, the precipitate may clog the pores if we use a filter paper that is too slow. A filter paper’s size is just its diameter. Filter paper comes in many sizes, including 4.25 cm, 7.0 cm, 11.0 cm, 12.5 cm, 15.0 cm, and 27.0 cm. Choose a size that fits comfortably into your funnel. For a typical 65-mm long-stem funnel, 11.0 cm and 12.5 cm filter paper are good choices. Because filter paper is hygroscopic, it is not easy to dry it to a constant weight. When accuracy is important, the filter paper is removed before we determine the precipitate’s mass. After transferring the precipitate and filter paper to a covered crucible, we heat the crucible to a temperature that coverts the paper to CO and H O , a process called . Igniting a poor quality filter paper leaves behind a residue of inorganic ash. For quantitative work, use a low-ash filter paper. This grade of filter paper is pretreated with a mixture of HCl and HF to remove inorganic materials. Quantitative filter paper typically has an ash content of less than 0.010% w/w. Gravity filtration is accomplished by folding the filter paper into a cone and placing it in a long-stem funnel (Figure 8.2.7
). To form a tight seal between the filter cone and the funnel, we dampen the paper with water or supernatant and press the paper to the wall of the funnel. When prepared properly, the funnel’s stem fills with the supernatant, increasing the rate of filtration. The precipitate is transferred to the filter in several steps. The first step is to decant the majority of the through the filter paper without transferring the precipitate (Figure 8.2.8
). This prevents the filter paper from clogging at the beginning of the filtration process. The precipitate is rinsed while it remains in its beaker, with the rinsings decanted through the filter paper. Finally, the precipitate is transferred onto the filter paper using a stream of rinse solution. Any precipitate that clings to the walls of the beaker is transferred using a rubber policeman (a flexible rubber spatula attached to the end of a glass stirring rod). An alternative method for filtering a precipitate is to use a filtering crucible. The most common option is a fritted-glass crucible that contains a porous glass disk filter. Fritted-glass crucibles are classified by their porosity: coarse (retaining particles larger than 40–60 μm), medium (retaining particles greater than 10–15 μm), and fine (retaining particles greater than 4–5.5 μm). Another type of filtering crucible is the Gooch crucible, which is a porcelain crucible with a perforated bottom. A glass fiber mat is placed in the crucible to retain the precipitate. For both types of crucibles, the pre- cipitate is transferred in the same manner described earlier for filter paper. Instead of using gravity, the supernatant is drawn through the crucible with the assistance of suction from a vacuum aspirator or pump (Figure 8.2.9
). Because the supernatant is rich with dissolved inert ions, we must remove residual traces of supernatant without incurring loss of analyte due to solubility. In many cases this simply involves the use of cold solvents or rinse solutions that contain organic solvents such as ethanol. The pH of the rinse solution is critical if the precipitate contains an acidic or a basic ion. When coagulation plays an important role in determining particle size, adding a volatile inert electrolyte to the rinse solution prevents the precipitate from reverting into smaller particles that might pass through the filter. This process of reverting to smaller particles is called . The volatile electrolyte is removed when drying the precipitate. In general, we can minimize the loss of analyte if we use several small portions of rinse solution instead of a single large volume. Testing the used rinse solution for the presence of an impurity is another way to guard against over-rinsing the precipitate. For example, if Cl is a residual ion in the supernatant, we can test for its presence using AgNO . After we collect a small portion of the rinse solution, we add a few drops of AgNO and look for the presence or absence of a precipitate of AgCl. If a precipitate forms, then we know Cl is present and continue to rinse the precipitate. Additional rinsing is not needed if the AgNO does not produce a precipitate. After separating the precipitate from its supernatant solution, we dry the precipitate to remove residual traces of rinse solution and to remove any volatile impurities. The temperature and method of drying depend on the method of filtration and the precipitate’s desired chemical form. Placing the precipitate in a laboratory oven and heating to a temperature of 110 C is sufficient to remove water and other easily volatilized impurities. Higher temperatures require a muffle furnace, a Bunsen burner, or a Meker burner, and are necessary if we need to decompose the precipitate before its weight is determined. Because filter paper absorbs moisture, we must remove it before we weigh the precipitate. This is accomplished by folding the filter paper over the precipitate and transferring both the filter paper and the precipitate to a porcelain or platinum crucible. Gentle heating first dries and then chars the filter paper. Once the paper begins to char, we slowly increase the temperature until there is no trace of the filter paper and any remaining carbon is oxidized to CO . Fritted-glass crucibles can not withstand high temperatures and are dried in an oven at a temperature below 200 C. The glass fiber mats used in Gooch crucibles can be heated to a maximum temperature of approximately 500 C. For a quantitative application, the final precipitate must have a well-defined composition. A precipitate that contains volatile ions or substantial amounts of hydrated water, usually is dried at a temperature that completely removes these volatile species. For example, one standard gravimetric method for the determination of magnesium involves its precipitation as MgNH PO •6H O. Unfortunately, this precipitate is difficult to dry at lower temperatures without losing an inconsistent amount of hydrated water and ammonia. Instead, the precipitate is dried at a temperature greater than 1000 C where it decomposes to magnesium pyrophosphate, Mg P O . An additional problem is encountered if the isolated solid is nonstoichiometric. For example, precipitating Mn as Mn(OH) and heating frequently produces a nonstoichiometric manganese oxide, MnO , where varies between one and two. In this case the nonstoichiometric product is the result of forming a mixture of oxides with different oxidation state of manganese. Other nonstoichiometric compounds form as a result of lattice defects in the crystal structure [Ward, R., ed., , American Chemical Society: Washington, D. C., 1963]. The best way to appreciate the theoretical and practical details discussed in this section is to carefully examine a typical precipitation gravimetric method. Although each method is unique, the determination of Mg in water and wastewater by precipitating MgNH PO • 6H O and isolating Mg P O provides an instructive example of a typical procedure. The description here is based on Method 3500-Mg D in , 19th Ed., American Public Health Asso- ciation: Washington, D. C., 1995. With the publication of the 20th Edition in 1998, this method is no longer listed as an approved method. Magnesium is precipitated as MgNH PO •6H O using (NH ) HPO as the precipitant. The precipitate’s solubility in a neutral solution is relatively high (0.0065 g/100 mL in pure water at 10 C), but it is much less soluble in the presence of dilute ammonia (0.0003 g/100 mL in 0.6 M NH ). Because the precipitant is not selective, a preliminary separation of Mg from potential interferents is necessary. Calcium, which is the most significant interferent, is removed by precipitating it as CaC O . The presence of excess ammonium salts from the precipitant, or from the addition of too much ammonia, leads to the formation of Mg(NH ) (PO ) , which forms Mg(PO ) after drying. The precipitate is isolated by gravity filtration, using a rinse solution of dilute ammonia. After filtering, the precipitate is converted to Mg P O and weighed. Transfer a sample that contains no more than 60 mg of Mg into a 600-mL beaker. Add 2–3 drops of methyl red indicator, and, if necessary, adjust the volume to 150 mL. Acidify the solution with 6 M HCl and add 10 mL of 30% w/v (NH ) HPO . After cooling and with constant stirring, add concentrated NH dropwise until the methyl red indicator turns yellow (pH > 6.3). After stirring for 5 min, add 5 mL of concentrated NH and continue to stir for an additional 10 min. Allow the resulting solution and precipitate to stand overnight. Isolate the precipitate by filtering through filter paper, rinsing with 5% v/v NH . Dissolve the precipitate in 50 mL of 10% v/v HCl and precipitate a second time following the same procedure. After filtering, carefully remove the filter paper by charring. Heat the precipitate at 500 C until the residue is white, and then bring the precipitate to constant weight at 1100 C. 1. Why does the procedure call for a sample that contains no more than 60 mg of Mg ? A 60-mg portion of Mg generates approximately 600 mg of MgNH PO •6H O, which is a substantial amount of precipitate. A larger quantity of precipitate is difficult to filter and difficult to rinse free of impurities. 2. Why is the solution acidified with HCl before we add the precipitant? The HCl ensures that MgNH PO • 6H O does not precipitate immediately upon adding the precipitant. Because \(\text{PO}_4^{3-}\) is a weak base, the precipitate is soluble in a strongly acidic solution. If we add the precipitant under neutral or basic conditions (that is, a high ), then the resulting precipitate will consist of smaller, less pure particles. Increasing the pH by adding base allows the precipitate to form under more favorable (that is, a low ) conditions. 3. Why is the acid–base indicator methyl red added to the solution? The indicator changes color at a pH of approximately 6.3, which indicates that there is sufficient NH to neutralize the HCl added at the beginning of the procedure. The amount of NH is crucial to this procedure. If we add insufficient NH , then the solution is too acidic, which increases the precipitate’s solubility and leads to a negative determinate error. If we add too much NH , the precipitate may contain traces of Mg(NH ) (PO ) , which, on drying, forms Mg(PO ) instead of Mg P O . This increases the mass of the ignited precipitate, and gives a positive determinate error. After adding enough NH to neutralize the HCl, we add an additional 5 mL of NH to complete the quantitative precipitation of MgNH PO • 6H O. 4. Explain why forming Mg(PO ) instead of Mg P O increases the precipitate’s mass. Each mole of Mg P O contains two moles of magnesium and each mole of Mg(PO ) contains only one mole of magnesium. A conservation of mass, therefore, requires that two moles of Mg(PO ) form in place of each mole of Mg P O . One mole of Mg P O weighs 222.6 g. Two moles of Mg(PO ) weigh 364.5 g. Any replacement of Mg P O with Mg(PO ) must increase the precipitate’s mass. 5. What additional steps, beyond those discussed in questions 2 and 3, help improve the precipitate’s purity? Two additional steps in the procedure help to form a precipitate that is free of impurities: digestion and reprecipitation. 6. Why is the precipitate rinsed with a solution of 5% v/v NH ? This is done for the same reason that the precipitation is carried out in an ammonical solution; using dilute ammonia minimizes solubility losses when we rinse the precipitate. Although no longer a common analytical technique, precipitation gravimetry still provides a reliable approach for assessing the accuracy of other methods of analysis, or for verifying the composition of standard reference materials. In this section we review the general application of precipitation gravimetry to the analysis of inorganic and organic compounds. Table 8.2.1
provides a summary of precipitation gravimetric methods for inorganic cations and anions. Several methods for the homogeneous generation of precipitants are shown in Table 8.2.2
. The majority of inorganic precipitants show poor selectivity for the analyte. Many organic precipitants, however, are selective for one or two inorganic ions. Table 8.2.3
lists examples of several common organic precipitants. Ba \(\text{SO}_4^{2-}\) Precipitation gravimetry continues to be listed as a standard method for the determination of \(\text{SO}_4^{2-}\) in water and wastewater analysis [Method 4500-SO42– C and Method 4500-SO42– D as published in , 20th Ed., American Public Health Association: Wash- ington, D. C., 1998]. Precipitation is carried out using BaCl in an acidic solution (adjusted with HCl to a pH of 4.5–5.0) to prevent the precipitation of BaCO or Ba (PO ) , and at a temperature near the solution’s boiling point. The precipitate is digested at 80–90 C for at least two hours. Ashless filter paper pulp is added to the precipitate to aid in its filtration. After filtering, the precipitate is ignited to constant weight at 800 C. Alternatively, the precipitate is filtered through a fine porosity fritted glass crucible (without adding filter paper pulp), and dried to constant weight at 105 C. This procedure is subject to a variety of errors, including occlusions of Ba(NO ) , BaCl , and alkali sulfates. Other standard methods for the determination of sulfate in water and wastewater include ion chromatography (see ), capillary ion electrophoresis (see ), turbidimetry (see ), and flow injection analysis (see ). Several organic functional groups or heteroatoms can be determined using precipitation gravimetric methods. Table 8.2.4
provides a summary of several representative examples. Note that the determination of alkoxy functional groups is an indirect analysis in which the functional group reacts with and excess of HI and the unreacted I determined by precipitating as AgCl. The stoichiometry of a precipitation reaction provides a mathematical relationship between the analyte and the precipitate. Because a precipitation gravimetric method may involve additional chemical reactions to bring the analyte into a different chemical form, knowing the stoichiometry of the precipitation reaction is not always sufficient. Even if you do not have a complete set of balanced chemical reactions, you can use a conservation of mass to deduce the mathematical relationship between the analyte and the precipitate. The following example demonstrates this approach for the direct analysis of a single analyte. To determine the amount of magnetite, Fe O , in an impure ore, a 1.5419-g sample is dissolved in concentrated HCl, resulting in a mixture of Fe and Fe . After adding HNO to oxidize Fe to Fe and diluting with water, Fe is precipitated as Fe(OH) using NH . Filtering, rinsing, and igniting the precipitate provides 0.8525 g of pure Fe O . Calculate the %w/w Fe O in the sample. A conservation of mass requires that the precipitate of Fe O contain all iron originally in the sample of ore. We know there are 2 moles of Fe per mole of Fe O (FW = 159.69 g/mol) and 3 moles of Fe per mole of Fe O (FW = 231.54 g/mol); thus \[0.8525 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} \times \frac{2 \ \mathrm{mol} \ \mathrm{Fe}}{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}} \times \frac{231.54 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} \nonumber\] The % w/w Fe O in the sample, therefore, is \[\frac{0.82405 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{1.5419 \ \mathrm{g} \ \text { sample }} \times 100=53.44 \% \nonumber\] A 0.7336-g sample of an alloy that contains copper and zinc is dissolved in 8 M HCl and diluted to 100 mL in a volumetric flask. In one analysis, the zinc in a 25.00-mL portion of the solution is precipitated as ZnNH PO , and isolated as Zn P O , yielding 0.1163 g. The copper in a separate 25.00-mL portion of the solution is treated to precipitate CuSCN, yielding 0.2383 g. Calculate the %w/w Zn and the %w/w Cu in the sample. A conservation of mass requires that all zinc in the alloy is found in the final product, Zn P O . We know there are 2 moles of Zn per mole of Zn P O ; thus \[0.1163 \ \mathrm{g} \ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7} \times \frac{2 \ \mathrm{mol} \ \mathrm{Zn}}{304.70 \ \mathrm{g}\ \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}} \times \frac{65.38 \ \mathrm{g} \ \mathrm{Zn}}{\mathrm{mol} \ \mathrm{Zn}}=0.04991 \ \mathrm{g} \ \mathrm{Zn}\nonumber\] This is the mass of Zn in 25% of the sample (a 25.00 mL portion of the 100.0 mL total volume). The %w/w Zn, therefore, is \[\frac{0.04991 \ \mathrm{g} \ \mathrm{Zn} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=27.21 \% \ \mathrm{w} / \mathrm{w} \mathrm{Zn} \nonumber\] For copper, we find that \[\begin{array}{c}{0.2383 \ \mathrm{g} \ \mathrm{CuSCN} \times \frac{1 \ \mathrm{mol} \ \mathrm{Zn}}{121.63 \ \mathrm{g} \ \mathrm{CuSCN}} \times \frac{63.55 \ \mathrm{g} \ \mathrm{Cu}}{\mathrm{mol} \ \mathrm{Cu}}=0.1245 \ \mathrm{g} \ \mathrm{Cu}} \\ {\frac{0.1245 \ \mathrm{g} \ \mathrm{Cu} \times 4}{0.7336 \ \mathrm{g} \text { sample }} \times 100=67.88 \% \ \mathrm{w} / \mathrm{w} \mathrm{Cu}}\end{array} \nonumber\] In Practice Exercise 8.2.2
the sample contains two analytes. Because we can precipitate each analyte selectively, finding their respective concentrations is a straightforward stoichiometric calculation. But what if we cannot separately precipitate the two analytes? To find the concentrations of both analytes, we still need to generate two precipitates, at least one of which must contain both analytes. Although this complicates the calculations, we can still use a conservation of mass to solve the problem. A 0.611-g sample of an alloy that contains Al and Mg is dissolved and treated to prevent interferences by the alloy’s other constituents. Aluminum and magnesium are precipitated using 8-hydroxyquinoline, which yields a mixed precipitate of Al(C H NO) and Mg(C H NO) that weighs 7.815 g. Igniting the precipitate converts it to a mixture of Al O and MgO that weighs 1.002 g. Calculate the %w/w Al and %w/w Mg in the alloy. The masses of the solids provide us with the following two equations. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{g} \ \mathrm{MgO}=1.002 \ \mathrm{g} \nonumber\] With two equations and four unknowns, we need two additional equations to solve the problem. A conservation of mass requires that all the aluminum in Al(C H NO) also is in Al O ; thus \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.43 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{101.96 \ \mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Al}_{2} \mathrm{O}_{3}} \nonumber\] \[\mathrm{g} \ \mathrm{Al}_{2} \mathrm{O}_{3}=0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \nonumber\] Using the same approach, a conservation of mass for magnesium gives \[\mathrm{g} \ \mathrm{MgO}=\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{40.304 \ \mathrm{g} \ \mathrm{MgO}}{\mathrm{mol} \ \mathrm{MgO}} \nonumber\] \[\mathrm{g} \ \mathrm{MgO}=0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2} \nonumber\] Substituting the equations for g MgO and g Al O into the equation for the combined weights of MgO and Al O leaves us with two equations and two unknowns. \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.815 \ \mathrm{g} \nonumber\] \[0.11096 \times \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}+ 0.12893 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=1.002 \ \mathrm{g} \nonumber\] Multiplying the first equation by 0.11096 and subtracting the second equation gives \[-0.01797 \times \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=-0.1348 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{2}=7.504 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}=7.815 \ \mathrm{g}-7.504 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}, \mathrm{H}_{6} \mathrm{NO}\right)_{2}=0.311 \ \mathrm{g} \nonumber\] Now we can finish the problem using the approach from . A conservation of mass requires that all the aluminum and magnesium in the original sample of Dow metal is in the precipitates of Al(C H NO) and the Mg(C H NO) . For aluminum, we find that \[0.311 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Al}}{459.45 \ \mathrm{g} \ \mathrm{Al}\left(\mathrm{C}_{9} \mathrm{H}_{6} \mathrm{NO}\right)_{3}} \times \frac{26.982 \ \mathrm{g} \ \mathrm{Al}}{\mathrm{mol} \ \mathrm{Al}}=0.01826 \ \mathrm{g} \ \mathrm{Al} \nonumber\] \[\frac{0.01826 \ \mathrm{g} \ \mathrm{Al}}{0.611 \ \mathrm{g} \text { sample }} \times 100=2.99 \% \mathrm{w} / \mathrm{w} \mathrm{Al} \nonumber\] and for magnesium we have \[7.504 \ \text{g Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mg}}{312.61 \ \mathrm{g} \ \mathrm{Mg}\left(\mathrm{C}_9 \mathrm{H}_{6} \mathrm{NO}\right)_{2}} \times \frac{24.305 \ \mathrm{g} \ \mathrm{Mg}}{\mathrm{mol} \ \mathrm{MgO}}=0.5834 \ \mathrm{g} \ \mathrm{Mg} \nonumber\] \[\frac{0.5834 \ \mathrm{g} \ \mathrm{Mg}}{0.611 \ \mathrm{g} \text { sample }} \times 100=95.5 \% \mathrm{w} / \mathrm{w} \mathrm{Mg} \nonumber\] A sample of a silicate rock that weighs 0.8143 g is brought into solution and treated to yield a 0.2692-g mixture of NaCl and KCl. The mixture of chloride salts is dissolved in a mixture of ethanol and water, and treated with HClO , precipitating 0.3314 g of KClO . What is the %w/w Na O in the silicate rock? The masses of the solids provide us with the following equations \[\mathrm{g} \ \mathrm{NaCl}+\mathrm{g} \ \mathrm{KCl}=0.2692 \ \mathrm{g} \nonumber\] \[\mathrm{g} \ \mathrm{KClO}_{4} = 0.3314 \ \mathrm{g} \nonumber\] With two equations are three unknowns—g NaCl, g KCl, and g KClO —we need one additional equation to solve the problem. A conservation of mass requires that all the potassium originally in the KCl ends up in the KClO ; thus \[\text{g KClO}_4 = \text{g KCl} \times \frac{1 \text{ mol Cl}}{74.55 \text{ g KCl}} \times \frac {138.55 \text{ g KClO}_4}{\text{mol Cl}} = 1.8585 \times \text{ g KCl} \nonumber\] Given the mass of KClO , we use the third equation to solve for the mass of KCl in the mixture of chloride salts \[\text{ g KCl} = \frac{\text{g KClO}_4}{1.8585} = \frac{0.3314 \text{ g}}{1.8585} = 0.1783 \text{ g KCl} \nonumber\] The mass of NaCl in the mixture of chloride salts, therefore, is \[\text{ g NaCl} = 0.2692 \text{ g} - \text{g KCl} = 0.2692 \text{ g} - 0.1783 \text{ g KCl} = 0.0909 \text{ g NaCl} \nonumber\] Finally, to report the %w/w Na O in the sample, we use a conservation of mass on sodium to determine the mass of Na O \[0.0909 \text{ g NaCl} \times \frac{1 \text{ mol Na}}{58.44 \text{ g NaCl}} \times \frac{61.98 \text{ g Na}_2\text{O}}{2 \text{ mol Na}} = 0.0482 \text{ g Na}_2\text{O} \nonumber\] giving the %w/w Na O as \[\frac{0.0482 \text{ g Na}_2\text{O}}{0.8143 \text{ g sample}} \times 100 = 5.92\% \text{ w/w Na}_2\text{O} \nonumber\] The previous problems are examples of direct methods of analysis because the precipitate contains the analyte. In an indirect analysis the precipitate forms as a result of a reaction with the analyte, but the analyte is not part of the precipitate. As shown by the following example, despite the additional complexity, we still can use conservation principles to organize our calculations. An impure sample of Na PO that weighs 0.1392 g is dissolved in 25 mL of water. A second solution that contains 50 mL of 3% w/v HgCl , 20 mL of 10% w/v sodium acetate, and 5 mL of glacial acetic acid is prepared. Adding the solution that contains the sample to the second solution oxidizes \(\text{PO}_3^{3-}\) to \(\text{PO}_4^{3-}\) and precipitates Hg Cl . After digesting, filtering, and rinsing the precipitate, 0.4320 g of Hg Cl is obtained. Report the purity of the original sample as % w/w Na PO . This is an example of an indirect analysis because the precipitate, Hg Cl , does not contain the analyte, Na PO . Although the stoichiometry of the reaction between Na PO and HgCl is given earlier in the chapter, let’s see how we can solve the problem using conservation principles. ( ) The reaction between Na PO and HgCl is an oxidation-reduction reaction in which phosphorous increases its oxidation state from +3 in Na PO to +5 in Na PO and in which mercury decreases its oxidation state from +2 in HgCl to +1 in Hg Cl . A redox reaction must obey a conservation of electrons because all the electrons released by the reducing agent, Na PO , must be accepted by the oxidizing agent, HgCl . Knowing this, we write the following stoichiometric conversion factors: \[\frac{2 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}} \text { and } \frac{1 \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \nonumber\] Now we are ready to solve the problem. First, we use a conservation of mass for mercury to convert the precipitate’s mass to the moles of HgCl . \[0.4320 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2} \times \frac{2 \ \mathrm{mol} \ \mathrm{Hg}}{472.09 \ \mathrm{g} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{HgCl}_{2}}{\mathrm{mol} \ \mathrm{Hg}}=1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \nonumber\] Next, we use the conservation of electrons to find the mass of Na PO . \[1.8302 \times 10^{-3} \ \mathrm{mol} \ \mathrm{HgCl}_{2} \times \frac{1 \ \mathrm{mol} \ e^{-}}{\mathrm{mol} \ \mathrm{HgCl}_{2}} \times \frac{1 \ \mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{2 \ \mathrm{mol} \ e^{-}} \times \frac{147.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}=0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] Finally, we calculate the %w/w Na PO in the sample. \[\frac{0.13538 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{0.1392 \ \mathrm{g} \text { sample }} \times 100=97.26 \% \mathrm{w} / \mathrm{w} \mathrm{Na}_{3} \mathrm{PO}_{3} \nonumber\] As you become comfortable using conservation principles, you will see ways to further simplify problems. For example, a conservation of electrons requires that the electrons released by Na PO end up in the product, Hg Cl , yielding the following stoichiometric conversion factor: \[\frac{2 \ \operatorname{mol} \ \mathrm{Na}_{3} \mathrm{PO}_{3}}{\mathrm{mol} \ \mathrm{Hg}_{2} \mathrm{Cl}_{2}} \nonumber\] This conversion factor provides a direct link between the mass of Hg Cl and the mass of Na PO . One approach for determining phosphate, \(\text{PO}_4^{3-}\), is to precipitate it as ammonium phosphomolybdate, (NH ) PO •12MoO . After we isolate the precipitate by filtration, we dissolve it in acid and precipitate and weigh the molybdate as PbMoO . Suppose we know that our sample is at least 12.5% Na PO and that we need to recover a minimum of 0.600 g of PbMoO ? What is the minimum amount of sample that we need for each analysis? To find the mass of (NH ) PO •12MoO that will produce 0.600 g of PbMoO , we first use a conservation of mass for molybdenum; thus \[0.600 \ \mathrm{g} \ \mathrm{PbMoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{Mo}}{351.2 \ \mathrm{g} \ \mathrm{PbMoO}_{3}} \times \frac{1876.59 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}}{12 \ \mathrm{mol} \ \mathrm{Mo}}= 0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \nonumber\] Next, to convert this mass of (NH ) PO •12MoO to a mass of Na PO , we use a conservation of mass on \(\text{PO}_4^{3-}\). \[0.2672 \ \mathrm{g} \ \left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \times \frac{1 \ \mathrm{mol} \ \mathrm{PO}_{4}^{3-}}{1876.59 \ \mathrm{g \ }\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3}} \times \frac{163.94 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}{\mathrm{mol} \ \mathrm{PO}_{4}^{3-}}=0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \nonumber\] Finally, we convert this mass of Na PO to the corresponding mass of sample. \[0.02334 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4} \times \frac{100 \ \mathrm{g} \text { sample }}{12.5 \ \mathrm{g} \ \mathrm{Na}_{3} \mathrm{PO}_{4}}=0.187 \ \mathrm{g} \text { sample } \nonumber\] A sample of 0.187 g is sufficient to guarantee that we recover a minimum of 0.600 g PbMoO . If a sample contains more than 12.5% Na PO , then a 0.187-g sample will produce more than 0.600 g of PbMoO . A precipitation reaction is a useful method for identifying inorganic and organic analytes. Because a qualitative analysis does not require quantitative measurements, the analytical signal is simply the observation that a precipitate forms. Although qualitative applications of precipitation gravimetry have been replaced by spectroscopic methods of analysis, they continue to find application in spot testing for the presence of specific analytes [Jungreis, E. ; 2nd Ed., Wiley: New York, 1997]. Any of the precipitants listed in , , and can be used for a qualitative analysis. The scale of operation for precipitation gravimetry is limited by the sensitivity of the balance and the availability of sample. To achieve an accuracy of ±0.1% using an analytical balance with a sensitivity of ±0.1 mg, we must isolate at least 100 mg of precipitate. As a consequence, precipitation gravimetry usually is limited to major or minor analytes, in macro or meso samples. The analysis of a trace level analyte or a micro sample requires a microanalytical balance. For a macro sample that contains a major analyte, a relative error of 0.1– 0.2% is achieved routinely. The principal limitations are solubility losses, impurities in the precipitate, and the loss of precipitate during handling. When it is difficult to obtain a precipitate that is free from impurities, it often is possible to determine an empirical relationship between the precipitate’s mass and the mass of the analyte by an appropriate calibration. The relative precision of precipitation gravimetry depends on the sample’s size and the precipitate’s mass. For a smaller amount of sample or precipitate, a relative precision of 1–2 ppt is obtained routinely. When working with larger amounts of sample or precipitate, the relative precision extends to several ppm. Few quantitative techniques can achieve this level of precision. For any precipitation gravimetric method we can write the following general equation to relate the signal (grams of precipitate) to the absolute amount of analyte in the sample \[\text { g precipitate }=k \times \mathrm{g} \text { analyte } \label{8.13}\] where , the method’s sensitivity, is determined by the stoichiometry between the precipitate and the analyte. Equation \ref{8.13} assumes we used a suitable blank to correct the signal for any contributions of the reagent to the precipitate’s mass. Consider, for example, the determination of Fe as Fe O . Using a conservation of mass for iron, the precipitate’s mass is \[\mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{\text{AW Fe}} \times \frac{\text{FW Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}} \nonumber\] and the value of is \[k=\frac{1}{2} \times \frac{\mathrm{FW} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{\mathrm{AW} \ \mathrm{Fe}} \label{8.14}\] As we can see from Equation \ref{8.14}, there are two ways to improve a method’s sensitivity. The most obvious way to improve sensitivity is to increase the ratio of the precipitate’s molar mass to that of the analyte. In other words, it helps to form a precipitate with the largest possible formula weight. A less obvious way to improve a method’s sensitivity is indicated by the term of 1/2 in Equation \ref{8.14}, which accounts for the stoichiometry between the analyte and precipitate. We can also improve sensitivity by forming a precipitate that contains fewer units of the analyte. Suppose you wish to determine the amount of iron in a sample. Which of the following compounds—FeO, Fe O , or Fe O —provides the greatest sensitivity? To determine which form has the greatest sensitivity, we use a conservation of mass for iron to find the relationship between the precipitate’s mass and the mass of iron. \[\begin{aligned} \mathrm{g} \ \mathrm{FeO} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{71.84 \ \mathrm{g} \ \mathrm{FeO}}{\mathrm{mol} \ \mathrm{Fe}}=1.286 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{159.69 \ \mathrm{g} \ \mathrm{Fe}_{2} \mathrm{O}_{3}}{2 \ \mathrm{mol} \ \mathrm{Fe}}=1.430 \times \mathrm{g} \ \mathrm{Fe} \\ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4} &=\mathrm{g} \ \mathrm{Fe} \times \frac{1 \ \mathrm{mol} \ \mathrm{Fe}}{55.85 \ \mathrm{g} \ \mathrm{Fe}} \times \frac{231.53 \ \mathrm{g} \ \mathrm{Fe}_{3} \mathrm{O}_{4}}{3 \ \mathrm{mol} \ \mathrm{Fe}}=1.382 \times \mathrm{g} \ \mathrm{Fe} \end{aligned} \nonumber\] Of the three choices, the greatest sensitivity is obtained with Fe O because it provides the largest value for . Due to the chemical nature of the precipitation process, precipitants usually are not selective for a single analyte. For example, silver is not a selective precipitant for chloride because it also forms precipitates with bromide and with iodide. Interferents often are a serious problem and must be considered if accurate results are to be obtained. Precipitation gravimetry is time intensive and rarely practical if you have a large number of samples to analyze; however, because much of the time invested in precipitation gravimetry does not require an analyst’s immediate supervision, it is a practical alternative when working with only a few samples. Equipment needs are few—beakers, filtering devices, ovens or burners, and balances—inexpensive, routinely available in most laboratories, and easy to maintain. | 59,802 | 3,361 |
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Often you will know the concentrations of reactants and products for a particular reaction and want to know whether the system is at equilibrium. If it is not, it is useful to predict how those concentrations will change as the reaction approaches equilibrium. A useful tool for making such predictions is the , . has the same mathematical form as the equilibrium-constant expression, but is a ratio of the actual concentrations (not a ratio of equilibrium concentrations). For example, suppose you are interested in the reaction \[2 \text{SO}_{2} (g) + \text{O}_{2} (g) \rightleftharpoons 2 \text{SO}_{3} (g) \nonumber \] \[ K_{\text{c}} = \frac{[\text{SO}_{3}]^{2}}{[\text{SO}_{2}]^{2} [\text{O}_{2}]} = 245 \text{ mol/L} \text{(at 1000 K)} \nonumber \] and you have added 0.10 mol of each gas to a container with volume 10.0 L. Is the system at equilibrium? If not, will the concentration of SO be greater than or less than 0.010 mol/L when equilibrium is reached? You can answer these questions by calculating and comparing it with . There are three possibilities: For the reaction given above, \[ Q = \frac{[\text{SO}_{3}]^{2}}{[\text{SO}_{2}]^{2} [\text{O}_{2}]} = \frac{(\frac{0.1 mol}{10 L})^{2}}{(\frac{0.1 mol}{10 L})^{2} (\frac{0.1 mol}{10 L})} = 100 \frac{\text{mol}}{\text{L}} \nonumber \] (In the expression for each actual concentration is enclosed in braces {curly brackets} in order to distinguish it from the equilibrium concentrations, which, in the expression, are enclosed in [square brackets].) In this case = 100 mol/L. This is less than , which has the value 245 mol/L. This implies that the concentrations of products are less than the equilibrium concentrations (and the concentrations of reactants are greater than the equilibrium concentrations). Therefore the reaction will proceed in the forward direction, producing more products, until the concentrations reach their equilibrium values. At 2300 K, the equilibrium constant, , is 1.7 x 10 for the reaction
\[ \text{N}_{2} (g) + \text{O}_{2} (g) \rightleftharpoons 2 \text{NO} (g) \nonumber \]
A mixture of the three gases at 2300 K has these concentrations, [N ] = 0.17 mol dm , [O ] = 0.17 mol dm , and [NO] = 0.034 mol dm .
(a) Is the system at equilibrium?
(b) In which direction must the reaction occur to reach equilibrium?
(c) What are the equilibrium concentrations of N , O , and NO? Use the known concentrations to calculate . Compare with to answer questions (a) and (b). Use an ICE table to answer part (c). \[Q = \frac{\{\text{NO\}}^{2}}{\{\text{N}_{2}\}\{\text{O}_{2}\}} = \frac{(0.034 \text{mol dm}^{-3})^{2}}{(0.17 \text{mol dm}^{-3})(0.17 \text{mol dm}^{-3})} = 4.0 \times 10^{-2} \nonumber \] Next, substitute the equilibrium concentrations into the expression and solve for . \[K_{\text{c}} = 1.7 \times {10^{ - 3}} = \frac{(0.034 - 2x)^{2}}{(0.17 + x)(0.1 + x)} \nonumber \] Now take the square root of both sides of this equation. This gives \(\sqrt {1.7 \times {10^{ - 3}}} = 0.0412 = \dfrac{0.034 - 2x}{0.17 + x}\) Multiplying both sides by 0.17 + gives \( 0.0070 + 0.412 x = 0.034 - 2 x \) \( 2 x + 0.0412 x =0.034 - 0.0070 \) \(x = \dfrac{0.0270}{2.0412} = 0.0132\) \( \text{[N}_{2} ] = \text{[O}_{2} ] = 0.17 + 0.013 = 0.183 \text{ mol dm}^{-3} \) \( \text{[NO] } = 0.034 - 2(0.0132) = 0.0076 \text{ mol dm}^{-3} \) Check the result by substituting these concentrations into the equilibrium constant expression. \[K_{\text{c}} = \frac{(0.0076)^{2}}{(0.18)(0.18)} = 1.8 \times {10^{ - 3}} \nonumber \] This agrees to two significant figures with the value of 1.7 x 10 . | 3,636 | 3,362 |
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The existence of coordination compounds with the same formula but different arrangements of the ligands was crucial in the development of coordination chemistry. Two or more compounds with the same formula but different arrangements of the atoms are called . Because isomers usually have different physical and chemical properties, it is important to know which isomer we are dealing with if more than one isomer is possible. As we will see, coordination compounds exhibit the same types of isomers as organic compounds, as well as several kinds of isomers that are unique. Isomers are compounds with the same molecular formula but different structural formulas and do not necessarily share similar properties. There are many different classes of isomers, like stereoisomers, enantiomers, and geometrical isomers. There are two main forms of isomerism: and (spatial isomerism). Isomers that contain the same number of atoms of each kind but differ in which atoms are bonded to one another are called structural isomers, which differ in structure or bond type. For inorganic complexes, there are three types of structural isomers: , and . Structural isomers, as their name implies, differ in their structure or bonding, which are separate from that differ in the spatial arrangement of the ligands are attached, but still have the bonding properties. The different chemical formulas in structural isomers are caused either by a difference in what ligands are bonded to the central atoms or how the individual ligands are bonded to the central atoms. When determining a structural isomer, you look at (1) the ligands that are bonded to the central metal and (2) which atom of the ligands attach to the central metal. Ionization isomers occur when a ligand that is bound to the metal center exchanges places with an anion or neutral molecule that was originally outside the coordination complex. The geometry of the central metal ion and the identity of other ligands are identical. For example, an octahedral isomer will have five ligands that are identical, but the sixth will differ. The non-matching ligand in one compound will be outside of the coordination sphere of the other compound. Because the anion or molecule outside the coordination sphere is different, the chemical properties of these isomers is different. A hydrate isomer is a specific kind of ionization isomer where a water molecule is one of the molecules that exchanges places. The difference between the ionization isomers can be view within the context of the ions generated when each are dissolved in solution. For example, when pentaamminebromocobalt(II) chloride is dissolved in water, \(\ce{Cl^{-}}\) ions are generated: \[\ce{CoBr(NH_3)_5Cl {(s)} \rightarrow CoBr(NH_3)^{+}5 (aq) + Cl^{-} (aq)} \] whereas when pentaamminechlorocobalt(II) bromide is dissolved, \(\ce{Br^{-}}\) ions are generated: \[\ce{CoCl(NH_3)_5Br {(s)} \rightarrow CoCl(NH_3)^{+}_{5} (aq) + Br^{-} (aq)}\] Notice that both anions are necessary to balance the charge of the complex, and that they differ in that one ion is directly attached to the central metal, but the other is not. A very similar type of isomerism results from replacement of a coordinated group by a solvent molecule ( ), which in the case of water is called . The best known example of this occurs for chromium chloride (\(\ce{CrCl_3 \cdot 6H_2O}\)) which may contain 4, 5, or 6 coordinated water molecules (assuming a of 6). The dot here is used essentially as an expression of ignorance to indicate that, though the parts of the molecule separated by the dot are bonded to one another in some fashion, the exact structural details of that interaction are not fully expressed in the resulting formula. Alfred Werner’s coordination theory indicates that several of the water molecules are actually bonded directly (via coordinate covalent bonds) to the central chromium ion. In fact, there are several possible compounds that use the brackets to signify bonding in the complex and the dots to signify "water molecules that are not bound to the central metal, but are part of the lattice: These isomers have very different chemical properties and on reaction with \(\ce{AgNO_3}\) to test for \(\ce{Cl^{-}}\) ions, would find 1, 2, and 3 \(\ce{Cl^{-}}\) ions in solution, respectively. Upon crystallization from water, many compounds incorporate water molecules in their crystalline frameworks. These "waters of crystallization" refers to water that is found in the crystalline framework of a metal complex or a salt, which is to the metal cation. In the first two hydrate isomers, there are water molecules that are artifacts of the crystallization and occur inside crystals. These waters of crystallization contribute to the total weight of water in a substance and are mostly present in a definite (stoichiometric) ratio. A compound with associated water of crystallization is known as a hydrate. The structure of hydrates can be quite elaborate, because of the existence of hydrogen bonds that define polymeric structures. For example, consider the aquo complex \(\ce{NiCl_2 \cdot 6H_2O}\) that consists of separated -[NiCl (H O) ] molecules linked more weakly to adjacent water molecules. Only four of the six water molecules in the formula are bound to the nickel (II) cation, and the remaining two are waters of crystallization as the crystal structure resolves (Figure \(\Page {3}\)). Water is particularly common solvent to be found in crystals because it is small and polar. But solvents can be found in some host crystals. Water is noteworthy because it is reactive, whereas other solvents such as benzene are considered to be chemically innocuous. Coordination isomerism occurs in compounds containing complex anionic and complex cationic parts and can be viewed as an interchange of some ligands from the cation to the anion. Hence, there are two complex compounds bound together, one with a negative charge and the other with a positive charge. In coordination isomers, the anion and cation complexes of a coordination compound exchange one or more ligands. For example, the \(\ce{[Zn(NH3)4] [Cu(Cl4)]}\) and \(\ce{[Cu(NH3)4] [Zn(Cl4)]}\) compounds are coordination isomers (Figure \(\Page {4}\)). What is the coordination isomer for the \(\ce{[Co(NH3)6] [Cr(CN)6]}\) compound? Coordination isomerism involves switching the metals between the cation and anion spheres. Hence the \(\ce{[Cr(NH3)6,Co(CN)6] }\) compound is a coordination isomer of \(\ce{[Co(NH3)6] [Cr(CN)6]}\). Linkage isomerism occurs with ligands that are capable of coordinating in more than one way. The best known cases involve the monodentate ligands: \(\ce{SCN^{-} / NCS^{-}}\) and \(\ce{NO_2^{-} / ONO^{-}}\). The only difference is what . The ligand(s) must have more than one donor atom, but bind to ion in only one place. For example, the (\(\ce{NO2^{-}}\)) ion is a ligand that can bind to the central atom through the nitrogen or the oxygen atom, but cannot bind to the central atom with both oxygen and nitrogen at once, in which case it would be called a rather than an . As with all structural isomers, the formula of the complex is unchanged for each isomer, but the properties may differ. The names used to specify the changed ligands are changed as well. For example, the \(\ce{NO2^{-}}\) ion is called when it binds with the \(\ce{N}\) atom and is called when it binds with the \(\ce{O}\) atom. The cationic cobalt complex [Co(NH ) (NO )]Cl exists in two separable linkage isomers of the complex ion: (NH ) (NO )] . When donation is from nitrogen to a metal center, the complex is known as a - complex and when donation is from one oxygen to a metal center, the complex is known as a - complex. An alternative formula structure to emphasize the different for the two isomers Another example of an ambidentate ligand is thiocyanate, \(\ce{SCN^{−}}\), which can attach to the transition metal at either the sulfur atom or the nitrogen atom. Such compounds give rise to linkage isomerism. Other ligands that give rise to linkage isomers include selenocyanate, \(\ce{SeCN^{−}}\) – isoselenocyanate, \(\ce{NCSe^{−}}\) and sulfite, \(\ce{SO3^{2−}}\). Are \(\ce{[FeCl5(NO2)]^{3–}}\) and \(\ce{[FeCl5(ONO)]^{3–}}\) complex ions linkage isomers of each other? Here, the difference is in how the ligand bonds to the metal. In the first isomer, the ligand bonds to the metal through an electron pair on the nitrogen. In the second isomer, the ligand bonds to the metal through an electron pair on one of the oxygen atoms. It's easier to see it: The existence of coordination compounds with the same formula but different arrangements of the ligands was crucial in the development of coordination chemistry. Two or more compounds with the same formula but different arrangements of the atoms are called . Because isomers usually have different physical and chemical properties, it is important to know which isomer we are dealing with if more than one isomer is possible. Recall that in many cases more than one structure is possible for organic compounds with the same molecular formula; examples discussed previously include n-butane versus isobutane and cis-2-butene versus trans-2-butene. As we will see, coordination compounds exhibit the same types of isomers as organic compounds, as well as several kinds of isomers that are unique. Metal complexes that differ only in which ligands are adjacent to one another ( ) or directly across from one another ( ) in the coordination sphere of the metal are called . They are most important for square planar and octahedral complexes. Because all vertices of a square are equivalent, it does not matter which vertex is occupied by the ligand B in a square planar MA B complex; hence only a single geometrical isomer is possible in this case (and in the analogous MAB case). All four structures shown here are chemically identical because they can be superimposed simply by rotating the complex in space: For an MA B complex, there are two possible isomers: either the A ligands can be adjacent to one another (cis), in which case the B ligands must also be cis, or the A ligands can be across from one another (trans), in which case the B ligands must also be trans. Even though it is possible to draw the cis isomer in four different ways and the trans isomer in two different ways, all members of each set are chemically equivalent: Because there is no way to convert the cis structure to the trans by rotating or flipping the molecule in space, they are fundamentally different arrangements of atoms in space. Probably the best-known examples of cis and trans isomers of an MA B square planar complex are cis-Pt(NH ) Cl , also known as cisplatin, and trans-Pt(NH ) Cl , which is actually toxic rather than therapeutic. The anticancer drug cisplatin and its inactive trans isomer. Cisplatin is especially effective against tumors of the reproductive organs, which primarily affect individuals in their 20s and were notoriously difficult to cure. For example, after being diagnosed with metastasized testicular cancer in 1991 and given only a 50% chance of survival, Lance Armstrong was cured by treatment with cisplatin. Square planar complexes that contain symmetrical bidentate ligands, such as [Pt(en) ] , have only one possible structure, in which curved lines linking the two N atoms indicate the ethylenediamine ligands: Draw the cis and trans isomers of the following compounds: Only one isomer of (tmeda)PtCl is possible [tmeda = (CH ) NCH CH N(CH ) ; both nitrogens connect to the platinum]. Draw this isomer and explain why the other isomer is not possible. Octahedral complexes also exhibit cis and trans isomers. Like square planar complexes, only one structure is possible for octahedral complexes in which only one ligand is different from the other five (MA B). Even though we usually draw an octahedron in a way that suggests that the four “in-plane” ligands are different from the two “axial” ligands, in fact all six vertices of an octahedron are equivalent. Consequently, no matter how we draw an MA B structure, it can be superimposed on any other representation simply by rotating the molecule in space. Two of the many possible orientations of an MA B structure are as follows: If two ligands in an octahedral complex are different from the other four, giving an MA B complex, two isomers are possible. The two B ligands can be cis or trans. Cis- and trans-[Co(NH ) Cl ]Cl are examples of this type of system: Replacing another A ligand by B gives an MA B complex for which there are also two possible isomers. In one, the three ligands of each kind occupy opposite triangular faces of the octahedron; this is called the fac isomer (for facial). In the other, the three ligands of each kind lie on what would be the meridian if the complex were viewed as a sphere; this is called the mer isomer (for meridional): Draw all the possible geometrical isomers for the complex [Co(H O) (ox)BrCl] , where ox is O CCO , which stands for oxalate. : formula of complex : structures of geometrical isomers This complex contains one bidentate ligand (oxalate), which can occupy only adjacent (cis) positions, and four monodentate ligands, two of which are identical (H O). The easiest way to attack the problem is to go through the various combinations of ligands systematically to determine which ligands can be trans. Thus either the water ligands can be trans to one another or the two halide ligands can be trans to one another, giving the two geometrical isomers shown here: In addition, two structures are possible in which one of the halides is trans to a water ligand. In the first, the chloride ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens. Exchanging the chloride and bromide ligands gives the other, in which the bromide ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens: This complex can therefore exist as four different geometrical isomers. Draw all the possible geometrical isomers for the complex [Cr(en) (CN) ] . Two geometrical isomers are possible: and . The existence of coordination compounds with the same formula but different arrangements of the ligands was crucial in the development of coordination chemistry. Two or more compounds with the same formula but different arrangements of the atoms are called . Many metal complexes form isomers, which are two or more compounds with the same formula but different arrangements of atoms. Structural isomers differ in which atoms are bonded to one another, while geometrical isomers differ only in the arrangement of ligands around the metal ion. Ligands adjacent to one another are cis, while ligands across from one another are trans. Table \(\Page {1}\) emphasizes the key differences of the three classes of structural isomers discussed below. The highlighted ions are the ions that switch or change somehow to make the type of structural isomer it is. , | 15,040 | 3,363 |
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If we have a system which is already in equilibrium, addition of an extra amount of one of the reactants or one of the products throws the system out of equilibrium. Either the forward or the reverse reaction will then occur in order to restore equilibrium conditions. We can easily tell which of these two possibilities will happen from Le Chatelier's principle. If we add more of one of the , the system will adjust in order to offset the gain in concentration of this component. The reaction will occur to a limited extent so that some of the added product can be consumed. Conversely, if one of the is added, the system will adjust by allowing the reaction to occur to some extent. In either case . We see this principle in operation in the case of the decomposition of HI at high temperatures:
The addition of H has increased the concentration of this component. Accordingly, Le Chatelier’s principle predicts that the system will achieve a new equilibrium in such a way as to reduce this concentration. The reverse reaction occurs to a limited extent. This not only reduces the concentration of H but the concentration of I as well. At the same time the concentration of HI is increased. The system finally ends up with the concentrations calculated in , namely, \[\frac{\text{ }\!\![\!\!\text{ H}_{\text{2}}]\text{ }\!\!]\!\!\text{ }\!\![\!\!\text{ I}_{\text{2}}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ HI }\!\!]\!\!\text{ }^{\text{2}}}=\frac{\text{0}\text{.1018 mol L}^{-\text{1}}\text{ }\times \text{ 0}\text{.001 82 mol L}^{-\text{1}}}{\text{(0}\text{.0963 mol L}^{-\text{1}}\text{)}^{\text{2}}\text{ }}=\text{0}\text{.02}=K_{c} \nonumber \] Le Chatelier's principle can also be applied to cases where one of the components is . In such a case the system responds by . Consider, for example, the ionization of the weak diprotic acid H S: Since H S is a weak acid, very few S ions are produced, but a much larger concentration of S ions can be obtained by adding a strong base. The base will consume most of the H O ions. As a result, more H S will react with H O in order to make up the deficiency of H O , and more S ions will also be produced. This trick of removing one of the products in order to increase the concentration of is often used by chemists, and also by living systems. Previously, we've investigated the effect of adding/subtracting a product/reactant in mathematical terms. The video below allows you to visually see the changes the equilibrium shifts that occur upon the addition of a reactant or product, courtesy of the North Caroline School of Science and Mathematics. When a mixture of 1 mol N and 3 mol H is brought to equilibrium over a catalyst at 773 K (500°C) and 10 atm (1.01 MPa), the mixture reacts to form NH according to the equation : As mentioned in Chaps. 3 and 12, NH is an important chemical because of its use in fertilizers. In the design of a Haber-process plant to manufacture ammonia, attempts are made to use as high a pressure and as low a temperature as possible. The pressure is usually of the order of 150 atm (15 MPa), while the temperature is not usually below 750 K. Although a lower temperature would give a higher yield, the reaction would go too slowly to be economical, at least with present-day catalysts. The discoverer of a better catalyst for this reaction would certainly become a millionaire over-night. Two of the equations in this article were created using help from . | 3,488 | 3,364 |
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The postulates of the kinetic molecular theory of gases ignore both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, however, all gases have nonzero molecular volumes. Furthermore, the molecules of real gases interact with one another in ways that depend on the structure of the molecules and therefore differ for each gaseous substance. In this section, we consider the properties of real gases and how and why they differ from the predictions of the ideal gas law. We also examine liquefaction, a key property of real gases that is not predicted by the kinetic molecular theory of gases. For an ideal gas, a plot of / versus gives a horizontal line with an intercept of 1 on the / axis. Real gases, however, show significant deviations from the behavior expected for an ideal gas, particularly at high pressures (part (a) in ). Only at relatively low pressures (less than 1 atm) do real gases approximate ideal gas behavior (part (b) in ). Real gases also approach ideal gas behavior more closely at higher temperatures, as shown in for N . Why do real gases behave so differently from ideal gases at high pressures and low temperatures? Under these conditions, the two basic assumptions behind the ideal gas law—namely, that gas molecules have negligible volume and that intermolecular interactions are negligible—are no longer valid. Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller ( ). As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of / is greater than the value predicted by the ideal gas law. Moreover, all molecules are attracted to one another by a combination of forces. These forces become particularly important for gases at low temperatures and high pressures, where intermolecular distances are shorter. Attractions between molecules reduce the number of collisions with the container wall, an effect that becomes more pronounced as the number of attractive interactions increases. Because the average distance between molecules decreases, the pressure exerted by the gas on the container wall decreases, and the observed pressure is than expected ( ). Thus as shown in , at low temperatures, the ratio of / is lower than predicted for an ideal gas, an effect that becomes particularly evident for complex gases and for simple gases at low temperatures. At very high pressures, the effect of nonzero molecular volume predominates. The competition between these effects is responsible for the minimum observed in the / versus plot for many gases. Nonzero molecular volume makes the actual volume than predicted at high pressures; intermolecular attractions make the pressure than predicted. At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate. Conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Eventually, a point is reached where the molecules can no longer overcome the intermolecular attractive forces, and the gas liquefies (condenses to a liquid). The Dutch physicist Johannes van der Waals (1837–1923; Nobel Prize in Physics, 1910) modified the ideal gas law to describe the behavior of real gases by explicitly including the effects of molecular size and intermolecular forces. In his description of gas behavior, the so-called van der Waals equation ,
\[ \left ( P+\dfrac{an^{2}}{V^{2}} \right )\left ( V-nb \right )=nRT \tag{11.1.1} \] and are empirical constants that are different for each gas. The values of and are listed in for several common gases. The pressure term + ( / ) corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law. Here, / represents the concentration of the gas ( / ) squared because it takes two particles to engage in the pairwise intermolecular interactions of the type shown in . The volume term − corrects for the volume occupied by the gaseous molecules. The correction for volume is negative, but the correction for pressure is positive to reflect the effect of each factor on and , respectively. Because nonzero molecular volumes produce a measured volume that is than that predicted by the ideal gas law, we must subtract the molecular volumes to obtain the actual volume available. Conversely, attractive intermolecular forces produce a pressure that is than that expected based on the ideal gas law, so the / term must be added to the measured pressure to correct for these effects. You are in charge of the manufacture of cylinders of compressed gas at a small company. Your company president would like to offer a 4.0 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you have on hand have a rupture pressure of 40 atm. Use both the ideal gas law and the van der Waals equation to calculate the pressure in a cylinder at 25°C. Is this cylinder likely to be safe against sudden rupture (which would be disastrous and certainly result in lawsuits because chlorine gas is highly toxic)? volume of cylinder, mass of compound, pressure, and temperature safety Use the molar mass of chlorine to calculate the amount of chlorine in the cylinder. Then calculate the pressure of the gas using the ideal gas law. Obtain and values for Cl from . Use the van der Waals equation to solve for the pressure of the gas. Based on the value obtained, predict whether the cylinder is likely to be safe against sudden rupture. We begin by calculating the amount of chlorine in the cylinder using the molar mass of chlorine (70.906 g/mol): \( \left ( 500 \; \cancel{g} \right )\left ( \dfrac{1 \; mol}{70.906 \; \cancel{g}}=7.05 mol \; Cl_{2} \right ) \) Using the ideal gas law and the temperature in kelvins (298 K), we calculate the pressure: \( P=\dfrac{nRT}{V}=\dfrac{\left ( 7.05 \; \cancel{mol} \right )\left [ 0.08206\left ( \cancel{L}\cdot atm \right )/\left ( \cancel{K}\cdot \cancel{mol} \right ) \right ]\left ( 298 \; \cancel{K} \right )}{4.0 \; \cancel {L}}=43 \; atm \) If chlorine behaves like an ideal gas, you have a real problem! Now let’s use the van der Waals equation with the and values for Cl from . Solving for gives \( P= \dfrac{nRT}{V-nb}-\dfrac{an^{2}}{V^{2}} \) \( =\dfrac{\left ( 7.05 \; \cancel{mol} \right )\left [ 0.08206\left ( \cancel{L}\cdot atm \right )/\left ( \cancel{K}\cdot \cancel{mol} \right ) \right ]\left ( 298 \; \cancel{K} \right )}{4.0 \; \cancel {L} - 7.05 \; \cancel{mol}\left ( 0.0542 \; \cancel{L}/\cancel{mol} \right )} - \dfrac{6.260^{2} \; \cancel{L^{2}}\cdot atm/\cancel{mol^{2}}\left ( 7.05 \; \cancel{mol^{2}} \right )}{4.0 \; \cancel{L^{2}}} \) \( = 47.7 \; atm - 19.4 \; atm = 28 \; atm \;(2\; significant \; figures) \) This pressure is well within the safety limits of the cylinder. The ideal gas law predicts a pressure 15 atm higher than that of the van der Waals equation. Exercise A 10.0 L cylinder contains 500 g of methane. Calculate its pressure to two significant figures at 27°C using the a. 77 atm; b. 67 atm Liquefaction of gases is the condensation of gases into a liquid form, which is neither anticipated nor explained by the kinetic molecular theory of gases. Both the theory and the ideal gas law predict that gases compressed to very high pressures and cooled to very low temperatures should still behave like gases, albeit cold, dense ones. As gases are compressed and cooled, however, they invariably condense to form liquids, although very low temperatures are needed to liquefy light elements such as helium (for He, 4.2 K at 1 atm pressure). Liquefaction can be viewed as an extreme deviation from ideal gas behavior. It occurs when the molecules of a gas are cooled to the point where they no longer possess sufficient kinetic energy to overcome intermolecular attractive forces. The precise combination of temperature and pressure needed to liquefy a gas depends strongly on its molar mass and structure, with heavier and more complex molecules usually liquefying at higher temperatures. In general, substances with large van der Waals coefficients are relatively easy to liquefy because large coefficients indicate relatively strong intermolecular attractive interactions. Conversely, small molecules with only light elements have small coefficients, indicating weak intermolecular interactions, and they are relatively difficult to liquefy. Gas liquefaction is used on a massive scale to separate O , N , Ar, Ne, Kr, and Xe. After a sample of air is liquefied, the mixture is warmed, and the gases are separated according to their boiling points. In , we will consider in more detail the nature of the intermolecular forces that allow gases to liquefy. A large value of indicates the presence of relatively strong intermolecular attractive interactions. The ultracold liquids formed from the liquefaction of gases are called cryogenic liquids , from the Greek , meaning “cold,” and , meaning “producing”). They have applications as refrigerants in both industry and biology. For example, under carefully controlled conditions, the very cold temperatures afforded by liquefied gases such as nitrogen (boiling point = 77 K at 1 atm) can preserve biological materials, such as semen for the artificial insemination of cows and other farm animals. These liquids can also be used in a specialized type of surgery called , which selectively destroys tissues with a minimal loss of blood by the use of extreme cold. Moreover, the liquefaction of gases is tremendously important in the storage and shipment of fossil fuels ( ). Liquefied natural gas (LNG) and liquefied petroleum gas (LPG) are liquefied forms of hydrocarbons produced from natural gas or petroleum reserves. LNG consists mostly of methane, with small amounts of heavier hydrocarbons; it is prepared by cooling natural gas to below about −162°C. It can be stored in double-walled, vacuum-insulated containers at or slightly above atmospheric pressure. Because LNG occupies only about 1/600 the volume of natural gas, it is easier and more economical to transport. LPG is typically a mixture of propane, propene, butane, and butenes and is primarily used as a fuel for home heating. It is also used as a feedstock for chemical plants and as an inexpensive and relatively nonpolluting fuel for some automobiles. No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Deviations from ideal gas behavior can be seen in plots of / versus at a given temperature; for an ideal gas, / versus = 1 under all conditions. At high pressures, most real gases exhibit larger / values than predicted by the ideal gas law, whereas at low pressures, most real gases exhibit / values close to those predicted by the ideal gas law. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the , which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces. If the temperature of a gas is decreased sufficiently, occurs, in which the gas condenses into a liquid form. Liquefied gases have many commercial applications, including the transport of large amounts of gases in small volumes and the uses of ultracold . : \( \left ( p+\dfrac{an^{2}}{V^{2}} \right )\left ( V-b \right )=nRT \) What factors cause deviations from ideal gas behavior? Use a sketch to explain your answer based on interactions at the molecular level. Explain the effect of nonzero atomic volume on the ideal gas law at high pressure. Draw a typical graph of volume versus 1/ for an ideal gas and a real gas. For an ideal gas, the product of pressure and volume should be constant, regardless of the pressure. Experimental data for methane, however, show that the value of decreases significantly over the pressure range 0 to 120 atm at 0°C. The decrease in over the same pressure range is much smaller at 100°C. Explain why decreases with increasing temperature. Why is the decrease less significant at higher temperatures. What is the effect of intermolecular forces on the liquefaction of a gas? At constant pressure and volume, does it become easier or harder to liquefy a gas as its temperature increases? Explain your reasoning. What is the effect of increasing the pressure on the liquefaction temperature? Describe qualitatively what and , the two empirical constants in the van der Waals equation, represent. In the van der Waals equation, why is the term that corrects for volume negative and the term that corrects for pressure positive? Why is / squared? Liquefaction of a gas depends strongly on two factors. What are they? As temperature is decreased, which gas will liquefy first—ammonia, methane, or carbon monoxide? Why? What is a cryogenic liquid? Describe three uses of cryogenic liquids. Air consists primarily of O , N , Ar, Ne, Kr, and Xe. Use the concepts discussed in this chapter to propose two methods by which air can be separated into its components. Which component of air will be isolated first? How can gas liquefaction facilitate the storage and transport of fossil fuels? What are potential drawbacks to these methods? The van der Waals constants for xenon are = 4.19 (L ·atm)/mol and = 0.0510 L/mol. If a 0.250 mol sample of xenon in a container with a volume of 3.65 L is cooled to −90°C, what is the pressure of the sample assuming ideal gas behavior? What would be the pressure under these conditions? The van der Waals constants for water vapor are = 5.46 (L ·atm)/mol and = 0.0305 L/mol. If a 20.0 g sample of water in a container with a volume of 5.0 L is heated to 120°C, what is the pressure of the sample assuming ideal gas behavior? What would be the pressure under these conditions? | 14,581 | 3,365 |
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Predicting the weather is tricky business. A thorough examination of a large amount of data is needed to make the daily forecast. Wind patterns, historical data, barometric pressure—these and many other data are fed into computers that then use a set of rules to predict what will happen based on past history. Some combinations of aqueous reactants result in the formation of a solid precipitate as a product. However, some combinations will not produce such a product. If solutions of sodium nitrate and ammonium chloride are mixed, no reaction occurs. One could write a molecular equation showing a double-replacement reaction, but both products, sodium chloride and ammonium nitrate, are soluble and would remain in the solution as ions. Every ion is a spectator ion and there is no net ionic equation at all. It is useful to be able to predict when a precipitate will occur in a reaction. To do so, you can use a set of guidelines called the (shown in Table \(\Page {1}\)). For practice using the solubility rules, predict if a precipitate will form when solutions of cesium bromide and lead (II) nitrate are mixed. \[\ce{Cs^+} \left( aq \right) + \ce{Br^-} \left( aq \right) + \ce{Pb^{2+}} \left( aq \right) + 2 \ce{NO_3^-} \left( aq \right) \rightarrow ?\nonumber \] The potential precipitates from a double-replacement reaction are cesium nitrate and lead (II) bromide. According to the solubility rules table, cesium nitrate is soluble because all compounds containing the nitrate ion, as well as all compounds containing the alkali metal ions, are soluble. Most compounds containing the bromide ion are soluble, but lead (II) is an exception. Therefore, the cesium and nitrate ions are spectator ions and the lead (II) bromide is a precipitate. The balanced net ionic reaction is: \[\ce{Pb^{2+}} \left( aq \right) + 2 \ce{Br^-} \left( aq \right) \rightarrow \ce{PbBr_2} \left( s \right)\nonumber \] | 1,931 | 3,366 |
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One way to report the heat absorbed or released by chemical reactions would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, starting from the elemental forms of each atom at 25 C and 1 atm pressure. (\(ΔH_f\)) is the enthalpy change for the formation of 1 mol of a compound from its component elements, such as the formation of carbon dioxide from carbon and oxygen. The formation of any chemical can be as a reaction from the corresponding elements: \[ \text{elements} \rightarrow \text{compound} \nonumber\] which in terms of the the Enthalpy of formation becomes \[\Delta H_{rxn} = \Delta H_{f} \label{7.8.1} \] For example, consider the combustion of carbon: \[ \ce{ C(s) + O2 (g) -> CO2 (g)} \nonumber\] then \[ \Delta H_{rxn} = \Delta H_{f}\left [CO_{2}\left ( g \right ) \right ] \nonumber \] The sign convention for Δ is the same as for any enthalpy change: \(ΔH_f < 0\) if heat is released when elements combine to form a compound and \(ΔH_f > 0\) if heat is absorbed. The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system. The magnitude of Δ for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution), the pressure of any gases present, and the temperature at which the reaction is carried out. To avoid confusion caused by differences in reaction conditions and ensure uniformity of data, the scientific community has selected a specific set of conditions under which enthalpy changes are measured. These standard conditions serve as a reference point for measuring differences in enthalpy, much as sea level is the reference point for measuring the height of a mountain or for reporting the altitude of an airplane. The standard conditions for which most thermochemical data are tabulated are a of 1 atmosphere (atm) for all gases and a of 1 M for all species in solution (1 mol/L). In addition, each pure substance must be in its standard state is usually its most stable form at a pressure of 1 atm at a specified temperature. We assume a temperature of 25°C (298 K) for all enthalpy changes given in this text, unless otherwise indicated. Enthalpies of formation measured under these conditions are called (\(ΔH^o_f\)) The standard enthalpy of formation of any element in its standard state is zero by definition. For example, although oxygen can exist as ozone (O ), atomic oxygen (O), and molecular oxygen (O ), O is the most stable form at 1 atm pressure and 25°C. Similarly, hydrogen is H (g), not atomic hydrogen (H). Graphite and diamond are both forms of elemental carbon, but because graphite is more stable at 1 atm pressure and 25°C, the standard state of carbon is graphite (Figure \(\Page {1}\)). Therefore, \(\ce{O2(g)}\), \(\ce{H2(g)}\), and graphite have \(ΔH^o_f\) values of zero. The standard enthalpy of formation of glucose from the elements at 25°C is the enthalpy change for the following reaction: \[ 6C\left (s, graphite \right ) + 6H_{2}\left (g \right ) + 3O_{2}\left (g \right ) \rightarrow C_{6}H_{12}O_{6}\left (s \right )\; \; \; \Delta H_{f}^{o} = - 1273.3 \; kJ \label{7.8.2} \] It is not possible to measure the value of \(ΔH^oo_f \) for glucose, −1273.3 kJ/mol, by simply mixing appropriate amounts of graphite, \(\ce{O2}\), and \(\ce{H2}\) and measuring the heat evolved as glucose is formed since the reaction shown in Equation \(\ref{7.8.2}\) does not occur at a measurable rate under any known conditions. Glucose is not unique; most compounds cannot be prepared by the chemical equations that define their standard enthalpies of formation. Instead, values of \(ΔH^oo_f \) are obtained using and standard enthalpy changes that have been measured for other reactions, such as combustion reactions. Values of \(ΔH^o_f\) for an extensive list of compounds are given in . Note that \(ΔH^o_f\) values are always reported in kilojoules per mole of the substance of interest. Also notice in that the standard enthalpy of formation of O (g) is zero because it is the most stable form of oxygen in its standard state. For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound. compound formula and phase. balanced chemical equation for its formation from elements in standard states Use to identify the standard state for each element. Write a chemical equation that describes the formation of the compound from the elements in their standard states and then balance it so that 1 mol of product is made. To calculate the standard enthalpy of formation of a compound, we must start with the elements in their standard states. The standard state of an element can be identified in : by a \(ΔH^o_f\) Hydrogen chloride contains one atom of hydrogen and one atom of chlorine. Because the standard states of elemental hydrogen and elemental chlorine are \(\ce{H2(g)}\) and \(\ce{Cl2(g)}\), respectively, the unbalanced chemical equation is \[\ce{H2(g) + Cl2(g) \rightarrow HCl(g)} \nonumber\] Fractional coefficients are required in this case because values are reported for of the product, \(\ce{HCl}\). Multiplying both \(\ce{H2(g)}\) and \(\ce{Cl2(g)}\) by 1/2 balances the equation: \[ \ce{1/2 H_{2} (g) + 1/2 Cl_{2} (g) \rightarrow HCl (g)} \nonumber\] The standard states of the elements in this compound are \(\ce{Mg(s)}\), \(\ce{C(s, graphite)}\), and \(\ce{O2(g)}\). The unbalanced chemical equation is thus \[\ce{Mg(s) + C (s, graphite) + O2 (g) \rightarrow MgCO3 (s)} \nonumber\] This equation can be balanced by inspection to give \[ \ce{Mg (s) + C (s, graphite ) + 3/2 O2 (g)\rightarrow MgCO3 (s)} \nonumber\] Palmitic acid, the major fat in meat and dairy products, contains hydrogen, carbon, and oxygen, so the unbalanced chemical equation for its formation from the elements in their standard states is as follows: \[\ce{C(s, graphite) + H2(g) + O2(g) \rightarrow CH3(CH2)14CO2H(s)} \nonumber\] There are 16 carbon atoms and 32 hydrogen atoms in 1 mol of palmitic acid, so the balanced chemical equation is \[\ce{16C (s, graphite) + 16 H2(g) + O2(g) -> CH3(CH2)14CO2H(s) } \nonumber\] For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound. \[ \ce{ Na (s) + 1/2 Cl2 (g) \rightarrow NaCl (s)} \nonumber \] \[ \ce{H_{2} (g) + 1/8 S8 (s) + 2O2 ( g) \rightarrow H2 SO4( l) } \nonumber\] \[\ce{2C(s) + O2(g) + 2H2(g) -> CH3CO2H(l)} \nonumber \] Definition of Heat of Formation Reactions: Tabulated values of standard enthalpies of formation can be used to calculate enthalpy changes for reaction involving substances whose \(\Delta{H_f^o}\) values are known. The standard enthalpy of reaction \(\Delta{H_{rxn}^o}\) is the enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard states. Consider the general reaction \[ aA + bB \rightarrow cC + dD \label{7.8.3}\] where \(A\), \(B\), \(C\), and \(D\) are chemical substances and \(a\), \(b\), \(c\), and \(d\) are their stoichiometric coefficients. The magnitude of \(ΔH^ο\) is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, minus the sum of the standard enthalpies of formation of the reactants, also multiplied by their coefficients: \[ \Delta H_{rxn}^{o} = \underbrace{ \left [c\Delta H_{f}^{o}\left ( C \right ) + d\Delta H_{f}^{o}\left ( D \right ) \right ] }_{\text{products} } - \underbrace{ \left [a\Delta H_{f}^{o}\left ( A \right ) + b\Delta H_{f}^{o}\left ( B \right ) \right ]}_{\text{reactants }} \label{7.8.4} \] More generally, we can write \[ \Delta H_{rxn}^{o} = \sum m\Delta H_{f}^{o}\left ( products \right ) - \sum n\Delta H_{f}^{o}\left ( reactants \right ) \label{7.8.5} \] where the symbol \(\sum\) means “sum of” and \(m\) and \(n\) are the stoichiometric coefficients of each of the products and the reactants, respectively. “Products minus reactants” summations such as Equation \(\ref{7.8.5}\) arise from the fact that enthalpy is a state function. Because many other thermochemical quantities are also state functions, “products minus reactants” summations are very common in chemistry; we will encounter many others in subsequent chapters. "Products minus reactants" summations are typical of state functions. To demonstrate the use of tabulated values, we will use them to calculate \(ΔH_{rxn}\) for the combustion of glucose, the reaction that provides energy for your brain: \[ \ce{ C6H12O6 (s) + 6O2 (g) \rightarrow 6CO2 (g) + 6H2O (l)} \label{7.8.6} \] Using Equation \(\ref{7.8.5}\), we write \[ \Delta H_{f}^{o} =\left \{ 6\Delta H_{f}^{o}\left [ CO_{2}\left ( g \right ) \right ] + 6\Delta H_{f}^{o}\left [ H_{2}O\left ( g \right ) \right ] \right \} - \left \{ \Delta H_{f}^{o}\left [ C_{6}H_{12}O_{6}\left ( s \right ) \right ] + 6\Delta H_{f}^{o}\left [ O_{2}\left ( g \right ) \right ] \right \} \label{7.8.7} \] From , the relevant values are , , and . Because O (g) is a pure element in its standard state, Inserting these values into Equation \(\ref{7.8.7}\) and changing the subscript to indicate that this is a combustion reaction, we obtain \[ \begin{align} \Delta H_{comb}^{o} &= \left [ 6\left ( -393.5 \; kJ/mol \right ) + 6 \left ( -285.8 \; kJ/mol \right ) \right ] - \left [-1273.3 + 6\left ( 0 \; kJ\;mol \right ) \right ] \label{7.8.8} \\[4pt] &= -2802.5 \; kJ/mol \end{align} \] As illustrated in Figure \(\Page {2}\), we can use Equation \(\ref{7.8.8}\) to calculate \(ΔH^ο_f\) for glucose because enthalpy is a state function. The figure shows two pathways from reactants (middle left) to products (bottom). The more direct pathway is the downward green arrow labeled \(ΔH^ο_{comb}\). The alternative hypothetical pathway consists of that convert the reactants to the elements in their standard states (upward purple arrow at left) and then convert the elements into the desired products (downward purple arrows at right). The reactions that convert the reactants to the elements are the reverse of the equations that define the \(ΔH^ο_f\) values of the reactants. Consequently, the enthalpy changes are \[ \begin{align} \Delta H_{1}^{o} &= \Delta H_{f}^{o} \left [ glucose \left ( s \right ) \right ] \nonumber \\[4pt] &= -1 \; \cancel{mol \; glucose}\left ( \dfrac{1273.3 \; kJ}{1 \; \cancel{mol \; glucose}} \right ) \nonumber \\[4pt] &= +1273.3 \; kJ \nonumber \\[4pt] \Delta H_{2}^{o} &= 6 \Delta H_{f}^{o} \left [ O_{2} \left ( g \right ) \right ] \nonumber \\[4pt] & =6 \; \cancel{mol \; O_{2}}\left ( \dfrac{0 \; kJ}{1 \; \cancel{mol \; O_{2}}} \right ) \nonumber \\[4pt] &= 0 \; kJ \end{align} \label{7.8.9} \] Recall that when we reverse a reaction, we must also reverse the of the accompanying enthalpy change (Equation \ref{7.8.4} since the products are now reactants and vice versa. The overall enthalpy change for conversion of the reactants (1 mol of glucose and 6 mol of O ) to the elements is therefore +1273.3 kJ. The reactions that convert the elements to final products (downward purple arrows in Figure \(\Page {2}\)) are identical to those used to define the values of the products. Consequently, the enthalpy changes (from \[ \begin{matrix} \Delta H_{3}^{o} = \Delta H_{f}^{o} \left [ CO_{2} \left ( g \right ) \right ] = 6 \; \cancel{mol \; CO_{2}}\left ( \dfrac{393.5 \; kJ}{1 \; \cancel{mol \; CO_{2}}} \right ) = -2361.0 \; kJ \\ \Delta H_{4}^{o} = 6 \Delta H_{f}^{o} \left [ H_{2}O \left ( l \right ) \right ] = 6 \; \cancel{mol \; H_{2}O}\left ( \dfrac{-285.8 \; kJ}{1 \; \cancel{mol \; H_{2}O}} \right ) = -1714.8 \; kJ \end{matrix} \] The overall enthalpy change for the conversion of the elements to products (6 mol of carbon dioxide and 6 mol of liquid water) is therefore −4075.8 kJ. Because enthalpy is a state function, the difference in enthalpy between an initial state and a final state can be computed using pathway that connects the two. Thus the enthalpy change for the combustion of glucose to carbon dioxide and water is the sum of the enthalpy changes for the conversion of glucose and oxygen to the elements (+1273.3 kJ) and for the conversion of the elements to carbon dioxide and water (−4075.8 kJ): \[ \Delta H_{comb}^{o} = +1273.3 \; kJ +\left ( -4075.8 \; kJ \right ) = -2802.5 \; kJ \label{7.8.10} \] This is the same result we obtained using the “products minus reactants” rule (Equation \(\ref{7.8.5}\)) and values. The two results must be the same because Equation \(\ref{7.8.10}\) is just a more compact way of describing the thermochemical cycle shown in Figure \(\Page {1}\). Long-chain fatty acids such as palmitic acid (\(\ce{CH3(CH2)14CO2H}\)) are one of the two major sources of energy in our diet (\(ΔH^o_f\) =−891.5 kJ/mol). Use the data in to calculate for the combustion of palmitic acid. Based on the energy released in combustion , which is the better fuel — glucose or palmitic acid? compound and \(ΔH^ο_{f}\) values \(ΔH^ο_{comb}\) per mole and per gram To determine the energy released by the combustion of palmitic acid, we need to calculate its \(ΔH^ο_f\). As always, the first requirement is a balanced chemical equation: \[C_{16}H_{32}O_{2(s)} + 23O_{2(g)} \rightarrow 16CO_{2(g)} + 16H_2O_{(l)} \nonumber \] Using Equation \(\ref{7.8.5}\) (“products minus reactants”) with values from (and omitting the physical states of the reactants and products to save space) gives \[ \begin{align*} \Delta H_{comb}^{o} &= \sum m \Delta H^o_f\left( {products} \right) - \sum n \Delta H^o_f \left( {reactants} \right) \\[4pt] &= \left [ 16\left ( -393.5 \; kJ/mol \; CO_{2} \right ) + 16\left ( -285.8 \; kJ/mol \; H_{2}O \; \right ) \right ] \\[4pt] & - \left [ -891.5 \; kJ/mol \; C_{16}H_{32}O_{2} + 23\left ( 0 \; kJ/mol \; O_{2} \; \right ) \right ] \\[4pt] &= -9977.3 \; kJ/mol \nonumber \end{align*} \] This is the energy released by the combustion of 1 mol of palmitic acid. The energy released by the combustion of 1 g of palmitic acid is \( \Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{9977.3 \; kJ}{\cancel{1 \; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{256.42 \; g} \right )= -38.910 \; kJ/g \nonumber \) As calculated in Equation \(\ref{7.8.8}\), \( of glucose is −2802.5 kJ/mol. The energy released by the combustion of 1 g of glucose is therefore \( \Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{-2802.5 \; kJ}{\cancel{1\; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{180.16\; g} \right ) = -15.556 \; kJ/g \nonumber \) The combustion of fats such as palmitic acid releases more than twice as much energy per gram as the combustion of sugars such as glucose. This is one reason many people try to minimize the fat content in their diets to lose weight. Use to calculate \(ΔH^o_{rxn}\) for the , which is used industrially on an enormous scale to obtain H (g): \[ \ce{ CO ( g ) + H2O (g ) -> CO2 (g) + H2 ( g )} \nonumber\] −41.2 kJ/mol We can also measure the enthalpy change for another reaction, such as a combustion reaction, and then use it to calculate a compound’s \(ΔH^ο_f\) which we cannot obtain otherwise. This procedure is illustrated in Example \(\Page {3}\). Beginning in 1923, [\(\ce{(C2H5)4Pb}\)] was used as an antiknock additive in gasoline in the United States. Its use was completely phased out in 1986 because of the health risks associated with chronic lead exposure. Tetraethyllead is a highly poisonous, colorless liquid that burns in air to give an orange flame with a green halo. The combustion products are \(\ce{CO2(g)}\), \(\ce{H2O(l)}\), and red \(\ce{PbO(s)}\). What is the standard enthalpy of formation of tetraethyllead, given that \(ΔH^ο_f\) is −19.29 kJ/g for the combustion of tetraethyllead and \(ΔH^ο_f\) of red PbO(s) is −219.0 kJ/mol? reactant, products, and \(ΔH^ο_{comb}\) values \(ΔH^ο_f\) of the reactants The balanced chemical equation for the combustion reaction is as follows: \[\ce{2(C2H5)4Pb(l) + 27O2(g) → 2PbO(s) + 16CO2(g) + 20H2O(l)} \nonumber\] Using Equation \(\ref{7.8.5}\) gives \[ \Delta H_{comb}^{o} = \left [ 2 \Delta H_{f}^{o}\left ( PbO \right ) + 16 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 20 \Delta H_{f}^{o}\left ( H_{2}O \right )\right ] - \left [2 \Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) + 27 \Delta H_{f}^{o}\left ( O_{2} \right ) \right ] \nonumber \] Solving for \(ΔH^o_f [\ce{(C2H5)4Pb}]\) give \[ \Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) = \Delta H_{f}^{o}\left ( PbO \right ) + 8 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 10 \Delta H_{f}^{o}\left ( H_{2}O \right ) - \dfrac{27}{2} \Delta H_{f}^{o}\left ( O_{2} \right ) - \dfrac{\Delta H_{comb}^{o}}{2} \nonumber \] The values of all terms other than \(ΔH^o_f [\ce{(C2H5)4Pb}]\) are given in . The magnitude of \(ΔH^o_{comb}\) is given in the problem in kilojoules per of tetraethyl lead. We must therefore multiply this value by the molar mass of tetraethyl lead (323.44 g/mol) to get \(ΔH^o_{comb}\) for 1 mol of tetraethyl lead: \[\begin{align*} \Delta H_{comb}^{o} &= \left ( \dfrac{-19.29 \; kJ}{\cancel{g}} \right )\left ( \dfrac{323.44 \; \cancel{g}}{mol} \right ) \\[4pt] &= -6329 \; kJ/mol \end{align*} \] Because the balanced chemical equation contains 2 mol of tetraethyllead, \(ΔH^o_{rxn}\) is \[\begin{align*} \Delta H_{rxn}^{o} &= 2 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb} \left ( \dfrac{-6329 \; kJ}{1 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb }} \right ) \\[4pt] &= -12,480 \; kJ \end{align*}\] Inserting the appropriate values into the equation for \(ΔH^o_f [\ce{(C2H5)4Pb}]\) \[ \begin{align*} \Delta H_{f}^{o} \left [ \left (C_{2}H_{4} \right )_{4}Pb \right ] & = \left [1 \; mol \;PbO \;\times 219.0 \;kJ/mol \right ]+\left [8 \; mol \;CO_{2} \times \left (-393.5 \; kJ/mol \right )\right ] +\left [10 \; mol \; H_{2}O \times \left ( -285.8 \; kJ/mol \right )\right ] + \left [-27/2 \; mol \; O_{2}) \times 0 \; kJ/mol \; O_{2}\right ] \left [12,480.2 \; kJ/mol \; \left ( C_{2}H_{5} \right )_{4}Pb \right ]\\[4pt]
&= -219.0 \; kJ -3148 \; kJ - 2858 kJ - 0 kJ + 6240 \; kJ = 15 kJ/mol \end{align*}\] Ammonium sulfate, \(\ce{(NH4)2SO4}\), is used as a fire retardant and wood preservative; it is prepared industrially by the highly exothermic reaction of gaseous ammonia with sulfuric acid: \[ \ce{2NH3(g) + H2SO4(aq) \rightarrow (NH4)2SO4(s)} \nonumber \] The value of \(ΔH^o_{rxn}\) is -179.4 \(\ce{H2SO4}\). kilojoules −1181 kJ/mol Calculating DH° using DHf°: The (\(ΔH_{f}\)) is the enthalpy change that accompanies the formation of a compound from its elements. (\(ΔH^o_{f}\)) are determined under : a pressure of 1 atm for gases and a concentration of 1 M for species in solution, with all pure substances present in their (their most stable forms at 1 atm pressure and the temperature of the measurement). The standard heat of formation of any element in its most stable form is defined to be zero. The (\(ΔH^o_{rxn}\)) can be calculated from the sum of the of the products (each multiplied by its stoichiometric coefficient) minus the sum of the standard enthalpies of formation of the reactants (each multiplied by its stoichiometric coefficient)—the “products minus reactants” rule. The (\(ΔH_{soln}\)) is the heat released or absorbed when a specified amount of a solute dissolves in a certain quantity of solvent at constant pressure. ( ) | 19,742 | 3,367 |
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Although chemists usually work with moles as units, occasionally it is helpful to refer to the actual number of atoms or molecules involved. When this is done, the symbol N is used. For example, in referring to 1 mol of mercury atoms, we could write \[n_{\text{Hg}} = 1 \text{mol} \nonumber \] and \[ N_{\text{Hg}} = 6.022\times 10^{23} \nonumber \] Notice that \(N_{Hg }\) is a unitless quantity, which requires the use of a conversion factor to obtain. This conversion factor involves the number of particles per unit amount of substance and is given the symbol \(N_A\) and called the . It is defined by the equation \[N_{\text{A}} = \dfrac{N}{n} \label{1} \] Since for any substance there are 6.022 × 10 particles per mole, \[\textit{N}_\text{A}=\dfrac{6.022\cdot10^{23}} {1\text{ mol}}=6.022\times 10^{23}\text{ mol}^{\text{–1}} \nonumber \] Calculate the number of O molecules in 0.189 mol O . Rearranging Equation \(\ref{1}\), we obtain \[N = n \times N_{\text{A}} = 0.189 \text{ mol} \times 6.022 \times 10^{23} \tfrac{1}{\text{ mol}} \ = 1.14 \times 10^{23} \nonumber \] Alternatively, we might include the identity of the particles involved: \[\begin{align*}\ce{N} &= \text{0.189 mol O}_{\text{2}}\cdot \left( \dfrac{6.022 \times 10^{23}\text{ O}_2\text{ molecules}} {\text{1 mol O}_2}\right) \\[4pt] &= 1.14\cdot10^{23}\text{ O}_{\text{2}}\text{ molecules}\end{align*} \nonumber \] Notice that Equation \(\ref{1}\), which defines the Avogadro constant, has the same form as the which defined density. The preceding example used the Avogadro constant as a conversion factor in the same way that density was used. As in previous examples, all that is necessary is to remember that number of particles and amount of substance are related by a conversion factor, the Avogadro constant. \[\large\text{Number of particles } \large\overset{\text{Avogadro constant}}{\longleftrightarrow} \large\text{amount of substance} \\ \quad \\ \large N \large\overset{\text{N}_{\text{A}}}{\longleftrightarrow} { \space} \large n\label{2} \] As long as the units cancel, is being used correctly. | 2,106 | 3,368 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/7%3A_Acids_and_Bases/7.09%3A_Acid_Solutions_that_Water_Contributes_p |
Unlike acids/bases, weak acids and weak bases do not completely dissociate (separate into ions) at equilibrium in water, so calculating the pH of these solutions requires consideration of a unique and . Although this is more difficult than calculating the pH of a strong acid or base solution, most biochemically important acids and bases are considered weak, and so it is very useful to understand how to calculate the pH of these substances. The same basic method can be used to determine the pH of aqueous solutions of many different weak acids and bases. An aqueous solution of a weak acid or base contains both the protonated and unprotonated forms of the compound, so an can be made and used to plug in concentrations into an . The ionization constant for the acid (K ) or base (K ) is a measure of how readily the acid donates protons or how readily a base accepts protons. Because you are calculating pH, you must solve for the unknown concentration of ions in solution at equilibrium. The first step in calculating the pH of an aqueous solution of any weak acid or base is to notice whether the initial concentration is high or low relative to 10 M (the concentration of hydronium and hydroxide ions in water due to the of water). If the concentration of the acid or base is very close to or less than 10 M, then the solution is considered dilute and additional steps must be taken to calculate pH. You must first be familiar with equilibrium constant expressions and how to write them for a chemical reaction. Then, by making an , you can find unknown concentration values that can be plugged into this equilibrium expression. To start, you must find the initial concentration of acetic acid in the vinegar. Assume that the vinegar is really just a solution of acetic acid in water, and that density = 1 g/mL. So if the vinegar is 3% acetic acid by mass and the molar mass of HC H O = 60.05 g/mol, then \[\dfrac{1.5L,vinegar}{} \times \dfrac{1000mL}{1L} \times \dfrac{1g}{1mL} \times \dfrac{3g,acetic acid}{100g,vinegar} \times \dfrac{1mol,acetic acid}{60.05g,acetic acid} = 0.75 \;mol\; HC_2H_3O_2 \nonumber \] Divide 0.75 mol by 1.5 L to get an initial concentration of 0.50 M. \[HC_2H_3O_{2(aq)} + H_2O_{(l)} \rightleftharpoons C_2H_3O^-_{2(aq)} + H_3O^+_{(aq)} \nonumber \] For every acetic acid molecule that dissociates, one acetate ion and one hydronium ion is produced. This can be represented by subtracting "x" from the original acetic acid concentration, and adding "x" to the original concentrations of the dissociated ions. You can create a modified equilibrium constant expression \[K_a = \dfrac{[C_2H_3O_2^-,H_3O^+]}{[HC_2H_3O_2]} \nonumber \] and then plug in the concentration values you found in the ICE table \[1.8 \times 10^{-5} = \dfrac{x^2}{0.5 - x} \nonumber \] so \[x^2 + (1.8 \times 10^{-5})x - (9 \times 10^{-6}) = 0 \nonumber \] then use the quadratic formula to calculate \[x = 0.0030\; M = [H_3O^+] \nonumber \] which can be plugged into the formula \[pH = -\log[H_3O^+] \nonumber \] \[-\log(0.0030) = pH = 2.5 \nonumber \] The same thing can be done for calculating the pH of a weak base. What is the pH of a \(7.0 \times 10^{-3}\) M NH solution? (pK = 4.74) \[NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH^+_{4(aq)} + OH^-_{(aq)} \nonumber \] Instead of K , you were given pK . So to get K pK = -log(K ) = 4.74 K = 10 = 1.8 x 10 Plug these values into the equilibrium expression to get \[K_b = \dfrac{[NH_4^+,OH^-]}{[NH_3]} = 1.8 \times 10^{-5} = \dfrac{x^2}{7 \times 10^{-3} - x} \nonumber \] and use the quadratic formula to find that \[x = 3.46 \times 10^{-4}\; M = [OH^-] \nonumber \] so \[pOH = -\log(3.46 \times 10^{-4}) = 3.46 \nonumber \] and in water at 25 degrees \[pH + pOH = 14 \nonumber \] Hence: \[14 - 3.46 = pH = 10.54 \nonumber \] Polyprotic acids have more than one proton to donate to water, and so they have more than one ionization constant (K , K , etc) that can be considered. Polyprotic bases take more than one proton from water, and also have more than one ionization constant (K , K , etc). Most often the first proton exchange is the only one that considerably affects pH. This is discussed more at the end of the first example. What is the pH of a grapefruit that contains 0.007 M citric acid solution (C H O )?
(K = 7.5 x 10 , K = 1.7 x 10 , K = 4.0 x 10 ) Make an ICE table for the first dissociation \[C_6H_8O_{7(aq)} + H_2O_{(l)} \rightleftharpoons C_6H_7O_7^- + H_3O^+_{(aq)} \nonumber \] \[K_{a1} = \dfrac{[C_6H_7O_7^-,H_3O^+]}{[C_6H_8O_7]} = 7.5 \times 10^{-4} = \dfrac{x^2}{0.007 - x} \nonumber \] and use the quadratic formula to find that \[x = 0.00195 \;M = [H_3O^+] \nonumber \] Then a second ICE table can be made for the second dissociation \[C_6H_7O^-_{7(aq)} + H_2O_{(l)} \rightleftharpoons C_6H_6O^{2-}_7 + H_3O^+_{(aq)} \nonumber \] Remember that, for the first dissociation, x = [H O ] = [C H O ], so you can plug in the first value of x in for the initial concentrations of C H O and H O . \[K_{a2} = \dfrac{[C_6H_6O_7^{2-},H_3O^+]}{[C_6H_7O_7^-]} = 1.7 \times 10^{-5} = \dfrac{(x)(0.00195 + x)}{0.00195- x} \nonumber \] and use the quadratic formula to find that \[x = 1.67 \times 10^{-5} \nonumber \] \[[H_3O^+] = 0.00195 + 1.67 \times 10^{-5} = 0.00197 \;M \nonumber \] \[-\log(0.00197) = pH = 2.71 \nonumber \] Note that if you ignored the addition of hydronium from the second dissociation, then [H O ] = 0.00195 M, and using this value to calculate pH still gives you the answer of 2.71. So even though you made two ICE tables (you could even make a third table for K ), the protons donated in the second dissociation were negligible compared to the first dissociation. So you can see that it is really only the first dissociation that affects pH. Most often this is the case, and only one ICE table is necessary. It is up to you how certain you want to be and how many ICE tables you want to make when you calculate these problems. What is the pH of a saturated solution of sodium carbonate (Na CO )?
(solubility in water is 21.6 g/100mL at room temperature and for carbonic acid, H CO , K = 4.5 x 10 , K = 4.7 x 10 First, you have to find the find the initial concentration of CO which can be found from \[\dfrac{21.6g,Na_2CO_3}{} \times \dfrac{1\;mol}{105.99\;g} = 0.204\; mol\; Na_2CO_3 = 0.204\; mol\; CO_3^{2-} \nonumber \] then divide 0.204 mol by 0.100 L to get 2.04 M CO Plug into an ICE table \[CO^{2-}_{3(aq)} + H_2O_{(l)} \rightleftharpoons HCO^-_{3(aq)} + OH^-_{(aq)} \nonumber \] But notice that the equilibrium constants are for carbonic acid. If you were considering the dissociation of carbonic acid, you would write the following expressions \[K_{a1} = \dfrac{[HCO_3^-,H_3O^+]}{[H_2CO_3]} = 4.5 \times 10^{-7} \nonumber \] for \[H_2CO_3 + H_2O \rightleftharpoons HCO_3^- + H_3O^+ \nonumber \] \[K_{a2} = \dfrac{[CO_3^{2-},H_3O^+]}{[HCO_3^-]} = 4.7 \times 10^{-11} \nonumber \] for \[HCO_3^- + H_2O \rightleftharpoons CO_3^{2-} + H_3O^+ \nonumber \] The second acid ionization constant corresponds to the first base ionization constant (because the base reactions go backwards). To convert the second acid ionization constant to the first base ionization constant, you use the equation \[K_a \times K_b = K_w = 10^{-14} \nonumber \] so that \[K_{a2} \times K_{b1} = 10^{-14} \nonumber \] \[K_{b1} = \dfrac{10^{-14}}{4.7 \times 10^{-11}} = 2.13 \times 10^{-4} \nonumber \] Use the same equation to convert the first acid ionization constant to the second base ionization constant \[K_{a1} \times K_{b2} = 10^{-14} \nonumber \] \[K_{b2} = \dfrac{10^{-14}}{4.5 \times 10^{-7}} = 2.22 \times 10^{-8} \nonumber \] The expressions for the protonation of carbonate are now known to be \[K_{b1} = \dfrac{[HCO_3^-,OH^-]}{[CO_3^{2-}]} = 2.13 \times 10^{-4} \nonumber \] for \[CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^- \nonumber \] \[K_{b2} = \dfrac{[H_2CO_3,OH^-]}{[HCO_3^-]} = 2.22 \times 10^{-8} \nonumber \] for \[HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^- \nonumber \] Plug the ICE tables values into the first equilibrium expression \[K_{b1} = \dfrac{x^2}{2.04 - x} = 2.13 \times 10^{-4} \nonumber \] and use the quadratic formula to solve \[x = 0.0207\; M = [OH^-] \nonumber \] You can ignore the second base ionization constant because it removes a negligible amount of protons from water. If you want to test this by making an ICE table, you should get that the total hydroxide concentration is 0.0207000222 M \approx 0.0207 M so pOH = -log(0.0207) = 1.68 pH + pOH = 14
14 - 1.68 = 'Dilute' refers to the concentration of the acid or base in water. If the concentration is close to or below 10 M, then you must consider the donation of hydronium ions from water as well as from your acid or base. This is done by making an ICE table to find the protonation of the acid or base, while also incorporating the . An average bee sting contains 5 micrograms of formic acid (\(HCO_2H\)). What is the pH of a 500 mL solution of formic acid? (pK = 3.75) First calculate the number of moles of formic acid was excreted. \[\dfrac{5.00 \mu g}{} \times \dfrac{1 g}{10^6 \mu g} \times \dfrac{1\;mol,HCO_2H}{46.03g,HCO_2H} = 1.09 \times 10^{-7} mol,HCO_2H \nonumber \] \[\dfrac{1.09 \times 10^{-7} mol}{0.500 L} = 2.17 \times 10^{-7}\; M \nonumber \] \[HCO_2H_{(aq)} + H_2O_{(l)} \rightleftharpoons CO_2H^-_{(aq)} + H_3O^+_{(aq)} \nonumber \] A second ICE table can be made for the autoionization of water \[2H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)} \nonumber \] Notice that total [H O ] = x + y \[pK_a = 3.75 \nonumber \] \[K_a = 10^{-3.75} = 1.78 \times 10^{-4} \nonumber \] So you can simultaneously solve both equations \[K_a = \dfrac{[CO_2H^-,H_3O^+]}{[HCO_2H]} = \dfrac{x(x + y)}{2.17 \times 10^{-7} - x} = 1.78 \times 10^{-4} \nonumber \] and \[K_w = [H_3O^+,OH^-] = (x + y)(y) = 10^{-14} \nonumber \] These calculations can be tricky, and it is very easy to make mistakes. It is usually easier to use variables to solve these problems instead of handling awkward numbers. For this problem use \[a = \dfrac{x(x+y)}{c-x} \nonumber \] and \[(x+y)(y) = w \nonumber \] \[ac-ax = x^2 +xy \nonumber \] and \[x=\dfrac{w}{y}-y \nonumber \] \[ac-a\left(\dfrac{w}{y}-y\right)=\left(\dfrac{w}{y}-y\right)^2+\left(\dfrac{w}{y}-y\right)y \nonumber \] \[ac-\dfrac{aw}{y}+ay=\dfrac{w^2}{y^2}-\dfrac{2wy}{y}+y^2+\dfrac{wy}{y}-y^2 \nonumber \] \[ac-\dfrac{aw}{y}+ay=\dfrac{w^2}{y^2}-2w+y^2+w-y^2 \nonumber \] \[ac-\dfrac{aw}{y}+ay=\dfrac{w^2}{y^2}-w \nonumber \] \[ac-\dfrac{aw}{y}+ay+w=\dfrac{w^2}{y^2} \nonumber \] \[acy^2-awy+ay^3+wy^2=w^2 \nonumber \] \[ay^3+(ac+w)y^2-awy-w^2=0 \nonumber \] so when we plug back in the values \[(1.78 \times 10^{-4})y^3+((1.78 \times 10^{-4})(2.17 \times 10^{-7})+(10^{-14})) y^2-(1.78 \times 10^{-4})(10^{-14})y-(10^{-14})^2=0 \nonumber \] \[(1.78 \times 10^{-4})y^3+(3.86 \times 10^{-11})y^2-(1.78 \times 10^{-18})y-10^{-28}=0 \nonumber \] and use a graphing calculator to find that \[y = 3.91 \times 10^{-8} \nonumber \] and \[x=\dfrac{w}{y}-y = \dfrac{10^{-14}}{3.91 \times 10^{-8}}-3.91 \times 10^{-8} = 2.17 \times 10^{-7} \nonumber \] and total \[[H_3O^+] = (x + y) = (2.17 \times 10^{-7} + 3.91 \times 10^{-8}) = 2.56 \times 10^{-7} \;M \nonumber \] so \[-\log[H_3O^+] = -\log(2.56 \times 10^{-7}) = pH = 6.59 \nonumber \] Compare this value to pH = 6.66, which is what would have been calculated if the autoprotonization of water was not considered. Buffer solutions resist pH change when more acid or base is added. They are made from a weak acid and its base or a weak base and its conjugate acid. The can be used to find the pH of a buffer solution, and is derived from the acid equilibrium expression. \[HA_{(aq)} + H_2O_{(l)} \rightleftharpoons A^-_{(aq)} + H_3O^+_{(aq)} \nonumber \] \[K_a = \dfrac{[A^-,H_3O^+]}{[HA]} \nonumber \] \[\log(K_a) = log\left( \dfrac{[A^-,H_3O^+]}{[HA]} \right) \nonumber \] \[\log(K_a) = \log[H_3O^+] + \log \left( \dfrac{[A^-]}{[HA]} \right) \nonumber \] \[-pK_a = -pH + \log \left( \dfrac{[A^-]}{[HA]} \right) \nonumber \] \[pH = pK_a + \log \left( \dfrac{[A^-]}{[HA]} \right) \nonumber \] A similar equation can be used for bases \[pOH = pK_b + \log \left( \dfrac{[HB^+]}{[B]} \right) \nonumber \] The pH of blood plasma is 7.40, and is maintained by a carbonic acid/hydrogen carbonate buffer system. What mass of sodium bicarbonate (NaHCO ) should be added to a one liter solution of 0.250 M H CO to maintain the solution at pH of 7.40? (pKa = 6.35) \[pH = pK_a + \log \left( \dfrac{[HCO_3^-]}{[H_2CO_3]} \right) \nonumber \] \[7.40 = 6.35 + \log \left( \dfrac{[HCO_3^-]}{[0.250]} \right) \nonumber \] \[1.05 = \log[HCO_3^-] -\log(0.250) \nonumber \] \[0.448 = \log[HCO_3^-] \nonumber \] \[[HCO_3^-] = 0.356\; M \nonumber \] \[(0.356\; M) \times (1\; L\; solution) = 0.356\; mol\; HCO_3^- \nonumber \] \[[0.356 mol\,HCO_3^{-} \times \dfrac{1 mol\,NaHCO_3}{1 mol\,HCO_3^-} \times \dfrac{84.01g}{1 mol,NaHCO_3} = 29.9 g \text{sodium bicarbonate} \nonumber \] It is also possible to use the Henderson-Hasselbalch equation to find pK , pH, or [HA] if the other variables are given or calculated. Also notice that because \(\dfrac{[A^-]}{[HA]}\) will cancel out the unit of volume, moles of \(HA\) and \(A^-\) can be used instead of molarity. | 13,299 | 3,369 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/20%3A_Electrochemistry/20.8%3A_Industrial_Electrolysis_Processes |
Electrolysis reactions are the basic foundations of today's modern industry. There are various elements, chemical compounds, and organic compounds that are only produced by electrolysis including aluminum, chlorine, and NaOH. Electrolysis is the process by which an electric current spurs an otherwise non-spontaneous reaction. The electrorefining process refines metals or compounds at a high purity for a low cost. The pure metal can coat an otherwise worthless object. Let's consider the electrorefining process of copper: At the anode, there is an impure piece of copper that has other metals such as Ag, Au, Pt, Sn, Bi, Sb, As, Fe, Ni, Co, and Zn. The copper in this impure ore is oxidized to form Cu at the anode, and moves through an aqueous sulfuric acid-Copper (II) sulfate solution into the cathode. When it reaches the cathode, the Cu is reduced to Cu. This whole process takes place at a fairly low voltage (about .15 to .30 V), so Ag, Au, and Pt are not oxidized at the anode, as their standard oxidation electrode potentials are -.800, -1.36 and -1.20 respectively; these unoxidized impurities turn into a mixture called anode mud, a sludge at the bottom of the tank. This sludge can be recovered and used in different processes. Unlike Ag, Au and Pt, the impurities of Sb, Bi and Sn in the ore are indeed oxidized at the anode, but they are precipitated as they form hydroxides and oxides. Finally, Fe, Ni, Co and Zn are oxidized as well, but they are dissolved in water. Therefore, the only solid we are left with is the pure solid copper plate at the cathode, which has a purity level of about 99.999%. The image below gives an outline about the fate of the main components of an impure iron ore. Electrosynthesis is the method of producing substances through electrolysis reactions. This is useful when reaction conditions must be carefully controlled. One example of electrosynthesis is that of MnO , Manganese dioxide. MnO occurs naturally in the form of the mineral pyrolusite, but this mineral is not easily used due to the nature of its size and lattice structure. However, MnO can be obtained a different way, through the electrolysis of MnSO in a sulfuric acid solution. \(\begin{align}
&\textrm{Oxidation: }
&&\mathrm{Mn^{2+} + 2H_2O \rightarrow MnO_2 + 4H^+ + 2e^-}
&&\mathrm{\hspace{12px}E^0_{MnO_2/H_2}=-1.23}\\
&\textrm{Reduction: }
&&\mathrm{2e^- + 2H^+ \rightarrow H_2}
&&\mathrm{{-E}^0_{H^+/H_2}= -0}\\
&\textrm{Overall: }
&&\mathrm{Mn^{2+} + 2H_2O \rightarrow MnO_2 + 2H^+ +H_2}
&&\mathrm{\hspace{12px}E^0_{MnO_2/H_2} -E^0_{H^+/H_2}= -1.23 - 0= -1.23}
\end{align}\) The commercial process for organic chemicals that is currently practiced on a scale comparable to that of inorganic chemicals and metals is the electrohydrodimerization of acrylonitrile to adiponitrile. \(\begin{align}
&\textrm{Anode: }
&&\mathrm{H_2O \rightarrow 2H^+ + \dfrac{1}{2} O_2 + 2e^-}\\
&\textrm{Cathode: }
&&\ce{2CH2=CHCN + 2H2O + 2e- \rightarrow NC(CH2)CN + 2OH-}\\
&\textrm{Overall: }
&&\underset{\textrm{acrylonitrile}} {\ce{2CH2=CHCN}} + H_2O \rightarrow \dfrac{1}{2} O_2 + \underset{\textrm{adiponitrile}} {\ce{ NC(CH2)4CN}} \end{align}\) The importance of adiponitrile is that it can be readily converted to other useful compounds. This process is the electrolysis of sodium chloride (NaCl) at an industrial level. We will begin by discussing the equation for the chlor-alkali process, followed by discussing three different types of the process: the diaphragm cell, the mercury cell and the membrane cell. We will begin the explanation of the chlor-alkali process by determining the reactions that occur during the electrolysis of NaCl. Because NaCl is in an aqueous solution, we also have to consider the electrolysis of water at both the anode and the cathode. Therefore, there are two possible reduction equations and two possible oxidation reactions. \begin{align}
& \mathrm{Na^+_{\large{(aq)}} + 2e^- \rightarrow Na_{\large{(s)}}}
&& \mathrm{E^0_{Na^+/Na}= -2.71\: V \label{1}}\\
& \mathrm{2 H_2O_{\large{(l)}} + 2e^- \rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}}}
&& \mathrm{E^0_{H_2O/H_2}= -0.83\: V \label{2}}
\end{align} \(\begin{align}
&\mathrm{2Cl^-_{\large{(aq)}} \rightarrow Cl_2 + 2e^-}
&&\mathrm{-E^0_{Cl_2/Cl^-}= -1.36\: V \label{3}}\\
&\mathrm{2H_2O_{\large{(l)}} \rightarrow O_2 + 4H^+ + 4e^-}
&&\mathrm{-E^0_{O_2/H_2O}= -1.23\: V \label{4}}
\end{align}\) As we can see, due to the very much more negative electrode potential, the reduction of sodium ions is much less likely to occur than the reduction of water, so we can assume that in the electrolysis of NaCl, the reduction that occurs is Equation \(\ref{2}\). Therefore, we should try to determine what the oxidation reaction that occurs is. Let's say we have Equation \(\ref{2}\) as the reduction and Equation \(\ref{3}\) as the oxidation. We would get: \(\begin{align}
&\textrm{Reduction: }
&&\mathrm{2 H_2O_{\large{(l)}} + 2e^- \rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}}}
&&\mathrm{\hspace{12px}E^0_{H_20/H_2}= -0.83\: V \label{2a}}\\
&\textrm{Oxidation: }
&&\mathrm{2Cl^-_{\large{(aq)}} \rightarrow Cl_2 + 2e^-} &&\mathrm{-E^0_{Cl_2/Cl^-}= -(1.36\: V) \label{3a}} \\
&\textrm{Overall: }
&&\mathrm{2 H_2O_{\large{(l)}} + 2Cl^-_{\large{(aq)}}}
&&\mathrm{\hspace{12px}E^0_{H_20/H_2} - E^0_{Cl_2/Cl^-}\label{5a}}\\
& &&\mathrm{\hspace{10px}\rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}} +Cl_2}
&&\mathrm{\hspace{22px} = -0.83 + (-1.36)= -2.19}
\end{align}\) Alternatively, we could also have Equation \(\ref{2}\) with \(\ref{4}\) \(\begin{align}
&\textrm{Reduction: }
&&\mathrm{2\,[2 H_2O_{\large{(l)}} + 2e^- \rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}}]}
&&\mathrm{\hspace{12px}E_{H_2O/H_2O}=-.83\: V \label{2b}}\\
&\textrm{Oxidation: }
&&\mathrm{2H_2O_{\large{(l)}} \rightarrow O_2 + 4H^+ + 4e^-}
&&\mathrm{-E_{O_2/H_20}^0= -(1.23\: V) \label{4b}}\\
&\textrm{Overall: }
&&\mathrm{2H_2O_{\large{(l)}} \rightarrow 2H_{2\large{(g)}} + O_{2\large{(g)}}}
&&\mathrm{\hspace{12px}E^0_{H_2O/H_2} - E^0_{O_2/H_20}\label{6b}}\\
& && && \mathrm{\hspace{22px}= -0.83\: V -(1.23\: V)= -2.06}
\end{align}\) At first glance it would appear as though Equation \(\ref{6b}\) would occur due to the smaller (less negative) electrode potential. However, O actually has a fairly large overpotential, so instead Cl is more likely to form, making Equation \(\ref{5a}\) the most probable outcome for the electrolysis of NaCl. Depending on the method used, there can be several different products produced through the chlor-alkali process. The value of these products is what makes the chlor-alkali process so important. The name comes from the two main products of the process, chlorine and the alkali, sodium hydroxide (NaOH). Therefore, one of the main uses of the chlor-alkali process is the production of NaOH. As described earlier, the equation for the chlor-alkali process, that is, the electrolysis of NaCl, is as follows: \(\begin{align}
&\textrm{Reduction: }
&&\mathrm{2 H_2O_{\large{(l)}} + 2e^- \rightarrow H_{2\large{(g)}} +2OH^-_{\large{(aq)}}}
&&\mathrm{E_{H_2O/H_2O}= -0.83\; V}\\
&\textrm{Oxidation: }
&&\mathrm{2Cl^-_{\large{(aq)}} \rightarrow Cl_2 + 2e^-}
&&\mathrm{E^0_{Cl_2/Cl^-}= -1.36\;V}\\
&\textrm{Overall: }
&&\mathrm{2Cl^- + 2H_2O_{\large{(l)}} \rightarrow 2 OH^- + H_{2\large{(g)}} +Cl_{2\large{(g)}}}
&&\mathrm{E^0_{H_2O/H_2}- E^0_{Cl/Cl_2^-}}\\
& && &&\hspace{12px}\mathrm{= -0.83\:V- (-1.36\: V)= -2.19\: V}
\end{align}\) **The chlor-alkali process often occurs in an apparatus called a diaphragm cell, which is illustrated in Figure \(\Page {1}\). If chlorine comes into contact with hydrogen, it produces a mixture which will explode violently on exposure to sunlight or heat. Hydrogen chloride gas would be produced. Obviously, the two gases need to be kept apart. To even further improve the purity of NaOH, a mercury cell can be used for the location of electrolysis, opposed to a diaphragm cell. In the mercury-cell process, also known as the Castner–Kellner process, a saturated brine solution floats on top of a thin layer of mercury. The mercury is the cathode, where sodium is produced and forms a sodium-mercury amalgam with the mercury. The amalgam is continuously drawn out of the cell and reacted with water which decomposes the amalgam into sodium hydroxide and mercury. The mercury is recycled into the electrolytic cell. Chlorine is produced at the anode and evaporates out of the cell. Mercury cells are being phased out due to concerns about mercury poisoning from mercury cell pollution A third way to make even more pure NaOH is to use a membrane cell. It is preferred over the diaphragm cell or mercury cell method because it uses the least amount of electric energy and produces the highest quality NaOH. For instance, it can produce NaOH with a degree of chlorine ion contamination of only 50 ppm. An ion-permeable membrane is used to separate the anode and cathode. Saturated brine is fed to the compartment. Current passed through the cell splits the sodium chloride into its constituent components, Na and Cl . Chloride ions are oxidized to chlorine gas at the anode. \[ 2Cl^-_{(aq)} \rightarrow 2e^- + Cl_{2(g)}\] The chlorine is purified by liquifying it under pressure. The oxygen stays as a gas when it is compressed at ordinary temperatures. The membrane passes Na ions to the cathode compartment. The hydrogen is produced at the nickel cathode: \[ 2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}\] The membrane is made from a polymer which only allows positive ions to pass through it. That means that the only the sodium ions from the sodium chloride solution can pass through the membrane - and not the chloride ions. The advantage of this is that the sodium hydroxide solution being formed in the right-hand compartment never gets contaminated with any sodium chloride solution. The sodium chloride solution being used has to be pure. If it contained any other metal ions, these would also pass through the membrane and so contaminate the sodium hydroxide solution. Jim Clark ( ) | 10,118 | 3,370 |
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Acids and bases have been known for a long time. When Robert characterized them in 1680, he noted that acids dissolve many substances, change the color of certain natural dyes (for example, they change litmus from blue to red), and lose these characteristic properties after coming into contact with alkalis (bases). In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO ), and interact with alkalis to form neutral substances. In 1815, Humphry contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions. Acids and bases are common solutions that exist everywhere. Almost every liquid that we encounter in our daily lives consists of acidic and basic properties, with the exception of water. They have completely different properties and are able to neutralize to form H O, which will be discussed later in a subsection. Acids and bases can be defined by their physical and chemical observations (Table \(\Page {1}\)). Acids and bases in aqueous solutions will conduct electricity because they contain dissolved ions. Therefore, acids and bases are . Strong acids and bases will be strong electrolytes. Weak acids and bases will be weak electrolytes. This affects the amount of conductivity. In 1884, the Swedish chemist Svante Arrhenius proposed two specific classifications of compounds, termed acids and bases. When dissolved in an aqueous solution, certain ions were released into the solution. The Arrhenius definition of acid-base reactions is a development of the "hydrogen theory of acids". It was used to provide a modern definition of acids and bases, and followed from Arrhenius's work with Friedrich Wilhelm Ostwald in establishing the presence of ions in aqueous solution in 1884. This led to Arrhenius receiving the Nobel Prize in Chemistry in 1903. An Arrhenius acid is a compound that increases the concentration of \(H^+\) ions that are present when added to water. These \(H^+\) ions form the hydronium ion (\(H_3O^+\)) when they combine with water molecules. This process is represented in a chemical equation by adding H O to the reactants side. \[ HCl_{(aq)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} \nonumber \] In this reaction, hydrochloric acid (\(HCl\)) dissociates into hydrogen (\(H^+\)) and chlorine (\(Cl^-\)) ions when dissolved in water, thereby releasing H ions into solution. Formation of the hydronium ion equation: \[ HCl_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)} + Cl^-_{(aq)} \nonumber \] The Arrhenius definitions of acidity and alkalinity are restricted to aqueous solutions and refer to the concentration of the solvated ions. Under this definition, pure \(H_2SO_4\) or \(HCl\) dissolved in toluene are not acidic, despite the fact that both of these acids will donate a proton to toluene. In addition, under the Arrhenius definition, a solution of sodium amide (\(NaNH_2\)) in liquid ammonia is not alkaline, despite the fact that the amide ion (\(NH^−_2\)) will readily deprotonate ammonia. Thus, the Arrhenius definition can only describe acids and bases in an aqueous environment. The Arrhenius definition can describe acids and bases in an aqueous environment. In chemistry, acids and bases have been defined differently by three sets of theories: One is the Arrhenius definition defined above, which revolves around the idea that acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (\(H^+\)) ions while bases produce hydroxide (\(OH^-\)) ions in solution. The other two definitions are discussed in detail alter in the chapter and include the Brønsted-Lowry definition the defines acids as substances that donate protons (\(H^+\)) whereas bases are substances that accept protons and the Lewis theory of acids and bases states that acids are electron pair acceptors while bases are electron pair donors. | 4,442 | 3,371 |
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In proposing his theory that octets can be completed by two atoms sharing electron pairs, Lewis provided scientists with the first description of covalent bonding. In this section, we expand on this and describe some of the properties of covalent bonds. The stability of a molecule is a function of the strength of the covalent bonds holding the atoms together. Triple bonds between like atoms are shorter than double bonds, and because more energy is required to completely break all three bonds than to completely break two, a triple bond is also stronger than a double bond. Similarly, double bonds between like atoms are stronger and shorter than single bonds. Bonds of the same order between atoms show a wide range of bond energies, however. Table \(\Page {1}\) lists the average values for some commonly encountered bonds. Although the values shown vary widely, we can observe four trends: Bonds between hydrogen and atoms in a given column in the periodic table are weaker down the column; bonds between like atoms usually become weaker down a column. Bond strengths increase as bond order , while bond distances . Bond energy is defined as the energy required to break a particular bond in a molecule in the gas phase. Its value depends on not only the identity of the bonded atoms but also their environment. Thus the bond energy of a C–H single bond is the same in all organic compounds. For example, the energy required to break a C–H bond in methane varies by as much as 25% depending on how many other bonds in the molecule have already been broken (Table \(\Page {2}\)); that is, the C–H bond energy depends on its molecular environment. Except for diatomic molecules, the bond energies listed in Table \(\Page {1}\) are values for all bonds of a given type in a range of molecules. Even so, they are not likely to differ from the actual value of a given bond by more than about 10%. We can estimate the enthalpy change for a chemical reaction by adding together the average energies of the bonds broken in the reactants and the average energies of the bonds formed in the products and then calculating the difference between the two. If the bonds formed in the products are stronger than those broken in the reactants, then energy will be released in the reaction (\(ΔH_{rxn} < 0\)): \[ ΔH_{rxn} \approx \sum{\text{(bond energies of bonds broken)}}−\sum{\text{(bond energies of bonds formed)}} \label{\(\Page {1}\)} \] The ≈ sign is used because we are adding together bond energies; hence this approach does not give exact values for Δ . Let’s consider the reaction of 1 mol of -heptane (C H ) with oxygen gas to give carbon dioxide and water. This is one reaction that occurs during the combustion of gasoline: \[\ce{CH3(CH2)5CH3(l) + 11 O2(g) \rightarrow 7 CO2(g) + 8 H2O(g)} \label{\(\Page {2}\)} \] In this reaction, 6 C–C bonds, 16 C–H bonds, and 11 O=O bonds are broken per mole of -heptane, while 14 C=O bonds (two for each CO ) and 16 O–H bonds (two for each H O) are formed. The energy changes can be tabulated as follows: The bonds in the products are stronger than the bonds in the reactants by about 4444 kJ/mol. This means that \(ΔH_{rxn}\) is approximately −4444 kJ/mol, and the reaction is highly exothermic (which is not too surprising for a combustion reaction). If we compare this approximation with the value obtained from measured \(ΔH_f^o\) values (\(ΔH_{rxn} = −481\;7 kJ/mol\)), we find a discrepancy of only about 8%, less than the 10% typically encountered. Chemists find this method useful for calculating approximate enthalpies of reaction for molecules whose actual \(ΔH^ο_f\) values are unknown. These approximations can be important for predicting whether a reaction is or —and to what degree. The compound (Research Development Explosive) is a more powerful explosive than dynamite and is used by the military. When detonated, it produces gaseous products and heat according to the following reaction. Use the approximate bond energies in Table \(\Page {1}\) to estimate the \(ΔH_{rxn}\) per mole of RDX. chemical reaction, structure of reactant, and Table \(\Page {1}\). \(ΔH_{rxn}\) per mole We must add together the energies of the bonds in the reactants and compare that quantity with the sum of the energies of the bonds in the products. A nitro group (–NO ) can be viewed as having one N–O single bond and one N=O double bond, as follows: In fact, however, both N–O distances are usually the same because of the presence of two equivalent resonance structures. We can organize our data by constructing a table: From Equation \(\Page {1}\), we have \[ \begin{align*} ΔH_{rxn} &\approx \sum{\text{(bond energies of bonds broken)}}−\sum{\text{(bond energies of bonds formed)}} \\[4pt] &= 7962 \; kJ/mol − 10,374 \; kJ/mol \\[4pt] &=−2412 \;kJ/mol \end{align*} \nonumber \] Thus this reaction is also highly exothermic The molecule -142b is a hydrochlorofluorocarbon that is used in place of chlorofluorocarbons (CFCs) such as the Freons and can be prepared by adding HCl to 1,1-difluoroethylene: Use tabulated bond energies to calculate \(ΔH_{rxn}\). −54 kJ/mol Bond Dissociation Energy (also referred to as Bond energy) is the enthalpy change (\(\Delta H\), heat input) required to break a bond (in 1 mole of a gaseous substance) What about when we have a compound which is not a diatomic molecule? Consider the dissociation of : There are four equivalent C-H bonds, thus we can that the dissociation energy for a single C-H bond would be: \[ \begin{align*} D(C-H) &= (1660/4)\, kJ/mol \\[4pt] &= 415 \,kJ/mol \end{align*} \nonumber \] The bond energy for a given bond is influenced by the rest of the molecule. However, this is a relatively small effect (suggesting that bonding electrons are localized between the bonding atoms). Thus, the bond energy for most bonds varies little from the average bonding energy for that type of bond Bond energy is always a value - it takes energy to break a covalent bond (conversely energy is released during bond formation) The more stable a molecule (i.e. the stronger the bonds) the less likely the molecule is to undergo a chemical reaction. If we know which bonds are broken and which bonds are made during a chemical reaction, we can estimate the enthalpy change of the reaction (\(\Delta H_{rxn}\)) even if we do not know the enthalpies of formation ((\(\Delta H_{f}^o\))for the reactants and products: \[\Delta H = \sum \text{bond energies of broken bonds} - \sum \text{bond energies of formed bonds} \label{8.8.3} \] What is the enthalpy of reaction between 1 mol of chlorine and 1 mol methane? We use Equation \ref{8.8.3}, which requires tabulating bonds broken and formed. \[\begin{align*} \Delta H &= [D(Cl-Cl) + D(C-H)] - [D(H-Cl)+D(C-Cl)] \\[4pt] &= [242 kJ + 413 kJ] - [431 kJ + 328 kJ] \\[4pt] &= -104 \,kJ \end{align*} \nonumber \] Thus, the reaction is (because the bonds in the products are stronger than the bonds in the reactants) What is the enthalpy of reaction for the combustion of 1 mol of ethane? We use Equation \ref{8.8.3}, which requires tabulating bonds broken and formed. \[\begin{align*} \Delta {H} &= [(6 \times 413) + (348) + (\frac{7}{2} \times 495)] - [(4 \times 799) + (6 \times 463)] \\[4pt] &= 4558 - 5974 \\[4pt] &= -1416\; kJ \end{align*} \nonumber \] Therefor the reaction is exothermic. As the number of bonds between two atoms increases, the bond grows shorter and stronger is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. In closely related compounds with bonds between the same kinds of atoms, the bond with the highest bond order is both the shortest and the strongest. In bonds with the same bond order between different atoms, trends are observed that, with few exceptions, result in the strongest single bonds being formed between the smallest atoms. Tabulated values of average bond energies can be used to calculate the enthalpy change of many chemical reactions. If the bonds in the products are stronger than those in the reactants, the reaction is exothermic and vice versa. The breakage and formation of bonds is similar to a relationship: you can either get married or divorced and it is more favorable to be married. | 8,358 | 3,372 |
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Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per liter per unit time, written as M/s, M/min, or M/h. To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses, perhaps plot the concentration as a function of time on a graph, and then calculate the change in the concentration per unit time. The progress of a simple reaction (A → B) is shown in the beakers are snapshots of the composition of the solution at 10 s intervals. The number of molecules of reactant (A) and product (B) are plotted as a function of time in the graph. Each point in the graph corresponds to one beaker in . The reaction rate is the change in the concentration of either the reactant or the product over a period of time. The concentration of A decreases with time, while the concentration of B increases with time. \[\textrm{rate}=\dfrac{\Delta [\textrm B]}{\Delta t}=-\dfrac{\Delta [\textrm A]}{\Delta t} \label{Eq1} \] Square brackets indicate molar concentrations, and the capital Greek delta (Δ) means “change in.” Because chemists follow the convention of expressing all reaction rates as positive numbers, however, a negative sign is inserted in front of Δ[A]/Δt to convert that expression to a positive number. The reaction rate calculated for the reaction A → B using Equation \(\ref{Eq1}\) is different for each interval (this is not true for every reaction, as shown below). A greater change occurs in [A] and [B] during the first 10 s interval, for example, than during the last, meaning that the reaction rate is greatest at first. Reaction rates generally decrease with time as reactant concentrations decrease. A Discussing Average Reaction Rates. Link: We can use Equation \(\ref{Eq1}\) to determine the reaction rate of hydrolysis of aspirin, probably the most commonly used drug in the world (more than 25,000,000 kg are produced annually worldwide). Aspirin (acetylsalicylic acid) reacts with water (such as water in body fluids) to give salicylic acid and acetic acid, as shown in Figure \(\Page {2}\). Because salicylic acid is the actual substance that relieves pain and reduces fever and inflammation, a great deal of research has focused on understanding this reaction and the factors that affect its rate. Data for the hydrolysis of a sample of aspirin are in Table \(\Page {1}\) and are shown in the graph in . The data in Table \(\Page {1}\) were obtained by removing samples of the reaction mixture at the indicated times and analyzing them for the concentrations of the reactant (aspirin) and one of the products (salicylic acid). The for a given time interval can be calculated from the concentrations of either the reactant or one of the products at the beginning of the interval (time = t ) and at the end of the interval (t ). Using salicylic acid, the reaction rate for the interval between t = 0 h and t = 2.0 h (recall that change is always calculated as final minus initial) is calculated as follows: The reaction rate can also be calculated from the concentrations of aspirin at the beginning and the end of the same interval, remembering to insert a negative sign, because its concentration decreases: If the reaction rate is calculated during the last interval given in Table \(\Page {1}\)(the interval between 200 h and 300 h after the start of the reaction), the reaction rate is significantly slower than it was during the first interval (t = 0–2.0 h): In the preceding example, the stoichiometric coefficients in the balanced chemical equation are the same for all reactants and products; that is, the reactants and products all have the coefficient 1. Consider a reaction in which the coefficients are not all the same, the fermentation of sucrose to ethanol and carbon dioxide: \[\underset{\textrm{sucrose}}{\mathrm{C_{12}H_{22}O_{11}(aq)}}+\mathrm{H_2O(l)}\rightarrow\mathrm{4C_2H_5OH(aq)}+4\mathrm{CO_2(g)} \label{Eq2} \] The coefficients indicate that the reaction produces four molecules of ethanol and four molecules of carbon dioxide for every one molecule of sucrose consumed. As before, the reaction rate can be found from the change in the concentration of any reactant or product. In this particular case, however, a chemist would probably use the concentration of either sucrose or ethanol because gases are usually measured as volumes and, as explained in , the volume of CO gas formed depends on the total volume of the solution being studied and the solubility of the gas in the solution, not just the concentration of sucrose. The coefficients in the balanced chemical equation tell us that the reaction rate at which ethanol is formed is always four times faster than the reaction rate at which sucrose is consumed: \[\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t}=-\dfrac{4\Delta[\textrm{sucrose}]}{\Delta t} \label{Eq3} \] The concentration of the reactant—in this case sucrose— with time, so the value of Δ[sucrose] is negative. Consequently, a minus sign is inserted in front of Δ[sucrose] in so the rate of change of the sucrose concentration is expressed as a positive value. Conversely, the ethanol concentration with time, so its rate of change is automatically expressed as a positive value. Often the reaction rate is expressed in terms of the reactant or product with the smallest coefficient in the balanced chemical equation. The smallest coefficient in the sucrose fermentation reaction ( ) corresponds to sucrose, so the reaction rate is generally defined as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm{sucrose}]}{\Delta t}=\dfrac{1}{4}\left (\dfrac{\Delta[\mathrm{C_2H_5OH}]}{\Delta t} \right ) \label{Eq4} \] Consider the thermal decomposition of gaseous N O to NO and O via the following equation: Write expressions for the reaction rate in terms of the rates of change in the concentrations of the reactant and each product with time. balanced chemical equation reaction rate expressions Because O has the smallest coefficient in the balanced chemical equation for the reaction, define the reaction rate as the rate of change in the concentration of O and write that expression. The balanced chemical equation shows that 2 mol of N O must decompose for each 1 mol of O produced and that 4 mol of NO are produced for every 1 mol of O produced. The molar ratios of O to N O and to NO are thus 1:2 and 1:4, respectively. This means that the rate of change of [N O ] and [NO ] must be divided by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate. For example, because NO is produced at four times the rate of O , the rate of production of NO is divided by 4. The reaction rate expressions are as follows: \(\textrm{rate}=\dfrac{\Delta[\mathrm O_2]}{\Delta t}=\dfrac{\Delta[\mathrm{NO_2}]}{4\Delta t}=-\dfrac{\Delta[\mathrm{N_2O_5}]}{2\Delta t}\) The is used in the manufacture of sulfuric acid. A key step in this process is the reaction of \(SO_2\) with \(O_2\) to produce \(SO_3\). \[2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber \] Write expressions for the reaction rate in terms of the rate of change of the concentration of each species. The of a reaction is the reaction rate at any given point in time. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate. Comparing this to calculus, the instantaneous rate of a reaction at a given time corresponds to the slope of a line tangent to the concentration-versus-time curve at that point—that is, the derivative of concentration with respect to time. The distinction between the instantaneous and average rates of a reaction is similar to the distinction between the actual speed of a car at any given time on a trip and the average speed of the car for the entire trip. Although the car may travel for an extended period at 65 mph on an interstate highway during a long trip, there may be times when it travels only 25 mph in construction zones or 0 mph if you stop for meals or gas. The average speed on the trip may be only 50 mph, whereas the instantaneous speed on the interstate at a given moment may be 65 mph. Whether the car can be stopped in time to avoid an accident depends on its instantaneous speed, not its average speed. There are important differences between the speed of a car during a trip and the speed of a chemical reaction, however. The speed of a car may vary unpredictably over the length of a trip, and the initial part of a trip is often one of the slowest. In a chemical reaction, the initial interval typically has the fastest rate (though this is not always the case), and the reaction rate generally changes smoothly over time. Chemical kinetics generally focuses on one particular instantaneous rate, which is the initial reaction rate, t = 0. Initial rates are determined by measuring the reaction rate at various times and then extrapolating a plot of rate versus time to t = 0. Using the reaction shown in Example \(\Page {1}\), calculate the reaction rate from the following data taken at 56°C: \[2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)} \nonumber \] balanced chemical equation and concentrations at specific times reaction rate Calculate the reaction rate in the interval between t = 240 s and t = 600 s. From Example \(\Page {1}\), the reaction rate can be evaluated using any of three expressions: Subtracting the initial concentration from the final concentration of N O and inserting the corresponding time interval into the rate expression for N O , Substituting actual values into the expression, Similarly, NO can be used to calculate the reaction rate: Allowing for experimental error, this is the same rate obtained using the data for N O . The data for O can also be used: Again, this is the same value obtained from the N O and NO data. Thus, the reaction rate does not depend on which reactant or product is used to measure it. Using the data in the following table, calculate the reaction rate of \(SO_2(g)\) with \(O_2(g)\) to give \(SO_3(g)\). \[2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)} \nonumber \] In this Module, the quantitative determination of a reaction rate is demonstrated. Reaction rates can be determined over particular time intervals or at a given point in time. A rate law describes the relationship between reactant rates and reactant concentrations. Reaction rates are reported as either the average rate over a period of time or as the instantaneous rate at a single time. Reaction rates can be determined over particular time intervals or at a given point in time. | 10,775 | 3,373 |
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It is possible to use kinetics studies of a chemical system, such as the effect of changes in reactant concentrations, to deduce events that occur on a microscopic scale, such as collisions between individual particles. Such studies have led to the collision model of chemical kinetics, which is a useful tool for understanding the behavior of reacting chemical species. The collision model explains why chemical reactions often occur more rapidly at higher temperatures. For example, the reaction rates of many reactions that occur at room temperature approximately double with a temperature increase of only 10°C. In this section, we will use the collision model to analyze this relationship between temperature and reaction rates. Before delving into the relationship between temperature and reaction rate, we must discuss three microscopic factors that influence the observed macroscopic reaction rates. Central to collision model is that a chemical reaction can occur only when the reactant molecules, atoms, or ions collide. Hence, the observed rate is influence by the frequency of collisions between the reactants. The is the average rate in which two reactants collide for a given system and is used to express the average number of collisions per unit of time in a defined system. While deriving the collisional frequency (\(Z_{AB}\)) between two species in a gas is , it is beyond the scope of this text and the equation for collisional frequency of \(A\) and \(B\) is the following: \[Z_{AB} = N_{A}N_{B}\left(r_{A} + r_{B}\right)^2\sqrt{ \dfrac{8\pi k_{B}T}{\mu_{AB}}} \label{freq} \] with The specifics of Equation \ref{freq} are not important for this conversation, but it is important to identify that \(Z_{AB}\) increases with increasing density (i.e., increasing \(N_A\) and \(N_B\)), with increasing reactant size ( . A Discussing Collision Theory of Kinetics: Previously, we discussed the kinetic molecular theory of gases, which showed that the average kinetic energy of the particles of a gas increases with increasing temperature. Because the speed of a particle is proportional to the square root of its kinetic energy, increasing the temperature will also increase the number of collisions between molecules per unit time. What the kinetic molecular theory of gases does not explain is why the reaction rate of most reactions approximately doubles with a 10°C temperature increase. This result is surprisingly large considering that a 10°C increase in the temperature of a gas from 300 K to 310 K increases the kinetic energy of the particles by only about 4%, leading to an increase in molecular speed of only about 2% and a correspondingly small increase in the number of bimolecular collisions per unit time. The collision model of chemical kinetics explains this behavior by introducing the concept of (\(E_a\)). We will define this concept using the reaction of \(\ce{NO}\) with ozone, which plays an important role in the depletion of ozone in the ozone layer: \[\ce{NO(g) + O_3(g) \rightarrow NO_2(g) + O_2(g)} \nonumber \] Increasing the temperature from 200 K to 350 K causes the rate constant for this particular reaction to increase by a factor of more than 10, whereas the increase in the frequency of bimolecular collisions over this temperature range is only 30%. Thus something other than an increase in the collision rate must be affecting the reaction rate. Experimental rate law for this reaction is \[\text{rate} = k [\ce{NO},\ce{O3}] \nonumber \] and is used to identify how the reaction rate (not the rate constant) vares with concentration. The rate constant, however, does vary with temperature. gure \(\Page {1}\) shows a plot of the rate constant of the reaction of \(\ce{NO}\) with \(\ce{O3}\) at various temperatures. The relationship is not linear but instead resembles the relationships seen in graphs of vapor pressure versus temperature (e.g, the ). In all three cases, the shape of the plots results from a distribution of kinetic energy over a population of particles (electrons in the case of conductivity; molecules in the case of vapor pressure; and molecules, atoms, or ions in the case of reaction rates). Only a fraction of the particles have sufficient energy to overcome an energy barrier. In the case of vapor pressure, particles must overcome an energy barrier to escape from the liquid phase to the gas phase. This barrier corresponds to the energy of the intermolecular forces that hold the molecules together in the liquid. In conductivity, the barrier is the energy gap between the filled and empty bands. In chemical reactions, the energy barrier corresponds to the amount of energy the particles must have to react when they collide. This energy threshold, called the , was first postulated in 1888 by the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry 1903). It is the minimum amount of energy needed for a reaction to occur. Reacting molecules must have enough energy to overcome electrostatic repulsion, and a minimum amount of energy is required to break chemical bonds so that new ones may be formed. Molecules that collide with less than the threshold energy bounce off one another chemically unchanged, with only their direction of travel and their speed altered by the collision. Molecules that are able to overcome the energy barrier are able to react and form an arrangement of atoms called the or the of the reaction. The activated complex is not a reaction intermediate; it does not last long enough to be detected readily. Any phenomenon that depends on the distribution of thermal energy in a population of particles has a nonlinear temperature dependence. We can graph the energy of a reaction by plotting the potential energy of the system as the reaction progresses. shows a plot for the NO–O system, in which the vertical axis is potential energy and the horizontal axis is the reaction coordinate, which indicates the progress of the reaction with time. The activated complex is shown in brackets with an asterisk. The overall change in potential energy for the reaction (\(ΔE\)) is negative, which means that the reaction releases energy. (In this case, \(ΔE\) is −200.8 kJ/mol.) To react, however, the molecules must overcome the energy barrier to reaction (\(E_a\) is 9.6 kJ/mol). That is, 9.6 kJ/mol must be put into the system as the activation energy. Below this threshold, the particles do not have enough energy for the reaction to occur. re \(\Page {3a}\) illustrates the general situation in which the products have a lower potential energy than the reactants. In contrast, F re \(\Page {3b}\) illustrates the case in which the products have a higher potential energy than the reactants, so the overall reaction requires an input of energy; that is, it is energetically uphill, and \(Δ > 0\). Although the energy changes that result from a reaction can be positive, negative, or even zero, in most cases an energy barrier must be overcome before a reaction can occur. This means that the activation energy is almost always positive; there is a class of reactions called barrierless reactions, but those are discussed elsewhere. For similar reactions under comparable conditions, the one with the smallest will occur most rapidly. Whereas \(ΔE\) is related to the tendency of a reaction to occur spontaneously, \(E_a\) gives us information about the reaction rate and how rapidly the reaction rate changes with temperature. For two similar reactions under comparable conditions, the reaction with the smallest \(E_a\) will occur more rapidly. Figure \(\Page {4}\) shows both the kinetic energy distributions and a potential energy diagram for a reaction. The shaded areas show that at the lower temperature (300 K), only a small fraction of molecules collide with kinetic energy greater than ; however, at the higher temperature (500 K) a much larger fraction of molecules collide with kinetic energy greater than . Consequently, the reaction rate is much slower at the lower temperature because only a relatively few molecules collide with enough energy to overcome the potential energy barrier. Discussing Transition State Theory: Even when the energy of collisions between two reactant species is greater than \(E_a\), most collisions do not produce a reaction. The probability of a reaction occurring depends not only on the collision energy but also on the spatial orientation of the molecules when they collide. For \(\ce{NO}\) and \(\ce{O3}\) to produce \(\ce{NO2}\) and \(\ce{O2}\), a terminal oxygen atom of \(\ce{O3}\) must collide with the nitrogen atom of \(\ce{NO}\) at an angle that allows \(\ce{O3}\) to transfer an oxygen atom to \(\ce{NO}\) to produce \(\ce{NO2}\) ( re \(\Page {4}\)). All other collisions produce no reaction. Because fewer than 1% of all possible orientations of \(\ce{NO}\) and \(\ce{O3}\) result in a reaction at kinetic energies greater than \(E_a\), most collisions of \(\ce{NO}\) and \(\ce{O3}\) are unproductive. The fraction of orientations that result in a reaction is called the (\(\rho\)) and its value can range from \(\rho=0\) (no orientations of molecules result in reaction) to \(\rho=1\) (all orientations result in reaction). The collision model explains why most collisions between molecules do not result in a chemical reaction. For example, nitrogen and oxygen molecules in a single liter of air at room temperature and 1 atm of pressure collide about 10 times per second. If every collision produced two molecules of \(\ce{NO}\), the atmosphere would have been converted to \(\ce{NO}\) and then \(\ce{NO2}\) a long time ago. Instead, in most collisions, the molecules simply bounce off one another without reacting, much as marbles bounce off each other when they collide. For an \(A + B\) elementary reaction, all three microscopic factors discussed above that affect the reaction rate can be summarized in a single relationship: \[\text{rate} = (\text{collision frequency}) \times (\text{steric factor}) \times (\text{fraction of collisions with } E > E_a ) \nonumber \] where \[\text{rate} = k[A,B] \label{14.5.2} \] Arrhenius used these relationships to arrive at an equation that relates the magnitude of the rate constant for a reaction to the temperature, the activation energy, and the constant, \(A\), called the : \[k=Ae^{-E_{\Large a}/RT} \label{14.5.3} \] The frequency factor is used to convert concentrations to collisions per second (scaled by the steric factor). Because the frequency of collisions depends on the temperature, \(A\) is actually not constant (Equation \ref{freq}). Instead, \(A\) increases slightly with temperature as the increased kinetic energy of molecules at higher temperatures causes them to move slightly faster and thus undergo more collisions per unit time. is known as the and summarizes the collision model of chemical kinetics, where \(T\) is the absolute temperature (in K) and is the ideal gas constant [8.314 J/(K·mol)]. \(E_a\) indicates the sensitivity of the reaction to changes in temperature. The reaction rate with a large \(E_a\) increases rapidly with increasing temperature, whereas the reaction rate with a smaller \(E_a\) increases much more slowly with increasing temperature. If we know the reaction rate at various temperatures, we can use the Arrhenius equation to calculate the activation energy. Taking the natural logarithm of both sides of , \[\begin{align} \ln k &=\ln A+\left(-\dfrac{E_{\textrm a}}{RT}\right) \\[4pt] &=\ln A+\left[\left(-\dfrac{E_{\textrm a}}{R}\right)\left(\dfrac{1}{T}\right)\right] \label{14.5.4} \end{align} \] is the equation of a straight line, \[y = mx + b \nonumber \] where \(y = \ln k\) and \(x = 1/T\). This means that a plot of \(\ln k\) versus \(1/T\) is a straight line with a slope of \(−E_a/R\) and an intercept of \(\ln A\). In fact, we need to measure the reaction rate at only two temperatures to estimate \(E_a\). Knowing the \(E_a\) at one temperature allows us to predict the reaction rate at other temperatures. This is important in cooking and food preservation, for example, as well as in controlling industrial reactions to prevent potential disasters. The procedure for determining \(E_a\) from reaction rates measured at several temperatures is illustrated in Example \(\Page {1}\). A Discussing The Arrhenius Equation: Many people believe that the rate of a tree cricket’s chirping is related to temperature. To see whether this is true, biologists have carried out accurate measurements of the rate of tree cricket chirping (\(f\)) as a function of temperature (\(T\)). Use the data in the following table, along with the graph of ln[chirping rate] versus \(1/T\) to calculate \(E_a\) for the biochemical reaction that controls cricket chirping. Then predict the chirping rate on a very hot evening, when the temperature is 308 K (35°C, or 95°F). chirping rate at various temperatures activation energy and chirping rate at specified temperature If cricket chirping is controlled by a reaction that obeys the Arrhenius equation, then a plot of \(\ln f\) versus \(1/T\) should give a straight line ( ). Also, the slope of the plot of \(\ln f\) versus \(1/T\) should be equal to \(−E_a/R\). We can use the two endpoints in to estimate the slope: \[\begin{align*}\textrm{slope}&=\dfrac{\Delta\ln f}{\Delta(1/T)} \\[4pt] &=\dfrac{5.30-4.37}{3.34\times10^{-3}\textrm{ K}^{-1}-3.48\times10^{-3}\textrm{ K}^{-1}} \\[4pt] &=\dfrac{0.93}{-0.14\times10^{-3}\textrm{ K}^{-1}} \\[4pt] &=-6.6\times10^3\textrm{ K}\end{align*} \nonumber \] A computer best-fit line through all the points has a slope of −6.67 × 10 K, so our estimate is very close. We now use it to solve for the activation energy: \[\begin{align*} E_{\textrm a} &=-(\textrm{slope})(R) \\[4pt] &=-(-6.6\times10^3\textrm{ K})\left(\dfrac{8.314 \textrm{ J}}{\mathrm{K\cdot mol}}\right)\left(\dfrac{\textrm{1 KJ}}{\textrm{1000 J}}\right) \\[4pt] &=\dfrac{\textrm{55 kJ}}{\textrm{mol}} \end{align*} \nonumber \] If the activation energy of a reaction and the rate constant at one temperature are known, then we can calculate the reaction rate at any other temperature. We can use to express the known rate constant (\(k_1\)) at the first temperature (\(T_1\)) as follows: \[\ln k_1=\ln A-\dfrac{E_{\textrm a}}{RT_1} \nonumber \] Similarly, we can express the unknown rate constant (\(k_2\)) at the second temperature (\(T_2\)) as follows: \[\ln k_2=\ln A-\dfrac{E_{\textrm a}}{RT_2} \nonumber \] These two equations contain four known quantities ( , , , and ) and two unknowns ( and ). We can eliminate by subtracting the first equation from the second: \[\begin{align*} \ln k_2-\ln k_1 &=\left(\ln A-\dfrac{E_{\textrm a}}{RT_2}\right)-\left(\ln A-\dfrac{E_{\textrm a}}{RT_1}\right) \\[4pt] &=-\dfrac{E_{\textrm a}}{RT_2}+\dfrac{E_{\textrm a}}{RT_1} \end{align*} \nonumber \] Then \[\ln \dfrac{k_2}{k_1}=\dfrac{E_{\textrm a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \nonumber \] To obtain the best prediction of chirping rate at 308 K ( ), we try to choose for and the measured rate constant and corresponding temperature in the data table that is closest to the best-fit line in the graph. Choosing data for = 296 K, where = 158, and using the \(E_a\) calculated previously, \[\begin{align*} \ln\dfrac{k_{T_2}}{k_{T_1}} &=\dfrac{E_{\textrm a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \\[4pt] &=\dfrac{55\textrm{ kJ/mol}}{8.314\textrm{ J}/(\mathrm{K\cdot mol})}\left(\dfrac{1000\textrm{ J}}{\textrm{1 kJ}}\right)\left(\dfrac{1}{296 \textrm{ K}}-\dfrac{1}{\textrm{308 K}}\right) \\[4pt] &=0.87 \end{align*} \nonumber \] Thus / = 2.4 and = (2.4)(158) = 380, and the chirping rate on a night when the temperature is 308 K is predicted to be 380 chirps per minute. The equation for the decomposition of \(\ce{NO2}\) to \(\ce{NO}\) and \(\ce{O2}\) is second order in \(\ce{NO2}\): \[\ce{2NO2(g) → 2NO(g) + O2(g)} \nonumber \] Data for the reaction rate as a function of temperature are listed in the following table. Calculate \(E_a\) for the reaction and the rate constant at 700 K. \(E_a\) = 114 kJ/mol; = 18,600 M ·s = 1.86 × 10 M ·s . What \(E_a\) results in a doubling of the reaction rate with a 10°C increase in temperature from 20° to 30°C? about 51 kJ/mol A Discussing Graphing Using the Arrhenius Equation: For a chemical reaction to occur, an energy threshold must be overcome, and the reacting species must also have the correct spatial orientation. The Arrhenius equation is \(k=Ae^{-E_{\Large a}/RT}\). A minimum energy (activation energy,v\(E_a\)) is required for a collision between molecules to result in a chemical reaction. Plots of potential energy for a system versus the reaction coordinate show an energy barrier that must be overcome for the reaction to occur. The arrangement of atoms at the highest point of this barrier is the activated complex, or transition state, of the reaction. At a given temperature, the higher the , the slower the reaction. The fraction of orientations that result in a reaction is the steric factor. The frequency factor, steric factor, and activation energy are related to the rate constant in the Arrhenius equation: \(k=Ae^{-E_{\Large a}/RT}\). A plot of the natural logarithm of versus 1/ is a straight line with a slope of − / . | 17,336 | 3,374 |
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We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously. As you have learned, the molecules of a gas are stationary but in constant and random motion. If someone opens a bottle of perfume in the next room, for example, you are likely to be aware of it soon. Your sense of smell relies on molecules of the aromatic substance coming into contact with specialized olfactory cells in your nasal passages, which contain specific receptors (protein molecules) that recognize the substance. How do the molecules responsible for the aroma get from the perfume bottle to your nose? You might think that they are blown by drafts, but, in fact, molecules can move from one place to another even in a draft-free environment. is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, , is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space. The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship , after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses Helium ( = 4.00 g/mol) effuses much more rapidly than ethylene oxide ( = 44.0 g/mol). Because helium is less dense than air, helium-filled balloons “float” at the end of a tethering string. Unfortunately, rubber balloons filled with helium soon lose their buoyancy along with much of their volume. In contrast, rubber balloons filled with air tend to retain their shape and volume for a much longer time. Because helium has a molar mass of 4.00 g/mol, whereas air has an average molar mass of about 29 g/mol, pure helium effuses through the microscopic pores in the rubber balloon \(\sqrt{\dfrac{29}{4.00}}=2.7\) times faster than air. For this reason, high-quality helium-filled balloons are usually made of Mylar, a dense, strong, opaque material with a high molecular mass that forms films that have many fewer pores than rubber. Hence, mylar balloons can retain their helium for days. At a given temperature, heavier molecules move more slowly than lighter molecules. During World War , scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of \(\ce{^{235}U}\). Naturally occurring uranium is only 0.720% \(\ce{^{235}U}\), whereas most of the rest (99.275%) is \(\ce{^{238}U}\), which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound \(UF_6\) (boiling point = 56°C). Given: isotopic content of naturally occurring uranium and atomic masses of U and U Asked for: ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure UF Strategy: A The first step is to calculate the molar mass of UF containing U and U. Luckily for the success of the separation method, fluorine consists of a single isotope of atomic mass 18.998. The molar mass of UF is The molar mass of UF is The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law using Equation \(\ref{10.8.1}\): \[\rm\dfrac{\text{rate }^{235}UF_6}{\text{rate }^{238}UF_6}=\sqrt{\dfrac{352.04\;g/mol}{349.03\;g/mol}}=1.0043 \nonumber \] To obtain 99.0% pure UF requires many steps. We can set up an equation that relates the initial and final purity to the number of times the separation process is repeated: In this case, 0.990 = (0.00720)(1.0043) , which can be rearranged to give \[1.0043^n=\dfrac{0.990}{0.00720}=137.50 \nonumber \] Taking the logarithm of both sides gives \[\begin{align} n\ln(1.0043)&=\ln(137.50) \\[4pt] n &=\dfrac{\ln(137.50)}{\ln(1.0043)} \\[4pt]&=1148 \end{align} \nonumber \] Thus at least a thousand effusion steps are necessary to obtain highly enriched U. Below is a small part of a system that is used to prepare enriched uranium on a large scale. Helium consists of two isotopes: He (natural abundance = 0.000134%) and He (natural abundance = 99.999866%). Their atomic masses are 3.01603 and 4.00260, respectively. Helium-3 has unique physical properties and is used in the study of ultralow temperatures. It is separated from the more abundant He by a process of gaseous effusion. ratio of effusion rates = 1.15200; one step gives 0.000154% He 96 steps Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses: \[KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{10.8.2} \] Multiplying both sides by 2 and rearranging give \[\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{10.8.3} \] Taking the square root of both sides gives \[\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{10.8.4} \] Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation \(\ref{10.8.4}\) shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy. Typically, gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour). The effect of molar mass on these speeds is dramatic, as illustrated in Figure \(\Page {3}\) for some common gases. Because all gases have the same average kinetic energy, according to the , molecules with lower masses, such as hydrogen and helium, have a wider distribution of speeds. The lightest gases have a wider distribution of speeds and the highest average speeds. Molecules with lower masses have a wider distribution of speeds and a higher average speed. Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 10 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line, as illustrated schematically in Figure \(\Page {4}\). The average distance traveled by a molecule between collisions is the . The denser the gas, the shorter the mean free path; conversely, as density decreases, the mean free path becomes longer because collisions occur less frequently. At 1 atm pressure and 25°C, for example, an oxygen or nitrogen molecule in the atmosphere travels only about 6.0 × 10 m (60 nm) between collisions. In the upper atmosphere at about 100 km altitude, where gas density is much lower, the mean free path is about 10 cm; in space between galaxies, it can be as long as 1 × 10 m (about 6 million miles). The the gas, the the mean free path. Calculate the rms speed of a sample -butene (C H ) at 20°C. compound and temperature rms speed Calculate the molar mass of cis-2-butene. Be certain that all quantities are expressed in the appropriate units and then use Equation 10.8.5 to calculate the rms speed of the gas. To use Equation 10.8.4, we need to calculate the molar mass of -2-butene and make sure that each quantity is expressed in the appropriate units. Butene is C H , so its molar mass is 56.11 g/mol. Thus \[\begin{align} u_{\rm rms} &= \sqrt{\dfrac{3RT}{M}} \\[4pt] &=\rm\sqrt{\dfrac{3\times8.3145\;\dfrac{J}{K\cdot mol}\times(20+273)\;K}{56.11\times10^{-3}\;kg}}\\[4pt] &=361\;m/s \end{align} \nonumber \] or approximately 810 mi/h. Calculate the rms speed of a sample of radon gas at 23°C. \(1.82 \times 10^2\; m/s\) (about 410 mi/h) The kinetic molecular theory of gases demonstrates how a successful theory can explain previously observed empirical relationships (laws) in an intuitively satisfying way. Unfortunately, the actual gases that we encounter are not “ideal,” although their behavior usually approximates that of an ideal gas. is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass ( ), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases. The of a molecule is the average distance it travels between collisions. | 9,655 | 3,375 |
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We begin with our understanding of the relationship between chemical behavior and atomic structure. That is, we assume the Periodic Law that the chemical and physical properties of the elements are periodic functions of atomic number. We further assume the structure of the atom as a massive, positively charged nucleus, whose size is much smaller than that of the atom as a whole, surrounded by a vast open space in which move negatively charged electrons. These electrons can be effectively partitioned into a core and a valence shell, and it is only the electrons in the valence shell which are significant to the chemical properties of the atom. The number of valence electrons in each atom is equal to the group number of that element in the Periodic Table. The atomic molecular theory is extremely useful in explaining what it means to form a compound from its component elements. That is, a compound consists of identical molecules, each comprised of the atoms of he component elements in a simple whole number ratio. However, the atomic molecular theory also opens up a wide range of new questions. We would like to know what atomic properties determine the number of atoms of each type which combine to form stable compounds. Why are some combinations observed and other combinations not observed? Some elements with very dissimilar atomic masses (for example, iodine and chlorine) form very similar chemical compounds, but other elements with very similar atomic masses (for example, oxygen and nitrogen) form very dissimilar compounds. What factors are responsible for the bonding properties of the elements in a similar group? In general, we need to know what forces hold atoms together in forming a molecule. We have developed a detailed understanding of the structure of the atom. Our task now is to apply this understanding to develop a similar level of detail about how atoms bond together to form molecules. To begin our analysis of chemical bonding, we define the of an atom by its tendencies to form molecules. The inert gases do not tend to combine with any other atoms. We thus assign their valence as 0, meaning that these atoms tend to form 0 bonds. Each halogen prefers to form molecules by combining with a single hydrogen atom (e.g. \(\ce{HF}\), \(\ce{HCl}\)). We thus assign their valence as 1, taking hydrogen to also have a valence of 1. What we mean by a valence of 1 is that these atoms prefer to bind to only one other atom. The valence of oxygen, sulfur, etc. is assigned as 2, since two hydrogens are required to satisfy bonding needs of these atoms. Nitrogen, phosphorus, etc. have a valence of 3, and carbon and silicon have a valence of 4. This concept also applies to elements just following the inert gases. Lithium, sodium, potassium, and rubidium bind with a single halogen atom. Therefore, they also have a valence of 1. Correspondingly, it is not surprising to find that, for example, the combination of two potassium atoms with a single oxygen atom forms a stable molecule, since oxygen's valence of 2 is satisfied by the two alkali atoms, each with valence 1. We can proceed in this manner to assign a valence to each element, by simply determining the number of atoms to which this element's atoms prefer to bind. In doing so, we discover that the periodic table is a representation of the valences of the elements: elements in the same group all share a common valence. The inert gases with a valence of 0 sit to one side of the table. Each inert gas is immediately preceded in the table by one of the halogens: fluorine precedes neon, chlorine precedes argon, bromine precedes krypton, and iodine precedes xenon. And each halogen has a valence of one. This "one step away, valence of one" pattern can be extended. The elements just prior to the halogens (oxygen, sulfur, selenium, tellurium) are each two steps away from the inert gases in the table, and each of these elements has a valence of two (e.g. \(\ce{H_2O}\), \(\ce{H_2S}\)). The elements just preceding these (nitrogen, phosphorus, antimony, arsenic) have valences of three (e.g. \(\ce{NH_3}\), \(\ce{PH_3}\)), and the elements before that (carbon and silicon most notably) have valences of four (\(\ce{CH_4}\), \(\ce{SiH_4}\)). The two groups of elements immediately after the inert gases, the alkali metals and the alkaline earths, have valences of one and two, respectively. Hence, for many elements in the periodic table, the valence of its atoms can be predicted from the number of steps the element is away from the nearest inert gas in the table. This systemization is quite remarkable and is very useful for remembering what molecules may be easily formed by a particular element. Next we discover that there is a pattern to the valences: for elements in groups 4 through 8 (e.g. carbon through neon), the valence of each atom the number of electrons in the valence shell in that atom always equals . For example, carbon has a valence of 4 and has 4 valence electrons, nitrogen has a valence of 3 and has 5 valence electrons, and oxygen has a valence of 2 and has 6 valence electrons. Hydrogen is an important special case with a single valence electron and a valence of 1. Interestingly, for each of these atoms, the number of bonds the atom forms is equal to the number of vacancies in its valence shell. To account for this pattern, we develop a model assuming that each atom attempts to bond to other atoms so as to completely fills its valence shell with electrons. For elements in groups 4 through 8, this means that each atom attempts to complete an "octet" of valence shell electrons. (Why atoms should behave this way is a question unanswered by this model.) Consider, for example, the combination of hydrogen and chlorine to form hydrogen chloride, \(\ce{HCl}\). The chlorine atom has seven valence electrons and seeks to add a single electron to complete an octet. Hence, chlorine has a valence of 1. Either hydrogen or chlorine could satisfy its valence by "taking" an electron from the other atom, but this would leave the second atom now needing two electrons to complete its valence shell. The only way for both atoms to complete their valence shells simultaneously is to two electrons. Each atom donates a single electron to the electron pair which is shared. It is this sharing of electrons that we refer to as a chemical bond, or more specifically, as a , so named because the bond acts to satisfy the valence of both atoms. The two atoms are thus held together by the need to share the electron pair. Many of the most important chemical fuels are compounds composed entirely of carbon and hydrogen, i.e. hydrocarbons. The smallest of these is methane, \(\ce{CH_4}\), a primary component of household natural gas. Other simple common fuels include ethane, \(\ce{C_2H_6}\), propane, \(\ce{C_3H_8}\), butane, \(\ce{C_4H_{10}}\), pentane, \(\ce{C_5H_{12}}\), hexane, \(\ce{C_6H_{14}}\), heptane, \(\ce{C_7H_{16}}\), and octane, \(\ce{C_8H_{18}}\). It is interesting to note that there is a consistency in these molecular formulae: in each case, the number of hydrogen atoms is two more than twice the number of carbon atoms, so that each compound has a molecular formula like \(\ce{C}_n \ce{H}_{2n+2}\). This suggests that there are strong similarities in the valences of the atoms involved which should be understandable in terms of our valence shell electron pair sharing model. In each molecule, the carbon atoms must be directly bonded together, since they cannot be joined together with a hydrogen atom. In the easiest example of ethane, the two carbon atoms are bonded together, and each carbon atom is in turn bonded to three hydrogen atoms. Thus, in this case, it is relatively apparent that the valence of each carbon atom is 4, just as in methane, since each is bonded to four other atoms. Therefore, by sharing an electron pair with each of the four atoms to which it is bonded, each carbon atom has a valence shell of eight electrons. In most other cases, it is not so trivial to determine which atoms are bonded to which, as there may be multiple possibilities which satisfy all atomic valences. Nor is it trivial, as the number of atoms and electrons increases, to determine whether each atom has an octet of electrons in its valence shell. We need a system of electron accounting which permits us to see these features more clearly. To this end, we adopt a standard notation for each atom which displays the number of valence electrons in the unbonded atom explicitly. In this notation, carbon and hydrogen look like Figure 6.1, representing the single valence electron in hydrogen and the four valence electrons in carbon. Using this notation, it is now relatively easy to represent the shared electron pairs and the carbon atom valence shell octets in methane and ethane. Linking bonded atoms together and pairing the valence shell electrons from each gives Figure 6.2. Recall that each shared pair of electrons represents a chemical bond. These are examples of what are called , after G. N. Lewis who first invented this notation. These structures reveal, at a glance, which atoms are bonded to which, i.e., the structural formula of the molecule. We can also easily count the number of valence shell electrons around each atom in the bonded molecule. Consistent with our model of the octet rule, each carbon atom has eight valence electrons and each hydrogen has two in the molecule. In a larger hydrocarbon, the structural formula of the molecule is generally not predictable from the number of carbon atoms and the number of hydrogen atoms, so the molecular structure must be given to deduce the Lewis structure and thus the arrangement of the electrons in the molecule. However, once given this information, it is straightforward to create a Lewis structure for molecules with the general molecular formula \(\ce{C}_n \ce{H}_{2n + 2}\) such as propane, butane, etc. For example, the Lewis structure for "normal" butane (with all carbons linked one after another) is found in Figure 6.3. It is important to note that there exist no hydrocarbons where the number of hydrogens exceeds two more than twice the number of carbons, for example, \(\ce{CH_5}\) does not exist, nor does \(\ce{C_2H_8}\). We correspondingly find that all attempts to draw Lewis structures which are consistent with the octet rule will fail for these molecules. Similarly, \(\ce{CH_3}\) and \(\ce{C_2H_5}\) are observed to be so extremely reactive that it is impossible to prepare stable quantities of either compound. Again we find that it is not possible to draw Lewis structures for these molecules which obey the octet rule. We conclude from these examples that, when it is possible to draw a Lewis structure in which each carbon has a complete octet of electrons in its valence shell, the corresponding molecule will be stable and the hydrocarbon compound will exist under ordinary conditions. After working a few examples, it is apparent that this always holds for compounds with molecular formula \(\ce{C}_n \ce{H}_{2n + 2}\). On the other hand, there are many stable hydrocarbon compounds with molecular formulae which do not fit the form \(\ce{C}_n \ce{H}_{2n + 2}\), particularly where the number of hydrogens is less than \(2n + 2\). In these compounds, the valences of the carbon atoms are not quite so obviously satisfied by electron pair sharing. For example, in ethene, \(\ce{C_2H_4}\) and acetylene, \(\ce{C_2H_2}\) there are not enough hydrogen atoms to permit each carbon atom to be bonded to four atoms each. In each molecule, the two carbon atoms must be bonded to one another. By simply arranging the electrons so that the carbon atoms share a single pair of electrons, we wind up with rather unsatisfying Lewis structures for ethene and acetylene, shown in Figure 6.4. Note that, in these structures, neither carbon atom has a complete octet of valence shell electrons. Moreover, these structures indicate that the carbon-carbon bonds in ethane, ethene, and acetylene should be very similar, since in each case a single pair of electrons is shared by the two carbons. However, these bonds are observed to be chemically and physically very different. First, we can compare the energy required to break each bond (the or ). We find that the carbon-carbon bond energy is \(347 \: \text{kJ}\) in \(\ce{C_2H_6}\), \(589 \: \text{kJ}\) in \(\ce{C_2H_4}\), and \(962 \: \text{kJ}\) in \(\ce{C_2H_2}\). Second, it is possible to observe the distance between the two carbon atoms, which is referred to as the . It is found that the carbon-carbon bond length is \(154 \: \text{pm}\) in \(\ce{C_2H_6}\), \(134 \: \text{pm}\) in \(\ce{C_2H_4}\), and \(120 \: \text{pm}\) in \(\ce{C_2H_2}\). (1 picometer \(= 1 \: \text{pm} = 10^{-12} \: \text{m}\)). These observations reveal clearly that the bonding between the carbon atoms in these three molecules must be very different. Note that the bond in ethene is about one and a half times as strong as the bond in ethane; this suggests that the two unpaired and unshared electrons in the ethene structure above are also paired and shared as a second bond between the two carbon atoms. Similarly, since the bond in acetylene is about two and a half times stronger than the bond in ethane, we can imagine that this results from the sharing of three pairs of electrons between the two carbon atoms. These assumptions produce the Lewis structures in Figure 6.5. These structures appear sensible from two regards. First, the trend in carbon-carbon bond strengths can be understood as arising from the increasing number of shared pairs of electrons. Second, each carbon atom has a complete octet of electrons. We refer to the two pairs of shared electrons in ethene as a and the three shared pairs in acetylene as a . We thus extend our model of valence shell electron pair sharing to conclude that carbon atoms can bond by sharing one, two, or three pairs of electrons as needed to complete an octet of electrons, and that the strength of the bond is greater when more pairs of electrons are shared. Moreover, the data above tell us that the carbon-carbon bond in acetylene is shorter than that in ethene, which is shorter than that in ethane. We conclude that triple bonds are shorter than double bonds which are shorter than single bonds. Many compounds composed primarily of carbon and hydrogen also contain some oxygen or nitrogen, or one or more of the halogens. We thus seek to extend our understanding of bonding and stability by developing Lewis structures involving these atoms. Recall that a nitrogen atom has a valence of 3 and has five valence electrons. In our notation, we could draw a structure in which each of the five electrons appears separately in a ring, similar to what we drew for \(\ce{C}\). However, this would imply that a nitrogen atom would generally form five bonds to pair its five valence electrons. Since the valence is actually 3, our notation should reflect this. One possibility looks like Figure 6.6. Note that this structure leaves three of the valence electrons "unpaired" and thus ready to join in a shared electron pair. The remaining two valence electrons are "paired", and this notation implies that they therefore are not generally available for sharing in a covalent bond. This notation is consistent with the available data, i.e. five valence electrons and a valence of 3. Pairing the two non-bonding electrons seems reasonable in analogy to the fact that electrons are paired in forming covalent bonds. Analogous structures can be drawn for oxygen, as well as for fluorine and the other halogens, as shown in Figure 6.7. With this notation in hand, we can now analyze structures for molecules including nitrogen, oxygen, and the halogens. The hydrides are the easiest, shown in Figure 6.8. Note that the octet rule is clearly obeyed for oxygen, nitrogen, and the halogens. At this point, it becomes very helpful to adopt one new convention: a pair of bonded electrons now be more easily represented in our Lewis structures by a straight line, rather than two dots. Double bonds and triple bonds are represented by double and triple straight lines between atoms. We will continue to show non-bonded electron pairs explicitly. As before, when analyzing Lewis structures for larger molecules, we must already know which atoms are bonded to which. For example, two very different compounds, ethanol and dimethyl ether, both have molecular formula \(\ce{C_2H_6O}\). In ethanol, the two carbon atoms are bonded together and the oxygen atom is attached to one of the two carbons; the hydrogens are arranged to complete the valences of the carbons and the oxygen, shown in Figure 6.9. This Lewis structure reveals not only that each carbon and oxygen atom has a completed octet of valence shell electrons but also that, in the stable molecule, there are four non-bonded electrons on the oxygen atom. Ethanol is an example of an . Alcohols can be easily recognized in Lewis structures by the \(\ce{C-O-H}\) group. The Lewis structures of all alcohols obey the octet rule. In dimethyl ether, the two carbons are each bonded to the oxygen, in the middle, shown in Figure 6.10. can be recognized in Lewis structures by the \(\ce{C-O-C}\) arrangement. Note that, in both ethanol and dimethyl ether, the octet rule is obeyed for all carbon and oxygen atoms. Therefore, it is not usually possible to predict the structural formula of a molecule from Lewis structures. We must know the molecular structure prior to determining the Lewis structure. Ethanol and dimethyl ether are examples of , molecules with the same molecular formula but different structural formulae. In general, isomers have rather different chemical and physical properties arising from their differences in molecular structures. A group of compounds called contain hydrogen, carbon, and nitrogen. The simplest amine is methyl amine, whose Lewis structure is shown in Figure 6.11. "Halogenated" hydrocarbons have been used extensively as refrigerants in air conditioning systems and refrigerators. These are the notorious "chlorofluorocarbons" or "CFCs" which have been implicated in the destruction of stratospheric ozone. Two of the more important CFCs include Freon 11, \(\ce{CFCl_3}\), and Freon 114, \(\ce{C_2F_4Cl_2}\), for which we can easily construct appropriate Lewis structures, shown in Figure 6.12. Finally, Lewis structures account for the stability of the diatomic form of the elemental halogens \(\ce{F_2}\), \(\ce{Cl_2}\), \(\ce{Br_2}\), and \(\ce{I_2}\). The single example of \(\ce{F_2}\) is sufficient, shown in Figure 6.13. We can conclude from these examples that molecules containing oxygen, nitrogen, and the halogens are expected to be stable when these atoms all have octets of electrons in their valence shells. The Lewis structure of each molecule reveals this character explicitly. On the other hand, there are many examples of common molecules with apparently unusual valences, including: carbon dioxide, \(\ce{CO_2}\), in which the carbon is bonded to only two atoms and each oxygen is only bonded to one; formaldehyde \(\ce{H_2CO}\); and hydrogen cyanide, \(\ce{HCN}\). Perhaps most conspicuously, we have yet to understand the bonding in two very important elemental diatomic molecules, \(\ce{O_2}\) and \(\ce{N_2}\), each of which has fewer atoms than the valence of either atom. We first analyze \(\ce{CO_2}\), noting that the bond strength of one of the \(\ce{C-O}\) bonds in carbon dioxide is \(532 \: \text{kJ}\), which is significantly greater than the bond strength of the \(\ce{C-O}\) bond in ethanol, \(358 \: \text{kJ}\). By analogy to the comparison of bond strengths in ethane to ethene, we can imagine that this difference in bond strengths results from double bonding in \(\ce{CO_2}\). Indeed, a Lewis structure of \(\ce{CO_2}\) in which only single electron pairs are shared (Figure 6.14) does not obey the octet rule, but one in which we pair and share the extra electrons reveals that double bonding permits the octet rule to be obeyed (Figure 6.15). A comparison of bond lengths is consistent with our reasoning: the single \(\ce{C-O}\) bond in ethanol is \(148 \: \text{pm}\), whereas the double bond in \(\ce{CO_2}\) is \(116 \: \text{pm}\). Knowing that oxygen atoms can double-bond, we can easily account for the structure of formaldehyde. The strength of the \(\ce{C-O}\) bond in \(\ce{H_2CO}\) is comparable to that in \(\ce{CO_2}\), consistent with the Lewis structure shown in Figure 6.16. What about nitrogen atoms? We can compare the strength of the \(\ce{C-N}\) bond in \(\ce{HCN}\), \(880 \: \text{kJ}\), to that in methyl amine, \(290 \: \text{kJ}\). This dramatic disparity again suggests the possibility of multiple bonding, and an appropriate Lewis structure for \(\ce{HCN}\) is shown in Figure 6.17. We can conclude that oxygen and nitrogen atoms, like carbon atoms, are capable of multiple bonding. Furthermore, our observations of oxygen and nitrogen reinforce our earlier deduction that multiple bonds are stronger than single bonds, and their bond lengths are shorter. As our final examples in this section, we consider molecules in which oxygen atoms are bonded to oxygen atoms. Oxygen-oxygen bonds appear primarily in two types of molecules. The first is simply the oxygen diatomic molecule, \(\ce{O_2}\), and the second are the peroxides, typified by hydrogen peroxide, \(\ce{H_2O_2}\). In a comparison of bond energies, we find that the strength of the \(\ce{O-O}\) bond in \(\ce{O_2}\) is \(499 \: \text{kJ}\) whereas the strength of the \(\ce{O-O}\) bond in \(\ce{H_2O_2}\) is \(142 \: \text{kJ}\). This is easily understood in a comparison of the Lewis structures of these molecules, showing that the peroxide bon is a single bond, whereas the \(\ce{O_2}\) bond is a double bond, shown in Figure 6.18. We conclude that an oxygen atom can satisfy its valence of 2 by forming two single bonds or by forming one double bond. In both cases, we can understand the stability of the resulting molecules by in terms of an octet of valence electrons. Before further developing our model of chemical bonding based on Lewis structures, we pause to consider the interpretation and importance of these structures. It is worth recalling that we have developed our model based on observations of the numbers of bonds formed by individual atoms and the number of valence electrons in each atom. In general, these structures are useful for predicting whether a molecule is expected to be stable under normal conditions. If we cannot draw a Lewis structure in which each carbon, oxygen, nitrogen, or halogen has an octet of valence electrons, then the corresponding molecule probably is not stable. Consideration of bond strengths and bond lengths enhances the model by revealing the presence of double and triple bonds in the Lewis structures of some molecules. At this point, however, we have observed no information regarding the geometries of molecules. For example, we have not considered the angles measured between bonds in molecules. Consequently, the Lewis structure model of chemical bonding does not at this level predict or interpret these bond angles. (This will be considered in Module 7.) Therefore, although the Lewis structure of methane is drawn as shown in Figure 6.19, this does imply that methane is a flat molecule, or that the angles between \(\ce{C-H}\) bonds in methane is \(90^\text{o}\). Rather, the structure simply reveals that the carbon atom has a complete octet of valence electrons in a methane molecule, that all bonds are single bonds, and that there are no non-bonding electrons. Similarly, one can write the Lewis structure for a water molecule in two apparently different ways, shown in Figure 6.20. However, it is very important to realize that these two structures are in the Lewis model, because both show that the oxygen atom as a complete octet of valence electrons, forms two single bonds with hydrogen atoms, and has two pairs of unshared electrons in its valence shell. In the same way, the two structures for Freon 114 shown in Figure 6.21 are also . These two drawings do not represent different structures or arrangements of the atoms in the bonds. Finally, we must keep in mind that we have drawn Lewis structures strictly as a convenient tool for our understanding of chemical bonding and molecular stability. It is based on commonly observed trends in valence, bonding, and bond strengths. These structures must not be mistaken as observations themselves, however. As we encounter addition experimental observations, we must be prepared to adapt our Lewis structure model to fit these observations, but we must never adapt our observations to fit the Lewis model. With these thoughts in mind, we turn to a set of molecules which challenge the limits of the Lewis model in describing molecular structures. First, we note that there are a variety of molecules for which atoms clearly must bond in such a way as to have more than eight valence electrons. A conspicuous example is \(\ce{SF_6}\), where the sulfur atom is bonded to six \(\ce{F}\) atoms. As such, the \(\ce{S}\) atom must have 12 valence shell electrons to form 6 covalent bonds. Similarly, the phosphorus atom in \(\ce{PCl_5}\) has 10 valence electrons in 5 covalent bonds, the \(\ce{Cl}\) atom in \(\ce{ClF_3}\) has 10 valence electrons in 3 covalent bonds and two lone pairs. We also observe the interesting compounds of the noble gas atoms, e.g. \(\ce{XeO_3}\), where the noble gas atom begins with eight valence electrons even before forming any bonds. In each of these cases, we note that the valence of the atoms \(\ce{S}\), \(\ce{P}\), \(\ce{Cl}\), and \(\ce{Xe}\) are normally 2, 3, 1, and 0, yet more bonds than this are formed. In such cases, it is not possible to draw Lewis structures in which \(\ce{S}\), \(\ce{P}\), \(\ce{Cl}\), and \(\ce{Xe}\) obey the octet rule. We refer to these molecules as "expanded valence" molecules, meaning that the valence of the central atom has expanded beyond the expected octet. There are also a variety of molecules for which there are too few electrons to provide an octet for every atom. Most notably, Boron and Aluminum, from Group III, display bonding behavior somewhat different than we have seen and thus less predictable from the model we have developed so far. These atoms have three valence shell electrons, so we might predict a valence of 5 on the basis of the octet rule. However, compounds in which boron or aluminum atoms form five bonds are never observed, so we must conclude that simple predictions based on the octet rule are not reliable for Group III. Consider first boron trifluoride, \(\ce{BF_3}\). The bonding is relatively simple to model with a Lewis structure (Figure 6.22) if we allow each valence shell electron in the boron atom to be shared in a covalent bond with each fluorine atom. Note that, in this structure, the boron atom has only six valence shell electrons, but the octet rule is obeyed by the fluorine atoms. We might conclude from this one example that boron atoms obey a sextet rule. However, boron will form a stable ion with hydrogen, \(\ce{BH_4^-}\), in which the boron atom does have a complete octet. In addition, \(\ce{BF_3}\) will react with ammonia, \(\ce{NH_3}\), to form a stable compound, \(\ce{NH_3BF_3}\), for which a Lewis structure can be drawn in which boron has a complete octet, shown in Figure 6.23. Compounds of aluminum follow similar trends. Aluminum trichloride, \(\ce{AlCl_3}\), aluminum hydride, \(\ce{AlH_3}\), and aluminum hydroxide, \(\ce{Al(OH)_3}\), all indicate a valence of 3 for aluminum, with six valence electrons in the bonded molecule. However, the stability of aluminum hydride ions, \(\ce{AlH_4^-}\), indicates that \(\ce{Al}\) can also support an octet of valence shell electrons as well. We conclude that, although the octet rule can still be of some utility in understanding the chemistry of Boron and Aluminum, the compounds of these elements are less predictable from the octet rule. This should not be disconcerting, however. The octet rule was developed on the basis of the observation that, for elements in Groups IV through VIII, the number of valence electrons plus the most common valence is equal to eight. Elements in Groups I, II, and III do not follow this observation most commonly. Another interesting challenge for the Lewis model we have developed is the set of molecules for which it is possible to draw more than one structure in agreement with the octet rule. A notable example is the nitric acid molecule, \(\ce{HNO_3}\), where all three oxygens are bonded to the nitrogen. Two structures can be drawn for nitric acid with nitrogen and all three oxygens obeying the octet rule. In each structure, of the oxygens not bonded to hydrogen, one shares a single bond with nitrogen while the other shares a double bond with nitrogen. These two structures are not identical, unlike the two freon structures in Figure 6.21, because the atoms are bonded differently in the two structures. Compounds with formulae of the form \(\ce{C}_n \ce{H}_{2n + 2}\) are often referred to as "saturated" hydrocarbons. Using Lewis structures, explain how and in what sense these molecules are "saturated". Molecules with formulae of the form \(\ce{C}_n \ce{H}_{2n + 1}\) (e.g. \(\ce{CH_3}\), \(\ce{C_2H_5}\)) are called "radicals" and are extremely reactive. Using Lewis structures, explain the reactivity of these molecules. State and explain the experimental evidence and reasoning which shows that multiple bonds are stronger and shorter than single bonds. Compare \(\ce{N_2}\) to \(\ce{H_4N_2}\). Predict which bond is stronger and explain why. Explain why the two Lewis structures for Freon 114, shown in Figure 6.21, are identical. Draw a Lewis Structure for an isomer of Freon 114, that is, another molecule with the same molecular formula as Freon 114 but a different structural formula. ; Chemistry) | 30,035 | 3,376 |
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The data we collect are characterized by their central tendency (where the values cluster), and their spread (the variation of individual values around the central value). We report our data’s central tendency by stating the mean or median, and our data’s spread using the range, standard deviation or variance. Our collection of data is subject to errors, including determinate errors that affect the data’s accuracy and indeterminate errors that affect its precision. A propagation of uncertainty allows us to estimate how these determinate and indeterminate errors affect our results. When we analyze a sample several times the distribution of the results is described by a probability distribution, two examples of which are the binomial distribution and the normal distribution. Knowing the type of distribution allows us to determine the probability of obtaining a particular range of results. For a normal distribution we express this range as a confidence interval. A statistical analysis allows us to determine whether our results are significantly different from known values, or from values obtained by other analysts, by other methods of analysis, or for other samples. We can use a -test to compare mean values and an -test to compare variances. To compare two sets of data you first must determine whether the data is paired or unpaired. For unpaired data you also must decide if you can pool the standard deviations. A decision about whether to retain an outlying value can be made using Dixon’s -test, Grubb’s test, or Chauvenet’s criterion. You should be sure to exercise caution if you decide to reject an outlier. Finally, the detection limit is a statistical statement about the smallest amount of analyte we can detect with confidence. A detection limit is not exact since its value depends on how willing we are to falsely report the analyte’s presence or absence in a sample. When reporting a detection limit you should clearly indicate how you arrived at its value. alternative hypothesis box plot confidence interval detection limit dot chart Grubb’s test kernel density plot mean method error one-tailed significance test paired t-test probability distribution range sample standard deviation tolerance type 1 error unpaired data bias central limit theorem constant determinate error determinate error error histogram limit of identification median normal distribution outlier personal error propagation of uncertainty repeatability sampling error standard error of the mean -test type 2 error variance binomial distribution Chauvenet’s criterion degrees of freedom Dixon’s -test -test indeterminate error limit of quantitation measurement error null hypothesis paired data population proportional determinate error reproducibility significance test Standard Reference Material two-tailed significance test uncertainty | 2,860 | 3,378 |
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The general idea is that there is a “natural” way to write each half-cell reaction, somehow showing the electron transfer that is characteristic of redox reactions. With inorganic reactions, it is common to show free electrons. With organic reactions, it is common to show species such as free hydrogen atoms or oxygen atoms. Importantly, these various ways to show electron transfer are easily related to each other. Abbreviations and conventions: We focus here on one chemical reaction, the oxidation of ethanol by dichromate. We will explore various ways of balancing it -- especially the organic part. The reaction is the oxidation of ethanol by dichromate, to yield ethanoic acid and Cr3+. The reaction occurs in acidic aqueous medium. The final, balanced equation is: \[3 CH_3CH_2OH + 2 Cr_2O_7^{2-} + 16H^+ \rightarrow 3 CH_3COOH + 4Cr^{3+} + 11\,H_2O \tag{1}\] So, how do we get it? The first step is to break the reaction into two parts: the oxidation half cell and the reduction half cell. We understand that the dichromate is oxidizing the organic compound -- and that the Cr itself is being reduced. We thus consider the two half cells separately, and later combine them. The inorganic half cell, the reduction of dichromate to Cr , is straightforward, using the usual methods learned in general chemistry. The Cr starts at ON = +6, and ends at ON = +3. Thus each Cr gains 3 electrons. We then balance out the H and O by using species present in the aqueous medium. The result is: \[Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \tag{2}\] This equation meets all the criteria for a balanced equation. It is balanced for each kind of atom, and for charge. The only thing about the equation that is “unusual” is showing the free electrons. But this is a half-cell reaction. The free electrons reflect a simple view of what happens to the Cr, and we understand that they will not appear in the final overall equation. The oxidation half reaction is the oxidation of ethanol to ethanoic acid.
Let’s look at this half reaction more carefully, trying to use good organic chemistry. Dealing with ON for carbon often is difficult. (An alert reader may counter that the ON in this particular example are actually not difficult at all. True, but they often are difficult with complex organic compounds. For the record, I show how to balance this equation using ON in Note 1.) Thus in organic chemistry, we often deal with oxidation and reduction of C by looking at changes in the number of H and O. Oxidation is the addition of O or removal of H. That statement refers to O and H atoms, just as the symbols indicate, not to ions. It is important to understand that, because oxidation and reduction deal with electron transfers. If we are considering a C atom getting oxidized (losing electrons), showing loss of an H atom is chemically logical, but showing loss of an H+ ion is not. The reaction being considered is actually the sum of two distinct reactions. I think it will be useful here to look at the two steps separately, since they are different in the H and O issues. (In practice, many people will be comfortable balancing the complete oxidation half-cell equation without considering the steps. This is addressed below, in Section B.6.) \[CH_3CH_2OH \rightarrow CH_3CHO + \, ??? \tag{3}\] (Equations that are incomplete or not balanced are marked with ???.) Inspection of Eqn 3 shows that this oxidation reaction involves loss of two hydrogen atoms. Thus we write: \[CH_3CH_2OH \rightarrow CH_3CHO + 2H \tag{4}\] This is a complete and balanced equation for this half reaction. It is also simple and chemically logical. The oxidation involves removing two H atoms, so we show 2 H atoms. \[CH_3CHO \rightarrow CH_3COOH ??? \tag{5}\] In this case, the oxidation involves a gain of one oxygen atom. Thus we write: \[CH_3CHO + O \rightarrow CH_3COOH \tag{6}\] This is a complete and balanced equation for this half reaction. Again, it is simple and chemically logical. We can also write this equation in another form. For the moment, this is “for fun” -- just to show two equivalent equations. However, it turns out that the form below will be convenient in the next section. Eqn 6 uses an O atom to reflect the electrons of this redox reaction. We can transform the equation to use H, instead of O. To do this, we recognize that: \[2H + O = H_2O \tag{7}\] Substituting Eqn 7 in Eqn 6 for O gives: \[CH_3CHO + H_2O \rightarrow CH_3COOH + 2H \tag{8}\] Both Eqn 6 and Eqn 8 are complete and balanced for this second oxidation step. One is no “better” than the other. The desired equation is the sum of Eqn 4 (first oxidation step) and Eqn 6 or 8 (second oxidation step). We choose Eqn 8 here for the second step, because the equation for the first step, Eqn 4, is in terms of H. The desired summed equation is: \[CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4 H \tag{9}\] Again, this is fully balanced for this double-oxidation half reaction. It is balanced for each kind of atom, and it is balanced for charge. Charge was not a difficulty, since all species were written with proper charge at the start. Eqn 9 is chemically logical, chemically instructive, and fully balanced. It is a good chemical equation -- for a half cell. It contains an unrealistic product, the H atom -- just as equations for common inorganic half cells typically contain electrons. The H is not the final product, but we can deal with that separately, when we combine half cells (Section B.7). Although I worked through the two oxidation steps above one by one, many will find it easy enough to balance the overall reaction in one step. The oxidation of CH CH OH to CH COOH requires one O atom and releases two H atoms. Just count the atoms. That leads to: CH CH OH + O → CH COOH + 2 H (10) That equation is logical and balanced. We then put it in a more convenient form by using Eqn 7; this leads to Eqn 9, as before. This section combines what was done above in Sections B.3, B.4 and B.5. It is fine to do it either way: going through the individual steps and then adding them, as in the earlier sections, or doing it all at once, as in this section. The purpose of presenting the individual steps earlier was to present one small step at a time, to make the logic clear. It is fine to do multiple steps together, if you
are comfortable doing so. We now have equations for two half-cell reactions. The reduction half cell is described in Eqn 2. The oxidation half cell is described in Eqn 9. The former shows electron transfer in terms of free electrons; the latter shows electron transfer in terms of hydrogen atoms. To combine these two half-cell equations, we need to get them in the same electron language. To do that, we re-write the H atom, H, as the sum of its parts: H = H + e (11) Substituting Eqn 11 in Eqn 9 gives CH CH OH + H O → CH COOH + 4 H + 4 e (12) To obtain the final equation for the overall reaction, we combine the reduction half cell and the oxidation half cell, as usual with redox equations. The reduction half cell is Eqn 2, and shows 6 e-. The form of the oxidation half cell that is convenient here is Eqn 12; it shows 4 e-. To get the electrons to balance out when the two are combined, we take 2 times Eqn 2, and add 3 times Eqn 12. The result is Eqn 1 -- the final desired equation. 3 CH CH OH + 2 Cr O + 16H → 3 CH COOH + 4 Cr + 11 H O (1) This final equation includes the H+ ion. That is fine here, since the reaction occurs in acidic aqueous medium. The choice of which species are acceptable depends on the specifics of the case at hand. However, free electrons and free hydrogen atoms generally do not belong in final equations for real processes. To help us visualize and balance redox reactions, we commonly divide them into two parts: one for oxidation and one for reduction. We recognize an oxidation half cell because there is a loss of electrons; equivalently, there may be a loss of hydrogen or gain of oxygen. We recognize a reduction half cell because there is a gain of electrons -- or a gain of hydrogen or loss of oxygen. These “half-cell reactions” are hypothetical, and we often write them with unrealistic species, such as free electrons or hydrogen atoms. These unrealistic species must disappear when we combine the two half cells into an equation for the complete redox reaction. However, the half-cell reactions are instructive to us, as well as useful in working out the complete equation.
For simple chemicals, we may write the redox half cell with free electrons, making use of the oxidation numbers. For example: \[Cr(VI) + 3 e^- \rightarrow Cr(III) \tag{13}\] For more complex chemicals, and particularly for organic chemicals, we may write the redox half cell with free hydrogen or oxygen atoms. For example: \[CH_2=CH_2 + 2 H \rightarrow CH_3CH_3 \tag{14}\] That is, redox half cells may show the formal species e-, H or O. In combining the half cells, we need to eliminate these formal intermediates. To do that, we recognize various relationships between them, all of which follow good chemical logic. We used two such relationships above: \[2 H + O = H_2O \tag{7}\] We used this relationship in Section B.4. In that case, we had two partial reactions, one showing H and one showing O. Eqn 7 allowed us to combine those into an equation showing only one electron form. \[H = H^+ + e^- \tag{11}\] We used this relationship in Section B.7. In that case, we had two partial reactions, one written with \(e^-\) and one written with H. Eqn 11 allowed us to combine those into a single equation -- in this case, one with all electron forms canceling out. Another relationship we might have used is: \[O + 2 e^- = O^{2-} \tag{15}\] Overall, Equations 7, 11 and 15 allow us to interconvert among three formal ways of representing electron intermediates: free electrons or hydrogen or oxygen atoms. Thus, we are free to write whichever of them seems most chemically appropriate for a particular half-cell; we can then combine half-cell equations no matter which electron form they show by using these relationships. Some additional relationships that might come up are briefly noted in Section F. From time to time the question comes up of whether we should teach -- or encourage -- a “shortcut”. This is a gray area, and needs to be considered in each particular case. However, as a general philosophical point, I prefer that we start by teaching things that are correct and logical. If shortcuts follow, that may be fine. Certainly, when students discover a shortcut on their own, they deserve a good explanation of why it works. And there may be times when we teach a shortcut or simplified method because it is not practical to be more rigorous. Examples here where this might be relevant include the breakdown of the oxidation to two steps vs one, and considering the half cells separately, rather than trying to balance the whole equation at once. Whether these are good examples is a matter of taste. As noted, this section is gray. The main point is to encourage the use of good chemical logic as much as possible. Shortcuts should be identified as such. Here is a possible alternative approach that was offered to me for balancing the organic half cell; the approach might be considered a shortcut. The presentation of this shortcut here is loosely based on a real discussion. The purpose here is to discuss the merits of one or another approach. Discussing alternative approaches should serve to highlight their strengths and weaknesses.
For simplicity, let’s consider step 1 of the oxidation reaction discussed above: Eqn 3. CH CH OH → CH CHO ??? (3) The suggested approach starts by noting that H+ is a likely product. Since 2 H are removed, we start by writing the following as our “trial equation”: CH CH OH → CH CHO + 2 H ??? (16) That achieves balance for H, but it is unbalanced for charge. To solve that problem, we now add electrons as needed to achieve charge balance: CH CH OH → CH CHO + 2 H+ + 2 e- (17) This is a complete and balanced equation for this half reaction. It is fully equivalent to Eqn 4 -- because 2 H = 2 H + 2 e- (by Eqn 11). Thus the first point is that this approach works; it gives a correct answer. The purpose of discussing it further, comparing it to what was done in Section B.3, is to explore the logic of the two approaches. Both approaches ultimately make use of electrons, in some form. In the approach of Section B.3, presented earlier, the electrons are considered from the start -- because this is a redox Balancing Organic Redox Reactions Page 8 equation, fundamentally involving electron transfer. The first equation written, Eqn 4, is chemically logical and balanced. Further manipulation of the equation may be done for convenience. In contrast, in the shortcut approach, the electrons that are so fundamental are ignored in the initial step, and added back only later to get the equation to work out. The first equation written, Eqn 16, is neither chemically logical nor balanced; further manipulation is required simply to get a meaningful equation. In the end, this comes out fine, but why not just do the chemistry logically in the first place? It is not obvious that the “shortcut” is any shorter, and it is certainly less logical. What, then, is its merit? In discussing the shortcut with one person, they seemed unsure how to explain what they had done, and unsure what to do in a case that did not ultimately produce H+. That discussion suggested that the shortcut was being used as a blind mechanical trick, with poor understanding of the chemistry. In contrast, the approach in Section B.3 is chemically logical right from the start. Then, it is meshed with whatever specific constraints are relevant to the problem at hand, such as relating H atoms to free electrons, using Eqn 11. Further, the approach is easily generalized to other cases -- simply by looking at what the electrons are doing. If anyone has an example where they think the proposed shortcut is particularly appropriate, please contribute. Discussing such examples could be instructive. We briefly note some additional cases. These were not relevant in our example above, but are relevant to some reactions. They can be seen as simple extensions to the story above. The first two reflect simple chemistry, and the final group considers biochemical cofactors that carry electrons (hydrogen). \[2 H = H_2 \tag{7}\] The hydride ion is \(H^-\) -- and a good source of electrons, as a reducing agent. For example, the reducing agent \(LiAlH_4\) can be thought of as formally containing \(Li^+ + Al^{3+} + 4 H^-\). The hydride ion is easily related to other hydrogen species, as desired. One such relationship is: \[H^- = H^+ + 2e^- \tag{18}\] In biochemistry, electrons are carried by cofactors, such as flavin adenine dinucleotide (FAD) or nicotinamide adenine dinucleotide (NAD). (The details of these cofactors are not of concern here.) We often show the electrons in the form of hydrogen atoms. Thus we commonly write simply FADH2 and FAD for the two forms of FAD -- with and without the two electrons (two hydrogen atoms). FAD + 2H = FADH (19) NAD is a bit more complicated, because NAD itself actually is an ion. The idea is the same, however: NAD + 2H = NADH + H (20) There is also a phosphorylated form of NAD, called NADP. However, the phosphate group has no direct role in carrying H, and the relationship is quite the same as for NAD: NADP + 2H = NADPH + H (21) This approach is easily extended to other electron (hydrogen) carriers, such as pyrroloquinoline quinone (PQQ): PQQ + 2H = PQQH (22) The particular equation at hand is actually easy enough to balance by the usual procedures using ON, as taught in general chemistry. For the record, we do that here. The basic description of the reaction is CH CH OH + Cr O → CH COOH + 2 Cr ??? (23) I have included a coefficient 2 for the Cr , since there must be 2 Cr for a single Cr O . For Cr: The Cr is reduced from +6 to +3, and there are 2 Cr; therefore, the equation as written above involves a 6 electron gain, total, for the two Cr atoms. For C: We assign the usual +1 for H and -2 for O. For CH CH OH , there are 6 H = 6+ and 1 O = 2-. Total of H and O is 4+, therefore the total ON for the two C atoms is 4-. Similarly, for CH COOH, the total ON for the two C atoms is 0. Thus the C atoms, together, lose 4 electrons. To balance the 6 e- gain of the Cr and the 4 e- loss of the C, we multiply the former by 2 and the latter by 3: 3 CH CH OH + 2 Cr O7 → 3 CH COOH + 4 Cr ??? (24) That takes care of the redox (electron transfer) part of the balancing. We now complete the balancing, using water and its common ions. There are 17 O on the left, 6 on the right. So we add 11 H O to the right to achieve O balance: 3 CH CH OH + 2 Cr O7 → 3 CH COOH + 4 Cr + 11 H O ??? (25) There are now 18 H on the left and 34 on the right. So we add 16 H + to the left to achieve H balance: 3 CH CH OH + 2 Cr O7 + 16 H+ → 3 CH COOH + 4 Cr + 11 H O (1) The usual checking shows that the equation is fully balanced, for each element and for charge. In fact, it is the correct final equation, Eqn 1. On my web page of practice quizzes for intro organic/biochem is one called “Quiz: Oxidation and reduction”. This is an exercise to help students see the equivalence of the two methods for dealing with redox of carbon compounds: assigning ON and counting H and O. Both methods focus on counting the electrons, the heart of the redox reaction. There are various ways to show the hydrogen atom. These include H, as I have done, or H· or [H]. It does not matter, so long as we understand what is meant -- in particular, that it is the neutral species, with one electron. (The dot on H· serves to remind us of that one electron. It also relates to the common electron dot formula for hydrogen gas, H:H.) The H atom is not a stable final product (or reactant) in such equations, and will not appear in the final equation for a complete real process (under ordinary conditions). However, it lets us focus on one step of a complex reaction, without worrying about the detail of where the hydrogen comes from or goes. Sometimes we use [H] to denote “reducing power”. The term may be used qualitatively or quantitatively. For example, the symbol [H] over the reaction arrow means to use a reducing agent, typically some source of H. In biochemistry, 2 H may well end up reducing NAD+ to NADH + H . This was introduced in But it is also common that we simply want to be able to count the amount of reducing power, without worrying about its form. My Metabolism handout, posted at the website for Intro Organic/Biochem, has a section on “Electron carriers (H carriers)”. It includes a discussion of how we show hydrogen atoms, with an emphasis on the biochemical usage. Similarly, the hydride ion, H-, might be shown as H:-. (The hydride ion was introduced in Section F.2.) This note follows from the preceding note, on the hydrogen atom. We might show the oxygen atom as O or [O]. I choose to show it simply as O. I suppose we might show it with dots, but that seems unwieldy in this case. >Robert Bruner ( ) The discussion here started with M. Farooq Wahab, then of the Chemistry Dept, University of Karachi, noting the difficulty of using ON with organic reactions. He also emphasized the importance of balancing such reactions in analytical chemistry, even though organic chemists are often rather casual about balancing equations. | 19,492 | 3,379 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/06%3A_Electron_Transfer/6.05%3A_Energy_Storage_and_Release |
Electron-transfer reactions play key roles in a great many biological processes, including collagen synthesis, steroid metabolism, the immune response, drug activation, neurotransmitter metabolism, nitrogen fixation, respiration, and photosynthesis. The latter two processes are of fundamental significance-they provide most of the energy that is required for the maintenance of life. From the point of view of global bioenergetics, aerobic respiration and photosynthesis are complementary processes (Figure 6.10). The oxygen that is evolved by photosynthetic organisms is consumed by aerobic microbes and animals. Similarly, the end products of aerobic respiratory metabolism (CO and H O) are the major nutritional requirements of photosynthetic organisms. The global C, H, and O cycles are thus largely due to aerobic respiration and photosynthesis. The extraction of energy from organic compounds, carried out by several catabolic pathways (e.g., the citric-acid cycle), involves the oxidation of these compounds to CO and H O with the concomitant production of water-soluble reductants (NADH and succinate). These reductants donate electrons to components of the mitochondrial electron-transfer chain, resulting in the reduction of oxygen to water: \[\frac{1}{2} O_{2} + NADH + H^{+} \rightarrow H_{2}O + NAD^{+} \tag{6.5}\] In aerobic organisms, the terminal oxidant is, of course, oxygen. However, some species of bacteria respire anaerobically and are able to use inorganic oxyanions (nitrate or sulfate) as terminal oxidants. The translocation of protons across the inner mitochondrial membrane accompanies the electron transfers that ultimately lead to the reduction of O ; these protons, in turn, activate ATP synthase, which catalyzes the phosphorylation of ADP to ATP (a process known as oxidative phosphorylation). Because the hydrolysis of ATP is very exoergonic (i.e., \(\Delta\)G < 0), the newly synthesized ATP is used as a molecular energy source to drive thermodynamically unfavorable reactions to completion. The rediscovery of cytochromes by Keilin in 1925 led him to propose that the reduction of O is linked to the oxidation of reduced substrates by a series of redox reactions, carried out by cellular components collectively referred to as the respiratory electron-transport chain. Progress toward a molecular understanding of these redox reactions has been painfully slow. Most of the components are multisubunit proteins that reside in the inner mitochondrial membrane (Figure 6.11). These proteins (Complexes I-IV) are quite difficult to purify with retention of properties, and they do not crystallize well. The components of the respiratory chain contain a variety of redox cofactors. Complex I (NADH-Q reductase; > 600 kDa) contains five iron-sulfur clusters and FMN. Complex II (succinate-Q reductase; 150 kDa) contains several iron-sulfur clusters, FAD (flavin adenine dinucleotide), and cytochrome b . Complex III (ubiquinol-cytochrome C reductase; 250 kDa) contains a [2Fe-2S] iron-sulfur center and cytochromes b , b , and c . Complex IV (cytochrome C oxidase; 200 kDa) contains at least two copper ions and cytochromes a and a ; Q denotes coenzyme Q, which may be bound to hydrophobic subunits of Complexes I, II, and/or III . Cytochrome c (cyt c in Figure 6.11) is a water-soluble protein (12.4 kDa) that is only peripherally associated with the inner mitochondrial membrane; it has been so thoroughly studied that it is generally regarded as the prime example of an electron transferase. More than 20 redox centers are involved in the electron-transport chain. Figure 6.12 depicts a simplified view of the flow of electrons from NADH to O via this series of electron carriers. Electron flow through Complexes I, III, and IV is associated with the release of relatively large amounts of energy, which is coupled to proton translocation by these complexes (and therefore ATP production). The redox potentials of the electron carriers thus appear to playa role in determining the pathway of electron flow through the electron-transport chain. Approximately 50 percent of the surface area of the inner mitochondrial membrane is lipid bilayer that is unoccupied by membrane proteins and through which these proteins, in principle, are free to diffuse laterally. Kinetic (laser photobleaching and fluorescence recovery) and ultrastructural (freeze-fracture electron microscopy) studies indicate that Complexes I-IV diffuse independently and laterally over the inner membrane, whereas cytochrome c diffuses in three dimensions (i.e., through the intramembrane space). Respiratory electron transport has been shown to be a diffusion-coupled kinetic process. The term "electron-transport chain" is thus somewhat misleading, because it implies a degree of structural order that does not exist beyond the level of a given protein complex. In view of these observations, why are all of the electron transfers associated with mitochondrial respiration required? For example, why is cytochrome c needed to shuttle electrons in Figures 6.11 and 6.12 when the cofactor reduction potentials of Complex III are more negative than those of Complex IV? Evidently, factors other than \(\Delta\)G° are of importance—these will be discussed in Sections III and IV. Photosynthesis could be viewed as the most fundamental bioenergetic process. Biological reactions are driven by an energy flux, with sunlight serving as the energy source. Photosynthesis is the process by which radiant solar energy is converted into chemical energy in the form of ATP and NADPH, which are then used in a series of enzymatic reactions to convert CO into organic compounds. The photosynthetic algae that appeared on Earth two million years ago released oxygen into the atmosphere and changed the environment from a reducing to an oxidizing one, setting the stage for the appearance of aerobically respiring organisms. Photosynthesis is initiated by the capture of solar energy, usually referred to as "light harvesting." A large number of organic pigments, including chlorophylls, carotenoids, phycoerythrin, and phycocyanin (in green plants and algae) are clustered together in pigment-protein complexes called photosystems. These pigments collectively absorb most of the sunlight reaching the Earth—their absorption spectra are displayed in Figure 6.13. Light is transformed into chemical energy in pigment-protein complexes called reaction centers. The concentration of reaction centers within a photosynthetic cell is too small to offer a suitable absorption cross section for sunlight. Hence, hundreds of these lightharvesting pigments function as molecular antennas; an x-ray structure of one subunit of a bacteriochlorophyll-protein complex is displayed in Figure 6.14. Absorption of a photon by an antenna pigment promotes the pigment into an electronically excited state, which can return to the ground state by a variety of relaxation processes, including fluorescence or resonance transfer of excitation energy to a nearby pigment at picosecond rates. As much as 100 ps may elapse between the photon absorption and the arrival of the light energy at a reaction center. During this time, the energy may "migrate" in a random-walk fashion among hundreds of pigments. The energy of the excited state is converted into electrochemical potential energy at the reaction center, which contains a primary electron donor P that transfers an electron to a nearby acceptor Al within the same protein (and P becomes oxidized to P ): \[P A_{1}A_{2}A_{3} \cdotp \cdotp \cdotp \xrightarrow{h_{\nu}} P^{\ast}A_{1}A_{2}A_{3} \cdotp \cdotp \cdotp \rightarrow P^{+} A_{1}^{-}A_{2}A_{3} \cdotp \cdotp \cdotp \tag{6.5}\] This is of paramount importance. The key problem is maintaining the charge separation, which involves minimization of the energy-wasting back reaction. Reaction centers contain an ordered array of secondary electron acceptors (A , A , A3•••) that optimize the \(\Delta\)G° that occurs at each step: \[P^{+} A_{1}^{-}A_{2}A_{3} \cdotp \cdotp \cdotp \rightarrow P^{+} A_{1}A_{2}^{-}A_{3} \cdotp \cdotp \cdotp \rightarrow P^{+} A_{1}A_{2}A_{3}^{-} \cdotp \cdotp \cdotp \tag{6.6}\] Thus, the back reaction is circumvented by optimizing forward electron transfers that rapidly remove electrons from A . As the acceptors are separated by greater and greater distances from P , the probability of the back electron transfer to P decreases. Put another way, the overlap of P and each acceptor orbital decreases in the order P /A > P /A > P /A . Photosynthetic bacteria contain only one type of reaction center (l00 kDa). The solution of the x-ray structure (at 2.9 Å resolution) of the reaction center was reported in 1984, providing conclusive proof that electrons can "tunnel" over 10-20 Å distances through protein interiors. The reaction-center protein contains many cofactors (Figure 6.15): two bacteriochlorophylls (BChl) in close proximity (the so-called "special pair"), two further bacteriochlorophylls that are spectroscopically identical, two bacteriopheophytins (BPh), two quinones (Q and Q ), and one iron center. (Q was lost during isolation of the reaction center and thus does not appear in Figure 6.15.) The reaction center contains an approximate two-fold rotation axis. Despite this strikingly high symmetry in the reaction center, one pathway of electron flow predominates, as the cartoon in Figure 6.16 indicates. | 9,468 | 3,381 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Chemistry_of_Cooking_(Rodriguez-Velazquez)/08%3A_Chocolate/8.04%3A_Definitions_and_Regulations_(ADD_US) |
The legislation for cocoa and chocolate products in Canada is found in Division 4 of the Food and Drug Regulations (FDR), under the Food and Drugs Act (FDA). The Canadian Food Inspection Agency (CFIA) is responsible for administering and enforcing the FDR and FDA. Here are some of the regulations governing cocoa and chocolate: The only sweetening agents permitted in chocolate in Canada are listed in Division 18 of the Food and Drug Regulations. Cocoa butter and sugar quantities are not defined in the regulations. Some semi-sweet chocolate may be sweeter than so-called sweet chocolate. And remember that bittersweet chocolate is not, as you might expect, sugarless. Only if the label states “unsweetened,” do you know that there is no sugar added. Products manufactured or imported into Canada that contain non-permitted ingredients (vegetable fats or oils, artificial sweeteners) cannot legally be called chocolate when sold in Canada. A non-standardized name such as “candy” must be used. Finally, lecithin, which is the most common emulsifying agent added to chocolate, is approved for use in chocolate in North America and Europe, but Canadian regulations state that no more than 1% can be added during the manufacturing process of chocolate. Emulsifiers like lecithin can help thin out melted chocolate so it flows evenly and smoothly. Because it is less expensive than cocoa butter at thinning chocolate, it can be used to help lower the cost. The lecithin used in chocolate is mainly derived from soy. Both GMO (genetically modified organism) and non-GMO soy lecithin are available. Check the manufacturer’s packaging and ingredient listing for the source of soy lecithin in your chocolate. | 1,715 | 3,382 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Mass_Spectrometry/The_M_1_Peak |
This page explains how the M+1 peak in a mass spectrum can be used to estimate the number of carbon atoms in an organic compound. If you had a complete (rather than a simplified) mass spectrum, you will find a small line 1 m/z unit to the right of the main molecular ion peak. This small peak is called the M+1 peak. The M+1 peak is caused by the presence of the C isotope in the molecule. C is a stable isotope of carbon - don't confuse it with the C isotope which is radioactive. Carbon-13 makes up 1.11% of all carbon atoms. If you had a simple compound like methane, CH , approximately 1 in every 100 of these molecules will contain carbon-13 rather than the more common carbon-12. That means that 1 in every 100 of the molecules will have a mass of 17 (13 + 4) rather than 16 (12 + 4). The mass spectrum will therefore have a line corresponding to the molecular ion [ CH ] as well as [ CH ] . The line at m/z = 17 will be much smaller than the line at m/z = 16 because the carbon-13 isotope is much less common. Statistically you will have a ratio of approximately 1 of the heavier ions to every 99 of the lighter ones. That's why the M+1 peak is much smaller than the M+ peak. Imagine a compound containing 2 carbon atoms. Either of them has an approximately 1 in 100 chance of being C. There's therefore a 2 in 100 chance of the molecule as a whole containing one C atom rather than a C atom - which leaves a 98 in 100 chance of both atoms being C. That means that the ratio of the height of the M+1 peak to the M+ peak will be approximately 2 : 98. That's pretty close to having an M+1 peak approximately 2% of the height of the M+ peak. If you measure the peak height of the M+1 peak as a percentage of the peak height of the M+ peak, that gives you the number of carbon atoms in the compound. We've just seen that a compound with 2 carbons will have an M+1 peak approximately 2% of the height of the M+ peak. Similarly, you could show that a compound with 3 carbons will have the M+1 peak at about 3% of the height of the M+ peak. The approximations we are making won't hold with more than 2 or 3 carbons. The proportion of carbon atoms which are C isn't 1% - it's 1.11%. And the approximation that a ratio of 2 : 98 is about 2% doesn't hold as the small number increases. Consider a molecule with 5 carbons in it. You could work out that 5.55 (5 x 1.11) molecules will contain 1 C to every 94.45 (100 - 5.55) which contain only C atoms. If you convert that to how tall the M+1 peak is as a percentage of the M+ peak, you get an answer of 5.9% (5.55/94.45 x 100). That's close enough to 6% that you might assume wrongly that there are 6 carbon atoms. Above 3 carbon atoms, then, you shouldn't really be making the approximation that the height of the M+1 peak as a percentage of the height of the M+ peak tells you the number of carbons - you will need to do some fiddly sums! Jim Clark ( ) | 2,923 | 3,383 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Synthesis_of_Alkenes/Alkenes_from_Dehydration_of_Alcohols |
One way to synthesize alkenes is by dehydration of alcohols, a process in which alcohols undergo or mechanisms to lose water and form a double bond. The dehydration reaction of alcohols to generate alkene proceeds by heating the alcohols in the presence of a strong acid, such as sulfuric or phosphoric acid, at high temperatures. The required range of reaction temperature decreases with increasing substitution of the hydroxy-containing carbon: If the reaction is not sufficiently heated, the alcohols do not dehydrate to form alkenes, but react with one another to form ethers (e.g., the ). Alcohols are amphoteric; they can act both as acid or base. The lone pair of electrons on oxygen atom makes the –OH group weakly basic. Oxygen can donate two electrons to an electron-deficient proton. Thus, in the presence of a strong acid, R—OH acts as a base and protonates into the very acidic alkyloxonium ion OH (The pKa value of a tertiary protonated alcohol can go as low as -3.8). This basic characteristic of alcohol is essential for its dehydration reaction with an acid to form alkenes. Different types of alcohols may dehydrate through a slightly different mechanism pathway. However, the general idea behind each dehydration reaction is that the –OH group in the alcohol donates two electrons to H from the acid reagent, forming an alkyloxonium ion. This ion acts as a very good leaving group which leaves to form a carbocation. The deprotonated acid (the nucleophile) then attacks the hydrogen adjacent to the carbocation and form a double bond. Primary alcohols undergo bimolecular elimination ( ) while secondary and tertiary alcohols undergo unimolecular elimination ( ). The relative reactivity of alcohols in dehydration reaction is ranked as the following Methanol < primary < secondary < tertiary Oxygen donates two electrons to a proton from sulfuric acid H SO , forming an alkyloxonium ion. Then the nucleophile HSO back-side attacks one adjacent hydrogen and the alkyloxonium ion leaves in a concerted process, making a double bond. Similarly to the reaction above, secondary and tertiary –OH protonate to form alkyloxonium ions. However, in this case the ion leaves first and forms a carbocation as the reaction intermediate. The water molecule (which is a stronger base than the HSO ion) then abstracts a proton from an adjacent carbon, forming a double bond. Notice in the mechanism below that the aleke formed depends on which proton is abstracted: the red arrows show formation of the more substituted 2-butene, while the blue arrows show formation of the less substituted 1-butene. Recall the general rule that more substituted alkenes are more stable than less substituted alkenes, and alkenes are more stable than alkenes. Therefore, the diastereomer of the 2-butene product is most abundant. Dehydration reaction of secondary alcohol: When more than one alkene product are possible, the favored product is usually the thermodynamically most stable alkene. More-substituted alkenes are favored over less-substituted ones; and trans-substituted alkenes are preferred compared to cis-substituted ones. Since the dehydration reaction of alcohol has a carbocation intermediate, hydride or alkyl shifts can occur which relocates the carbocation to a more stable position. The dehydrated products therefore are a mixture of alkenes, with and without . Tertiary cation is more stable than secondary cation, which in turn is more stable than primary cation due to a phenomenon known as hyperconjugation, where the interaction between the filled orbitals of neighboring carbons and the singly occupied p orbital in the carbocation stabilizes the positive charge in carbocation. Similarly, when there is no hydride available for hydride shifting, an alkyl group can take its bonding electrons and swap place with an adjacent cation, a process known as alkyl shift. Test your understanding by predicting what product(s) will be formed in each of the following reactions: 1. Did you notice the reaction temperature? It is only 25°, which is much lower than the required temperature of 170°C for dehydration of primary alcohol. This reaction will not produce any alkene but will form ether. 2. . Notice that the reactant is a secondary -OH group, which will form a relatively unstable secondary carbocation in the intermediate. Thus hydride shift from an adjacent hydrogen will occur to make the carbocation tertiary, which is much more stable. The products are a mixture of alkenes that are formed with or without carbocation rearrangement (A number of products are formed faster than hydride shift can occur). | 4,660 | 3,386 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/08%3A_Phase_Equilibrium/8.10%3A_Cooling_Curves |
The method that is used to map the phase boundaries on a phase diagram is to measure the rate of cooling for a sample of known composition. The rate of cooling will change as the sample (or some portion of it) begins to undergo a phase change. These “breaks” will appear as changes in slope in the temperature-time curve. Consider a binary mixture for which the phase diagram is as shown in Figure \(\Page {1A}\). A cooling curve for a sample that begins at the temperature and composition given by point a is shown in Figure \(\Page {1B}\). As the sample cools from point a, the temperature will decrease at a rate determined by the sample composition, and the geometry of the experiment (for example, one expects more rapid cooling is the sample has more surface area exposed to the cooler surroundings) and the temperature difference between the sample and the surroundings. When the temperature reaches that at point b, some solid compound B will begin to form. This will lead to a slowing of the cooling due to the exothermic nature of solid formation. But also, the composition of the liquid will change, becoming richer in compound A as B is removed from the liquid phase in the form of a solid. This will continue until the liquid attains the composition at the eutectic point (point c in the diagram.) When the temperature reaches that at point c, both compounds A and B will solidify, and the composition of the liquid phase will remain constant. As such, the temperature will stop changing, creating what is called the . Once all of the material has solidified (at the time indicated by point c’), the cooling will continue at a rate determined by the heat capacities of the two solids A and B, the composition, and (of course) the geometry of the experimental set up. By measuring cooling curves for samples of varying composition, one can map the entire phase diagram. | 1,894 | 3,389 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/Ylide_Addition |
Sometimes, nucleophiles adding to a carbonyl do not follow the normal reactivity patterns that have been common so far. This is often the case with the addition of ylides. Ylides are compounds that are often depicted with a positive charge one one atom and a negative charge on the next atom. They are examples of zwitterions, compounds that contain both positive and negative charges within the same molecule. What distinguishes them from other zwitterions is the proximity of the opposite charges. The classic example of an ylide addition to a carbonyl is the , which involves the addition of a phosphorus ylide to an aldehyde or ketone. Rather than producing an alcohol, the reaction produces and alkene. The reaction is driven by formation of a phosphorus oxide "side product". This is a special case. The phosphorus-oxygen bond is strong enough to change the course of this reaction away from the normal pattern, and it isn't something you would have been able to predict based on related reactions. An ylide is an example of a molecular compound that contains both a positive and a negative formal charge on two adjacent atoms. The charges are right beside each other: in this case, there is a positive charge on the phosphorus and a negative charge on the carbon. Phosphorus ylides are made one charge at a time. A phosphonium ion must first be assembled, containing the positive charge on phosphorus. This event occurs via a nucleophilic substitution reaction, in which a phosphorus nucleophile displaces a halogen from an alkyl halide. Show, with reaction arrows, formation of the three alkyltriphenyl phosphonium bromide salts shown below. In most cases, the source of the phosphorus is triphenylphosphine. Triphenylphosphine is used for several practical reasons. First of all, it is a solid, so it is easy to weigh out the right amount of it and add it to a reaction. Secondly, organophosphorus compounds are often very toxic and smelly, but triphenylphosphine is less offensive. Thirdly, in the Wittig reaction, the original phosphorus compound is eventually discarded as waste, and the more useful alkene is kept. Since the phosphorus part doesn't matter that much, the most convenient possible phosphine is generally used. However, there are other variations of this reaction that use other phosphorus compounds. Once the phosphonium salt has been made, the phosphorus ylide can then be obtained via deprotonation of a phosphonium ion. The hydrogens on a carbon next to a phosphorus cation are a little bit acidic because of the positive charge on the phosphorus. One of these hydrogens is easily removed via adddition of a very strong base such as sodium hydride. Show, with reaction arrows, formation of ylides from the three alkyltriphenyl phosphonium bromide salts shown above in Problem CO18.1. The reaction of a phosphorus ylide with a carbonyl compound does begin like other nucleophilic additions. The ylide donates its nucleophilic lone pair to the carbonyl and the carbonyl pi bond breaks. However, the strong P-O bond then takes over the reaction. To begin, a lone pair on the resulting alkoxide ion is donated to the positively charged phosphonium ion. Wait! That violates one of our mechanistic rules. Usually, we don't have an atom donate to a positively charged atom that already has an octet; if we do so, the atom will have too many electrons. However, the octet rule doesn't strictly apply to sulfur and phosphorus. These atoms are larger than second-row atoms like nitrogen and oxygen, and they are often observed to "exceed the octet rule". Sulfur and phosphorus are frequently observed with trigonal bipyramidal or octahedral molecular geometries, meaning they may have up to 12 electrons in their valence shells. So go ahead! Donate a pair of electrons to the phosphorus. It can't help itself, because of the strength of the P-O bond that forms. Show the products of the reactions of each of the ylides you made in Problem CO18.2. with the following electrophiles: a) butanal b) benzaldehyde c) 4-methylpentanal This is when things really get interesting. It turns out that one P-O bond just isn't enough. The phosphorus is so oxophilic that it takes the oxygen atom all to itself, pulling it right out of the molecule. It probably doesn't hurt that the four-membered ring is pretty strained, so it is motivated to decompose (but be careful: there are plenty of stable four- and even three-membered rings in nature). The arrows shown in the decomposition of the four-membered ring (called a betaine) are just meant to keep track of electrons; there isn't a true nucleophile and electrophile in this step. Instead this step may resemble a pericyclic reaction, which is covered in another section. Exactly how to draw the P=O bond is debatable. There isn't much doubt that it is a double bond; it is stronger and shorter than a P-O single bond. However, quantum mechanical calculations indicate that the phosphorus can't form a pi bond. This double bond is different than other double bonds you have seen. For that reason, some people prefer to draw this compound as an ylide, too, with a positive charge on the phosphorus, a single bond, and a negative charge on the oxygen. The phosphorus oxide compound forms, leaving behind an alkene. Alkenes are very common in nature, and this reaction has frequently been used to make interesting alkene-containing compounds for further use or study. The juvenile hormone of the cecropia moth caterpillar (JH-1, below) is a regulatory hormone used to control the organism's development by preventing it from pupating until conditions are right. Sulfur ylides are also good nucleophiles for aldehydes and ketones. However, the unusual stability of the phosphorus-oxygen bond does not have a similar analogue in sulfur chemistry. Sulfur ylides are formed in a manner very similar to phosphorus ylides. Show, with arrows, the mechanism for formation of the sulfur ylide above. Once formed, sulfur ylides react with aldehydes or ketones. Like phosphorus ylides, the reaction starts out just like any other nucleophile, but a second step takes a very different direction. Epoxides are formed in these reactions, and the original sulfur compound (a thioether) is regenerated. Show, with arrows, the mechanism for the epoxide-forming reaction above. , | 6,333 | 3,390 |
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The heat changes which accompany a chemical reaction are caused largely by changes in the electronic energy of the molecules. If we restrict our attention to , and hence to fairly simple molecules, we can go quite a long way toward predicting whether a reaction will be exothermic by considering the bonds which are broken and made in the course of the reaction. In order to do this we must first become familiar with the idea of a bond enthalpy. we point out that when a chemical bond forms, negative charges move closer to positive charges than before, and so there is a lowering of the energy of the molecule relative to the atoms from which it was made. This means that energy is required to break a molecule into its constituent atoms. The D of a diatomic molecule X—Y is the enthalpy change for the (usually hypothetical) process: \[\ce{XY(g) \rightarrow X(g) + Y(g)} \nonumber \] \[\Delta H^{o} (298 K) = D_{x-y} \nonumber \] We have already used the term to describe this quantity, though strictly speaking the bond energy is a measure of Δ rather than Δ . As we have already seen, Δ and Δ are nearly equal, and so either term may be used. As an example, let us consider the bond enthalpy for carbon monoxide. It is possible to establish the thermochemical equation \[\ce{CO(g) \rightarrow C(g) + O(g)} \nonumber \] \[\Delta H^{o}(298 K) = 1073 kJ mol^{-1}\label{3} \] Accordingly we can write \[\ce{C_{C\equiv O}= 1073 kJ mol^{-1}} \nonumber \] even though the process to which Eq. \(\ref{3}\) corresponds is hypothetical: Neither carbon nor oxygen exists as a monatomic gas at 298 K. For triatomic and polyatomic molecules, the bond enthalpy is usually defined as a mean. In the case of water, for instance, we have \[\ce{H_{2}O(g) \rightarrow 2H(g) + O(g)} \nonumber \] \[\Delta H^{o} (298 K) = 927.2 \text{kJ mol}^{-1} \nonumber \] Since it requires 927.2 kJ to break open O—H bonds, we take this value as the mean bond enthalpy and write \[D_{O-H} = 463.6 \text{kJ mol}^{-1} \nonumber \] In methanol, CH OH,however, a value of 427 kJ mol for the O—H bond enthalpy fits the experimental data better. In other words the strength of the O—H varies somewhat from compound to compound. Because of this fact, we must expect to obtain only approximate results, accurate only to about ± 50 kJ mol , from the use of bond enthalpies. Bond enthalpies for both single and multiple bonds are given in Table \(\Page {1}\). \(\Page {1}\) Average Bond Energies/kJ mol . As an example of how a table of bond enthalpies can he used to predict the Δ value for a reaction, let us take the simple case \[\ce{H_{2}(g) + F_{2}(g) \rightarrow 2HF(g)}\label{9} \] 298 K, 1 atm
We can regard this reaction as occurring in two stages (Figure \(\Page {1}\) ). In the first stage all the reactant molecules are broken up into atoms: \[\ce{H_{2}(g) + F_{2}(g) \rightarrow 2H(g) + F(g)}\label{10} \] 298 K, 1 atm For this stage \[\Delta H_I = H_{H-H}+ D_{F-F}\label{11} \] since 1 mol H and 1 mol F have been dissociated. In the second stage the H and F atoms are reconstituted to form HF molecules: \[\ce{2H(g) + 2F(g) \rightarrow 2HF(g)} \nonumber \] 298 K, 1 atm For which \[\Delta H_{II} = – 2D_{H-F} \nonumber \] where a negative sign is necessary since this stage corresponds to the of dissociation. Since Eq. \(\ref{9}\) corresponds to the sum of Equations \(\ref{10}\) and \(\ref{11}\), allows us to add Δ values: We can work this same trick of subdividing a reaction into a bond-breaking stage followed by a bond-making stage for the general case of any gaseous reaction. In the first stage all the bonds joining the atoms in the reactant molecules are broken and a set of gaseous atoms results. For this stage \[\Delta H_{I} = \sum_{\text{bonds broken}} D \nonumber \] The enthalpy change is the sum of the bond enthalpies for all bonds broken. In the second stage these gaseous atoms are reconstituted into the product molecules. For this second stage therefore \[\Delta H_{II} = – \sum_{\text{bonds formed}} D \nonumber \] where the negative sign is necessary because the of bond breaking is occurring in this stage. The total enthalpy change for the reaction at standard pressure is thus \[\Delta H^{o} = \Delta H_{I} + \Delta H_{II} \nonumber \] or \[\Delta H^{o} = \sum D \text{(bond broken)} – \sum D \text{(bond formed)} \nonumber \] The use of this equation is illustrated in the next example. Using Table \(\Page {1}\) calculate the value of Δ °(298 K) for the reaction \[\ce{CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)} \nonumber \] It is best to sketch the molecules and their bonds in order to make sure that none are missed. Thus \(\Delta H^{o} = \sum D \text{(bond broken)} – \sum D \text{(bond formed)}\) \(= 4 D_{C\bond{-}H} + 2 D_{D\bond{=}D} - 2 D_{C\bond{=}O} - 4 D_{O\bond{-}H}\) \(= (4 * 416 + 2 * 498 – 2 * 803 – 4 * 467) \text{kJ mol}^{-1}\) \(= – 814 \text{kJ mol}^{-1}\) | 5,061 | 3,391 |
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Most of the solids discussed so far have been molecules or ions with low molecular masses, ranging from tens to hundreds of atomic mass units. Many of the molecular materials in consumer goods today, however, have very high molecular masses, ranging from thousands to millions of atomic mass units, and are formed from a carefully controlled series of reactions that produce giant molecules called polymers (from the Greek and , meaning “many parts”). Polymers are used in corrective eye lenses, plastic containers, clothing and textiles, and medical implant devices, among many other uses. They consist of basic structural units called monomers , which are repeated many times in each molecule. As shown schematically in , polymerization is the process by which monomers are connected into chains or networks by covalent bonds. Polymers can form via a , in which two monomer molecules are joined by a new covalent bond and a small molecule such as water is eliminated, or by an , a variant of a condensation reaction in which the components of a species AB are added to adjacent atoms of a multiple bond. (For more information about condensation and addition reactions, see .) Many people confuse the terms and . Plastic is the property of a material that allows it to be molded into almost any shape. Although many plastics are polymers, many polymers are not plastics. In this section, we introduce the reactions that produce naturally occurring and synthetic polymers. Polymers are formed via condensation or addition reactions. Polymers that occur naturally are crucial components of all organisms and form the fabric of our lives. Hair, silk, skin, feathers, muscle, and connective tissue are all primarily composed of proteins, the most familiar kind of naturally occurring, or biological, polymer. The monomers of many biological polymers are the amino acids each called an . The residues are linked together by amide bonds, also called peptide bonds, via a condensation reaction where H O is eliminated: In the above equation, R represents an alkyl or aryl group, or hydrogen, depending on the amino acid. We write the structural formula of the product with the free amino group on the left (the ) and the free carboxylate group on the right (the ). For example, the structural formula for the product formed from the amino acids glycine and valine (glycyl-valine) is as follows: The most important difference between synthetic and naturally occurring polymers is that the former usually contain very few different monomers, whereas biological polymers can have as many as 20 different kinds of amino acid residues arranged in many different orders. Chains with less than about 50 amino acid residues are called peptides , whereas those with more than about 50 amino acid residues are called proteins . Many proteins are enzymes , which are catalysts that increase the rate of a biological reaction. Synthetic polymers usually contain only a few different monomers, whereas biological polymers can have many kinds of monomers, such as amino acids arranged in different orders. Many small peptides have potent physiological activities. The , for example, are powerful, naturally occurring painkillers found in the brain. Other important peptides are the hormones vasopressin and oxytocin. Although their structures and amino acid sequences are similar, vasopressin is a blood pressure regulator, whereas oxytocin induces labor in pregnant women and milk production in nursing mothers. Oxytocin was the first biologically active peptide to be prepared in the laboratory by Vincent du Vigneaud (1901–1978), who was awarded the Nobel Prize in Chemistry in 1955. Many of the synthetic polymers we use, such as plastics and rubbers, have commercial advantages over naturally occurring polymers because they can be produced inexpensively. Moreover, many synthetic polymers are actually more desirable than their natural counterparts because scientists can select monomer units to tailor the physical properties of the resulting polymer for particular purposes. For example, in many applications, wood has been replaced by plastics that are more durable, lighter, and easier to shape and maintain. Polymers are also increasingly used in engineering applications where weight reduction and corrosion resistance are required. Steel rods used to support concrete structures, for example, are often coated with a polymeric material when the structures are near ocean environments where steel is vulnerable to corrosion (For more information on corrosion, see .) In fact, the use of polymers in engineering applications is a very active area of research. Probably the best-known example of a synthetic polymer is ( ). Its monomers are linked by amide bonds (which are called peptide bonds in biological polymers), so its physical properties are similar to those of some proteins because of their common structural unit—the amide group. Nylon is easily drawn into silky fibers that are more than a hundred times longer than they are wide and can be woven into fabrics. Nylon fibers are so light and strong that during World War II, all available nylon was commandeered for use in parachutes, ropes, and other military items. With polymer chains that are fully extended and run parallel to the fiber axis, nylon fibers resist stretching, just like naturally occurring silk fibers, although the structures of nylon and silk are otherwise different. Replacing the flexible –CH – units in nylon by aromatic rings produces a stiffer and stronger polymer, such as the very strong polymer known as Kevlar. Kevlar fibers are so strong and rigid that they are used in lightweight army helmets, bulletproof vests, and even sailboat and canoe hulls, all of which contain multiple layers of Kevlar fabric. Not all synthetic polymers are linked by amide bonds—for example, contain monomers that are linked by ester bonds. Polyesters are sold under trade names such as Dacron, Kodel, and Fortrel, which are used in clothing, and Mylar, which is used in magnetic tape, helium-filled balloons, and high-tech sails for sailboats. Although the fibers are flexible, properly prepared Mylar films are almost as strong as steel. Polymers based on skeletons with only carbon are all synthetic. Most of these are formed from ethylene (CH =CH ), a two-carbon building block, and its derivatives. The relative lengths of the chains and any branches control the properties of polyethylene. For example, higher numbers of branches produce a softer, more flexible, lower-melting-point polymer called low-density polyethylene (LDPE), whereas high-density polyethylene (HDPE) contains few branches. Substances such as glass that melt at relatively low temperatures can also be formed into fibers, producing . Because most synthetic fibers are neither soluble nor low melting, multistep processes are required to manufacture them and form them into objects. Graphite fibers are formed by heating a precursor polymer at high temperatures to decompose it, a process called pyrolysis . The usual precursor for graphite is polyacrylonitrile, better known by its trade name—Orlon. A similar approach is used to prepare fibers of silicon carbide using an organosilicon precursor such as polydimethylsilane {[–(CH ) Si–] }. A new type of fiber consisting of carbon nanotubes, hollow cylinders of carbon just one atom thick, is lightweight, strong, and impact resistant. Its performance has been compared to that of Kevlar, and it is being considered for use in body armor, flexible solar panels, and bombproof trash bins, among other uses. Because there are no good polymer precursors for elemental boron or boron nitride, these fibers have to be prepared by time-consuming and costly indirect methods. Even though boron fibers are about eight times stronger than metallic aluminum and 10% lighter, they are significantly more expensive. Consequently, unless an application requires boron’s greater resistance to oxidation, these fibers cannot compete with less costly graphite fibers. Polyethylene is used in a wide variety of products, including beach balls and the hard plastic bottles used to store solutions in a chemistry laboratory. Which of these products is formed from the more highly branched polyethylene? type of polymer application Determine whether the polymer is LDPE, which is used in applications that require flexibility, or HDPE, which is used for its strength and rigidity. A highly branched polymer is less dense and less rigid than a relatively unbranched polymer. Thus hard, strong polyethylene objects such as bottles are made of HDPE with relatively few branches. In contrast, a beach ball must be flexible so it can be inflated. It is therefore made of highly branched LDPE. Exercise Which products are manufactured from LDPE and which from HPDE? are giant molecules that consist of long chains of units called connected by covalent bonds. is the process of linking monomers together to form a polymer. is the property of a material that allows it to be molded. Biological polymers formed from amino acid residues are called or , depending on their size. are proteins that catalyze a biological reaction. A particle that is more than a hundred times longer than it is wide is a , which can be formed by a high-temperature decomposition reaction called . How are amino acids and proteins related to monomers and polymers? Draw the general structure of an amide bond linking two amino acid residues. Although proteins and synthetic polymers (such as nylon) both contain amide bonds, different terms are used to describe the two types of polymer. Compare and contrast the terminology used for the | 9,727 | 3,392 |
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Convert the temperatures indicated to complete the following table Make a graph representing the potential energy of a harmonic oscillator as a function of displacement from equilibrium. On the same graph, include a function describing the kinetic energy as a function of displacement from equilibrium as well as the total energy of the system. Calculate the work required to move a 3.2 kg mass 10.0 m against a resistive force of 9.80 N. Calculate the work needed for a 22.4 L sample of gas to expand to 44.8 L against a constant external pressure of 0.500 atm. If the internal and external pressure of an expanding gas are equal at all points along the entire expansion pathway, the expansion is called “reversible.” Calculate the work of a reversible expansion for 1.00 mol of an ideal gas expanding from 22.4 L at 273 K to a final volume of 44.8 L. | 864 | 3,393 |
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Because binary ionic compounds are confined mainly to and elements on the one hand and group VI and VII elements on the other, we find that they consist mainly of ions having an electronic structure which is the same as that of a noble gas. In calcium fluoride, for example, the calcium atom has lost electrons in order to achieve the electronic structure of argon, and thus has a charge of +2: 1 2 2 3 3 → 1 2 2 3 3 + By contrast, a fluorine atom needs to acquire but electron in order to achieve a neon structure. The resulting fluoride ion has a charge of –1: The outermost shell of each of these ions has the electron configuration , where is 3 for Ca and 2 for F . Such an noble-gas electron configuration is encountered quite often. It is called an because it contains eight electrons. In a crystal of calcium fluoride, the Ca and F ions are packed together in the lattice shown below. Careful study of the diagram shows that each F ion is surrounded by Ca ions, while each Ca ion has F ions as nearest neighbors. Thus there must be twice as many F ions as Ca ions in the entire crystal lattice. Only a small portion of the lattice is shown, but if it were extended indefinitely in all directions, you could verify the ratio of two F for every Ca . This ratio makes sense if you consider that two F ions (each with a –1 charge) are needed to balance the +2 charge of each Ca ion, making the net charge on the crystal zero. The formula for calcium fluoride is thus CaF . Newcomers to chemistry often have difficulty in deciding what the formula of an ionic compound will be. A convenient method for doing this is to regard the compound as being formed from its atoms and to use Lewis diagrams. The can then be applied. Each atom must lose or gain electrons in order to achieve an octet. Furthermore, all electrons lost by one kind of atom must be gained by the other. An exception to the octet rule occurs in the case of the three ions having the He 1 structure, that is, H , Li and Be . In these cases rather than eight electrons are needed in the outermost shell to comply with the rule. Find the formula of the ionic compound formed from O and Al. We first write down Lewis diagrams for each atom involved: We now see that each O atom needs 2 electrons to make up an octet, while each Al atom has 3 electrons to donate. In order that the same number of electrons would be donated as accepted, we need 2 Al atoms (2 × 3 donated) and 3 O atoms (3 × 2 accepted). The whole process is then The resultant oxide consists of aluminum ions, Al , and oxide ions, O , in the ratio of 2:3. The formula is Al O . | 2,668 | 3,394 |
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The anti-inflammatory, analgesic, and antipyretic drugs are a heterogeneous group of compounds, often chemically unrelated (although most of them are organic acids), which nevertheless share certain therapeutic actions and side effects. The prototype is aspirin; hence these compounds are often referred to as aspirin-like drugs. All aspirin-like drugs are antipyretic, analgesic, and anti-inflammatory, but there are important differences in their activities. For example, acetaminophen is antipyretic and analgesic but is only weakly anti-inflammatory. The reason for the differences are not clear; variations in the sensitivity of enzymes in the target tissues may be important. When employed as analgesics, these drugs are usually effective only against pain of low-to-moderate intensity, particularly that associated with inflammation. Aspirin drugs do not change the perception of sensory modalities other than pain. The type of pain is important; chronic postoperative pain or pain arising from inflammation is particularly well controlled by aspirin-like drugs, whereas pain arising from the hollow viscera is usually not relieved. As antipyretics, aspirin-like drugs reduce the body temperature in feverish states. Although all such drugs are antipyretics and analgesics, some are not suitable for either routine or prolonged use because of toxicity; phenylbutaxone is an example. This class of drugs finds its chief clinical application as anti inflammatory agents in the treatment of musculoskelatal disorders, such as rheumatoid arthritis, osteoarthritis, and ankylosing spondylitis. In general, aspirin-like drugs provide only symptomatic relief from the pain and inflammation associated with the disease and do not arrest the progression of pathological injury. There has been substantial progress in elucidating the mechanism of action of aspirin-like drugs, and it is now possible to understand why such heterogeneous agents have the same basic therapeutic activities and often the same side effects. Indeed, their therapeutic activity appears to depend to a large extent upon the inhibition of a defined biochemical pathways responsible for the biosynthesis of prostaglandins (see figure below) and related autacoids. Aspirin-like drugs inhibit the conversion of arachidonic acid to the unstable endoperoxide intermediate, PGG2, which is catalyzed by the cyclooxygenase. Individual agents have differing modes of inhibitory activity on the cyclooxygenase. Aspirin itself acetylates a serine at the active site of the enzyme. Platelets are especially susceptible to this action because (unlike most other cells) they are incapable of regenerating the enzyme, presumably because they have little or no capacity for protein biosynthesis. In practical terms this means that a single dose of aspirin will inhibit the platelet cyclooxygenase for the life of the platelet (8 to 10 days); in man a dose as small as 40 mg per day is sufficient to produce this effect. In contrast to aspirin, salicylic acid has no acetylating capacity and is almost inactive against cyclooxygenase . Nevertheless, it is as active as aspirin in reducing the synthesis of prostaglandins . The basis of this action and, thus, of the anti-inflammatory effect of salicylic acid is not clearly understood. Since aspirin is rapidly hydrolyzed to salicylic acid in vivo (half-life in human plasma, approximately 15 minutes), the acetylated and nonacetylated species probably act as pharmacologically distinct entities. Most of the other common aspirin-like drugs are irreversible inhibitors of the cyclooxygenase, although there are some exceptions. For indomethacin, the mode of inhibition is particularly complex and probably involves a site on the enzyme different from that which is acetylated by aspirin. Prostaglandins are associated particularly with the development of pain that accompanies injury or inflammation. Large doses of PGE2 or PGF2 , given to women by injection to induce abortion, cause intense local pain. Prostaglandins can also cause headache and vascular pain when infused intravenously in man. While the doses of prostaglandins required to elicit pain are high in comparison with the concentrations expected , induction of hyperalgesia occurs when minute amounts of PGE1 are given intradermally to man. Furthermore, in experiments in man where separate infusions of PGE1, bradykinin, or histamine caused no pain, marked pain was experienced when PGE1 was added to bradykinin or histamine. When PGE1 was infused with histamine, itching was also noted. The hypothalamus regulates the set point at which body temperature is maintained. In fever, this set point is elevated, and aspirin-like drugs promote its return to normal. These drugs do not influence body temperature when it is elevated by such factors as exercise or increases in the surrounding temperature. Fever may be a result of infection, tissue damage, inflammation, graft rejection, malignancy, or other disease states. A multitude of microorganisms can cause fever. There is evidence that bacterial endotoxins act by stimulating the biosynthesis and release by neutrophils and other cells of an endogenous pyrogen, a protein with a molecular weight of 10,000 to 20,000. The current view is that the endogenous pyrogen passes from the general circulation into the central nervous system, where it acts upon discrete sites within the brain, especially the preoptic hypothalamic area. There is evidence that the resultant elevation of body temperature is mediated by the release of prostaglandins and that aspirin-like drugs suppress the effects of endogenous pyrogen by inhibiting the synthesis of these substances. The evidence includes the ability of prostaglandins, especially PGE2, to produce fever when infused into the cerebral ventricles or when infected into the hypothalamus. Fever is a frequent side effect of prostaglandins when they are administered to a women as abortifacients. Moreover some studies have demonstrated an increase in prostaglandin-like substances in the cerebrospinal fluid when endogenous pyrogen is injected intravenously. The fever produced by the administration of pyrogen, but not that by prostaglandins, is reduced by aspirin-like drugs. Aspirin is very useful, but it has many side effects and therefore must be used carefully. Like most powerful drugs, an overdose of aspirin or salicylates can be fatal. If a child or adult takes an overdose of aspirin, induce vomiting to empty the unabsorbed medication from the stomach (if the person is still awake and conscious). Obtain emergency medical care right away.
The most common side effects of aspirin are heartburn and other symptoms of stomach irritation such as indigestion, pain, nausea, and vomiting. The stomach irritation may lead to bleeding from the stomach, which may cause black stools. These symptoms may be reduced by taking aspirin with meals, with an antacid, with a glass of milk, or by taking enteric-coated or timed-release aspirin. Also, it is best not to take aspirin with alcohol or coffee (or other beverages containing caffeine, such as tea or cocoa and many soft drinks). Alcohol and caffeine make the stomach more sensitive to irritation. The non aspirin salicylate preparations sometimes are less irritating to the stomach and may be substituted for aspirin by your doctor.
A few people develop asthma, hay fever, nasal congestion, or hives from aspirin or non-steroid anti-inflammatory drugs (NSAIDs). These people should never take aspirin, nor should people who have active stomach or duodenal ulcers. Anyone who has ever had a peptic ulcer should be very careful about taking aspirin because it can lead to a recurrence.
Aspirin is known to interfere with the action of the platelets. As a result, some people who take a lot of aspirin experience easy bruising of the skin. Therefore, people who have major bleeding problems should not take aspirin. Also, keep in mind that aspirin should not be taken for 10-14 days before surgery (including surgery in the mouth) to avoid excessive bleeding during or after the operation. These side effects probably depend on aspirin-like drugs' ability to block endogenous prostaglandin biosynthesis. Platelet function appears to be disturbed because aspirin-like drugs prevent the formation by the platelets of thrombozane A2 (TXA2), a potent aggregating agent. This accounts for the tendency of these drugs to increase the bleeding time. Aspirin increases oxygen consumption by the body, increasing carbon dioxide production-an effect that stimulates respiration. Therefore, overdose with aspirin is often characterized by marked increases in respiratory rate, which cause the overdosed individual to appear to pant. This occurrence results in other, severe, metabolic consequences. Prolongation of gestation by aspirin-like drugs has been demonstrated in both experimental animals and the human female. Furthermore, prostaglandins of the E and F series are potent uterotropic agents , and their biosynthesis by the uterus increases dramatically in the hours before parturition. It is thus hypothesized that prostaglandins play a major role in the initiation and progression of labor and delivery.
High doses of salicylate may cause ringing in the ears and slight deafness. Sometimes, however, these symptoms indicate mild overdose, which could become more serious.
Aspirin and NSAIDs sometimes affect the normal function of the kidneys and aspirin-like drugs promote the retention of salt and water by reducing the prostaglandin-induced inhibition of both the reabsorption of chloride and the action of antidiuretic hormone. This may cause edema in some patients with arthritis who are treated with an aspirin-like drug.
Recent reports have said there could be a link between the use of aspirin and the development of . is a rare but possibly fatal disease seen most often in children and teenagers. It usually affects those recovering from chicken pox or a viral illness such as the flu. These reports have raised concern in pediatricians (doctors who specialize in treating children) and parents of children with arthritis who need to take large doses of aspirin to control their disease. In the U.S., about 10 to 20 thousand tons of aspirin are consumed each year; it is our most popular analgesic. Aspirin is one of the most effective analgesic, antipyretic, and anti-inflammatory agents. Chemical structure Acetaminophen is an effective alternative to aspirin as an analgesic and antipyretic agent. However, its anti-inflammatory effect is minor and not clinically useful. It is commonly felt that acetaminophen may have fewer side effects than aspirin, but it should be noted that an acute overdose may produce severe or even fatal liver damage. Acetaminophen does not inhibit platelet aggregation and therefore is not useful for preventing vascular clotting. Side effects are usually fewer than those of aspirin; the drug produces less gastric distress and less ringing in the ears. However, as stated previously, overdose can lead to severe damage of the liver. Acetaminophen has been proved to be a reasonable substitute for aspirin when analgesic or antipyretic effectiveness is desired, especially in patients who cannot tolerate aspirin. This might include patients with peptic ulcer disease of gastric distress or those in whom the anticoagulant action of aspirin might be undesirable.
Aspirin is often combined with acetaminophen in a single tablet for relief of arthritis and other painful conditions. Sometimes other drugs such as caffeine, an antihistamine, nasal drying agents, and sedatives are also added. Although some of these preparations may have special uses for certain acute conditions such as a cold or a headache, they should not be taken for a chronic (long-term) form of arthritis. If a combination is required, each drug should be prescribed separately. The dose of each should be adjusted individually to achieve the greatest benefit with the fewest side effects. Researchers attribute the pain-relieving activity of acetaminophen to the drug's ability to elevate the pain threshold, although the precise mechanisms involved in this process have not been clearly identified. The antipyretic, or fever-reducing, effect of acetaminophen is far better understood. Research shows that the drug inhibits the action of fever-producing agents on the heat-regulating centers of the brain by blocking the formation and release of prostaglandins in the central nervous system. However, unlike aspirin and other NSAIDs, acetaminophen has no significant effect on the prostaglandins involved in other body processes. Despite claims to the contrary, stomach upset and hepatic toxicity are statistically as much a problem with acetaminophen as with aspirin-like drugs. Acetaminophen is normally metabolized in the liver and kidney by P450 enzymes. No toxicity is observed with therapeutic doses, however, after ingestion of large quantities (>2,000 mg/kg), a highly reactive metabolite, N-acetyl-p-benzoquinoneimine, is generated (see figure below). This species is electrophilic intermediate which is conjugated with glutathion to a non-toxic compound. Overdosing depletes glutithione and N-acetyl-p-benzoquinone reacts with nucleophilic portions (sulfhdryl groups) of critical liver cell protein. This results in cellular dysfunction and hepatic and renal toxicity. Antidote treatment consists of amino acid supplements to replenish glutathione. The P450 metabolizing enzymes differ somewhat in character between the liver and kidney. Factors that enhance renal toxicity include chronic liver disease, possibly gender, concurrent renal insults, and conditions that alter the activity of P450-metabolizing enzyme systems. Other aspirin-like drugs include diflunisal, phenylbutazone, apazone, indomethacine, sulindac, fenamates, tolmetin, ibuprofen (see figure below), and piroxicam. Gold is not of course an aspirin-like drug. However, its end effect is similar to aspirin, so it will be briefly considered here. Gold in elemental form has been employed for centuries as an antipruritic (anti itch medication) to relieve the itching palm. At present, gold treatment includes different forms of gold salts used to treat rheumatoid arthritis and related diseases. In some people, it helps relieve joint pain and stiffness, reduce swelling and bone damage, and reduce the chance of joint deformity and disability. The significant preparations of gold are all compounds in which the gold is attached to sulfur. The three prominent drugs are aurothioglucose, auranofin, and gold sodium thiomalate. It takes months for gold compounds to leave the body. This means that side effects to gold therapy may take some time to resolve. Sometimes side effects even appear after the last gold injection. Rash and a metallic taste in the mouth are side effects of gold injections that may not seem serious at first. However, they are early warning signs for more serious reactions. If either of these side effects develop, the health care provider should be contacted promptly.
Some side effects may cause multiple symptoms, not all of which may occur. Side effects with multiple symptoms are: The term opiate refers to any natural or synthetic drug that exerts actions upon the body similar to those induced by morphine, the major pain-relieving agent obtained from the opium poppy (Papaver somniferum). They were so highly regarded in the nineteenth century as remedies for pain, anxiety, cough, and diarrhea that some physicians referred to them as G.O.M.- `God's Own Medicine'. Opiates interact with what appear to be several closely related receptors, and they share some of the properties of certain naturally occurring peptides, the enkephalins, endorphins, and dynorphins. The term opium refers to the crude resinous extract obtained from the opium poppy. Crude opium contains a wide variety of ingredients, including morphine and codeine, both of which are widely used in medicine. The bulk of the ingredients of opium, however, consists of such organic substances as resins, oils, sugars, and proteins that account for more than 75 % of the weight of the opium but exert little pharmacological activity. Morphine is the major pain relieving drug found in opium, being approximately 10% of the crude exudate. Codeine is structurally close to morphine (see figs below), although it is much less potent and amounts to only 0.5% of the opium extract. Heroin does not occur naturally but is a semisynthetic derivative produced by a chemical modification of morphine that increases the potency (see figs. below). It takes only 3 mg. of heroin to produce the same analgesic effect as 10 mg of morphine. However, at these equally effective doses, it may be difficult to distinguish between the effects of the two compounds.
of morphine Studies of the binding of opioid drugs and peptides to specific sites in brain and other organs have suggested the existence of perhaps as many as eight types of receptors. In the CNS, there is reasonably firm evidence for four major categories of receptors, designated , and . To add confusion, there may well be subtypes of each of these receptors. Although there is considerable variation in binding characteristics and anatomical distribution among different species, inferences have been drawn from data that attempt to relate pharmacological effects to interactions with a particular constellation of receptors. For example, analgesia has been associated with both and receptors, while dysphoria or psychotomimetic (alteration of behavior or personality) effects have been ascribed to receptors; based primarily on their localization in limbic regions of the brain, receptors are thought to be involved in alterations of affective behavior. The actions of opioid drugs that are currently available have usually been interpreted with respect to the participation of only three types of receptors - and ; at each, a given agent may act as an agonist, a partial agonist, or an antagonist ( ). The receptor is thought to meduate supraspinal analgesia, respiratory depression, euphoria, and physical dependance; the receptor, spinal analgesia, miosis, and sedation; the receptor, dysphoria, hallucinations, and respiratory and vasomotor stimulation. It has been observed that opioids can selectively inhibit certain excitatory inputs to identified neurons. For example, the iontophoretic administration (the induction of an ionized substance through intact skin by the application of a direct current) of morphine into the substantia gelatinosa suppresses the discharge of spinal neurons in lamina IV of the dorsal horn that is evoked by noxious stimuli (e.g. heat) without changing responses to other inputs. While a postsynaptic action at discrete dedritic sites cannot be excluded, these findings suggest that opioids selectively inhibit the release of excitatory transmitters from terminals of nerves carrying pain related stimuli. In other situations, postsynaptic actions of opioids appear to be important. For example, application of opioids to neurons in the locus ceruleus reduces both spontaneous discharge and responses evoked by noxious stimuli. However, excitation of the neurons by antidromic stimulation (i.e. causing the neurons to fire backwards) is also suppressed, and the cells are hyperpolarized by the drugs. Opioids have been observed to inhibit prostaglandin-induced increases in the accumulation of cyclic AMP in in brain tissue. Of potential relevance to mechanisms that underlie the phenomena of tolerance and withdrawal, the responses to prostaglandins recover in the continued presence of opioids. Opioid-induced analgesia is due to actions of several sites within the CNS and involves several systems of neurotransmitters. Although opioids do not alter the threshold or responsivity of afferent nerve endings to noxious stimulation or impair the conduction of the nerve impulses along peripheral nerves, they may decrease conduction of impulses of primary afferent fibers when they enter the spinal cord and decrease activity in other sensory endings. There are opioid binding sites ( receptors) on the terminal axons of primary afferents within laminae I and II (substantia gelatinosa) of the spinal cord and in the spinal nucleus of the trigeminal nerve. Morphine-like drugs acting at this site are thought to decrease the release of neurotransmitters, such as substance P, that mediate transmission of pain impulses. High doses of opioids can produce muscular rigidity in man, and both opioids and endogenous peptides cause catalepsy, circling, and stereotypical behavior in rats and other animals. These effects are probably related to actions at opioid receptors in the substania nigra and striatum, and involve interactions with both dopaminergic and GABA-ergic neurons. The mechanism by which opioids produce euphoria, tranquility, and other alterations of mood remains unsettled. Microinjections of opioids into the ventral tegmentum activate dopaminergic neurons that project to the nucleus accumbens. Animals will work to receive such injections, and activation (or disinhibition) of these neurons has been postulated to be a critical element in the reinforcing effects of opioids and opioid-induced euphoria. However, the administration of dopaminergic antagonists does not consistently prevent these reinforcing effects. The neural systems that mediate opioid reinforcement in the ventral tegmentum appear to be distinct from those involved in the classical manifestations of physical dependence and analgesia. . Morphine exerts a narcotic action manifested by analgesia, drowsiness, changes in mood, and mental clouding. The major medical action of morphine sought in the CNS is analgesia, which may usually be induced by doses below those that cause other effects on the CNS, such as sedation or respiratory depression. The relief of pain by morphine-like opioids is relatively selective, in that other sensory modalities (touch, vibration, vision, hearing, etc.) are not inhibited. Patients frequently report that the pain is still present but that they feel more comfortable. Continuous dull pain is relieved more effectively than sharp intermittent pain, but with sufficient amounts of morphine it is possible to relieve even the severe pain associated with renal or biliary colic. In fact, its analgesic action appears to result not from a decrease of pain impulses into the CNS but from an altered perception of the painful stimuli. . A second major action of morphine-like drugs is to depress respiration through interaction with receptors located in the brainstem. At high doses, respiration may become so slow and irregular that life is threatened. In man, death from morphine poisoning is nearly always due to respiratory arrest. The primary mechanism of respiratory depression by morphine involves a reduction in the responsiveness of the brain stem respiratory centers to increases in carbon dioxide tension (P ). High concentrations of opioid receptors, as well, as endogenous peptides, are found in the medullary areas believed to be important in ventilatory control. Respiratory depression is mediated by a subpopulation of receptors ( 1), distinct from those that are involved in the production of analgesia ( 2). Thus, a 'pure' 1-opioid agonist could theoretically produce analgesia with little respiratory depression. . Opiates suppress the "cough center" which is also located in the brainstem, the medulla. Such an action is thought to underlie the use of opiate narcotics as cough suppressants. Codeine appears to be particularly effective in this action and is widely used for this purpose. . The opiates have been used for centuries for the relief of diarrhea and for the treatment of dysentery, and these uses were developed long before these agents were used as analgesics or euphoiants. Opiates appear to exert their effect on the gastrointestinal tract primarily in the intestine, where peristaltic movements, which normally propel food down the intestine, are markedly diminished. Also, the tone of the intestine is greatly increased to the point where almost complete spastic paralysis of movement occurs. This combination of decreased propulsion and increased tone leads to a marked decrease in the movement of food through the intestine. This stasis is followed by a dehydration of the feces, which hardens the stool and further retards the advance of material. All these effects contribute to the constipating properties of opiates. Indeed, nothing more effective has yet been developed for treating sever diarrhea. Naloxone, when administered to normal individuals, produces no analgesia, euphoria, or respiratory depression. However, it rapidly precipitates withdrawal in narcotic-dependent individuals. Naloxone antagonizes the actions of morphine at all its receptors; however its affinity for receptors is generally more than ten fold higher than for or receptors. The uses of naloxone include the reversal of the respiratory depression that follows acute narcotic intoxication and the reversal of narcotic-induced respiratory depression in newborns of mothers who have received narcotics. The use of naloxone is limited by a short duration of action and the necessity of parenteral route of administration. Naltrexone became clinically available in 1985 as a new narcotic antagonist. Its actions resemble those of naloxone, but naltrexone is well is well absorbed orally and is long acting, necessitating only a dose of 50 to 100 mg. Therefore, it is useful in narcotic treatment programs where it is desired to maintain an individual on chronic therapy with a narcotic antagonist. In individuals taking naltrexone, subsequent injection of an opiate will produce little or no effect. naltrexone appears to be particularly effective for the treatment of narcotic dependence in addicts who have more to gain by being drug-free rather than drug dependant. | 26,101 | 3,396 |
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One technique we can use to measure the amount of heat involved in a chemical or physical process is known as . Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The change in temperature of the measuring part of the calorimeter is converted into the amount of heat (since the previous calibration was used to establish its heat capacity). The measurement of heat transfer using this approach requires the definition of a (the substance or substances undergoing the chemical or physical change) and its (the other components of the measurement apparatus that serve to either provide heat to the system or absorb heat from the system). Knowledge of the heat capacity of the surroundings, and careful measurements of the masses of the system and surroundings and their temperatures before and after the process allows one to calculate the heat transferred as described in this section. A is a device used to measure the amount of heat involved in a chemical or physical process. The thermal energy change accompanying a chemical reaction is responsible for the change in temperature that takes place in a calorimeter. If the reaction releases heat ( < 0), then heat is absorbed by the calorimeter ( > 0) and its temperature increases. Conversely, if the reaction absorbs heat ( > 0), then heat is transferred from the calorimeter to the system ( < 0) and the temperature of the calorimeter decreases. In both cases, . The heat capacity of the calorimeter or of the reaction mixture may be used to calculate the amount of heat released or absorbed by the chemical reaction. The amount of heat released or absorbed per gram or mole of reactant can then be calculated from the mass of the reactants. Because Δ is defined as the heat flow at constant pressure, measurements made using a constant-pressure calorimeter give Δ values directly. This device is particularly well suited to studying reactions carried out in solution at a constant atmospheric pressure. A “student” version, called a (Figure \(\Page {2}\)), is often encountered in general chemistry laboratories. Commercial calorimeters operate on the same principle, but they can be used with smaller volumes of solution, have better thermal insulation, and can detect a change in temperature as small as several millionths of a degree (10 °C). Before we practice calorimetry problems involving chemical reactions, consider a simpler example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W. The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature—that is, when they reach thermal equilibrium (Figure \(\Page {4}\)). If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either the calorimeter or the calorimeter’s surroundings. Under these ideal circumstances, the net heat change is zero: \[q_\mathrm{\,substance\: M} + q_\mathrm{\,substance\: W}=0 \label{7.3.1}\] This relationship can be rearranged to show that the heat gained by substance M is equal to the heat lost by substance W: \[q_\mathrm{\,substance\: M}=-q_\mathrm{\,substance\: W} \label{7.3.2}\] A 360-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water was measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron ( ), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings). The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = −heat taken in by water, or: \[q_\ce{rebar}=−q_\ce{water} \label{5.3.3}\] Since we know how heat is related to other measurable quantities, we have: \[(c×m×ΔT)_\ce{rebar}=−(c×m×ΔT)_\ce{water} \label{5.3.4}\] Letting f = final and i = initial, in expanded form, this becomes: \[c_\ce{rebar}×m_\ce{rebar}×(T_\mathrm{f,rebar}−T_\mathrm{i,rebar})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \label{5.3.5}\] The density of water is 1.0 g/mL, so 425 mL of water = 425 g. Noting that the final temperature of both the rebar and water is 42.7 °C, substituting known values yields: \[\mathrm{(0.449\:J/g\: °C)(360g)(42.7°C−\mathit T_\mathrm{i,rebar})=(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)} \label{5.3.6a}\] \[\mathrm{\mathit T_{i,rebar}=\dfrac{(4.184\:J/g\: °C)(425\:g)(42.7°C−24.0°C)}{(0.449\:J/g\: °C)(360\:g)}+42.7°C} \label{5.3.6b}\] Solving this gives Ti,rebar= 248 °C, so the initial temperature of the rebar was 248 °C. A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water. The initial temperature of the copper was 335.6 °C. A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water initially at 22.6 °C. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature. : The final temperature (reached by both copper and water) is 38.8 °C. This method can also be used to determine other quantities, such as the specific heat of an unknown metal. A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to determine the specific heat of the metal. Use this result to identify the metal. Assuming perfect heat transfer, heat given off by metal = −heat taken in by water, or: \[q_\ce{metal}=−q_\ce{water} \label{5.3.7}\] In expanded form, this is: \[c_\ce{metal}×m_\ce{metal}×(T_\mathrm{f,metal}−T_\mathrm{i, metal})=−c_\ce{water}×m_\ce{water}×(T_\mathrm{f,water}−T_\mathrm{i,water}) \label{5.3.8}\] Noting that since the metal was submerged in boiling water, its initial temperature was 100.0 °C; and that for water, 60.0 mL = 60.0 g; we have: \[\mathrm{(\mathit c_{metal})(59.7\:g)(28.5°C−100.0°C)=−(4.18\:J/g\: °C)(60.0\:g)(28.5°C−22.0°C)} \label{5.3.9}\] Solving this: \[\mathrm{\mathit c_{metal}=\dfrac{−(4.184\:J/g\: °C)(60.0\:g)(6.5°C)}{(59.7\:g)(−71.5°C)}=0.38\:J/g\: °C}\] Comparing this with values in , our experimental specific heat is closest to the value for copper (0.39 J/g °C), so we identify the metal as copper. A 92.9-g piece of a silver/gray metal is heated to 178.0 °C, and then quickly transferred into 75.0 mL of water initially at 24.0 °C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7 °C. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal). \(c_{metal}= 0.13 \;J/g\; °C\) This specific heat is close to that of either gold or lead. It would be difficult to determine which metal this was based solely on the numerical values. However, the observation that the metal is silver/gray in addition to the value for the specific heat indicates that the metal is lead. When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have been discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the surroundings. Because energy is neither created nor destroyed during a chemical reaction, there is no overall energy change during the reaction. The heat produced or consumed in the reaction (the “system”), , plus the heat absorbed or lost by the solution (the “surroundings”), , must add up to zero: \[q_\ce{reaction}+q_\ce{solution}=0\ \label{ \(\Page {3}\)}\] This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution: \[q_\ce{reaction}=−q_\ce{solution} \label{\(\Page {4}\)}\] This concept lies at the heart of all calorimetry problems and calculations. Because the heat released or absorbed at constant pressure is equal to Δ , the relationship between heat and Δ is \[ \Delta H_{rxn}=q_{rxn}=-q_{calorimater}=-mC_{s} \Delta T \label{\(\Page {5}\)} \] The use of a constant-pressure calorimeter is illustrated in Example \(\Page {3}\). When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range averages 0.9969 g/cm . What is Δ (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water. mass of substance, volume of solvent, and initial and final temperatures Δ To calculate Δ , we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is \[ \left (100.0 \; mL\; H2O \right ) \left ( 0.9969 \; g/ \cancel{mL} \right )+ 5.03 \; g \; KOH=104.72 \; g \] The temperature change is (34.7°C − 23.0°C) = +11.7°C. Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus \[ q_{calorimater}=mC_{s} \Delta T =\left ( 104.72 \; \cancel{g} \right ) \left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right )\left ( 11.7 \; ^{o}C \right )=5130 \; J =5.13 \; lJ \] The temperature of the solution increased because heat was absorbed by the solution ( > 0). Where did this heat come from? It was released by KOH dissolving in water. From Equation \(\Page {1}\), we see that This experiment tells us that dissolving 5.03 g of KOH in water is accompanied by the of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of KOH in water must be exothermic. The last step is to use the molar mass of KOH to calculate Δ —the heat released when dissolving 1 mol of KOH: \[ \Delta H_{soln}= \left ( \dfrac{5.13 \; kJ}{5.03 \; \cancel{g}} \right )\left ( \dfrac{56.11 \; \cancel{g}}{1 \; mol} \right )=-57.2 \; kJ/mol \] A coffee-cup calorimeter contains 50.0 mL of distilled water at 22.7°C. Solid ammonium bromide (3.14 g) is added and the solution is stirred, giving a final temperature of 20.3°C. Using the same assumptions as in Example \(\Page {3}\), find Δ for NH Br (in kilojoules per mole). 16.6 kJ/mol Conservation of Energy: Coffee Cup Calorimetry: Constant-pressure calorimeters are not very well suited for studying reactions in which one or more of the reactants is a gas, such as a combustion reaction. The enthalpy changes that accompany combustion reactions are therefore measured using a constant-volume calorimeter, such as the bomb calorimeter shown schematically in Figure \(\Page {3}\)). The reactant is placed in a steel cup inside a steel vessel with a fixed volume (the “bomb”). The bomb is then sealed, filled with excess oxygen gas, and placed inside an insulated container that holds a known amount of water. Because combustion reactions are exothermic, the temperature of the bath and the calorimeter increases during combustion. If the heat capacity of the bomb and the mass of water are known, the heat released can be calculated. Because the volume of the system (the inside of the bomb) is fixed, the combustion reaction occurs under conditions in which the volume, but not the pressure, is constant. The heat released by a reaction carried out at constant volume is identical to the change in (ΔU) rather than the enthalpy change (Δ ); ΔU is related to Δ by an expression that depends on the change in the number of moles of gas during the reaction. The difference between the heat flow measured at constant volume and the enthalpy change is usually quite small, however (on the order of a few percent). Assuming that Δ < Δ , the relationship between the measured temperature change and Δ is given in Equation \ref{7.3.6}, where is the total heat capacity of the steel bomb and the water surrounding it: \[ \Delta H_{comb} < q_{comb} = q_{calorimater} = C_{bomb} \Delta T \label{7.3.6}\] To measure the heat capacity of the calorimeter, we first burn a carefully weighed mass of a standard compound whose enthalpy of combustion is accurately known. Benzoic acid (C H CO H) is often used for this purpose because it is a crystalline solid that can be obtained in high purity. The combustion of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat per gram (i.e., its Δ = −26.38 kJ/g). This value and the measured increase in temperature of the calorimeter can be used in Equation \ref{5.42} to determine . The use of a bomb calorimeter to measure the Δ of a substance is illustrated in Example \(\Page {4}\). The combustion of 0.579 g of benzoic acid in a bomb calorimeter caused a 2.08°C increase in the temperature of the calorimeter. The chamber was then emptied and recharged with 1.732 g of glucose and excess oxygen. Ignition of the glucose resulted in a temperature increase of 3.64°C. What is the Δ of glucose? mass and Δ for combustion of standard and sample Δ of glucose The first step is to use Equation \(\Page {2}\) and the information obtained from the combustion of benzoic acid to calculate . We are given Δ , and we can calculate from the mass of benzoic acid: \[ q_{comb} = \left ( 0.579 \; \cancel{g} \right )\left ( -26.38 \; kJ/\cancel{g} \right ) = - 15.3 \; kJ \] From Equation \(\Page {2}\), \[ -C_{bomb} = \dfrac{q_{comb}}{\Delta T} = \dfrac{-15.3 \; kJ}{2.08 \; ^{o}C} =- 7.34 \; kJ/^{o}C \] According to the strategy, we can now use the heat capacity of the bomb to calculate the amount of heat released during the combustion of glucose: \[ q_{comb}=-C_{bomb}\Delta T = \left ( -7.34 \; kJ/^{o}C \right )\left ( 3.64 \; ^{o}C \right )=- 26.7 \; kJ \] Because the combustion of 1.732 g of glucose released 26.7 kJ of energy, the Δ of glucose is \[ \Delta H_{comb}=\left ( \dfrac{-26.7 \; kJ}{1.732 \; \cancel{g}} \right )\left ( \dfrac{180.16 \; \cancel{g}}{mol} \right )=-2780 \; kJ/mol =2.78 \times 10^{3} \; kJ/mol\] This result is in good agreement (< 1% error) with the value of Δ = −2803 kJ/mol that calculated using enthalpies of formation. When 2.123 g of benzoic acid is ignited in a bomb calorimeter, a temperature increase of 4.75°C is observed. When 1.932 g of methylhydrazine (CH NHNH ) is ignited in the same calorimeter, the temperature increase is 4.64°C. Calculate the Δ of methylhydrazine, the fuel used in the maneuvering jets of the US space shuttle. −1.30 × 10 kJ/mol Conservation of Energy: Bomb Calorimetry: is the set of techniques used to measure enthalpy changes during chemical processes. It uses devices called , which measure the change in temperature when a chemical reaction is carried out. The magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system. The of an object is the amount of energy needed to raise its temperature by 1°C; its units are joules per degree Celsius. The of a substance is the amount of energy needed to raise the temperature of 1 g of the substance by 1°C, and the is the amount of energy needed to raise the temperature of 1 mol of a substance by 1°C. Liquid water has one of the highest specific heats known. Heat flow measurements can be made with either a , which gives Δ values directly, or a , which operates at constant volume and is particularly useful for measuring enthalpies of combustion. Thermal energy itself cannot be measured easily, but the temperature change caused by the flow of thermal energy between objects or substances can be measured. Calorimetry describes a set of techniques employed to measure enthalpy changes in chemical processes using devices called . To have any meaning, the quantity that is actually measured in a calorimetric experiment, the change in the temperature of the device, must be related to the heat evolved or consumed in a chemical reaction. We begin this section by explaining how the flow of thermal energy affects the temperature of an object. ). | 16,967 | 3,397 |
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Substances consisting entirely of single-bonded carbon and hydrogen atoms and lacking functional groups are called alkanes. There are three basic types of structures that classify the alkanes: (1) linear straight-chain alkanes, (2) branched alkanes, and (3) . In the straight-chain alkanes, each carbon is bound to its two neighbors and to two hydrogen atoms. (In red) Exceptions are the two terminal carbon nuclei, which are bound to only one carbon atom and three hydrogen atoms. (In blue) - - - - ) . Methane (n=1) is the first member of the homologous series of the alkanes, followed by ethane (n=2) and so forth. - = CH - CH Branched alkanes are derived from the straight-chain alkanes system by removing one of the hydrogen atoms from a methylene group (-CH -) and replacing it with an alkyl group. Straight-chained and branched alkanes follow the same general formula: C H . The smallest branched alkane is 2-methylpropane or isobutane (Pictures shown above). 2-methylpropane has the same molecular formula as butane (C H ) but with a different connectivity, resulting in a different structure. These 2 compounds form a pair of ( , Greeks, equal). Isomers are compounds with the same molecular formula but different structural formulae. For higher alkane homologs (n > 4), more than two isomers are possible. For example, pentane has three possible isomers in which one is a linear straight-chain alkane and two are branched alkanes. When branched, the nomenclature can be different because of common and IUPAC names. | 1,562 | 3,398 |
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What does it mean to say an equation is balanced? Why is it important for an equation to be balanced? An equation is balanced when the same number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter. Consider molecular, complete ionic, and net ionic equations. Balance the following equations: Balance the following equations: Write a balanced molecular equation describing each of the following chemical reactions. Write a balanced equation describing each of the following chemical reactions. Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen. Fill in the blank with a single chemical formula for a covalent compound that will balance the equation: Aqueous hydrogen fluoride (hydrofluoric acid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride will also react with sand (silicon dioxide). A novel process for obtaining magnesium from sea water involves several reactions. Write a balanced chemical equation for each step of the process. From the balanced molecular equations, write the complete ionic and net ionic equations for the following: Use the following equations to answer the next five questions: a.) \(H_2O (solid) → H_2O(liquid)\) b.) c.) \(\ce{2CH3OH}(g)+\ce{3O2}(g)\rightarrow \ce{2CO2}(g)+\ce{4H2O}(g)\) d.) Indicate what type, or types, of reaction each of the following represents: oxidation-reduction (addition); acid-base (neutralization); oxidation-reduction (combustion) < Indicate what type, or types, of reaction each of the following represents: Silver can be separated from gold because silver dissolves in nitric acid while gold does not. Is the dissolution of silver in nitric acid an acid-base reaction or an oxidation-reduction reaction? Explain your answer. It is an oxidation-reduction reaction because the oxidation state of the silver changes during the reaction. Determine the oxidation states of the elements in the following compounds: Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides. H +1, P +5, O −2; Al +3, H +1, O −2; Se +4, O −2; K +1, N +3, O −2; In +3, S −2; P +3, O −2 Determine the oxidation states of the elements in the compounds listed. None of the oxygen-containing compounds are peroxides or superoxides. Classify the following as acid-base reactions or oxidation-reduction reactions: acid-base; oxidation-reduction: Na is oxidized, H is reduced; oxidation-reduction: Mg is oxidized, Cl is reduced; acid-base; oxidation-reduction: P is oxidized, O is reduced; acid-base Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the oxidizing and reducing agents in each of the following equations: Complete and balance the following acid-base equations: Complete and balance the following acid-base equations: Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms. Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms. Complete and balance the equations for the following acid-base neutralization reactions. If water is used as a solvent, write the reactants and products as aqueous ions. In some cases, there may be more than one correct answer, depending on the amounts of reactants used. When heated to 700–800 °C, diamonds, which are pure carbon, are oxidized by atmospheric oxygen. (They burn!) Write the balanced equation for this reaction. The military has experimented with lasers that produce very intense light when fluorine combines explosively with hydrogen. What is the balanced equation for this reaction? \(\ce{H2}(g)+\ce{F2}(g)\rightarrow \ce{2HF}(g)\) Write the molecular, total ionic, and net ionic equations for the following reactions: Great Lakes Chemical Company produces bromine, Br , from bromide salts such as NaBr, in Arkansas brine by treating the brine with chlorine gas. Write a balanced equation for the reaction of NaBr with Cl . \(\ce{2NaBr}(aq)+\ce{Cl2}(g)\rightarrow \ce{2NaCl}(aq)+\ce{Br2}(l)\) In a common experiment in the general chemistry laboratory, magnesium metal is heated in air to produce MgO. MgO is a white solid, but in these experiments it often looks gray, due to small amounts of Mg N , a compound formed as some of the magnesium reacts with nitrogen. Write a balanced equation for each reaction. Lithium hydroxide may be used to absorb carbon dioxide in enclosed environments, such as manned spacecraft and submarines. Write an equation for the reaction that involves 2 mol of LiOH per 1 mol of CO . (Hint: Water is one of the products.) \(\ce{2LiOH}(aq)+\ce{CO2}(g)\rightarrow \ce{Li2CO3}(aq)+\ce{H2O}(l)\) Calcium propionate is sometimes added to bread to retard spoilage. This compound can be prepared by the reaction of calcium carbonate, CaCO , with propionic acid, C H CO H, which has properties similar to those of acetic acid. Write the balanced equation for the formation of calcium propionate. Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas: Copper(II) sulfide is oxidized by molecular oxygen to produce gaseous sulfur trioxide and solid copper(II) oxide. The gaseous product then reacts with liquid water to produce liquid hydrogen sulfate as the only product. Write the two equations which represent these reactions. Write balanced chemical equations for the reactions used to prepare each of the following compounds from the given starting material(s). In some cases, additional reactants may be required. Calcium cyclamate Ca(C H NHSO ) is an artificial sweetener used in many countries around the world but is banned in the United States. It can be purified industrially by converting it to the barium salt through reaction of the acid C H NHSO H with barium carbonate, treatment with sulfuric acid (barium sulfate is very insoluble), and then neutralization with calcium hydroxide. Write the balanced equations for these reactions. Complete and balance each of the following half-reactions (steps 2–5 in half-reaction method): Complete and balance each of the following half-reactions (steps 2–5 in half-reaction method): Balance each of the following equations according to the half-reaction method: Balance each of the following equations according to the half-reaction method: Balance each of the following equations according to the half-reaction method: Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following: Determine the number of moles and the mass requested for each reaction in . 0.435 mol Na, 0.217 mol Cl , 15.4 g Cl ; 0.005780 mol HgO, 2.890 × 10 mol O , 9.248 × 10 g O ; 8.00 mol NaNO , 6.8 × 10 g NaNO ; 1665 mol CO , 73.3 kg CO ; 18.86 mol CuO, 2.330 kg CuCO ; 0.4580 mol C H Br , 86.05 g C H Br Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following: Determine the number of moles and the mass requested for each reaction in . 0.0686 mol Mg, 1.67 g Mg; 2.701 × 10 mol O , 0.08644 g O ; 6.43 mol MgCO , 542 g MgCO 713 mol H O, 12.8 kg H O; 16.31 mol BaO , 2762 g BaO ; 0.207 mol C H , 5.81 g C H H is produced by the reaction of 118.5 mL of a 0.8775-M solution of H PO according to the following equation: \(\ce{2Cr + 2H3PO4 \rightarrow 3H2 + 2CrPO4}\). a.) b.) 1. \(118.5\: mL\times \dfrac{1\: L}{1000\: mL} = 0.1185\: L\) 2. \(0.1185\: L \times \dfrac{0.8775\: moles\: \ce{H3PO4}}{1\: L} = 0.1040\: moles\: \ce{H3PO4}\) 3. \(0.1040\: moles\: \ce{H3PO4} \times \dfrac{3\: moles\:\ce{H_2}}{2\: moles\: \ce{H3PO4}} = 0.1560\: moles\: \ce{H2}\) 4. \(0.1560\: moles\: \ce{H2} \times \dfrac{2.02 g}{1\: mole} = 0.3151g\: \ce{H2}\) Gallium chloride is formed by the reaction of 2.6 L of a 1.44 M solution of HCl according to the following equation: \(\ce{2Ga + 6HCl \rightarrow 2GaCl3 + 3H2}\). \(\mathrm{volume\: HCl\: solution \rightarrow mol\: HCl \rightarrow mol\: GaCl_3}\); 1.25 mol GaCl , 2.2 × 10 g GaCl I is produced by the reaction of 0.4235 mol of CuCl according to the following equation: \(\ce{2CuCl2 + 4KI \rightarrow 2CuI + 4KCl + I2}\). Silver is often extracted from ores as K[Ag(CN) ] and then recovered by the reaction \(\ce{2K[Ag(CN)2]}(aq)+\ce{Zn}(s)\rightarrow \ce{2Ag}(s)+\ce{Zn(CN)2}(aq)+\ce{2KCN}(aq)\) 5.337 × 10 molecules; 10.41 g Zn(CN) What mass of silver oxide, Ag O, is required to produce 25.0 g of silver sulfadiazine, AgC H N SO , from the reaction of silver oxide and sulfadiazine? \(\ce{2C10H10N4SO2 + Ag2O \rightarrow 2AgC10H9N4SO2 + H2O}\) Carborundum is silicon carbide, SiC, a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, SiO , with carbon at high temperature. Carbon monoxide, CO, is the other product of this reaction. Write the balanced equation for the reaction, and calculate how much SiO is required to produce 3.00 kg of SiC. \(\ce{SiO2 + 3C \rightarrow SiC + 2CO}\), 4.50 kg SiO Automotive air bags inflate when a sample of sodium azide, NaN , is very rapidly decomposed. \(\ce{2NaN3}(s) \rightarrow \ce{2Na}(s) + \ce{3N2}(g)\) What mass of sodium azide is required to produce 2.6 ft (73.6 L) of nitrogen gas with a density of 1.25 g/L? 142g NaN Urea, CO(NH ) , is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the CO produced by combustion of 1.00×10 kg of carbon followed by the reaction? \[\ce{CO2}(g)+\ce{2NH3}(g)\rightarrow \ce{CO(NH2)2}(s)+\ce{H2O}(l)\] 5.00 × 10 kg In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na CO was quickly spread on the area and CO was released by the reaction. Was sufficient Na CO used to neutralize all of the acid? A compact car gets 37.5 miles per gallon on the highway. If gasoline contains 84.2% carbon by mass and has a density of 0.8205 g/mL, determine the mass of carbon dioxide produced during a 500-mile trip (3.785 liters per gallon). 1.28 × 10 g CO What volume of a 0.750 M solution of hydrochloric acid, a solution of HCl, can be prepared from the HCl produced by the reaction of 25.0 g of NaCl with an excess of sulfuric acid? \[\ce{NaCl}(s)+\ce{H2SO4}(l)\rightarrow \ce{HCl}(g)+\ce{NaHSO4}(s)\] What volume of a 0.2089 KI solution contains enough KI to react exactly with the Cu(NO ) in 43.88 mL of a 0.3842 solution of Cu(NO ) ? \[\ce{2Cu(NO3)2 + 4KI \rightarrow 2CuI + I2 + 4KNO3}\] 161.40 mL KI solution A mordant is a substance that combines with a dye to produce a stable fixed color in a dyed fabric. Calcium acetate is used as a mordant. It is prepared by the reaction of acetic acid with calcium hydroxide. \[\ce{2CH3CO2H + Ca(OH)2 \rightarrow Ca(CH3CO2)2 + 2H2O}\] What mass of Ca(OH) is required to react with the acetic acid in 25.0 mL of a solution having a density of 1.065 g/mL and containing 58.0% acetic acid by mass? The toxic pigment called white lead, Pb (OH) (CO ) , has been replaced in white paints by rutile, TiO . How much rutile (g) can be prepared from 379 g of an ore that contains 88.3% ilmenite (FeTiO ) by mass? \[\ce{2FeTiO3 + 4HCl + Cl2 \rightarrow 2FeCl3 + 2TiO2 + 2H2O}\] 176 g TiO The following quantities are placed in a container: 1.5 × 10 atoms of hydrogen, 1.0 mol of sulfur, and 88.0 g of diatomic oxygen. What is the limiting reactant in a reaction that produces sodium chloride from 8 g of sodium and 8 g of diatomic chlorine? The limiting reactant is Cl . Which of the postulates of Dalton's atomic theory explains why we can calculate a theoretical yield for a chemical reaction? A student isolated 25 g of a compound following a procedure that would theoretically yield 81 g. What was his percent yield? \(\mathrm{Percent\: yield = 31\%}\) A sample of 0.53 g of carbon dioxide was obtained by heating 1.31 g of calcium carbonate. What is the percent yield for this reaction? \[\ce{CaCO3}(s)\rightarrow \ce{CaO}(s)+\ce{CO2}(s)\] Freon-12, CCl F , is prepared from CCl by reaction with HF. The other product of this reaction is HCl. Outline the steps needed to determine the percent yield of a reaction that produces 12.5 g of CCl F from 32.9 g of CCl . Freon-12 has been banned and is no longer used as a refrigerant because it catalyzes the decomposition of ozone and has a very long lifetime in the atmosphere. Determine the percent yield. \(\ce{g\: CCl4\rightarrow mol\: CCl4\rightarrow mol\: CCl2F2 \rightarrow g\: CCl2F2}, \mathrm{\:percent\: yield=48.3\%}\) Citric acid, C H O , a component of jams, jellies, and fruity soft drinks, is prepared industrially via fermentation of sucrose by the mold . The equation representing this reaction is \[\ce{C12H22O11 + H2O + 3O2 \rightarrow 2C6H8O7 + 4H2O}\] What mass of citric acid is produced from exactly 1 metric ton (1.000 × 10 kg) of sucrose if the yield is 92.30%? Toluene, C H CH , is oxidized by air under carefully controlled conditions to benzoic acid, C H CO H, which is used to prepare the food preservative sodium benzoate, C H CO Na. What is the percent yield of a reaction that converts 1.000 kg of toluene to 1.21 kg of benzoic acid? \[\ce{2C6H5CH3 + 3O2 \rightarrow 2C6H5CO2H + 2H2O}\] \(\mathrm{percent\: yield=91.3\%}\) In a laboratory experiment, the reaction of 3.0 mol of H with 2.0 mol of I produced 1.0 mol of HI. Determine the theoretical yield in grams and the percent yield for this reaction. Outline the steps needed to solve the following problem, then do the calculations. Ether, (C H ) O, which was originally used as an anesthetic but has been replaced by safer and more effective medications, is prepared by the reaction of ethanol with sulfuric acid. 2C H OH + H SO ⟶ (C H ) + H SO ·H O What is the percent yield of ether if 1.17 L (d = 0.7134 g/mL) is isolated from the reaction of 1.500 L of C H OH (d = 0.7894 g/mL)? Convert mass of ethanol to moles of ethanol; relate the moles of ethanol to the moles of ether produced using the stoichiometry of the balanced equation. Convert moles of ether to grams; divide the actual grams of ether (determined through the density) by the theoretical mass to determine the percent yield; 87.6% Outline the steps needed to determine the limiting reactant when 30.0 g of propane, C H , is burned with 75.0 g of oxygen. \[\mathrm{percent\: yield=\dfrac{0.8347\:\cancel{g}}{0.9525\:\cancel{g}}\times 100\%=87.6\%}\] Determine the limiting reactant. Outline the steps needed to determine the limiting reactant when 0.50 g of Cr and 0.75 g of H PO react according to the following chemical equation? \[\ce{2Cr + 2H3PO4 \rightarrow 2CrPO4 + 3H2}\] Determine the limiting reactant. The conversion needed is \(\ce{mol\: Cr \rightarrow mol\: H2PO4}\). Then compare the amount of Cr to the amount of acid present. Cr is the limiting reactant. What is the limiting reactant when 1.50 g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advanced batteries, according to the following unbalanced equation? \[\ce{Li + N2 \rightarrow Li3N}\] \[\ce{6Li} + \ce{N2} \rightarrow \: \ce{2Li3N}\] \[1.50g\: \ce{Li} \times \dfrac{1\: mole\: \ce{Li}}{6.94g\: \ce{Li}} \times\dfrac{2\: mole\: \ce{Li3N}}{6\:mole\: \ce{Li}} = 0.0720\: moles\: \ce{Li3N}\] \[1.50g\: \ce{N2} \times \dfrac{1\: mole\: \ce{N2}}{28.02g\: \ce{N2}} \times\dfrac{2\: mole\: \ce{Li3N}}{1\:mole\: \ce{N2}} = 0.107\: moles\: \ce{Li3N}\] \(\ce{Li}\) is the limiting reactant Uranium can be isolated from its ores by dissolving it as UO (NO ) , then separating it as solid UO (C O )·3H O. Addition of 0.4031 g of sodium oxalate, Na C O , to a solution containing 1.481 g of uranyl nitrate, UO (NO ) , yields 1.073 g of solid UO (C O )·3H O. \[\ce{Na2C2O4 + UO2(NO3)2 + 3H2O ⟶ UO2(C2O4)·3H2O + 2NaNO3}\] Determine the limiting reactant and the percent yield of this reaction. Na C O is the limiting reactant. percent yield = 86.6% How many molecules of C H Cl can be prepared from 15 C H molecules and 8 Cl molecules? How many molecules of the sweetener saccharin can be prepared from 30 C atoms, 25 H atoms, 12 O atoms, 8 S atoms, and 14 N atoms? Only four molecules can be made. The phosphorus pentoxide used to produce phosphoric acid for cola soft drinks is prepared by burning phosphorus in oxygen. Would you agree to buy 1 trillion (1,000,000,000,000) gold atoms for $5? Explain why or why not. Find the current price of gold at \(\mathrm{(1\: troy\: ounce=31.1\: g)}\) This amount cannot be weighted by ordinary balances and is worthless. What volume of 0.0105-M HBr solution is be required to titrate 125 mL of a 0.0100- Ca(OH) solution? \[\ce{Ca(OH)2}(aq)+\ce{2HBr}(aq) \rightarrow \ce{CaBr2}(aq)+\ce{2H2O}(l)\] Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 NaOH to reach the end point. If we assume that the acidity of the rain is due to the presence of sulfuric acid, what was the concentration of sulfuric acid in this sample of rain? 3.4 × 10 H SO What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503 AgNO requires 20.22 mL of the AgNO solution to reach the end point? \[\ce{AgNO3}(aq)+\ce{NaCl}(aq)\rightarrow \ce{AgCl}(s)+\ce{NaNO3}(aq)\] In a common medical laboratory determination of the concentration of free chloride ion in blood serum, a serum sample is titrated with a Hg(NO ) solution. \[\ce{2Cl-}(aq)+\ce{Hg(NO3)2}(aq)\rightarrow \ce{2NO3-}(aq)+\ce{HgCl2}(s)\] What is the Cl concentration in a 0.25-mL sample of normal serum that requires 1.46 mL of 5.25 × 10 Hg(NO ) ( ) to reach the end point? 9.6 × 10 Cl Potatoes can be peeled commercially by soaking them in a 3-M to 6-M solution of sodium hydroxide, then removing the loosened skins by spraying them with water. Does a sodium hydroxide solution have a suitable concentration if titration of 12.00 mL of the solution requires 30.6 mL of 1.65 M HCI to reach the end point? A sample of gallium bromide, GaBr , weighing 0.165 g was dissolved in water and treated with silver nitrate, AgNO , resulting in the precipitation of 0.299 g AgBr. Use these data to compute the %Ga (by mass) GaBr . 22.4% The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon and hydrogen. A 3.000-mg sample of naphthalene burns to give 10.3 mg of CO . Determine its empirical and molecular formulas. A 0.025-g sample of a compound composed of boron and hydrogen, with a molecular mass of ~28 amu, burns spontaneously when exposed to air, producing 0.063 g of B O . What are the empirical and molecular formulas of the compound. The empirical formula is BH . The molecular formula is B H . Sodium bicarbonate (baking soda), NaHCO , can be purified by dissolving it in hot water (60 °C), filtering to remove insoluble impurities, cooling to 0 °C to precipitate solid NaHCO , and then filtering to remove the solid, leaving soluble impurities in solution. Any NaHCO that remains in solution is not recovered. The solubility of NaHCO in hot water of 60 °C is 164 g L. Its solubility in cold water of 0 °C is 69 g/L. What is the percent yield of NaHCO when it is purified by this method? What volume of 0.600 HCl is required to react completely with 2.50 g of sodium hydrogen carbonate? \[\ce{NaHCO3}(aq)+\ce{HCl}(aq)\rightarrow \ce{NaCl}(aq)+\ce{CO2}(g)+\ce{H2O}(l)\] 49.6 mL What volume of 0.08892 HNO is required to react completely with 0.2352 g of potassium hydrogen phosphate? \[\ce{2HNO3}(aq)+\ce{K2HPO4}(aq)\rightarrow \ce{H2PO4}(aq)+\ce{2KNO3}(aq)\] What volume of a 0.3300- solution of sodium hydroxide would be required to titrate 15.00 mL of 0.1500 oxalic acid? \[\ce{C2O4H2}(aq)+\ce{2NaOH}(aq)\rightarrow \ce{Na2C2O4}(aq)+\ce{2H2O}(l)\] 13.64 mL What volume of a 0.00945- solution of potassium hydroxide would be required to titrate 50.00 mL of a sample of acid rain with a H SO concentration of 1.23 × 10 . \[\ce{H2SO4}(aq)+\ce{2KOH}(aq)\rightarrow \ce{K2SO4}(aq)+\ce{2H2O}(l)\] 1.30 mL A sample of solid calcium hydroxide, Ca(OH) , is allowed to stand in water until a saturated solution is formed. A titration of 75.00 mL of this solution with 5.00 × 10 HCl requires 36.6 mL of the acid to reach the end point. \[\ce{Ca(OH)2}(aq)+\ce{2HCl}(aq)\rightarrow \ce{CaCl2}(aq)+\ce{2H2O}(l)\] What is the molarity? 1.22 What mass of Ca(OH) will react with 25.0 g of propionic acid to form the preservative calcium propionate according to the equation? How many milliliters of a 0.1500- solution of KOH will be required to titrate 40.00 mL of a 0.0656- solution of H PO ? \[\ce{H3PO4}(aq)+\ce{2KOH}(aq)\rightarrow \ce{K2HPO4}(aq)+\ce{2H2O}(l)\] 34.99 mL KOH Potassium acid phthalate, KHC H O , or KHP, is used in many laboratories, including general chemistry laboratories, to standardize solutions of base. KHP is one of only a few stable solid acids that can be dried by warming and weighed. A 0.3420-g sample of KHC H O reacts with 35.73 mL of a NaOH solution in a titration. What is the molar concentration of the NaOH? \[\ce{KHC6H4O4}(aq)+\ce{NaOH}(aq)\rightarrow \ce{KNaC6H4O4}(aq)+\ce{H2O}(aq)\] The reaction of WCl with Al at ~400 °C gives black crystals of a compound containing only tungsten and chlorine. A sample of this compound, when reduced with hydrogen, gives 0.2232 g of tungsten metal and hydrogen chloride, which is absorbed in water. Titration of the hydrochloric acid thus produced requires 46.2 mL of 0.1051 NaOH to reach the end point. What is the empirical formula of the black tungsten chloride? The empirical formula is WCl . | 21,963 | 3,400 |
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This page explains the relationship between the arrangement of the ions in a typical ionic solid like sodium chloride and its physical properties - melting point, boiling point, brittleness, solubility and electrical behavior. It also explains why cesium chloride has a different structure from sodium chloride even though sodium and cesium are both in Group 1 of the Periodic Table. Sodium chloride is taken as a typical ionic compound. Compounds like this consist of a giant (endlessly repeating) lattice of ions. So sodium chloride (and any other ionic compound) is described as having a giant ionic structure. You should be clear that giant in this context does not just mean very large. It means that you can't state exactly how many ions there are. There could be billions of sodium ions and chloride ions packed together, or trillions, or whatever - it simply depends how big the crystal is. That is different from, say, a water molecule which always contains exactly 2 hydrogen atoms and one oxygen atom - never more and never less. A small representative bit of a sodium chloride lattice looks like this: If you look at the diagram carefully, you will see that the sodium ions and chloride ions alternate with each other in each of the three dimensions. This diagram is easy enough to draw with a computer, but extremely difficult to draw convincingly by hand. We normally draw an "exploded" version which looks like this: Only those ions joined by lines are actually touching each other. The sodium ion in the center is being touched by 6 chloride ions. By chance we might just as well have centered the diagram around a chloride ion - that, of course, would be touched by 6 sodium ions. Sodium chloride is described as being 6:6-coordinated. This diagram represents only a tiny part of the whole sodium chloride crystal; the pattern repeats in this way over countless ions. Draw a perfect square: Now draw an identical square behind this one and offset a bit. You might have to practice a bit to get the placement of the two squares right. If you get it wrong, the ions get all tangled up with each other in your final diagram. Turn this into a perfect cube by joining the squares together: Now the tricky bit! Subdivide this big cube into 8 small cubes by joining the mid point of each edge to the mid point of the edge opposite it. To complete the process you will also have to join the mid point of each face (easily found once you've joined the edges) to the mid point of the opposite face. Now all you have to do is put the ions in. Use different colors or different sizes for the two different ions, and don't forget a key. It does not matter whether you end up with a sodium ion or a chloride ion in the center of the cube - all that matters is that they alternate in all three dimensions. You should be able to draw a perfectly adequate free-hand sketch of this in under two minutes - less than one minute if you're not too fussy! The more attraction there is between the positive and negative ions, the more energy is released. The more energy that is released, the more energetically stable the structure becomes. That means that to gain maximum stability, you need the maximum number of attractions. So why does each ion surround itself with 6 ions of the opposite charge? That represents the maximum number of chloride ions that you can fit around a central sodium ion before the chloride ions start touching each other. If they start touching, you introduce repulsions into the crystal which makes it less stable. We'll look first at the arrangement of the ions and then talk about why the structures of sodium chloride and cesium chloride are different afterwards. Imagine a layer of chloride ions as shown below. The individual chloride ions aren't touching each other. That's really important - if they were touching, there would be repulsion. Now let's place a similarly arranged layer of cesium ions on top of these. Notice that the cesium ions aren't touching each other either, but that each cesium ion is resting on four chloride ions from the layer below. Now let's put another layer of chloride ions on, exactly the same as the first layer. Again, the chloride ions in this layer are NOT touching those in the bottom layer - otherwise you are introducing repulsion. Since we are looking directly down on the structure, you can't see the bottom layer of chloride ions any more, of course. If you now think about a cesium ion sandwiched between the two layers of chloride ions, it is touching four chloride ions in the bottom layer, and another four in the top one. Each cesium ion is touched by eight chloride ions. We say that it is 8-coordinated. If we added another layer of cesium ions, you could similarly work out that each chloride ion was touching eight cesium ions. The chloride ions are also 8-coordinated. Overall, then, cesium chloride is 8:8-coordinated. The final diagram in this sequence takes a slightly tilted view of the structure so that you can see how the layers build up. These diagrams are quite difficult to draw without it looking as if ions of the same charge are touching each other. They aren't! Diagrams of ionic crystals are usually simplified to show the most basic unit of the repeating pattern. For cesium chloride, you could, for example, draw a simple diagram showing the arrangement of the chloride ions around each cesium ion: By reversing the colors (green chloride ion in the center, and orange cesium ions surrounding it), you would have an exactly equivalent diagram for the arrangement of cesium ions around each chloride ion. When attractions are set up between two ions of opposite charges, energy is released. The more energy that can be released, the more stable the system becomes. That means that the more contact there is between negative and positive ions, the more stable the crystal should become. If you can surround a positive ion like cesium with eight chloride ions rather than just six (and vice versa for the chloride ions), then you should have a more stable crystal. So why does not sodium chloride do the same thing? Look again at the last diagram: Now imagine what would happen if you replaced the cesium ion with the smaller sodium ion. Sodium ions are, of course, smaller than cesium ions because they have fewer layers of electrons around them. You still have to keep the chloride ions in contact with the sodium. The effect of this would be that the whole arrangement would shrink, bringing the chloride ions into contact with each other - and that introduces repulsion. Any gain in attractions because you have eight chlorides around the sodium rather than six is more than countered by the new repulsions between the chloride ions themselves. When sodium chloride is 6:6-coordinated, there are no such repulsions - and so that is the best way for it to organize itself. Which structure a simple 1:1 compound like NaCl or CsCl crystallizes in depends on the radius ratio of the positive and the negative ions. If the radius of the positive ion is bigger than 73% of that of the negative ion, then 8:8-coordination is possible. Less than that (down to 41%) then you get 6:6-coordination. In CsCl, the cesium ion is about 93% of the size of the chloride ion - so is easily within the range where 8:8-coordination is possible. But with NaCl, the sodium ion is only about 52% of the size of the chloride ion. That puts it in the range where you get 6:6-coordination. This is also typical of ionic solids. The attractions between the solvent molecules and the ions are not big enough to overcome the attractions holding the crystal together. Solid sodium chloride does not conduct electricity, because there are no electrons which are free to move. When it melts, sodium chloride undergoes electrolysis, which involves conduction of electricity because of the movement and discharge of the ions. In the process, sodium and chlorine are produced. This is a chemical change rather than a physical process. The positive sodium ions move towards the negatively charged electrode (the cathode). When they get there, each sodium ion picks up an electron from the electrode to form a sodium atom. These float to the top of the melt as molten sodium metal. (And assuming you are doing this open to the air, this immediately catches fire and burns with an orange flame.) \[ Na^+ + e^- \rightarrow Na\] The movement of electrons from the cathode onto the sodium ions leaves spaces on the cathode. The power source (the battery or whatever) moves electrons along the wire in the external circuit to fill those spaces. That flow of electrons would be seen as an electric current (the external circuit is all the rest of the circuit apart from the molten sodium chloride.) Meanwhile, chloride ions are attracted to the positive electrode (the anode). When they get there, each chloride ion loses an electron to the anode to form an atom. These then pair up to make chlorine molecules. Chlorine gas is produced. Overall, the change is . . . \[ 2Cl^- \rightarrow Cl_2 + 2e^-\] The new electrons deposited on the anode are pumped off around the external circuit by the power source, eventually ending up on the cathode where they will be transferred to sodium ions. Molten sodium chloride conducts electricity because of the movement of the ions in the melt, and the discharge of the ions at the electrodes. Both of these have to happen if you are to get electrons flowing in the external circuit. In solid sodium chloride, of course, that ion movement ca not happen and that stops any possibility of any current flow in the circuit. Jim Clark ( ) | 9,640 | 3,401 |
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This page explains how to write the formula for an organic compound given its name - and vice versa. It covers alkanes, cycloalkanes, alkenes, simple compounds containing halogens, alcohols, aldehydes and ketones. There are two skills you have to develop in this area: The first of these is more important (and also easier!) than the second. In an exam, if you can't write a formula for a given compound, you aren't going to know what the examiner is talking about and could lose lots of marks. However, you might only be asked to write a name for a given formula once in a whole exam - in which case you only risk 1 mark. So, we're going to look mainly at how you decode names and turn them into formulae. In the process you will also pick up tips about how to produce names yourself. In the early stages of an organic chemistry course people frequently get confused and daunted by the names because they try to do too much at once. Don't try to read all these pages in one go. Just go as far as the compounds you are interested in at the moment and ignore the rest. Come back to them as they arise during the natural flow of your course. A modern organic name is simply a code. Each part of the name gives you some useful information about the compound. For example, to understand the name 2-methylpropan-1-ol you need to take the name to pieces. The prop in the middle tells you how many carbon atoms there are in the longest chain (in this case, 3). The an which follows the "prop" tells you that there aren't any carbon-carbon double bonds. The other two parts of the name tell you about interesting things which are happening on the first and second carbon atom in the chain. Any name you are likely to come across can be broken up in this same way. You will need to remember the codes for the number of carbon atoms in a chain up to 6 carbons. There is no easy way around this - you have got to learn them. If you don't do this properly, you won't be able to name anything! Whether or not the compound contains a carbon-carbon double bond is shown by the two letters immediately after the code for the chain length. For example, butane means four carbons in a chain with no double bond. Propene means three carbons in a chain with a double bond between two of the carbons. Compounds like methane, CH , and ethane, CH CH , are members of a family of compounds called alkanes. If you remove a hydrogen atom from one of these you get an alkyl group. For example: These groups must, of course, always be attached to something else. Write the structural formula for 2-methylpentane. Start decoding the name from the bit that counts the number of carbon atoms in the longest chain - pent counts 5 carbons. Are there any carbon-carbon double bonds? No - an tells you there aren't any. Now draw this carbon skeleton: Put a methyl group on the number 2 carbon atom: Does it matter which end you start counting from? No - if you counted from the other end, you would draw the next structure. That's exactly the same as the first one, except that it has been flipped over. Finally, all you have to do is to put in the correct number of hydrogen atoms on each carbon so that each carbon is forming four bonds. Write the structural formula for 2,3-dimethylbutane. Start with the carbon backbone. There are 4 carbons in the longest chain (but) with no carbon-carbon double bonds (an). This time there are two methyl groups (di) on the number 2 and number 3 carbon atoms. Completing the formula by filling in the hydrogen atoms gives: Write the structural formula for 2,2-dimethylbutane. This is exactly like the last example, except that both methyl groups are on the same carbon atom. Notice that the name shows this by using 2,2- as well as di. The structure is worked out as before: Write the structural formula for 3-ethyl-2-methylhexane. hexan shows a 6 carbon chain with no carbon-carbon double bonds. This time there are two different alkyl groups attached - an ethyl group on the number 3 carbon atom and a methyl group on number 2. Filling in the hydrogen atoms gives: How do you know what order to write the different alkyl groups at the beginning of the name? The convention is that you write them in alphabetical order - hence ethyl comes before methyl which in turn comes before propyl. In a cycloalkane the carbon atoms are joined up in a ring - hence cyclo. Write the structural formula for cyclohexane. hexan shows 6 carbons with no carbon-carbon double bonds. cyclo shows that they are in a ring. Drawing the ring and putting in the correct number of hydrogens to satisfy the bonding requirements of the carbons gives: Write the structural formula for propene. prop counts 3 carbon atoms in the longest chain. en tells you that there is a carbon-carbon double bond. That means that the carbon skeleton looks like this: Putting in the hydrogens gives you: Write the structural formula for but-1-ene. "but" counts 4 carbon atoms in the longest chain and en tells you that there is a carbon-carbon double bond. The number in the name tells you where the double bond starts. No number was necessary in the propene example above because the double bond has to start on one of the end carbon atoms. In the case of butene, though, the double bond could either be at the end of the chain or in the middle - and so the name has to code for the its position. The carbon skeleton is: And the full structure is: Incidentally, you might equally well have decided that the right-hand carbon was the number 1 carbon, and drawn the structure as: Write the structural formula for 3-methylhex-2-ene. The longest chain has got 6 carbon atoms (hex) with a double bond starting on the second one (-2-en). But this time there is a methyl group attached to the chain on the number 3 carbon atom, giving you the underlying structure: Adding the hydrogens gives the final structure: Be very careful to count the bonds around each carbon atom when you put the hydrogens in. It would be very easy this time to make the mistake of writing an H after the third carbon - but that would give that carbon a total of 5 bonds. Write the structural formula for 1,1,1-trichloroethane. This is a two carbon chain (eth) with no double bonds (an). There are three chlorine atoms all on the first carbon atom. Write the structural formula for 2-bromo-2-methylpropane. First sort out the carbon skeleton. It's a three carbon chain with no double bonds and a methyl group on the second carbon atom. Draw the bromine atom which is also on the second carbon. And finally put the hydrogen atoms in. Notice that the whole of the hydrocarbon part of the name is written together - as methylpropane - before you start adding anything else on to the name. Write the structural formula for 1-iodo-3-methylpent-2-ene. This time the longest chain has 5 carbons (pent), but has a double bond starting on the number 2 carbon. There is also a methyl group on the number 3 carbon. Now draw the iodine on the number 1 carbon. Giving a final structure: All alcohols contain an -OH group. This is shown in a name by the ending . Write the structural formula for methanol. This is a one carbon chain with no carbon-carbon double bond (obviously!). The ending shows it's an alcohol and so contains an -OH group. Add example text here. Write the structural formula for 2-methylpropan-1-ol. The carbon skeleton is a 3 carbon chain with no carbon-carbon double bonds, but a methyl group on the number 2 carbon. The -OH group is attached to the number 1 carbon. The structure is therefore: Write the structural formula for ethane-1,2-diol. This is a two carbon chain with no double bond. The diol shows 2 -OH groups, one on each carbon atom. All aldehydes contain the group: If you are going to write this in a condensed form, you write it as -CHO - never as -COH, because that looks like an alcohol. The names of aldehydes end in al. Write the structural formula for propanal. This is a 3 carbon chain with no carbon-carbon double bonds. The al ending shows the presence of the -CHO group. The carbon in that group counts as one of the chain. Write the structural formula for 2-methylpentanal. This time there are 5 carbons in the longest chain, including the one in the -CHO group. There aren't any carbon-carbon double bonds. A methyl group is attached to the number 2 carbon. Notice that in aldehydes, the carbon in the -CHO group is always counted as the number 1 carbon. Ketones contain a carbon-oxygen double bond just like aldehydes, but this time it's in the middle of a carbon chain. There isn't a hydrogen atom attached to the group as there is in aldehydes. Ketones are shown by the ending one. This is a 3 carbon chain with no carbon-carbon double bond. The carbon-oxygen double bond has to be in the middle of the chain and so must be on the number 2 carbon. Ketones are often written in this way to emphasize the carbon-oxygen double bond. Write the structural formula for pentan-3-one. This time the position of the carbon-oxygen double bond has to be stated because there is more than one possibility. It's on the third carbon of a 5 carbon chain with no carbon-carbon double bonds. If it was on the second carbon, it would be pentan-2-one. This could equally well be written: | 9,253 | 3,402 |
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Earlier in the course we performed stoichiometric calculations with chemical reactions using quantities of moles and mass (typically in grams). These same principles can be applied to chemical reactions involving gases except that we first have to convert volumes of gases into moles. Hydrogen gas reacts with oxygen gas to produce water vapor via the following balanced chemical equation: 2H + O -> 2H O If the temperature is 320 K and pressure is 1.34 atm, what volume of oxygen is required to produce 65.0 g of water? Strategy: Since we are given the temperature and pressure, to find the volume of oxygen using the ideal gas law we need to first calculate the moles of oxygen. To find the moles of oxygen required, we can first calculate the moles of water in 65.0g. \[n_{water}=\frac{m_{water}}{mm_{water}}=\frac{\text{65.0 g}}{\text{18.0g/mol}}=\text{3.61 mol of water}\] From the balanced chemical equation, we can see that 1 equivalent of oxygen produces 2 equivalents of water. We can therefore write the following ratio: 1 mol O : 2 mol H O We can now solve for the amount of oxygen: \[(\text{3.61 mol } H_{2}O)\times(\frac{ \text{1 mol } O_{2}}{\text{2 mol } H_{2}O })= \text{1.81 mol } O_{2}\] Finally, now that we know how many moles of oxygen are required, we can calculate the volume of the oxygen using the ideal gas law and the temperature&pressure provided in question: \[PV=nRT\] \[V=\frac{nRT}{P}\] \[{V = \rm \frac{1.81 mol\ \cdot 0.08206 \frac{L atm}{mol K}\cdot 320K}{1.34 atm }}\] \[V = 35.5 \rm L\] | 1,544 | 3,403 |
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Michael Faraday discovered in 1833 that there is always a simple relationship between the amount of substance produced or consumed at an electrode during and the quantity of electrical charge which passes through the cell. For example, the half-equation \[\text{Ag}^{+} + e^{-} \rightarrow \text{Ag} \nonumber \] tells us that when 1 mol Ag is plated out as 1 mol Ag, 1 mol must be supplied from the cathode. Since the negative charge on a single electron is known to be 1.6022 × 10 C, we can multiply by the Avogadro constant to obtain the charge per mole of electrons. This quantity is called the , symbol : F = 1.6022 × 10 C × 6.0221 × 10 mol = 9.649 × 10 C mol Thus in the case of Eq. (1), 96 490 C would have to pass through the cathode in order to deposit 1 mol Ag. For any electrolysis the electrical charge passing through an electrode is related to the amount of electrons – by \[\text{F}=\frac{Q}{n_{e^{-}}} \nonumber \] Thus serves as a conversion factor between – and . Calculate the quantity of electrical charge needed to plate 1.386mol Cr from an acidic solution of K Cr O according to half-equation \[\ce{H2Cr2O7}(aq) + \text{12H}^{+}(aq) + \text{12}e^{-} \rightarrow \ce{2Cr}(s) + \ce{7H2O}(l)\label{3} \]. According to Eq. \(\ref{3}\), 12 mol is required to plate 2 mol Cr, giving us a stoichiometric ratio ( /Cr). Then the Faraday constant can be used to find the quantity of charge. In road-map form \(\xrightarrow{S\text{(}e^{-}\text{/Cr)}}\) – \(\xrightarrow{F}\) = 1.386 mol Cr × \(\frac{\text{12 mol }e^{-}}{\text{2 mol Cr}}\) × \(\frac{\text{9}\text{.649 }\times \text{ 10}^{\text{4}}\text{ C}}{\text{ 1 mol }e^{-}}\) = 8.024 × 10 C Often the electrical current rather than the quantity of electrical charge is measured in an electrolysis experiment. Since a is defined as the quantity of charge which passes a fixed point in an electrical circuit when a current of one ampere flows for one second, the charge in coulombs can be calculated by multiplying the measured current (in amperes) by the time (in seconds) during which it flows: \[\text{Q} = \text{It} \nonumber \] In this equation represents current and represents time. If you remember that coulomb = 1 ampere × 1 second 1 C = 1 A s you can adjust the time units to obtain the correct result. Hydrogen peroxide, H O , can be manufactured by electrolysis of cold concentrated sulfuric acid. The reaction at the anode is \[\ce{2H2SO4 -> H2S2O8 + 2H}^{+} + \text{2}e^{-}\label{5} \] When the resultant peroxydisulfuric acid, H S O , is boiled at reduced pressure, it decomposes: \[\ce{2H2O + H2S2O8 -> 2H2SO4 + H2O2}\label{6} \] Calculate the mass of hydrogen peroxide produced if a current of 0.893 flows for 1 h. he product of current and time gives us the quantity of electricity, . Knowing this we easily calculate the amount of electrons, –. From half-equation \(\ref{5}\) we can then find the amount of peroxydisulfuric acid. Equation \(\ref{6}\) then leads to and finally to . The road map to describe this logic is as follows: \[I\xrightarrow{t}Q\xrightarrow{F}n_{e^{-}}\xrightarrow{S_{e}}n_{\text{H}_{\text{2}}\text{S}_{\text{2}}\text{O}_{\text{8}}}\xrightarrow{S}n_{\text{H}_{\text{2}}\text{O}_{\text{2}}}\xrightarrow{M}m_{\text{H}_{\text{2}}\text{O}_{\text{2}}} \nonumber \] so that \[m_{\text{H}_{\text{2}}\text{O}_{\text{2}}}=\text{0}\text{.893 A }\times \text{ 3600 s }\times \text{ }\frac{\text{1 mol }e^{-}}{\text{96 490 C}}\text{ }\times \text{ }\frac{\text{1 mol H}_{\text{2}}\text{S}_{\text{2}}\text{O}_{\text{8}}}{\text{2 mol }e^{-}} \nonumber \] \[=\frac{\text{1 mol H}_{\text{2}}\text{O}_{\text{2}}}{\text{1 mol H}_{\text{2}}\text{S}_{\text{2}}\text{O}_{\text{8}}}\text{ }\times \text{ }\frac{\text{34}\text{.01 g H}_{\text{2}}\text{O}_{\text{2}}}{\text{1 mol H}_{\text{2}}\text{O}_{\text{2}}} \nonumber \] = 05666 \(\frac{\text{A s}}{\text{C}}\) × g H O = 0.5666 g H O | 3,932 | 3,404 |
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Our inability to locate an exactly may seem rather strange, but it arises whether we think in terms of or of particles. Suppose an experiment is to be done to locate a billiard ball moving across a pool table whose surface is hidden under a black cloth. One way to do this would be to try to bounce a second billiard ball off the first one (Figure \(\Page {1}\). When a hit was made and the second ball emerged from under the cloth, we would have a pretty good idea of where the first ball was. The only trouble with the experiment is that the position and speed of the first ball would almost certainly be changed by the collision. To lessen this effect, a table-tennis ball could be substituted for the second billiard ball—its smaller mass would produce a much smaller change in the motion of the first ball. Clearly, the lighter and more delicate the “probe” we use to try to locate the first ball, the less our measurement will affect it. The best way to locate the first billiard ball and determine its speed would be to remove the cover from the table so it could be seen. In this case, however, something is still “bouncing” off the first billiard ball (Figure 1 ). If we are to see the ball, particles of visible light, or , must strike the ball and be reflected to our eyes. Since each photon is very small and has very little energy by comparison with that needed to change the motion of the billiard ball, looking at a ball is an excellent means of observing it without changing its position or speed. But to observe an electron is quite another story, since the mass of an electron is far smaller than that of a billiard ball. Anything (such as a photon of light) which can be bounced off an electron in such a way as to locate it precisely would have far more energy than would be required to change the path of the electron. Hence it would be impossible to predict the electron’s future speed or position from the experiment. The idea that it is impossible to determine accurately the location and the speed of any particle as small as an electron is called the . It was first proposed in 1927 by Werner Heisenberg (1901 to 1976). According to the uncertainty principle, even if we draw an analogy between and a billiard ball ( ), it will be impossible to determine both the electron’s exact position in the box and its exact speed. Since depends on speed (½ ) and assigns exact values of kinetic energy to the electron in the box, the speed can be calculated accurately. This means that determination of the electron’s position will be very inexact. The Heisenberg Uncertainty principle is stated mathematically as \[\Delta X \Delta P \ge \frac{\hbar}{2} \nonumber \] where \(ΔX\) is the uncertainty in the position, ΔP is the uncertainty in the momentum, and \(\hbar\) is Planck's Constant divided by \(2π\). It will be possible to talk about the that the electron is at a specific location, but there will also be some probability of finding it somewhere else in the box. Since it is impossible to know precisely where the electron is at a given instant, the question, “How does it get from one place to another?” is pointless. There is a finite probability that it was at the other place to begin with! It is possible to be quantitative about the of finding a “billiard-ball” electron at a given location, however. Shortly after the uncertainty principle was proposed, the German physicist Max Born (1882 to 1969) suggested that the intensity of the electron wave at any position in the box was proportional to the probability of finding the electron (as a particle) at that same position. Thus if we can determine the shapes of the waves to be associated with an electron, we can also determine the relative probability of its being located at one point as opposed to another. The wave and particle models for the electron are thus connected to and reinforce each other. Niels Bohr suggested the term to describe their relationship. It does no good to ask, “Is the electron a wave or a particle?” Both are ways of drawing an analogy between the microscopic world and macroscopic things whose behavior we understand. Both are useful in our thinking, and they are complementary rather than mutually exclusive. A graphic way of indicating the probability of finding the electron at a particular location is by the density of shading or stippling along the length of the box. This has been done in Figure 2 for the same three electron waves previously illustrated in . Notice that the density of dots is large wherever the electron wave is large.(This would correspond in the guitar-string analogy to places where the string was vibrating quite far from its rest position.) Where the electron wave is small (near the ends of the box in all three cases and at the nodes indicated in the figure), there are only a small number of dots. (A is a place where the intensity of the wave is zero, that is, in the guitar-string analogy, where the string has not moved from its rest position.) If the electron is thought of as a wave occupying all parts of the box at once, we can speak of an which has greater or lesser density in various parts of the box. There will be a greater quantity of negative charge in a region of high density (a region where there is a greater concentration of dots) than in one of low density. In an atom or molecule, according to the uncertainty principle, the best we can do is indicate in various regions—we cannot locate the precise position of the electron. Therefore electron dot-density diagrams, such as the ones shown in Figure 2, give a realistic and useful picture of the behavior of electrons in atoms. In such a diagram the . We will encounter electron dot-density diagrams quite often throughout this book. These have all been generated by a computer from accurate mathematical descriptions of the atom or molecule under discussion. | 5,923 | 3,405 |
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Both enthalpy and the internal energy are often described as . This means that they depend only on the of the system, i.e., on its pressure, temperature, composition, and amount of substance, but not on its previous history. Thus any solution of NaCl at 25°C and 1 bar (100 kPa) which contains a mixture of 1 mol NaCland 50 mol H O has the same internal energy and the same enthalpy as any other solution with the same specifications. It does not matter whether the solution was prepared by simply dissolving NaCl( ) in H O, by reacting NaOH( ) with HCl( ), or by some more exotic method. The fact that the internal energy and the enthalpy are both state functions has an important corollary. It means that when a system undergoes any change whatever, then the alteration in its enthalpy (or its internal energy) depends only on the initial state of the system and its final state. The initial value of the enthalpy will be , and the final value will be . No matter what pathway we employ to get to state 2, we will always end up with the value for the enthalpy. The enthalpy change Δ = – will thus be independent of the path used to travel from state 1 to state 2. This corollary is of course the basis of . The change in enthalpy for a given chemical process is the same whether we produce that change in one or in several steps. | 1,354 | 3,406 |
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refers to the increase in the average temperature of the Earth’s atmosphere due to elevated greenhouse gas concentrations, heightening the greenhouse effect. includes both global warming driven by human-induced emissions of greenhouse gases and the resulting large-scale shifts in weather patterns. Though there have been previous periods of climatic change, since the mid-20th century humans have had an unprecedented impact on Earth's climate system and caused change on a global scale. The largest driver of warming is the emission of gases that create a greenhouse effect, of which more than 90% are carbon dioxide (CO
2) and methane. Fossil fuel burning (coal, oil, and natural gas) for energy consumption is the main source of these emissions, with additional contributions from agriculture, deforestation, and manufacturing. The human cause of climate change is not disputed by any scientific body of national or international standing. Temperature rise is accelerated or tempered by climate feedbacks, such as loss of sunlight-reflecting snow and ice cover, increased water vapour (a greenhouse gas itself), and changes to land and ocean carbon sinks. The is the process by which radiation from a planet's atmosphere warms the planet's surface to a temperature above what it would be without this atmosphere (Figure \(\Page {2}\)). Radiatively active gases (i.e. greenhouse gases) in a planet's atmosphere radiate energy in all directions. Part of this radiation is directed towards the surface, thus warming it. The intensity of downward radiation – that is, the strength of the greenhouse effect – depends on the amount of greenhouse gases that the atmosphere contains. The temperature rises until the intensity of upward radiation from the surface, thus cooling it, balances the downward flow of energy. Earth's natural greenhouse effect is critical to supporting life and initially was a precursor to life moving out of the ocean onto land. Human activities, mainly the burning of fossil fuels and clear cutting of forests, have increased the greenhouse effect and caused global warming. The term is a slight misnomer in the sense that physical greenhouses warm via a different mechanism. The greenhouse effect as an atmospheric mechanism functions through radiative heat loss while a traditional greenhouse as a built structure blocks convective heat loss. The result, however, is an increase in temperature in both cases. The greenhouse effect of solar radiation on the Earth's surface caused by emission of greenhouse gases. A (sometimes abbreviated ) is a gas that absorbs and emits radiant energy within the thermal infrared range, causing the greenhouse effect. It’s important to realize that (H O) is also a greenhouse gas. While humans have little direct impact on water vapor concentrations in the atmosphere, it is still an essential component of the natural greenhouse effect that occurs in our atmosphere. The four major categories of greenhouse gases that have been impacted by humans the most will be discussed in detail below. See Table \(\Page {1}\) for a numeric comparison of these greenhouse gases. • Carbon dioxide, CO • Methane, CH • Nitrous oxide, N O • Synthetic , including (HFCs), (PFCs), and (SF ) Carbon dioxide (CO ) is the greenhouse gas responsible for most of the human-caused climate change in our atmosphere. It has the highest concentration in the atmosphere of any of the greenhouse gases that we’ll discuss here. Remember that CO is a direct product of both combustion and cellular respiration, causing it to be produced in great quantities both naturally and anthropogenically. Any time biomass or fossil fuels are burned, CO is released. Major anthropogenic sources include: electricity production from coal-fired and natural gas power plants, transportation, and industry (Chapter 4). To get an idea of how CO concentration has changed over time, watch this video compiled by the National Oceanic and Atmospheric Administration (NOAA): . This video contains atmospheric CO concentrations measured directly, dating back to 1958, as well as atmospheric CO concentrations measured indirectly from ice core data, dating back to 800,000 BCE. By 1990, a quantity of over seven billion tons of carbon (equivalent to 26 billion tons of carbon dioxide when the weight of the oxygen atoms are also considered) was being emitted into the atmosphere every year, much of it from industrialized nations. Similar to the action of the naturally existing greenhouse gases, any additional greenhouse gases leads to an increase in the surface temperature of the Earth. While CO is produced by aerobic cellular respiration, gases such as CH and N O are often the products of anaerobic metabolisms. Agriculture is a major contributor to CH emissions. In addition to anaerobic bacteria, methane is also a significant component of natural gas, and is commonly emitted through the mining and use of natural gas and petroleum, in addition to coal mining. Finally, landfills contribute significantly to CH emissions, as the waste put into the landfill largely undergoes anaerobic decomposition as it is buried under many layers of trash and soil. Natural sources of CH include swamps and wetlands, and volcanoes. The vast majority of N O production by humans comes from agricultural land management. While some N O is naturally emitted to the atmosphere from soil as part of the nitrogen cycle, human changes in land management, largely due to agricultural practices, have greatly increased N O emissions. Some N O is also emitted from transportation and industry. Due to their relatively high concentrations in the atmosphere compared to synthetic gases, CO , CH , and N O, are responsible for most of the human-caused global climate change over the past century. Figure \(\Page {2}\) shows the increases in all three gases since 1750. Ice core data shows us that the atmospheric CO concentration never exceeded 300 ppm before the industrial revolution. As of early 2015, the current atmospheric CO concentration is 400 ppm. One class of greenhouse gas chemicals that has no natural sources is the fluorinated gases. These include HFCs, PFCs, and SF6, among others. Because these are synthetic chemicals that are only created by humans, these gases were essentially non-existent before the industrial revolution. These synthetic gases are used for a wide variety of applications, from refrigerants to semiconductor manufacturing, and propellants to fire retardants. They tend to have a long lifetime in the atmosphere, as seen in Table \(\Page {1}\). Some of these chemicals, as well as the older (CFCs), have been phased out by international environmental legislation under the Montreal Protocol. Due to their long lifespan, many of these now banned CFCs remain in the atmosphere. Newer chemical replacements, such as HFCs, provide many of the same industrial applications, but unfortunately have their own environmental consequences. Just as greenhouse gases differ in their sources and their residence time in the atmosphere, they also differ in their ability to produce the greenhouse effect. This is measured by the , or GWP, of each greenhouse gas. The GWP of a greenhouse gas is based on its ability to absorb and scatter energy, as well as its lifetime in the atmosphere. Since CO is the most prevalent greenhouse gas, all other greenhouse gases are measured relative to it. As the reference point, CO always has a GWP of 1. Note the very high GWP values of the synthetic fluorinated gases in Table \(\Page {1}\). This is largely due to their very long residence time in the atmosphere. Also note the higher GWP values for CH and N O compared to CO . In addition to greenhouse gases, other manmade changes may be forcing climate change. Increases in near-surface ozone from internal combustion engines, aerosols such as carbon black, mineral dust and aviation-induced exhaust are acting to raise the surface temperature. This primarily occurs due to a decrease in the of light-colored surfaces by the darker-colored carbon black, soot, dust, or particulate matter. As you know, it is more comfortable to wear a white shirt on a hot summer day than a black shirt. Why is this? Because the lighter-colored material bounces more solar radiation back toward space than the darker-colored material does, allowing it to stay cooler. The darker-colored material absorbs more solar radiation, increasing its temperature. Just as the white shirt has a higher albedo than the black shirt, light-colored objects in nature (such as snow) have a higher albedo than dark-colored objects (such as soot or dust). As humans increase the amount of carbon black, soot, dust, and particulates in the atmosphere, we decrease the albedo of light-colored surfaces, causing them to absorb more solar radiation and become warmer than they would without human influence. An example of this can be seen in the snow on Figure \(\Page {3}\). We will only discuss some of the consequences of climate change in this section, including changes in temperature, precipitation, ocean level, and ocean acidity. There are many more changes that have been seen, and are projected to continue in the future. These include: changes in the amount and distribution of ice and snow, changes in seasonality, ecosystem shift, and habitat changes of plant and animal populations, in addition to others. For more information about these consequence of climate change, visit this site: . Temperature and are the two most direct impacts on the Earth’s climate due to climate change. By now, you should already understand why an increase in greenhouse gas levels in the atmosphere causes an increase in temperature. But why does it also impact precipitation patterns? As you already know, water vapor is an important component of the Earth’s atmosphere . As the air in the troposphere warms and cools, the amount of water vapor that it holds changes dramatically. Here in Georgia, we have very hot and humid summers. The high summer humidity in this region is possible due to the increased capability warm air has to hold water vapor. Simply put, warmer air can hold more water than cooler air. As air cools, its ability to hold water vapor decreases, and any excess water will leave the air as liquid water. A great example of this is the formation of dew on surfaces overnight. During the day, the temperature is warmer than it is at night, and the air has a relatively high holding capacity for water vapor. When the sun sets, the air cools, decreasing its capacity to hold water vapor. That extra water must go somewhere, and it does that by accumulating on surfaces. Similarly, when warm and cool air fronts collide, the chances for rain and thunderstorms increase. Furthermore, an increase in temperature enhances evaporation occurring at the Earth’s surface. This increased evaporation leads to greater concentrations of water vapor in the atmosphere which can lead to increased precipitation. The change in temperature that we have already seen in the Earth’s average atmospheric temperature is relatively small (about1.2 °C, according to Figure 7.5.1). However, as with many of the aspects of climate change, the potential for greater changes increases dramatically as time progresses in the future. This can be seen in Figure \(\Page {4}\), which displays a model of the predicted temperature increase. Notice that these changes occur relatively rapidly, and are not uniform across the globe. What might be some of the reasons for this? Changes in precipitation occur due to a variety of factors, including changes in atmospheric water vapor content due to changing temperature, as discussed above. Also at play is the heightened rate of water on Earth’s surface under warmer temperatures. More evaporation leads to more precipitation. Finally, shifts in wind patterns impact the distribution of precipitation events. As you can see in Figure \(\Page {5}\), there are some areas of the globe that are expected to have an increase in precipitation, while others are expected to have a dramatic decrease. Some major population centers projected to have a moderate to severe precipitation increase include (population estimates of the metropolitan area given in parentheses): New York, United States (20.1 million); Bogotá, Colombia (12.1 m.); and Manila, Philippines (11.9 m.). What sort of challenges might these cities face in the future as they deal with this change in their climate? In contrast, many more major metropolitan areas are projected to have a moderate to severe precipitation decrease (droughts) by the end of the 21st century. These include Delhi, India (21.8 m.); Lagos, Nigeria (21 m.); São Paulo, Brazil (20.9 m.); Kolkata, India (14.6 m.); Istanbul, Turkey (14.4 m.); Los Angeles, United States (13.3 m.); Rio de Janeiro, Brazil (12 m.); Paris, France (12 m.); and Lahore, Pakistan (11.3 m.). The largest challenge that these areas are likely to face is a dwindling water supply for drinking and agriculture. See Chapter 8 for more detail on challenges faced by societies to supply clean, reliable water to their populations and farms. Additional challenges may be felt by all areas of the world with regard to changes in the seasonality or timing of precipitation, as well the form in which precipitation falls (e.g., mist or downpour; rain, ice, or snow). All of these factors affect the availability of soil water for plants, the flow of rivers and streams, and the overall accessibility of water worldwide. Furthermore, scientists predict an increase in the number and severity of storms as climate change progresses. For a full discussion of the potential impacts of this, see the assigned article. Sea level rise While we know that water continuously cycles around the world, and that the overall quantity of water on Earth will not change due to global climate change, the distribution of this water is changing. In particular, oceans are increasing in volume while land ice stores (such as ) are decreasing. This contributes to an increase in sea level worldwide (Figure \(\Page {6}\)). From the data in Figure \(\Page {6}\), we see that sea level has increased at an average of 0.06 inches (0.15 cm) per year over the time period shown above. Most of this rise, however, has occurred within the most recent decades. The rate of increase has gone up to between 0.11 to 0.14 inches (0.28 to 0.36 cm) per year since 1993. There are two forces causing sea level to rise, both caused by climate change. First, the increased global temperature has caused increased ice melting in many regions of the globe. Melting (such as the glacier shown in Figure \(\Page {7}\)) contributes to sea level rise because water that used to be stored in ice sitting on top of land becomes running water which reaches the ocean through We also observe melting (see ence/indicators/index.html for data and figures). Sea ice, such as the ice that covers the arctic regions of the Northern Hemisphere, has no land underneath it. When it melts, the water stays in the same locations, and the overall sea level does not change. The second factor that influences sea level rise is a phenomenon called thermal expansion. Due to the physical properties of water, as water warms, its density decreases. A less dense substance will have fewer molecules in a given area than a more dense substance (see Chapter 1 supplemental material). This means that as the overall temperature of the oceans increases due to global climate change, the same amount of water molecules will now occupy a slightly larger volume. This may not seem significant, but considering the 1.3 billion trillion liters (264 billion gallons) of water in the ocean, even a small change in density can have large effects on sea level as a whole. Scientists have already documented sea level rise in some areas of the world, including one familiar to most of us: the Southeastern United States. Figure \(\Page {8}\) depicts the measured land area lost due to increasing sea level since 1996. Note that the Southeast (defined here as the Atlantic coast of North Carolina south to Florida) is particularly susceptible to land area loss due to the gently sloping nature of our coastline. Moving northward into the Mid-Atlantic States (defined here as Virginia north to Long Island, New York), coastal habitats tend to have a steeper geography, which protects against some losses. While the ecological effects of sea level rise remain in the United States, we don’t project any catastrophic loss of life, property, or livelihood for some time. This is, in part, due to large investments that we have made in infrastructure to protect our cities and farmlands. This is not the case in many areas of the world. For a discussion of the impacts of sea level rise on less industrialized nations of Bangladesh, Maldives, Kiribati, and Fiji, review the required article reading. Dissolved CO is essential for many organisms, including shell-building animals and other organisms that form a hard coating on their exterior (e.g., shellfish, corals, Haptophyte algae). This hard coating is built out of aragonite, a mineral form of the molecule calcium carbonate, CaCO . These organisms rely on the formation of carbonate ions (see Chapter 1 supplemental material for information on ions), CO , from dissolved CO , through a natural, chemical reaction that occurs. This takes place through a chain-reaction equation, where bicarbonate (HCO ) is formed as an intermediate, and hydrogen ions (H ) are generated (equations \(\Page {1}\) and \(\Page {2}\)). \[CO_{2} + H_{2}O \leftrightarrow H^{+} + HCO_{3}^{-} \nonumber \] \[HCO_{3}^{-} \leftrightarrow H^{+} + CO_{3}^{2-} \nonumber \] To have a better visualization of this process, follow along with the interactive graphic at: . As you can see, both equations \(\Page {1}\) and \(\Page {2}\) each produce one H . This is significant to water chemistry because an increase in H+ concentration means a decrease in the pH of the water. You can see in Figure \(\Page {9}\) that a lower pH means that the liquid is more acidic. As shown in the interactive graphic, an increase in CO in the atmosphere causes additional CO to be dissolved in the ocean. This means that more CO in the atmosphere leads to more acidic ocean environments. Unfortunately for shell-building animals, the buildup of H in the more acidic ocean environment blocks the absorption of calcium and CO , and makes the formation of aragonite more difficult. An aragonite deficit is already being documented in many of the world’s oceans, as shown in Figure \(\Page {10}\). The increasing acidity of the world’s oceans is resulting in habitat changes across the globe. This is only expected to worsen as atmospheric CO levels continue to increase. Many organisms, including the corals that are the foundation species of the beautiful coral reefs, are very sensitive to changes in ocean pH. Scientists have documented cases of ecosystem destruction through coral bleaching, caused by the effects of climate change including ocean acidification and increased temperature. For more information, visit the NOAA Coral Reef Conservation Program website: coralreef.noaa.gov/threats/climate/. While the situation surrounding global climate change is in serious need of our attention, it is important to realize that many scientists, leaders, and concerned citizens are making solutions to climate change part of their life’s work. The two solutions to the problems caused by climate change are and , and we will likely need a combination of both in order to prosper in the future. We know that climate change is already occurring, as we can see and feel the effects of it. For this reason, it is essential to also adapt to our changing environment. This means that we must change our behaviors in response to the changing environment around us. Some adaptation strategies are discussed in the required article reading. Adaption strategies will vary greatly by region, depending on the largest specific impacts in that area. For example, in the city of Delhi, India, a dramatic decrease in rainfall is projected over the next century (Figure 7.6.2). This city will likely need to implement policies and practices relating to conservation of water, for example: rainwater harvesting, water re-use, and increased irrigation efficiency. Rain-limited cities near oceans, such as Los Angeles, California may choose to use desalination to provide drinking water to their citizens. involves taking the salt out of seawater to make it potable (Chapter 14). Cities with low elevations near oceans may need to implement adaptation strategies to rising sea levels, from seawalls and levees to relocation of citizens. One adaptations strategy gaining use is the creation or conservation of , which provide natural protection against storm surges and flooding. In general, a strategy to mitigate climate change is one that reduces the amount of greenhouse gases in the atmosphere or prevents additional emissions. Mitigations strategies attempt to “fix” the problems caused by climate change. Governmental regulations regarding fuel efficiency of vehicles is one example of an institutionalized mitigation strategy already in place in the United States and in many other countries around the world. Unlike some other countries, there are no or charges on burning fossil fuels in the United States. This is another governmental mitigation strategy that has been shown to be effective in many countries including India, Japan, France, Costa Rica, Canada, and the United Kingdom. In addition to government measures and incentives, technology can also be harnessed to mitigate climate change. One strategy for this is the use of (CCS). Through CCS, 80-90% of the CO that would have been emitted to the atmosphere from sources such as a coal-fired power plant is instead captured and then stored deep beneath the Earth’s surface. The CO is often injected and sequestered hundreds of miles underground into porous rock formations sealed below an impermeable layer, where it is stored permanently (Figure \(\Page {11}\)). Scientists are also looking into the use of soils and vegetation for carbon storage potential. Proper management of soil and forest ecosystems has been shown to create additional carbon sinks for atmospheric carbon, reducing the overall atmospheric CO burden. Increasing soil carbon further benefits communities by providing better-quality soil for agriculture and cultivation. Technologies related to alternative energy sources (Chapter 15) mitigate climate change by providing people with energy not derived from the combustion of fossil fuels. Finally, simple activities such as energy conservation, choosing to walk or bike instead of driving, and disposing of waste properly are activities that, when done by large numbers of people, actively mitigate climate change by preventing carbon emissions. Take a moment to identify ways that you personally can be involved in the mitigation of or adaptation to climate change. What changes can you make in your own life to prevent excess carbon emissions? Similar to your ecological footprint, which you should have already calculated in lab, you can also calculate your carbon footprint. Use the EPA’s carbon footprint calculator to do so, and investigate the Reduce Your Emissions section to find ways to decrease your carbon footprint. Figure \(\Page {12}\) list various technologies and approaches that companies and individuals can adapt to reduce greenhouse gases. Technologies related to alternative energy sources mitigate climate change by providing people with energy not derived from the combustion of fossil fuels. Finally, simple activities such as energy conservation, choosing to walk or bike instead of driving, and disposing of waste properly are activities that, when done by large numbers of people, actively mitigate climate change by preventing carbon emissions. ( ) | 24,091 | 3,407 |
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In there is almost always a difference in energy, and hence in enthalpy, between the reactants and the products. The for dissociation of N O for example, is This can happen if some N O in the mixture dissociates, since the resulting absorption of energy will produce a cooling effect. We would therefore expect that by raising the temperature of the equilibrium mixture, we would shift the equilibrium in favor of dissociation. Indeed that raising the temperature from 200 to 600 K changes an equilibrium mixture which is almost pure N O into an equilibrium mixture which is almost pure NO In the general case, if we raise the temperature of any mixture of species which are in chemical equilibrium with each other, Le Chatelier's principle tells us that we will shift the equilibrium in the direction of those species with the higher energy. Thus, if the reaction is , as in the dissociation just discussed, raising the temperature will swing the equilibrium toward the , and the value of the equilibrium constant will increase with temperature. Conversely, if the reaction is , a rise in temperature will favor the , and will get smaller as the temperature increases. We can also turn the argument around. If we find a reaction for which increases with temperature, we know immediately that the reaction must be endothermic. Conversely, if decreases as temperature increases, the reaction must be exothermic. | 1,443 | 3,408 |
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Although all the elements in this group form compounds in which their oxidation state of +5 equals the group number, their other properties vary considerably. Nitrogen is clearly nonmetallic and consists of diatomic triply-bonded N molecules. Phosphorus, also a nonmetal, exists as tetrahedral P molecules (Figure ) in the vapor and the white allotropic form of the solid. On standing, white phosphorus slowly changes to the red allotrope, whose structure is shown in Figure . The most stable form of the element is black phosphorus, which has a layer structure (Figure ). Black phosphorus can be made by heating the white form with a mercury catalyst for 8 days at 220 to 370°C. Arsenic is a semimetal and consists of As molecules in the gas phase. When As ( ) is condensed to a solid, three allotropes may form. The most stable of these is metallic arsenic, in which each arsenic atom has three nearest neighbors, with three more arsenic atoms somewhat farther away. Antimony, also a semimetal, has two allotropes, the more stable one being metallic, like arsenic. In the case of bismuth, only the metallic form occurs. Note that for all the group VA elements the 8 – rule is followed. The number of bonds or nearest neighbors for each atom is 8 minus the group number in , P , and even in the metallic forms of As, Sb, and Bi. The table summarizes the atomic properties of the group VA elements. Overall, the trends are what we would expect, based on our experience with previous groups. These elements exhibit a much wider variety of oxidation states, however, especially in the case of nitrogen. This element forms compounds in which it has every possible oxidation number from –3 (the group number minus 8) to +5 (the group number). As in previous groups, the oxidation state in which the pair of electrons is not used for bonding becomes more prominent toward the bottom of the periodic table. There are a few compounds, Bi(NO ) , for example, in which discrete Bi ions are present. Electron Configuration Usual Oxidation State Radius/pm Ionic (Charge) (3-)171 - - - Density/ g cm Electro- negativity Melting Point (in °C) -210 The most important compounds of the group VA elements are those of nitrogen and phosphorus. Both elements are essential to all living organisms, and both are progressively removed from soil when plants are cultivated and crops harvested. According to Liebig’s law of the minimum, an insufficient supply of either element can limit plant growth and reduce crop yields, and so these elements are important components of fertilizer. More recently both elements have been implicated in several kinds of pollution problems. As we discuss the properties of nitrogen and phosphorus compounds, their effects on food production and environmental degradation will also be discussed. The importance of nitrogen fertilizer was first recognized over a century ago. By the late 1800s the only major ore of nitrogen, Chile saltpeter, NaNO , was being mined in Chile and shipped to Europe for application to agricultural land, but the supply was obviously limited. Most nitrogen at the earth’s surface is in the form of N ( ), which makes up 78 percent of the atmosphere by volume (or by amount of substance). Therefore chemists began to look for ways of obtaining nitrogen compounds directly from the atmosphere. Any process which does this is called . Nitrogen fixation can occur naturally when an electrical discharge (lightning) heats air to a high temperature. The following reaction occurs: \[\text{N}_2(g) + \text{O}_2(g) \rightarrow \text{2NO}(g) \label{1} \] The nitrogen monoxide (nitric oxide) formed can react further at ordinary temperatures, producing the brown gas, nitrogen dioxide: \[\text{2NO}(g) + \text{O}_2(g) \rightarrow \text{2NO}_2(g)\label{2} \] The Lewis diagrams for these and other important nitrogen compounds are shown in Figure . From the figure you can see that both NO and NO have an odd number of electrons and violate the octet rule. In such a case it is common for two molecules to combine (dimerize) by pairing their odd electrons. In the case of NO , dimerization occurs below room temperature, producing colorless dinitrogen tetroxide: \[\text{2NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) \nonumber \] At room temperature, however, the NO and N O are in equilibrium, as evidenced by the brown color of the mixture. NO dimerizes only at very low temperatures in the solid state. The first industrial nitrogen fixation was done by mimicking nature. Reaction \(\ref{1}\) was carried out in a plant near Niagara Falls, where hydroelectric generation provided inexpensive power to support an electric arc. NO was further oxidized to NO which was dissolved in H O to convert it to nitric acid, HNO : \[\text{3NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow \text{2H}^{+}(aq) + \text{2NO}_3^{-}(aq) + \text{NO}(g)\label{4} \] Note that NO is not the acid anhydride of HNO . This reaction involves, dis-proportionation of NO (which contains N in the +4 oxidation state) to form HNO in the + 5 state) and NO (N in the + 2 state). The NO can be recycled by reoxidizing it to NO , and so it was not wasted. The HNO produced in Equation (12.5) was neutralized with NaOH to make a substitute for Chile saltpeter: \[\text{NaOH}(aq) + \text{HNO}_3(aq) \rightarrow \text{NaNO}_3(aq) + \text{H}_2\text{O}(l) \nonumber \] Fixation of nitrogen by the electric-arc process used a great deal of energy and was rather expensive. Other methods were designed to replace it, and the most successful of these is the , which is the major one used today. Nitrogen is reacted with hydrogen at a high temperature and extremely high pressure over a catalyst consisting of iron and aluminum oxide: \[\text{N}_2(g) + \text{3H}_2(g) \xrightleftharpoons[\text{1000atm}]{\text{450}{}^\circ \text{C}} \text{2NH}_3(g) \nonumber \] The ammonia produced by the Haber process is used directly as a fertilizer. It can be liquefied under pressure and injected through special nozzles about a foot under the soil surface. This prevents loss of gaseous ammonia which would otherwise irritate the nose, throat, and lungs of anyone near a fertilized field. You are probably familiar with the odor of ammonia since it is the most common weak base encountered in the chemical laboratory. Prior to the recent development of underground injection techniques, most ammonia was converted to ammonium nitrate for fertilizer use: \[\text{NH}_3(aq) + \text{HNO}_3(aq) \rightarrow \text{NH}_4\text{NO}_3(aq)\label{7} \] Except for ammonia, ammonium nitrate contains a greater mass fraction of nitrogen than any other compound of comparable cost. Ammonium nitrate manufacture requires that half the ammonia produced in the Haber process be converted to nitric acid. The first step is oxidation of ammonia over a catalyst of platinum metal: \[\text{4NH}_3(g) + \text{5O}_2(g) \xrightarrow[\text{800}{}^\circ \text{C}]{\text{Pt}} \text{4NO}(g) + {6H}_2\text{O}(g) \nonumber \] This is called the Ostwald process. It is followed by Eqs. \(\ref{2}\) and \(\ref{4}\), yielding nitric acid, which can be combined with ammonia (Equation \(\ref{7}\)). Nitric acid and nitrates have commercial applications other than fertilizer production. Because NO is a strong oxidizing agent, it reacts vigorously with substances whose elements are in low oxidation states. One example of this is black powder, which consists of charcoal (carbon), sulfur, an potassium nitrate, KNO (saltpeter or nitre). During the American revolution, for example, both armies had numerous persons whose job was to find caves in which the relatively soluble KNO had been deposited as water evaporated. A second example is nitroglycerin which contains carbon and hydrogen in low oxidation states as well as nitrate. Still another example of an explosive nitrate is NH NO , which contains nitrogen in its maximum and minimum oxidation state. NH NO decomposes as follows: \[\text{NH}_4\text{NO}_3(s) \xrightarrow[\text{shock}]{\text{heat or}} \text{N}_2\text{O}(g) + \text{2H}_2\text{O}(g) \nonumber \] ΔH = –37 kJ mol The reaction is exothermic and produces 3 mol of gaseous products for every mole of solid reactant. This causes a tremendous increase in pressure, and, if the reaction is rapid, an explosion. The compound dinitrogen monoxide (nitrous oxide or laughing gas), produced by decomposition of NH NO , is a third important oxide of nitrogen (in addition to NO and NO ). N O is produced during microbial decomposition of organic matter containing nitrogen. Because it is quite unreactive, it is the second most-concentrated nitrogen-containing substance in the atmosphere (after N ). It is used commercially as an anesthetic, is mildly intoxicating, and is poisonous in large doses. The other two important oxides of nitrogen, NO and NO , play a major role in an air-pollution problem known as (or Los Angeles smog). NO is formed by Equation \(\ref{1}\) in automobile engines and other high-temperature combustion processes. At normal temperatures NO is oxidized to NO (Equation \(\ref{2}\)). Both these oxides are free radicals and are rather reactive. Moreover, brown-colored NO absorbs sunlight, and the energy of the absorbed photons breaks a nitrogen-oxygen bond:
The oxygen atoms produced are highly reactive. They combine with hydrocarbon molecules (from evaporated or unburned gasoline) to form aldehydes, ketones, and a number of other compounds which form an almost fog-like cloud and irritate the eyes, throat, and lungs. Photochemical smog is especially bad in cities like Los Angeles and Denver which have lots of sunshine and automobile traffic, but its effects have been observed in every large city in the United States. As in the case of carbon and silicon, there are major differences between the chemistries of nitrogen and phosphorus. The concentrations of phosphorus compounds in the earth’s atmosphere are so small as to be negligible, but phosphorus is more abundant than nitrogen in the solid crust. Here it is found as phosphate rock, which is mainly hydroxyapatite, Ca (PO ) (OH) , or fluorapatite, Ca (PO ) F . (These are the same substances involved in the discussion of dental decay in the section on group IVA elements.) Phosphate rock is quite insoluble, and hence its phosphate ions cannot be assimilated by plants. Production of phosphate fertilizer requires treatment of apatite acid. This protonates the PO ions, converting them to H PO , whose calcium salt is much more soluble: \[\text{Ca}_{10}(\text{PO}_4)_6(\text{OH})_2 + \text{7H}_2\text{SO}_4 + \text{H}_2\text{O} \rightarrow \text{3Ca(H}_2\text{PO}_4)_2•\text{H}_2\text{O} + \text{7CaSO}_4 \nonumber \] \[\text{Ca}_{10}(\text{PO}_4)_6(\text{OH})_2 + \text{14H}_3\text{PO}_4 \rightarrow \text{10Ca(H}_2\text{PO}_4)_2 + \text{2H}_2\text{O} \nonumber \] The compound Ca(H PO )•H O is known as superphosphate, and Ca(H PO ) is called triple superphosphate. The phosphoric acid, H PO , used to make triple superphosphate is also obtained from phosphate rock. The first step is a reduction with carbon (coke) and silicon dioxide in an electric furnace: \[\text{2Ca}_{10}(\text{PO}_4)_6\text{(OH)}_2 + \text{18SiO}_2 + \text{30C} \rightarrow \text{3P}_4 + \text{30CO} + \text{2Ca(OH)}_2 + \text{18CaSiO}_3 \nonumber \] The phosphorus obtained this way is then oxidized to phosphorus pentoxide: \[\text{P}_4(s) + \text{5O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s) \nonumber \] (The name phosphorus pentoxide for P O comes from the empirical formula P O of this compound.) Phosphorus pentoxide is the acid anhydride of phosphoric acid:. \[\text{P}_4\text{O}_6(s) + \text{6H}_2\text{O}(l) \rightarrow \text{4H}_3\text{PO}_3(aq) \nonumber \] Although not a very strong acid, phosphoric acid is triprotic. Therefore, 1 mol of this acid can transfer 3 mol of protons to a strong base. There is another oxide of phosphorus, P O , which involves the + 3 oxidation state, corresponding to use of the 3p , but not the 3s , electrons for bonding. P O is the acid anhydride of phosphorous acid, H PO : \[\text{P}_4\text{O}_6(s) + \text{6H}_2\text{O}(l) \rightarrow \text{4H}_3\text{PO}_3(aq) \nonumber \] Phosphorous acid is quite weak, and, contrary to what its formula might suggest, can only donate two protons. This is apparently because its Lewis structure is
Only the two protons bonded to highly electronegative oxygen atoms are expected to be acidic. Another major commercial use of phosphates is in laundry detergents. The problem of precipitation of soap by hard-water ions such as Ca was mentioned in the section on alkaline earth metals. This can be prevented, and the cleaning power of synthetic detergents can be improved, by adding phosphates. The compound usually used is sodium tripolyphosphate, whose anion is a condensation polymer of hydrogen phosphate and dihydrogen phosphate ions:
The use of phosphates in detergents is responsible in part for an environmental problem known as , or premature, aging of bodies of water. Over a period of many thousands of years, a lake or other body of water will slowly accumulate essential nutrient elements such as nitrogen or phosphorus because their compounds dissolve in streams that feed the lake. As the water becomes richer in nutrients, more plants and microorganisms can grow. Some of the organic matter which remains when these organisms die precipitates to the bottom of the lake and is not decomposed. Eventually the lake fills up with debris, becoming a swamp, and finally dry land. This process of eutrophication can be greatly accelerated by human input of nutrients such as nitrogen or phosphorus fertilizers, or phosphates from detergents. Since reduction in the use of detergent phosphates would appear to have the least negative effects—people’s clothes might not look as clean—many have suggested that prohibiting or limiting phosphate content is the way to solve the problem. Many states have passed laws implementing such limitations or bans. | 14,038 | 3,410 |
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You learned that the integrated rate law for each common type of reaction (zeroth, first, or second order in a single reactant) can be plotted as a straight line. Using these plots offers an alternative to the methods described for showing how reactant concentration changes with time and determining reaction order. We will illustrate the use of these graphs by considering the thermal decomposition of NO gas at elevated temperatures, which occurs according to the following reaction: \[(\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \label{14.26}\] Experimental data for this reaction at 330°C are listed in ; they are provided as [NO ], ln[NO ], and 1/[NO ] versus time to correspond to the integrated rate laws for zeroth-, first-, and second-order reactions, respectively. The actual concentrations of NO are plotted versus time in part (a) in . Because the plot of [NO ] versus is not a straight line, we know the reaction is not zeroth order in NO . A plot of ln[NO ] versus (part (b) in ) shows us that the reaction is not first order in NO because a first-order reaction would give a straight line. Having eliminated zeroth-order and first-order behavior, we construct a plot of 1/[NO ] versus (part (c) in ). This plot is a straight line, indicating that the reaction is second order in NO . We have just determined the reaction order using data from a single experiment by plotting the concentration of the reactant as a function of time. Because of the characteristic shapes of the lines shown in \(\Page {2}\), the graphs can be used to determine the reaction order of an unknown reaction. In contrast, the method of initial rates required multiple experiments at different NO concentrations as well as accurate initial rates of reaction, which can be difficult to obtain for rapid reactions. Dinitrogen pentoxide (N O ) decomposes to NO and O at relatively low temperatures in the following reaction: 2N O (soln) → 4NO (soln) + O (g) This reaction is carried out in a CCl solution at 45°C. The concentrations of N O as a function of time are listed in the following table, together with the natural logarithms and reciprocal N O concentrations. Plot a graph of the concentration versus , ln concentration versus , and 1/concentration versus and then determine the rate law and calculate the rate constant. balanced chemical equation, reaction times, and concentrations graph of data, rate law, and rate constant Use the data in the table to separately plot concentration, the natural logarithm of the concentration, and the reciprocal of the concentration (the vertical axis) versus time (the horizontal axis). Compare the graphs with those in to determine the reaction order. Write the rate law for the reaction. Using the appropriate data from the table and the linear graph corresponding to the rate law for the reaction, calculate the slope of the plotted line to obtain the rate constant for the reaction. Here are plots of [N O ] versus , ln[N O ] versus , and 1/[N O ] versus : The plot of ln[N O ] versus gives a straight line, whereas the plots of [N O ] versus and 1/[N O ] versus do not. This means that the decomposition of N O is first order in [N O ]. The rate law for the reaction is therefore rate = [N O ] Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus for a first-order reaction is − . We can calculate the slope using any two points that lie on the line in the plot of ln[N O ] versus . Using the points for = 0 and 3000 s, \[\textrm{slope}=\dfrac{\ln[\mathrm{N_2O_5}]_{3000}-\ln[\mathrm{N_2O_5}]_0}{3000\textrm{ s}-0\textrm{ s}}=\dfrac{(-4.756)-(-3.310)}{3000\textrm{ s}}=-4.820\times10^{-4}\textrm{ s}^{-1}\] Thus = 4.820 × 10 s . 1,3-Butadiene (CH =CH—CH=CH ; C H ) is a volatile and reactive organic molecule used in the production of rubber. Above room temperature, it reacts slowly to form products. Concentrations of C H as a function of time at 326°C are listed in the following table along with ln[C H ] and the reciprocal concentrations. Graph the data as concentration versus , ln concentration versus , and 1/concentration versus . Then determine the reaction order in C H , the rate law, and the rate constant for the reaction. second order in C H ; rate = [C H ] ; = 1.3 × 10 M ·s For a zeroth-order reaction, a plot of the concentration of any reactant versus time is a straight line with a slope of − . For a first-order reaction, a plot of the natural logarithm of the concentration of a reactant versus time is a straight line with a slope of − . For a second-order reaction, a plot of the inverse of the concentration of a reactant versus time is a straight line with a slope of . | 4,807 | 3,411 |
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When matter absorbs or releases heat energy, we cannot always explain this energy change on in terms of a speeding up or a slowing down of molecular motion as we do for . This is particularly true when the heat change accompanies a chemical change. Here we must consider changes in the kinetic and potential energies of in the atoms and molecules involved, that is, changes in the . As a simple example of a chemical change, let us consider an exothermic reaction involving only one kind of atom, the decomposition of ozone, O : \[\ce{2O_{3}(g) \rightarrow 3O_2(g)} \,\,\, 25°C\label{1} \] O is a which occurs in very low but very important concentrations in the upper atmosphere. It can be produced in somewhat higher concentrations in the laboratory by using an electric spark discharge and can then be concentrated and purified. The result is a blue gas which is dangerously unstable and liable to explode without warning. Let us now suppose that we have a pure sample of 2 mol O ( ) in a closed container at 25°C and are able to measure the quantity of heat evolved when it subsequently explodes to form O gas according to Eq. \(\ref{1}\) (see Figure \(\Page {1}\) ). It is found that 287.9 kJ is released. This energy heats up the surroundings. Where do these 287.9 kJ come from? Certainly not entirely from the translational kinetic energy of the molecules. To begin with we had 2 mol O at 25°C and at the end 3 mol O at 25°C. Since the translational kinetic energy of any gas is / , this corresponds to an increase of / (3 mol – 2 mol) = / × 1 mol × 8.314 J K mol × 298 K = 3.72 kJ. After the reaction the translational energy of the molecules is higher, and so heat should have been absorbed, not given off. In any case 3.73 kJ is only 1.3 percent of the total heat change. The changes in rotational and vibrational energy are even smaller, accounting for a decrease in energy of the substance in the container by only 0.88 kJ. From the standpoint of , the most important thing that happens as 2 mol O is converted into 3 mol O is rearrangement of the valence electrons so that they are closer to positively charged O nuclei. that the closer electrons are to nuclei, the lower their total energy. Thus three O molecules have less electronic energy than the two O molecules from which they were formed. The remaining energy first appears as kinetic energy of the O molecules. Immediately after the reaction the O is at a very high temperature. Eventually this energy finds its way to the surroundings, and the O cools to room temperature. A detailed summary of the various energy changes which occur when O reacts to form O is given in Table \(\Page {1}\). The important message of this table is that 99 percent of the energy change is attributable to the change in electronic energy. This is a typical figure for gaseous reactions. What makes a gaseous reaction exothermic or endothermic is the . Changes in the energies of molecular motion can usually be neglected by comparison. * Final value – initial value. † There is no experimental means of determining the initial or final value of the electronic energy—only the change in electronic energy can be measured. Highly accurate calculations of electronic energy from wave-mechanical theory require complicated mathematics end a great deal of computer power. Therefore we have represented the initial electronic energy by x. The sum of all the different kinds of energy which the molecules of a substance can possess is called the and given the symbol . (The symbol also widely used.) In a gas we can regard the internal energy as the sum of the electronic, translational, rotational, and vibrational energies. In the case of liquids and solids the molecules are closer together, and we must include the potential energy due to their interactions with each other. (intermolecular forces) attract one molecule to others. In addition, the motion of one molecule now affects its neighbors, and we can no longer subdivide the energy into neat categories as in the case of a gas. Equation \(\ref{2}\) tells us how to detect and measure changes in the internal energy of a system. If we carry out any process in a closed container the volume remains constant), the quantity of heat absorbed by the system equals the increase in internal energy. \[q_{V} = \triangle U\label{2} \] A convenient device for making such measurements is a bomb calorimeter (Figure \(\Page {2}\) ), which contains a steel-walled vessel (bomb) with a screw-on gas-tight lid. In the bomb can be placed a weighed sample of a combustible substance together with O ( ) at about 3 MPa (30 atm) pressure. When the substance is ignited by momentarily passing electrical current through a heating wire, the heat energy released by its combustion raises the temperature of water surrounding the bomb. Measurement of the change in temperature of the water permits calculation of (and thus Δ ), provided the heat capacity of the calorimeter is known. The heat capacity can be determined as in or by igniting a substance for which Δ is already known. When 0.7943g of glucose, C H O , is ignited in a bomb calorimeter, the temperature rise is found to be 1.841 K. The heat capacity of the calorimeter is 6.746 kJ K . Find Δ for the reaction \(C_{6}H_{12}O_{6}(s) + 6O_{2}(g) \rightarrow 6CO_{2}(g) + 6H_{2}O(l)\) under the prevailing conditions. The heat energy absorbed by the calorimeter in increasing its temperature by 1.841 K is given by \[q = C \triangle T = 6.745 \text{kJ K^{-1} }* 1.841 \text{K} = 12.42 \text{kJ}\) \nonumber \] Since this heat energy was released by the reaction system, we must regard it as negative. Accordingly, \[q_{V} = –12.42 \text{kJ} = \triangle U \nonumber \] We need now only to calculate the change in internal energy per mole, that is, Δ . Now \[n_{\text{glucose}}=\text{0}\text{.7953 g }\times \text{ }\frac{\text{1 mol}}{\text{180}\text{.16 g}}=\text{4}\text{.409 }\times \text{ 10}^{-\text{3}}\text{ mol} \nonumber \] Thus \[\triangle U_{m}= \frac{-12.42 kJ}{4.409} \times 10^{-3} \text{mol}= -2817 \text{kJ mol}^{-1} \nonumber \] Ed Vitz (Kutztown University), (University of | 6,207 | 3,412 |
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In absorption spectroscopy a beam of electromagnetic radiation passes through a sample. Much of the radiation passes through the sample without a loss in intensity. At selected wavelengths, however, the radiation’s intensity is attenuated. This process of attenuation is called absorption. There are two general requirements for an analyte’s absorption of electromagnetic radiation. First, there must be a mechanism by which the radiation’s electric field or magnetic field interacts with the analyte. For ultraviolet and visible radiation, absorption of a photon changes the energy of the analyte’s valence electrons. A bond’s vibrational energy is altered by the absorption of infrared radiation. provides a list of the types of atomic and molecular transitions associated with different types of electromagnetic radiation. The second requirement is that the photon’s energy, \(h \nu\), must exactly equal the difference in energy, \(\Delta E\), between two of the analyte’s quantized energy states. shows a simplified view of a photon’s absorption, which is useful because it emphasizes that the photon’s energy must match the difference in energy between a lower-energy state and a higher-energy state. What is missing, however, is information about what types of energy states are involved, which transitions between energy states are likely to occur, and the appearance of the resulting spectrum. We can use the energy level diagram in Figure 10.2.1
to explain an absorbance spectrum. The lines labeled and represent the analyte’s ground (lowest) electronic state and its first electronic excited state. Superimposed on each electronic energy level is a series of lines representing vibrational energy levels. The energy of infrared radiation produces a change in a molecule’s or a polyatomic ion’s vibrational energy, but is not sufficient to effect a change in its electronic energy. As shown in Figure 10.2.1
, vibrational energy levels are quantized; that is, a molecule or polyatomic ion has only certain, discrete vibrational energies. The energy for an allowed vibrational mode, \(E_{\nu}\), is \[E_{\nu}=\nu+\frac{1}{2} h \nu_{0} \nonumber\] where \(\nu\) is the vibrational quantum number, which has values of 0, 1, 2, ..., and \(\nu_0\) is the bond’s fundamental vibrational frequency. The value of \(\nu_0\), which is determined by the bond’s strength and by the mass at each end of the bond, is a characteristic property of a bond. For example, a carbon-carbon single bond (C–C) absorbs infrared radiation at a lower energy than a carbon-carbon double bond (C=C) because a single bond is weaker than a double bond. At room temperature most molecules are in their ground vibrational state (\(\nu = 0\)) . A transition from the ground vibrational state to the first vibrational excited state (\(\nu = 1\)) requires absorption of a photon with an energy of \(h \nu_0\). Transitions in which \(\Delta \nu = \pm 1\) give rise to the fundamental absorption lines. Weaker absorption lines, called overtones, result from transitions in which \(\Delta \nu\) is ±2 or ±3. The number of possible normal vibrational modes for a linear molecule is 3 – 5, and for a non-linear molecule is 3 – 6, where is the number of atoms in the molecule. Not surprisingly, infrared spectra often show a considerable number of absorption bands. Even a relatively simple molecule, such as ethanol (C H O), for example, has \(3 \times 9 - 6\), or 21 possible normal modes of vibration, although not all of these vibrational modes give rise to an absorption. The IR spectrum for ethanol is shown in Figure 10.2.2
. Why does a non-linear molecule have 3 – 6 vibrational modes? Consider a molecule of methane, CH . Each of methane’s five atoms can move in one of three directions ( , , and ) for a total of \(5 \times 3 = 15\) different ways in which the molecule’s atoms can move. A molecule can move in three ways: it can move from one place to another, which we call translational motion; it can rotate around an axis, which we call rotational motion; and its bonds can stretch and bend, which we call vibrational motion. Because the entire molecule can move in the , , and directions, three of methane’s 15 different motions are translational. In addition, the molecule can rotate about its , , and axes, accounting for three additional forms of motion. This leaves 15 – 3 – 3 = 9 vibrational modes. A linear molecule, such as CO , has 3 – 5 vibrational modes because it can rotate around only two axes. The valence electrons in organic molecules and polyatomic ions, such as \(\text{CO}_3^{2-}\), occupy quantized sigma bonding (\(\sigma\)), pi bonding (\(\pi\)), and non-bonding ( ) molecular orbitals (MOs). Unoccupied sigma antibonding (\(\sigma^*\)) and pi antibonding (\(\pi^*\)) molecular orbitals are slightly higher in energy. Because the difference in energy between the highest-energy occupied MOs and the lowest-energy unoccupied MOs corresponds to ultraviolet and visible radiation, absorption of a photon is possible. Four types of transitions between quantized energy levels account for most molecular UV/Vis spectra. Table 10.2.1
lists the approximate wavelength ranges for these transitions, as well as a partial list of bonds, functional groups, or molecules responsible for these transitions. Of these transitions, the most important are \(n \rightarrow \pi^*\) and \(\pi \rightarrow \pi^*\) because they involve important functional groups that are characteristic of many analytes and because the wavelengths are easily accessible. The bonds and functional groups that give rise to the absorption of ultraviolet and visible radiation are called . Many transition metal ions, such as Cu and Co , form colorful solutions because the metal ion absorbs visible light. The transitions that give rise to this absorption are valence electrons in the metal ion’s -orbitals. For a free metal ion, the five -orbitals are of equal energy. In the presence of a complexing ligand or solvent molecule, however, the -orbitals split into two or more groups that differ in energy. For example, in an octahedral complex of \(\text{Cu(H}_2\text{O)}_6^{2+}\) the six water molecules perturb the -orbitals into the two groups shown in Figure 10.2.3
. The resulting \(d \rightarrow d\) transitions for transition metal ions are relatively weak. A more important source of UV/Vis absorption for inorganic metal–ligand complexes is charge transfer, in which absorption of a photon produces an excited state in which there is transfer of an electron from the metal, , to the ligand, . \[M-L+h \nu \rightarrow\left(M^{+}-L^{-}\right)^{*} \nonumber\] Charge-transfer absorption is important because it produces very large absorbances. One important example of a charge-transfer complex is that of -phenanthroline with Fe , the UV/Vis spectrum for which is shown in Figure 10.2.4
. Charge-transfer absorption in which an electron moves from the ligand to the metal also is possible. Why is a larger absorbance desirable? An analytical method is more sensitive if a smaller concentration of analyte gives a larger signal. Comparing the IR spectrum in Figure 10.2.2
to the UV/Vis spectrum in Figure 10.2.4
shows us that UV/Vis absorption bands are often significantly broader than those for IR absorption. We can use Figure 10.2.1
to explain why this is true. When a species absorbs UV/Vis radiation, the transition between electronic energy levels may also include a transition between vibrational energy levels. The result is a number of closely spaced absorption bands that merge together to form a single broad absorption band. The energy of ultraviolet and visible electromagnetic radiation is sufficient to cause a change in an atom’s valence electron configuration. Sodium, for example, has a single valence electron in its 3 atomic orbital. As shown in Figure 10.2.5
, unoccupied, higher energy atomic orbitals also exist. The valence shell energy level diagram in Figure 10.2.5
might strike you as odd because it shows that the 3 orbitals are split into two groups of slightly different energy. The reasons for this splitting are unimportant in the context of our treatment of atomic absorption. For further information about the reasons for this splitting, consult the chapter’s additional resources. Absorption of a photon is accompanied by the excitation of an electron from a lower-energy atomic orbital to an atomic orbital of higher energy. Not all possible transitions between atomic orbitals are allowed. For sodium the only allowed transitions are those in which there is a change of ±1 in the orbital quantum number ( ); thus transitions from \(s \rightarrow p\) orbitals are allowed, but transitions from \(s \rightarrow s\) and from \(s \rightarrow d\) orbitals are forbidden. The atomic absorption spectrum for Na is shown in Figure 10.2.6
, and is typical of that found for most atoms. The most obvious feature of this spectrum is that it consists of a small number of discrete absorption lines that correspond to transitions between the ground state (the 3 atomic orbital) and the 3 and the 4 atomic orbitals. Absorption from excited states, such as the \(3p \rightarrow 4s\) and the \(3p \rightarrow 3d\) transitions included in Figure 10.2.5
, are too weak to detect. Because an excited state’s lifetime is short—an excited state atom typically returns to a lower energy state in 10 to 10 seconds—an atom in the exited state is likely to return to the ground state before it has an opportunity to absorb a photon. Another feature of the atomic absorption spectrum in Figure 10.2.6
is the narrow width of the absorption lines, which is a consequence of the fixed difference in energy between the ground state and the excited state, and the lack of vibrational and rotational energy levels. Natural line widths for atomic absorption, which are governed by the uncertainty principle, are approximately 10 nm. Other contributions to broadening increase this line width to approximately 10 nm. As light passes through a sample, its power decreases as some of it is absorbed. This attenuation of radiation is described quantitatively by two separate, but related terms: transmittance and absorbance. As shown in Figure 10.2.7
a, transmittance is the ratio of the source radiation’s power as it exits the sample, , to that incident on the sample, . \[T=\frac{P_{\mathrm{T}}}{P_{0}} \label{10.1}\] Multiplying the transmittance by 100 gives the percent transmittance, % , which varies between 100% (no absorption) and 0% (complete absorption). All methods of detecting photons—including the human eye and modern photoelectric transducers—measure the transmittance of electromagnetic radiation. Equation \ref{10.1} does not distinguish between different mechanisms that prevent a photon emitted by the source from reaching the detector. In addition to absorption by the analyte, several additional phenomena contribute to the attenuation of radiation, including reflection and absorption by the sample’s container, absorption by other components in the sample’s matrix, and the scattering of radiation. To compensate for this loss of the radiation’s power, we use a method blank. As shown in Figure 10.2.7
b, we redefine as the power exiting the method blank. An alternative method for expressing the attenuation of electromagnetic radiation is absorbance, , which we define as \[A=-\log T=-\log \frac{P_{\mathrm{T}}}{P_{0}} \label{10.2}\] Absorbance is the more common unit for expressing the attenuation of radiation because it is a linear function of the analyte’s concentration. We will show that this is true in the next section when we introduce Beer’s law. A sample has a percent transmittance of 50%. What is its absorbance? A percent transmittance of 50.0% is the same as a transmittance of 0.500. Substituting into Equation \ref{10.2} gives \[A=-\log T=-\log (0.500)=0.301 \nonumber\] What is the % for a sample if its absorbance is 1.27? To find the transmittance, , we begin by noting that \[A=1.27=-\log T \nonumber\] Solving for \[\begin{array}{c}{-1.27=\log T} \\ {10^{-1.27}=T}\end{array} \nonumber\] gives a transmittance of 0.054, or a %T of 5.4%. Equation \ref{10.1} has an important consequence for atomic absorption. As we learned from , atomic absorption lines are very narrow. Even with a high quality monochromator, the effective bandwidth for a continuum source is \(100-1000 \times\) greater than the width of an atomic absorption line. As a result, little radiation from a continuum source is absorbed when it passes through a sample of atoms; because ≈ the measured absorbance effectively is zero. For this reason, atomic absorption requires that we use a line source instead of a continuum source. When monochromatic electromagnetic radiation passes through an infinitesimally thin layer of sample of thickness , it experiences a decrease in its power of (Figure 10.2.8
). This fractional decrease in power is proportional to the sample’s thickness and to the analyte’s concentration, ; thus \[-\frac{d P}{P}=\alpha C d x \label{10.3}\] where is the power incident on the thin layer of sample and \(\alpha\) is a proportionality constant. Integrating the left side of Equation \ref{10.3} over the sample’s full thickness \[-\int_{P=P_0}^{P=P_t} \frac{d P}{P}=\alpha C \int_{x=0}^{x=b} d x \nonumber\] \[\ln \frac{P_{0}}{P_T}=\alpha b C \nonumber\] converting from ln to log, and substituting into Equation \ref{10.2}, gives \[A=a b C \label{10.4}\] where is the analyte’s with units of cm conc . If we express the concentration using molarity, then we replace with the , \(\varepsilon\), which has units of cm M . \[A=\varepsilon b C \label{10.5}\] The absorptivity and the molar absorptivity are proportional to the probability that the analyte absorbs a photon of a given energy. As a result, values for both and \(\varepsilon\) depend on the wavelength of the absorbed photon. A \(5.00 \times 10^{-4}\) M solution of analyte is placed in a sample cell that has a pathlength of 1.00 cm. At a wavelength of 490 nm, the solution’s absorbance is 0.338. What is the analyte’s molar absorptivity at this wavelength? Solving Equation \ref{10.5} for \(\epsilon\) and making appropriate substitutions gives \[\varepsilon=\frac{A}{b C}=\frac{0.338}{(1.00 \ \mathrm{cm})\left(5.00 \times 10^{-4} \ \mathrm{M}\right)}=676 \ \mathrm{cm}^{-1} \ \mathrm{M}^{-1} \nonumber\] A solution of the analyte from Example 10.2.2
has an absorbance of 0.228 in a 1.00-cm sample cell. What is the analyte’s concentration? Making appropriate substitutions into Beer’s law \[A=0.228=\varepsilon b C=\left(676 \ \mathrm{M}^{-1} \ \mathrm{cm}^{-1}\right)(1 \ \mathrm{cm}) C \nonumber\] and solving for gives a concentration of \(3.37 \times 10^{-4}\) M. Equation \ref{10.4} and Equation \ref{10.5}, which establish the linear relationship between absorbance and concentration, are known as Beer’s law. Calibration curves based on Beer’s law are common in quantitative analyses. As is often the case, the formulation of a law is more complicated than its name suggests. This is the case, for example, with Beer’s law, which also is known as the Beer-Lambert law or the Beer-Lambert-Bouguer law. Pierre Bouguer, in 1729, and Johann Lambert, in 1760, noted that the transmittance of light decreases exponentially with an increase in the sample’s thickness. \[T \propto e^{-b} \nonumber\] Later, in 1852, August Beer noted that the transmittance of light decreases exponentially as the concentration of the absorbing species increases. \[T \propto e^{-C} \nonumber\] Together, and when written in terms of absorbance instead of transmittance, these two relationships make up what we know as Beer’s law. We can extend Beer’s law to a sample that contains several absorbing components. If there are no interactions between the components, then the individual absorbances, , are additive. For a two-component mixture of analyte’s and , the total absorbance, , is \[A_{tot}=A_{X}+A_{Y}=\varepsilon_{X} b C_{X}+\varepsilon_{Y} b C_{Y} \nonumber\] Generalizing, the absorbance for a mixture of components, , is \[A_{m i x}=\sum_{i=1}^{n} A_{i}=\sum_{i=1}^{n} \varepsilon_{i} b C_{i} \label{10.6}\] Beer’s law suggests that a plot of absorbance vs. concentration—we will call this a Beer’s law plot—is a straight line with a -intercept of zero and a slope of or \(\varepsilon b\). In some cases a Beer’s law plot deviates from this ideal behavior (see Figure 10.2.9
), and such deviations from linearity are divided into three categories: fundamental, chemical, and instrumental. Beer’s law is a limiting law that is valid only for low concentrations of analyte. There are two contributions to this fundamental limitation to Beer’s law. At higher concentrations the individual particles of analyte no longer are independent of each other. The resulting interaction between particles of analyte may change the analyte’s absorptivity. A second contribution is that an analyte’s absorptivity depends on the solution’s refractive index. Because a solution’s refractive index varies with the analyte’s concentration, values of and \(\varepsilon\) may change. For sufficiently low concentrations of analyte, the refractive index essentially is constant and a Beer’s law plot is linear. A chemical deviation from Beer’s law may occur if the analyte is involved in an equilibrium reaction. Consider, for example, the weak acid, HA. To construct a Beer’s law plot we prepare a series of standard solutions—each of which contains a known total concentration of HA—and then measure each solution’s absorbance at the same wavelength. Because HA is a weak acid, it is in equilibrium with its conjugate weak base, A . In the equations that follow, the conjugate weak base A is written as A as it is easy to mistake the symbol for anionic charge as a minus sign. \[\mathrm{HA}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{A}^{-}(a q) \nonumber\] If both HA and A absorb at the selected wavelength, then Beer’s law is \[A=\varepsilon_{\mathrm{HA}} b C_{\mathrm{HA}}+\varepsilon_{\mathrm{A}} b C_{\mathrm{A}} \label{10.7}\] Because the weak acid’s total concentration, , is \[C_{\mathrm{total}}=C_{\mathrm{HA}}+C_{\mathrm{A}} \nonumber\] we can write the concentrations of HA and A as \[C_{\mathrm{HA}}=\alpha_{\mathrm{HA}} C_{\mathrm{total}} \label{10.8}\] \[C_{\text{A}} = (1 - \alpha_\text{HA})C_\text{total} \label{10.9}\] where \(\alpha_\text{HA}\) is the fraction of weak acid present as HA. Substituting Equation \ref{10.8} and Equation \ref{10.9} into Equation \ref{10.7} and rearranging, gives \[A=\left(\varepsilon_{\mathrm{HA}} \alpha_{\mathrm{HA}}+\varepsilon_{\mathrm{A}}-\varepsilon_{\mathrm{A}} \alpha_{\mathrm{A}}\right) b C_{\mathrm{total}} \label{10.10}\] To obtain a linear Beer’s law plot, we must satisfy one of two conditions. If \(\varepsilon_\text{HA}\) and \(\varepsilon_{\text{A}}\) have the same value at the selected wavelength, then Equation \ref{10.10} simplifies to \[A = \varepsilon_{\text{A}}bC_\text{total} = \varepsilon_\text{HA}bC_\text{total} \nonumber\] Alternatively, if \(\alpha_\text{HA}\) has the same value for all standard solutions, then each term within the parentheses of Equation \ref{10.10} is constant—which we replace with —and a linear calibration curve is obtained at any wavelength. \[A=k b C_{\mathrm{total}} \nonumber\] Because HA is a weak acid, the value of \(\alpha_\text{HA}\) varies with pH. To hold \(\alpha_\text{HA}\) constant we buffer each standard solution to the same pH. Depending on the relative values of \(\alpha_\text{HA}\) and \(\alpha_{\text{A}}\), the calibration curve has a positive or a negative deviation from Beer’s law if we do not buffer the standards to the same pH. There are two principal instrumental limitations to Beer’s law. The first limitation is that Beer’s law assumes that radiation reaching the sample is of a single wavelength—that is, it assumes a purely monochromatic source of radiation. As shown in , even the best wavelength selector passes radiation with a small, but finite effective bandwidth. Polychromatic radiation always gives a negative deviation from Beer’s law, but the effect is smaller if the value of \(\varepsilon\) essentially is constant over the wavelength range passed by the wavelength selector. For this reason, as shown in Figure 10.2.10
, it is better to make absorbance measurements at the top of a broad absorption peak. In addition, the deviation from Beer’s law is less serious if the source’s effective bandwidth is less than one-tenth of the absorbing species’ natural bandwidth [(a) Strong, F. C., III , , 16A–34A; Gilbert, D. D. , , A278–A281]. When measurements must be made on a slope, linearity is improved by using a narrower effective bandwidth. is the second contribution to instrumental deviations from Beer’s law. Stray radiation arises from imperfections in the wavelength selector that allow light to enter the instrument and to reach the detector without passing through the sample. Stray radiation adds an additional contribution, , to the radiant power that reaches the detector; thus \[A=-\log \frac{P_{\mathrm{T}}+P_{\text { stray }}}{P_{0}+P_{\text { stray }}} \nonumber\] For a small concentration of analyte, is significantly smaller than and , and the absorbance is unaffected by the stray radiation. For higher concentrations of analyte, less light passes through the sample and and become similar in magnitude. This results is an absorbance that is smaller than expected, and a negative deviation from Beer’s law. | 21,688 | 3,413 |
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\[[H_2O\:(acidulated)\rightleftharpoons H^+\,(aq)+OH]^-\times4\] At cathode: \[[H^+\,(aq)+e^-\rightarrow\dfrac{1}{2}H_2(g)]\times4\] At anode: \[4OH^-\,(aq)\rightarrow O_2+2H_2O + 4e^-\] Net reaction: \[2H_2O \xrightarrow{\large{electrolysis}} 2H_2\,(g) + O_2\,(g)\] Oxygen can thus be obtained from acidified water by its electrolysis. 2Li(s) + D (g) → 2LiD(s) 4LiD(s) + AlCl (soln) → LiAlD (s) + 3LiCl(soln) CO(g) + H O(g) → CO (g) + H (g) Calculate ΔG° for this reaction at 298 K and determine the temperature at which the reaction changes from spontaneous to nonspontaneous (or vice versa). | 612 | 3,416 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.05%3A_Hydrogen_and_Hydroxide_Ions |
If you refer to , you will see that pure water does conduct some electrical current, albeit much less than-even the weak electrolytes listed there. This is because water itself is a very weak electrolyte. It ionizes to hydrogen ions and hydroxide ions to an extremely small extent: \[\text{H}_{2}\text{O}(l) \rightleftharpoons \text{H}^{+}(aq) + \text{OH}^{-}(aq)\label{1} \] Careful measurements show that at 25°C the concentrations of H ( ) and OH ( ) are each 1.005 × 10 mol dm . At higher temperatures more H ( ) and OH ( ) are produced while at lower temperatures less ionization of water occurs. Nevertheless, in pure water the concentration of H ( ) always equals the concentration of OH ( ). Dissolving acids or bases in water can change the concentrations of both H ( ) and OH ( ), causing them to differ from one another. The special case of a solution in which these two concentrations remain equal is called a A hydrogen ion, H , is a hydrogen atom which has lost its single electron; that is, a hydrogen ion is just a proton. Because a proton is only about one ten-thousandth as big as an average atom or ion, water dipoles can approach very close to a hydrogen ion in solution. Consequently the proton can exert a very strong attractive force on a lone pair of electrons in a water molecule—strong enough to form a coordinate covalent bond: The H O formed in this way is called a (on the left in the figure below). All three of its O―H bonds are exactly the same, and the ion has a pyramidal structure as predicted by VSEPR theory (1 ). To emphasize the fact that a proton cannot exist by itself in aqueous solution, Eq. (1 ) is often rewritten as \[\text{2H}_{2}\text{O}(l) \rightleftharpoons \text{H}_{3}\text{O}^{+}(aq) + \text{OH}^{-}(aq) \nonumber \] Like other ions in aqueous solution, both hydronium and hydroxide ions are hydrated. Moreover, hydrogen bonds are involved in attracting water molecules to hydronium and hydroxide ions. In both cases three water molecules appear to be rather tightly held, giving formulas H O(H O) (or H O ) and HO(H O) (or H O ). Possible structures for the hydrated hydronium and hydroxide ions are shown in Figure \(\Page {1}\). Hydrogen bonding of hydronium and hydroxide ions to water molecules accounts rather nicely for the unusually large electrical currents observed for some electrolytes containing H and OH. The case of the hydronium ion is illustrated in Figure \(\Page {2}\). When a hydronium ion collides with one end of a hydrogen-bonded chain of water molecules, a different hydronium ion can be released at the other end. Only a slight movement of six protons and a rearrangement of covalent and hydrogen bonds is needed. In effect a hydronium ion can almost instantaneously “jump” the length of several water molecules. It need not elbow its way through a crowd as other ions must. The same is true of aqueous hydroxide ions. Since both ions move faster, they can transfer more electrical charge per unit time, that is, more current. | 3,026 | 3,417 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/10%3A_Spectroscopic_Methods/10.08%3A_Spectroscopy_Based_on_Scattering |
The blue color of the sky during the day and the red color of the sun at sunset are the result of light scattered by small particles of dust, molecules of water, and other gases in the atmosphere. The efficiency of a photon’s scattering depends on its wavelength. We see the sky as blue during the day because violet and blue light scatter to a greater extent than other, longer wavelengths of light. For the same reason, the sun appears red at sunset because red light is less efficiently scattered and is more likely to pass through the atmosphere than other wavelengths of light. The scattering of radiation has been studied since the late 1800s, with applications beginning soon thereafter. The earliest quantitative applications of scattering, which date from the early 1900s, used the elastic scattering of light by colloidal suspensions to determine the concentration of colloidal particles. If we send a focused, monochromatic beam of radiation with a wavelength \(\lambda\) through a medium of particles with dimensions \(< 1.5 \lambda\), the radiation scatters in all directions. For example, visible radiation of 500 nm is scattered by particles as large as 750 nm in the longest dimension. Two general categories of scattering are recognized. In elastic scattering, radiation is first absorbed by the particles and then emitted without undergoing a change in the radiation’s energy. When the radiation emerges with a change in energy, the scattering is inelastic. Only elastic scattering is considered in this text. Elastic scattering is divided into two types: Rayleigh, or small-particle scattering, and large-particle scattering. Rayleigh scattering occurs when the scattering particle’s largest dimension is less than 5% of the radiation’s wavelength. The intensity of the scattered radiation is proportional to its frequency to the fourth power, \(\nu^4\)—which accounts for the greater scattering of blue light than red light—and is distributed symmetrically (Figure 10.8.1
a). For larger particles, scattering increases in the forward direction and decreases in the backward direction as the result of constructive and destructive interferences (Figure 10.8.1
b). Turbidimetry and nephelometry are two techniques that rely on the elastic scattering of radiation by a suspension of colloidal particles. In the detector is placed in line with the source and the decrease in the radiation’s transmitted power is measured. In the scattered radiation is measured at an angle of 90 to the source. The similarity of turbidimetry to absorbance spectroscopy and of nephelometry to fluorescence spectroscopy is evident in the instrumental designs shown in Figure 10.8.2
. In fact, we can use a UV/Vis spectrophotometer for turbidimetry and we can use a spectrofluorometer for nephelometry. When developing a scattering method the choice between using turbidimetry or using nephelometry is determined by two factors. The most important consideration is the intensity of the scattered radiation relative to the intensity of the source’s radiation. If the solution contains a small concentration of scattering particles, then the intensity of the transmitted radiation, , is approximately the same as the intensity of the source’s radiation, . As we learned earlier in the section on molecular absorption, there is substantial uncertainty in determining a small difference between two intense signals. For this reason, nephelometry is a more appropriate choice for a sample that contains few scattering particles. Turbidimetry is a better choice when the sample contains a high concentration of scattering particles. A second consideration in choosing between turbidimetry and nephelometry is the size of the scattering particles. For nephelometry, the intensity of scattered radiation at 90 increases when the particles are small and Rayleigh scattering is in effect. For larger particles, as shown in Figure 10.8.1
, the intensity of scattering decreases at 90 . When using an ultraviolet or a visible source of radiation, the optimum particle size is 0.1–1 μm. The size of the scattering particles is less important for turbidimetry where the signal is the relative decrease in transmitted radiation. In fact, turbidimetric measurements are feasible even when the size of the scattering particles results in an increase in reflection and refraction, although a linear relationship between the signal and the concentration of scattering particles may no longer hold. For turbidimetry the measured transmittance, , is the ratio of the intensity of source radiation transmitted by the sample, , to the intensity of source radiation transmitted by a blank, . \[T=\frac{I_{\mathrm{T}}}{I_{0}} \nonumber\] The relationship between transmittance and the concentration of the scattering particles is similar to that given by Beer’s law \[-\log T=k b C \label{10.1}\] where is the concentration of the scattering particles in mass per unit volume (w/v), is the pathlength, and is a constant that depends on several factors, including the size and shape of the scattering particles and the wavelength of the source radiation. The exact relationship is established by a calibration curve prepared using a series of standards that contain known concentrations of analyte. As with Beer’s law, Equation \ref{10.1} may show appreciable deviations from linearity. For nephelometry the relationship between the intensity of scattered radiation, , and the concentration of scattering particles is \[I_{\mathrm{s}}=k I_{0} C \label{10.2}\] where is an empirical constant for the system and is the intensity of the source radiation. The value of is determined from a calibration curve prepared using a series of standards that contain known concentrations of analyte. The choice of wavelength is based primarily on the need to minimize potential interferences. For turbidimetry, where the incident radiation is transmitted through the sample, a monochromator or filter allow us to avoid wavelengths that are absorbed instead of scattered by the sample. For nephelometry, the absorption of incident radiation is not a problem unless it induces fluorescence from the sample. With a nonfluorescent sample there is no need for wavelength selection and a source of white light may be used as the incident radiation. For both techniques, other considerations in choosing a wavelength including the intensity of scattering, the trans- ducer’s sensitivity (many common photon transducers are more sensitive to radiation at 400 nm than at 600 nm), and the source’s intensity. Although Equation \ref{10.1} and Equation \ref{10.2} relate scattering to the concentration of the scattering particles, the intensity of scattered radiation also is influenced by the size and the shape of the scattering particles. Samples that contain the same number of scattering particles may show significantly different values for –log or depending on the average diameter of the particles. For a quantitative analysis, therefore, it is necessary to maintain a uniform distribution of particle sizes throughout the sample and between samples and standards. Most turbidimetric and nephelometric methods rely on precipitation reaction to form the scattering particles. As we learned in , a precipitate’s properties, including particle size, are determined by the conditions under which it forms. To maintain a reproducible distribution of particle sizes between samples and standards, it is necessary to control parameters such as the concentration of reagents, the order of adding reagents, the pH and temperature, the agitation or stirring rate, the ionic strength, and the time between the precipitate’s initial formation and the measurement of transmittance or scattering. In many cases a surface-active agent—such as glycerol, gelatin, or dextrin—is added to stabilize the precipitate in a colloidal state and to prevent the coagulation of the particles. Turbidimetry and nephelometry are used to determine the clarity of water. The primary standard for measuring clarity is formazin, an easily prepared, stable polymer suspension (Figure 10.8.3
) [Hach, C. C.; Bryant, M. “Turbidity Standards,” Technical Information Series, Booklet No. 12, Hach Company: Loveland, CO, 1995]. A stock standard of formazin is prepared by combining a 1g/100mL solution of hydrazine sulfate, N H •H SO , with a 10 g/100 mL solution of hexamethylenetetramine to produce a suspension of particles that is defined as 4000 nephelometric turbidity units (NTU). A set of external standards with NTUs between 0 and 40 is prepared by diluting the stock standard. This method is readily adapted to the analysis of the clarity of orange juice, beer, and maple syrup. A number of inorganic cations and anions are determined by precipitating them under well-defined conditions. The transmittance or scattering of light, as defined by Equation \ref{10.1} or Equation \ref{10.2}, is proportional to the concentration of the scattering particles, which, in turn, is related by the stoichiometry of the precipitation reaction to the analyte’s concentration. Several examples of analytes determined in this way are listed in Table 10.8.1
. The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical analytical method. Although each method is unique, the following description of the determination of sulfate in water provides an instructive example of a typical procedure. The description here is based on Method 4500–SO42––C in , American Public Health Association: Washington, D. C. 20th Ed., 1998. Adding BaCl to an acidified sample precipitates \(\text{SO}_4^{2-}\) as BaSO . The concentration of \(\text{SO}_4^{2-}\) is determined either by turbidimetry or by nephelometry using an incident source of radiation of 420 nm. External standards that contain know concentrations of \(\text{SO}_4^{2-}\) are used to standardize the method. Transfer a 100-mL sample to a 250-mL Erlenmeyer flask along with 20.00 mL of an appropriate buffer. For a sample that contains more than 10 mg \(\text{SO}_4^{2-}\)/L, the buffer’s composition is 30 g of MgCl •6H O, 5 g of CH COONa•3H O, 1.0 g of KNO , and 20 mL of glacial CH COOH per liter. The buffer for a sample that contain less than 10 mg \(\text{SO}_4^{2-}\)/L is the same except for the addition of 0.111 g of Na SO per L. Place the sample and the buffer on a magnetic stirrer operated at the same speed for all samples and standards. Add a spoonful of 20–30 mesh BaCl , using a measuring spoon with a capacity of 0.2–0.3 mL, to precipitate the \(\text{SO}_4^{2-}\) as BaSO4. Begin timing when the BaCl is added and stir the suspension for 60 ± 2 s. When the stirring is complete, allow the solution to sit without stirring for 5.0± 0.5 min before measuring its transmittance or its scattering. Prepare a calibration curve over the range 0–40 mg \(\text{SO}_4^{2-}\)/L by diluting a stock standard that is 100-mg \(\text{SO}_4^{2-}\)/L. Treat each standard using the procedure described above for the sample. Prepare a calibration curve and use it to determine the amount of sulfate in the sample. 1. What is the purpose of the buffer? If the precipitate’s particles are too small, is too small to measure reliably. Because rapid precipitation favors the formation of micro-crystalline particles of BaSO , we use conditions that favor the precipitate’s growth over the nucleation of new particles. The buffer’s high ionic strength and its acidity favor the precipitate’s growth and prevent the formation of microcrystalline BaSO . 2. Why is it important to use the same stirring rate and time for the samples and standards? How fast and how long we stir the sample after we add BaCl influences the size of the precipitate’s particles. 3. Many natural waters have a slight color due to the presence of humic and fulvic acids, and may contain suspended matter (Figure 10.8.4
). Explain why these might interfere with the analysis for sulfate. For each interferent, suggest a way to minimize its effect on the analysis. Suspended matter in a sample contributes to scattering and, therefore, results in a positive determinate error. We can eliminate this interference by filtering the sample prior to its analysis. A sample that is colored may absorb some of the source’s radiation, leading to a positive determinate error. We can compensate for this interference by taking a sample through the analysis without adding BaCl . Because no precipitate forms, we use the transmittance of this sample blank to correct for the interference. 4. Why is Na SO added to the buffer for samples that contain less than 10 mg \(\text{SO}_4^{2-}\)/L? The uncertainty in a calibration curve is smallest near its center. If a sample has a high concentration of \(\text{SO}_4^{2-}\), we can dilute it so that its concentration falls near the middle of the calibration curve. For a sample with a small concentration of \(\text{SO}_4^{2-}\), the buffer increases the concentration of sulfate by \[\begin{array}{c}{\frac{0.111 \ \mathrm{g} \ \mathrm{Na}_{2} \mathrm{SO}_{4}}{\mathrm{L}} \times \frac{96.06 \ \mathrm{g} \ \mathrm{SO}_{4}^{2-}}{142.04 \ \mathrm{g} \ \mathrm{Na}_{2} \mathrm{SO}_{4}} \times} \\ {\qquad \frac{1000 \ \mathrm{mg}}{\mathrm{g}} \times \frac{20.00 \ \mathrm{mL}}{250.0 \ \mathrm{mL}}=6.00 \ \mathrm{mg} \ \mathrm{SO}_{4}^{2-} / \mathrm{L}}\end{array} \nonumber\] After using the calibration curve to determine the amount of sulfate in the sample as analyzed, we subtract 6.00 mg \(\text{SO}_4^{2-}\)/L to determine the amount of sulfate in the original sample. To evaluate the method described in Representative Method 10.8.1, a series of external standard was prepared and analyzed, providing the results shown in the following table. Analysis of a 100.0-mL sample of a surface water gives a transmittance of 0.538. What is the concentration of sulfate in the sample? Linear regression of –log versus concentration of \(\text{SO}_4^{2-}\) gives the calibration curve shown below, which has the following calibration equation. \[-\log T=-1.04 \times 10^{-5}+0.0190 \times \frac{\mathrm{mg} \ \mathrm{SO}_{4}^{2-}}{\mathrm{L}} \nonumber\] Substituting the sample’s transmittance into the calibration curve’s equation gives the concentration of sulfate in sample as 14.2 mg \(\text{SO}_4^{2-}\)/L. | 14,384 | 3,418 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.06%3A_Reaction_Mechanisms |
One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the mechanism of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism. In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water: \[\ce{2C8H18 (l) + 25O2(g) -> 16CO2 (g) + 18H2O(g)} \label{14.6.1} \] For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an , involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the of the reaction. The sum of the individual steps, or elementary reactions, in the mechanism must give the balanced chemical equation for the overall reaction. The overall sequence of elementary reactions is the mechanism of the reaction. To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide. \[\ce{NO2(g) + (g) -> NO(g) + CO2 (g)} \label{14.6.2} \] From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of \(\ce{NO2}\) with a molecule of \(\ce{CO}\) that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows: \[rate = k[\ce{NO2}]^2 \label{14.6.3} \] The fact that the reaction is second order in \([\ce{NO2}]\) and independent of \([\ce{CO}]\) tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be \[rate = k[\ce{NO2},\ce{CO}]. \nonumber \] The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2: According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The \(\ce{NO3}\) molecule is an in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step. The sum of the elementary reactions in a reaction mechanism give the overall balanced chemical equation of the reaction. The of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as ; if there are two reactant molecules, it is ; and if there are three reactant molecules (a relatively rare situation), it is . Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!) Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order of the elementary reaction is the same as its molecularity ( \(\Page {1}\)). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction (A → products) is \[rate = k[A]. \nonumber \] For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in e \(\Page {1}\). For a bimolecular elementary reaction of the form A + B → products, the general rate law is \[rate = k[A,B]. \nonumber \] For elementary reactions, the order of the elementary reaction is the same as its molecularity. In contrast, the rate law be determined from the balanced chemical equation for the overall reaction (unless it is a single step mechanism and is therefore also an elementary step). Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we cannot obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the slowest overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the , that must give the experimentally determined rate law for the overall reaction.This statement is true if one step is substantially slower than all the others, typically by a factor of 10 or more. If two or more slow steps have comparable rates, the experimentally determined rate laws can become complex. Our discussion is limited to reactions in which one step can be identified as being substantially slower than any other. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions. Look at the rate laws for each elementary reaction in our example as well as for the overall reaction. The experimentally determined rate law for the reaction of \(NO_2\) with \(CO\) is the same as the predicted rate law for step 1. This tells us that the first elementary reaction is the rate-determining step, so \(k\) for the overall reaction must equal \(k_1\). That is, NO is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2. Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect. In an alternative mechanism for the reaction of \(\ce{NO2}\) with \(\ce{CO}\) with \(\ce{N2O4}\) appearing as an intermediate. Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate = [NO ] )? elementary reactions rate law for each elementary reaction and overall rate law The rate law for step 1 is rate = [NO ] ; for step 2, it is rate = [N O ,CO]. If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate = [NO ] . This is the same as the experimentally determined rate law. Hence this mechanism, with N O as an intermediate, and the one described previously, with NO as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO and N O , directly. Iodine monochloride (\(\ce{ICl}\)) reacts with \(\ce{H2}\) as follows: \[\ce{2ICl(l) + H_2(g) \rightarrow 2HCl(g) + I_2(s)} \nonumber \] The experimentally determined rate law is \(rate = k[\ce{ICl},\ce{H2}]\). Write a two-step mechanism for this reaction using only bimolecular elementary reactions and show that it is consistent with the experimental rate law. (Hint: \(\ce{HI}\) is an intermediate.) This mechanism is consistent with the experimental rate law if the first step is the rate-determining step. Assume the reaction between \(\ce{NO}\) and \(\ce{H_2}\) occurs via a three-step process: Write the rate law for each elementary reaction, write the balanced chemical equation for the overall reaction, and identify the rate-determining step. Is the rate law for the rate-determining step consistent with the experimentally derived rate law for the overall reaction: \[\text{rate} = k[\ce{NO}]^2[\ce{H_2}]? \tag{observed} \] The overall reaction is then \[\ce{2NO(g) + 2H_2(g) -> N2(g) + 2H2O(g)} \nonumber \] Reaction Mechanism (Slow step followed by fast step): A balanced chemical reaction does not necessarily reveal either the individual elementary reactions by which a reaction occurs or its rate law. A reaction mechanism is the microscopic path by which reactants are transformed into products. Each step is an elementary reaction. Species that are formed in one step and consumed in another are intermediates. Each elementary reaction can be described in terms of its , the number of molecules that collide in that step. The slowest step in a reaction mechanism is the rate-determining step. | 9,514 | 3,419 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.14%3A_Dalton's_Law_of_Partial_Pressures |
The atmosphere of Venus is markedly different from that of Earth. The gases in the Venusian atmosphere are \(96.5\%\) carbon dioxide and \(3\%\) nitrogen. The atmospheric pressure on Venus is roughly 92 times that of Earth, so the amount of nitrogen on Venus contributes pressure well over \(2700 \: \text{mm} \: \ce{Hg}\). And there is no oxygen present, so humans could not breathe there. Not that anyone would want to go to Venus—the surface temperature is usually over \(460^\text{o} \text{C}\). Gas pressure results from collisions between gas particles and the inside walls of their container. If more gas is added to a rigid container, the gas pressure increases. The identities of the two gases do not matter. John Dalton, the English chemist who proposed the atomic theory, also studied mixtures of gases. He found that each gas in a mixture exerts a pressure independently of every other gas in the mixture. For example, our atmosphere is composed of about \(78\%\) nitrogen and \(21\%\) oxygen, with smaller amounts of several other gases making up the rest. Since nitrogen makes up \(78\%\) of the gas particles in a given sample of air, it exerts \(78\%\) of the pressure. If the overall atmospheric pressure is \(1.00 \: \text{atm}\), then the pressure of just the nitrogen in the air is \(0.78 \: \text{atm}\). The pressure of the oxygen in the air is \(0.21 \: \text{atm}\). The of a gas is the contribution that gas makes to the total pressure when the gas is part of a mixture. The partial pressure of nitrogen is represented by \(P_{N_2}\). states that the total pressure of a mixture of gases is equal to the sum of all of the partial pressures of the component gases. Dalton's law can be expressed with the following equation: \[P_\text{total} = P_1 + P_2 + P_3 + \cdots\nonumber \] The figure below shows two gases that are in separate, equal-sized containers at the same temperature and pressure. Each exerts a different pressure, \(P_1\) and \(P_2\), reflective of the number of particles in the container. On the right, the two gases are combined into the same container, with no volume change. The total pressure of the gas mixture is equal to the sum of the individual pressures. If \(P_1 = 300 \: \text{mm} \: \ce{Hg}\) and \(P_2 = 500 \: \text{mm} \: \ce{Hg}\), then \(P_\text{total} = 800 \: \text{mm} \: \ce{Hg}\). | 2,370 | 3,420 |
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