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https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/09%3A_Titrimetric_Methods/9.08%3A_Chapter_Summary_and_Key_Terms |
In a titrimetric method of analysis, the volume of titrant that reacts stoichiometrically with a titrand provides quantitative information about the amount of analyte in a sample. The volume of titrant that corresponds to this stoichiometric reaction is called the equivalence point. Experimentally we determine the titration’s end point using an indicator that changes color near the equivalence point. Alternatively, we can locate the end point by monitoring a property of the titrand’s solution—absorbance, potential, and temperature are typical examples—that changes as the titration progresses. In either case, an accurate result requires that the end point closely match the equivalence point. Knowing the shape of a titration curve is critical to evaluating the feasibility of a titrimetric method. Many titrations are direct, in which the analyte participates in the titration as the titrand or the titrant. Other titration strategies are possible when a direct reaction between the analyte and titrant is not feasible. In a back titration a reagent is added in excess to a solution that contains the analyte. When the reaction between the reagent and the analyte is complete, the amount of excess reagent is determined by a titration. In a displacement titration the analyte displaces a reagent, usually from a complex, and the amount of displaced reagent is determined by an appropriate titration. Titrimetric methods have been developed using acid–base, complexation, oxidation–reduction, and precipitation reactions. Acid–base titrations use a strong acid or a strong base as a titrant. The most common titrant for a complexation titration is EDTA. Because of their stability against air oxidation, most redox titrations use an oxidizing agent as a titrant. Titrations with reducing agents also are possible. Precipitation titrations often involve Ag as either the analyte or titrant. acid–base titration argentometric titration auxiliary oxidizing agent buret direct titration equivalence point Gran plot Kjeldahl analysis Mohr method redox indicator symmetric equivalence point titrant titrimetry acidity asymmetric equivalence point auxiliary reducing agent complexation titration displacement titration Fajans method indicator leveling potentiometric titration redox titration thermometric titration titration curve Volhard method alkalinity auxiliary complexing agent back titration conditional formation constant end point formal potential Jones reductor metallochromic indicator precipitation titration spectrophotometric titration titrand titration error Walden reductor | 2,603 | 3,691 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Geochemistry_(Lower)/03%3A_The_Atmosphere/3.01%3A_Structure_and_Composition_of_the_Atmosphere |
Life as we know it on the Earth is entirely dependent on the tenuous layer of gas that clings to the surface of the globe, adding about 1% to its diameter and an insignificant amount to its total mass. And yet the atmosphere serves as the earth’s window and protective shield, as a medium for the transport of heat and water, and as source and sink for exchange of carbon, oxygen, and nitrogen with the biosphere. The atmosphere acts as a compressible fluid tied to the earth by gravitation; as a receptor of solar energy and a thermal reservoir, it constitutes the working fluid of a heat engine that transports and redistributes matter and energy over the entire globe. The atmosphere is also a major temporary repository of a number of chemical elements that move in a cyclic manner between the hydrosphere, atmosphere, and the upper lithosphere. Finally, the atmosphere is a site for a large variety of complex photochemically initiated reactions involving both natural and anthropogenic substances. On the scale of cubic meters the air is a homogeneous mixture of its constituent gases, but on a larger scale the atmosphere is anything but uniform. Variations of temperature, pressure, and moisture content in the layers of air near the earth’s surface give rise to the dynamic effects we know as the weather. Although the density of the atmosphere decreases without limit with increasing height, for most practical purposes one can roughly place its upper boundary at about 500 km. However, half the mass of the atmosphere lies within 5 km, and 99.99% within 80 km of the surface. The average atmospheric pressure at sea level is 1.01 x 10 pascals, or 1010 millibars. A 1-cm cross section of the earth’s surface supports a column weighing 1030 g; the total mass of the atmosphere is about 5.27 x 10 g. About 80% of the mass of the atmosphere resides in the first 10 km; this well-mixed region of fairly uniform composition is known as the . The gases ozone, water vapor, and carbon dioxide are only minor components of the atmosphere, but they exert a huge effect on the Earth by absorbing radiation in the ranges indicated by the shading. Ozone in the upper atmosphere filters out the ultraviolet light below about 360 nm that is destructive of life. O , H O, CO and CH are "greenhouse" gases that trap some of the heat absorbed from the Sun and prevent it from re-radiating back into space. We commonly think of gas molecules as moving about in a completely random manner, but the Earth’s gravitational field causes downward motions to be very slightly favored so that the molecules in any thin layer of the air collide more frequently with those in the layer below. This gives rise to a pressure gradient that is the most predictable and well known structural characteristic of the atmosphere. This gradient is described by an exponential law which predicts that the atmospheric pressure should decrease by 50% for every 6 km increase in altitude. This law also predicts that the composition of a gas mixture will change with altitude, the lower-molecular weight components being increasingly favored at higher altitudes. However, this gravitational fractionation effect is completely obliterated below about 160 km owing to turbulence and convective flows (winds). The atmosphere is divided vertically into several major regions which are distinguished by the sign of the temperature gradient. In the lowermost region, the troposphere, the temperature falls with increasing altitude. The major source of heat input into this part of the atmosphere is long-wave radiation from the earth’s surface, while the major loss is radiation into space. At higher elevations the temperature begins to rise with altitude as we move into a region in which heat is produced by exothermic chemical reactions, mainly the decomposition of ozone that is formed photochemically from dioxygen in the stratosphere. At still higher elevations the ozone gives out and the temperature begins to drop; this is the mesosphere, which is finally replaced by the thermosphere which consists largely of a plasma (gaseous ions). This outer section of the atmosphere which extends indefinitely to perhaps 2000 km is heated by absorption of intense u.v. radiation from the Sun and also from the solar wind, a continual rain of electrons, protons, and other particles emitted from the Sun’s surface. Except for water vapor, whose atmospheric abundance varies from practically zero up to 4%, the fractions of the major atmospheric components N , O , and Ar are remarkably uniform below about 100 km. At greater heights, diffusion becomes the principal transport process, and the lighter gases become relatively more abundant. In addition, photochemical processes result in the formation of new species whose high reactivities would preclude their existence in significant concentrations at the higher pressures found at lower elevations. The atmospheric gases fall into three abundance categories: major, minor, and trace. Nitrogen, the most abundant component, has accumulated over time as a result of its geochemical inertness; a very small fraction of it passes into the other phases as a result of biological activity and natural fixation by lightning. It is believed that denitrifying bacteria in marine sediments may provide the major route for the return of N to the atmosphere. Oxygen is almost entirely of biological origin, and cycles through the hydrosphere, the biosphere, and sedimentary rocks. Argon consists mainly of Ar which is a decay product of K in the mantle and crust. The most abundant of the minor gases aside from water vapor is carbon dioxide, about which more will be said below. Next in abundance are neon and helium. Helium is a decay product of radioactive elements in the earth, but neon and the other inert gases are primordial, and have probably been present in their present relative abundances since the earth’s formation. Two of the minor gases, ozone and carbon monoxide, have abundances that vary with time and location. A variable abundance implies an imbalance between the rates of formation and removal. In the case of carbon monoxide, whose major source is anthropogenic (a small amount is produced by biological action), the variance is probably due largely to localized differences in fuel consumption, particularly in internal combustion engines. The nature of the carbon monoxide sink (removal mechanism) is not entirely clear; it may be partly microbial. Ozone is formed by the reaction of O with oxygen atoms produced photochemically. As a consequence the abundance of ozone varies with the time of day, the concentration of O atoms from other sources (photochemical smog, for example), and particularly with altitude; at 30 km, the ozone concentration reaches a maximum of 12 ppm. The concentration of atmospheric carbon dioxide, while fairly uniform globally, is increasing at a rate of 0.2-0.7% per year as a result of fossil fuel burning. The present CO content of the atmosphere is about 129 10 g. Most of the CO , however, is of natural origin, and represents the smallest part of the total carbonate reservoir that includes oceanic CO , HCO , and carbonate sediments. The latter contain about 600 times as much CO as the atmosphere, and the oceans contain about 50 times as much. These relative amounts are controlled by the rates of the reactions that interconvert the various forms of carbonate. The surface conditions on the earth are sensitively dependent on the atmospheric CO concentration. This is due mainly to the strong infrared absorption of CO , which promotes the absorption and trapping of solar heat (see below). Since CO acts as an acid in aqueous solution, the pH of the oceans is also dependent on the concentration of CO in the atmosphere; it has been estimated that if only 1% of the carbonate presently in sediments were still in the atmosphere, the pH of the oceans would be 5.9, instead of the present 8.2. The amount of energy (the solar flux) impinging on the outer part of the atmosphere is 1367 watts m . About 30% of this is reflected or scattered back into space by clouds, dust, and the atmospheric gas molecules themselves, and by the earth’s surface. About 19% of the radiation is absorbed by clouds or the atmosphere (mainly by and O , but not CO ), leaving 51% of the incident energy available for absorption by the earth’s surface. If one takes into account the uneven illumination of the earth’s surface and the small flux of internal heat to the surface, the assumption of thermal equilibrium requires that the earth emit about 240 watts m . This corresponds to the power that would be emitted by a black body at 255 K, or –18°C, which is the average temperature of the atmosphere at an altitude of 5 km. The observed mean global surface temperature of the earth is 13°C, and is presumably the temperature required to maintain thermal equilibrium between the earth and the atmosphere. The energy radiated by the earth has a longer wavelength (maximum 12 ) than the incident radiation. Most gases absorb radiation in this range quite efficiently, including those gases such as CO and N O that do not absorb the incident radiation. The energy absorbed by atmospheric gases is re-radiated in all directions; some of it therefore escapes into space, but a portion returns to the earth and is reabsorbed, thus raising its temperature.This is commonly called the greenhouse effect. If the amount of an infrared-absorbing gas such as carbon dioxide increases, a larger fraction of the incident solar radiation is trapped, and the mean temperature of the earth will increase. Any significant increase in the temperature of the oceans would increase the atmospheric concentrations of both water and CO , producing the possibility of a runaway process that would be catastrophic from a human perspective. Fossil fuel combustion and deforestation during the last two hundred years have increased the atmospheric CO concentration by 25%, and this increase is continuing. The same combustion processes responsible for the increasing atmospheric CO concentration also introduce considerable quantities of particulate materials into the upper atmosphere. The effect of these would be to scatter more of the incoming solar radiation, reducing the amount that reaches and heats the earth’s surface. The extent to which this process counteracts the greenhouse effect is still a matter of controversy; all that is known for sure is that the average temperature of the Earth is increasing. Carbon dioxide is not the only atmospheric gas of anthropogenic origin that can affect the heat balance of the earth; other examples are SO and N O. Nitrous oxide is of particular interest, since its abundance is fairly high, and is increasing at a rate of about 0.5% per year. It is produced mainly by bacteria, and much of the increase is probably connected with introduction of increased nitrate into the environment through agricultural fertilization and sewage disposal. Besides being a strong infrared absorber, N O is photochemically active, and can react with ozone. Any significant depletion of the ozone content of the upper atmosphere would permit more ultraviolet radiation to reach the earth. This would have numerous deleterious effects on present life forms, as well as contributing to a temperature increase. The warming effect attributed to anthropogenic additions of greenhouse gases to the atmosphere is estimated to be about 2 watts per m , or about 1.5% of the 150 watts per m trapped by clouds and atmospheric gases. This is a relatively large perturbation compared to the maximum variation in solar output of 0.5 watts per m that has been observed during the past century. Continuation of greenhouse gas emission at present levels for another century could increase the atmospheric warming effect by 6-8 watts per m . A less-appreciated side effect of the increase in atmospheric carbon dioxide (and of other plant nutrients such as nitrates) may be reduction in plant species diversity by selectively encouraging the growth of species which are ordinarly held in check by other species that are able to grow well with fewer nutrients. This effect, for which there is already some evidence, could be especially pronounced when the competing species utilize the C and C photosynthetic pathways that differ in their sensitivity to CO . ) | 12,439 | 3,692 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Averill_et_al./15.E%3A_Chemical_Equilibrium_(Exercises) |
" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact 1. ♦ The total concentrations of dissolved Al in a soil sample represent the sum of “free” Al and bound forms of Al that are stable enough to be considered definite chemical species. The distribution of aluminum among its possible chemical forms can be described using equilibrium constants such as the following: a. Write an equilibrium equation for each expression. b. Which anion has the highest affinity for Al : OH , SO , or F ? Explain your reasoning. c. A 1.0 M solution of Al is mixed with a 1.0 M solution of each of the anions. Which mixture has the lowest Al concentration? 2. Many hydroxy acids form lactones (cyclic esters) that contain a 5- or 6-membered ring. Common hydroxy acids found in nature are glycolic acid, a constituent of cane sugar juice; lactic acid, which has the characteristic odor and taste of sour milk; and citric acid, found in fruit juices. The general reaction for lactone formation can be written as follows: Use the information in the following table to calculate the equilibrium constant for lactone formation for each hydroxy acid given and determine which ring size is most stable. 3. ♦ Phosphorus pentachloride, an important reagent in organic chemistry for converting alcohols to alkyl chlorides (ROH → RCl), is hydrolyzed in water to form phosphoric acid and hydrogen chloride. In the gaseous state, however, PCl can decompose at 250°C, according to PCl (g)⇌PCl (g)+Cl (g), for which K = 0.0420. a. Are products or reactants favored in the decomposition of PCl (g)? b. If a 2.00 L flask containing 104.1 g PCl is heated to 250°C, what is the equilibrium concentration of each species in this reaction? c. What effect would an increase in pressure have on the equilibrium position? Why? d. If a 1.00 × 10 L vessel containing 2.00 × 10 kg of PCl with a constant chlorine pressure of 2.00 atm is allowed to reach equilibrium, how many kilograms of PCl are produced? What is the percent yield of PCl ? 4. ♦ Carbon disulfide (CS ) is used in the manufacture of rayon and in electronic vacuum tubes. However, extreme caution must be used when handling CS in its gaseous state because it is extremely toxic and can cause fatal convulsions. Chronic toxicity is marked by psychic disturbances and tremors. CS is used to synthesize H S at elevated T via the following reaction: CS (g)+4H (g)⇌CH (g)+2H S(g) K=3.3×10 a. If the equilibrium concentration of methane in this reaction is 2.5 × 10 M and the initial concentration of each reactant is 0.1635 M, what is the concentration of H S at equilibrium? b. Exposure to CS concentrations greater than 300 ppm for several hours can start to produce adverse effects. After working for several hours in a laboratory that contains large quantities of CS , you notice that the fume hoods were off and there was not enough ventilation to remove any CS vapor. Given the equilibrium CS (l)⇌CS (g), where T = 20°C and K = 0.391, determine whether you are in any danger. 5. ♦ Chloral hydrate, a sedative commonly referred to as “knockout drops,” is in equilibrium with trichloroacetaldehyde in highly concentrated aqueous solutions: The equilibrium constant for this reaction as written is 3 × 10 . Are the products or the reactants favored? Write an equilibrium expression for this reaction. How could you drive this reaction to completion? 6. Hydrogen cyanide is commercially produced in the United States by the following reaction: , where HCN is continuously removed from the system. This reaction is carried out at approximately 1100°C in the presence of a catalyst; however, the high temperature causes other reactions to occur. Why is it necessary to run this reaction at such an elevated temperature? Does the presence of the catalyst affect the equilibrium position? 7. ♦ Hemoglobin transports oxygen from the lungs to the capillaries and consists of four subunits, each capable of binding a single molecule of O . In the lungs, PO is relatively high (100 mmHg), so hemoglobin becomes nearly saturated with O . In the tissues, however, PO is relatively low (40 mmHg), so hemoglobin releases about half of its bound oxygen. Myoglobin, a protein in muscle, resembles a single subunit of hemoglobin. The plots show the percent O saturation versus PO for hemoglobin and myoglobin. Based on these plots, which molecule has the higher affinity for oxygen? What advantage does hemoglobin have over myoglobin as the oxygen transporter? Why is it advantageous to have myoglobin in muscle tissue? Use equilibrium to explain why it is more difficult to exercise at high altitudes where the partial pressure of oxygen is lower. 8. ♦ Sodium sulfate is widely used in the recycling industry as well as in the detergent and glass industries. This compound combines with H SO via Na SO +H SO ⇌2NaHSO . Sodium hydrogen sulfate is used as a cleaning agent because it is water soluble and acidic. a. Write an expression for K for this reaction. b. Relate this equilibrium constant to the equilibrium constant for the related reaction: 2Na SO +2H SO ⇌4NaHSO . c. The dissolution of Na SO in water produces the equilibrium reaction SO +H O ⇌HSO +OH where K = 8.33 × 10 . 11. What is the concentration of OH in a solution formed from the dissolution of 1.00 g of sodium sulfate to make 150.0 mL of aqueous solution? Neglect the autoionization of water in your answer. 9. ♦ One of the Venera orbiter satellites measured S concentrations at the surface of Venus. The resulting thermochemical data suggest that S formation at the planet’s surface occurs via the following equilibrium reaction: 4CO +2SO ⇌4CO +S Write an expression for K for this reaction and then relate this expression to those for the following reactions: a. 2CO + SO ⇌2CO +12S b. CO(g) +12SO ⇌CO +14S c. At 450°C, the equilibrium pressure of CO is 85.0 atm, SO is 1.0 atm, CO is 1.0 atm, and S is 3.0 × 10 atm. What are K and K at this temperature? What is the concentration of S ? 10. ♦ Until the early part of the 20th century, commercial production of sulfuric acid was carried out by the “lead-chamber” process, in which SO was oxidized to H SO in a lead-lined room. This process may be summarized by the following sequence of reactions: 1) NO + NO +2H SO ⇌2NOHSO +H O K 2) 2NOHSO +SO +2H O ⇌3H SO +2NO K a. Write the equilibrium constant expressions for reactions 1 and 2 and the sum of the reactions (reaction 3). b. Show that K = K × K . c. If insufficient water is added in reaction 2 such that the reaction becomes NOHSO +12SO +H O ⇌32H SO +NO ,does K still equal K × K ? d. Based on part c, write the equilibrium constant expression for K . 11. Phosgene (carbonic dichloride, COCl ) is a colorless, highly toxic gas with an odor similar to that of moldy hay. Used as a lethal gas in war, phosgene can be immediately fatal; inhalation can cause either pneumonia or pulmonary edema. For the equilibrium reaction COCl ⇌CO +Cl , K is 0.680 at −10°C. If the initial pressure of COCl is 0.681 atm, what is the partial pressure of each component of this equilibrium system? Is the formation of products or reactant favored in this reaction? 12. ♦ British bituminous coal has a high sulfur content and produces much smoke when burned. In 1952, burning of this coal in London led to elevated levels of smog containing high concentrations of sulfur dioxide, a lung irritant, and more than 4000 people died. Sulfur dioxide emissions can be converted to SO and ultimately to H SO , which is the cause of acid rain. The initial reaction is 2SO +O ⇌2SO , for which K = 44. a. Given this K , are product or reactants favored in this reaction? b. What is the partial pressure of each species under equilibrium conditions if the initial pressure of each reactant is 0.50 atm? c. Would an increase in pressure favor the formation of product or reactants? Why? 13. Oxyhemoglobin is the oxygenated form of hemoglobin, the oxygen-carrying pigment of red blood cells. Hemoglobin is built from α and β protein chains. Assembly of the oxygenated (oxy) and deoxygenated (deoxy) β-chains has been studied with the following results: Is it more likely that hemoglobin β chains assemble in an oxygenated or deoxygenated state? Explain your answer. 14. ♦ Inorganic weathering reactions can turn silicate rocks, such as diopside (CaMgSi O ), to carbonate via the following reaction: Write an expression for the equilibrium constant. Although this reaction occurs on both Earth and Venus, the high surface temperature of Venus causes the reaction to be driven in one direction on that planet. Predict whether high temperatures will drive the reaction to the right or the left and then justify your answer. The estimated partial pressure of carbon dioxide on Venus is 85 atm due to the dense Venusian atmosphere. How does this pressure influence the reaction? 15. Silicon and its inorganic compounds are widely used to manufacture textile glass fibers, cement, ceramic products, and synthetic fillers. Two of the most important industrially utilized silicon halides are SiCl and SiHCl , formed by reaction of elemental silicon with HCl at temperatures greater than 300°C: Which of these two reactions is favored by increasing [HCl]? by decreasing the volume of the system? 16. ♦ The first step in the utilization of glucose in humans is the conversion of glucose to glucose-6-phosphate via the transfer of a phosphate group from ATP (adenosine triphosphate), which produces glucose-6-phosphate and ADP (adenosine diphosphate): a. Is the formation of products or reactants favored in this reaction? b. Would K increase, decrease, or remain the same if the glucose concentration were doubled? c. If −RT ln K = −RT′ ln K', what would K be if the temperature were decreased to 0°C? d. Is the formation of products favored by an increase or a decrease in the temperature of the system? 17. In the presence of O , the carbon atoms of glucose can be fully oxidized to CO with a release of free energy almost 20 times greater than that possible under conditions in which O is not present. In many animal cells, the TCA cycle (tricarboxylic acid cycle) is the second stage in the complete oxidation of glucose. One reaction in the TCA cycle is the conversion of citrate to isocitrate, for which K = 0.08 in the forward direction. Speculate why the cycle continues despite this unfavorable value of K. What happens if the citrate concentration increases? 18. ♦ Soil is an open system, subject to natural inputs and outputs that may change its chemical composition. Aqueous-phase, adsorbed, and solid-phase forms of Al(III) are of critical importance in controlling the acidity of soils, although industrial effluents, such as sulfur and nitrogen oxide gases, and fertilizers containing nitrogen can also have a large effect. Dissolution of the mineral gibbsite, which contains Al in the form Al(OH) , occurs in soil according to the following reaction: When gibbsite is in a highly crystalline state, K = 9.35 for this reaction at 298 K. In the microcrystalline state, K = 8.11. Is this change consistent with the increased surface area of the microcrystalline state? Answers 1. a. b. c. 2. 3. a. reactant b. [Cl2] = [PCl3] = 0.0836 M; [PCl5] = 0.166 M c. increasing pressure favors reactant (PCl5) d. 1.59 × 103 kg; 52.5% 4. a. b. 5. Products are favored; K = [chloral hydrate] / [Cl3CHO,H2O] ; high concentrations of water will favor chloral hydrate formation. 6. 7. 8 a. b. c. 9. K = {[CO ] [S ] } / {[CO] [SO ] } a.K' = {[CO ] [S ] } / {[CO] [SO ] } = K b.K'' = {[CO ] [S ] } / {[CO] [SO ] } = K c.. K = 1.6; K = 93; [S ] = 5.1 × 10 M 10. a. b. c. d. 11. P = PCl = 0.421 atm; P = 0.260 atm; reactants are slightly favored. 12. a. b. c. 13. 14. 15. Both reactions are favored by increasing [HCl] and decreasing volume. 16. a. b. c. d. 17. 18, 19. | 12,021 | 3,694 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Electrochemistry_and_Thermodynamics |
The connection between cell potential, Gibbs energy and constant equilibrium are directly related in the following multi-part equation: \[ \Delta G^o= -RT\ln K_{eq} = -nFE^o_{cell} \] ∆G is the change of Gibbs (free) energy for a system and ∆G° is the Gibbs energy change for a system under standard conditions (1 atm, 298K). On an energy diagram, ∆G can be represented as: Where ∆G is the difference in the energy between reactants and products. In addition ∆G is unaffected by external factors that change the kinetics of the reaction. For example if E (activation energy) were to decrease in the presence of a catalyst or the kinetic energy of molecules increases due to a rise in temperature, the ∆G value would remain the same. E° is the electromotive force (also called cell voltage or cell potential) between two half-cells. The greater the E° of a reaction the greater the driving force of electrons through the system, the more likely the reaction will proceed (more spontaneous). E° is measured in volts (V). The overall voltage of the cell = the half-cell potential of the reduction reaction + the half-cell potential of the oxidation reaction. To simplify, \[E^o_{cell} = E^o_{reduction} - E^o_{oxidation} \label{1}\] or \[E^o_{cell} = E^o_{cathode} - E^o_{anode} \label{2}\] The potential of an oxidation reduction (loss of electron) is the negative of the potential for a reduction potential (gain of electron). Most tables only record the standard reduction half-reactions as To find the standard oxidation potential, simply reverse the sign of the standard reduction potential. The more positive reduction potential of reduction reactions are more spontaneous. When viewing a cell reduction potential table, the higher the cell is on the table, the higher potential it has as an oxidizing agent. Eº cell is the standard state cell potential, which means that the value was determined under standard states. The standard states include a concentration of 1 Molar (mole per liter) and an atmospheric pressure of 1. Similar to the standard state cell potential, Eº , the E is the non-standard state cell potential, which means that it is not determined under a concentration of 1 M and pressure of 1 atm. The two are closely related in the sense that the standard cell potential is used to calculate for the cell potential in many cases. \[ E_{cell}= E^o_{cell} -\dfrac{RT}{nF} \ln\; Q \label{3}\] Other simplified forms of the equation that we typically see: \[ E_{cell}= E^o_{cell} -\dfrac{0.0257}{n} \ln \; Q \label{4}\] or in terms of \(\log_{10}\) (base 10) instead of the natural logarithm (base e) \[ E_{cell}= E^o_{cell} - \dfrac{0.0592}{n} \log_{10}\; Q \label{5}\] Both equations applies when the temperature is 25ºC. Deviations from 25ºC requires the use of the original equation. Essentially, Eº is E at standard conditions What is the value of \(E_{cell}\) for the voltaic cell below: \[\ce{Pt(s)} | \ce{Fe^{2+}}(0.1\, M), \ce{Fe^{3+}}(0.2\, M) || \ce{Ag^{+}} (0.1\, M)| \ce{Ag(s)} \nonumber\] To use the , we need to establish \(E^o_{cell}\) and the reaction to which the cell diagram corresponds so that the form of the reaction quotient (Q) can be revealed. Once we have determined the form of the Nernst equation, we can insert the concentration of the species. Solve: \[ \begin{align*} E^o_{cell} &= E^o_{SRP}(cathode - E^o_{SRP}(anode)\\[4pt] &= E^o(Ag^{+}/Ag) - E^o(Fe^{3+}/Fe^{2+}) \\[4pt] &= 0.800\,V - 0.771\,V \\[4pt] &= 0.029\, V \end{align*}\] Now to determine \(E_{cell}\) for the reaction \[\ce{Fe^{2+}}(0.1\,M) + \ce{Ag^{+}}(1.0\,M) → \ce{Fe^{3+}}(0.20\,M) + \ce{Ag(s)} \nonumber\] Use the Nernst equation \[\begin{align*} E_{cell} &= 0.029\,V - \left(\dfrac{0.0592\,V}{1}\right) \log \dfrac{[Fe^{3+}]}{[Fe^{2+},Ag]} \\[4pt] &=0.029\,V - 0.0592\,V \log \dfrac{0.2}{(0.1)(0.1)} \\[4pt] &=-0.048\,V \end{align*}\] Note that this reaction is spontaneous (positive \(E_{cell}\)) as written under standard conditions, but is non-spontaneous (negative \(E_{cell}\)) under the non-standard conditions of question. \(K\) is the equilibrium constant of a general reaction \[ aA + bB \leftrightharpoons cC + dD \label{6} \] and is expressed by the reaction quotient: \[ K_c= \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{7}\] Given \(K = 2.81 \times 10^{-16}\) for a following reaction \[\ce{Cu^{2+}(aq) + Ag(s) \rightleftharpoons Cu(s) + 2Ag^{+}} \nonumber\] Find ∆G. Use the following formula: \[\begin{align*} ∆G &=-RT\ln K \\[4pt] &= (8.314) (298\,K) \ln (2.81 \times 10^{-16}) \\[4pt] &= -8.87 \times 10^5 J/mole \\[4pt] &= 8.871 \,kJ/mol \end{align*}\] The relationship between \(∆G\), \(K\), and \(E^°_{cell}\) can be represented by the following diagram. where \(E^o_{cell}\) can be calculated using the following formula: \[E^o_{cell} = E^o_{cathode} – E^o_{anode} = E^o_{Reduction} – E^o_{Oxidation} \label{8}\] Find the \(E^o_{cell}\) for the following coupled half-reactions 1. Determine the cathode and anode in the reaction \[\ce{Zn(s) <=> Zn^{2+}(aq) + 2e^{-}} \nonumber\] Anode, Oxidation half-reaction (since \(\ce{Zn(s)}\) increase oxidation state from 0 to +2) \[\ce{Cu^{2+}(aq) + 2e^{-} <=> Cu(s)} \nonumber\] Cathode, Reduction half-rection (since \(\ce{Cu^{2+}(aq)}\) decreases oxidation state from +2 to 0) 2. Determine the \(E^o_{cell}\) values using the standard reduction potential table ( ) \[\ce{Zn(s) <=> Zn^{2+}(aq) + 2e^{-}} \quad\quad E_{SRP} = -0.763 \nonumber\] \[\ce{Cu^{2+}(aq) + 2e^{-} <=> Cu(s)} \quad\quad E_{SRP}=+0.340 \nonumber\] 3. Use \[\begin{align*} E^°_{cell} &= E^°_{SRP}(\text{cathode}) - E^o_{SRP}(\text{anode}) \\[4pt] &= 0.340 \,V - (-0.763\, V) \\[4pt] &= 1.103\, V \end{align*}\] Find ∆G for the following reaction: \[\ce{2Al(s) + 3Br2(l) <=> 2Al^{3+}}(aq, 1.0\,M) + \ce{6Br^{-}}(aq, 1.0\,M) \nonumber\] Separate the reaction into its two half reactions \[\ce{2Al(s) <=> 2Al^{3+}(aq)} \nonumber\] \[\ce{3Br2(l) <=> 6Br^{-}(aq, 1\,M)} \nonumber\] Balance the half equations using O, H, and charge using e- 2Al ↔ 2Al +6e 6e + 3Br ↔ 6Br From the balanced half reactions, we can conclude the number of moles of e for use later in the calculation of ∆G. Determine the E° values using the standard reduction potentials, using the E° cell table. 2Al ↔ Al +6e- -1.676V 3Br + 6e ↔ 6Br +1.065V : Determine = 1.065 - (-1.676) Once E° cell has be calculated and the number of moles of electrons have been determined, we can use = (-6 mol e )(96485 C/mol e )(2.741 V) = -1586kJ This equation can be used to calculate E° cell given K or K given \(E^o_{cell}\). If T=298 K, the RT is a constant then the following equation can be used: Given the \(E^°_{cell}\) for the reaction \[\ce{Cu(s) + 2H^{+}(aq) \rightleftharpoons Cu^{2+}(aq) +H2(g)} \nonumber\] is -0.34V, find the equilibrium constant (\(K\)) for the reaction. Split into two half reaction \[\ce{Cu(s) <=> Cu^{2+}(aq)} \nonumber\] \[\ce{2H^{+}(aq) <=> H2(g)} \nonumber\] Balance the half reactions with charges to determine n Cu ↔ Cu + 2e 2H +2e ↔ H From the example above, \(E^°_{cell} = -0.34\,V\) \[\begin{align*} -0.34 &= \left(\dfrac{0.025693}{2} \right) \ln K \\[4pt] K &= \exp \left(\dfrac{-0.34 \times 2}{0.025693}\right) \\[4pt] K &= 3.19 \times 10^{-12} \end{align*}\] Given the following reaction determine \(∆G\), \(K\), and \(E^o_{cell}\) for the following reaction at standard conditions? Is this reaction spontaneous? \[\ce{Mn^{2+}(aq) + K(s) <=> MnO2(s) + K(aq)} \nonumber\] : Separate and balance the half reactions. Label which one is reduction and which one is oxidation. Find the corresponding \(E^o\) values for the half reactions. \[\ce{MnO2(s) + 4H^{+}(aq) + 2e^{-} <=> Mn^{2+}(aq) + 2H2O(l)} \nonumber\] Reduction with \(E_{SRP}\) of +1.23V \[\ce{K^{+}(aq) + e^{-} <=> K(s)} \nonumber\] Oxidation with \(E_{SRP}\) of -2.92V Write net balanced reaction in acidic solution, and determine the E° cell. \[\begin{align*} E^o_{cell} &= E^o_{SRP}(cathode) - E^{°}_{SRP} (anode) \\[4pt] &= +1.23\,V - (-2.92\,V) = 4.15\,V \end{align*}\] Mn + 2K + 2H O ↔ + 4H + K E° cell = 4.15V Find ∆G for the reaction. Use ∆G = -nFE° = -2 e x 96485 C x 4.15 = -800.60kJ Therefore, since \(E^o_{cell}\) is positive and \(∆G\) is negative, this reaction is spontaneous. Find \(E^o_{cell}\) for \(\ce{2Br^{-}(aq) + I2(s) <=> Br2(s) + 2I^{-}(aq)}\) +0.530 V Find \(E^o\) for \(\ce{Sn(s) <=> Sn^{2+}(aq) + 2e^{-}}\) +0.137 V Find \(E^o\) cell for \(\ce{Zn(s) | Zn^{2+}(aq) || Cr^{3+}(aq), Cr^{2+}(aq)}\) +0.339 V Find ∆G for the following combined half reactions: \[\ce{F2(g) + 2e^{-} <=> 2F^{-}(aq)} \nonumber\] \[\ce{Li^{+}(aq) + e^{-} <=> Li(s)} \nonumber\] +1139.68 kJ Find the equilibrium constant (\(K\)) for the following reaction: \[\ce{Zn(s) + 2H^{+}(aq) <=> Zn^{2+}(aq) + H2(g)} \nonumber\] Find \(E^o_{cell}\) first! 6.4 x 10 Find \(E^o_{cell}\) for the given reaction at standard conditions: \[\ce{Cu^{+}(aq) + e^{-} <=> Cu(s)} \nonumber\] \[\ce{I2(s) + 2e^{-} <=> 2I^{-}(aq)} \nonumber\] +0.195 V | 8,986 | 3,698 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity_of_Alpha_Hydrogens/Alpha_Halogenation |
A carbonyl containing compound with \(\alpha\) hydrogens can undergo a substitution reaction with halogens. This reaction comes about because of the tendency of carbonyl compounds to form enolates in basic condition and enols in acidic condition. In these cases even weak bases, such as the hydroxide anion, is sufficient enough to cause the reaction to occur because it is not necessary for a complete conversion to the enolate. For this reaction Cl , Br or I can be used as the halogens. General reaction Under acidic conditions the reaction occurs thought the formation of an enol which then reacts with the halogen. Step 1: Protonation of the carbonyl Step 2: Enol formation Step 3: S 2 attack Step 4: Deprotonation Under basic conditions the enolate forms and then reacts with the halogen. Note! This is base promoted and not base catalyzed because an entire equivalent of base is required. Step 1: Enolate formation Step 2: S 2 attack The fact that an electronegative halogen is placed on an α carbon means that the product of a base promoted α halogenation is actually more reactive than the starting material. The electron withdrawing effect of the halogen makes the α carbon even more acidic and therefor promotes further reaction. Because of this multiple halogenations can occur. This effect is exploited in the haloform reaction discussed later. If a monohalo product is required then acidic conditions are usually used. Methyl ketones typically undergo halogenation three times to give a trihalo ketone due to the increased reactivity of the halogenated product as discussed above. This trihalomethyl group is an effective leaving group due to the three electron withdrawing halogens and can be cleaved by a hydroxide anion to effect the haloform reaction. The product of this reaction is a carboxylate and a haloform molecule (CHCl , CHBr , CHI ). Overall the haloform reaction represents an effective method for the conversion of methyl ketones to carboxylic acids. Typically, this reaction is performed using iodine because the subsequent iodoform (CHI ) is a bright yellow precipitate which is easily filtered off. 1) Formation of the trihalo species 2) Nulceophilic attack on the carbonyl carbon 3) Removal of the leaving group 4) Deprotonation Please draw the products of the following reactions | 2,334 | 3,699 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Exemplars/Commercial_Galvanic_Cells |
Because galvanic cells can be self-contained and portable, they can be used as batteries and fuel cells. A is a galvanic cell (or a series of galvanic cells) that contains all the reactants needed to produce electricity. In contrast, a is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. In this section, we describe the chemistry behind some of the more common types of batteries and fuel cells. There are two basic kinds of batteries: disposable, or primary, batteries, in which the electrode reactions are effectively irreversible and which cannot be recharged; and rechargeable, or secondary, batteries, which form an insoluble product that adheres to the electrodes. These batteries can be recharged by applying an electrical potential in the reverse direction. The recharging process temporarily converts a rechargeable battery from a galvanic cell to an electrolytic cell. Batteries are cleverly engineered devices that are based on the same fundamental laws as galvanic cells. The major difference between batteries and the galvanic cells we have previously described is that commercial batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. The use of highly concentrated or solid reactants has another beneficial effect: the concentrations of the reactants and the products do not change greatly as the battery is discharged; consequently, the output voltage remains remarkably constant during the discharge process. This behavior is in contrast to that of the Zn/Cu cell, whose output decreases logarithmically as the reaction proceeds (Figure \(\Page {1}\)). When a battery consists of more than one galvanic cell, the cells are usually connected in series—that is, with the positive (+) terminal of one cell connected to the negative (−) terminal of the next, and so forth. The overall voltage of the battery is therefore the sum of the voltages of the individual cells. The major difference between batteries and the galvanic cells is that commercial typically batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass. An obvious exception is the standard car battery which used solution phase chemistry. The dry cell, by far the most common type of battery, is used in flashlights, electronic devices such as the Walkman and Game Boy, and many other devices. Although the dry cell was patented in 1866 by the French chemist Georges Leclanché and more than 5 billion such cells are sold every year, the details of its electrode chemistry are still not completely understood. In spite of its name, the is actually a “wet cell”: the electrolyte is an acidic water-based paste containing \(MnO_2\), \(NH_4Cl\), \(ZnCl_2\), graphite, and starch (part (a) in Figure \(\Page {1}\)). The half-reactions at the anode and the cathode can be summarized as follows: \[\ce{2MnO2(s) + 2NH^{+}4(aq) + 2e^{−} -> Mn2O3(s) + 2NH3(aq) + H2O(l)} \nonumber \] \[\ce{Zn(s) -> Zn^{2+}(aq) + 2e^{−}} \nonumber \] The \(\ce{Zn^{2+}}\) ions formed by the oxidation of \(\ce{Zn(s)}\) at the anode react with \(\ce{NH_3}\) formed at the cathode and \(\ce{Cl^{−}}\) ions present in solution, so the overall cell reaction is as follows: \[\ce{2MnO2(s) + 2NH4Cl(aq) + Zn(s) -> Mn2O3(s) + Zn(NH3)2Cl2(s) + H2O(l)} \label{Eq3} \] The dry cell produces about 1.55 V and is inexpensive to manufacture. It is not, however, very efficient in producing electrical energy because only the relatively small fraction of the \(\ce{MnO2}\) that is near the cathode is actually reduced and only a small fraction of the zinc cathode is actually consumed as the cell discharges. In addition, dry cells have a limited shelf life because the \(\ce{Zn}\) anode reacts spontaneously with \(\ce{NH4Cl}\) in the electrolyte, causing the case to corrode and allowing the contents to leak out. The is essentially a Leclanché cell adapted to operate under alkaline, or basic, conditions. The half-reactions that occur in an alkaline battery are as follows: \[\ce{2MnO2(s) + H2O(l) + 2e^{−} -> Mn2O3(s) + 2OH^{−}(aq)} \nonumber \] \[\ce{Zn(s) + 2OH^{−}(aq) -> ZnO(s) + H2O(l) + 2e^{−}} \nonumber \] \[\ce{Zn(s) + 2MnO2(s) -> ZnO(s) + Mn2O3(s)} \nonumber \] This battery also produces about 1.5 V, but it has a longer shelf life and more constant output voltage as the cell is discharged than the Leclanché dry cell. Although the alkaline battery is more expensive to produce than the Leclanché dry cell, the improved performance makes this battery more cost-effective. Although some of the small button batteries used to power watches, calculators, and cameras are miniature alkaline cells, most are based on a completely different chemistry. In these "button" batteries, the anode is a zinc–mercury amalgam rather than pure zinc, and the cathode uses either \(\ce{HgO}\) or \(\ce{Ag2O}\) as the oxidant rather than \(\ce{MnO2}\) in Figure \(\Page {1b}\)). The cathode, anode and overall reactions and cell output for these two types of button batteries are as follows (two half-reactions occur at the anode, but the overall oxidation half-reaction is shown): The major advantages of the mercury and silver cells are their reliability and their high output-to-mass ratio. These factors make them ideal for applications where small size is crucial, as in cameras and hearing aids. The disadvantages are the expense and the environmental problems caused by the disposal of heavy metals, such as \(\ce{Hg}\) and \(\ce{Ag}\). None of the batteries described above is actually “dry.” They all contain small amounts of liquid water, which adds significant mass and causes potential corrosion problems. Consequently, substantial effort has been expended to develop water-free batteries. One of the few commercially successful water-free batteries is the . The anode is lithium metal, and the cathode is a solid complex of \(I_2\). Separating them is a layer of solid \(LiI\), which acts as the electrolyte by allowing the diffusion of Li ions. The electrode reactions are as follows: \[I_{2(s)} + 2e^− \rightarrow {2I^-}_{(LiI)}\label{Eq11} \] \[2Li_{(s)} \rightarrow 2Li^+_{(LiI)} + 2e^− \label{Eq12} \] \[2Li_{(s)}+ I_{2(s)} \rightarrow 2LiI_{(s)} \label{Eq12a} \] with \(E_{cell} = 3.5 \, V\) As shown in part (c) in Figure \(\Page {1}\), a typical lithium–iodine battery consists of two cells separated by a nickel metal mesh that collects charge from the anode. Because of the high internal resistance caused by the solid electrolyte, only a low current can be drawn. Nonetheless, such batteries have proven to be long-lived (up to 10 yr) and reliable. They are therefore used in applications where frequent replacement is difficult or undesirable, such as in cardiac pacemakers and other medical implants and in computers for memory protection. These batteries are also used in security transmitters and smoke alarms. Other batteries based on lithium anodes and solid electrolytes are under development, using \(TiS_2\), for example, for the cathode. Dry cells, button batteries, and lithium–iodine batteries are disposable and cannot be recharged once they are discharged. Rechargeable batteries, in contrast, offer significant economic and environmental advantages because they can be recharged and discharged numerous times. As a result, manufacturing and disposal costs drop dramatically for a given number of hours of battery usage. Two common rechargeable batteries are the nickel–cadmium battery and the lead–acid battery, which we describe next. The , or NiCad, battery is used in small electrical appliances and devices like drills, portable vacuum cleaners, and AM/FM digital tuners. It is a water-based cell with a cadmium anode and a highly oxidized nickel cathode that is usually described as the nickel(III) oxo-hydroxide, NiO(OH). As shown in Figure \(\Page {2}\), the design maximizes the surface area of the electrodes and minimizes the distance between them, which decreases internal resistance and makes a rather high discharge current possible. The electrode reactions during the discharge of a \(NiCad\) battery are as follows: \[2NiO(OH)_{(s)} + 2H_2O_{(l)} + 2e^− \rightarrow 2Ni(OH)_{2(s)} + 2OH^-_{(aq)} \label{Eq13} \] \[Cd_{(s)} + 2OH^-_{(aq)} \rightarrow Cd(OH)_{2(s)} + 2e^- \label{Eq14} \] \[Cd_{(s)} + 2NiO(OH)_{(s)} + 2H_2O_{(l)} \rightarrow Cd(OH)_{2(s)} + 2Ni(OH)_{2(s)} \label{Eq15} \] \(E_{cell} = 1.4 V\) Because the products of the discharge half-reactions are solids that adhere to the electrodes [Cd(OH) and 2Ni(OH) ], the overall reaction is readily reversed when the cell is recharged. Although NiCad cells are lightweight, rechargeable, and high capacity, they have certain disadvantages. For example, they tend to lose capacity quickly if not allowed to discharge fully before recharging, they do not store well for long periods when fully charged, and they present significant environmental and disposal problems because of the toxicity of cadmium. A variation on the NiCad battery is the nickel–metal hydride battery (NiMH) used in hybrid automobiles, wireless communication devices, and mobile computing. The overall chemical equation for this type of battery is as follows: \[NiO(OH)_{(s)} + \rightarrow Ni(OH)_{2(s)} + M_{(s)} \label{Eq16} \] The NiMH battery has a 30%–40% improvement in capacity over the NiCad battery; it is more environmentally friendly so storage, transportation, and disposal are not subject to environmental control; and it is not as sensitive to recharging memory. It is, however, subject to a 50% greater self-discharge rate, a limited service life, and higher maintenance, and it is more expensive than the NiCad battery. Directive 2006/66/EC of the European Union prohibits the placing on the market of portable batteries that contain more than 0.002% of cadmium by weight. The aim of this directive was to improve "the environmental performance of batteries and accumulators" The is used to provide the starting power in virtually every automobile and marine engine on the market. Marine and car batteries typically consist of multiple cells connected in series. The total voltage generated by the battery is the potential per cell (E° ) times the number of cells. As shown in Figure \(\Page {3}\), the anode of each cell in a lead storage battery is a plate or grid of spongy lead metal, and the cathode is a similar grid containing powdered lead dioxide (\(PbO_2\)). The electrolyte is usually an approximately 37% solution (by mass) of sulfuric acid in water, with a density of 1.28 g/mL (about 4.5 M \(H_2SO_4\)). Because the redox active species are solids, there is no need to separate the electrodes. The electrode reactions in each cell during discharge are as follows: \[PbO_{2(s)} + ^−_{4(aq)} + 3H^+_{(aq)} + 2e^− \rightarrow PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq17} \] with \(E^°_{cathode} = 1.685 \; V\) \[Pb_{(s)} + HSO^−_{4(aq)} \rightarrow PbSO_{4(s) }+ H^+_{(aq)} + 2e^−\label{Eq18} \] with \(E^°_{anode} = −0.356 \; V\) \[Pb_{(s)} + PbO_{2(s)} + 2HSO^−_{4(aq)} + 2H^+_{(aq)} \rightarrow 2PbSO_{4(s)} + 2H_2O_{(l)} \label{Eq19} \] and \(E^°_{cell} = 2.041 \; V\) As the cell is discharged, a powder of \(PbSO_4\) forms on the electrodes. Moreover, sulfuric acid is consumed and water is produced, decreasing the density of the electrolyte and providing a convenient way of monitoring the status of a battery by simply measuring the density of the electrolyte. This is often done with the use of a hydrometer. When an external voltage in excess of 2.04 V per cell is applied to a lead–acid battery, the electrode reactions reverse, and \(PbSO_4\) is converted back to metallic lead and \(PbO_2\). If the battery is recharged too vigorously, however, of water can occur: \[ 2H_2O_{(l)} \rightarrow 2H_{2(g)} +O_{2 (g)} \label{EqX} \] This results in the evolution of potentially explosive hydrogen gas. The gas bubbles formed in this way can dislodge some of the \(PbSO_4\) or \(PbO_2\) particles from the grids, allowing them to fall to the bottom of the cell, where they can build up and cause an internal short circuit. Thus the recharging process must be carefully monitored to optimize the life of the battery. With proper care, however, a lead–acid battery can be discharged and recharged thousands of times. In automobiles, the alternator supplies the electric current that causes the discharge reaction to reverse. A fuel cell is a galvanic cell that requires a constant external supply of reactants because the products of the reaction are continuously removed. Unlike a battery, it does not store chemical or electrical energy; a fuel cell allows electrical energy to be extracted directly from a chemical reaction. In principle, this should be a more efficient process than, for example, burning the fuel to drive an internal combustion engine that turns a generator, which is typically less than 40% efficient, and in fact, the efficiency of a fuel cell is generally between 40% and 60%. Unfortunately, significant cost and reliability problems have hindered the wide-scale adoption of fuel cells. In practice, their use has been restricted to applications in which mass may be a significant cost factor, such as manned space vehicles. These space vehicles use a hydrogen/oxygen fuel cell that requires a continuous input of H (g) and O (g), as illustrated in Figure \(\Page {4}\). The electrode reactions are as follows: \[O_{2(g)} + 4H^+ + 4e^− \rightarrow 2H_2O_{(g)} \label{Eq20} \] \[2H_{2(g)} \rightarrow 4H^+ + 4e^− \label{Eq21} \] \[2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(g)} \label{Eq22} \] The overall reaction represents an essentially pollution-free conversion of hydrogen and oxygen to water, which in space vehicles is then collected and used. Although this type of fuel cell should produce 1.23 V under standard conditions, in practice the device achieves only about 0.9 V. One of the major barriers to achieving greater efficiency is the fact that the four-electron reduction of \(O_2 (g)\) at the cathode is intrinsically rather slow, which limits current that can be achieved. All major automobile manufacturers have major research programs involving fuel cells: one of the most important goals is the development of a better catalyst for the reduction of \(O_2 (g)\). Commercial batteries are galvanic cells that use solids or pastes as reactants to maximize the electrical output per unit mass. A battery is a contained unit that produces electricity, whereas a fuel cell is a galvanic cell that requires a constant external supply of one or more reactants to generate electricity. One type of battery is the Leclanché dry cell, which contains an electrolyte in an acidic water-based paste. This battery is called an alkaline battery when adapted to operate under alkaline conditions. Button batteries have a high output-to-mass ratio; lithium–iodine batteries consist of a solid electrolyte; the nickel–cadmium (NiCad) battery is rechargeable; and the lead–acid battery, which is also rechargeable, does not require the electrodes to be in separate compartments. A fuel cell requires an external supply of reactants as the products of the reaction are continuously removed. In a fuel cell, energy is not stored; electrical energy is provided by a chemical reaction. | 15,362 | 3,700 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Brown_et_al./01.E%3A_Matter_and_Measurement_(Exercises) |
. In addition to these individual basis; please contact Please be sure you are familiar with the topics discussed in Essential Skills 1 ( ) before proceeding to the Conceptual Problems. What is the difference between mass and weight? Is the mass of an object on Earth the same as the mass of the same object on Jupiter? Why or why not? Is it accurate to say that a substance with a mass of 1 kg weighs 2.2 lb? Why or why not? What factor must be considered when reporting the weight of an object as opposed to its mass? Construct a table with the headings “Solid,” “Liquid,” and “Gas.” For any given substance, state what you expect for each of the following: Classify each substance as homogeneous or heterogeneous and explain your reasoning. Classify each substance as homogeneous or heterogeneous and explain your reasoning. Classify each substance as a pure substance or a mixture and explain your reasoning. Classify each substance as a pure substance or a mixture. Classify each substance as an element or a compound. Classify each substance as an element or a compound. What techniques could be used to separate each of the following? What techniques could be used to separate each of the following? Match each separation technique in (a) with the physical/chemical property that each takes advantage of in (b). The following figures illustrate the arrangement of atoms in some samples of matter. Which figures are related by a physical change? By a chemical change? Classify each statement as an extensive property or an intensive property. Classify each statement as a physical property or a chemical property. NUMERICAL PROBLEMS Please be sure you are familiar with the topics discussed in Essential Skills 1 (Section 1.9) before proceeding to the Numerical Problems. If a person weighs 176 lb on Earth, what is his or her mass on Mars, where the force of gravity is 37% of that on Earth? If a person weighs 135 lb on Earth, what is his or her mass on Jupiter, where the force of gravity is 236% of that on Earth? Calculate the volume of 10.00 g of each element and then arrange the elements in order of decreasing volume. The numbers in parentheses are densities. Given 15.00 g of each element, calculate the volume of each and then arrange the elements in order of increasing volume. The numbers in parentheses are densities. A silver bar has dimensions of 10.00 cm × 4.00 cm × 1.50 cm, and the density of silver is 10.49 g/cm3. What is the mass of the bar? Platinum has a density of 21.45 g/cm3. What is the mass of a platinum bar measuring 3.00 cm × 1.50 cm × 0.500 cm? Complete the following table. Gold has a density of 19.30 g/cm3. If a person who weighs 85.00 kg (1 kg = 1000 g) were given his or her weight in gold, what volume (in cm3) would the gold occupy? Are we justified in using the SI unit of mass for the person’s weight in this case? An irregularly shaped piece of magnesium with a mass of 11.81 g was dropped into a graduated cylinder partially filled with water. The magnesium displaced 6.80 mL of water. What is the density of magnesium? The density of copper is 8.92 g/cm3. If a 10.00 g sample is placed in a graduated cylinder that contains 15.0 mL of water, what is the total volume that would be occupied? At 20°C, the density of fresh water is 0.9982 kg/m3, and the density of seawater is 1.025 kg/m3. Will a ship float higher in fresh water or in seawater? Explain your reasoning. AnswerS 1. Unlike weight, mass does not depend on location. The mass of the person is therefore the same on Earth and Mars: 176 lb ÷ 2.2 lb/kg = 80 kg. 3. a. Cu: 1.12 cm b. Ca: 6.49 cm c. Ti: 2.22 cm d. Ir: 0.4376 cm Volume decreases: Ca > Ti > Cu > Ir 5. 629 g 9. 1.74 g/cm 1. Milk turns sour. This is a ________________ 2. HCl being a strong acid is a __________, Wood sawed in two is ___________ 3. CuSO is dissolved in water 4. Aluminum Phosphate has a density of 2.566 g/cm3 5. Which of the following are examples of matter? 6. The formation of gas bubbles is a sign of what type of change? 7. True or False: Bread rising is a physical property. 8. True or False: Dicing potatoes is a physical change. 9. Is sunlight matter? 10. The mass of lead is a _____________property. 1)chemical change 2) chemical property, physical change 3) physical change 4) physical property 5) All of the above 6) chemical 7) False 8) True 9) No 10) physical property | 4,404 | 3,703 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Geochemistry_(Lower)/04%3A_The_Biosphere/4.01%3A_Chemistry_and_Energetics_of_the_Life_Process |
The biosphere comprises the various regions near the earth’s surface that contain and are dynamically affected by the metabolic activity of the approximately 1.5 million animal species and 0.5 million plant species that are presently known and are still being discovered at a rate of about 10,000 per year. The biosphere is the youngest of the dynamical systems of the earth, having had its genesis about 2 billion years ago. It is also the one that has most profoundly affected the other major environmental systems, particularly the atmosphere and the hydrosphere. About a third of the chemical elements cycle through living organisms, which are responsible for massive deposits of silicon, iron, manganese, sulfur, and carbon. Large quantities of methane and nitrous oxide are introduced into the atmosphere by bacterial action, and plants alone inject about 400,000 tons of volatile substances (including some metals) into the atmosphere annually. It has been suggested that biological activity might be responsible for the deficiency of hydrogen on Earth, compared to its very high relative abundance in the solar system as a whole. Bacteria capable of reducing hydrogen compounds into H transform this element into a form in which it can escape from the earth; such bacteria may have been especially active in the reducing atmosphere of the early planet. A second mechamism might be the microbial production of methane, which presently injects about 10 T of CH into the atmosphere each year. Some of this reaches the stratosphere, where it is oxidized to CO and H O. The water vapor is photolyzed to H , which escapes into space. This may be the major mechanism by which water vapor (and thus hydrogen) is transported to the upper atmosphere; the low temperature of the upper atmosphere causes most of the water originating at lower levels to condense before it can migrate to the top of the atmosphere. The increase in the abundance of atmospheric oxygen from its initial value of essentially zero has without question been the most important single effect of life on earth, and the time scale of this increase parallels the development of life forms from their most primitive stages up to the appearance of the first land animals about 0.5 billion years ago. All life processes involve the uptake and storage of energy, and its subsequent orderly release in small steps during the metabolic process. This energy is taken up in the combination of ADP with inorganic phosphate to form ATP, in which form the energy is stored and eventually delivered to sites where it can provide the free energy needed for driving non-spontaneous reactions such as protein and carbohydrate synthesis. ADP + PO → ATP ΔG° = +30 kJ The three main metabolic processes are glycolysis, respiration, and photosynthesis. The first two of these extract free energy from glucose by breaking it up into smaller, more thermodynamically stable fragments. Photosynthesis reverses this process by capturing the energy of sunlight into ATP which then drives the buildup of glucose from CO . As its name implies, this most primitive (and least efficient) of all metabolic processes is based on the breakdown of a sugar into fragments having a smaller total free energy. Thus the 6-carbon sugar glucose can be broken down into two 3-carbon lactic acid units, or into three 2-carbon ethanol units. C H O → 2 CH CHOHCOOH Δ = –197 kJ In this process, two molecules of ATP are produced, thus capturing 61 kJ of free energy. Since the standard free energy of glucose (with respect to its elements) is –2870 kJ, this represents an overall efficiency of about 2 percent. The net reaction of glycolysis is essentially a rearrangement of the atoms initially present in the energy source. This is a form of fermentation, which is defined as the enzymatic breakdown of organic molecules in which other organic compounds serve as electron acceptors. Since there is no need for an external oxidizing or reducing agent, there is no change in the oxidation state of the environment. When the enzymatic degradation of organic molecules is accompanied by transfer of electrons to an external (and usually better) electron acceptor, the process is known as respiration. The overall reaction of respiration is the oxidation of glucose to carbon dioxide: C H O + 6 O → 6 CO + 6 H O Δ = –2380 kJ In this process, 36 molecules of ADP are converted into ATP, thus capturing 1100 kJ of free energy: an efficiency of 38 percent. The oxidizing agent (electron sink) need not be oxygen; some bacteria reduce nitrates to NO or to N , and sulfates or sulfur to H S. These metabolic products can have far-reaching localized environmental effects, particularly if hydrogen ions are involved. Falling down the respiratory ladder presents a succinct picture of oxidation-reduction and the role of non-O electron sinks in biological energy capture. The energy of sunlight is trapped in the form of an intermediate which is able to deliver electrons to successively lower free-energy levels through the mediation of various molecules (mainly cytochromes) comprising an electron-transport chain. The free energy thus gained is utilized in part to reduce CO to glucose, which is then available to supply metabolic energy by glycolysis or respiration. In green plants and eukaryotic algae, the source of hydrogen is water, the net reaction being 6 CO + 6 H O → C H O + 6 O Δ = +2830 kJ For every CO molecule fixed in this way, 469 kJ of free energy must be supplied. Red light of 680 nm wavelength has an energy of 168 kJ/mol; this implies that about three photons must be absorbed for every carbon atom taken up, but experiment indicates that about ten seem to be required. utilizes the energy of red light to add hydrogen (from H O) and electrons (from the O in H O) to CO , reducing it to . is the reverse of this process; electrons are removed from the carbohydrate (food is "burned") in small steps, each one releasing a small amount of energy. Some of this energy is liberated as heat, but part of it is used to add a phosphate group to ADP, converting it into ATP. (adenosine triphosphate) is to an organism's energy needs as money is to our material needs; it circulates to wherever it is required in order to bring about energy-requiring reactions or to make muscle cells contract. Each increment of energy given up by ATP converts it back to ADP and phosphate, ready to repeat the cycle. are able to operate in both modes during the daylight hours, reverting to respiration-only at night. carry out only the right side of the cycle and thus require as source of carbohydrate (" ") either directly (by eating plants), or indirectly by eating other animals that eat plants. There are many kinds of photosynthetic bacteria, but with one exception (cyanobacteria) they are incapable of using water as a source of hydrogen for reducing carbon dioxide. Instead, they consume hydrogen sulfide or other reduced sulfur compounds, organic molecules, or elemental hydrogen itself, excreting the reducing agent in an oxidized state. Green plants, cyanobacteria, green filamentous bacteria and the purple nonsulfur bacteria utilize glucose by respiration during periods of darkness, while the green sulfur bacteria and the purple sulfur bacteria are strictly anaerobic. | 7,339 | 3,704 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Case_Studies%3A_Proteins/Membrane_Transport |
Membrane transport is essential for cellular life. As cells proceed through their life cycle, a vast amount of exchange is necessary to maintain function. Transport may involve the incorporation of biological molecules and the discharge of waste products that are necessary for normal function. Membrane transport refers to the movement of particles (solute) across or through a membranous barrier. These membranous barriers, in the case of the cell for example, consist of a phospholipid bilayer. The phospholipids orient themselves in such a way so that the hydrophilic (polar) heads are nearest the extracellular and intracellular mediums, and the hydrophobic (non-polar) tails align between the two hydrophilic head groups. Membrane transport is dependent upon the permeability of the membrane, transmembrane solute concentration, and the size and charge of the solute. Solute particles can traverse the membrane via three mechanisms: passive, facilitated, and active transport. Some of these transport mechanisms require the input of energy and use of a transmembrane protein, whereas other mechanisms do not incorporate secondary molecules. Passive transport is the simplest method of transport and is dependent upon the concentration gradient, and the size and charge of the solute. In passive transport, small uncharged solute particles diffuse across the membrane until both sides of the membrane have reached an equilibrium that is similar in concentration. The direction of solute travel is indicative of the concentration of that particular particle on each side of the membrane. (1) (2) (3) Figure 1. Passive diffusion of O and CO across a membrane over time 1-3. Based on the thermodynamics of the system, particles will move from an area of high concentration to an area of low concentration in order to increase the entropy of the cell. Additionally, this particle movement will occur spontaneously as the free energy (Gibbs free energy; ∆G) of the system is negative. Where: The properties of the membrane must also be considered when determining the rate of flow of the substrate. Darcy’s Law can be used to determine flow rate. Where: Where: Membrane Substrate 300 Escherichia coli Glycerol 25 Glucose 1 Lactose The diffusion of small charged particles, on the other hand, across a membrane is dependent upon the charge and transmembrane concentration of the solute. Again, however, the direction of solute travel is indicative of the thermodynamics of the system. Particles will travel from an area of high to low concentration, as well as travel so that the electrical potential across the membrane is diminished. As a result of this movement, the entropy of the system has increased. Passive transport is independent of membrane proteins and the catabolism of biological molecules for energy. This energy deficient process commonly occurs in the blood brain barrier as specific molecules, such as sodium thiopental (Figure 2. Sodium Thiopental Structure at right), can diffuse across the membrane. Sodium thiopental is a barbiturate frequently used in the methods of lethal injection. Sodium thiopental is a negatively charged particle and proceeds across the blood brain barrier to neuronal synaptic clefts. Sodium thiopental is an agonist of γ-aminobutyric acid (GABA), which acts as a neurotransmitter inhibitor. Sodium thiopental acts an anesthetic to cause unconsciousness. Additionally, passive diffusion occurs across the placenta as all solute particles are exchanged between mother and fetus. Placental physiology juxtaposes maternal and fetal capillaries in order to exchange solute particles, such as oxygen and carbon dioxide gas. These uncharged molecules proceed across maternal and fetal capillary membranes in the direction from high to low concentration. This spontaneous process occurs in accord with the aforementioned properties. Despite the cellular system, passive diffusion across the membranes of all biological cells is in accord with the characteristics outlined above. Facilitated diffusion, not to be confused with simple diffusion, is a form of passive transport mediated by transport proteins imbedded within the cellular membrane. Facilitated diffusion allows the passage of lipophobic molecules through the cell membrane’s lipid bilayer. Just as in passive transport, molecules, particles, and ions travel freely across the cellular membrane from high concentration to low concentration in an attempt to achieve equilibrium and thereby increase the entropy of the system. Also like passive transport, the Gibbs Free Energy of the system is negative, allowing the particle movement to be spontaneous. Facilitated diffusion, however, uses channel proteins to facilitate solute movement. (1) (2) (3) (4) Figure 3. Facilitated Diffusion via channel protein across a membrane proceeding images 1-4. Channel proteins are pores immersed in the lipid bilayer membrane and are the hallmark of facilitated diffusion. All channel proteins have two things in common: While all channel proteins have an inherent selectivity filter, others have additional gating. Gating is a response to a predetermined trigger that allows the channel protein to undergo a conformational change. This action subsequently causes another conformational change that either opens or closes the channel, allowing or disallowing its specific particle to pass. Channel proteins can be physically or chemically modulated through a number of different mechanisms. Voltage gated protein channels play a particularly important role in excitable neuronal and muscle tissues. Ligand gated channel proteins are activated in response to the binding of a ligand. Typically, ligand binding occurs at an allosteric binding site independent of the channel protein’s pore. The binding of a ligand at the allosteric binding site causes a conformational change in the structure of the channel protein, subsequently causing an influx or efflux of ions. Release of the ligand allows the channel protein to return to its original shape. Structurally, ligand gated channel proteins generally differ from other channels due to the presence of an additional protein domain that serves as the allosteric binding site. The prototypical example of ligand gating is the nicotinic acetylcholine receptor located on the postsynaptic side of the neuromuscular junction. Channel proteins may be gated in less common instances by methods such as light activation, mechanical activation, or secondary messanger activation. Light activated protein channels contain a photoswitch through which a photon causes a conformational change in the channel protein causing it to open or close. Only one such protein channel exists naturally. Mechanically activated protein channels open or close in response to a mechanical stimulus and are vital to the touch, hearing, and balance sensations in human. Ligand-gated protein channels are typically linked to second messanger gating. Second messenger gating functions stepwise in that a neurotransmitter binds to a channel protein receptor which, in turn, reveals an active site to which the conformation-changing ligand binds. Active transport, simply put, is the movement of particles through a transport protein from low concentration to high concentration at the expense of metabolic energy. The most common energy source used by cells is adenosine triphosphate or ATP, though other sources such as light energy or the energy stored in an electrochemical gradient are also utilized. In the case of ATP, energy is chemically harvested through hydrolysis. ATP hydrolysis in turn causes a conformational change in the transport protein which allows mechanical movement of the particle in question. Active transport systems are, therefore, energy-coupling devices as chemical and mechanical processes are linked to achieve particle movement. Active transport is classified as either Primary Active Transport or Secondary Active Transport. Figure 4 (displayed above) displays a ribbon structure of a commonly depicted ABC Vitamin B importer active transport protein. Primary active transport uses the energy found in ATP, photons, and electrochemical gradients directly in the transport of molecules from low concentration to high concentration across the cellular membrane. The enzyme-catalyzed hydrolysis reaction removing a phosphate from ATP, thereby forming ADP, causes a conformational change in the transport protein allowing particles to influx or efflux. Enzymes catalyzing ATP-driven primary active transport are called ATPases. _____________________________________________________________________________________ Figure 5. Primary active transport, with the use of ATP, is depicted above progressing left to right and top to bottom. The most universal example of ATP hydrolysis driving primary active transport in cells is the sodium-potassium pump. The sodium-potassium pump is responsible for controlling both sodium and potassium concentrations inside the cell. The sodium-potassium pump is extremely important in maintaining the cell’s resting potential. Figure 6. Ribbon structure of a sodium-potassium ATPase pump. An electrochemical gradient has two components: 1) an electrical component caused by charge difference on either side of the cellular membrane and 2) a chemical component resulting from differing concentrations of ions across the cellular membrane. The electrochemical gradient is generated by the presence of a proton (H ) gradient. A proton gradient is an interconvertible form of energy that can ultimately be used by the transport protein to move particles across the cellular membrane. A quintessential example of electrochemical gradient energy in primary active transport is the mitochondrial electron transport chain (ETC). The ETC uses the energy produced from the reduction of NADH to NAD to create a proton gradient by pumping protons into the inner mitochondrial space. The energy stored in a photon, the basic unit of light, is used to generate a proton gradient through a process similar to that found in electrochemical gradients. The stepwise passing of electrons in an electron transport chain reduces a molecule like NADH and ultimately generates a proton gradient. Plant photosynthesis is an example of primary active transport using photon energy. Chlorophyll absorbs a photon of light and consequently loses an electron which it passes pheophytin causing a subsequent electron transport chain. This ETC ultimately ends in the reduction of NADH to NAD creating a proton gradient across the chloroplast membrane. Secondary active transport achieves an identical result as primary active transport in that particles are moved from low concentration to high concentration at the expense of energy. Secondary active transport, however, functions independent of direct ATP coupling. Rather, the electrochemical energy generated from pumping ions out of the cell is used. Secondary active transport is classified as either symporter of antiporter. Symport secondary active transport uses a downhill movement of one particle to transport another particle against its concentration gradient. Symports move both particles in the same direction through a transmembrane transport protein. A common symport example is SGLT1, a glucose symport. SGLT1 tranports one glucose molecule into the cell for every two sodium ions transported into the cell. The SGLT1 symport is located throughout the body, particularly in the nephron of the kidney. Antiport secondary active transport moves two or more different particles across the cellular membrane in opposite directions. Antiport secondary active transport moves one particle down its concentration gradient and uses the energy generated from that process to move another particle up its concentration gradient. The sodium-calcium exchanger found throughout humans in excitable cells is a simple and common example of an antiport. Three sodium ions travel down their concentration gradient in exchange for one calcium ion. (1 ) Pardee, A. Membrane transport proteins: proteins that appear to be parts of membrane transport systems are being isolated and characterized. , , , 632-637. DOI: 10.1126/science.162.3854.632 (2 ) Garrett, R .; Grisham, C. , 4 ed.; Brooks/Cole: Boston, . (3 ) Stein, W. D. ; Stein, W. D., Eds, Academic Press: New York, . (4 ) Alberty, R. Standard Gibbs free energy, enthalpy, and entropy changes as a function of pH and pMg for several reactions involving adenosine phosphates. , , , 3290-3302. (5 ) Pal, L.; Joyce, M.; Fleming, P. A simple method for calculation of the permeability coefficient of porous media. , , , 10-16. (6 ) Walter, A.; Gutknecht, J. Permeability of small nonelectrolytes through lipid bilayer membranes. , , , 207-217. DOI:10.1007/BF01870127 (7 ) Nikaido, H.; Vaara, M. Molecular basis of bacterial outer membrane permeability. , , , 1-32. (8 ) Fischer, H.; Gottschlich, R.; Seelig, A. Blood-brain barrier permeation: molecular parameters governing passive diffusion. 1998, , 201-211. DOI: 10.1007/s002329900434 (9 ) Jevtovic-Todorovic, V.; Wozniak, D.; Powell, S.; Olney, J. Propofol and sodium thiopental protect against MK-801-induced neuronal necrosis in the posterior cingulated/retrosplenial cortex. , , , 185-189. (10 ) Sibley, C.; Boyd, R. Control of transfer across the mature placenta. , , , 382-435. (11 ) Reynolds, L.; Redmer, D. Utero-placental vascular development and placental function. , , , 1839-1851. (12 ) Cusslera, E.; Arisa, R.; Bhown, A. On the limits of facilitated diffusion. , , , 149-164. DOI:10.1016/S0376-7388(00)85094-2 (13 ) Hille, B. , 3 ed. Sinauer Associates: Sunderland, MA, . (14 ) Hellgren M.; Sandberg L.; Edholm O. A comparison between two prokaryotic potassium channels (KirBac1.1 and KcsA) in a molecular dynamics (MD) simulation study. , 120, 1–9. DOI: 10.1016/j.bpc.2005.10.002 (15 ) Coulomb, C-A. Recherches théoriques et expérimentales sur la force de torsion et sur l'élasticité des fils de metal. , , 229-269. (16 ) Catterall, W. Structure and function of voltage-gated ion channels. , , , 493-531. DOI: 10.1016/0166-2236(93)90193-P (17 ) Sheng, M.; Pak, D. Ligand-gated ion channel interactions with cytoskeletal and signaling proteins. , , , 755-78. DOI:10.1146/annurev.physiol. 62.1.755 (18 ) Purves, D.; Augustine, G.; Fitzpatrick, D.; Hall, W.; LaMantia, A.; McNamara, J.; White, L. , 4 ed. Sinauer Associates: Sunderland, MA, . (19 ) Banghart, M.; Volgraf, M.; Trauner, D. Engineering light-gated ion channels. , , , 15129–15141. DOI: 10.1021/bi0618058 (20 ) Pivetti, C.; Yen, M.; Miller, S.; Busch, W.; Tseng, Y.; Booth, I.; Saier, M. Two Families of Mechanosensitive Channel Proteins. , , , 66-85. DOI: 10.1128/MMBR.67.1.66-85.2003 (21^) Hodgkin, A.; Keynes, R. Active transport of cations in giant axons from sepia and loligo. , , , 28-60. PMC1365754 (22^) Nicholls, D. The influence of respiration and ATP hydrolysis on the proton-electrochemical gradient across the inner membrane of rat-liver mitochondria as determined by ion distribution. , , , 305-315. DOI: 10.1111/j.1432-1033.1974. tb03899.x (23 ) Mandel, L. Primary active sodium transport, oxygen consumption, and ATP: Coupling and regulation. , , , 3-9. DOI: 10.1038/ki.1986.2 (24 ) Zilberstein, D.; Schuldiner, S.; Padan, E. Proton electrochemical gradient in cells and its relation to active transport of lactose. , , , 669-673. DOI: 10.1021/bi00571a018 (25^) Karp, G. ,5 ed. John Wiley & Sons: Hoboken, NJ, . (26^) Raven, P.; Evert, R.; Eichhorn, S. , 7 ed. W.H. Freeman and Company Publishers: New York, . (27 ) Pâun, A.; Pâun, G. The power of communication: P systems with symport/antiport. , , , 295-305. DOI: 10.1007/BF03037362 (28^) Wright, E.; Hirayama B.; Loo D. Active sugar transport in health and disease. , , , 32–43. DOI: 10.1111/j.1365-2796.2006.01746.x (29 ) Yu, S.; Choi, D. Na+–Ca2+ exchange currents in cortical neurons: concomitant forward and reverse operation and effect of glutamate. , , 1273–81. | 16,203 | 3,705 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Addition_Reactions_of_Dienes |
Addition reactions of isolated dienes proceed more or less as expected from the behavior of simple alkenes. Thus, if one molar equivalent of 1,5-hexadiene is treated with one equivalent of bromine a mixture of 5,6-dibromo-1-hexene, 1,2,5,6-tetrabromohexane and unreacted diene is obtained, with the dibromo compound being the major product (about 50%). Similar reactions of conjugated dienes, on the other hand, often give unexpected products. The addition of bromine to 1,3-butadiene is an example. As shown below, a roughly 50:50 mixture of 3,4-dibromo-1-butene (the expected product) and 1,4-dibromo-2-butene (chiefly the E-isomer) is obtained. The latter compound is remarkable in that the remaining double bond is found in a location where there was no double bond in the reactant. This interesting relocation requires an explanation. The expected addition product from reactions of this kind is the result of , i.e. bonding to the adjacent carbons of a double bond. The unexpected product comes from , i.e. bonding at the terminal carbon atoms of a conjugated diene with a shift of the remaining double bond to the 2,3-location. These numbers refer to the four carbons of the conjugated diene and are not IUPAC nomenclature numbers. Product compositions are often temperature dependent, as the addition of HBr to 1,3-butadiene demonstrates. Bonding of an electrophilic atom or group to one of the end carbon atoms of a conjugated diene (#1) generates an allyl cation intermediate. Such cations are stabilized by charge delocalization, and it is this delocalization that accounts for the 1,4-addition product produced in such addition reactions. As shown in the diagram, the positive charge is distributed over carbons #2 and #4 so it is at these sites that the nucleophilic component bonds. Note that resonance stabilization of the allyl cation is greater than comparable stabilization of 1,3-butadiene, because charge is delocalized in the former, but created and separated in the latter. An explanation for the temperature influence is shown in the following energy diagram for the addition of HBr to 1,3-butadiene. The initial step in which a proton bonds to carbon #1 is the , as indicated by the large activation energy (light gray arrow). The second faster step is the , and there are two reaction paths (colored blue for 1,2-addition and magenta for 1,4-addition). The 1,2-addition has a smaller activation energy than 1,4-addition, but the 1,4-product is more stable than the 1,2-product. At low temperatures, the products are formed irreversibly and reflect the relative rates of the two competing reactions. This is termed . At higher temperatures, equilibrium is established between the products, and the 1,4-product dominates. | 2,763 | 3,706 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Case_Studies%3A_Proteins/Sickle_Cell_Anemia |
The incorrect amino acid sequence in a protein may lead to fatal consequences. For example, the inherited disease, sickle cell anemia, results from a single incorrect amino acid at the 6th position of the beta - protein chain out of 146. Hemoglobin consists of four protein chains - two beta and two alpha. See the graphic on the left for the sequences. This one alteration of the sequence of amino acids in hemoglobin changes its molecular geometry and hence its ability to carry oxygen and its solubility characteristics. The red blood cells change into a sickled shape instead of the normal round shape, become trapped in the small blood capillaries, and cause a great deal of pain. | 703 | 3,707 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Theory |
Crystal field theory (CFT) describes the breaking of orbital degeneracy in due to the presence of . CFT qualitatively describes the strength of the metal-ligand bonds. Based on the strength of the metal-ligand bonds, the energy of the system is altered. This may lead to a change in as well as . This theory was developed by Hans Bethe and John Hasbrouck van Vleck. In Crystal Field Theory, it is assumed that the ions are (a simplification). When applied to alkali metal ions containing a symmetric sphere of charge, calculations of bond energies are generally quite successful. The approach taken uses classical potential energy equations that take into account the attractive and repulsive interactions between charged particles (that is, Coulomb's Law interactions). \[E \propto \dfrac{q_1 q_2}{r}\] with This approach leads to the correct prediction that large cations of low charge, such as \(K^+\) and \(Na^+\), should form few coordination compounds. For transition metal cations that contain varying numbers of d electrons in orbitals that are NOT spherically symmetric, however, the situation is quite different. The shapes and occupations of these d-orbitals then become important in building an accurate description of the bond energy and properties of the transition metal compound. When examining a single transition metal ion, the five have the same energy (Figure \(\Page {1}\)). When ligands approach the metal ion, some experience more opposition from the -orbital electrons than others based on the geometric structure of the molecule. Since ligands approach from different directions, not all orbitals interact directly. These interactions, however, create a splitting due to the electrostatic environment. For example, consider a molecule with octahedral geometry. Ligands approach the metal ion along the \(x\), \(y\), and \(z\) axes. Therefore, the electrons in the \(d_{x^2-y^2}\) orbitals (which lie along these axes) experience greater repulsion. It requires more energy to have an electron in these orbitals than it would to put an electron in one of the other orbitals. This causes a splitting in the energy levels of the -orbitals. This is known as . For octahedral complexes, crystal field splitting is denoted by \(\Delta_o\) (or \(\Delta_{oct}\)). The energies of the \(d_{z^2}\) and \(d_{x^2-y^2}\) orbitals increase due to greater interactions with the ligands. The \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\) According to the , electrons are filled from lower to higher energy orbitals (Figure \(\Page {1}\)). For the octahedral case above, this corresponds to the d , d , and d orbitals. Following , electrons are filled in order to have the highest number of unpaired electrons. For example, if one had a d complex, there would be three unpaired electrons. If one were to add an electron, however, it has the ability to fill a higher energy orbital ( or ) e called . As mentioned above, CFT is based primarily on symmetry of ligands around a central metal/ion and how this anisotropic (properties depending on direction) ligand field affects the metal's atomic orbitals; the energies of which may increase, decrease or not be affected at all. Once the ligands' electrons interact with the electrons of the -orbitals, the electrostatic interactions cause the energy levels of the d-orbital to fluctuate depending on the orientation and the nature of the ligands. For example, the and the strength of the ligands determine splitting; the higher the oxidation state or the stronger the ligand, the larger the splitting. Ligands are classified as strong or weak based on the spectrochemical series: I < Br < Cl < CN < F < OH < ox < NO < H O < SC < EDTA < NH < en < O < CN Note that SCN and NO - ligands are represented twice in the above spectrochemical series since there are two different Lewis base sites (e.g., free electron pairs to share) on each ligand (e.g., for the SCN ligand, the electron pair on the sulfur or the nitrogen can form the to a metal). The specific atom that binds in such ligands is underlined. In addition to octahedral complexes, two common geometries observed are that of tetrahedral and square planar. These complexes differ from the octahedral complexes in that the orbital levels are raised in energy due to the interference with electrons from ligands. For the tetrahedral complex, the d , d , and d orbitals are raised in energy while the d , d orbitals are lowered. For the square planar complexes, there is greatest interaction with the d orbital and therefore it has higher energy. The next orbital with the greatest interaction is d , followed below by d . The orbitals with the lowest energy are the d and d orbitals. There is a large energy separation between the d orbital and the d and d orbitals, meaning that the crystal field splitting energy is large. We find that the square planar complexes have the greatest crystal field splitting energy compared to all the other complexes. This means that most square planar complexes are low spin, strong field ligands. To understand CFT, one must understand the description of the lobes: In an , there are six ligands attached to the central transition metal. The d-orbital splits into two different levels (Figure \(\Page {4}\)). The bottom three energy levels are named \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\) (collectively referred to as \(t_{2g}\)). The two upper energy levels are named \(d_{x^²-y^²}\), and \(d_{z^²}\) (collectively referred to as \(e_g\)). The reason they split is because of the electrostatic interactions between the electrons of the ligand and the lobes of the d-orbital. In an octahedral, the electrons are attracted to the axes. Any orbital that has a lobe on the axes moves to a higher energy level. This means that in an octahedral, the energy levels of \(e_g\) are higher (0.6∆ ) while \(t_{2g}\) is lower (0.4∆ ). \(t_{2g}\) \(e_g\) In a tetrahedral complex, there are four ligands attached to the central metal. The d orbitals also split into two different energy levels. The top three consist of the \(d_{xy}\), \(d_{xz}\), and \(d_{yz}\) orbitals. The bottom two consist of the \(d_{x^2-y^2}\) and \(d_{z^2}\) orbitals. The reason for this is due to poor orbital overlap between the metal and the ligand orbitals. The orbitals are directed on the axes, while the ligands are not. The difference in the splitting energy is tetrahedral splitting constant (\(\Delta_{t}\)), which less than (\(\Delta_{o}\)) for the same ligands: \[\Delta_{t} = 0.44\,\Delta_o \label{1}\] Consequentially, \(\Delta_{t}\) is typically smaller than the , so tetrahedral complexes are usually . In a square planar, there are four ligands as well. However, the difference is that the electrons of the ligands are only attracted to the \(xy\) plane. Any orbital in the xy plane has a higher energy level (Figure \(\Page {6}\)). There are four different energy levels for the square planar (from the highest energy level to the lowest energy level): d , d , d , and both d and d . The splitting energy (from highest orbital to lowest orbital) is \(\Delta_{sp}\) and tends to be larger then \(\Delta_{o}\) \[\Delta_{sp} = 1.74\,\Delta_o \label{2}\] Moreover, \(\Delta_{sp}\) is also larger than the pairing energy, so the square planar complexes are usually complexes. For the complex ion [Fe(Cl) ] determine the number of d electrons for Fe, sketch the d-orbital energy levels and the distribution of d electrons among them, list the number of lone electrons, and label whether the complex is paramagnetic or diamagnetic. A complex absorbs at 545 nm. What is the respective octahedral crystal field splitting (\(\Delta_o\))? What is the color of the complex? \[\begin{align*} \Delta_t &= \dfrac{hc}{\lambda}\\[4pt] &= \dfrac{ (6.626 \times 10^{-34} J \cdot s)(3 \times 10^8 m/s)}{545 \times 10^{-9} m} \\[4pt] &=3.65 \times 10^{-19}\; J \end{align*}\] However, the splitting (\(\Delta_t\)) is ~4/9 that of the splitting (\(\Delta_o\)). \[\begin{align*} \Delta_t &= 0.44\Delta_o \\[4pt] \Delta_o &= \dfrac{\Delta_t}{0.44} \\[4pt] &= \dfrac{3.65 \times 10^{-19} J}{0.44} \\[4pt] &= 8.30 \times 10^{-18}J \end{align*}\] This is the energy needed to promote electron in complex. Often the crystal field splitting is given per mole, which requires this number to be multiplied by Avogadro's Number (\(6.022 \times 10^{23}\)). This complex appears red, since it absorbs in the complementary green color (determined via the color wheel). For each of the following, sketch the d-orbital energy levels and the distribution of d electrons among them, state the geometry, list the number of d-electrons, list the number of lone electrons, and label whether they are paramagnetic or dimagnetic: 1. octahedral, 2, 2, paramagnetic 2. tetrahedral, 8, 2, paramagnetic (see 3. octahedral, 6, 4, paramagnetic, high spin 4. octahedral, 6, 0, diamagnetic, low spin 5. True | 8,952 | 3,709 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Quantifying_Nature/Units_of_Measure/Unit_Conversions |
In the field of science, the metric system is used in performing measurements. The metric system is actually easier to use than the English system, as you will see shortly. The metric system uses prefixes to indicate the magnitude of a measured quantity. The prefix itself gives the conversion factor. You should memorize some of the common prefixes, as you will be using them on a regular basis. Common prefixes are shown below: Suppose you wanted to convert the mass of a \(250\; mg\) aspirin tablet to grams. Start with what you know and let the conversion factor units decide how to set up the problem. If a unit to be converted is in the numerator, that unit must be in the denominator of the conversion factor in order for it to cancel. Notice how the units cancel to give grams. the conversion factor numerator is shown as \(1 \times 10^{-3}\) because on most calculators, it must be entered in this fashion, not as just 10 . If you don't know how to use the scientific notation on your calculator, try to find out as soon as possible. Look in your calculator's manual, or ask someone who knows. Also, notice how the unit, mg is assigned the value of 1, and the prefix, milli-, is applied to the gram unit. In other words, \(1\, mg\) literally means \(1 \times 10^{-3}\, g\). Next, let's try a more involved conversion. Suppose you wanted to convert 250 mg to kg. You may or may not know a direct, one-step conversion. In fact, the better method (foolproof) to do the conversion would be to go to the base unit first, and then to the final unit you want. In other words, convert the milligrams to grams and then go to kilograms: The world's ocean is estimated to contain \(\mathrm{1.4 \times 10^9\; km^3}\) of water. \(\mathrm{1.4e9 \left (\dfrac{1000\: m}{1\: km} \right )^3 \left (\dfrac{10\: dm}{1\:m} \right )^3\\
= 1.4e21\: dm^3 \left (\dfrac{1\: L}{1\: dm^3} \right )\\
= 1.5e21\: L \left (\dfrac{1.1\: kg}{1\: L} \right )\\
= 1.5e21\: kg \left (\dfrac{1000\: g}{1\: kg} \right )\left (\dfrac{1\: mol}{18\: g} \right )\\
= 8.3e22\: mol\: H_2O \left (\dfrac{2\: mol\: H\: atoms}{1\: mol\: H_2O} \right )\\
= 1.7e23\: mol\: H \left (\dfrac{6.02e23\: atoms}{1\: mol} \right )\\
= 5.0e46\: H\: atoms}\) In this example, a quantity has been converted from a unit for volume into other units of volume, weight, amount in moles, and number of atoms. Every factor used for the unit conversion is a unity. The numerator and denominator represent the same quantity in different ways. Even in this simple example, several concepts such as the quantity in moles, Avogadro's number, and specific density (or specific gravity) have been applied in the conversion. If you have not learned these concepts, you may have difficulty in understanding some of the conversion processes. Identify what you do not know and find out in your text or from a resource. A typical city speed for automobiles is 50 km/hr. Some years ago, most people believed that 10 seconds to dash a 100 meter race was the lowest limit. Which speed is faster, 50 km/hr or 10 m/s? For comparison, the two speeds must be expressed in the same unit. Let's convert 50 km/hr to m/s. \[ \mathrm{50 \;\dfrac{\cancel{km}}{\cancel{hr}} \left(\dfrac{1000\; m}{1\; \cancel{km}}\right) \left(\dfrac{1\; \cancel{hr}}{60\;\cancel{min}}\right) \left(\dfrac{1\;\cancel{min}}{ 60\; s}\right) =13.89\; m/s} \] Thus, 50 km/hr is faster. Note: a different unit can be selected for the comparison (e.g., miles/hour) but the result will be the same (test this out if interested). The speed of a typhoon is reported to be 100 m/s. What is the speed in km/hr and in miles per hour? These conversions are accomplished in the same way as metric - metric conversions. The only difference is the conversion factor used. It would be a good idea to memorize a few conversion factors involving converting mass, volume, length and temperature. Here are a few useful conversion factors. All of the above conversions are to three significant figures, except length, which is an exact number. As before, let the units help you set up the conversion. Suppose you wanted to convert mass of my \(23\, lb\) cat to kilograms. One can quickly see that this conversion is not achieved in one step. The pound units will be converted to grams, and then from grams to kilograms. Let the units help you set up the problem: \[ \dfrac{23 \, lb}{1} \times \dfrac{454\,g}{1 \, lb} \times \dfrac{1 \, kg}{ 1 \times 10^3 \, g} = 10 \, kg\] Let's try a conversion which looks "intimidating", but actually uses the same basic concepts we have already examined. Suppose you wish to convert pressure of 14 lb/in to g/cm . When setting up the conversion, worry about one unit at a time, for example, convert the pound units to gram units, first: Next, convert in to cm . Set up the conversion without the exponent first, using the conversion factor, 1 in = 2.54 cm. Since we need in and cm , raise everything to the second power: Notice how the units cancel to the units sought. Always check your units because they indicate whether or not the problem has been set up correctly. Mr. Smart is ready for a T-bone steak. He went to market A and found the price to be 4.99 dollars per kilograms. He drove out of town to a roadside market, which sells at 2.29 per pound. Which price is better for Mr. Smart? To help Mr. Smart, we have to know that 1.0 kg is equivalent to 2.206531 lb or 1 lb = 453.2 g. By the way, are these the same? \[ \mathrm{4.99\; \dfrac{$}{\cancel{kg}} \left( \dfrac{1\; \cancel{kg}}{2.206532\; lb} \right) = 2.26468 \;\dfrac{$}{lb}}\] Of course, with the money system in Canada, there is no point quoting the price as detailed as it is given above. This brings about the significant digit issue, and the quantization. The price is therefore 2.26 $/lb, better for Mr. Smart than the price of 2.29 $/lb.
Converting a quantity into SI units.
To convert temperature from one scale to another scale.
Converting two quantities.
Determine the costs per unit common volume. 6486 m
Convert quantities into SI units. | 6,090 | 3,710 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Synthesis_of_Alcohols/The_Manufacture_of_Alcohols |
This page looks at the manufacture of alcohols by the direct hydration of alkenes, concentrating mainly on the hydration of ethene to make ethanol. It then compares that method with making ethanol by fermentation. Ethanol is manufactured by reacting ethene with steam. The catalyst used is solid silicon dioxide coated with phosphoric(V) acid. The reaction is reversible. Only 5% of the ethene is converted into ethanol at each pass through the reactor. By removing the ethanol from the equilibrium mixture and recycling the ethene, it is possible to achieve an overall 95% conversion. A flow scheme for the reaction looks like this: Some - but not all - other alcohols can be made by similar reactions. The catalyst used and the reaction conditions will vary from alcohol to alcohol. The reason that there is a problem with some alcohols is well illustrated with trying to make an alcohol from propene, CH CH=CH . In principle, there are two different alcohols which might be formed: You might expect to get either propan-1-ol or propan-2-ol depending on which way around the water adds to the double bond. In practice what you get is propan-2-ol. If you add a molecule H-X across a carbon-carbon double bond, the hydrogen nearly always gets attached to the carbon with the most hydrogens on it already - in this case the CH rather than the CH. The effect of this is that there are bound to be some alcohols which it is impossible to make by reacting alkenes with steam because the addition would be the wrong way around. This method only applies to ethanol and you cannot make any other alcohol this way. The starting material for the process varies widely, but will normally be some form of starchy plant material such as maize (US: corn), wheat, barley or potatoes. is a complex carbohydrate, and other carbohydrates can also be used - for example, in the lab sucrose (sugar) is normally used to produce ethanol. Industrially, this wouldn't make sense. It would be silly to refine sugar if all you were going to use it for was fermentation. There is no reason why you should not start from the original sugar cane, though. The first step is to break complex carbohydrates into simpler ones. For example, if you were starting from starch in grains like wheat or barley, the grain is heated with hot water to extract the starch and then warmed with malt. Malt is germinated barley which contains enzymes which break the starch into a simpler carbohydrate called maltose, \(C_{12}H_{22}O_{11}\). Maltose has the same molecular formula as sucrose but contains two glucose units joined together, whereas contains one glucose and one fructose unit. Yeast is then added and the mixture is kept warm (say 35°C) for perhaps several days until fermentation is complete. Air is kept out of the mixture to prevent oxidation of the ethanol produced to ethanoic acid (vinegar). Enzymes in the yeast first convert carbohydrates like maltose or sucrose into even simpler ones like glucose and fructose, both \(C_6H_{12}O_6\), and then convert these in turn into ethanol and carbon dioxide. You can show these changes as simple chemical equations, but the biochemistry of the reactions is much, much more complicated than this suggests. \[ C_{12}H_{22}O_{11} + H_2O \longrightarrow 2C_6H_{12}O_6 \] \[ C_6H_{12}O_6 \longrightarrow 2CH_3CH_2OH + 2CO2\] Yeast is killed by ethanol concentrations in excess of about 15%, and that limits the purity of the ethanol that can be produced. The ethanol is separated from the mixture by fractional distillation to give 96% pure ethanol. For theoretical reasons (minimum boiling point ), it is impossible to remove the last 4% of water by fractional distillation. Jim Clark ( ) | 3,719 | 3,711 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/20%3A_Carbohydrates/20.06%3A_Glycosides |
Although abundant quantities of glucose and fructose are found in the free state, they and less common sugars occur widely in plants and animals combined with various hydroxy compounds. The bonding is through oxygen to the carbonyl carbon, as in the \(\alpha\)- and \(\beta\)-methylglucosides discussed in , to give acetal or ketal structures. These substances are sometimes simply called , but it is desirable to specify that the bonding is through oxygen by using the name \(\ce{O}\)-glycoside. Hydrolysis of an \(\ce{O}\)-glycoside gives the sugar and the hydroxy compound, called the component of the glycoside. A specific example is glucovanillin, which can be isolated from the green fruit pods of vanilla, a climbing orchid cultivated in several tropical countries. Hydrolysis gives glucose and the aglycone, vanillin, which is the principal ingredient of vanilla flavoring. As the vanilla pods mature, a natural hydrolysis reaction proceeds to the extent that the pods may be covered with small crystals of vanillin. The configurations of glycosides are designated by the same convention used for the sugar anomers. Thus if a glycoside of a \(D\) sugar has the \(D\) configuration at the anomeric carbon, it is designated as the \(\alpha\)-\(D\)-glycoside, and if it has the \(L\) configuration it is called the \(\beta\)-\(D\)-glycoside (see ). If the sugar involved in glycoside formation is glucose, the derivative is a ; if fructose, a ; if galactose, a ; and so on. When the hydroxy compound, or aglycone, is another sugar, then the glycoside is a , and if the sugar is already a dissacharide, the glycoside is a , and so on. Among the natural products that occur as glycosides (most commonly as \(\beta\)-\(D\)-glucosides) are many plant pigments (the anthocyanins), the flavorings vanillin and amygdalin, and many steroids (such as the cardiac glycosides and saponins). The structures of some of these substances will be discussed in later chapters. Not all glycosides are \(\ce{O}\)-glycosides. A group of \(\ce{N}\)-glycosides of special biological importance are derived from heterocyclic nitrogen bases and \(D\)-ribose and 2-deoxy-\(D\)-ribose. They commonly are known as , or more specifically, as and ; the \(\ce{N}\)-glycoside linkage is always \(\beta\): The \(\ce{N}\)-glycoside of \(D\)-ribose and the nitrogen heterocycle, adenine, is : A is a phosphate ester of a nucleoside. The hydroxyl group at the \(\ce{C_5}\) position of the pentose is the most common site of esterification. The nucleotides of adenosine are ATP, ADP, and AMP ( ). A is a combination of two nucleosides through a common phosphate ester link. Familiar examples are \(\ce{NAD}^\oplus\), \(\ce{NADH}\), \(\ce{FAD}\), and \(\ce{FADH_2}\) ( ). Polynucleotides are polymers of nucleosides linked through phosphate ester bonds. Polynucleotides also are called nucleic acids (RNA and DNA) and are the genetic material of cells, as will be discussed in . and (1977) | 2,996 | 3,712 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Protein_Structure/The_Structure_of_Proteins |
This page explains how amino acids combine to make proteins and what is meant by the primary, secondary and tertiary structures of proteins. Quaternary structure isn't covered. It only applies to proteins consisting of more than one polypeptide chain. In chemistry, if you were to draw the structure of a general 2-amino acid, you would probably draw it like this: However, for drawing the structures of proteins, we usually twist it so that the "R" group sticks out at the side. It is much easier to see what is happening if you do that. That means that the two simplest amino acids, glycine and alanine, would be shown as: Glycine and alanine can combine together with the elimination of a molecule of water to produce a dipeptide. It is possible for this to happen in one of two different ways - so you might get two different dipeptides. Either: Or: In each case, the linkage shown in blue in the structure of the dipeptide is known as a peptide link. In chemistry, this would also be known as an amide link, but since we are now in the realms of biochemistry and biology, we'll use their terms. If you joined three amino acids together, you would get a tripeptide. If you joined lots and lots together (as in a protein chain), you get a polypeptide. A protein chain will have somewhere in the range of 50 to 2000 amino acid residues. You have to use this term because strictly speaking a peptide chain isn't made up of amino acids. When the amino acids combine together, a water molecule is lost. The peptide chain is made up from what is left after the water is lost - in other words, is made up of amino acid residues. By convention, when you are drawing peptide chains, the -NH group which hasn't been converted into a peptide link is written at the left-hand end. The unchanged -COOH group is written at the right-hand end. The end of the peptide chain with the -NH group is known as the N-terminal, and the end with the -COOH group is the C-terminal. A protein chain (with the N-terminal on the left) will therefore look like this: The "R" groups come from the 20 amino acids which occur in proteins. The peptide chain is known as the backbone, and the "R" groups are known as side chains. Now there's a problem! The term "primary structure" is used in two different ways. At its simplest, the term is used to describe the order of the amino acids joined together to make the protein. In other words, if you replaced the "R" groups in the last diagram by real groups you would have the primary structure of a particular protein. This primary structure is usually shown using abbreviations for the amino acid residues. These abbreviations commonly consist of three letters or one letter. Using three letter abbreviations, a bit of a protein chain might be represented by, for example: If you look carefully, you will spot the abbreviations for glycine (Gly) and alanine (Ala) amongst the others. If you followed the protein chain all the way to its left-hand end, you would find an amino acid residue with an unattached -NH group. The N-terminal is always written on the left of a diagram for a protein's primary structure - whether you draw it in full or use these abbreviations. The wider definition of primary structure includes all the features of a protein which are a result of covalent bonds. Obviously, all the peptide links are made of covalent bonds, so that isn't a problem. But there is an additional feature in proteins which is also covalently bound. It involves the amino acid cysteine. If two cysteine side chains end up next to each other because of folding in the peptide chain, they can react to form a sulfur bridge. This is another covalent link and so some people count it as a part of the primary structure of the protein. Because of the way sulphur bridges affect the way the protein folds, other people count this as a part of the tertiary structure (see below). This is obviously a potential source of confusion! Within the long protein chains there are regions in which the chains are organised into regular structures known as alpha-helices (alpha-helixes) and beta-pleated sheets. These are the secondary structures in proteins. These secondary structures are held together by hydrogen bonds. These form as shown in the diagram between one of the lone pairs on an oxygen atom and the hydrogen attached to a nitrogen atom: Although the hydrogen bonds are always between C=O and H-N groups, the exact pattern of them is different in an alpha-helix and a beta-pleated sheet. When you get to them below, take some time to make sure you see how the two different arrangements works. In an alpha-helix, the protein chain is coiled like a loosely-coiled spring. The "alpha" means that if you look down the length of the spring, the coiling is happening in a clockwise direction as it goes away from you. The next diagram shows how the alpha-helix is held together by hydrogen bonds. This is a very simplified diagram, missing out lots of atoms. We'll talk it through in some detail after you have had a look at it. What's wrong with the diagram? Two things: First of all, only the atoms on the parts of the coils facing you are shown. If you try to show all the atoms, the whole thing gets so complicated that it is virtually impossible to understand what is going on. Secondly, I have made no attempt whatsoever to get the bond angles right. I have deliberately drawn all of the bonds in the backbone of the chain as if they lie along the spiral. In truth they stick out all over the place. Again, if you draw it properly it is virtually impossible to see the spiral. So, what do you need to notice? Notice that all the "R" groups are sticking out sideways from the main helix. Notice the regular arrangement of the hydrogen bonds. All the N-H groups are pointing upwards, and all the C=O groups pointing downwards. Each of them is involved in a hydrogen bond. And finally, although you can't see it from this incomplete diagram, each complete turn of the spiral has 3.6 (approximately) amino acid residues in it. If you had a whole number of amino acid residues per turn, each group would have an identical group underneath it on the turn below. Hydrogen bonding can't happen under those circumstances. Each turn has 3 complete amino acid residues and two atoms from the next one. That means that each turn is offset from the ones above and below, such that the N-H and C=O groups are brought into line with each other. In a beta-pleated sheet, the chains are folded so that they lie alongside each other. The next diagram shows what is known as an "anti-parallel" sheet. All that means is that next-door chains are heading in opposite directions. Given the way this particular folding happens, that would seem to be inevitable. It isn't, in fact, inevitable! It is possible to have some much more complicated folding so that next-door chains are actually heading in the same direction. The folded chains are again held together by hydrogen bonds involving exactly the same groups as in the alpha-helix. The tertiary structure of a protein is a description of the way the whole chain (including the secondary structures) folds itself into its final 3-dimensional shape. This is often simplified into models like the following one for the enzyme dihydrofolate reductase. Enzymes are, of course, based on proteins. The model shows the alpha-helices in the secondary structure as coils of "ribbon". The beta-pleated sheets are shown as flat bits of ribbon ending in an arrow head. The bits of the protein chain which are just random coils and loops are shown as bits of "string". The color coding in the model helps you to track your way around the structure - going through the spectrum from dark blue to end up at red. You will also notice that this particular model has two other molecules locked into it (shown as ordinary molecular models). These are the two molecules whose reaction this enzyme catalyses. The tertiary structure of a protein is held together by interactions between the the side chains - the "R" groups. There are several ways this can happen. Some amino acids (such as aspartic acid and glutamic acid) contain an extra -COOH group. Some amino acids (such as lysine) contain an extra -NH group. You can get a transfer of a hydrogen ion from the -COOH to the -NH group to form zwitterions just as in simple amino acids. You could obviously get an ionic bond between the negative and the positive group if the chains folded in such a way that they were close to each other. Notice that we are now talking about hydrogen bonds between side groups - not between groups actually in the backbone of the chain. Lots of amino acids contain groups in the side chains which have a hydrogen atom attached to either an oxygen or a nitrogen atom. This is a classic situation where hydrogen bonding can occur. For example, the amino acid serine contains an -OH group in the side chain. You could have a hydrogen bond set up between two serine residues in different parts of a folded chain. You could easily imagine similar hydrogen bonding involving -OH groups, or -COOH groups, or -CONH groups, or -NH groups in various combinations - although you would have to be careful to remember that a -COOH group and an -NH group would form a zwitterion and produce stronger ionic bonding instead of hydrogen bonds. Several amino acids have quite large hydrocarbon groups in their side chains. A few examples are shown below. Temporary fluctuating dipoles in one of these groups could induce opposite dipoles in another group on a nearby folded chain. The dispersion forces set up would be enough to hold the folded structure together. Sulfur bridges which form between two cysteine residues have already been discussed under primary structures. Wherever you choose to place them doesn't affect how they are formed! | 9,866 | 3,714 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_OpenStax/01%3A_Essential_Ideas_of_Chemistry_(Exercises) |
Explain how you could experimentally determine whether the outside temperature is higher or lower than 0 °C (32 °F) without using a thermometer. Place a glass of water outside. It will freeze if the temperature is below 0 °C. Identify each of the following statements as being most similar to a hypothesis, a law, or a theory. Explain your reasoning. Identify each of the following statements as being most similar to a hypothesis, a law, or a theory. Explain your reasoning. (a) law (states a consistently observed phenomenon, can be used for prediction); (b) theory (a widely accepted explanation of the behavior of matter); (c) hypothesis (a tentative explanation, can be investigated by experimentation) Identify each of the underlined items as a part of either the macroscopic domain, the microscopic domain, or the symbolic domain of chemistry. For any in the symbolic domain, indicate whether they are symbols for a macroscopic or a microscopic feature. Identify each of the underlined items as a part of either the macroscopic domain, the microscopic domain, or the symbolic domain of chemistry. For those in the symbolic domain, indicate whether they are symbols for a macroscopic or a microscopic feature. (a) symbolic, microscopic; (b) macroscopic; (c) symbolic, macroscopic; (d) microscopic According to one theory, the pressure of a gas increases as its volume decreases because the molecules in the gas have to move a shorter distance to hit the walls of the container. Does this theory follow a macroscopic or microscopic description of chemical behavior? Explain your answer. The amount of heat required to melt 2 lbs of ice is twice the amount of heat required to melt 1 lb of ice. Is this observation a macroscopic or microscopic description of chemical behavior? Explain your answer. Macroscopic. The heat required is determined from macroscopic properties. 2 Liquids can change their shape (flow); solids can’t. Gases can undergo large volume changes as pressure changes; liquids do not. Gases flow and change volume; solids do not. 4.The mixture can have a variety of compositions; a pure substance has a definite composition. Both have the same composition from point to point. 6 Molecules of elements contain only one type of atom; molecules of compounds contain two or more types of atoms. They are similar in that both are comprised of two or more atoms chemically bonded together. 8. Answers will vary. Sample answer: Gatorade contains water, sugar, dextrose, citric acid, salt, sodium chloride, monopotassium phosphate, and sucrose acetate isobutyrate. 11. (a) element; (b) element; (c) compound; (d) mixture, (e) compound; (f) compound; (g) compound; (h) mixture 12. In each case, a molecule consists of two or more combined atoms. They differ in that the types of atoms change from one substance to the next. 14. Gasoline (a mixture of compounds), oxygen, and to a lesser extent, nitrogen are consumed. Carbon dioxide and water are the principal products. Carbon monoxide and nitrogen oxides are produced in lesser amounts. 16. (a) Increased as it would have combined with oxygen in the air thus increasing the amount of matter and therefore the mass. (b) 0.9 g 18. (a) 200.0 g; (b) The mass of the container and contents would decrease as carbon dioxide is a gaseous product and would leave the container. (c) 102.3 g Classify the six underlined properties in the following paragraph as chemical or physical: Fluorine is a pale yellow gas that reacts with most substances. The free element melts at −220 °C and boils at −188 °C. Finely divided metals burn in fluorine with a bright flame. Nineteen grams of fluorine will react with 1.0 gram of hydrogen. Classify each of the following changes as physical or chemical: (a) physical; (b) chemical; (c) chemical; (d) physical; (e) physical Classify each of the following changes as physical or chemical: The volume of a sample of oxygen gas changed from 10 mL to 11 mL as the temperature changed. Is this a chemical or physical change? physical A 2.0-liter volume of hydrogen gas combined with 1.0 liter of oxygen gas to produce 2.0 liters of water vapor. Does oxygen undergo a chemical or physical change? Explain the difference between extensive properties and intensive properties. The value of an extensive property depends upon the amount of matter being considered, whereas the value of an intensive property is the same regardless of the amount of matter being considered. Identify the following properties as either extensive or intensive. The density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (V). \(\mathrm{density=\dfrac{mass}{volume}}\) \(\mathrm{d=\dfrac{m}{V}}\) Considering that mass and volume are both extensive properties, explain why their ratio, density, is intensive. Being extensive properties, both mass and volume are directly proportional to the amount of substance under study. Dividing one extensive property by another will in effect “cancel” this dependence on amount, yielding a ratio that is independent of amount (an intensive property). Is one liter about an ounce, a pint, a quart, or a gallon? Is a meter about an inch, a foot, a yard, or a mile? about a yard Indicate the SI base units or derived units that are appropriate for the following measurements: Indicate the SI base units or derived units that are appropriate for the following measurements: (a) kilograms; (b) meters; (c) kilometers/second; (d) kilograms/cubic meter; (e) kelvin; (f) square meters; (g) cubic meters Give the name and symbol of the prefixes used with SI units to indicate multiplication by the following exact quantities. Give the name of the prefix and the quantity indicated by the following symbols that are used with SI base units. (a) centi-, \(\times\) 10 ; (b) deci-, \(\times\) 10 ; (c) Giga-, \(\times\) 10 ; (d) kilo-, \(\times\) 10 ; (e) milli-, \(\times\) 10 ; (f) nano-, \(\times\) 10 ; (g) pico-, \(\times\) 10 ; (h) tera-, \(\times\) 10 A large piece of jewelry has a mass of 132.6 g. A graduated cylinder initially contains 48.6 mL water. When the jewelry is submerged in the graduated cylinder, the total volume increases to 61.2 mL. Visit this and select the Same Volume Blocks. (a) 8.00 kg, 5.00 L, 1.60 kg/L; (b) 2.00 kg, 5.00 L, 0.400 kg/L; (c) red < green < blue < yellow; (d) If the volumes are the same, then the density is directly proportional to the mass. Visit this and select Custom Blocks and then My Block. Visit this and select Mystery Blocks. (a) (b) Answer is one of the following. A/yellow: mass = 65.14 kg, volume = 3.38 L, density = 19.3 kg/L, likely identity = gold. B/blue: mass = 0.64 kg, volume = 1.00 L, density = 0.64 kg/L, likely identity = apple. C/green: mass = 4.08 kg, volume = 5.83 L, density = 0.700 kg/L, likely identity = gasoline. D/red: mass = 3.10 kg, volume = 3.38 L, density = 0.920 kg/L, likely identity = ice; and E/purple: mass = 3.53 kg, volume = 1.00 L, density = 3.53 kg/L, likely identity = diamond. (c) B/blue/apple (0.64 kg/L) < C/green/gasoline (0.700 kg/L) < D/red/ice (0.920 kg/L) < E/purple/diamond (3.53 kg/L) < A/yellow/gold (19.3 kg/L) Express each of the following numbers in scientific notation with correct significant figures: Express each of the following numbers in exponential notation with correct significant figures: (a) 7.04 \(\times\) 10 ; (b) 3.344 \(\times\) 10 ; (c) 5.479 \(\times\) 10 ; (d) 2.2086 \(\times\) 10 ; (e) 1.00000 \(\times\) 10 ; (f) 6.51 \(\times\) 10 ; (g) 7.157 \(\times\) 10 Indicate whether each of the following can be determined exactly or must be measured with some degree of uncertainty: Indicate whether each of the following can be determined exactly or must be measured with some degree of uncertainty: (a) exact; (b) exact; (c) uncertain; (d) exact; (e) uncertain; (f) uncertain How many significant figures are contained in each of the following measurements? How many significant figures are contained in each of the following measurements? (a) two; (b) three; (c) five; (d) four; (e) six; (f) two; (g) five The following quantities were reported on the labels of commercial products. Determine the number of significant figures in each. Round off each of the following numbers to two significant figures: (a) 0.44; (b) 9.0; (c) 27; (d) 140; (e) 1.5 \(\times\) 10 ; (f) 0.44 Round off each of the following numbers to two significant figures: Perform the following calculations and report each answer with the correct number of significant figures. (a) 2.15 \(\times\) 10 ; (b) 4.2 \(\times\) 10 ; (c) 2.08; (d) 0.19; (e) 27,440; (f) 43.0 Perform the following calculations and report each answer with the correct number of significant figures. Consider the results of the archery contest shown in this figure. Classify the following sets of measurements as accurate, precise, both, or neither. Write conversion factors (as ratios) for the number of: (a) \(\mathrm{\dfrac{1.0936\: yd}{1\: m}}\); (b) \(\mathrm{\dfrac{0.94635\: L}{1\: qt}}\); (c) \(\mathrm{\dfrac{2.2046\: lb}{1\: kg}}\) Write conversion factors (as ratios) for the number of: The label on a soft drink bottle gives the volume in two units: 2.0 L and 67.6 fl oz. Use this information to derive a conversion factor between the English and metric units. How many significant figures can you justify in your conversion factor? \(\mathrm{\dfrac{2.0\: L}{67.6\: fl\: oz}=\dfrac{0.030\: L}{1\: fl\: oz}}\) Only two significant figures are justified. The label on a box of cereal gives the mass of cereal in two units: 978 grams and 34.5 oz. Use this information to find a conversion factor between the English and metric units. How many significant figures can you justify in your conversion factor? Soccer is played with a round ball having a circumference between 27 and 28 in. and a weight between 14 and 16 oz. What are these specifications in units of centimeters and grams? 68–71 cm; 400–450 g A woman's basketball has a circumference between 28.5 and 29.0 inches and a maximum weight of 20 ounces (two significant figures). What are these specifications in units of centimeters and grams? How many milliliters of a soft drink are contained in a 12.0-oz can? 355 mL A barrel of oil is exactly 42 gal. How many liters of oil are in a barrel? The diameter of a red blood cell is about 3 \(\times\) 10 in. What is its diameter in centimeters? 8 \(\times\) 10 cm The distance between the centers of the two oxygen atoms in an oxygen molecule is 1.21 \(\times\) 10 cm. What is this distance in inches? Is a 197-lb weight lifter light enough to compete in a class limited to those weighing 90 kg or less? yes; weight = 89.4 kg A very good 197-lb weight lifter lifted 192 kg in a move called the clean and jerk. What was the mass of the weight lifted in pounds? Many medical laboratory tests are run using 5.0 μL blood serum. What is this volume in milliliters? 5.0 \(\times\) 10 mL If an aspirin tablet contains 325 mg aspirin, how many grams of aspirin does it contain? Use scientific (exponential) notation to express the following quantities in terms of the SI base units in : (a) 1.3 \(\times\) 10 kg; (b) 2.32 \(\times\) 10 kg; (c) 5.23 \(\times\) 10 m; (d) 8.63 \(\times\) 10 kg; (e) 3.76 \(\times\) 10 m; (f) 5.4 \(\times\) 10 m; (g) 1 \(\times\) 10 s; (h) 2.7 \(\times\) 10 s; (i) 1.5 \(\times\) 10 K Complete the following conversions between SI units. Gasoline is sold by the liter in many countries. How many liters are required to fill a 12.0-gal gas tank? 45.4 L Milk is sold by the liter in many countries. What is the volume of exactly 1/2 gal of milk in liters? A long ton is defined as exactly 2240 lb. What is this mass in kilograms? 1.0160 \(\times\) 10 kg Make the conversion indicated in each of the following: Make the conversion indicated in each of the following: Many chemistry conferences have held a 50-Trillion Angstrom Run (two significant figures). How long is this run in kilometers and in miles? (1 Å = 1 \(\times\) 10 m) A chemist’s 50-Trillion Angstrom Run (see ) would be an archeologist’s 10,900 cubit run. How long is one cubit in meters and in feet? (1 Å = 1 \(\times\) 10 cm) 0.46 m; 1.5 ft/cubit The gas tank of a certain luxury automobile holds 22.3 gallons according to the owner’s manual. If the density of gasoline is 0.8206 g/mL, determine the mass in kilograms and pounds of the fuel in a full tank. As an instructor is preparing for an experiment, he requires 225 g phosphoric acid. The only container readily available is a 150-mL Erlenmeyer flask. Is it large enough to contain the acid, whose density is 1.83 g/mL? Yes, the acid's volume is 123 mL. To prepare for a laboratory period, a student lab assistant needs 125 g of a compound. A bottle containing 1/4 lb is available. Did the student have enough of the compound? A chemistry student is 159 cm tall and weighs 45.8 kg. What is her height in inches and weight in pounds? 62.6 in (about 5 ft 3 in.) and 101 lb In a recent Grand Prix, the winner completed the race with an average speed of 229.8 km/h. What was his speed in miles per hour, meters per second, and feet per second? Solve these problems about lumber dimensions. (a) To describe to a European how houses are constructed in the US, the dimensions of “two-by-four” lumber must be converted into metric units. The thickness \(\times\) width \(\times\) length dimensions are 1.50 in. \(\times\) 3.50 in. \(\times\) 8.00 ft in the US. What are the dimensions in cm \(\times\) cm \(\times\) m? (b) This lumber can be used as vertical studs, which are typically placed 16.0 in. apart. What is that distance in centimeters? (a) 3.81 cm \(\times\) 8.89 cm \(\times\) 2.44 m; (b) 40.6 cm The mercury content of a stream was believed to be above the minimum considered safe—1 part per billion (ppb) by weight. An analysis indicated that the concentration was 0.68 parts per billion. What quantity of mercury in grams was present in 15.0 L of the water, the density of which is 0.998 g/ml? \(\mathrm{\left(1\: ppb\: Hg=\dfrac{1\: ng\: Hg}{1\: g\: water}\right)}\) Calculate the density of aluminum if 27.6 cm has a mass of 74.6 g. 2.70 g/cm Osmium is one of the densest elements known. What is its density if 2.72 g has a volume of 0.121 cm ? Calculate these masses. (a) What is the mass of 6.00 cm of mercury, density = 13.5939 g/cm ? (b) What is the mass of 25.0 mL octane, density = 0.702 g/cm ? (a) 81.6 g; (b) 17.6 g Calculate these masses. Calculate these volumes. (a) 5.1 mL; (b) 37 L Calculate these volumes. Convert the boiling temperature of gold, 2966 °C, into degrees Fahrenheit and kelvin. 5371 °F, 3239 K Convert the temperature of scalding water, 54 °C, into degrees Fahrenheit and kelvin. Convert the temperature of the coldest area in a freezer, −10 °F, to degrees Celsius and kelvin. −23 °C, 250 K Convert the temperature of dry ice, −77 °C, into degrees Fahrenheit and kelvin. Convert the boiling temperature of liquid ammonia, −28.1 °F, into degrees Celsius and kelvin. −33.4 °C, 239.8 K The label on a pressurized can of spray disinfectant warns against heating the can above 130 °F. What are the corresponding temperatures on the Celsius and kelvin temperature scales? The weather in Europe was unusually warm during the summer of 1995. The TV news reported temperatures as high as 45 °C. What was the temperature on the Fahrenheit scale? 113 °F ). | 15,399 | 3,717 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Nomenclature/Omitting_numbers_in_Nomenclature |
In naming organic chemicals, the nomenclature must specify the location of all special features. This is commonly done using "locants" such as numbers, as in 2-bromobutane. There are some cases where the number is omitted. However, students are sometimes confused about when they can omit a number. This page summarizes the cases where the number, most often a 1, can be omitted, and discusses a common misconception about when it can be omitted. The general rule is that the locant (number) can be omitted when there is no ambiguity; that is, when only one location is possible. It cannot be omitted by making any assumption that a certain location is more likely. The examples below are presented without showing the structures. All are simple compounds, and you should be able to draw them. In fact, this page will probably work best if you draw the structures for yourself as you go, so you can think about whether or not a number is required. The cases where the number can be omitted can be considered in three classes: The aldehyde and carboxylic acid functional groups can occur only at the end of a chain. When one of these is the main functional group, its position is as "1". Thus butanal (aldehyde) and butanoic acid (carboxylic acid) have the indicated functional group at the 1-position. Similarly, butanedial and butanedioic acid have one functional group at each end; there is no other possibility, and the numbers are omitted. All positions on a ring are equivalent. If there is one substituent, that substituent the "1" position. Thus bromocyclobutane and bromobenzene do not need numbers. There is only one possible compound of each name, and the bromo position is defined as "1". This applies to single ring compounds, where all positions on the ring are equivalent. There are several chemicals, generally small ones, where the number for a substituent is commonly omitted, even though they do not quite fit one of the specific exceptions discussed above. However, they do fit the general criterion that the number is not needed, because there is no other choice. A list of some of those special cases follows. I encourage you to draw the structures, and be sure you agree that only one structure is possible in each case: Sometimes people think that the 1 is a "default" number. That is, they think that the number 1 is implied if no number is given. Using this logic, they might give octanol or octene as (incorrect) names for 1-octanol or 1-octene. As you read the cases above where the number is omitted, it is important to realize that it is omitted because there is no other possibility, not because there is a default number. That is, one never "assumes" that a group is at "1". A chemical name conveys information about the structure of the chemical; a name can be thought of as an "instruction manual" for how to draw the structure. When we first started discussing the nomenclature of organic compounds, we made the point that there are two types of wrong names. One type of incorrect name fails to provide the correct information. Another type of incorrect name may provide the information, but not follow the official (IUPAC) "rules" for doing so. As an example, consider the simple chemical 1-chloropropane. If someone names this chloropropane, they have failed to communicate what compound they mean (since the chloro could be at either the 1-position or the 2-position). If someone names it 3-chloropropane, it is quite clear what they mean, even though they have violated the naming rules (give the substituent the lowest possible number). The first type of error is much more important, as it means the primary goal of naming -- communication of the structure -- has failed. How is this relevant to the discussion of omitting numbers? Well, you can do more harm by omitting a number that is needed than by giving one that is not needed. If it doubt, include the number. Omit the number only when it is perfectly clear what the meaning is. | 3,991 | 3,718 |
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The simplest example of a macromolecular solid is . Crystals of diamond contain only carbon atoms, and these are linked to each other by covalent bonds in a giant three-dimensional network, as shown below. Note how each carbon atom is surrounded tetrahedrally by four bonds. Such a network of carbon atoms extends throughout the crystal so that the whole diamond is one extremely large covalently bonded entity, i.e., a macromolecule. Because strong covalent bonds, rather than London forces or dipole forces, hold the carbon atoms together in this crystal, it takes a great deal of energy to separate them. Accordingly, diamond has an extremely high melting point, 3550°C—much higher than any ionic solid. Diamond is also the hardest substance known. Each carbon atom is held firmly in its place from all sides and is thus very difficult to displace or remove. Carbon also exists in a second, more familiar, crystalline form called , whose crystal structure is also shown in part of the figure. You use graphite every time you write with a pencil. (Pencil leads consist of C, not Pb!) The structure of graphite consists of flat layers. In each layer the carbon atoms are arranged in a regular hexagonal array. We can regard each layer as a large number of benzene rings fused together to form a gigantic honeycomb. All carbon-carbon bonds in this honeycomb are equivalent and intermediate in character between a single and a double bond. While there are strong covalent bonds between the carbon atoms in a given plane, only weak London forces attract the planes together. The various layers can therefore slide past each other quite easily. When a pencil lead rubs across paper, the planes slide past each other and thin plates of crystal are left behind on the paper. These sliding plates also make graphite useful as a lubricant. When an element can exist in more than one crystalline form, as carbon can in diamond and graphite, each form is said to be an . Other elements, such as sulfur and phosphorus, also form allotropes. | 2,047 | 3,719 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Spectroscopy/Infrared_Spectroscopy/Infrared_spectroscopy_2 |
The covalent bonds in molecules are not rigid sticks or rods, such as found in molecular model kits, but are more like stiff springs that can be stretched and bent. In addition to the facile of groups about single bonds, molecules experience a wide variety of motions, characteristic of their component atoms. Consequently, virtually all organic compounds will absorb infrared radiation that corresponds in energy to these vibrations. Infrared spectrometers, similar in principle to the UV-Visible spectrometer, permit chemists to obtain absorption spectra of compounds that are a unique reflection of their molecular structure. Below are some characteristic vibrational modes of organic molecules: The energy of molecular vibration is rather than continuous, meaning that a molecule can only stretch and bend at certain 'allowed' frequencies. If a molecule is exposed to electromagnetic radiation that matches the frequency of one of its vibrational modes, it will in most cases absorb energy from the radiation and jump to a higher vibrational energy state - what this means is that the of the vibration will increase, but the vibrational will remain the same. The difference in energy between the two vibrational states is equal to the energy associated with the wavelength of radiation that was absorbed. It turns out that it is the region of the electromagnetic spectrum which contains frequencies corresponding to the vibrational frequencies of organic bonds. Let's take 2-hexanone as an example. Picture the carbonyl bond of the ketone group as a spring. This spring is constantly bouncing back and forth, stretching and compressing, pushing the carbon and oxygen atoms further apart and then pulling them together. This is the of the carbonyl bond. In the space of one second, the spring 'bounces' back and forth 5.15 x 10 times - in other words, the ground-state frequency of carbonyl stretching for a the ketone group is about 5.15 x 10 Hz. If our ketone sample is irradiated with infrared light, the carbonyl bond will specifically absorb light with this same frequency, which by equations 4.1 and 4.2 corresponds to a wavelength of 5.83 x 10 m and an energy of 4.91 kcal/mol. When the carbonyl bond absorbs this energy, it jumps up to an excited vibrational state. The value of E - the energy difference between the low energy (ground) and high energy (excited) vibrational states - is equal to 4.91 kcal/mol, the same as the energy associated with the absorbed light frequency. The molecule does not remain in its excited vibrational state for very long, but quickly releases energy to the surrounding environment in form of heat, and returns to the ground state. With an instrument called an infrared spectrophotometer, we can 'see' this vibrational transition. In the spectrophotometer, infrared light with frequencies ranging from about 10 to 10 Hz is passed though our sample of cyclohexane. Most frequencies pass right through the sample and are recorded by a detector on the other side. Our 5.15 x 10 Hz carbonyl stretching frequency, however, is absorbed by the 2-hexanone sample, and so the detector records that the intensity of this frequency, after having passed through the sample, is something less than 100% of its initial intensity. The vibrations of a 2-hexanone molecule are not, of course, limited to the simple stretching of the carbonyl bond. The various carbon-carbon bonds also stretch and bend, as do the carbon-hydrogen bonds, and all of these vibrational modes also absorb different frequencies of infrared light. The power of infrared spectroscopy arises from the observation that . The carbonyl bond in a ketone, as we saw with our 2-hexanone example, typically absorbs in the range of 5.11 - 5.18 x 10 Hz, depending on the molecule. The carbon-carbon triple bond of an alkyne, on the other hand, absorbs in the range 6.30 - 6.80 x 10 Hz. The technique is therefore very useful as a means of identifying which functional groups are present in a molecule of interest. If we pass infrared light through an unknown sample and find that it absorbs in the carbonyl frequency range but not in the alkyne range, we can infer that the molecule contains a carbonyl group but not an alkyne. Some bonds absorb infrared light more strongly than others, and some bonds do not absorb at all. . Such vibrations are said to be . In general, the greater the polarity of the bond, the stronger its IR absorption. The carbonyl bond is very polar, and absorbs very strongly. The carbon-carbon triple bond in most alkynes, in contrast, is much less polar, and thus a stretching vibration does not result in a large change in the overall dipole moment of the molecule. Alkyne groups absorb rather weakly compared to carbonyls. Some kinds of vibrations are . The stretching vibrations of completely symmetrical double and triple bonds, for example, do not result in a change in dipole moment, and therefore do not result in any absorption of light (but other bonds and vibrational modes in these molecules absorb IR light). Now, let's look at some actual output from IR spectroscopy experiments. Below is the IR spectrum for 2-hexanone. There are a number of things that need to be explained in order for you to understand what it is that we are looking at. On the horizontal axis we see IR wavelengths expressed in terms of a unit called (cm ), which tells us how many waves fit into one centimeter. On the vertical axis we see ‘ ’, which tells us how strongly light was absorbed at each frequency (100% transmittance means no absorption occurred at that frequency). The solid line traces the values of % transmittance for every wavelength – the ‘peaks’ (which are actually pointing down) show regions of strong absorption. For some reason, it is typical in IR spectroscopy to report wavenumber values rather than wavelength (in meters) or frequency (in Hz). The ‘upside down’ vertical axis, with absorbance peaks pointing down rather than up, is also a curious convention in IR spectroscopy. We wouldn’t want to make things too easy for you! The key absorption peak in this spectrum is that from the carbonyl double bond, at 1716 cm (corresponding to a wavelength of 5.86 mm, a frequency of 5.15 x 10 Hz, and a E value of 4.91 kcal/mol). Notice how strong this peak is, relative to the others on the spectrum: . Within that range, carboxylic acids, esters, ketones, and aldehydes tend to absorb in the shorter wavelength end (1700-1750 cm-1), while conjugated unsaturated ketones and amides tend to absorb on the longer wavelength end (1650-1700 cm ). The jagged peak at approximately 2900-3000 cm is characteristic of tetrahedral carbon-hydrogen bonds. This peak is not terribly useful, as just about every organic molecule that you will have occasion to analyze has these bonds. Nevertheless, it can serve as a familiar reference point to orient yourself in a spectrum. You will notice that there are many additional peaks in this spectrum in the longer-wavelength 400 -1400 cm region. This part of the spectrum is called the . While it is usually very difficult to pick out any specific functional group identifications from this region, it does, nevertheless, contain valuable information. The reason for this is suggested by the name: just like a human fingerprint, the pattern of absorbance peaks in the fingerprint region is unique to every molecule, meaning that the data from an unknown sample can be compared to the IR spectra of known standards in order to make a positive identification. In the mid-1990's, for example, several paintings were identified as forgeries because scientists were able to identify the IR footprint region of red and yellow pigment compounds that would not have been available to the artist who supposedly created the painting (for more details see Chemical and Engineering News, Sept 10, 2007, p. 28). Now, let’s take a look at the IR spectrum for 1-hexanol. As you can see, the carbonyl peak is gone, and in its place is a very broad ‘mountain’ centered at about 3400 cm . This signal is characteristic of the O-H stretching mode of alcohols, and is a dead giveaway for the presence of an alcohol group. The breadth of this signal is a consequence of hydrogen bonding between molecules. In the spectrum of octanoic acid we see, as expected, the characteristic carbonyl peak, this time at 1709 cm . We also see a low, broad absorbance band that looks like an alcohol, except that it is displaced slightly to the right (long-wavelength) side of the spectrum, causing it to overlap to some degree with the C-H region. This is the characteristic carboxylic acid O-H single bond stretching absorbance. The spectrum for 1-octene shows two peaks that are characteristic of alkenes: the one at 1642 cm is due to stretching of the carbon-carbon double bond, and the one at 3079 cm-1 is due to stretching of the s bond between the alkene carbons and their attached hydrogens. Alkynes have characteristic IR absorbance peaks in the range of 2100-2250 cm due to stretching of the carbon-carbon triple bond, and terminal alkenes can be identified by their absorbance at about 3300 cm-1, due to stretching of the bond between the sp-hybridized carbon and the terminal hydrogen. It is possible to identify other functional groups such as amines and ethers, but the characteristic peaks for these groups are considerably more subtle and/or variable, and often are overlapped with peaks from the fingerprint region. For this reason, we will limit our discussion here to the most easily recognized functional groups, which are summarized in this table. As you can imagine, obtaining an IR spectrum for a compound will not allow us to figure out the complete structure of even a simple molecule, unless we happen to have a reference spectrum for comparison. In conjunction with other analytical methods, however, IR spectroscopy can prove to be a very valuable tool, given the information it provides about the presence or absence of key functional groups. IR can also be a quick and convenient way for a chemist to check to see if a reaction has proceeded as planned. If we were to run a reaction in which we wished to convert cyclohexanone to cyclohexanol, for example, a quick comparison of the IR spectra of starting compound and product would tell us if we had successfully converted the ketone group to an alcohol. To illustrate the usefulness of infrared absorption spectra, examples for five C H O isomers are presented below their corresponding structural formulas. Try to associate each spectrum with one of the isomers in the row above it.
| 10,650 | 3,720 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Lipids/Glycerides/Phosphoglycerides_or_Phospholipids |
Phospholipids are similar to the triglycerides with a couple of exceptions. Phospholglycerides are esters of only two fatty acids, phosphoric acid and a trifunctional alcohol - glycerol (IUPAC name is 1,2,3-propantriol). The fatty acids are attached to the glycerol at the 1 and 2 positions on glycerol through ester bonds. There may be a variety of fatty acids, both saturated and unsatured, in the phospholipids. The third oxygen on glycerol is bonded to phosphoric acid through a bond (oxygen-phosphorus double bond oxygen). In addition, there is usually a complex amino alcohol also attached to the phosphate through a second phosphate ester bond. The complex amino alcohols include choline, ethanolamine, and the amino acid-serine. The properties of a phospholipid are characterized by the properties of the fatty acid chain and the phosphate/amino alcohol. The long hydrocarbon chains of the fatty acids are of course non-polar. The phosphate group has a negatively charged oxygen and a positively charged nitrogen to make this group ionic. In addition there are other oxygen of the ester groups, which make on whole end of the molecule strongly ionic and polar. Phospholipids are major components in the of cell membranes. There are two common phospholipids: Lecithin is probably the most common phospholipid. It is found in egg yolks, wheat germ, and soybeans. Lecithin is extracted from soy beans for use as an emulsifying agent in foods. Lecithin is an emulsifier because it has both polar and non-polar properties, which enable it to cause the mixing of other fats and oils with water components. See more discussion on this property in . Lecithin is also a major component in the lipid bilayers of cell membranes. Lecithin contains the ammonium salt of choline joined to the phosphate by an ester linkage. The nitrogen has a positive charge, just as in the ammonium ion. In choline, the nitrogen has the positive charge and has four methyl groups attached. Cephalins are phosphoglycerides that contain ehtanolamine or the amino acid serine attached to the phosphate group through phosphate ester bonds. A variety of fatty acids make up the rest of the molecule. Cephalins are found in most cell membranes, particularly in brain tissues. They also iimportant in the blood clotting process as they are found in blood platelets. Note: The MEP coloration of the electrostatic potential does not show a strong red color for the phosphate-amino alcohol portion of the molecule as it should to show the strong polar property of that group. | 2,567 | 3,721 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.4%3A_Molecular_Orbital_Theory |
For almost every covalent molecule that exists, we can now draw the Lewis structure, predict the electron-pair geometry, predict the molecular geometry, and come close to predicting bond angles. However, one of the most important molecules we know, the oxygen molecule O , presents a problem with respect to its Lewis structure. We would write the following Lewis structure for O : This electronic structure adheres to all the rules governing Lewis theory. There is an O=O double bond, and each oxygen atom has eight electrons around it. However, this picture is at odds with the magnetic behavior of oxygen. By itself, O is not magnetic, but it is attracted to magnetic fields. Thus, when we pour liquid oxygen past a strong magnet, it collects between the poles of the magnet and defies gravity. Such attraction to a magnetic field is called , and it arises in molecules that have unpaired electrons. And yet, the Lewis structure of O indicates that all electrons are paired. How do we account for this discrepancy? Magnetic susceptibility measures the force experienced by a substance in a magnetic field. When we compare the weight of a sample to the weight measured in a magnetic field (Figure \(\Page {1}\)), paramagnetic samples that are attracted to the magnet will appear heavier because of the force exerted by the magnetic field. We can calculate the number of unpaired electrons based on the increase in weight. Experiments show that each O molecule has two unpaired electrons. The Lewis-structure model does not predict the presence of these two unpaired electrons. Unlike oxygen, the apparent weight of most molecules decreases slightly in the presence of an inhomogeneous magnetic field. Materials in which all of the electrons are paired are and weakly repel a magnetic field. Paramagnetic and diamagnetic materials do not act as permanent magnets. Only in the presence of an applied magnetic field do they demonstrate attraction or repulsion. Molecular orbital theory (MO theory) provides an explanation of chemical bonding that accounts for the paramagnetism of the oxygen molecule. It also explains the bonding in a number of other molecules, such as violations of the octet rule and more molecules with more complicated bonding (beyond the scope of this text) that are difficult to describe with Lewis structures. Additionally, it provides a model for describing the energies of electrons in a molecule and the probable location of these electrons. Unlike valence bond theory, which uses hybrid orbitals that are assigned to one specific atom, MO theory uses the combination of atomic orbitals to yield molecular orbitals that are delocalized over the entire molecule rather than being localized on its constituent atoms. MO theory also helps us understand why some substances are electrical conductors, others are semiconductors, and still others are insulators. Table \(\Page {1}\) summarizes the main points of the two complementary bonding theories. Both theories provide different, useful ways of describing molecular structure. describes the distribution of electrons in molecules in much the same way that the distribution of electrons in atoms is described using atomic orbitals. Using quantum mechanics, the behavior of an electron in a molecule is still described by a wave function, , analogous to the behavior in an atom. Just like electrons around isolated atoms, electrons around atoms in molecules are limited to discrete (quantized) energies. The region of space in which a valence electron in a molecule is likely to be found is called a . Like an atomic orbital, a molecular orbital is full when it contains two electrons with opposite spin. We will consider the molecular orbitals in molecules composed of two identical atoms (H or Cl , for example). Such molecules are called . In these diatomic molecules, several types of molecular orbitals occur. The mathematical process of combining atomic orbitals to generate molecular orbitals is called the . The wave function describes the wavelike properties of an electron. Molecular orbitals are combinations of atomic orbital wave functions. Combining waves can lead to constructive interference, in which peaks line up with peaks, or destructive interference, in which peaks line up with troughs (Figure \(\Page {2}\)). In orbitals, the waves are three dimensional, and they combine with in-phase waves producing regions with a higher probability of electron density and out-of-phase waves producing nodes, or regions of no electron density. There are two types of molecular orbitals that can form from the overlap of two atomic orbitals on adjacent atoms. The two types are illustrated in Figure \(\Page {3}\). The in-phase combination produces a lower energy (read as "sigma-s") in which most of the electron density is directly between the nuclei. The out-of-phase addition (which can also be thought of as subtracting the wave functions) produces a higher energy \(σ^∗_s\) m (read as "sigma-s-star") molecular orbital in which there is a node between the nuclei. The asterisk signifies that the orbital is an antibonding orbital. Electrons in a σ orbital are attracted by both nuclei at the same time and are more stable (of lower energy) than they would be in the isolated atoms. Adding electrons to these orbitals creates a force that holds the two nuclei together, so we call these orbitals . Electrons in the \(σ^∗_s\) orbitals are located well away from the region between the two nuclei. The attractive force between the nuclei and these electrons pulls the two nuclei apart. Hence, these orbitals are called . Electrons fill the lower-energy bonding orbital before the higher-energy antibonding orbital, just as they fill lower-energy atomic orbitals before they fill higher-energy atomic orbitals. In orbitals, the wave function gives rise to two lobes with opposite phases, analogous to how a two-dimensional wave has both parts above and below the average. We indicate the phases by shading the orbital lobes different colors. When orbital lobes of the same phase overlap, constructive wave interference increases the electron density. When regions of opposite phase overlap, the destructive wave interference decreases electron density and creates nodes. When orbitals overlap end to end, they create σ and σ* orbitals (Figure \(\Page {4}\)). If two atoms are located along the -axis in a Cartesian coordinate system, the two orbitals overlap end to end and form σ (bonding) and \(σ^∗_{px}\) (antibonding) (read as "sigma-p-x" and "sigma-p-x star," respectively). Just as with -orbital overlap, the asterisk indicates the orbital with a node between the nuclei, which is a higher-energy, antibonding orbital. The side-by-side overlap of two orbitals gives rise to a and a \( , as shown in Figure \(\Page {5}\). In valence bond theory, we describe π bonds as containing a nodal plane containing the internuclear axis and perpendicular to the lobes of the orbitals, with electron density on either side of the node. In molecular orbital theory, we describe the π orbital by this same shape, and a π bond exists when this orbital contains electrons. Electrons in this orbital interact with both nuclei and help hold the two atoms together, making it a bonding orbital. For the out-of-phase combination, there are two nodal planes created, one along the internuclear axis and a perpendicular one between the nuclei. In the molecular orbitals of diatomic molecules, each atom also has two sets of orbitals oriented side by side ( and ), so these four atomic orbitals combine pairwise to create two π orbitals and two \(π^*\) orbitals. The \(π_{py}\) and \(π^∗_{py}\) orbitals are oriented at right angles to the \(π_{pz}\) and \(π^∗_{pz}\) orbitals. Except for their orientation, the π and π orbitals are identical and have the same energy; they are . The \(π^∗_{py}\) and \(π^∗_{pz}\) antibonding orbitals are also degenerate and identical except for their orientation. A total of six molecular orbitals results from the combination of the six atomic orbitals in two atoms: \(σ_{px}\) and \(σ^∗_{px}\), \(π_{py}\) and \(π^∗_{py}\), \(π_{pz}\) and \(π^∗_{pz}\). Predict what type (if any) of molecular orbital would result from adding the wave functions so each pair of orbitals shown overlap. The orbitals are all similar in energy. Label the molecular orbital shown as \(σ\) or \(π\), bonding or antibonding and indicate where the node occurs. The orbital is located along the internuclear axis, so it is a \(σ\) orbital. There is a node bisecting the internuclear axis, so it is an antibonding orbital. While the descriptions of bonding described in this chapter involve many theoretical concepts, they also have many practical, real-world applications. For example, drug design is an important field that uses our understanding of chemical bonding to develop pharmaceuticals. This interdisciplinary area of study uses biology (understanding diseases and how they operate) to identify specific targets, such as a binding site that is involved in a disease pathway. By modeling the structures of the binding site and potential drugs, computational chemists can predict which structures can fit together and how effectively they will bind (Figure \(\Page {6}\)). Thousands of potential candidates can be narrowed down to a few of the most promising candidates. These candidate molecules are then carefully tested to determine side effects, how effectively they can be transported through the body, and other factors. Dozens of important new pharmaceuticals have been discovered with the aid of computational chemistry, and new research projects are underway. The relative energy levels of atomic and molecular orbitals are typically shown in a (Figure \(\Page {7}\)). For a diatomic molecule, the atomic orbitals of one atom are shown on the left, and those of the other atom are shown on the right. Each horizontal line represents one orbital that can hold two electrons. The molecular orbitals formed by the combination of the atomic orbitals are shown in the center. Dashed lines show which of the atomic orbitals combine to form the molecular orbitals. For each pair of atomic orbitals that combine, one lower-energy (bonding) molecular orbital and one higher-energy (antibonding) orbital result. Thus we can see that combining the six 2 atomic orbitals results in three bonding orbitals (one σ and two π) and three antibonding orbitals (one σ* and two π*). We predict the distribution of electrons in these molecular orbitals by filling the orbitals in the same way that we fill atomic orbitals, by the Aufbau principle. Lower-energy orbitals fill first, electrons spread out among degenerate orbitals before pairing, and each orbital can hold a maximum of two electrons with opposite spins (Figure \(\Page {7}\)). Just as we write electron configurations for atoms, we can write the molecular electronic configuration by listing the orbitals with superscripts indicating the number of electrons present. For clarity, we place parentheses around molecular orbitals with the same energy. In this case, each orbital is at a different energy, so parentheses separate each orbital. Thus we would expect a diatomic molecule or ion containing seven electrons (such as \(\ce{Be2+}\)) would have the molecular electron configuration \((σ_{1s})^2(σ^∗_{1s})^2(σ_{2s})^2(σ^∗_{2s})^1\). It is common to omit the core electrons from molecular orbital diagrams and configurations and include only the valence electrons. The filled molecular orbital diagram shows the number of electrons in both bonding and antibonding molecular orbitals. The net contribution of the electrons to the bond strength of a molecule is identified by determining the that results from the filling of the molecular orbitals by electrons. When using Lewis structures to describe the distribution of electrons in molecules, we define bond order as the number of bonding pairs of electrons between two atoms. Thus a single bond has a bond order of 1, a double bond has a bond order of 2, and a triple bond has a bond order of 3. We define bond order differently when we use the molecular orbital description of the distribution of electrons, but the resulting bond order is usually the same. The MO technique is more accurate and can handle cases when the Lewis structure method fails, but both methods describe the same phenomenon. In the molecular orbital model, an electron contributes to a bonding interaction if it occupies a bonding orbital and it contributes to an antibonding interaction if it occupies an antibonding orbital. The bond order is calculated by subtracting the destabilizing (antibonding) electrons from the stabilizing (bonding) electrons. Since a bond consists of two electrons, we divide by two to get the bond order. We can determine bond order with the following equation: \[\textrm{bond order}=\dfrac{(\textrm{number of bonding electrons})−(\textrm{number of antibonding electrons})}{2} \nonumber \] The order of a covalent bond is a guide to its strength; a bond between two given atoms becomes stronger as the bond order increases. If the distribution of electrons in the molecular orbitals between two atoms is such that the resulting bond would have a bond order of zero, a stable bond does not form. We next look at some specific examples of MO diagrams and bond orders. A dihydrogen molecule (H ) forms from two hydrogen atoms. When the atomic orbitals of the two atoms combine, the electrons occupy the molecular orbital of lowest energy, the σ bonding orbital. A dihydrogen molecule, H , readily forms because the energy of a H molecule is lower than that of two H atoms. The σ orbital that contains both electrons is lower in energy than either of the two 1 atomic orbitals. A molecular orbital can hold two electrons, so both electrons in the H molecule are in the σ bonding orbital; the electron configuration is \((σ_{1s})^2\). We represent this configuration by a molecular orbital energy diagram (Figure \(\Page {8}\)) in which a single upward arrow indicates one electron in an orbital, and two (upward and downward) arrows indicate two electrons of opposite spin. A dihydrogen molecule contains two bonding electrons and no antibonding electrons so we have \[\ce{bond\: order\: in\: H2}=\dfrac{(2−0)}{2}=1 \nonumber \] Because the bond order for the H–H bond is equal to 1, the bond is a single bond. A helium atom has two electrons, both of which are in its 1 orbital. Two helium atoms do not combine to form a dihelium molecule, He , with four electrons, because the stabilizing effect of the two electrons in the lower-energy bonding orbital would be offset by the destabilizing effect of the two electrons in the higher-energy antibonding molecular orbital. We would write the hypothetical electron configuration of He as \((σ_{1s})^2(σ^∗_{1s})^2\) as in Figure \(\Page {9}\). The net energy change would be zero, so there is no driving force for helium atoms to form the diatomic molecule. In fact, helium exists as discrete atoms rather than as diatomic molecules. The bond order in a hypothetical dihelium molecule would be zero. \[\ce{bond\: order\: in\: He2}=\dfrac{(2−2)}{2}=0 \nonumber \] A bond order of zero indicates that no bond is formed between two atoms. Eight possible homonuclear diatomic molecules might be formed by the atoms of the second period of the periodic table: Li , Be , B , C , N , O , F , and Ne . However, we can predict that the Be molecule and the Ne molecule would not be stable. We can see this by a consideration of the molecular electron configurations (Table \(\Page {1}\)). We predict valence molecular orbital electron configurations just as we predict electron configurations of atoms. Valence electrons are assigned to valence molecular orbitals with the lowest possible energies. Consistent with Hund’s rule, whenever there are two or more degenerate molecular orbitals, electrons fill each orbital of that type singly before any pairing of electrons takes place. As we saw in valence bond theory, σ bonds are generally more stable than π bonds formed from degenerate atomic orbitals. Similarly, in molecular orbital theory, σ orbitals are usually more stable than π orbitals. However, this is not always the case. The MOs for the valence orbitals of the second period are shown in Figure \(\Page {10}\). Looking at Ne molecular orbitals, we see that the order is consistent with the generic diagram shown in the previous section. However, for atoms with three or fewer electrons in the orbitals (Li through N) we observe a different pattern, in which the σ orbital is higher in energy than the π set. Obtain the molecular orbital diagram for a homonuclear diatomic ion by adding or subtracting electrons from the diagram for the neutral molecule. This switch in orbital ordering occurs because of a phenomenon called . s-p mixing does not create new orbitals; it merely influences the energies of the existing molecular orbitals. The σ wavefunction mathematically combines with the σ wavefunction, with the result that the σ orbital becomes more stable, and the σ orbital becomes less stable (Figure \(\Page {11}\)). Similarly, the antibonding orbitals also undergo s-p mixing, with the σ becoming more stable and the σ becoming less stable. s-p mixing occurs when the and orbitals have similar energies. The energy difference between 2 and 2 orbitals in O, F, and Ne is greater than that in Li, Be, B, C, and N. Because of this, O , F , and Ne exhibit negligible s-p mixing (not sufficient to change the energy ordering), and their MO diagrams follow the normal pattern, as shown in Figure \(\Page {7}\). All of the other period 2 diatomic molecules do have s-p mixing, which leads to the pattern where the σ orbital is raised above the π set. Using the MO diagrams shown in Figure \(\Page {11}\), we can add in the electrons and determine the molecular electron configuration and bond order for each of the diatomic molecules. As shown in Table \(\Page {1}\), Be and Ne molecules would have a bond order of 0, and these molecules do not exist. The combination of two lithium atoms to form a lithium molecule, Li , is analogous to the formation of H , but the atomic orbitals involved are the valence 2 orbitals. Each of the two lithium atoms has one valence electron. Hence, we have two valence electrons available for the σ bonding molecular orbital. Because both valence electrons would be in a bonding orbital, we would predict the Li molecule to be stable. The molecule is, in fact, present in an appreciable concentration in lithium vapor at temperatures near the boiling point of the element. All of the other molecules in Table \(\Page {1}\) with a bond order greater than zero are also known. The O molecule has enough electrons to half fill the \((π^∗_{2py},\:π^∗_{2pz})\) level. We expect the two electrons that occupy these two degenerate orbitals to be unpaired, and this molecular electronic configuration for O is in accord with the fact that the oxygen molecule has two unpaired electrons ( Figure \(\Page {10}\)). The presence of two unpaired electrons has proved to be difficult to explain using Lewis structures, but the molecular orbital theory explains it quite well. In fact, the unpaired electrons of the oxygen molecule provide a strong piece of support for the molecular orbital theory. When two identical atomic orbitals on different atoms combine, two molecular orbitals result (e.g., \(H_2\) in Figure \(\Page {8}\)). The bonding orbital is lower in energy than the original atomic orbitals because the atomic orbitals are in-phase in the molecular orbital. The antibonding orbital is higher in energy than the original atomic orbitals because the atomic orbitals are out-of-phase. In a solid, similar things happen, but on a much larger scale. Remember that even in a small sample there are a huge number of atoms (typically > 10 atoms), and therefore a huge number of atomic orbitals that may be combined into molecular orbitals. When valence atomic orbitals, all of the same energy and each containing one (1) electron, are combined, /2 (filled) bonding orbitals and /2 (empty) antibonding orbitals will result. Each bonding orbital will show an energy lowering as the atomic orbitals are in-phase, but each of the bonding orbitals will be a little different and have slightly different energies. The antibonding orbitals will show an increase in energy as the atomic orbitals are out-of-phase, but each of the antibonding orbitals will also be a little different and have slightly different energies. The allowed energy levels for all the bonding orbitals are so close together that they form a band, called the valence band. Likewise, all the antibonding orbitals are very close together and form a band, called the conduction band. Figure \(\Page {12}\)) shows the bands for three important classes of materials: insulators, semiconductors, and conductors. In order to conduct electricity, electrons must move from the filled valence band to the empty conduction band where they can move throughout the solid. The size of the band gap, or the energy difference between the top of the valence band and the bottom of the conduction band, determines how easy it is to move electrons between the bands. Only a small amount of energy is required in a conductor because the band gap is very small. This small energy difference is “easy” to overcome, so they are good conductors of electricity. In an insulator, the band gap is so “large” that very few electrons move into the conduction band; as a result, insulators are poor conductors of electricity. Semiconductors conduct electricity when “moderate” amounts of energy are provided to move electrons out of the valence band and into the conduction band. Semiconductors, such as silicon, are found in many electronics. Semiconductors are used in devices such as computers, smartphones, and solar cells. Solar cells produce electricity when light provides the energy to move electrons out of the valence band. The electricity that is generated may then be used to power a light or tool, or it can be stored for later use by charging a battery. As of December 2014, up to 46% of the energy in sunlight could be converted into electricity using solar cells. Draw the molecular orbital diagram for the oxygen molecule, O . From this diagram, calculate the bond order for O . How does this diagram account for the paramagnetism of O ? We draw a molecular orbital energy diagram similar to that shown in Figure \(\Page {7}\). Each oxygen atom contributes six electrons, so the diagram appears as shown in Figure \(\Page {7}\). We calculate the bond order as \[\ce{O2}=\dfrac{(8−4)}{2}=2 \nonumber \] Oxygen's paramagnetism is explained by the presence of two unpaired electrons in the (π , π )* molecular orbitals. The main component of air is N . From the molecular orbital diagram of N , predict its bond order and whether it is diamagnetic or paramagnetic. N has a bond order of 3 and is diamagnetic. Give the molecular orbital configuration for the valence electrons in \(\ce{C2^2-}\). Will this ion be stable? Looking at the appropriate MO diagram, we see that the π orbitals are lower in energy than the σ orbital. The valence electron configuration for C is \((σ_{2s})^2(σ^∗_{2s})^2(π_{2py},\:π_{2pz})^4\). Adding two more electrons to generate the \(\ce{C2^2-}\) anion will give a valence electron configuration of \((σ_{2s})^2(σ^∗_{2s})^2(π_{2py},\:π_{2pz})^4(σ_{2px})^2\) Since this has six more bonding electrons than antibonding, the bond order will be 3, and the ion should be stable. How many unpaired electrons would be present on a \(\ce{Be2^2-}\) ion? Would it be paramagnetic or diamagnetic? two, paramagnetic Molecular orbital (MO) theory describes the behavior of electrons in a molecule in terms of combinations of the atomic wave functions. The resulting molecular orbitals may extend over all the atoms in the molecule. Bonding molecular orbitals are formed by in-phase combinations of atomic wave functions, and electrons in these orbitals stabilize a molecule. Antibonding molecular orbitals result from out-of-phase combinations of atomic wave functions and electrons in these orbitals make a molecule less stable. Molecular orbitals located along an internuclear axis are called σ MOs. They can be formed from orbitals or from orbitals oriented in an end-to-end fashion. Molecular orbitals formed from orbitals oriented in a side-by-side fashion have electron density on opposite sides of the internuclear axis and are called π orbitals. We can describe the electronic structure of diatomic molecules by applying molecular orbital theory to the valence electrons of the atoms. Electrons fill molecular orbitals following the same rules that apply to filling atomic orbitals; Hund’s rule and the Aufbau principle tell us that lower-energy orbitals will fill first, electrons will spread out before they pair up, and each orbital can hold a maximum of two electrons with opposite spins. Materials with unpaired electrons are paramagnetic and attracted to a magnetic field, while those with all-paired electrons are diamagnetic and repelled by a magnetic field. Correctly predicting the magnetic properties of molecules is in advantage of molecular orbital theory over Lewis structures and valence bond theory. | 25,490 | 3,722 |
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Most of us have little difficulty distinguishing living organisms from inanimate matter. The former are capable of reproducing nearly exact copies of themselves; they can appropriate both matter and energy from their surroundings, moving, growing, and repairing damage caused by external factors; and groups of them evolve and adapt in response to long-term environmental changes. On a macroscopic scale the differences are sufficiently striking that early philosophers and scientists postulated the existence of a vital force without which living organisms would be inanimate. It was thought that organic compounds could only be manufactured in living organisms, and chemistry was divided into the subfields of inorganic and organic on this basis. This subdivision persists today, but the definition of organic has changed in response to the discovery of numerous ways to make organic compounds from inorganic starting materials. As seen in the sections on organic compounds, organic chemistry now means “chemistry of compounds containing carbon.” No restriction is placed on the origin of the compounds, and hundreds of thousands of organic compounds which are foreign to all living systems have been produced in laboratories around the world. Indeed, concern about the effects of some of these synthetic substances on the environment has led to yet another definition of organic. The general public now takes it to mean “free of substances produced as a result of human activities.” Just as the division between organic and inorganic chemistry has become more arbitrary with the advance of knowledge, the distinction between life and nonlife has also blurred. Living organisms are made up of atoms and molecules which follow the same chemical principles as any other set of atoms and molecules. Yet there is a difference-these atoms, molecules, and even groups of molecules are organized to a much greater degree than in any of the cases we have discussed. Above a certain level of complexity a collection of chemicals begins to exhibit most of the behavior patterns that we associate with life. A virus, for example, may consist of fewer than 100 associated large molecules. It is the structures of these molecules and the ways in which they are associated that determine a virus’ behavior and make it appear to be on the threshold of life. Biochemistry is the study of chemical elements found in living systems, and how these elements combine to form molecules and collections of molecules which carry out the biological functions and behaviors that we associate with life. Our treatment must of necessity be brief, but even if it were not, the complexity of biochemical systems would insure that it would be incomplete. Much of modern chemistry, both inorganic and organic, involves the extension to complex biological systems of principles and facts gleaned from studies of more general chemical behavior. More than 99 percent of the atoms in most living organisms are H, O, N, or C. These are the smallest atoms which can form one, two, three, and four covalent bonds, respectively, and they are especially suited to make up the more than 10 different, but often related, kinds of molecules estimated to exist in the biosphere. Each such molecule has a specific function to perform in a specific organism, and its molecular structure is very important in determining how that function is carried out. Many biological molecules are formed by condensation polymerization from small building-block molecules. Reversal of such condensations (hydrolysis) breaks large molecules down into the building blocks again, allowing them to be used by another organism. Examples of condensation reactions are the formation of lipids from fatty acids and glycerol, the formation of cellulose and starch from glucose, the formation of proteins from amino acids, and the formation of nucleic acids from ribose or deoxyribose, phosphate, and nitrogenous bases. Lipids may be divided into two categories: nonpolar (hydrophobic) and polar (hydrophilic). Both types consist of long hydrocarbon chains, but polar lipids have electrically charged or hydrogen-bonding groups at one end. Lipid bilayers, in which the hydrophilic ends of polar lipids contact an aqueous phase while the hydrophobic tails intertwine, are important components of cell walls and other membranes. Nonpolar substances often dissolve in the hydrophobic portions of lipid tissue and may be concentrated along ecological food chains, a process referred to as bioamplification. Carbohydrates provide a storehouse for solar energy absorbed during photosynthesis. Simple sugars usually contain five or six C atoms in a ring and a large number of OH groups. Disaccharides are formed from a condensation reaction between two simple sugars. Polysaccharides, such as cellulose and starch, are condensation polymers of simple sugars. Their structures and chemical reactivities are very dependent on the exact structure of the simple sugar from which they are made. Fibrous proteins, in which polypeptide chains are arranged parallel to one another, are fundamental components of structural tissues such as tendons, hair, etc. Globular proteins, on the other hand, have compact, nearly spherical structures in which the polypeptide chain folds back on itself. Enzymes, antibodies, hormones, and hemoglobin are examples of globular proteins. Proteins are made by polymerization of 20 different amino acids. The order of amino acid side chains along the polymer backbone constitutes primary protein structure. Secondary structure involves hydrogen bonding to form α-helix or pleated-sheet structures. The intricate folding of the polypeptide chain in a globular protein is referred to as tertiary structure. Some proteins (hemoglobin, for example) have quaternary structure— several polypeptide chains are nested together. Nucleic acids, such as DNA and RNA and formed from nitrogenous bases, sugars and phosphate, constitute a blueprint and a mechanism for synthesizing useful proteins. Codons, each consisting of a sequence of three nitrogenous bases along a nucleic acid polymer chain, indicate which amino acid goes where. DNA also has secondary structure, the double helix. When cells divide, the double-stranded DNA molecule can replicate itself, and complementary base pairing insures that each new cell will contain identical DNA. During protein synthesis, information is transcribed from DNA to mRNA and then translated from the mRNA code into a protein. In the translation process, tRNA molecules bonded to specific amino acids, base pair their anticodon sequence with a codon sequence on the mRNA. The ribosome, itself constituted of RNA and protein performs the catalytic activity of synthesizing the protein. In this way the primary protein structure is determined, and secondary and higher order structures then follow directly. | 6,901 | 3,723 |
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The experimental facts described in the preceding section can be accounted for by assuming that any atom is made up of three kinds of . Some important properties of the three kinds of subatomic particles are listed in the following table. Experimental evidence for the existence of the neutron was first correctly interpreted in 1932 by James Chadwick (1891 to 1974), a discovery for which he was awarded the Nobel Prize in 1935. The modern picture of a helium atom, which is made up of two electrons, two protons, and two neutrons, is shown below. Because each proton and each neutron has more than 1800 times the mass of an electron, nearly all the mass of the helium atom is accounted for by the nucleus. This agrees with Rutherford’s interpretation of the Geiger-Marsden experiment. The number of units of positive charge on the nucleus is usually about half the number of units of mass because about half the nuclear particles are uncharged neutrons. The two electrons move about rapidly, occupying all the volume of the atom outside the nucleus. Their negative charge neutralizes the positive charge of the two protons, producing a neutral or uncharged atom. The protons and neutrons in the nucleus of an atom such as helium are held very tightly by strong nuclear forces. It is very difficult either to separate the nuclear particles or to add extra ones. The electrons, on the other hand, are held to the atom by their electrostatic attraction for the positively charged protons in the nucleus. This force is strong, but not so strong that an atom cannot lose or gain electrons. When the number of electrons is not the same as the number of protons, an atom has a net electric charge and is called an . The \(α\) particles emitted by radioactive elements consist of two protons and two neutrons tightly bound together. Thus an \(α\) particle is the same as a helium nucleus; that is, a helium atom that has lost its two electrons or a helium ion whose charge is . When particles are emitted into a closed container, they slowly pick up electrons from their surroundings, and eventually the container becomes filled with helium. The structure of any atom may be specified by indicating how many electrons, protons, and neutrons it contains. The number of protons is the same as the number of electrons and is given by the atomic number . Instead of directly specifying how many neutrons are present, we use the \(A\). This is the total number of particles in the nucleus; hence \[\begin{align*}A &= \text{number of protons} + \text{number of neutrons}\\[4pt] &= Z + N \end{align*} \nonumber \] where \(N\) represents the number of neutrons. To symbolize a particular atom, the mass number and atomic number are written as a superscript and subscript preceding the chemical symbol (\(\ce{Sy}\)) as follows: \[{}_{Z}^{A}\text{Sy} \nonumber \] The helium atom, whose structure was represented above, has 2 protons and 2 electrons ( = 2)as well as 2 neutrons. Hence = 2 + 2 = 4, and the atom is represented by \[{}_{2}^{4}\text{He} \nonumber \] In the case of an ion the positive or negative charge is indicated as a superscript to the right of the chemical symbol. Thus a helium atom which had lost two electrons (a helium ion with two more protons than electrons) would be written as \[{}_{2}^{4}\text{He}^{\text{2+}} \nonumber \] How many electrons, protons, and neutrons are there in each of the atoms represented below? For an atom the number of electrons equals the number of protons and is given by . For an ion the atomic number gives the number of protons, but the number of electrons must be determined from the charge. Thus The number of neutrons can be obtained by subtracting the number of protons ( ) from the total number of particles in the nucleus ( ): | 3,794 | 3,724 |
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In chemistry we are interested not only in those changes occurring in a closed container at constant volume but also in those occurring in an open container at constant (i.e., atmospheric) When a change occurs at constant pressure, there is another energy factor we must consider in addition to the heat absorbed and the change in . This is the which the system does as its volume expands against the external pressure. In order to understand the nature and magnitude of this expansion work, let us consider the simple example illustrated in Figure \(\Page {1}\). Here a sample of oxygen gas is heated at a constant pressure from an initial temperature to a final temperature by means of an electrical heating coil. The is confined in a cylinder by a piston, and the is maintained by placing a weight of the correct magnitude on top of the piston. The whole apparatus is maintained in a vacuum so that there is no atmospheric pressure to consider in addition to the effect of the weight and piston. When the heating coil in this apparatus is switched on, the temperature rises and the gas expands in compliance with Charles’ law, lifting the piston and weight in the process. Energy must he supplied from the heating coil not only to increase the energy of the oxygen molecules but also to . In other words, \[q_{P} = \triangle U + w_{exp}\label{1} \] where To calculate the magnitude of this expansion work, we begin with the definition of pressure: \[P=\frac{\text{force}}{\text{area}}=\frac{F}{A} \nonumber \] or where is the total farce exerted by the weight and piston on the oxygen gas and is the area of the piston. As the gas sample is heated, the volume increases from an initial value to a final value , while the piston and its weight move from a height to a height . The work done is thus given by the expression \(w_{exp} = \text{force exerted × distance moved}\) \(= (P * A) * (h_{2} – h_{1})\) \(= P * (A * h_{2} – A * h_{1})\) However, since the volume of a cylinder is the area of the base times the height, \(A * h_1 = V_1\) initial volume of gas and \(A * h_2 = V_2\) final volume Thus \(w_{exp} = P * (V_2 – V_1)\) or \[w_{exp} = P \triangle V\label{2} \] Inserting this value for into Eq. \(\ref{1}\), we obtain the final result \[q_{P} = \triangle U + P \triangle V\label{3} \] It is important to realize that the expansion work Δ does not depend on our sample being in the apparatus of Figure \(\Page {1}\) in which there is an obvious gain in the potential energy of a weight. If instead of a weight we allowed the atmosphere to exert a pressure on the gas, the result would still be Δ . In this second case, instead of lifting a visible weight, the expanding gas would push back the atmosphere and hence be lifting invisible air molecules. The work done, and hence the , would still be Δ . This simple example of an expanding gas helps us to see what is involved in the general case of a chemical or physical change occurring at constant pressure. In any such case the heat energy absorbed by the system, , will always exactly account for the increase in internal energy Δ and the expansion work Δ . In other words the relationship \(q_{P} = \triangle U + P \triangle V\) is valid in the general case. When 1 mol liquid H O is boiled at 100°C and 101.3 kPa (1.000 atm) pressure, its volume expands from 19.8 cm in the liquid state to 30.16 dm³ in the gaseous state. The heat energy absorbed by the vaporization process is found experimentally to be 40.67 kJ. Calculate the increase in internal energy Δ of H O. We must first calculate the expansion work P ΔV. \(w_{exp} = P * (V_2 – V_1)\) \(= 101.3 \text{kPa} * (30.16 \text{dm}^3– 0.0198 \text{dm}^3) = 3053 \text{kPa dm}^3\) , we see that 1Pa × 1 dm = 1 J, and we have Also, since we have : As we see in the sections on and , the vaporization of a liquid is always an endothermic process. Since the molecules attract each other, energy must be supplied to separate them as vaporization occurs. However, not all the energy supplied when a liquid boils goes to increasing the potential energy of the molecules. A significant proportion is needed to increase the potential energy of the air as well. It is far easier to carry out most chemical reactions in a container open to the atmosphere than in a closed system like a bomb calorimeter. However, when a reaction is carried out at constant atmospheric pressure, it is necessary to measure P, and (as in Example \(\Page {1}\) ) as well as in order to calculate the change in from the equation Δ = – Δ . These extra measurements are a nuisance, and they can be avoided by introducing a new quantity which is related to the internal energy but also includes the potential energy of the atmosphere. This quantity is also used in the sections on . It is called the enthalpy, symbol , and is defined by \[H = U + PV\label{4} \] From Eq. \(\ref{4}\) we can see that the enthalpy is always larger than the internal energy by a quantity . This extra energy is added to take account of the fact that whenever a body of volume is introduced into the atmosphere, the potential energy of the atmosphere is increased by [by the same argument used to derive Eq. \(\ref{2}\)]. The enthalpy is thus of the form Enthalpy = internal energy + potential energy of atmosphere When a system undergoes a chemical or physical change at constant pressure, the change in enthalpy is given by Change in enthalpy = change in internal energy + change in potential energy of atmosphere The change in enthalpy thus includes the energy changes for which heat energy must be supplied from the surroundings. In more formal language \[\triangle H = \triangle U + \triangle (PV) = \triangle U + (P_2V_2 – P_1V_1) \nonumber \] If we consider conditions of constant pressure, = = , and \[\triangle H = \triangle U + (PV_2 – PV_1) = \triangle U + P(V_2 – V_1) = \triangle U + P \triangle V \nonumber \] but from Eq. \(\ref{3}\) \[q_P = \triangle U + P \triangle V \nonumber \] Thus In other words, . The change in enthalpy Δ can be obtained from a single experimental measurement: the heat energy absorbed at constant pressure. | 6,219 | 3,725 |
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The laws that describe the behavior of gases were well established long before anyone had developed a coherent model of the properties of gases. In this section, we introduce a theory that describes why gases behave the way they do. The theory we introduce can also be used to derive laws such as the ideal gas law from fundamental principles and the properties of individual particles. The kinetic molecular theory of gases explains the laws that describe the behavior of gases. Developed during the mid-19th century by several physicists, including the Austrian Ludwig Boltzmann (1844–1906), the German Rudolf Clausius (1822–1888), and the Englishman James Clerk Maxwell (1831–1879), who is also known for his contributions to electricity and magnetism, this theory is based on the properties of individual particles as defined for an ideal gas and the fundamental concepts of physics. Thus the kinetic molecular theory of gases provides a molecular explanation for observations that led to the development of the ideal gas law. The kinetic molecular theory of gases is based on the following five postulates: Although the molecules of real gases have nonzero volumes and exert both attractive and repulsive forces on one another, for the moment we will focus on how the kinetic molecular theory of gases relates to the properties of gases we have been discussing. In the following sections, we explain how this theory must be modified to account for the behavior of real gases. Postulates 1 and 4 state that gas molecules are in constant motion and collide frequently with the walls of their containers. The collision of molecules with their container walls results in a (impulse) from molecules to the walls (Figure \(\Page {2}\)). The to the wall perpendicular to \(x\) axis as a molecule with an initial velocity \(u_x\) in \(x\) direction hits is expressed as: \[\Delta p_x=2mu_x \label{10.7.1} \] The , a number of collisions of the molecules to the wall per unit area and per second, increases with the molecular speed and the number of molecules per unit volume. \[f\propto (u_x) \times \Big(\dfrac{N}{V}\Big) \label{10.7.2} \] The pressure the gas exerts on the wall is expressed as the product of impulse and the collision frequency. \[P\propto (2mu_x)\times(u_x)\times\Big(\dfrac{N}{V}\Big)\propto \Big(\dfrac{N}{V}\Big)mu_x^2 \label{10.7.3} \] At any instant, however, the molecules in a gas sample are traveling at different speed. Therefore, we must replace \(u_x^2\) in the expression above with the average value of \(u_x^2\), which is denoted by \(\overline{u_x^2}\). The overbar designates the average value over all molecules. The exact expression for pressure is given as : \[P=\dfrac{N}{V}m\overline{u_x^2} \label{10.7.4} \] Finally, we must consider that there is nothing special about \(x\) direction. We should expect that \[\overline{u_x^2}= \overline{u_y^2}=\overline{u_z^2}=\dfrac{1}{3}\overline{u^2}. \nonumber \] Here the quantity \(\overline{u^2}\) is called the defined as the average value of square-speed (\(u^2\)) over all molecules. Since \[u^2=u_x^2+u_y^2+u_z^2 \nonumber \] for each molecule, then \[\overline{u^2}=\overline{u_x^2}+\overline{u_y^2}+\overline{u_z^2}. \nonumber \] By substituting \(\dfrac{1}{3}\overline{u^2}\) for \(\overline{u_x^2}\) in the expression above, we can get the final expression for the pressure: \[P=\dfrac{1}{3}\dfrac{N}{V}m\overline{u^2} \label{10.7.5} \] Because volumes and intermolecular interactions are negligible, postulates 2 and 3 state that all gaseous particles behave identically, regardless of the chemical nature of their component molecules. This is the essence of the ideal gas law, which treats all gases as collections of particles that are identical in all respects except mass. Postulate 2 also explains why it is relatively easy to compress a gas; you simply decrease the distance between the gas molecules. Postulate 5 provides a molecular explanation for the temperature of a gas. Postulate 5 refers to the kinetic energy of the molecules of a gas \((\overline{e_K})\), which can be represented as \((T)\) \[\overline{e_K}=\dfrac{1}{2}m\overline{u^2}=\dfrac{3}{2}\dfrac{R}{N_A}T \label{10.7.6} \] where \(N_A\) is the Avogadro's constant. The total translational kinetic energy of 1 mole of molecules can be obtained by multiplying the equation by \(N_A\): \[N_A\overline{e_K}=\dfrac{1}{2}M\overline{u^2}=\dfrac{3}{2}RT \label{10.7.7} \] where \(M\) is the molar mass of the gas molecules and is related to the molecular mass by \(M=N_Am\). By rearranging the equation, we can get the relationship between the root-mean square speed (\(u_{\rm rms}\)) and the temperature. The rms speed (\(u_{\rm rms}\)) is the square root of the sum of the squared speeds divided by the number of particles: \[u_{\rm rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \label{10.7.8} \] where \(N\) is the number of particles and \(u_i\) is the speed of particle \(i\). The relationship between \(u_{\rm rms}\) and the temperature is given by: \[u_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \label{10.7.9} \] In Equation \(\ref{10.7.9}\), \(u_{\rm rms}\) has units of meters per second; consequently, the units of molar mass \(M\) are kilograms per mole, temperature \(T\) is expressed in kelvins, and the ideal gas constant \(R\) has the value 8.3145 J/(K•mol). Equation \(\ref{10.7.9}\) shows that \(u_{\rm rms}\) of a gas is proportional to the square root of its Kelvin temperature and inversely proportional to the square root of its molar mass. The root mean-square speed of a gas increase with increasing temperature. At a given temperature, heavier gas molecules have slower speeds than do lighter ones. The rms speed and the average speed do not differ greatly (typically by less than 10%). The distinction is important, however, because the rms speed is the speed of a gas particle that has average kinetic energy. Particles of different gases at the same temperature have the same average kinetic energy, not the same average speed. In contrast, the most probable speed (vp) is the speed at which the greatest number of particles is moving. If the average kinetic energy of the particles of a gas increases linearly with increasing temperature, then Equation \(\ref{10.7.8}\) tells us that the rms speed must also increase with temperature because the mass of the particles is constant. At higher temperatures, therefore, the molecules of a gas move more rapidly than at lower temperatures, and vp increases. At a given temperature, all gaseous particles have the same average kinetic energy but not the same average speed. The speeds of eight particles were found to be 1.0, 4.0, 4.0, 6.0, 6.0, 6.0, 8.0, and 10.0 m/s. Calculate their average speed (\(v_{\rm av}\)) root mean square speed (\(v_{\rm rms}\)), and most probable speed (\(v_{\rm m}\)). particle speeds average speed (\(v_{\rm av}\)), root mean square speed (\(v_{\rm rms}\)), and most probable speed (\(v_{\rm m}\)) Use Equation \(\ref{10.7.6}\) to calculate the average speed and Equation \(\ref{10.7.8}\) to calculate the rms speed. Find the most probable speed by determining the speed at which the greatest number of particles is moving. The average speed is the sum of the speeds divided by the number of particles: \[v_{\rm av}=\rm\dfrac{(1.0+4.0+4.0+6.0+6.0+6.0+8.0+10.0)\;m/s}{8}=5.6\;m/s \nonumber \] The rms speed is the square root of the sum of the squared speeds divided by the number of particles: \[v_{\rm rms}=\rm\sqrt{\dfrac{(1.0^2+4.0^2+4.0^2+6.0^2+6.0^2+6.0^2+8.0^2+10.0^2)\;m^2/s^2}{8}}=6.2\;m/s \nonumber \] The most probable speed is the speed at which the greatest number of particles is moving. Of the eight particles, three have speeds of 6.0 m/s, two have speeds of 4.0 m/s, and the other three particles have different speeds. Hence \(v_{\rm m}=6.0\) m/s. The \(v_{\rm rms}\) of the particles, which is related to the average kinetic energy, is greater than their average speed. At any given time, what fraction of the molecules in a particular sample has a given speed? Some of the molecules will be moving more slowly than average, and some will be moving faster than average, but how many in each situation? Answers to questions such as these can have a substantial effect on the amount of product formed during a chemical reaction. This problem was solved mathematically by Maxwell in 1866; he used statistical analysis to obtain an equation that describes the distribution of molecular speeds at a given temperature. Typical curves showing the distributions of speeds of molecules at several temperatures are displayed in Figure \(\Page {3}\). Increasing the temperature has two effects. First, the peak of the curve moves to the right because the most probable speed increases. Second, the curve becomes broader because of the increased spread of the speeds. Thus increased temperature increases the of the most probable speed but decreases the relative number of molecules that have that speed. Although the mathematics behind curves such as those in Figure \(\Page {3}\) were first worked out by Maxwell, the curves are almost universally referred to as Boltzmann distributions, after one of the other major figures responsible for the kinetic molecular theory of gases. We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously. The temperature of a 4.75 L container of N gas is increased from 0°C to 117°C. What is the qualitative effect of this change on the temperatures and volume effect of increase in temperature Use the relationships among pressure, volume, and temperature to predict the qualitative effect of an increase in the temperature of the gas. A sample of helium gas is confined in a cylinder with a gas-tight sliding piston. The initial volume is 1.34 L, and the temperature is 22°C. The piston is moved to allow the gas to expand to 2.12 L at constant temperature. What is the qualitative effect of this change on the no change no change no change no change decreases decreases decreases The behavior of ideal gases is explained by the . Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same . The actual values of speed and kinetic energy are not the same for all particles of a gas but are given by a , in which some molecules have higher or lower speeds (and kinetic energies) than average. | 10,663 | 3,727 |
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You are expected to be able to define and explain the significance of terms identified in . The , expressed as , is essentially a statement of the law of conservation of energy. The significance of this law is that it tells us that any proposed process that would violate this condition can be dismissed as impossible, without even inquiring further into the details of the process. For simple mechanical operations on macroscopic objects, the First Law, conservation of energy, is all we usually need to determine such things as how many joules of energy is required to lift a weight or to boil some water, how many grams of glucose you must metabolize in order to climb a hill, or how much fuel your car needs to drive a given distance. But if you think about it, there are a number of "simple mechanical operations" that never occur, even though they would not violate energy conservation. What do all these scenarios that conform to the First Law but are nevertheless never seen to occur have in common? In every case, energy becomes less spread out, less "diluted". In the first two examples, thermal energy (dispersed) gets concentrated into organized kinetic energy of a macroscopic object— a book, a propeller. In the third case, the thermal energy gets concentrated into a smaller volume as the gas contracts. The second law of thermodynamics says in effect, that the extent to which any natural process can occur is limited by the dilution of thermal energy (increase in entropy) that accompanies it, and once the change has occurred, it can never be un-done without spreading even more energy around. This is one of the most profound laws of nature, and should be a part of every educated person's world view. It is unfortunate that this law is so widely misrepresented as simply ordaining the increase in "disorder". A more brief statement of the Second Law (for those who know the meaning of "entropy") is The more formal and historical ways of stating the Second Law will be presented farther below after we introduce the topic of heat engines. It is also worth knowing this important consequence of the Second Law: Just because the energy is “there” does not mean it will be available to do anything useful. We explained how processes that take place spontaneously always proceed in a direction that leads to the spreading and sharing of thermal energy. Because all natural processes lead to the spreading and sharing of thermal energy, and because entropy is a measure of the extent to which energy is dispersed in the world, it follows that: In any spontaneous macroscopic change, the entropy of the world increases. All natural processes that allow the free exchange of thermal energy amongst chemically-significant numbers of particles are accompanied by a spreading or “dilution” of energy that leaves the world forever changed. In other words, . At first sight, this might seem to be inconsistent with our observations of very common instances in which there is a clear decrease in entropy, such as the freezing of a liquid, the formation of a precipitate, or the growth of an organism. ... but It is important to understand that the criterion for spontaneous change is the entropy change of the system and the surroundings— that is, of the “world”, which we denote by Δ : \[ΔS_{total} = ΔS_{system} + ΔS_{surroundings} \label{23.1}\] : \[ΔS_{surroundings} = \dfrac{q_{surr}}{ T} \label{23.2}\] Thus the freezing of water is accompanied by a flow of heat (the heat of fusion) into the surroundings, causing ΔS to increase. At temperatures below the freezing point, this increase more than offsets the decrease in the entropy of the water itself, so ΔSworld exceeds zero and the process is spontaneous. The problem example below works this out in detail for a specific example. Note that it does not matter whether the change in the system occurs reversibly or irreversibly; as mentioned previously, it is always possible to define an alternative (irreversible) pathway in which the amount of heat exchanged with the surroundings is the same as ; because Δ is a state function, the entropy change of the surroundings will have the same value as for the unrealizable reversible pathway. Examples of such processes, , are the free expansion of an ideal gas into a vacuum, and the mixing of two ideal gases. In practice, almost all processes involving mixing and diffusion can be regarded as driven exclusively by the entropy increase of the system. Most processes involving chemical and phase changes involve the exchange of heat with the surroundings, so their tendency to occur cannot always be predicted by focusing attention on the system alone. Further, owing to the – term in Δ , the spontaneity of all such processes will depend on the temperature, as we illustrated for the dissociation of H previously. As a quantitative example, let us consider the . We know that liquid water will spontaneously change into ice when the temperature drops below 0°C at 1 atm pressure. Since the entropy of the solid is less than that of the liquid, we know the entropy of the water (the system here) will decrease on freezing. The amount of decrease is found by dividing the heat of fusion of ice by the temperature for the reversible pathway, which occurs at the normal freezing point: \[ΔS_{system} = \dfrac{-6000 \; J/mol}{273 \;K} = -21.978 \; J/mol\] If the process is actually carried at 0°C, then the heat of fusion is transferred to the surroundings at the same temperature, and the entropy of the surroundings increases by \[ΔS_{surroundings} = \dfrac{6000 \; J/mol}{273 \;K} = 21.979\; J/mol\] so that Δ = 0. Under these conditions the process can proceed in either direction (freezing or melting) without affecting the entropy of the world; this means that both ice and liquid water can be present simultaneously without any change occurring; the system is said to be in . Suppose now that the water is supercooled to –1°C before it freezes. The entropy change of the water still corresponds to the reversible value = (–6000J)/(273K). The entropy change of the surroundings, however, is now given by \[ΔS_{surroundings} = \dfrac{6000 \; J/mol}{273 \;K} = 22.059\; J/mol\] The total entropy change is now \[ΔS_{total} = (–21.978 + 22.059) J;\ K^{–1} mol^{–1} = +0.081\; J \;K^{–1} mol^{–1}\] indicating that the process can now occur (“is spontaneous”) only in the one direction. Why did we use 273 K when evaluating Δ and 272 K for calculating Δ ? In the latter case it is possible to formulate a reversible pathway by which heat can be transferred to the surroundings at any temperature. Δ , however, is a state function of water, and will vary with temperature only slightly. Note that in order to actually freeze water, it must be cooled to very slightly below its normal freezing point, a condition known as supercooling. Freezing of supercooled water is of course an irreversible process (once it starts, it cannot be stopped except by raising the temperature by a finite amount), and the positive value of \(ΔS_{total}\) tells us that this process will occur spontaneously at temperatures below 273 K. Under these conditions, the process is driven by the entropy increase of the surroundings resulting from flow of the heat of fusion of water into the surroundings. The principle that thermal energy (and the molecules carrying it) tends to spread out is based on simple statistics. It must be remembered, however, that the laws of probability have meaningful application only to systems made up of large numbers of independent actors. If you trap a hundred flies in a bottle, they will generally distribute themselves more or less uniformly throughout the container; if there are only four flies, however, it is quite likely that all of them will occasionally be located in one particular half of the bottle. Similarly, you can trust with complete certainty that the spontaneous movement of half the molecules of the air to one side of the room you now occupy will not occur, even though the molecules are moving randomly and independently. On the other hand, if we consider a box whose dimensions are only a few molecular diameters, then we would expect that the random and short-term displacement of the small number of particles it contains to one side of the box would occur quite frequently. This is, in fact, the cause of the blueness of the sky: random fluctuations in the air density over tiny volumes of space whose dimensions are comparable with the wavelength of light results in selective scattering of the shorter wavelengths, so that blue light is scattered out, leaving the red light for the enjoyment of sunset-watchers to the east. This refers to the irregular zig-zag-like movement of extremely small particles such as plant pollen when they are suspended in a drop of liquid. Any such particle is continually being buffeted by the thermal motions of the surrounding liquid molecules. If size of the particle is very large compared to that the the liquid molecules, the forces that result from collisions of these molecules with the particle will cancel out and the particle remains undisturbed. If the particle is very small, however (perhaps only a thousand times larger than a molecule of the liquid), then the chances that it will undergo sufficiently more hits from one direction than from another during a brief interval of time become significant. In these two examples, the entropy of the system decreases without any compensating flow of heat into the surroundings, leading to a net (but only temporary) decrease in the entropy of the world. This does not represent a failure of the Second Law, however, because no one has ever devised a way to extract useful work from these processes. The Industrial Revolution of the 19th century was largely driven by the invention of the steam engine. The first major use of such engines was to pump water out of mines, whose flooding from natural seepage seriously limited the depths to which they could be driven, and thus the availability of the metal ores that were essential to the expansion of industrial activities. The steam engine is a type of heat engine, a device that converts heat, provided by burning a fuel, into mechanical work, typically delivered through the motion of a piston in opposition to an opposing force. An engine is therefore an energy conversion device in which, ideally, every joule of heat released by combustion of the fuel could be extracted as work at the output shaft; such an engine would operate at 100 percent efficiency. However, engineers of the time were perplexed to find that the efficiencies of steam engines were rather low (usually around 20%), with most of the heat being exhausted uselessly to the environment. Everyone understood that an efficiency exceeding 100% would be impossible (that would violate conservation of energy, and thus the First Law), but it was not clear why efficiencies could not rise significantly beyond the small values observed even as mechanical designs improved The answer was found by a young French engineer, Sadi Carnot, who in 1824 published an analysis of an idealized heat engine that is generally considered to be the foundation of the science of thermodynamics— notwithstanding the fact that Carnot still accepted the belief that heat is a fluid-like substance called “caloric”. We will not replicate his analysis here (this is normally done in more advanced courses in physical chemistry), but will simply state his conclusion in his own [translated] words: "The production of motive power is then due in steam-engines not to an actual consumption of caloric, but to its from a warm body to a cold body...the production of heat alone is not sufficient to give birth to the impelling power: it is necessary that there should also be cold; without it, the heat would be useless. The ultimate attainable efficiency of any heat engine will depend on the temperatures at which heat is supplied to and removed from it." The left side of the figure represents a generalized heat engine into which a quantity of heat , extracted from a source or “reservoir” at temperature is partly converted into work . The remainder of the heat is exhausted to a reservoir at a lower temperature . In practice, would be the temperature of the steam in a steam engine, or the temperature of the combustion mixture in an internal combustion or turbine engine. The low temperature reservoir is ordinarily that of the local environment. The efficiency ε ( ) of a heat engine is the fraction of the heat abstracted from the high temperature reservoir that can be converted into work: \[ ε = \dfrac{w}{q_H} \label{3.3}\] Carnot’s crucial finding (for which he would certainly have deserved a Nobel prize if these had existed at the time) is that the efficiency is proportional to the "distance'' in temperature that the heat can “fall” as it passes through the engine: \[ ε = 1 - \dfrac{T_L}{T_H} \label{3.4}\] This is illustrated graphically in the right half of the figure just above, in which the efficiency is simply the fraction of the “complete” fall (in temperature) to absolute zero (arrow ) that the heat undergoes in the engine (arrow .) Clearly, the only way to attain 100% efficiency would be to set the temperature of the exhaust reservoir to 0°K, which would be impossible. For most terrestrial heat engines, T is just the temperature of the environment, normally around 300 K, so the only practical way to improve the efficiency is to make T as high as possible. This is the reason that high pressure (superheated) steam is favored in commercial thermal power plants. The highest temperatures (and the greatest operating efficiencies) are obtained in gas turbine engines. However, as operating temperatures rise, the costs of dealing with higher steam pressures and the ability of materials such as turbine blades to withstand high temperatures become significant factors, placing an upper limit of around 600K on T , thus imposing a maximum of around 50 percent efficiency on thermal power generation. For , in which safety considerations require lower steam pressures, the efficiency is lower. One consequence of this is that a larger fraction of the heat is exhausted to the environment, which may result in greater harm to aquatic organisms when the cooling water is returned to a stream or estuary. Several proposals have been made for building a heat engine that makes use of the temperature differential between the surface waters of the ocean and cooler waters that, being more dense, reside at greater depth. If the exhaust temperature is 5°C, what is the maximum amount of work that could be extracted from 1000 L of surface water at 10°C? (The specific heat capacity of water is 4.184 J g K .) The amount of heat (q ) that must be extracted to cool the water by 5 K is (4.184 J g K )(10 g)(5 K) = 2.09 10 J. The ideal thermodynamic efficiency is given by \[ 1 -\dfrac{278 \;K}{283\; K} = 0.018\] The amount of work that could be done would be \[(0.018)(2.09 \times 10^7 \;J) = 3.7 \times 10^6 \;J\] : It may be only 1.8% efficient, but it’s free! Few toys illustrate as many principles of physical science as this popular device that has been around for many years. At first glance it might appear to be a perpetual motion machine, but it's really just a simple heat engine. Modern "dippy birds" (as they are sometimes called) utilize dichloromethane as the working fluid. This liquid boils at 39° C, and therefore has a rather high vapor pressure at room temperature. The liquid (to which a dye is often added for dramatic effect) is stored in a reservoir at the bottom of the bird. The bird's beak is covered with felt which, when momentarily dipped in water, creates a cooling effect as the water evaporates. This causes some of the CH Cl vapor to condense in the head, reducing the pressure inside the device, causing more liquid to boil off and re-condense in the head. The redistribution of fluid upsets the balance, causing the bird to dip its beak back into the water. Once the head fills with liquid, it drains back into the bottom, tipping the bird upright to repeat the cycle. We will leave it to you to relate this to the heat engine diagram above by identifying the heat source and sink, and estimate the thermodynamic efficiency of the engine. If a heat engine is run “in reverse” by performing work on it (that is, changing “work out” to “work in” in Fig 8), it becomes a device for transporting heat against a thermal gradient. Refrigerators and air conditioners are the most commonly-encountered heat pumps. A heat pump can also be used to heat the interior of a building. In this application, the low temperature reservoir can be a heat exchanger buried in the earth or immersed in a well. In this application heat pumps are more efficient than furnaces or electric heating, but the capital cost is rather high. It was the above observation by Carnot that eventually led to the formulation of the near the end of the 19th Century. One statement of this law (by Kelvin and Planck) is as follows: It is impossible for a connected to a reservoir to produce a in the surroundings. To help you understand this statement and how it applies to heat engines, consider the schematic heat engine in the figure in which a working fluid (combustion gases or steam) expands against the restraining force of a weight that is mechanically linked to the piston. From a thermodynamic perspective, the working fluid is the system and everything else is surroundings. Expansion of the fluid occurs when it absorbs heat from the surroundings; return of the system to its initial state requires that the surrounding do work on the system. Now re-read the above statement of the Second Law, paying special attention to the italicized phrases which are explained below: Note carefully that the Second Law applies only to a — isothermal expansion of a gas against a non-zero pressure always does work on the surroundings, but an engine must repeat this process continually; to do so it must be returned to its initial state at the end of every cycle. When operating isothermally, the work – it does on the surroundings in the expansion step (power stroke) is nullified by the work + the surroundings must do on the system in order to complete the cycle. The Second Law can also be stated in an alternative way: Thus the Second Law does allow an engine to convert heat into work, but only if “other changes” (transfer of a portion of the heat directly to the surroundings) are allowed. And since heat can only flow spontaneously from a source at a higher temperature to a sink at a lower temperature, the impossibility of isothermal conversion of heat into work is implied. | 18,805 | 3,728 |
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We often write transition-metal ions in aqueous solution with symbols such as Cr , Cu , and Fe as though they were monatomic, but this is far from being the case. These ions are actually hydrated in solution and can be regarded as complex ions. Thus, for example, the grayish-violet color of many chromium(III) salts when dissolved in H O is due to the species [Cr(H O) ] rather than to a bare Cr ion. The same color is evident in many crystalline solids such as [Cr(H O) ]Cl which are known to contain the Cr ion surrounded octahedrally by six H O molecules. In much the same way the blue color of many solutions of copper(II) salts can be attributed to the species [Cu(H O) ] and the pale violet color of some solutions of iron(III) salts to the [Fe(H O) ] ion. Because [Fe(H O) ] is capable of donating a proton, the conjugate base, [Fe(H O) OH] is generally present when Fe is dissolved in water. This imparts a yellow color to the solution. Figure \(\Page {1}\) shows examples of colored ion complexes in aqueous solution. Figure \(\Page {1}\) Examples of colored aqueous transition metal complexes Not all salts of transition-metal ions yield the hydrated ion when dissolved in H O. Figure \(\Page {2}\) compares three aqueous copper complexes. When CuCl is dissolved in H O, a beautiful green color due mainly to the complex [CuCl (H O) ] is produced. This is obviously different from the sky-blue color of [Cu(H O) ] which is obtained when Copper(II) sulfate or copper(II) nitrate are dissolved. This is because the Cl ion is a stronger Lewis base with respect to the Cu ion than is H O. Thus, if there is a competition between H O and Cl to bond as a ligand to Cu , the Cl ion will usually win out over the H O. Figure \(\Page {2}\) The Different Colored Copper Chloride Complexes The superior strength of the Cl as a Lewis base is easily demonstrated by adding Cl ions to a sky-blue solution of copper(II) sulfate. A green color immediately appears due to the formation of chloro complexes: \[\begin{align*} \ce{[Cu(H2O)4]^{2+}} + \ce{Cl^{-}} &\rightleftharpoons \ce{[Cu(H2O)3Cl]^{+}} + \ce{H2O} \label{1} \\[4pt] \ce{[Cu(H2O)3Cl]^{+}} + \ce{Cl^{-}} &\rightleftharpoons \ce{[Cu(H2O)2Cl2]} + \ce{H2O} \end{align*} \] If a large excess of Cl ion is added, the solution changes color again from green to yellow. This is because of even further displacement of H O ligands by Cl ligands: \[ \begin{align*} \ce{[Cu(H2O)2Cl2]} + \ce{Cl^{-}} &\rightleftharpoons \ce{[Cu(H2O)2Cl3]^{-}} + \ce{H2O} \\[4pt] \ce{[Cu(H2O)2Cl3]^{-}} + \ce{Cl^{-}} &\rightleftharpoons \ce{[CuCl4]^{2-}} + \ce{H2O} \end{align*} \nonumber \] Under favorable circumstances yellow crystals of salts like Cs [CuCl ], containing the complex ion CuCl can be obtained from these solutions. Because they might very possibly form complexes with it, one must be careful about what ions are added to a solution containing hydrated transition-metal ions. Not only the chloride ions, but the other halide ions are liable to complex, and the same is true of species like NH and CN . These ligands differ quite a lot in their affinity for a particular metal ion, but the rules governing this situation are not simple. One finds, for instance, that although NH will complex very readily with Cu it has little or no affinity for Fe . In other words, a ligand which is a strong Lewis base with respect to one metal ion is not necessarily a strong base with respect to another. There are some ions, however, which almost always function as very weak Lewis bases. The perchlorate ion, ClO in particular, forms almost no complexes. The nitrate ion, NO , and sulfate ion, SO , only occasionally form complexes. The addition of ligands to a solution in order to form a highly colored complex is often used to detect the presence or absence of a given metal in solution. The deep blue color of [Cu(NH ) ] produced when excess NH is added to solution of Cu(II) salts is a case in point. This can be seen in the following video, where a aqueous solution of ammonia is added to a copper sulfate solution: The initial copper sulfate solution is sky blue, due to the [Cu(H O) ] complex. When ammonia is added, a precipitate of Cu(OH) ( ) is formed. as it settles to the bottom, it can be seen that the remaining solution is a dark blue, due to the [Cu(NH ) ] complex formed by copper with ammonia. Other well-known color reactions are the blood-red complex formed between Fe(III) ions and the thiocyanate ion, SCN , as well as the pink-red complex of Ni(II) with dimethylglyoxime. While most of the reactions we have been describing are very fast and occur just as quickly as the solutions are mixed, this is not always the case. With certain types of complexes, ligand substitution is quite a slow process. For example, if Cl ions are added to a solution containing [Cr(H O) ] ions, it is a few days before the grayish-violet color of the original ion is replaced by the green color of the chloro complexes [Cr(H O ) Cl] and [Cr(H O) Cl] . Alternatively the solution may he heated, in which case the green color will usually appear within 10 min. The reaction is thus a slow reaction with a high activation energy. Ligand substitution reactions of other Cr(III) complexes behave similarly. In consequence Cr(III) complexes are said to be , as opposed to a complex like Fe(H O) which swaps ligands very quickly and is said to be . Other examples of inert complexes are those of Co(III), Pt(IV), and Pt(II). Almost all the compounds which were used to establish the nature and the geometry of coordination compounds were inert rather than labile. There is very little point in trying to prepare cis and trans isomers of a labile complex, for example, because either will quickly react to form an equilibrium mixture of the cis and trans forms. A final complication in dealing with aqueous solutions of transition-metal complexes is their acid-base behavior. are capable of donating protons to water and acting as weak acids. Most hydrated ions with a charge of + 3, like Al and Fe behave similarly and are about as strong as acetic acid. The hydrated Hg(II) ion is also noticeably acidic in this way. Perhaps the most obvious of these cationic acids is the hydrated Fe(III) ion. When most Fe(III) salts are dissolved in water, the color of the solution is yellow or brown, though the Fe(H2O) ion itself is pale violet. The yellow color is due to the conjugate base produced by the loss of a proton. The equilibrium involved is \[ \text{[Fe(H}_{2} \text{O)}_{6} \text{]}^{3+} + \text{H}_{2} \text{O} \rightleftharpoons \text{[Fe(H}_{2} \text{O)}_{5} \text{OH]}^{2+} + \text{H}_{3} \text{O}^{+} \nonumber \] If solutions of Fe(III) salts are acidified with perchloric acid or nitric acid, the brown base is protonated and the yellow color disappears from the solution entirely. | 6,879 | 3,729 |
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Rather than drawing a complete diagram like the figures in the , it is convenient to specify a galvanic cell in shorthand form. The two cells we have just described would be written as \[\ce{Zn} | \ce{Zn^{2+}(1 \,M)}\, || \, \ce{Cu^{2+}(1\,M)} | \ce{Cu}\label{1} \] and \[\ce{Pt(s)} | \ce{Cl_{2}(g)} | \ce{Cl^{–}(1\, M)}\, ||\, \ce{Fe^{2+}(1 \,M), Fe^{3+}(1\, M)} | \ce{Pt(s)} \label{2} \] The components of the cell are written in order, starting with the left-hand and moving across the salt bridge to the right. A single vertical line indicates a phase boundary, such as that between the solid \(\ce{Zn}\) electrode and \(\ce{Zn^{2+}(aq)}\), or between \(\ce{Cl2(g)}\) and \(\ce{Cl^{–}(aq)}\). The double vertical line represents a salt bridge. Spectator ions, like \(\ce{SO4^{2–}(aq)}\) in the \(\ce{Zn–Cu}\) cell, are usually omitted. By convention, the electrode written to the of the salt bridge in this cell notation is always taken to be the , and the associated half-equation is always written as an . The right-hand electrode is therefore always the , and the half-equation is always written as a . This is easy to remember, because reading from left to right gives anode and cathode in alphabetical order. The corresponding to a given shorthand description is obtained by summing the half-equations after multiplying by any factors necessary to equalize the number of lost at the anode with the number gained at the cathode. Unless otherwise stated, standard conditions of 1 are usually implied. However, can be created. Write the half-equations and cell reactions for each of the following cells: Since the half-equation at the left electrode is taken by convention to be oxidation, we have \[\ce{Ag} \rightarrow \ce{Ag^{+} + }e^– \nonumber \] At the right-hand electrode, then, we must have a reduction: \[\ce{2H^{+} + 2}e^{–} \ce{\rightarrow H_{2}} \nonumber \] Multiplying the first half-equation by 2 and summing gives the cell reaction \[\ce{2Ag + 2H^{+} \rightarrow 2Ag^{+} + H_{2} (g)} \nonumber \] Following the same procedure as in part a, we obtain \[\ce{2Cr^{3+} + 7H_{2}O \rightarrow Cr_{2}O_{7}^{2–} + 14H^{+} + 6}e^{-} \nonumber \] Left electrode \[\ce{(Br_{2}(l) + 2}e^– \ce{\rightarrow 2Br^{–}) \times 3} \nonumber \] Right electrode \[\ce{2Cr^{3+} + 7H_{2}O + 3Br_{2}(l) \rightarrow Cr_{2}O_{7}^{2–} + 14H^{+} + 6Br^{-}} \nonumber \] Cell reaction Note: were used in arriving at the above results. Describe in shorthand notation a galvanic cell for which the cell reaction is \[\ce{Cu(s) + 2Fe^{3+}(aq) \rightarrow Cu^{2+}(aq) + 2Fe^{2+}(aq)} \nonumber \] First divide the cell reaction into half-equations: Then write the oxidation as the left-hand electrode and the reduction on the right: \[\ce{Cu} \mid \ce{Cu^{2+}} \parallel \ce{Fe^{2+}, Fe^{3+}} \mid \ce{Pt} \nonumber \] (Since both Fe and Fe are in solution, a Pt electrode is used.) The conventions we have developed can be used to decide whether the cell reaction is actually spontaneous. If it is, an oxidation will release electrons to the external circuit at the left-hand electrode. If a voltmeter is placed in the circuit, these electrons will make its left-hand terminal negative. Since the right-hand electrode corresponds to reduction, electrons will be withdrawn from the right-hand terminal of the voltmeter. This is shown for the Zn–Cu cell in . You can readily confirm that the spontaneous cell reaction ( ) corresponds to the shorthand cell notation of Equation \(\ref{1}\). For the cell shown in , the shorthand notation is \[\ce{Pt(s)} | \ce{Cl_2(g)} \mid \ce{Cl^{–}(1 M)} \parallel \ce{Fe^{2+}(1 M), Fe^{3+}(1 M)} \mid \ce{Pt(s)} \nonumber \] According to the conventions we have just developed, this corresponds to the cell reaction \[\ce{2Cl^{–}(aq) + 2Fe^{3+}(aq) \rightarrow Cl_2(g) + 2Fe^{2+}(aq)} \nonumber \] But this is the of the spontaneous cell reaction we described before ( ). Since the cell reaction is , electrons will not be forced into the external circuit at the left-hand electrode, and they will not be withdrawn at the right. In fact the reverse will actually occur. Thus if a voltmeter is connected to this cell, its right-hand terminal will become more negative and its left-hand terminal will become more positive. This is shown in . In general, if a galvanic cell is connected to a voltmeter, the electrode connected to the negative terminal of the meter must be the anode. If our shorthand cell notation shows that electrode on the left, then the corresponding cell reaction must be spontaneous. Electrons will be released by the oxidation half-equation on the left and accepted by the reduction on the right. If, on the other hand, the voltmeter shows that the right-hand electrode is releasing electrons, then we must have written our shorthand notation . This means that the reverse of the cell reaction obtained by our rules must actually be occurring, and it is that reverse reaction which is spontaneous. Thus by simply observing which electrode in the cell releases electrons and which accepts them, that is, by finding which electrode is negative and which positive, we can determine whether the cell reaction is spontaneous. | 5,206 | 3,731 |
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You may have wondered why we have been so careful to define atomic weights and isotopic weights as of masses. The reason will be clearer once the most important and accurate experimental technique by which isotopic weights are measured has been described. This technique, called , has developed from the experiments with cathode-ray tubes mentioned earlier in this chapter. It depends on the fact that an electrically charged particle passing through a magnetic field of constant strength moves in a circular path. The radius of such a path is directly proportional to the mass and the speed of the particle, and inversely proportional to the charge . Thus the greater the mass or speed of the particle, the greater the radius of its path. The greater the charge, the smaller the radius. In a mass spectrometer, as seen below, atoms or molecules in the gaseous phase are bombarded by a beam of electrons. Occasionally one of these electrons will strike another electron in a particular atom, and both electrons will have enough energy to escape the attraction of the positive nucleus. This leaves behind a positive ion since the atom now has one more proton than it has electrons. For example, \[\ce{^{12}_6C} +\ce{e^-} \text{ (high speed electron)} \rightarrow \ce{^{12}_6C^+} + \ce{2 e^-} \nonumber \] Once positive ions are produced in a mass spectrometer, they are accelerated by the attraction of a negative electrode and pass through a slit. This produces a narrow beam of ions traveling parallel to one another. The beam then passes through electric and magnetic fields. The fields deflect away all ions except those traveling at a certain speed. The beam of ions is then passed between the poles of a large electromagnet. Since the speed and charge are the same for all ions, the radii of their paths depend only on their masses. For different ions of masses and \[\frac{r_{\text{2}}}{r_{\text{1}}}=\frac{m_{\text{2}}}{m_{\text{1}}} \nonumber \] and the ratio of masses may be obtained by measuring the ratio of radii, The paths of the ions are determined either by a photographic plate (which darkens where the ions strike it, as in Figure \(\Page {1}\)) or a metal plate connected to a galvanometer (a device which detects the electric current due to the beam of charged ions). When a sample of carbon is vaporized in a mass spectrometer, two lines are observed on the photographic plate. The darker line is 27.454 cm, and the other is 29.749 cm from the entrance slit. Determine the relative atomic masses (isotopic weights) of the two isotopes of carbon. Since the distance from the entrance slit to the line on the photographic plate is twice the radius of the circular path of the ions, we have \[\frac{m_{\text{2}}}{m_{\text{1}}}=\frac{r_{\text{2}}}{r_{\text{1}}}=\frac{2r_{\text{2}}}{2r_{\text{1}}}=\frac{\text{29}\text{.749 cm}}{\text{27}\text{.454 cm}}=\text{1}\text{.083 59} \nonumber \] Thus = 1.083 . If we assume that the darker mark on the photographic plate is produced because there are a greater number of 12 ions than of the less common13 , then may be equated with the relative mass of and may be assigned a value of 12.000 000 exactly. The isotopic weight of is then \[ m_2 = (1.083 59)(12.000 000) = 13.0031 \nonumber \] Notice that in mass spectrometry all that is required is that the charge and speed of the two ions whose relative masses are to be determined be . If the mass of an individual ion were to be measured accurately, its actual speed upon entering the magnetic field and the exact magnitude of its electric charge would have to be known very accurately. Therefore it is easier to measure the ratio of two masses than to determine a single absolute mass, and so atomic weights are reported as pure numbers. | 3,786 | 3,732 |
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The elements of group IIIA show considerable variability in properties from top to bottom of the periodic table. B is a semimetal or metalloid, and the element has a covalent network structure in which icosahedrons of boron atoms (Figure ) are linked together. None of its compounds contain B ions—covalent bonding is the rule there as well. Many B compounds, especially those containing O, are similar to those of Si (another diagonal relationship) instead of Al. In elemental Al the atoms are closest packed. Al is definitely metallic, but like Be, to which it is related diagonally, it forms many compounds with extensive covalent character. The other three elements in the group, Ga, In, and Tl, are also metals, but their chemistry is affected to some extent by the fact that they follow the transition elements in the periodic system (Table \(\Page {1}\)). The presence of a filled subshell, as opposed to only and subshells in B and Al, introduces some new atomic properties. Some properties of the group IIIA elements are shown in Table \(\Page {2}\). In this case the trends are not quite what might have been expected on the basis of previous experience with the periodic table. Both ionization energy and electronegativity decrease significantly from B to Al, but as one proceeds on down the group, these atomic properties change very little. Indeed electronegativity increases from Ga to In to Tl. There is also a break in the expected steady increase in atomic radius down the group. Al and Ga have nearly identical radii, and the ionic radii of A1 and Ga differ by only 8 pm. In the case of Tl the most common oxidation state is +1,corresponding to loss of only the 6 valence electron, rather than an oxidation state of +3, which would entail loss of the 6 pair as well. Many of these anomalies can be understood if we look more closely at the electron configurations of Ga, In, and Tl. In each case a subshell has been filled between the pair and the final valence electron. The 4 subshell also intervenes in the building-up process for Tl. Since and electrons are not as efficient as or electrons at screening nuclear charge, the valence electrons of Ga, In, and Tl are more strongly attracted, the atoms are smaller than would be expected, and ionization energies are unusually large. Another factor affecting the chemistry of the group IIIA elements is the relative sizes of the first, second, and third ionization energies. These are roughly in the ratio 1:3:4.5 for all elements, and the large increase from first to second ionization energy becomes more pronounced toward the bottom of the group. In the case of Tl, whose large radius prevents close approach and strong bonding to other atoms, the energy required to unpair or ionize the 6 pair of electrons and use them to form bonds is often too large to be compensated by the bond energies of the two additional bonds. Consequently Tl is the more common oxidation state, rather than Tl . This reluctance of the 6 pair of electrons to be used in bonding is called the inert-pair effect. It also affects the chemistry of Hg, Pb, and Bi, elements which are adjacent to Tl in the sixth period. Unlike groups IA and IIA, none of the group IIIA elements react directly with hydrogen to form hydrides. The halides of B, Al, and Ga will react with sodium hydride, however, to form tetrahydro anions: \[\text{4NaH} + \text{BF}_3 \rightarrow \text{NaBH}_4 + \text{3NaF} \nonumber \] The compounds NaBH , NaAlH , and NaGaH do not contain H ions. Instead the hydrogens are covalently bonded to the group IIIA atom: All these anions are tetrahedral, as would be predicted by VSEPR theory. The tetrahydroaluminate and tetrahydrogallate ions react readily with H O, splitting the H O molecule so that the H ends up with another H to form H , while the OH ends up with the group IIIA atom: \[\text{AlH}_4^{-}(aq) + \text{4HOH}(l) \rightarrow \text{Al(OH)}_3(s) + \text{OH}^{-}(aq) + \text{4H}_2(g) \nonumber \] Splitting of an H O molecule (hydrolysis) by these compounds is similar to the hydrolysis of esters. The hydrolysis of AlH is violently explosive, making it necessary to handle NaAlH in a dry environment. The anion in NaBH by contrast, involves bonds with greater covalent character and is much less readily hydrolyzed. Only B forms a wide variety of hydride compounds. The simplest of these is diborane, B H , a volatile, readily hydrolyzed compound which may decompose explosively. Diborane is made as follows: \[\ce{3NaBH_{4} + 4BF_{3} \rightarrow 3NaBF_{2} + 2B_{2}H_{6}} \nonumber \] Diborane is electron deficient—if you try to draw a Lewis diagram, you will soon find it to be impossible—and for a number of years theoretical chemists were mystified about what held the atoms together. The current picture of bonding in diborane is shown in the Figure and the accompanying Jmol Applet. Each B is assumed to be surrounded by four hybrid orbitals. Four of the eight hybrids from the two B’s overlap the 1 orbitals of four H’s, forming four B—H bonds, all of which lie in a plane. The other four hybrids overlap between the B’s, forming two banana bonds, a form we described when discussing the double bonds of ethene. Each of the remaining two H nuclei is embedded in the middle of one of the banana bonds. In this way the two electrons in each banana bond hold three nuclei (two B’s and one H) together, and the bond is called a . Below is a 3D Jmol model of diborane. We've added a molecular electrostatic potential surface function, as well as molecular orbitals. Let us look at the MEP options first. a) All three show that the boron atoms are surrounded by negative charge, the free hydrogen atoms are neutral, and the two hydrogen atoms in the banana bonds are electron deficient. This picture makes sense in terms of our discussion on diborane structure. First, the two banana bond hydrogen atoms are sharing an electron pair not just with one boron atom, but with two. As you can see in the dot density diagram, the electron density for this bond is pulled away from the hydrogen atoms. The negative charge on boron arises from their being four, instead of three, bonds to the atom. b)This orbital spans the region between the hydrogen atom and borons. Since orbitals contain only 2 electrons, this means that there is a strong bonding presence of only two electrons shared between 3 atoms (rather than 2 individual bonds of 2 electrons each). All the group IIIA elements form trihalides, although Tl does so reluctantly, preferring to remain in the +1 oxidation state. The general reaction is \[\text{2M}(s) + \text{3X}_2 \rightarrow \text{2MX}_3 \nonumber \] M = B, Al, Ga, In, Tl X = F, Cl, Br, I Below is a video example of one such reaction, that of Al and Br : In this video, aluminum foil is cut into smaller squares, and placed in a watch glass with liquid bromine, also giving off gaseous bromine vapor fumes. Initially, no reaction occurs, and the aluminum foil has a protective oxide coating. Soon, white fumes start to come off, signaling the start of the reaction. Since the reaction is exothermic, more bromine fumes also emerge. The aluminum and bromine react violently, causing flashes of light. After the reaction has occurred, a fan is used to remove the fumes, so that the AlBr product may be seen. The overall reaction is: \[\text{2Al}(s) + \text{3Br}_2 \rightarrow \text{2AlBr}_3 \nonumber \] All the boron halides consist of discrete BX molecules, which are electron deficient. Below is a 3D model of one of these halides,BCl . As predicted by VSEPR theory the molecule has a trigonal planar structure. On this model, Molecular Electrostatic Potential Surface options have been provided. First, set the minimum cutoff to -0.0100 and the maximum cutoff to 0.0100. Again, any of the three surface potential options are useful, but in this case, "MEP on Isopotential Surface" is most informative. First, all three chlorides have a partial negative charge. The boron halides are strong Lewis acids because each can accept an electron pair to complete an octet on B. Looking at the positive isopotential surface around the boron atom, it is clear how open each side of the molecule is to accepting an electron pair. The bromides and iodides of Al, Ga, In, and Tl are primarily covalent, and two MX molecules usually combine to form a dimer (dimer means two units): In the dimer structure a lone electron pair from a Br atom in one AlBr unit has been donated to the Al of the other AlBr unit, and vice versa. This forms two coordinate covalent bonds to hold the dimer together. (These bonds are shown as arrows.) The fluorides of Al, Ga, In, and Tl are primarily ionic. Only the larger, more easily polarizable Br and I ions can be distorted enough by the small M ions to give mainly covalent bonds. Most of the chlorides are border-line cases. This is reflected in the melting points: AlF , 1291°C AlCl , 190°C Al Br , 97.5°C Al I , 191°C Solid AlCl consists of A1 and Cl ions, but in the liquid and gas phases Al Br dimers predominate. The oxides and oxyanions of B and Al are the main compounds of commercial importance. Borax, Na B O •10H O, is the principal ore of B, and Al is obtained from bauxite, Al O • H O. Bauxite is an example of a . It contains an indeterminate amount of water (xH O) in the crystal lattice. Although Al is the most abundant metal in the earth’s crust, most of it is found in complicated aluminosilicates (feldspars). It is extremely difficult to convert the feldspars to Al metal, and any Si which remains as an impurity greatly degrades the strength and other properties of Al. Consequently the metal is obtained from bauxite, of which there is only a limited supply. The first step in the recovery of Al is called the . This depends on the fact that Al O is ; that is, it can behave as either an acid or a base. If it behaves as a base, it can react with, and dissolve in, acidic solutions. If it behaves as an acid, it can dissolve in basic solutions. The amphoteric behavior of Al O comes about because of the presence of A1 ions and O ions. Al has a very small radius and a large ionic charge, and so it strongly attracts H O molecules—so strongly that the Al(H O) ion can donate protons. (Other cations which do this were described in Sec. 11.3.) Oxide ion, of course, is a good proton acceptor and a strong base. The Bayer process makes use of the acidic behavior of Al O by dissolving it in strong base: \[\text{Al}_2\text{O}_3(s) + \text{3H}_2\text{O}(l) + \text{2OH}^{-}(aq) \rightarrow \text{2Al(OH)}_4^{-}(aq) \nonumber \] The oxides of most metals are strongly basic and do not dissolve, while the oxides of nonmetals such as Si are acidic and more soluble in base than Al O . Thus a fairly pure sample of aluminum hydroxide, Al(OH) , can be obtained by filtering off the basic oxides and acidifying the solution slightly. CO , an acidic oxide, is used for this purpose: \[\text{CO}_2(g) + \text{Al(OH)}_4^{-}(aq) \rightarrow \text{Al(OH)}_3(s) + \text{HCO}_3^{-}(aq) \nonumber \] The Al(OH) is then heated to drive off H O: \[\text{2Al(OH)}_3(s) \xrightarrow{\Delta } \text{Al}_2\text{O}_3(s) + \text{3H}_2{O}(g) \nonumber \] And Al is obtained by the : \[\text{Al}_2\text{O}_3(l) \xrightarrow{\text{electrolysis}} \text{2Al}(l) + \text{3H}_2\text{O}(g) \nonumber \] This process is described in greater detail in the section on aluminum production. The last step in aluminum production requires tremendous quantities of electrical energy. This, as well as the scarcity of its ore, makes aluminum more expensive than iron, the only metal which is more widely used. Nevertheless, aluminum has several advantages over iron. One is its considerably lower density, making possible alloys of comparable strength with considerably lower mass. A second advantage of aluminum is the coating of Al O which forms on its surface when it is exposed to air. This coating sticks to the surface and is insoluble in neutral solutions, and so it prevents further oxidation. Iron, by contrast, forms a hydrous oxide (rust) which flakes off easily, exposing additional metal to the air. Thus under normal environmental conditions aluminum will last much longer than iron. This is a problem when aluminum beverage cans litter roadsides, but in general it makes aluminum a very favorable candidate for recycling. | 12,419 | 3,734 |
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The thermochemical quantities that you probably encounter most often are the caloric values of food. Food supplies the raw materials that your body needs to replace cells and the energy that keeps those cells functioning. About 80% of this energy is released as heat to maintain your body temperature at a sustainable level to keep you alive. The nutritional Calorie (with a capital C) that you see on food labels is equal to 1 kcal (kilocalorie). The caloric content of food is determined from its enthalpy of combustion (Δ ) per gram, as measured in a bomb calorimeter, using the general reaction \[ \ce{food + excess\ O2 (g) -> CO2 (g) + H2O (l) + N2 (g)} \label{5.8.1} \] There are two important differences, however, between the caloric values reported for foods and the Δ of the same foods burned in a calorimeter. First, the Δ described in joules (or kilojoules) are negative for all substances that can be burned. In contrast, the caloric content of a food is always expressed as a number because it is . Therefore, \[ \mathrm{caloric\ content} = - \Delta H_\mathrm{comb} \label{5.8.2} \] Second, when foods are burned in a calorimeter, any nitrogen they contain (largely from proteins, which are rich in nitrogen) is transformed to N . In the body, however, nitrogen from foods is converted to urea [(H N) C=O], rather than N before it is excreted. The Δ of urea measured by bomb calorimetry is −632.0 kJ/mol. Consequently, the enthalpy change measured by calorimetry for any nitrogen-containing food is greater than the amount of energy the body would obtain from it. The difference in the values is equal to the Δ of urea multiplied by the number of moles of urea formed when the food is broken down. This point is illustrated schematically in the following equations: \[ \ce{food + excess\ O2 (g) ->[\Delta H_{1} < 0] CO2 (g) + H2O (l) + \cancel{(H2N)2C=O (s) }} \nonumber \] \[\ce{ \cancel{(H2N)2C=O (s)} +3/2 O2 (g) ->[\Delta H_{2} = 632.0 \mathrm{kJ/mol}] CO2 (g) + 2H2O (l) + N2 (g)} \nonumber \] which adds up to
\[ \ce{food + excess\ O2 (g) ->[\Delta H_{3}=\Delta H_{1}+\Delta H_{2} < 0] 2 CO2 (g) + 3 H2O (l) + N2 (g)} \nonumber \] All three Δ values are negative, and, by Hess’s law, Δ = Δ + Δ . The magnitude of Δ must be less than Δ , the calorimetrically measured Δ for a food. By producing urea rather than N , therefore, humans are excreting some of the energy that was stored in their food. Because of their different chemical compositions, foods vary widely in caloric content. As we saw , for instance, a fatty acid such as palmitic acid produces about 39 kJ/g during combustion, while a sugar such as glucose produces 15.6 kJ/g. Fatty acids and sugars are the building blocks of fats and carbohydrates, respectively, two of the major sources of energy in the diet. Nutritionists typically assign average values of 38 kJ/g (about 9 Cal/g) and 17 kJ/g (about 4 Cal/g) for fats and carbohydrates, respectively, although the actual values for specific foods vary because of differences in composition. Proteins, the third major source of calories in the diet, vary as well. Proteins are composed of amino acids, which have the following general structure: In addition to their amine and carboxylic acid components, amino acids may contain a wide range of other functional groups: R can be hydrogen (–H); an alkyl group (e.g., –CH ); an aryl group (e.g., –CH C H ); or a substituted alkyl group that contains an amine, an alcohol, or a carboxylic acid (Figure \(\Page {1}\)). Of the 20 naturally occurring amino acids, 10 are required in the human diet; these 10 are called because our bodies are unable to synthesize them from other compounds. Because R can be any of several different groups, each amino acid has a different value of Δ . Proteins are usually estimated to have an average Δ of 17 kJ/g (about 4 Cal/g). Calculate the amount of available energy obtained from the biological oxidation of 1.000 g of alanine (an amino acid). Remember that the nitrogen-containing product is urea, not N , so biological oxidation of alanine will yield energy than will combustion. The value of Δ for alanine is −1577 kJ/mol. amino acid and Δ per mole caloric content per gram The actual energy available biologically from alanine is less than its Δ because of the production of urea rather than N . We know the Δ values for alanine and urea, so we can use Hess’s law to calculate Δ for the oxidation of alanine to CO , H O, and urea. We begin by writing balanced chemical equations for (1) the oxidation of alanine to CO , H O, and urea; (2) the combustion of urea; and (3) the combustion of alanine. Because alanine contains only a single nitrogen atom, whereas urea and N each contain two nitrogen atoms, it is easier to balance Equations 1 and 3 if we write them for the oxidation of 2 mol of alanine: \[ \left ( 1 \right )\; \; 2C_{3}H_{7}NO_{2}\left ( s \right )+6O_{2}\left ( g \right ) \rightarrow 5CO_{2}\left ( g \right )+5H_{2}O\left ( l \right )+\left ( H_{2}N \right )_{2}C=O\left ( s \right ) \nonumber \] \[ \left ( 2 \right ) \; \; \left( H_{2}N \right )_{2}C=O\left ( s \right ) + \dfrac{3}{2}O_{2}\left ( g \right ) \rightarrow CO_{2}\left ( g \right )+2H_{2}O\left ( l \right )+ N_{2}\left ( g \right ) \nonumber \] \[ \left ( 3 \right ) \; \; \left ( 1 \right )\; \; 2C_{3}H_{7}NO_{2}\left ( s \right )+\dfrac{15}{2}O_{2}\left ( g \right )\rightarrow 6CO_{2}\left ( g \right )+7H_{2}O\left ( l \right )+N_{2}\left ( g \right ) \nonumber \] Adding Equations 1 and 2 and canceling urea from both sides give the overall chemical equation directly: \[ \left ( 1 \right )\; \; 2C_{3}H_{7}NO_{2}\left ( s \right )+6O_{2}\left ( g \right ) \rightarrow 5CO_{2}\left ( g \right )+5H_{2}O\left ( l \right )+\cancel{\left ( H_{2}N \right )_{2}C=O\left ( s \right )} \nonumber \] \[ \cancel{ \left ( 2 \right ) \; \; \left( H_{2}N \right )_{2}C=O\left ( s \right )} + \dfrac{3}{2}O_{2}\left ( g \right ) \rightarrow CO_{2}\left ( g \right )+2H_{2}O\left ( l \right )+ N_{2}\left ( g \right ) \nonumber \] \[ \left ( 3 \right ) \; \; \left ( 1 \right )\; \; 2C_{3}H_{7}NO_{2}\left ( s \right )+\dfrac{15}{2}O_{2}\left ( g \right )\rightarrow 6CO_{2}\left ( g \right )+7H_{2}O\left ( l \right )+N_{2}\left ( g \right ) \nonumber \] By Hess’s law, Δ = Δ + Δ . We know that Δ = 2Δ (alanine), Δ = Δ (urea), and Δ = 2Δ (alanine → urea). Rearranging and substituting the appropriate values gives \[ \Delta {H_{1}} = \Delta {H_{3}} - \Delta {H_{2}} \nonumber \] \[ =2\left ( -1577 \; kJ/mol \right )-\left ( -632.0 \; kJ/mol \right ) \nonumber \]= -2522 \; kJ/\left ( 2 \;mol\; analine \right ) \nonumber \] Thus Δ (alanine → urea) = −2522 kJ/(2 mol of alanine) = −1261 kJ/mol of alanine. Oxidation of alanine to urea rather than to nitrogen therefore results in about a 20% decrease in the amount of energy released (−1261 kJ/mol versus −1577 kJ/mol). The energy released per gram by the biological oxidation of alanine is \[ \left (\dfrac{-1261 \; kJ}{1 \; \cancel{mol}} \right )\left (\dfrac{1 \; \cancel{mol}}{89.094 \; g} \right )= -14.15 \; kJ/g \nonumber \] This is equal to −3.382 Cal/g. Calculate the energy released per gram from the oxidation of valine (an amino acid) to CO , H O, and urea. Report your answer to three significant figures. The value of Δ for valine is −2922 kJ/mol. −22.2 kJ/g (−5.31 Cal/g) The reported caloric content of foods does not include Δ for those components that are not digested, such as fiber. Moreover, meats and fruits are 50%−70% water, which cannot be oxidized by O to obtain energy. So water contains no calories. Some foods contain large amounts of fiber, which is primarily composed of sugars. Although fiber can be burned in a calorimeter just like glucose to give carbon dioxide, water, and heat, humans lack the enzymes needed to break fiber down into smaller molecules that can be oxidized. Hence fiber also does not contribute to the caloric content of food. We can determine the caloric content of foods in two ways. The most precise method is to dry a carefully weighed sample and carry out a combustion reaction in a bomb calorimeter. The more typical approach, however, is to analyze the food for protein, carbohydrate, fat, water, and “minerals” (everything that doesn’t burn) and then calculate the caloric content using the average values for each component that produces energy (9 Cal/g for fats, 4 Cal/g for carbohydrates and proteins, and 0 Cal/g for water and minerals). An example of this approach is shown in Table \(\Page {1}\) for a slice of roast beef. The compositions and caloric contents of some common foods are given in Table \(\Page {2}\). Because the Calorie represents such a large amount of energy, a few of them go a long way. An average 73 kg (160 lb) person needs about 67 Cal/h (1600 Cal/day) to fuel the basic biochemical processes that keep that person alive. This energy is required to maintain body temperature, keep the heart beating, power the muscles used for breathing, carry out chemical reactions in cells, and send the nerve impulses that control those automatic functions. Physical activity increases the amount of energy required but not by as much as many of us hope (Table \(\Page {2}\)). A moderately active individual requires about 2500−3000 Cal/day; athletes or others engaged in strenuous activity can burn 4000 Cal/day. Any excess caloric intake is stored by the body for future use, usually in the form of fat, which is the most compact way to store energy. When more energy is needed than the diet supplies, stored fuels are mobilized and oxidized. We usually exhaust the supply of stored carbohydrates before turning to fats, which accounts in part for the popularity of low-carbohydrate diets. What is the minimum number of Calories expended by a 72.6 kg person who climbs a 30-story building? (Assume each flight of stairs is 14 ft high.) How many grams of glucose are required to supply this amount of energy? (The energy released during the combustion of glucose was calculated in Example 5.5.4). mass, height, and energy released by combustion of glucose calories expended and mass of glucose needed The energy needed to climb the stairs equals the difference between the person’s potential energy ( ) at the top of the building and at ground level. Recall that = . Because and are given in non-SI units, we must convert them to kilograms and meters, respectively \[ PE = \left ( 72.6 \; kg \right )\left ( 9.81 \; m/s^{2} \right )\left ( 128 m \right ) = 8.55 × 10^{4} \left ( kg \cdot m^{2}/s^{2} \right ) = 91.2 kJ \nonumber \] To convert to Calories, we divide by 4.184 kJ/kcal: \[ PE = \left ( 91.2 \; \cancel{kJ} \right ) \left ( \dfrac{1 \; kcal}{4.184 \; \cancel{kJ}} \right )=21.8 \;kcal = 21.8 \; Cal \nonumber \] Because the combustion of glucose produces 15.6 kJ/g (Example 5), the mass of glucose needed to supply 85.5 kJ of energy is \[ PE = \left ( 91.2 \; \cancel{kJ} \right )\left ( \dfrac{1 \; g \; glucose}{15.6 \; \cancel{kJ}} \right )=5.85\; g \; glucose \nonumber \] This mass corresponds to only about a teaspoonful of sugar! Because the body is only about 30% efficient in using the energy in glucose, the actual amount of glucose required would be higher: (100%/30%) × 5.85 g = 19.5 g. Nonetheless, this calculation illustrates the difficulty many people have in trying to lose weight by exercise alone. Calculate how many times a 160 lb person would have to climb the tallest building in the United States, the 110-story Willis Tower in Chicago, to burn off 1.0 lb of stored fat. Assume that each story of the building is 14 ft high and use a calorie content of 9.0 kcal/g of fat. About 55 times The calculations in Example 5.8.2 ignore various factors, such as how fast the person is climbing. Although the rate is irrelevant in calculating the change in potential energy, it is very relevant to the amount of energy actually required to ascend the stairs. The calculations also ignore the fact that the body’s conversion of chemical energy to mechanical work is significantly less than 100% efficient. According to the average energy expended for various activities listed in Table 5.8.3, a person must run more than 4.5 h at 10 mph or bicycle for 6 h at 13 mph to burn off 1 lb of fat (1.0 lb × 454 g/lb × 9.0 Cal/g = 4100 Cal). But if a person rides a bicycle at 13 mph for only 1 h per day 6 days a week, that person will burn off 50 lb of fat in the course of a year (assuming, of course, the cyclist doesn’t increase his or her intake of calories to compensate for the exercise). Thermochemical concepts can be applied to determine the actual energy available in food. The nutritional is equivalent to 1 kcal (4.184 kJ). The caloric content of a food is its Δ per gram. The combustion of nitrogen-containing substances produces N (g), but the biological oxidation of such substances produces urea. Hence the actual energy available from nitrogen-containing substances, such as proteins, is less than the Δ of urea multiplied by the number of moles of urea produced. The typical caloric contents for food are 9 Cal/g for fats, 4 Cal/g for carbohydrates and proteins, and 0 Cal/g for water and minerals. | 13,237 | 3,735 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/13%3A_Air/13.05%3A_13.4__Automobile_Emissions |
or flue gas is emitted as a result of the combustion of fuels such as natural gas, gasoline, petrol, biodiesel blends, diesel fuel, fuel oil, or coal. It is a major component of motor vehicle emissions (and from stationary internal combustion engines), which can also include: Motor vehicle emissions contribute to air pollution and are a major ingredient in the creation of smog in some large cities. A 2013 study by MIT indicates that 53,000 early deaths occur per year in the United States alone because of vehicle emissions. Carbon monoxide and nitrogen oxides are the two main pollutant gases from automobile emissions. Ozone is a result of the reaction between nitrogen oxides and volatile organic compounds (VOCs). ) is a colorless, odorless gas that can be harmful when inhaled in large amounts. CO is released when something is burned. The greatest sources of CO to outdoor air are cars, trucks and other vehicles or machinery that burn fossil fuels. A variety of items in your home such as unvented kerosene and gas space heaters, leaking chimneys and furnaces, and gas stoves also release CO and can affect air quality indoors. Breathing air with a high concentration of CO reduces the amount of oxygen that can be transported in the blood stream to critical organs like the heart and brain. At very high levels, which are possible indoors or in other enclosed environments, CO can cause dizziness, confusion, unconsciousness and death. Very high levels of CO are not likely to occur outdoors. However, when CO levels are elevated outdoors, they can be of particular concern for people with some types of heart disease. These people already have a reduced ability for getting oxygenated blood to their hearts in situations where the heart needs more oxygen than usual. They are especially vulnerable to the effects of CO when exercising or under increased stress. In these situations, short-term exposure to elevated CO may result in reduced oxygen to the heart accompanied by chest pain also known as angina. is one of a group of highly reactive gasses known as "oxides of nitrogen," or "nitrogen oxides (NO )." Other nitrogen oxides include nitrous acid and nitric acid. NO is a yellowish-brown to reddish-brown foul-smelling gas that is a major contributor to smog and acid rain. Nitrogen oxides result when atmospheric nitrogen and oxygen react at the high temperatures created by combustion engines. Most emissions in the U.S. result from combustion in vehicle engines, electrical utility, and industrial combustion. NO primarily gets in the air from the burning of fuel. NO forms from emissions from cars, trucks and buses, power plants, and off-road equipment. Breathing air with a high concentration of NO can irritate airways in the human respiratory system. Such exposures over short periods can aggravate respiratory diseases, particularly asthma, leading to respiratory symptoms (such as coughing, wheezing or difficulty breathing), hospital admissions and visits to emergency rooms. Longer exposures to elevated concentrations of NO may contribute to the development of asthma and potentially increase susceptibility to respiratory infections. People with asthma, as well as children and the elderly are generally at greater risk for the health effects of NO . NO along with other NO reacts with other chemicals in the air to form both particulate matter and ozone. Both of these are also harmful when inhaled due to effects on the respiratory system. NO and other NO interact with water, oxygen and other chemicals in the atmosphere to form acid rain. Acid rain harms sensitive ecosystems such as lakes and forests. The nitrate particles that result from NO make the air hazy and difficult to see though. This affects the many national parks that we visit for the view. NO in the atmosphere contributes to nutrient pollution in coastal waters. is a colorless gas with a slightly sweet odor that is not emitted directly into the air, but is created by the interaction of sunlight, heat, oxides of nitrogen (NO ) and volatile organic compounds (VOCs). Ozone is likely to reach unhealthy levels on hot sunny days in urban environments. Emissions from industrial facilities and electric utilities, motor vehicle exhaust, gasoline vapors, and chemical solvents are some of the major sources of NO and VOCs. Ozone is a gas composed of three atoms of oxygen (O ). Ozone occurs both in the Earth's upper atmosphere and at ground level. Ozone can be good or bad, depending on where it is found. Called stratospheric ozone, good ozone occurs naturally in the upper atmosphere, where it forms a protective layer that shields us from the sun's harmful ultraviolet rays. This beneficial ozone has been partially destroyed by manmade chemicals, causing what is sometimes called a "hole in the ozone." The good news is, this hole is diminishing. Ozone at ground level is a harmful air pollutant, because of its effects on people and the environment, and it is the main ingredient in “smog." Tropospheric, or ground level ozone, is not emitted directly into the air, but is created by chemical reactions between oxides of nitrogen (NOx) and volatile organic compounds (VOC). This happens when pollutants emitted by cars, power plants, industrial boilers, refineries, chemical plants, and other sources chemically react in the presence of sunlight. Ozone is most likely to reach unhealthy levels on hot sunny days in urban environments, but can still reach high levels during colder months. Ozone can also be transported long distances by wind, so even rural areas can experience high ozone levels. Ozone in the air we breathe can harm our health especially on hot sunny days when ozone can reach unhealthy levels. People most at risk from breathing air containing ozone include people with asthma, children, older adults, and people who are active outdoors, especially outdoor workers. Ozone affects sensitive vegetation and ecosystems, including forests, parks, wildlife refuges and wilderness areas. In particular, ozone harms sensitive vegetation during the growing season. ( ) | 6,135 | 3,737 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/03%3A__The_Vocabulary_of_Analytical_Chemistry/3.08%3A_Problems |
and in the presence of both glycolic acid and ascorbic acid (AA), the signal is
\[S_{samp,2} = k_\text{GA} C_\text{GA} + k_\text{AA} C_\text{AA} \nonumber\]
When the concentration of glycolic acid is \(1.0 \times 10^{-4} \text{ M}\) and the concentration of ascorbic acid is \(1.0 \times 10^{-5} \text{ M}\), the ratio of their signals is
\[\frac {S_{samp,2}} {S_{samp,1}} = 1.44 \nonumber\]
(a) Using the ratio of the two signals, determine the value of the selectivity ratio .
(b) Is the method more selective toward glycolic acid or ascorbic acid?
(c) If the concentration of ascorbic acid is \(1.0 \times 10^{-5} \text{ M}\), what is the smallest concentration of glycolic acid that can be determined such that the error introduced by failing to account for the signal from ascorbic acid is less than 1%? (d) What is the largest concentration of ascorbic acid that may be present if a concentration of \(1.12 \times 10^{-6} \text{ M}\) hypoxanthine is to be determined within 1.0%? | 1,064 | 3,739 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/19%3A_Nuclear_Chemistry/19.02%3A_Naturally_Occurring_Radioactivity |
When we discuss natural radioactivity in the sections under The Structure of Atoms, we mention the properties of \(\alpha\) particles, \(\beta\) particles, and \(\gamma\) rays, but little attention is paid to the atoms which are left behind when one of these forms of radiation is emitted. Now we consider the subject of radioactivity in more detail. An alpha particle corresponds to a helium nucleus. It consists of two protons and two neutrons, and so it has a mass number of 4 and an atomic number (nuclear charge) of 2. From a chemical point of view we would write it as He , indicating its lack of two electrons with the superscript 2+. In writing a nuclear reaction, though, it is unnecessary to specify the charge, because the presence or absence of electrons around the nucleus is usually unimportant. For these purposes an α particle is indicated as \({}_{\text{2}}^{\text{4}}\text{He}\) or \({}_{\text{2}}^{\text{4}}\alpha\label{1}\). In certain nuclei a particles are produced by combination of two protons and two neutrons which are then emitted. An example of naturally occurring emission of an α particle is the disintegration of one of the isotopes of uranium, \({}_{\text{92}}^{\text{238}}\text{U}\). The equation for this process is \[{}_{\text{92}}^{\text{238}}\text{U }\to \text{ }{}_{\text{90}}^{\text{234}}\text{Th + }{}_{\text{2}}^{\text{4}}\text{He} \label{2} \] Note that if we sum the mass numbers on each side of a , such as Equation \(\ref{2}\), the total is the same. That is, 238 on the left equals 234 + 4 on the right. Similarly, the atomic numbers (subscripts) must also balance (92 on the left and 90 + 2 on the right). This is a general rule which must be followed in writing any nuclear reaction. The total number of nucleons (i.e., protons and neutrons) remains unchanged, and electrical charge is neither created nor destroyed in the process. When a nucleus emits an a particle, its atomic number is reduced by 2 and it becomes the nucleus of an element two places earlier in the periodic table. That one element could transmute into another in this fashion was first demonstrated by Rutherford and Soddy in 1902. It caused a tremendous stir in the scientific circles of the day since it quite clearly contradicted Dalton’s hypothesis that atoms are immutable. It gave Rutherford, who was then working at McGill University in Canada, an international reputation. The type of nucleus that will spontaneously emit an a particle is fairly restricted. The mass number is usually greater than 209 and the atomic number greater than 82. In addition the nucleus must have a lower ratio of neutrons to protons than normal. The emission of an a particle raises the neutron/proton ratio as illustrated by the nuclear equation \[{}_{\text{84}}^{\text{210}}\text{Po }\to \text{ }{}_{\text{82}}^{\text{206}}\text{Pb + }{}_{\text{2}}^{\text{4}}\text{He} \nonumber \] The nucleus of \({}_{\text{84}}^{\text{210}}\text{Po}\) contains 210 – 84 = 126 neutrons and 84 protons, giving a ratio of 126:84 = 1.500. This is increased to 124:82 = 1.512 by the α-emission. A beta particle is an electron which has been emitted from an atomic nucleus. It has a very small mass and is therefore assigned a mass number of 0. The β particle has a negative electrical charge, and so its nuclear charge is taken to be –1. Thus it is given the symbol \({}_{-\text{1}}^{\text{0}}e\) or \({}_{-\text{1}}^{\text{0}}\beta\) in a nuclear equation. Two examples of unstable nuclei which emit \(β\) particles are \[\ce{ _{90}^{234}Th -> _{91}^{234}Pa + _{-1}^{0}e} \nonumber \] and \[{}_{\text{6}}^{\text{14}}\text{C }\to \text{ }{}_{\text{7}}^{\text{14}}\text{N + }{}_{-\text{1}}^{\text{0}}e \nonumber \] Note that both of these equations accord with the conservation of mass number and atomic number, showing again that both the total number of nucleons and the total electrical charge remain unchanged. We can consider a β particle emitted from a nucleus to result from the transformation of a neutron into a proton and an electron according to the reaction \[{}_{\text{0}}^{\text{1}}n\text{ }\to \text{ }{}_{\text{1}}^{\text{1}}p\text{ + }{}_{-\text{1}}^{\text{0}}e \nonumber \] (Indeed, free neutrons unattached to any nucleus soon decay in this way.) Thus when a nucleus emits a β particle, the nuclear charge rises by 1 while the mass number is unaltered. Therefore the disintegration of a nucleus by β decay produces the nucleus of an element one place in the periodic table than the original element. β decay is a very common form of radioactive disintegration and, unlike α decay, is found among both heavy and light nuclei. Nuclei which disintegrate in this way usually have a neutron/proton ratio which is higher than normal. When a β particle is lost, a neutron is replaced by a proton and the neutron/proton ratio decreases. For example, in the decay process \[{}_{\text{82}}^{\text{206}}\text{Pb }\to \text{ }{}_{\text{83}}^{\text{206}}\text{Bi + }{}_{-\text{1}}^{\text{0}}e \nonumber \] the neutron/proton ratio changes from 1.561 to 1.530. Gamma rays correspond to electromagnetic radiation similar to light waves or radiowaves except that they have an extremely short wavelength—about a picometer. Because of the wave-particle duality we can also think of them as particles or photons having the same velocity as light and an extremely high energy. Since they have zero charge and are not nucleons, they are denoted in nuclear equations by the symbol \({}_{\text{0}}^{\text{0}}\gamma\) or, more simply, \(\gamma\). Virtually all nuclear reactions are accompanied by the emission of γ rays. This is because the occurrence of a nuclear transformation usually leaves the resultant nucleus in an unstable high-energy state. The nucleus then loses energy in the form of a γ-ray photon as it adopts a lower-energy more stable form. Usually these two processes succeed each other so rapidly that they cannot be distinguished. Thus when U decays by \(α\) emission, it also emits a \(γ\) ray. This is actually a two-stage process. In the first stage a high-energy (or excited) form of Th is produced: \[{}_{\text{92}}^{\text{238}}\text{U }\to \text{ }{}_{\text{90}}^{\text{234}}\text{Th* + }{}_{\text{2}}^{\text{4}}\text{He} \label{eq7} \] This excited nucleus-then emits a \(γ\) ray: \[{}_{\text{90}}^{\text{234}}\text{Th* }\to \text{ }{}_{\text{90}}^{\text{234}}\text{Th} + {}_{\text{0}}^{\text{0}}\gamma \label{eq8} \] Usually when a nuclear reaction is written, the \(γ\) ray is omitted. Thus Equations \ref{eq7} and \ref{eq8} are usually combined to give \[{}_{\text{92}}^{\text{238}}\text{U }\to \text{ }{}_{\text{90}}^{\text{234}}\text{Th} + {}_{\text{2}}^{\text{4}}\text{He} \nonumber \] | 6,721 | 3,740 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/15%3A_Quality_Assurance/15.07%3A_Chapter_Summary_and_Key_Terms |
Few analyses are so straightforward that high quality results are obtained with ease. Good analytical work requires careful planning and an attention to detail. Creating and maintaining a quality assurance program is one way to help ensure the quality of analytical results. Quality assurance programs usually include elements of quality control and quality assessment. Quality control encompasses all activities used to bring a system into statistical control. The most important facet of quality control is written documentation, including statements of good laboratory practices, good measurement practices, standard operating procedures, and protocols for a specific purpose. Quality assessment includes the statistical tools used to determine whether an analysis is in a state of statistical control, and, if possible, to suggest why an analysis has drifted out of statistical control. Among the tools included in quality assessment are the analysis of duplicate samples, the analysis of blanks, the analysis of standards, and the analysis of spike recoveries. Another important quality assessment tool, which provides an ongoing evaluation of an analysis, is a control chart. A control chart plots a property, such as a spike recovery, as a function of time. Results that exceed warning and control limits, or unusual patterns of results indicate that an analysis is no longer under statistical control. control chart
good laboratory practices proficiency standard quality assurance program spike recovery trip blank duplicate samples good measurement practices protocol for a specific purpose quality control standard operations procedure field blank method blank quality assessment reagent blank statistical control | 1,748 | 3,743 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/16%3A_Chemistry_of_Benzene_-_Electrophilic_Aromatic_Substitution |
When you have completed Chapter 16, you should be able to In the preceding chapter, you studied the concept of aromaticity and spent considerable time on the theoretical aspects of the chemistry of aromatic compounds. In this chapter, you will begin to study the chemical reactions of aromatic compounds, focusing in particular on electrophilic aromatic substitution, and to a lesser extent on nucleophilic aromatic substitution. We will discuss, in detail, the mechanism of electrophilic substitution, paying particular attention to the factors that determine both the rate and position of substitution in those aromatic compounds which already have one or more substituents present in the aromatic ring. When we discuss nucleophilic aromatic substitution, you will see that it can be achieved by two different mechanisms, one of which involves the formation of an unusual looking intermediate, benzyne. You will also see how alkyl and acyl groups can be introduced on to an aromatic ring; how, once introduced, alkyl groups can be converted to carboxyl groups; and how bromine can be introduced to the alkyl side chain of alkylbenzene. The latter reaction is particularly useful because the benzylic bromide so produced undergoes the reactions of a typical alkyl bromide, thus providing us with a synthetic route to a large variety of compounds.
| 1,368 | 3,744 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Vicinal_Syn_Dihydroxylation |
oxidizes alkenes to give glycols through syn addition. A glycol, also known as a vicinal diol, is a compound with two -OH groups on adjacent carbons. The reaction with \(OsO_4\) is a concerted process that has a cyclic intermediate and no rearrangements. Vicinal syn dihydroxylation complements the epoxide-hydrolysis sequence which constitutes an dihydroxylation of an alkene. When an alkene reacts with osmium tetroxide, stereocenters can form in the glycol product. Cis alkenes give products and trans alkenes give . \(OsO_4\) is formed slowly when osmium powder reacts with gasoues \(O_2\) at ambient temperature. Reaction of bulk solid requires heating to 400 °C: \[Os_{(s)} + 2O_{2\;(g)} \rightarrow OS_4\] Since Osmium tetroxide is expensive and highly toxic, the reaction with alkenes has been modified. Catalytic amounts of OsO and stoichiometric amounts of an oxidizing agent such as hydrogen peroxide are now used to eliminate some hazards. Also, an older reagent that was used instead of OsO was potassium permanganate, \(KMnO_4\). Although syn diols will result from the reaction of KMnO and an alkene, potassium permanganate is less useful since it gives poor yields of the product because of . Example: Dihydroxylation of 1-ethyl-1-cycloheptene Antitumor drugs have been formed by using dihydroxylation. This method has been applied to the enantioselective synthesis of ovalicin, which is a class of fungal-derived products called antiangiogenesis agents. These antitumor products can cut off the blood supply to solid tumors. A derivative of ovalicin, TNP-470, is chemically stable, nontoxic, and noninflammatory. TNP-470 has been used in research to determine its effectiveness in treating cancer of the breast, brain, cervix, liver, and prostate. Questions: 1. Give the major product. 2. What is the product in the dihydroxylation of (Z)-3-hexene? 3. What is the product in the dihydroxylation of (E)-3-hexene? 4. Draw the intermediate of this reaction. 5. Fill in the missing reactants, reagents, and product. Answers: 1. A syn-1,2-ethanediol is formed. There is no stereocenter in this particular reaction. The OH groups are on the same side. 2. Meso-3,4-hexanediol is formed. There are 2 stereocenters in this reaction. 3. A racemic mixture of 3,4-hexanediol is formed. There are 2 stereocenters in both products. 4. A cyclic osmic ester is formed. 5. reaction is needed in the first box to form the cyclohexene. The second box needs a reagent to reduce the intermediate cyclic ester (not shown). The third box has the product: 1,2-cyclohexanediol. | 2,598 | 3,745 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity_of_Alpha_Hydrogens/Claisen_Reactions |
Because esters can contain \(\alpha\) hydrogens they can undergo a condensation reaction similar to the aldol reaction called a . In a fashion similar to the aldol, one ester acts as a nucleophile while a second ester acts as the electrophile. During the reaction a new carbon-carbon bond is formed; the product is a β-keto ester. A major difference with the aldol reaction is the fact that hydroxide cannot be used as a base because it could possibly react with the ester. Instead, an alkoxide version of the alcohol used to synthesize the ester is used to prevent transesterification side products. Basic reaction Going from reactants to products simply 1) Enolate formation 2) Nucleophilic attack 3) Removal of leaving group A diester can undergo an intramolecular reaction called a Dieckmann condensation. Claisen condensations between different ester reactants are called reactions. Crossed Claisen reactions in which both reactants can serve as donors and acceptors generally give complex mixtures. Because of this most Crossed Claisen reactions are usually not performed unless one reactant has no alpha hydrogens. ) | 1,147 | 3,746 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.04%3A_Heat_Capacity_and_Specific_Heat |
If a swimming pool and wading pool, both full of water at the same temperature, were subjected to the same input of heat energy, the wading pool would certainly rise in temperature more quickly than the swimming pool. The heat capacity of an object depends both on its mass and its chemical composition. Because of its much larger mass, the swimming pool of water has a larger heat capacity than the wading pool. Different substances respond to heat in different ways. If a metal chair sits in the bright sun on a hot day, it may become quite hot to the touch. An equal mass of water under the same sun exposure will not become nearly as hot. This means that water has a high (the amount of heat required to raise the temperature of an object by \(1^\text{o} \text{C}\)). Water is very resistant to changes in temperature, while metals generally are not. The of a substance is the amount of energy required to raise the temperature of 1 gram of the substance by \(1^\text{o} \text{C}\). The table below lists the specific heats of some common substances. The symbol for specific heat is \(c_p\), with the \(p\) subscript referring to the fact that specific heats are measured at constant pressure. The units for specific heat can either be joules per gram per degree \(\left( \text{J/g}^\text{o} \text{C} \right)\) or calories per gram per degree \(\left( \text{cal/g}^\text{o} \text{C} \right)\). This text will use \(\text{J/g}^\text{o} \text{C}\) for specific heat. Notice that water has a very high specific heat compared to most other substances. Water is commonly used as a coolant for machinery because it is able to absorb large quantities of heat (see table above). Coastal climates are much more moderate than inland climates because of the presence of the ocean. Water in lakes or oceans absorbs heat from the air on hot days and releases it back into the air on cool days. | 1,909 | 3,747 |
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We would expect the NMR spectrum of two spin-coupled protons, A & B, to display a pair of doublets. However, if the ratio of Δν to J (both in Hz) decreases to less than 10 a significant distortion of this expected pattern will take place, as shown in the following diagram. By four examples of this behavior will be presented, starting with a compound in which Δν/J = 7.9 and followed by cases having smaller ratios. The fourth example, 2-methyl-2-phenyl-1,3-propanediol is especially interesting. Here, the two identical methylene groups have a pair of diastereotopic protons (H & H ). Because these hydrogens have similar but different chemical shifts, the expected doublet pair experiences a second order distortion. Second order splitting distortions are not limited to single proton pairs. The following spectrum of 2-dimethylaminoethyl acetate shows what appears to be a well behaved pair of triplets coming from adjacent methylene groups. Such cases are often deceptively simple, and become very complex when Δν/J is less than 10. will display a simple example. For additional examples of splitting patterns Here. | 1,141 | 3,749 |
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A solution consists of two or more substances dissolved in a liquid form. Not to get confused with a , which is heterogeneous--multiple substances exist in varying structures-- solutions are homogenous, which means that atoms of the solute are evenly dispersed throughout the solvent (ex. water, ethanol). Think of it as comparing a cup of (dissolved) sugar water and a cup of water with lego blocks in it. The solute is the substance dissolved in the solution, and the solvent is the substance doing the dissolving. ex. A solution of NaCl in water A mixture of lego blocks and water Note: All solutions are mixtures but not all mixtures are solutions. Water (H O) is the most common solvent, used for dissolving many compounds or brewing coffee. Other common solvents include turpentine (a paint thinner), acetone (nail polish remover), and ethanol (used in some perfumes). Such solvents usually contain carbon and are called organic solvents. Solutions with water as the solvent are called aqueous solutions; they have special properties that are covered . Different chemical compounds dissolve in solutes in varying degrees. Some compounds, such as the strong acid hydrochloric acid (HCl), dissociate completely in solution into ions. Others, like the weak base ammonia (NH ), only partly dissociate. Yet other compounds like alcohol do not dissociate at all and remain compounds. Laboratory reactions often involve acids and bases, which are covered in more detail . Concentration is the measure of the amount of solute in a certain amount of solvent. Knowing the concentration of a solution is important determining the strength of an acid or base (pH), among other things. When there is so much solute present in a concentration that it no longer dissolves, the solution is saturated. Scientists often use to measure concentration. molarity = moles/Liter Since reaction stoichiometry relies on molar ratios, molarity is the main measurement for concentration. A less common unit for concentration is called . Molality = moles/Kg of solvent Scientists sometimes use molality to measure concentration because liquid volumes change slightly based on the temperature and pressure. Mass, however, stays the same and can be measured accurately using a balance. Commercial concentrated products are usually expressed in mass percent; such as commercial concentrated sulfuric acid, which is 93-98% H SO by mass in water (Hill, Petrucci 116). Solutions used in the laboratory are usually made from either solid solutes (often salts) or stock solutions. To make a solution from solid solutes, first calculate how many moles of solute are in the desired solutions (using the molarity). Calculate the amount of solid you need in grams using the moles needed and the molar mass of the solute and weight out the needed amount. Transfer the solute to a container (preferably a volumetric flask, which most accurately measures volume of solution labeled on the flask) and add a small amount of solvent. Mix thoroughly to dissolve the solute. Once the solute has dissolved, add the remaining solvent to make the solution of the desired volume and mix thoroughly. For example, to make 0.5 Liters of 0.5 molar NaCl: 1. Multiply the concentration (0.5 mols/Liters) by the volume of solution you want (0.5 Liters) to find the moles of NaCl you need. 0.5 moles/Liter * 0.5 Liters = 0.25 moles of NaCl 2. Multiply the moles of NaCl by its molar mass (58.44 g/mol) to find the grams of solute needed. (0.25 moles NaCl)*(58.44 grams/mole) = 14.61 grams of NaCl Making a solution of a certain concentration from a stock solution is called a dilution. When diluting a solution, keep in mind that adding a solvent to a solution changes the concentration of the solution, but not the amount of solute already present. The equation to use is simply M and V are the concentration and volume of the original (stock) solution to dilute; M and V are the desired concentration and volume of the final solution. For reactions that take place in solutions: 1. 2. 3. 4. 5. | 4,147 | 3,751 |
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Fuels cells generate electricity from an electrochemical reaction in which oxygen and a fuel combine to form water. Fuel cells work by converting the chemical energy found in the fuel, e.g. hydrogen gas, into electrical energy (electricity). In a fuel cell, the fuel is oxidized at the anode which yields electrons that flow through an external circuit doing electrical work, before an oxidant (often oxygen) is reduced by these electrons at the cathode. \[O_{2(g)} + \underset{\text{Fuel}}{H_{2(g)}} \rightarrow 2H_2O_{(l)}\] The resulting electricity can be used in a variety of ways, including: powering motor vehicles, electrical devices, and airplanes. Fuel cells also produce heat as a by-product which is finding increasing use in heating homes, especially in Japan. With its multifunctional products and by-products, fuel cells are rapidly becoming the hot ticket alternative source of energy as we move into a greener and more progressive era. A hydrogen fuel cell enables hydrogen to be combined electrochemically with oxygen to produce electricity, water, and heat. One fuel cell alone only produces a small amount of power. However, grouping the individual fuel cells together creates a fuel cell stack (as shown below in the diagram). When delivered to a fuel cell engine, fuel cell stacks create enough energy to power buses and other vehicles, and have even been used to power spacecraft. Each fuel cell in the stack contains two electrodes - a positive cathode (where the reduction occurs) and a negative anode (where the oxidation occurs). The energy-producing reactions take place at the surfaces of the electrodes. Each individual pair of electrodes is separated by an electrolyte (either in solid or liquid form). This electrolyte carries electrically charged particles (ions) between the electrodes. The rate of a given reaction can be increased with the help of a catalyst such as platinum or nickel. The power (\(P\)) produced by a fuel cell is equal to the voltage (\(V\)) multiplied by the current (\(I\)) at which it is being operated. This is measured in watts (W). \[P= IV\] There are a number of factors that reduce the efficiency in fuel cells, which can be broadly categorized as reversible, irreversible and fuel utilization losses. 'Reversible' losses correspond to losses that underpin the deviation between the standard electrode potential of the full electrochemical cell and the actual operating open-circuit voltage (OCV). 'Irreversible' losses describe the various contributions from different components of the fuel cell to the loss of voltage with respect to OCV, as the current drawn from the cell/stack is increased. This includes an activation barrier that must be overcome at low currents, which can represent a loss of ~200 mV. This results from the reactions requiring energy to overcome a threshold before occurring, despite a thermodynamic driving force. This is especially relevant in the case of the reduction of oxygen at the cathode. Further losses are experienced in the intermediate regime as a result of ohmic resistance (both ionic and electronic) and due to mass transport limitations at high current loads. Particularly in the case of PEMFCs, the flows of waste water and unspent fuel occur at rates that exceed the capabilities of the fuel cell's physical capabilities and not all fuel that enters at the inlet is utilized. This also results in a drop of overall efficiency. Since the efficiency of a fuel cell is directly proportional to the power generated, its efficiency is almost proportional to its voltage. MCFC's use molten carbonate salts as their electrolyte. This fuel cell has a high electrical efficiency of 60 %. These cells operate at about 600 degrees Celsius. The generated power varies, and some units have been built with outputs as high as 100 MW. Because of the high temperatures, these cells are not generally used in the home. In this type of fuel cell, there is a gas diffusion electrode, a zinc anode separated by electrolyte and a mechanical separator. This fuel cell's anode and cathode are made with specks of platinum on a carbon and silicon carbide matrix that supports the phosphoric acid electrolyte. PAFC's are commonly used in large commercial vehicles i.e. buses and were the first fuel cells to be commercialized. This fuel cell uses a polymeric membrane as the electrolyte along with platinum-activated, carbon-based electrodes. PEMFC's can function at relatively low temperatures and are therefore commonly used in scenarios This fuel cell is similar to the Proton Exchange Membrane Fuel Cell; however, instead of using gaseous hydrogen as the fuel, liquid methanol is used. Fuel cells create little to no environmentally damaging emissions: generally, the only byproduct in a hydrogen fuel cell, typically found in automobiles, is water. Fuel cells are being used all over the globe. Here are some examples: 1. Answer: A) Solid Oxide Fuel Cell (SOFC) Explanation: As explained by the table in the 'Different Types of Fuel Cells' section, the Solid Oxide Fuel Cell (SOFC) is the most efficient with an efficiency of 50-65%. 2. Answer: Water OR Heat Explanation: Both water and heat are by-products of an active hydrogen fuel cell. 3. Answer: United Technologies Corporation Explanation: United Technologies Corporation was the first company to commercialize fuel cells. 4. Answer: It serves as a bridge between anode and cathode. Explanation: An electrolyte serves a bridge that connects the anode and cathode parts of a fuel cell. This is because of the ionic nature of salts in solution. As a voltage is applied, these ions align either at the cathode or at the anode according to their charge. This creates a bridge that the voltage can cross. 5. Answer: No (very little) carbon-dioxide emissions OR Very high efficiency Explanation: The very efficient nature of fuel cells relative to standard combustion engines and other similar devices is highly valued in today's society because of the rising cost of energy. The relatively low emissions produced by fuel cells is also beneficial to the environment. | 6,140 | 3,752 |
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The picture of electrons "orbiting" the nucleus like planets around the sun remains an enduring one, not only in popular images of the atom but also in the minds of many of us who know better. The proposal, first made in 1913, that the centrifugal force of the revolving electron just exactly balances the attractive force of the nucleus (in analogy with the centrifugal force of the moon in its orbit exactly counteracting the pull of the Earth's gravity) is a nice picture, but is simply untenable. One origin for this hypothesis that suggests this perspective is plausible is the similarity of the gravity and Coulombic interactions. The expression for the force of gravity between two masses (Newton's Law of gravity) is \[F_{gravity} \propto \dfrac{m_1m_2}{r^2}\label{1}\] with The expression for the Coulomb force between two charged species is \[F_{Coulomb} \propto \dfrac{q_1q_2}{r^2}\label{2}\] with However, an electron, unlike a planet or a satellite, is electrically charged, and it has been known since the mid-19th century that an electric charge that undergoes acceleration (changes velocity and direction) will emit electromagnetic radiation, losing energy in the process. A revolving electron would transform the atom into a miniature radio station, the energy output of which would be at the cost of the potential energy of the electron; according to classical mechanics, the electron would simply spiral into the nucleus and the atom would collapse. By the 1920's, it became clear that a tiny object such as the electron cannot be treated as a classical particle having a definite position and velocity. The best we can do is specify the probability of its manifesting itself at any point in space. If you had a magic camera that could take a sequence of pictures of the electron in the 1s orbital of a hydrogen atom, and could combine the resulting dots in a single image, you would see something like this. Clearly, the electron is more likely to be found the closer we move toward the nucleus. This is confirmed by this plot which shows the quantity of electron charge per unit volume of space at various distances from the nucleus. This is known as a probability density plot. The per unit volume of space part is very important here; as we consider radii closer to the nucleus, these volumes become very small, so the number of electrons per unit volume increases very rapidly. In this view, it appears as if the electron does fall into the nucleus! According to classical mechanics, the electron would simply spiral into the nucleus and the atom would collapse. Quantum mechanics is a different story. As you know, the potential energy of an electron becomes more negative as it moves toward the attractive field of the nucleus; in fact, it approaches negative infinity. However, because the total energy remains constant (a hydrogen atom, sitting peacefully by itself, will neither lose nor acquire energy), the loss in potential energy is compensated for by an increase in the electron's kinetic energy (sometimes referred to in this context as "confinement" energy) which determines its momentum and its effective velocity. So as the electron approaches the tiny volume of space occupied by the nucleus, its potential energy dives down toward minus-infinity, and its kinetic energy (momentum and velocity) shoots up toward positive-infinity. This "battle of the infinities" cannot be won by either side, so a compromise is reached in which theory tells us that the fall in potential energy is just twice the kinetic energy, and the electron dances at an average distance that corresponds to the Bohr radius. There is still one thing wrong with this picture; according to the (a better term would be "indeterminacy"), a particle as tiny as the electron cannot be regarded as having either a definite location or momentum. The Heisenberg principle says that either the location or the momentum of a quantum particle such as the electron can be known as precisely as desired, but as one of these quantities is specified more precisely, the value of the other becomes increasingly indeterminate. It is important to understand that this is not simply a matter of observational difficulty, but rather a fundamental property of nature. What this means is that within the tiny confines of the atom, the electron cannot really be regarded as a "particle" having a definite energy and location, so it is somewhat misleading to talk about the electron "falling into" the nucleus. Arthur Eddington, a famous physicist, once suggested, not entirely in jest, that a better description of the electron would be "wavicle"! We can, however, talk about where the electron has the highest probability of manifesting itself— that is, where the maximum negative charge will be found. This is just the curve labeled "probability density"; its steep climb as we approach the nucleus shows unambiguously that the electron is most likely to be found in the tiny volume element at the nucleus. But wait! Did we not just say that this does not happen? What we are forgetting here is that as we move out from the nucleus, the number of these small volume elements situated along any radius increases very rapidly with \(r\), going up by a factor of \(4πr^2\). So the probability of finding the electron somewhere on a given radius circle is found by multiplying the probability density by \(4πr^2\). This yields the curve you have probably seen elsewhere, known as the , that is shown on the right side of the above diagram. The peak of the radial probability for principal quantum number \(n = 1\) corresponds to the Bohr radius. To sum up, the probability density and radial probability plots express two different things: the first shows the electron density at any single point in the atom, while the second, which is generally more useful to us, tells us the the relative electron density summed over all points on a circle of given radius. ) | 5,965 | 3,753 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Properties_of_Alkynes/Spectroscopy_of_the_Alkynes |
Unlike alkenyl hydrogens, alkynyl hydrogens give rise to hydrogens, or relatively high field chemical shifts for H-NMR when subjected to an external magnetic field. This can be explained by the cylindrical \(pi\) cloud around the carbon-carbon triple bond. Long range coupling is also observed in the alkynes. Infrared spectroscopy is a useful complement to NMR data, and displays characteristic peaks for terminal and internal alkynes. As discussed before, a carbon-carbon triple bond is the functional characteristic of the alkynes, and protons, or hydrogens, bound to these sp-hybridized carbon atoms resonate at ? = 1.7-3.1 ppm. For example, in the NMR spectrum of 3,3-dimethyl-1-butyne, the terminal hydrogen of the alkyne appears at ? = 2.06 ppm. 3,3-dimethyl-1-butyne. The H-NMR spectrum of 3,3-dimethyl-1-butyne shows a high field signal due to the alkynyl hydrogen on the terminal alkyne. This high field position suggests a relatively shielded hydrogen, which can be explained by the cylindrical electron cloud around the axis of the molecule. When the two ? bonds are subjected to an external magnetic field, these ? electrons will enter into a motion that results in this strong shielding effect with high field chemical shifts, something that is absent in the alkenes. This electron cloud can be seen in the figure below. Another important aspect to keep in mind for alkyne NMR is the splitting of the peaks, or spin spin coupling. In 1-pentyne, for example, the terminal hydrogen is split by the hydrogens across the triple bond, even though it is separated from them by three carbons. As indicated in the figure below, the three peaks for the terminal hydrogen can be explained by this long range coupling, where the -CH group adjacent the sp carbon splits the terminal hydrogen. Infrared Spectroscopy can be helpful in identifying terminal and internal alkynes. Terminal Alkyne: Internal Alkyne: 3-chloro-1-propyne 4,4-dichloro-2-pentyne Internal alkynes, with its sp carbons attached to other carbons, will show weak bands for its triple bond at the 2100-2260 cm region. However, this stretch is relatively weak, and is sometimes not present at all if the internal alkyne is symmetrical. In these cases, the IR spectrum loses its value as a helpful tool. Terminal alkynes, where the sp carbon is attached to a hydrogen, will show bands on the IR spectrum for both its alkynyl hydrogen and its triple bond. The C-H stretch on the terminal alkyne tends to appear as a strong, narrow band in the 3260-3330 cm region while the triple bond shows a weak peak at 2100-2260 cm . Additional C-H bends will appear between 610-700 cm . The figure below shows these different characteristic stretches for internal and terminal alkynes. Based on the NMR and IR provided, what is the structure of C H ? Answer: The degree of unsaturation is 2, indicating that the structure contains an alkyne, a triple bond. Because we don't see a peak at the 2100-2260 cm range on the IR spectrum as expected, and a C-H stretch for an internal alkyne at 3260-3330 cm is also absent, we can assume that a symmetrical internal alkyne is present. The NMR spectrum is more useful for this problem, and indicates that there are two equivalent -CH groups and two equivalent -CH groups. With this information, we get a molecule that looks like this: hexyne Therefore, using both the IR and NMR spectrum, we get hexyne. | 3,431 | 3,755 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Properties_of_Alkynes/Acidity_of_Terminal_Alkynes |
Alkanes are undoubtedly the weakest commonly encountered in organic chemistry. It is difficult to measure such weak acids, but estimates put the pK of ethane at about 48. Hybridizing the carbon so as to increase the s-character of the C-H increases the acidity, with the greatest change occurring for the sp-C-H groups found in terminal alkynes. Thus, the pK of ethene is estimated at 44, and the pK of ethyne (acetylene) is found to be 25, making it 10 times stronger an acid than ethane. This increase in acidity permits the isolation of insoluble silver and copper salts of such compounds. Despite the dramatic increase in acidity of terminal alkynes relative to other hydrocarbons, they are still very weak acids, especially when compared with water, which is roughly a billion times more acidic. If we wish to prepare nucleophilic salts of terminal alkynes for use in synthesis, it will therefore be necessary to use a much stronger base than hydroxide (or ethoxide) anion. Such a base is sodium amide (NaNH ), discussed above, and its reactions with terminal alkynes may be conducted in liquid ammonia or ether as solvents. The products of this acid-base reaction are ammonia and a sodium acetylide salt. Because the acetylide anion is a powerful nucleophile it may displace halide ions from 1º-alkyl halides to give a more highly substituted alkyne as a product (S 2 reaction). This synthesis application is described in the following equations. The first two equations show how acetylene can be converted to propyne; the last two equations present a synthesis of 2-pentyne from propyne. Because RC≡C: Na is a very strong base (roughly a billion times stronger than NaOH), its use as a nucleophile in S 2 reactions is limited to 1º-alkyl halides; 2º and 3º-alkyl halides undergo elimination by an E2 mechanism. The enhanced acidity of terminal alkynes relative to alkanes also leads to metal exchange reactions when these compounds are treated with organolithium or Grignard reagents. This exchange, shown below in equation 1, can be interpreted as an acid-base reaction which, as expected, proceeds in the direction of the weaker acid and the weaker base. This factor clearly limits the usefulness of Grignard or lithium reagents when a terminal triple bond is present, as in equation 2. The acidity of terminal alkynes also plays a role in product determination when vicinal (or geminal) dihalides undergo base induced bis-elimination reactions. The following example illustrates eliminations of this kind starting from 1,2-dibromopentane, prepared from 1-pentene by addition of bromine. The initial elimination presumably forms 1-bromo-1-pentene, since base attack at the more acidic and less hindered 1 º-carbon should be favored. The second elimination then produces 1-pentyne. If the very strong base sodium amide is used, the terminal alkyne is trapped as its sodium salt, from which it may be released by mild acid treatment. However, if the weaker base KOH is used for the elimination, the terminal alkyne salt is not formed, or is formed reversibly, and the initially generated 1-pentyne rearranges to the more stable 2-pentyne via an allene intermediate. In the case of non-terminal alkynes, sodium and potassium amide, and related strong bases from 1 º-amines, are able to abstract protons from carbon atoms adjacent to the triple bond. The resulting allenic carbanions undergo rapid proton transfer equilibria, leading to the relatively stable terminal alkyne conjugate base. This isomerization may be used to prepare longer chain 1-alkynes, as shown in the following conversion of 3-heptyne to 1-heptyne. The R and R' substituents on the allenic intermediate range from propyl to hydrogen, as the proton transfers proceed. | 3,763 | 3,756 |
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In we showed that a covalent bond results from an overlap of atomic orbitals—usually one orbital from each of two bonded atoms. Maximum bond strength is achieved when maximum overlap occurs. When we try to integrate this idea of orbital overlap with VSEPR theory, however, a problem arises. For example, the 2 and 2 atomic orbitals for a C atom are either spherically symmetrical (2 ) or dumbbell shaped at angles of 90° to each other (2 2 ). The VSEPR theory predicts that the C―Cl bonds in CCl are oriented at angles of 109.5° from one another and that all four C―Cl bonds are equivalent. If each C―Cl bond is formed by overlap of Cl orbitals with C 2 and 2 orbitals, it is hard to understand how four equivalent bonds can be formed. It is also difficult to see why the angles between bonds are 109.5° rather than the 90° angle between orbitals To understand molecular shapes, we first need to consider a simple case, that of beryllium chloride. As seen in , VSEPR theory predicts a linear BeCl molecule with a Cl—Be—Cl angle of 180°, in agreement with experiments. However, if we associate two valence electron pairs with a beryllium atom and place them in the lowest energy orbitals, we obtain the configuration 2 22. (Note that since the electrons must be in order to form bonds, they do not obey Hund’s rule.) The electron density distribution of such a configuration is shown in Figure 1 . The 2 cloud is indicated in red, and the 2 cloud in gray. Figure \(\Page {1}\): Electron-density distribution for the valence electron configuration 2s 2p . Use the buttons at the bottom (you may need to scroll down) to toggle orbitals off and on. (a) In the top image, color coded to shows s (red) and p (gray) electron densities. (b) Color coded to show the two different sp hybrid orbital electron densities. (Computer generated.) An important aspect of Figure \(\Page {1}\) the fact that the density of electron probability is greater along the axis than in any other direction. This can be seen more clearly in Figure \(\Page {1}\) where the dots have been color coded to indicate one electron pair to the left and one to the right of the nucleus. In this case we should certainly predict that chlorine atoms bonded to the beryllium through these electron pairs would lie on a straight line, the axis. Each of the two orbitals whose electron densities are shown in Figure \(\Page {1}\) is called an . The word indicates that each orbital is derived from two or more of the atomic orbitals discussed in , and the designation indicates that a single and a single orbital contributed to each hybrid. Careful comparison of the and electron densities with those of the two hybrid orbitals will reveal another important fact. For every dot in Figure \(\Page {1}\) , there is a corresponding dot (in the same location) in Figure \(\Page {1}\) . That is, the overall electron density (due to the four electrons occupying two orbitals) is exactly the same in both cases. We have not created something new with the two hybrids. Rather, we are looking at the same electron density, but we have color coded it to emphasize its concentration along the axis. In actual fact all electrons are identical—we cannot distinguish one from another experimentally. Labeling one electron cloud as and another as is an aid to our thinking, just as color-coding one -hybrid electron density red and the other gray is, but it is the which determines the experimentally observable molecular geometry. In other words, it does not matter whether we think of the total electron cloud as being formed from and orbitals or from hybrids, but it does matter . It is much easier to think of two Be—Cl bonds separated by 180° in terms of hybrids than in terms of separate and orbitals. By contrast it is much easier to explain the periodic table by using and orbitals rather than hybrids. Since both correspond to the same physical reality, we can use whichever approach suits us best. When beryllium forms a linear molecule such as beryllium chloride, it is not the hybrids themselves that form the two bonds but rather an overlap between each of these orbitals and some orbital on each other atom. The situation is shown schematically in Figure \(\Page {2}\). As a result of each orbital overlap, there is a concentration of electron density between two nuclei. This pulls the nuclei together and forms a covalent bond. Somewhat more complex hybrid orbitals are found in BCl , where the boron is surrounded by electron pairs in a trigonal arrangement. If we place these three electron pairs in the valence shell of boron, one 2 and two 2 orbitals (2 and 2 , for example) will be filled, giving the electron configuration This configuration is illustrated in Figure \(\Page {3}\) as a dot-density diagram in which each electron pair is represented in a different color. Immediately below this diagram is another (Figure \(\Page {3}\) ) which is dot-for-dot the same as the upper diagram but with a different color-coding. The electron density distribution is the same in both diagrams, but in the bottom diagram the three electron pairs are distributed in a trigonal arrangement. Because electrons are indistinguishable, and because both the upper and the lower diagrams in Figure \(\Page {3}\) correspond to an identical total electron density, we are entitled to use either formulation when it suits our purposes. To explain the trigonal geometry of BCl and similar molecules, the hybrid picture is obviously more suitable than the and picture. The three electron-pair bonds usually formed by boron result from an overlap of each of these three hybrids with a suitable orbital in each other atom. \(\Page {3}\) Electron-density distribution for the valence electron configuration 2s 2p 2p . (a) Color coded to show 2s (black), 2p (green), and 2p (blue) electron densities; (b) color coded to show electron densities of three sp hybrids at 120° angles. (Computer-generated.) In addition to the two hybrids just considered, a third combination of s and orbitals, called , is possible. As the name suggests, hybrids are obtained by combining an orbital with orbitals ( , and ). Suppose we have two electrons, two electrons, two electrons, and two electrons, as shown by the boundary-surface diagrams in Figure \(\Page {3}\). When these are all arranged around the nucleus, the total electron cloud is essentially spherical. The boundary-surface diagrams in Figure \(\Page {4}\) show a slightly “bumpy” surface, but we must remember that electron clouds are fuzzy and do not stop suddenly at the boundary surface shown. When this fuzziness is taken into account, the four atomic orbitals blend into each other perfectly to form an exactly spherical shape. This blending of and orbitals is much the same as that discussed in the previous cases of and hybrids. It can be clearly seen in two dimensions in Figure \(\Page {1}\) or Figure \(\Page {3}\) . The total electron probability cloud in Figure \(\Page {4}\) can be subdivided in a different way―into four equivalent hybrid orbitals, each occupied by two electrons. Each of these four hybrid orbitals has a similar appearance to each of the and hybrids encountered previously, but the hybrids are arranged around the nucleus. The four orbitals, each occupied by two electrons, also appear bumpy in a boundary-surface diagram, but when the fuzziness of the electron clouds is taken into account, the result is a spherical electron cloud equivalent in every way to an configuration. We are thus equally entitled to look at an octet of electrons in terms of four hybrids, each doubly occupied, or in terms of one and three orbitals, each doubly occupied. Certainly, nature cannot tell the difference! According to the VSEPR theory, if an atom has an octet of electrons in its valence shell, these electrons are arranged tetrahedrally in pairs around it. Since hybrids also correspond to a tetrahedral geometry, they are the obvious choice for a wave-mechanical description of the octets we find in Lewis structures. Consider, for example, a molecule of methane, CH , whose Lewis structure is Also shown in this figure are four other simple molecules whose Lewis structures are In each case the octets in the Lewis structure can be translated into hybrids. These hybrids can either overlap with the 1 orbitals of H or with hybrids in other atoms. Alternatively they can form lone pairs, in which case no bond is formed and no overlap is necessary. Suggest which hybrid orbitals should be used to describe the bonding of the central atom in the following molecules: In making decisions on which hybrids need to be used, we must first decide on the of the molecule from VSEPR theory. This in turn is best derived from a Lewis structure. The Lewis structures of these four molecules are Hybrid orbitals can also be used to describe those shapes which occur when there is more than an octet of electrons in an atom’s valence shell. The combination of a orbital with an orbital and three orbitals yield a set of five . These hybrids are directed toward the corners of a trigonal bipyramid. An octahedral arrangement is possible if one more orbital is included. Then , all at 90° to one another, are formed. Note that as soon as we get beyond an octet (whose eight electrons fill all the and orbitals), the inclusion of orbitals is mandatory. This can only occur for atoms in the third row of the periodic table and below. Thus bonding in PCl involves hybrid orbitals which include a P 3 orbital. Bonding in SF uses two S 3 orbitals in hybrids. | 9,754 | 3,757 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/12%3A_Chemistry_of_the_Representative_Elements/12.05%3A_Group_IVA |
Near the middle of the periodic table there is greatest variability of properties among elements of the same group. This is certainly true of group IVA, which contains carbon, a nonmetal, silicon and germanium, both semi- metals, and tin and lead, which are definitely metallic. Elemental carbon exists in two allotropic forms, diamond and graphite, whose structures are shown below. Electron Configuration Usual Oxidation State Radius/pm Ionic (M ) - - - 122 Density/ g cm Electro- negativity Melting Point (in °C) 3550 Some properties of the group IVA elements are summarized in the table. As in the case of group IIIA, there is a large decrease in ionization energy and electronegativity from carbon to silicon, but little change farther down the group. This occurs for the same reason in both groups, namely, that elements farther down the group have filled subshells. Note also that ionization energies, especially the third and fourth, are rather large. Formation of true +4 ions is very difficult, and in their +4 oxidation states all group IVA elements form predominantly covalent bonds. The +2 oxidation state, corresponding to use of the , but not the , electrons for bonding, occurs for all elements. It is most important in the case of tin and especially lead, the latter having an inert pair like that of thallium. In the +4 oxidation state lead is a rather strong oxidizing agent, gaining two electrons (6 ) and being reduced to the +2 state. Carbon’s ability to form strong bonds with other carbon atoms and the tremendous variety of organic compounds have already been discussed extensively in the section on organic compounds. You may want to review the subsections dealing with hydrocarbons and the other organic compounds. The most important inorganic carbon compounds are carbon monoxide and carbon dioxide. Both are produced by combustion of any fuel containing carbon: \(\text{CO} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO}_2\) (2) The triple bond in is the strongest chemical bond known, and contains two double bonds, and so both molecules are quite stable. Equations (1) and (2) occur stepwise when a fuel is burned, and the strong bond makes Eq. (2) slow unless the temperature is rather high. If there is insufficient O or if the products of combustion are cooled rapidly, significant quantities of CO can be produced. This is precisely what happens in an automobile engine, and the exhaust contains between 3 and 4% CO unless pollution controls have been installed. CO is about 200 times better than O at bonding to hemoglobin, the protein which transports O through the bloodstream from the lungs to the tissues. Consequently a small concentration of CO in the air you breathe can inhibit transport of O to the brain, causing drowsiness, loss of consciousness, and death. (After a few minutes of breathing undiluted auto exhaust, more than half your hemoglobin will be incapable of transporting O , and you will faint.) CO in automobile exhaust can be used to put animals to sleep. Because CO is colorless and odorless, your senses cannot detect it, and people must constantly be cautioned not to run cars in garages or other enclosed spaces. With the large number of cars and the great number of miles driven, it is important to limit CO emissions from automobiles. In the early 1970s new EPA standards led to the adoption of catalytic converters, which convert the poisonous CO into CO . Implementation and increasing effectiveness of these converters has caused CO levels to drop since the 1970s, despite the increase in automobiles on the road . Like the organic compounds of carbon, the oxygen compounds of silicon which make up most of the earth’s crust have already been described. These substances illustrate a major contrast between the chemistry of carbon and silicon. The latter element does form a few compounds, called silanes, which are analogous to the alkanes, but the Si—Si bonds in silanes are much weaker than Si—O bonds. Consequently the silanes combine readily with oxygen from air, forming Si—O—Si linkages. Unlike the alkanes, which must be ignited with a spark or a match before they will burn, silanes catch fire of their own accord in air: \(\text{2Si}_4\text{H}_10 + \text{13O}_2 \rightarrow \text{4SiO}_2 + \text{5H}_2\text{O}\) Another important group of silicon compounds is the . These polymeric substances contain Si—O—Si linkages and may be thought of as derived from silicon dioxide, SiO . To make silicones, one must first reduce silicon dioxide to silicon. This can be done using carbon as the reducing agent in a high-temperature furnace: \(\text{SiO}_2(s) + \text{2C}(s) \xrightarrow{\text{3000}{}^\circ \text{C}} \text{Si}(l) + \text{2CO}(g)\) The silicon is then reacted with chloromethane: \(\text{Si}(s) + \text{2CH}_3\text{Cl}(g) \xrightarrow[\text{Cu catalyst}]{\text{300}{}^\circ \text{C}} (\text{CH}_3)_2\text{SiCl}_2(g)\) The dichlorodimethylsilane obtained in this reaction polymerizes when treated with water: Besides the metals themselves, some tin and lead compounds are of commercial importance. Tin(II) fluoride (stannous fluoride), SnF , is added to some toothpastes to inhibit dental caries. Tooth decay involves dissolving of dental enamel [mainly Ca (PO ) (OH) ] in acids synthesized by bacteria in the mouth. Fluoride ions from SnF inhibit decay by transforming tooth surfaces into Ca (PO ) F , which is less soluble in acid: \(\text{Ca}_{10}(\text{PO}_4)_6\text{(OH)}_2 + \text{SnF}_2 \rightarrow \text{Ca}_{10}(\text{PO}_4)_6\text{F}_2 + \text{Sn(OH)}_2\) Since F is a weaker base than OH , the F compound has less tendency to react with acids. Note that when tin or lead are in the +2 oxidation state and are combined with a highly electronegative element like fluorine, the compounds formed are rather ionic. Lead is found in two main commercial applications. One, the lead-acid storage battery is used to start cars and power golf carts. The other is the lead found in automobile fuel. In the +4 oxidation state lead forms primarily covalent compounds and bonds strongly to carbon. The compound tetraethyllead may be synthesized by reacting with a sodium-lead alloy: \(\text{4NaPb} + \text{4CH}_3\text{CH}_2\text{Cl} \rightarrow (\text{CH}_3\text{CH}_2)\text{Pb} + \text{4Pb} + \text{4NaCl}\) Sodium dissolved in the lead makes the latter more reactive. Tetraethyl-lead prevents gasoline from igniting too soon or burning unevenly in an automobile engine, circumstances which cause the engine to “knock” or “ping.” This is where the term leaded gasoline comes from. A major problem connected to using tetraethyl-lead is the introduction of lead into the atmosphere. Lead is toxic, and thus use of TEL as an antiknock agent has been phased out in favor of other agents less dangerous to public health. | 6,829 | 3,758 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.06%3A_Acid-Base_Reactions |
Early in the history of chemistry it was noted that aqueous solutions of a number of substances behaved very similarly, although the substances themselves did not at first seem to be related. Solutions were classified as if they had the following characteristics: sour taste; ability to dissolve metals such as Zn, Mg, or Fe; ability to release a gas from solid limestone (CaCO ) or other carbonates; ability to change the color of certain dyes (litmus paper turns red in the presence of acid). Another group of substances called can also be distinguished by the properties of their aqueous solutions. These are bitter taste, slippery or soapy feel, and the ability to change the color of certain dyes (litmus paper turns blue in base). Most important of all, acids and bases appear to be opposites. Any acid can counteract or the properties of a base. Similarly any base can neutralize an acid. The video below demonstrates one of the ways acids and bases are determined (indicators - often colorful as seen below) and shows an example of a neutralization reaction. \[\text{HCl}(aq) + \text{NaOH}(aq) \rightleftharpoons \text{Na}^{+}(aq) + \text{Cl }^{-}(aq) + \text{H}_{2}\text{O}(l) \nonumber \] | 1,216 | 3,759 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.16%3A_Higher-Order_Structure |
Several of the amino acid side chains are difficult to fit into either the α helix or β-sheet types of structure. An obvious example is proline, in which the R group is a ring and includes nitrogen bonded to the α carbon. When proline is involved in a peptide bond, all hydrogens are gone from the nitrogen, leaving no site for hydrogen bonding. Another problem involves side chains having the same charge, such as those of lysine and arginine or glutamic acid and aspartic acid, which repel one another considerably. When they occur close together, these groups may also destabilize an α helix or β sheet structure. The presence of several groups of the type just mentioned allows a protein chain to bend sharply instead of laying flat or curling regularly into a spiral. Such sharp bends connect sections of α helix or β sheet structures in the globular proteins. They allow the polypeptide chain to curl back upon itself, folding the protein into a very compact, nearly spherical shape. The nature of this folding is referred to as , since it involves a third organizational level above the primary amino acid sequence and the secondary α helix or β sheet. An excellent example of tertiary structure in a globular protein is provided by the three-dimensional view of sperm-whale myoglobin in Figure \(\Page {1}\). The molecular backbone consists of eight relatively straight segments of a helix. The longest of these contain 23 amino acids, the shortest just 7. Fewer than 50 of myoglobin's 153 amino acids are found in the bends between the helixes, but all the proline residues fall into this category, as do many of the others which destabilize the α helix or β sheet. (Although only the α helix is prominent in the structure of myoglobin, other globular proteins are found to have regions consisting of β sheets.) Several other important generalizations may be made about the structure of myoglobin. Even in regions where the chain twists into an α helix, nearly all the nonpolar R groups point toward the interior of the molecule. Here they crowd so closely together that only four water molecules can squeeze their way in. The outside surface of the protein, however, contains all the polar R groups. These interact strongly with the many water molecules which normally surround myoglobin in muscle tissue. Although myoglobins isolated from a variety of mammals differ slightly in their primary structure, they all seem to have nearly the same overall molecular shape. Apparently some of the amino acids are much more important than others in determining the bend points and other crucial features of tertiary structure. Substitutions at less-important positions do not cause great variations in the ability of the protein to carry out its biological function. Finally, Figure \(\Page {1}\) shows clearly that myoglobin contains a prosthetic group. In this case it is the heme group, which contains an iron atom surrounded by four nitrogens in a flat ring structure known as a porphyrin. The porphyrin ring is not covalently bonded to the protein chain, but rather fits snugly into a “pocket” surrounded by several segments of helix. A nitrogen atom in a histidine side chain on one helix does form a coordinate covalent bond with iron in the heme group, but apart from this the prosthetic group is positioned solely by the way the protein chain is folded. The iron in the heme group marks the active site at which the oxygen-storage function of myoglobin is accomplished. Iron, like other first-row transition-metal ions, ordinarily forms six bonds directed toward the corners of an octahedron. In myoglobin only five bonds to iron are found. The sixth position (opposite the histidine nitrogen) can be occupied by an oxygen molecule, providing a convenient storage site. If the concentration of oxygen near the protein falls, this binding is reversed, releasing oxygen to replenish the supply. If an iron ion can bond to an oxygen molecule, you may wonder why the complicated porphyrin and protein structures are necessary. It is to of the iron in the heme group from the iron(II) to the iron(III) oxidation state. Because the oxygen molecule needs to gain electrons to be oxidized while iron(II) can only supply one, no electron transfer occurs. If the iron(II) atom in heme were not surrounded by the protein chain, a water molecule would be able to take part as an intermediate in the electron-transfer mechanism and reduction of the oxygen could occur. As matters stand, though, the oxygen molecule cannot be reduced until it is released by the myoglobin. Thus the combined effects of the tertiary structure of the protein, the prosthetic group, and a specialized active site allow the myoglobin to fulfill its biological function of storing oxygen molecules until needed, without allowing them to be reduced during storage. In some proteins there is yet a fourth level of organization, labeled . This may be illustrated by hemoglobin, whose function as an oxygen carrier in the bloodstream is well known. Hemoglobin consists essentially of four myoglobin molecules packed together in a single unit. Four separate polypeptide chains are each folded as in myoglobin and then nested together. The way these four subunits fit together constitutes the quaternary structure. Like the other type of structures, quaternary structure contributes to the function of a protein. In the case of hemoglobin, an oxygen molecule attached to one subunit causes slight shifts in tertiary and quaternary structure which make it easier for other oxygen molecules to bond to the other subunits. Consequently hemoglobin in the lungs can be loaded with its full complement of four oxygen molecules rather easily, a factor which increases its efficiency in carrying oxygen to body tissues. The converse is also true—loss of one oxygen molecule causes slight structural rearrangements which allow the remaining three to depart more readily. | 5,954 | 3,761 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Solids/Metal_Structures |
Copper is the most common mentioned metal that has the fcc structure. This element is one of the noble metals. It has been widely used for door knobs and other tools, and has been widely recognized. Yet, most noble metals (\(\ce{Cu}\), \(\ce{Ag}\), \(\ce{Au}\), \(\ce{Ni}\), \(\ce{Pd}\), \(\ce{Pt}\), \(\ce{Rh}\), and \(\ce{Ir}\)) have fcc type structures. Among the group 2 elements, only \(\ce{Ca}\) and \(\ce{Sr}\) have the fcc structure, whereas \(\ce{Be}\) and \(\ce{Mg}\) have hcp structures. So do \(\ce{Zn}\), \(\ce{Cd}\), \(\ce{Sc}\), \(\ce{Y}\), \(\ce{Lu}\), \(\ce{Ti}\), \(\ce{Zr}\), \(\ce{Hf}\), \(\ce{Tc}\), \(\ce{Re}\), \(\ce{Ru}\), \(\ce{Os}\) and most rare earth elements. These are mentioned to bring your attention to these two common types of structures, and you are encouraged to at least be able to give a few examples for each type. Amazingly, the hcp and fcc structure are very similar in many aspects, but nature knows best. The structures adopted by various metals occur by their design. When crystallization takes place, the atoms arrange themselves according to their structure types. | 1,132 | 3,762 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/16%3A_Acids_and_Bases/16.5%3A_Weak_Acids_and_Weak_Bases |
The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For example, the general equation for the ionization of a weak acid in water, where HA is the parent acid and A− is its conjugate base, is as follows: \[HA_{(aq)}+H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+A^−_{(aq)} \label{16.5.1}\] The equilibrium constant for this dissociation is as follows: \[K=\dfrac{[H_3O^+,A^−]}{[H_2O,HA]} \label{16.5.2}\] As we noted earlier, the concentration of water is essentially constant for all reactions in aqueous solution, so \([H_2O]\) in Equation \ref{16.5.2} can be incorporated into a new quantity, the acid ionization constant (\(K_a\)), also called the acid dissociation constant: \[K_a=K[H_2O]=\dfrac{[H_3O^+,A^−]}{[HA]} \label{16.5.3}\] Thus the numerical values of K and \(K_a\) differ by the concentration of water (55.3 M). Again, for simplicity, \(H_3O^+\) can be written as \(H^+\) in Equation \(\ref{16.5.3}\). Keep in mind, though, that free \(H^+\) does not exist in aqueous solutions and that a proton is transferred to \(H_2O\) in all acid ionization reactions to form \(H^3O^+\). The larger the \(K_a\), the stronger the acid and the higher the \(H^+\) concentration at equilibrium.Like all equilibrium constants, acid–base ionization constants are actually measured in terms of the activities of \(H^+\) or \(OH^−\), thus making them . The values of \(K_a\) for a number of common acids are given in Table \(\Page {1}\). Weak bases react with water to produce the hydroxide ion, as shown in the following general equation, where B is the parent base and BH+ is its conjugate acid: \[\ce{B(aq) + H2O(l) <=> BH^{+}(aq) + OH^{−} (aq)} \label{16.5.4}\] The equilibrium constant for this reaction is the (\(K_b\)), also called the base dissociation constant: \[K_b=K[H_2O]=\dfrac{[BH^+,OH^−]}{[B]} \label{16.5.5}\] Once again, the concentration of water is constant, so it does not appear in the equilibrium constant expression; instead, it is included in the \(K_b\). The larger the \(K_b\), the stronger the base and the higher the \(OH^−\) concentration at equilibrium. The values of \(K_b\) for a number of common weak bases are given in Table \(\Page {2}\). There is a simple relationship between the magnitude of \(K_a\) for an acid and \(K_b\) for its conjugate base. Consider, for example, the ionization of hydrocyanic acid (\(HCN\)) in water to produce an acidic solution, and the reaction of \(CN^−\) with water to produce a basic solution: \[HCN_{(aq)} \rightleftharpoons H^+_{(aq)}+CN^−_{(aq)} \label{16.5.6}\] \[CN^−_{(aq)}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+HCN_{(aq)} \label{16.5.7}\] The equilibrium constant expression for the ionization of HCN is as follows: \[K_a=\dfrac{[H^+,CN^−]}{[HCN]} \label{16.5.8}\] The corresponding expression for the reaction of cyanide with water is as follows: \[K_b=\dfrac{[OH^−,HCN]}{[CN^−]} \label{16.5.9}\] If we add Equations \(\ref{16.5.6}\) and \(\ref{16.5.7}\), we obtain the following (recall that the equilibrium constant for the sum of two reactions is the product of the equilibrium constants for the individual reactions): \[\cancel{HCN_{(aq)}} \rightleftharpoons H^+_{(aq)}+\cancel{CN^−_{(aq)}} \;\;\; K_a=[H^+]\cancel{[CN^−]}/\cancel{[HCN]}\] \[\cancel{CN^−_{(aq)}}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+\cancel{HCN_{(aq)}} \;\;\; K_b=[OH^−]\cancel{[HCN]}/\cancel{[CN^−]}\] \[H_2O_{(l)} \rightleftharpoons H^+_{(aq)}+OH^−_{(aq)} \;\;\; K=K_a \times K_b=[H^+,OH^−]\] In this case, the sum of the reactions described by \(K_a\) and \(K_b\) is the equation for the autoionization of water, and the product of the two equilibrium constants is \(K_w\): \[K_aK_b = K_w \label{16.5.10}\] Thus if we know either \(K_a\) for an acid or \(K_b\) for its conjugate base, we can calculate the other equilibrium constant for any conjugate acid–base pair. Just as with \(pH\), \(pOH\), and \(pK_w\), we can use negative logarithms to avoid exponential notation in writing acid and base ionization constants, by defining \(pK_a\) as follows: \[pKa = −\log_{10}K_a \label{16.5.11}\] \[K_a=10^{−pK_a} \label{16.5.12}\] and \(pK_b\) as \[pK_b = −\log_{10}K_b \label{16.5.13}\] \[K_b=10^{−pK_b} \label{16.5.14}\] Similarly, Equation \ref{16.5.10}, which expresses the relationship between \(K_a\) and \(K_b\), can be written in logarithmic form as follows: \[pK_a + pK_b = pK_w \label{16.5.15}\] At 25°C, this becomes \[pK_a + pK_b = 14.00 \label{16.5.16}\] The values of \(pK_a\) and \(pK_b\) are given for several common acids and bases in Table \(\Page {1}\) and Table \(\Page {2}\), respectively, and a more extensive set of data is provided in Tables E1 and E2. Because of the use of negative logarithms, smaller values of \(pK_a\) correspond to larger acid ionization constants and hence stronger acids. For example, nitrous acid (\(HNO_2\)), with a \(pK_a\) of 3.25, is about a 1000 times stronger acid than hydrocyanic acid (HCN), with a \(pK_a\) of 9.21. Conversely, smaller values of \(pK_b\) correspond to larger base ionization constants and hence stronger bases. The relative strengths of some common acids and their conjugate bases are shown graphically in Figure 16.5. The conjugate acid–base pairs are listed in order (from top to bottom) of increasing acid strength, which corresponds to decreasing values of \(pK_a\). This order corresponds to decreasing strength of the conjugate base or increasing values of \(pK_b\). At the bottom left of Figure \(\Page {2}\) are the common strong acids; at the top right are the most common strong bases. Notice the inverse relationship between the strength of the parent acid and the strength of the conjugate base. Thus the conjugate base of a strong acid is a very weak base, and the conjugate base of a very weak acid is a strong base. The conjugate base of a strong acid is a weak base and vice versa. We can use the relative strengths of acids and bases to predict the direction of an acid–base reaction by following a single rule: an acid–base equilibrium always favors the side with the weaker acid and base, as indicated by these arrows: \[\text{stronger acid + stronger base} \ce{ <=>>} \text{weaker acid + weaker base} \nonumber\] In an acid–base reaction, the proton always reacts with the stronger base. For example, hydrochloric acid is a strong acid that ionizes essentially completely in dilute aqueous solution to produce \(H_3O^+\) and \(Cl^−\); only negligible amounts of \(HCl\) molecules remain undissociated. Hence the ionization equilibrium lies virtually all the way to the right, as represented by a single arrow: \[HCl_{(aq)} + H_2O_{(l)} \rightarrow \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)} \label{16.5.17}\] In contrast, acetic acid is a weak acid, and water is a weak base. Consequently, aqueous solutions of acetic acid contain mostly acetic acid molecules in equilibrium with a small concentration of \(H_3O^+\) and acetate ions, and the ionization equilibrium lies far to the left, as represented by these arrows: \[ \ce{ CH_3CO_2H_{(aq)} + H_2O_{(l)} <<=> H_3O^+_{(aq)} + CH_3CO_{2(aq)}^- } \nonumber\] Similarly, in the reaction of ammonia with water, the hydroxide ion is a strong base, and ammonia is a weak base, whereas the ammonium ion is a stronger acid than water. Hence this equilibrium also lies to the left: \[H_2O_{(l)} + NH_{3(aq)} \ce{ <<=>} NH^+_{4(aq)} + OH^-_{(aq)} \nonumber\] All acid–base equilibria favor the side with the weaker acid and base. Thus the proton is bound to the stronger base. : \(pK_a\) and \(K_b\) : corresponding \(K_b\) and \(pK_b\), \(K_a\) and \(pK_a\) : The constants \(K_a\) and \(K_b\) are related as shown in Equation \ref{16.5.10}. The \(pK_a\) and \(pK_b\) for an acid and its conjugate base are related as shown in Equation \ref{16.5.15} and Equation \ref{16.5.16}. Use the relationships \(pK = −\log K\) and \(K = 10{−pK}\) (Equations \ref{16.5.11} and \ref{16.5.13}) to convert between \(K_a\) and \(pK_a\) or \(K_b\) and \(pK_b\). : We are given the \(pK_a\) for butyric acid and asked to calculate the \(K_b\) and the \(pK_b\) for its conjugate base, the butyrate ion. Because the \(pK_a\) value cited is for a temperature of 25°C, we can use Equation \ref{16.5.16}: \(pK_a\) + \(pK_b\) = pKw = 14.00. Substituting the \(pK_a\) and solving for the \(pK_b\), \[\begin{align*} 4.83 + pK_b &=14.00 \\[4pt] pK_b &=14.00−4.83 \\[4pt] &=9.17 \end{align*}\] Because \(pK_b = −\log K_b\), \(K_b\) is \(10^{−9.17} = 6.8 \times 10^{−10}\). In this case, we are given \(K_b\) for a base (dimethylamine) and asked to calculate \(K_a\) and \(pK_a\) for its conjugate acid, the dimethylammonium ion. Because the initial quantity given is \(K_b\) rather than \(pK_b\), we can use Equation \ref{16.5.10}: \(K_aK_b = K_w\). Substituting the values of \(K_b\) and \(K_w\) at 25°C and solving for \(K_a\), \[ \begin{align*} K_a(5.4 \times 10^{−4}) &=1.01 \times 10^{−14} \\[4pt] K_a &=1.9 \times 10^{−11} \end{align*}\] Because \(pK_a\) = −log \(K_a\), we have \(pK_a = −\log(1.9 \times 10^{−11}) = 10.72\). We could also have converted \(K_b\) to \(pK_b\) to obtain the same answer: \[ \begin{align*} pK_b &=−\log(5.4 \times 10^{−4}) \\[4pt] &=3.27 \\[10pt] pKa + pK_b &=14.00 \\[4pt] pK_a &=10.73 \\ K_a &=10^{−pK_a} \\[4pt] &=10^{−10.73} \\[4pt] &=1.9 \times 10^{−11} \end{align*}\] If we are given any one of these four quantities for an acid or a base (\(K_a\), \(pK_a\), \(K_b\), or \(pK_b\)), we can calculate the other three. Lactic acid (\(CH_3CH(OH)CO_2H\)) is responsible for the pungent taste and smell of sour milk; it is also thought to produce soreness in fatigued muscles. Its \(pK_a\) is 3.86 at 25°C. Calculate \(K_a\) for lactic acid and \(pK_b\) and \(K_b\) for the lactate ion. \(K_a = 1.4 \times 10^{−4}\) for lactic acid; \(pK_b\) = 10.14 and \(K_b = 7.2 \times 10^{−11}\) for the lactate ion You will notice in Table \(\Page {1}\) that acids like \(H_2SO_4\) and \(HNO_3\) lie above the hydronium ion, meaning that they have \(pK_a\) values less than zero and are stronger acids than the \(H_3O^+\) ion. Recall that the acidic proton in virtually all oxoacids is bonded to one of the oxygen atoms of the oxoanion. Thus nitric acid should properly be written as \(HONO_2\). Unfortunately, however, the formulas of oxoacids are almost always written with hydrogen on the left and oxygen on the right, giving \(HNO_3\) instead. In fact, all six of the common strong acids that we first encountered in Chapter 4 have \(pK_a\) values less than zero, which means that they have a greater tendency to lose a proton than does the \(H_3O^+\) ion. Conversely, the conjugate bases of these strong acids are weaker bases than water. Consequently, the proton-transfer equilibria for these strong acids lie far to the right, and adding any of the common strong acids to water results in an essentially stoichiometric reaction of the acid with water to form a solution of the \(H_3O^+\) ion and the conjugate base of the acid. Although \(K_a\) for \(HI\) is about 108 greater than \(K_a\) for \(HNO_3\), the reaction of either \(HI\) or \(HNO_3\) with water gives an essentially stoichiometric solution of \(H_3O^+\) and I− or \(NO_3^−\). In fact, a 0.1 M aqueous solution of any strong acid actually contains 0.1 M \(H_3O^+\), regardless of the identity of the strong acid. This phenomenon is called the leveling effect: any species that is a stronger acid than the conjugate acid of water (\(H_3O^+\)) is leveled to the strength of \(H_3O^+\) in aqueous solution because \(H_3O^+\) is the strongest acid that can exist in equilibrium with water. Consequently, it is impossible to distinguish between the strengths of acids such as HI and HNO3 in aqueous solution, and an alternative approach must be used to determine their relative acid strengths. One method is to use a solvent such as anhydrous acetic acid. Because acetic acid is a stronger acid than water, it must also be a weaker base, with a lesser tendency to accept a proton than \(H_2O\). Measurements of the conductivity of 0.1 M solutions of both HI and \(HNO_3\) in acetic acid show that HI is completely dissociated, but \(HNO_3\) is only partially dissociated and behaves like a weak acid in this solvent. This result clearly tells us that HI is a stronger acid than \(HNO_3\). The relative order of acid strengths and approximate \(K_a\) and \(pK_a\) values for the strong acids at the top of Table \(\Page {1}\) were determined using measurements like this and different nonaqueous solvents. In , \(H_3O^+\) is the strongest acid and \(OH^−\) is the strongest base that can exist in equilibrium with \(H_2O\). The leveling effect applies to solutions of strong bases as well: In aqueous solution, any base stronger than \(\ce{OH^{−}}\) is leveled to the strength of \(\ce{OH^{−}}\) because \(\ce{OH^{−}}\) is the strongest base that can exist in equilibrium with water. Salts such as \(\ce{K_2O}\), \(\ce{NaOCH3}\) (sodium methoxide), and \(\ce{NaNH2}\) (sodamide, or sodium amide), whose anions are the conjugate bases of species that would lie below water in Table \(\Page {2}\), are all strong bases that react essentially completely (and often violently) with water, accepting a proton to give a solution of \(\ce{OH^{−}}\) and the corresponding cation: \[\ce{K2O(s) + H2O(l) -> 2OH^{−}(aq) + 2K^{+} (aq)} \nonumber\] \[\ce{NaOCH3(s) + H2O(l) -> OH^{−}(aq) + Na^{+} (aq) + CH3OH(aq)} \nonumber\] \[\ce{NaNH2(s) + H2O(l) -> OH^{−}(aq) + Na^{+} (aq) + NH3(aq)} \nonumber\] Other examples that you may encounter are potassium hydride (\(KH\)) and organometallic compounds such as methyl lithium (\(\ce{CH3Li}\)). Calculating the pH of Weak Acids and Weak Bases: Two species that differ by only a proton constitute a conjugate acid–base pair. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For an aqueous solution of a weak acid, the dissociation constant is called the acid ionization constant (Ka). Similarly, the equilibrium constant for the reaction of a weak base with water is the base ionization constant (Kb). For any conjugate acid–base pair, \(K_aK_b = K_w\). Smaller values of \(pK_a\) correspond to larger acid ionization constants and hence stronger acids. Conversely, smaller values of \(pK_b\) correspond to larger base ionization constants and hence stronger bases. At 25°C, \(pK_a + pK_b = 14.00\). Acid–base reactions always proceed in the direction that produces the weaker acid–base pair. No acid stronger than \(H_3O^+\) and no base stronger than \(OH^−\) can exist in aqueous solution, leading to the phenomenon known as the leveling effect. | 14,716 | 3,763 |
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The simplest kind of is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form 2A → products. A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer). The differential rate law for the simplest second-order reaction in which 2A → products is as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2 \label{14.4.8}\] Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M ·s ). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s). For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time: \[\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \label{14.4.9}\] Because has the form of an algebraic equation for a straight line, = + , with = 1/[A] and = 1/[A] , a plot of 1/[A] versus for a simple second-order reaction is a straight line with a slope of and an intercept of 1/[A] . Second-order reactions generally have the form 2A → products or A + B → products. Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO to NO and O and the decomposition of HI to I and H . Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture. Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows: For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively. The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law ( ) or the integrated rate law ( ). To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in . From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7: \[\dfrac{5.0\times10^{-5}\textrm{ M/min}}{1.8\times10^{-5}\textrm{ M/min}}=2.8\hspace{5mm}\textrm{and}\hspace{5mm}\dfrac{3.4\times10^{-3}\textrm{ M}}{2.0\times10^{-3} \textrm{ M}}=1.7\] Because (1.7) = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration. rate ∝ [monomer] This means that the reaction is second order in the monomer. Using and the data from any row in , we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following: \[\begin{align}\textrm{rate}&=k[\textrm A]^2 \\8.0\times10^{-5}\textrm{ M/min}&=k(4.4\times10^{-3}\textrm{ M})^2 \\4.1 \textrm{ M}^{-1}\cdot \textrm{min}^{-1}&=k\end{align}\] We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in part (a) in . The measurements show that the concentration of the monomer (initially 5.4 × 10 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus should be a straight line, as shown in part (b) in . Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, = 4.1 M ·min , which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally. For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders. At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen. \[\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)}\] Experimental data for the reaction at 300°C and four initial concentrations of NO are listed in the following table: Determine the reaction order and the rate constant. balanced chemical equation, initial concentrations, and initial rates reaction order and rate constant We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10 ) ÷ (1.35 × 10 ) = 4.0], which means that the reaction rate is proportional to [NO ] . Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO ] . This behavior is characteristic of a second-order reaction. We have rate = [NO ] . We can calculate the rate constant ( ) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following: When the highly reactive species HO forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows: \[2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber\] The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table: Determine the reaction order and the rate constant. second order in HO ; = 1.4 × 10 M ·s If a plot of reactant concentration versus time is linear, but a plot of 1/reaction concentration versus time is linear, then the reaction is second order. If a flask that initially contains 0.056 M NO is heated at 300°C, what will be the concentration of NO after 1.0 h? How long will it take for the concentration of NO to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction ( ) and the rate constant calculated above. balanced chemical equation, rate constant, time interval, and initial concentration final concentration and time required to reach specified concentration We know and [NO ] , and we are asked to determine [NO ] at = 1 h (3600 s). Substituting the appropriate values into , Thus [NO ] = 5.1 × 10 M. In this case, we know and [NO ] , and we are asked to calculate at what time [NO ] = 0.1[NO ] = 0.1(0.056 M) = 0.0056 M. To do this, we solve for , using the concentrations given. \[t=\dfrac{(1/[\mathrm{NO_2}])-(1/[\mathrm{NO_2}]_0)}{k}=\dfrac{(1/0.0056 \textrm{ M})-(1/0.056\textrm{ M})}{0.54 \;\mathrm{M^{-1}\cdot s^{-1}}}=3.0\times10^2\textrm{ s}=5.0\textrm{ min} \nonumber\] NO decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min. In the previous exercise, you calculated the rate constant for the decomposition of HO as = 1.4 × 10 M ·s . This high rate constant means that HO decomposes rapidly under the reaction conditions given in the problem. In fact, the HO molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO , calculate the concentration of HO that remains after 1.0 h at 25°C. How long will it take for 90% of the HO to decompose? Use the integrated rate law for a second-order reaction ( ) and the rate constant calculated in the exercise in Example \(\Page {3}\). 2.0 × 10 M; 6.4 × 10 s In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form \(A + B \rightarrow products\), in which the reaction is first order in \(A\) and first order in \(B\). The differential rate law for this reaction is as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=-\dfrac{\Delta[\textrm B]}{\Delta t}=k[\textrm A,\textrm B] \label{14.4.10}\] Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant. Second-Order Integrated Rate Law Equation: | 9,758 | 3,764 |
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Nucleic acids were first isolated from cell nuclei (hence the name) in 1870. Since then they have been found in other portions of cells as well, especially the ribosomes, which are the sites of protein synthesis. Most nucleic acids are extremely long-chain polymers—some forms of DNA have molecular weights greater than 10 . Nucleic acids are made up from three distinct structural units. These are: . There are five of these bases. All are shown in Figure \(\Page {2}\). Three of them, adenine, guanine, and cytosine, are common to both DNA and RNA. Thymine occurs only in DNA, and uracil only in RNA. . H PO provides the unit that holds the various segments of the nucleic acid chain to each other. The combination of a sugar and a nitrogenous base is called a . A typical example of a nucleoside is adenosine, derived by the condensation of ribose and adenine: | 882 | 3,765 |
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Our contemporary society requires the constant expenditure of huge amounts of energy to heat our homes, provide telephone and cable service, transport us from one location to another, provide light when it is dark outside, and run the machinery that manufactures material goods. The United States alone consumes almost 10 kJ per person per day, which is about 100 times the normal required energy content of the human diet. This figure is about 30% of the world’s total energy usage, although only about 5% of the total population of the world lives in the United States. In contrast, the average energy consumption elsewhere in the world is about 10 kJ per person per day, although actual values vary widely depending on a country’s level of industrialization. In this section, we describe various sources of energy and their impact on the environment. According to the law of conservation of energy, energy can never actually be “consumed”; it can only be changed from one form to another. What consumed on a huge scale, however, are resources that can be readily converted to a form of energy that is useful for doing work. Energy that is not used to perform work is either stored as potential energy for future use or transferred to the surroundings as heat. A major reason for the huge consumption of energy by our society is the low of most machines in transforming stored energy into work. Efficiency can be defined as the ratio of useful work accomplished to energy expended. Automobiles, for example, are only about 20% efficient in converting the energy stored in gasoline to mechanical work; the rest of the energy is released as heat, either emitted in the exhaust or produced by friction in bearings and tires. The production of electricity by coal- or oil-powered steam turbines (Figure \(\Page {1}\)) is can be more than 50% efficient. In general, it is more efficient to use primary sources of energy directly (such as natural gas or oil) than to transform them to a secondary source such as electricity prior to their use. For example, if a furnace is well maintained, heating a house with natural gas is about 70% efficient. In contrast, burning the natural gas in a remote power plant, converting it to electricity, transmitting it long distances through wires, and heating the house by electric baseboard heaters have an overall efficiency of less than 35%. The total expenditure of energy in the world each year is about 3 × 10 kJ. 80% of this energy is provided by the combustion of fossil fuels: oil, coal, and natural gas (the sources of the energy consumed in the United States in 2019 are shown in Figure \(\Page {2}\)). Natural gas and petroleum are the preferred fuels because many of the products derived from them are gases or liquids that are readily transported, stored, and burned. Natural gas and petroleum are derived from the remains of marine creatures that died hundreds of millions of years ago and were buried beneath layers of sediment. As the sediment turned to rock, the tremendous heat and pressure inside Earth transformed the organic components of the buried sea creatures to petroleum and natural gas. Coal is a complex solid material derived primarily from plants that died and were buried hundreds of millions of years ago and were subsequently subjected to high temperatures and pressures. Because plants contain large amounts of , derived from linked glucose units, the structure of coal is more complex than that of petroleum (Figure \(\Page {3}\)). In particular, coal contains a large number of oxygen atoms that link parts of the structure together, in addition to the basic framework of carbon–carbon bonds. It is impossible to draw a single structure for coal; however, because of the prevalence of rings of carbon atoms (due to the original high cellulose content), coal is more similar to an aromatic hydrocarbon than an aliphatic one. There are four distinct classes of coal (Table \(\Page {1}\)); their hydrogen and oxygen contents depend on the length of time the coal has been buried and the pressures and temperatures to which it has been subjected. Lignite, with a hydrogen:carbon ratio of about 1.0 and a high oxygen content, has the lowest Δ . Anthracite, in contrast, with a hydrogen:carbon ratio of about 0.5 and the lowest oxygen content, has the highest Δ and is the highest grade of coal. The most abundant form in the United States is bituminous coal, which has a high sulfur content because of the presence of small particles of pyrite (FeS ). The combustion of coal releases the sulfur in FeS as SO , which is a major contributor to acid rain. Table \(\Page {2}\) compares the Δ per gram of oil, natural gas, and coal with those of selected organic compounds. Peat, a precursor to coal, is the partially decayed remains of plants that grow in swampy areas. It is removed from the ground in the form of soggy bricks of mud that will not burn until they have been dried. Even though peat is a smoky, poor-burning fuel that gives off relatively little heat, humans have burned it since ancient times (Figure \(\Page {4}\)). If a peat bog were buried under many layers of sediment for a few million years, the peat could eventually be compressed and heated enough to become lignite, the lowest grade of coal; given enough time and heat, lignite would eventually become anthracite, a much better fuel. Oil and natural gas resources are limited. Current estimates suggest that the known reserves of petroleum will be exhausted in about 60 years, and supplies of natural gas are estimated to run out in about 120 years. Coal, on the other hand, is relatively abundant, making up more than 90% of the world’s fossil fuel reserves. As a solid, coal is much more difficult to mine and ship than petroleum (a liquid) or natural gas. Consequently, more than 75% of the coal produced each year is simply burned in power plants to produce electricity. A great deal of current research focuses on developing methods to convert coal to gaseous fuels ( gasification) or liquid fuels (coal liquefaction). In the most common approach to coal gasification, coal reacts with steam to produce a mixture of CO and H2 known as synthesis gas, or syngas:Because coal is 70%–90% carbon by mass, it is approximated as C in Equation \(\ref{7.9.1}\). \[\mathrm{C_{(s)} +H_2O_{(g)} → CO_{(g)}+H_{2(g)}} \;\;\; ΔH= \mathrm{131\: kJ} \label{7.9.1}\] Converting coal to syngas removes any sulfur present and produces a clean-burning mixture of gases. Syngas is also used as a reactant to produce methane and methanol. A promising approach is to convert coal directly to methane through a series of reactions: \(\mathrm{2C(s)+2H_2O(g)→\cancel{2CO(g)}+\cancel{2H_2(g)}}\hspace{20px}ΔH_1= \mathrm{262\: kJ}\\
\mathrm{\cancel{CO(g)}+\cancel{H_2O(g)}→CO_2(g)+\cancel{H_2(g)}}\hspace{20px}ΔH_2=\mathrm{−41\: kJ}\\
\mathrm{\cancel{CO(g)}+\cancel{3H_2(g)}→CH_4(g)+\cancel{H_2O(g)}}\hspace{20px}ΔH_3=\mathrm{−206\: kJ}\\
\overline{\mathrm{Overall:\hspace{10px}2C(s)+2H_2O(g)→CH_4(g)+CO_2(g)}\hspace{20px}ΔH_\ce{comb}= \mathrm{15\: kJ}}\hspace{40px}\label{7.9.2}\) Burning a small amount of coal or methane provides the energy consumed by these reactions. Unfortunately, methane produced by this process is currently significantly more expensive than natural gas. As supplies of natural gas become depleted, however, this coal-based process may well become competitive in cost. Similarly, the techniques available for converting coal to liquid fuels are not yet economically competitive with the production of liquid fuels from petroleum. Current approaches to coal liquefaction use a catalyst to break the complex network structure of coal into more manageable fragments. The products are then treated with hydrogen (from syngas or other sources) under high pressure to produce a liquid more like petroleum. Subsequent distillation, cracking, and reforming can be used to create products similar to those obtained from petroleum. The total yield of liquid fuels is about 5.5 bbl of crude liquid per ton of coal (1 bbl is 42 gal or 160 L). Although the economics of coal liquefaction are currently even less attractive than for coal gasification, liquid fuels based on coal are likely to become economically competitive as supplies of petroleum are consumed. If bituminous coal is converted to methane by the process in Equation \(\ref{7.9.1}\), what is the ratio of the Δ of the methane produced to the enthalpy of the coal consumed to produce the methane? (Note that 1 mol of CH is produced for every 2 mol of carbon in coal.) chemical reaction and Δ (Table \(\Page {2}\)) ratio of Δ of methane produced to coal consumed Write a balanced chemical equation for the conversion of coal to methane. Referring to Table \(\Page {2}\), calculate the Δ of methane and carbon. Calculate the ratio of the energy released by combustion of the methane to the energy released by combustion of the carbon. The balanced chemical equation for the conversion of coal to methane is as follows: \[\ce{2C (s) + 2H2O(g) → CH4(g) + CO2(g)} \nonumber\] Thus 1 mol of methane is produced for every 2 mol of carbon consumed. The Δ of 1 mol of methane is The Δ of 2 mol of carbon (as coal) is B The ratio of the energy released from the combustion of methane to the energy released from the combustion of carbon is The energy released from the combustion of the product (methane) is 131% of that of the reactant (coal). The fuel value of coal is actually increased by the process! How is this possible when the law of conservation of energy states that energy cannot be created? The reaction consumes 2 mol of water (\(ΔH^\circ_\ce{f}=\mathrm{−285.8\: kJ/mol}\)) but produces only 1 mol of CO2 (\(ΔH^\circ_\ce{f}=\mathrm{−393.5\: kJ/mol}\)). Part of the difference in potential energy between the two (approximately 180 kJ/mol) is stored in CH4 and can be released during combustion. Using the data in Table \(\Page {2}\), calculate the mass of hydrogen necessary to provide as much energy during combustion as 1 bbl of crude oil (density approximately 0.75 g/mL). 36 kg Even if carbon-based fuels could be burned with 100% efficiency, producing only CO (g) and H O(g), doing so could still potentially damage the environment when carried out on the vast scale required by an industrial society. The amount of CO released is so large and is increasing so rapidly that it is apparently overwhelming the natural ability of the planet to remove CO from the atmosphere. In turn, the elevated levels of CO are thought to be affecting the temperature of the planet through a mechanism known as the greenhouse effect. As you will see, there is little doubt that atmospheric CO levels are increasing, and the major reason for this increase is the combustion of fossil fuels. There is substantially less agreement, however, on whether the increased CO levels are responsible for a significant increase in temperature. Figure \(\Page {5}\) illustrates the global carbon cycle, the distribution and flow of carbon on Earth. Normally, the fate of atmospheric CO is to either (1) dissolve in the oceans and eventually precipitate as carbonate rocks or (2) be taken up by plants. The rate of uptake of CO by the ocean is limited by its surface area and the rate at which gases dissolve, which are approximately constant. The rate of uptake of CO by plants, representing about 60 billion metric tons of carbon per year, partly depends on how much of Earth’s surface is covered by vegetation. Unfortunately, the rapid deforestation for agriculture is reducing the overall amount of vegetation, and about 60 billion metric tons of carbon are released annually as CO from animal respiration and plant decay. The amount of carbon released as CO every year by fossil fuel combustion is estimated to be about 5.5 billion metric tons. The net result is a system that is slightly out of balance, experiencing a slow but steady increase in atmospheric CO levels (Figure \(\Page {6}\)). As a result, average CO levels have increased by about 30% since 1850. Most of Earth’s carbon is found in the crust, where it is stored as calcium and magnesium carbonate in sedimentary rocks. The oceans also contain a large reservoir of carbon, primarily as the bicarbonate ion (HCO ). Green plants consume about 60 billion metric tons of carbon per year as CO during photosynthesis, and about the same amount of carbon is released as CO annually from animal and plant respiration and decay. The combustion of fossil fuels releases about 5.5 billion metric tons of carbon per year as CO . The increasing levels of atmospheric CO are of concern because CO absorbs thermal energy radiated by the Earth, as do other gases such as water vapor, methane, and chlorofluorocarbons. Collectively, these substances are called greenhouse gases; they mimic the effect of a greenhouse by trapping thermal energy in the Earth’s atmosphere, a phenomenon known as the greenhouse effect (Figure \(\Page {7}\)). Venus is an example of a planet that has a runaway greenhouse effect. The atmosphere of Venus is about 95 times denser than that of Earth and contains about 95% CO . Because Venus is closer to the sun, it also receives more solar radiation than Earth does. The result of increased solar radiation and high CO levels is an average surface temperature of about 450°C, which is hot enough to melt lead. Data such as those in Figure Figure \(\Page {6}\) indicate that atmospheric levels of greenhouse gases have increased dramatically over the past 100 years, and it seems clear that the heavy use of fossil fuels by industry is largely responsible. It is not clear, however, how large an increase in temperature ( ) may result from a continued increase in the levels of these gases. Estimates of the effects of doubling the preindustrial levels of CO range from a 0°C to a 4.5°C increase in the average temperature of Earth’s surface, which is currently about 14.4°C. Even small increases, however, could cause major perturbations in our planet’s delicately balanced systems. For example, an increase of 5°C in Earth’s average surface temperature could cause extensive melting of glaciers and the Antarctic ice cap. It has been suggested that the resulting rise in sea levels could flood highly populated coastal areas, such as New York City, Calcutta, Tokyo, Rio de Janeiro, and Sydney. An analysis conducted in 2009 by leading climate researchers from the US National Oceanic and Atmospheric Administration, Switzerland, and France shows that CO in the atmosphere will remain near peak levels far longer than other greenhouse gases, which dissipate more quickly. The study predicts a rise in sea levels of approximately 3 ft by the year 3000, excluding the rise from melting glaciers and polar ice caps. According to the analysis, southwestern North America, the Mediterranean, and southern Africa are projected to face droughts comparable to that of the Dust Bowl of the 1930s as a result of global climate changes. The increase in CO levels is only one of many trends that can affect Earth’s temperature. In fact, geologic evidence shows that the average temperature of Earth has fluctuated significantly over the past 400,000 years, with a series of glacial periods (during which the temperature was 10°C–15°C than it is now and large glaciers covered much of the globe) interspersed with relatively short, warm interglacial periods (Figure \(\Page {8}\)). Although average temperatures appear to have increased by 0.5°C in the last century, the statistical significance of this increase is open to question, as is the existence of a cause-and-effect relationship between the temperature change and CO levels. Despite the lack of incontrovertible scientific evidence, however, many people believe that we should take steps now to limit CO emissions and explore alternative sources of energy, such as solar energy, geothermal energy from volcanic steam, and nuclear energy, to avoid even the possibility of creating major perturbations in Earth’s environment. In 2010, international delegates met in Cancún, Mexico, and agreed on a broad array of measures that would advance climate protection. These included the development of low-carbon technologies, providing a framework to reduce deforestation, and aiding countries in assessing their own vulnerabilities. They avoided, however, contentious issues of assigning emissions reductions commitments. A student at UCLA decided to fly home to New York for Christmas. The round trip was 4500 air miles, and part of the cost of her ticket went to buy the 100 gal of jet fuel necessary to transport her and her baggage. Assuming that jet fuel is primarily -dodecane (C H ) with a density of 0.75 g/mL, how much energy was expended and how many tons of \(\ce{CO2}\) were emitted into the upper atmosphere to get her home and back? volume and density of reactant in combustion reaction energy expended and mass of CO emitted A After writing a balanced chemical equation for the reaction, calculate \(ΔH^\circ_\ce{comb}\) B Determine the number of moles of dodecane in 100 gal by using the density and molar mass of dodecane and the appropriate conversion factors. C Obtain the amount of energy expended by multiplying \(ΔH^\circ_\ce{comb}\) by the number of moles of dodecane. Calculate the amount of CO2 emitted in tons by using mole ratios from the balanced chemical equation and the appropriate conversion factors. A We first need to write a balanced chemical equation for the reaction: \[\ce{2C12H26 (l) + 37O2(g) -> 24CO2(g) + 26H2O(l)} \nonumber\] We can calculate \(ΔH^\circ_\ce{comb}\) using the \(ΔH^\circ_\ce{f}\) values corresponding to each substance in the specified phase (phases are not shown for simplicity): \[\begin{align*}
ΔH^\circ_\ce{comb}&=ΣmΔH^\circ_\ce{f}(\ce{products})−ΣnΔH^\circ_\ce{f}(\ce{reactants})\\
&= [24ΔH^\circ_\ce{f}(\ce{CO2}) + 26ΔH^\circ_\ce{f}(\ce{H2O})]−[37ΔH^\circ_\ce{f}(\ce{O2}) + 2ΔH^\circ_\ce{f}(\ce{C12H26})]\\
&= \mathrm{[24(−393.5\: kJ/mol\: CO_2) + 26(−285.8\: kJ/mol\: H_2O)]} \\
&\:\:\:\:\:\mathrm{−[37(0\: kJ/mol\: O_2) + 2(−350.9\: kJ/mol\: C_{12}H_{26})]}\\
&=\mathrm{−16,173.0\: kJ}
\end{align*}\] According to the balanced chemical equation for the reaction, this value is \(ΔH^\circ_\ce{comb}\) for the combustion of 2 mol of n-dodecane. So we must divide by 2 to obtain \(ΔH^\circ_\ce{comb}\) per mole of n-dodecane: \[ΔH^\circ_\ce{comb}=\mathrm{−8,086.5\; kJ/mol\; C_{12}H_{26}} \nonumber\] B The number of moles of dodecane in 100 gal can be calculated as follows, using density, molar mass, and appropriate conversion factors: \(\mathrm{100\: \cancel{gal} \left( \dfrac{3.785 \cancel{L}}{1 \cancel{gal}}\right )\left(\dfrac{1000 \cancel{mL}}{\cancel{L}}\right)\left(\dfrac{0.75 \cancel{g}}{\cancel{mL}}\right)\left(\dfrac{1\: mol}{170.34 \cancel{g}}\right)= 1.7×10^3\: mol\: C_{12}H_{26}}\) C The total energy released is \(ΔH^\circ_\ce{comb}= \mathrm{(−8086.5\: kJ/\cancel{mol}) (1.7×10^3 \cancel{mol}) =−1.4×10^7\: kJ}\) From the balanced chemical equation for the reaction, we see that each mole of dodecane forms 12 mol of \(\ce{CO2}\) upon combustion. Hence the amount of \(\ce{CO2}\) emitted is \(\mathrm{1.7×10^3 \cancel{mol\: C_{12}H_{26}}\left(\dfrac{\dfrac{24}{2} \cancel{mol\: CO_2}}{1 \cancel{mol\: C_{12}H_{26}}}\right)\left(\dfrac{44.0 \cancel{g}}{1 \cancel{mol\: CO_2}}\right)\left(\dfrac{1 \cancel{lb}}{454 \cancel{g}}\right)\left(\dfrac{1\: tn}{2000 \cancel{lb}}\right)= 0.99\: tn}\) Suppose the student in Example \(\Page {2}\) couldn’t afford the plane fare, so she decided to drive home instead. Assume that the round-trip distance by road was 5572 miles, her fuel consumption averaged 31 mpg, and her fuel was pure isooctane (C H , density = 0.6919 g/mL). How much energy was expended and how many tons of CO were produced during her trip? 2.2 × 10 kJ; 1.6 tons of CO (about twice as much as is released by flying) Thermochemical concepts can be used to calculate the efficiency of various forms of fuel, which can then be applied to environmental issues. More than 80% of the energy used by modern society (about 3 × 10 kJ/yr) is from the combustion of fossil fuels. Because of their availability, ease of transport, and facile conversion to convenient fuels, natural gas and petroleum are currently the preferred fuels. Supplies of , a complex solid material derived from plants that lived long ago, are much greater, but the difficulty in transporting and burning a solid makes it less attractive as a fuel. Coal releases the smallest amount of energy per gram of any fossil fuel, and natural gas the greatest amount. The combustion of fossil fuels releases large amounts of CO that upset the balance of the and result in a steady increase in atmospheric CO levels. Because CO is a , which absorbs heat before it can be radiated from Earth into space, CO in the atmosphere can result in increased surface temperatures (the ). The temperature increases caused by increased CO levels because of human activities are, however, superimposed on much larger variations in Earth’s temperature that have produced phenomena such as the ice ages and are still poorly understood. | 21,250 | 3,766 |
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Molecules or ions which can either donate or accept a proton, depending on their circumstances, are called . The most important amphiprotic species is water itself. When an acid donates a proton to water, the water molecule is a proton acceptor, and hence a base. Conversely, when a base reacts with water, a water molecule donates a proton, and hence acts as an acid. Another important group of amphiprotic species is the . Each amino acid molecule contains an acidic carboxyl group and a basic amino group. In fact the amino acids usually exist in (German for “double ion”) form, where the proton has transferred from the carboxyl to the amino group. In the case of glycine, for example, the zwitterion is Write equations to show the amphiprotic behavior of (a) H PO and (b) H O. To make an amphiprotic species behave as an acid requires a fairly good proton acceptor. Conversely, to make it behave as a base requires a proton donor. Base: \(\text{H}_2 \text{PO}_4^- + \text{H}_3 \text{O}^+ \rightarrow \text{H}_3 \text{PO}_4 + \text{H}_2 \text{O}\) Base: \( \text{H}_2\text{O} + \text{H}_2\text{SO}_4 \rightarrow \text{H}_3\text{O}^+ + \text{HSO}_4^- \) | 1,214 | 3,768 |
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The focus of this chapter is on the interaction of ultraviolet, visible, and infrared radiation with matter. Because these techniques use optical materials to disperse and focus the radiation, they often are identified as optical spectroscopies. For convenience we will use the simpler term spectroscopy in place of optical ; however, you should understand we will consider only a limited piece of what is a much broader area of analytical techniques. Despite the difference in instrumentation, all spectroscopic techniques share several common features. Before we consider individual examples in greater detail, let’s take a moment to consider some of these similarities. As you work through the chapter, this overview will help you focus on the similarities between different spectroscopic methods of analysis. You will find it easier to understand a new analytical method when you can see its relationship to other similar methods. —light—is a form of energy whose behavior is described by the properties of both waves and particles. Some properties of electromagnetic radiation, such as its refraction when it passes from one medium to another (Figure 10.1.1
), are explained best when we describe light as a wave. Other properties, such as absorption and emission, are better described by treating light as a particle. The exact nature of electromagnetic radiation remains unclear, as it has since the development of quantum mechanics in the first quarter of the 20th century [Home, D.; Gribbin, J. 1991, 2 Nov. 30–33]. Nevertheless, this dual model of wave and particle behavior provide a useful description for electromagnetic radiation. Electromagnetic radiation consists of oscillating electric and magnetic fields that propagate through space along a linear path and with a constant velocity. In a vacuum, electromagnetic radiation travels at the speed of light, , which is \(2.99792 \times 10^8\) m/s. When electromagnetic radiation moves through a medium other than a vacuum, its velocity, , is less than the speed of light in a vacuum. The difference between and is sufficiently small (<0.1%) that the speed of light to three significant figures, \(3.00 \times 10^8\) m/s, is accurate enough for most purposes. The oscillations in the electric field and the magnetic field are perpendicular to each other and to the direction of the wave’s propagation. Figure 10.1.2
shows an example of plane-polarized electromagnetic radiation, which consists of a single oscillating electric field and a single oscillating magnetic field. An electromagnetic wave is characterized by several fundamental properties, including its velocity, amplitude, frequency, phase angle, polarization, and direction of propagation [Ball, D. W. , , 24–25]. For example, the amplitude of the oscillating electric field at any point along the propagating wave is \[A_{t}=A_{e} \sin (2 \pi \nu t+\Phi) \nonumber\] where is the magnitude of the electric field at time , is the electric field’s maximum , \(\nu\) is the wave’s —the number of oscillations in the electric field per unit time—and \(\Phi\) is a that accounts for the fact that need not have a value of zero at = 0. The identical equation for the magnetic field is \[A_{t}=A_{m} \sin (2 \pi \nu t+\Phi) \nonumber\] where is the magnetic field’s maximum amplitude. Other properties also are useful for characterizing the wave behavior of electromagnetic radiation. The , \(\lambda\), is defined as the distance between successive maxima (see Figure 10.1.2
). For ultraviolet and visible electromagnetic radiation the wavelength usually is expressed in nanometers (1 nm = 10 m), and for infrared radiation it is expressed in microns (1 mm = 10 m). The relationship between wavelength and frequency is \[\lambda = \frac {c} {\nu} \nonumber\] Another unit useful unit is the , \(\overline{\nu}\), which is the reciprocal of wavelength \[\overline{\nu} = \frac {1} {\lambda} \nonumber\] Wavenumbers frequently are used to characterize infrared radiation, with the units given in cm . When electromagnetic radiation moves between different media—for example, when it moves from air into water—its frequency, \(\nu\), remains constant. Because its velocity depends upon the medium in which it is traveling, the electromagnetic radiation’s wavelength, \(\lambda\), changes. If we replace the speed of light in a vacuum, , with its speed in the medium, \(v\), then the wavelength is \[\lambda = \frac {v} {\nu} \nonumber\] This change in wavelength as light passes between two media explains the refraction of electromagnetic radiation shown in Figure 10.1.1
. In 1817, Josef Fraunhofer studied the spectrum of solar radiation, observing a continuous spectrum with numerous dark lines. Fraunhofer labeled the most prominent of the dark lines with letters. In 1859, Gustav Kirchhoff showed that the D line in the sun’s spectrum was due to the absorption of solar radiation by sodium atoms. The wavelength of the sodium D line is 589 nm. What are the frequency and the wavenumber for this line? The frequency and wavenumber of the sodium D line are \[\nu=\frac{c}{\lambda}=\frac{3.00 \times 10^{8} \ \mathrm{m} / \mathrm{s}}{589 \times 10^{-9} \ \mathrm{m}}=5.09 \times 10^{14} \ \mathrm{s}^{-1} \nonumber\] \[\overline{\nu}=\frac{1}{\lambda}=\frac{1}{589 \times 10^{-9} \ \mathrm{m}} \times \frac{1 \ \mathrm{m}}{100 \ \mathrm{cm}}=1.70 \times 10^{4} \ \mathrm{cm}^{-1} \nonumber\] Another historically important series of spectral lines is the Balmer series of emission lines from hydrogen. One of its lines has a wavelength of 656.3 nm. What are the frequency and the wavenumber for this line? The frequency and wavenumber for the line are \[\nu=\frac{c}{\lambda}=\frac{3.00 \times 10^{8} \ \mathrm{m} / \mathrm{s}}{656.3 \times 10^{-9} \ \mathrm{m}}=4.57 \times 10^{14} \ \mathrm{s}^{-1} \nonumber\] \[\overline{\nu}=\frac{1}{\lambda}=\frac{1}{656.3 \times 10^{-9} \ \mathrm{m}} \times \frac{1 \ \mathrm{m}}{100 \ \mathrm{cm}}=1.524 \times 10^{4} \ \mathrm{cm}^{-1} \nonumber\] When matter absorbs electromagnetic radiation it undergoes a change in energy. The interaction between matter and electromagnetic radiation is easiest to understand if we assume that radiation consists of a beam of energetic particles called photons. When a is absorbed by a sample it is “destroyed” and its energy acquired by the sample [Ball, D. W. , 20–21]. The energy of a photon, in joules, is related to its frequency, wavelength, and wavenumber by the following equalities \[E=h \nu=\frac{h c}{\lambda}=h c \overline{\nu} \nonumber\] where is Planck’s constant, which has a value of \(6.626 \times 10^{-34}\) Js. What is the energy of a photon from the sodium D line at 589 nm? The photon’s energy is \[E=\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \ \mathrm{Js}\right)\left(3.00 \times 10^{8} \ \mathrm{m} / \mathrm{s}\right)}{589 \times 10^{-7} \ \mathrm{m}}=3.37 \times 10^{-19} \ \mathrm{J} \nonumber\] What is the energy of a photon for the Balmer line at a wavelength of 656.3 nm? The photon’s energy is \[E=\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \ \mathrm{Js}\right)\left(3.00 \times 10^{8} \ \mathrm{m} / \mathrm{s}\right)}{656.3 \times 10^{-9} \ \mathrm{m}}=3.03 \times 10^{-19} \ \mathrm{J} \nonumber\] The frequency and the wavelength of electromagnetic radiation vary over many orders of magnitude. For convenience, we divide electromagnetic radiation into different regions—the —based on the type of atomic or molecular transitions that gives rise to the absorption or emission of photons (Figure 10.1.3
). The boundaries between the regions of the electromagnetic spectrum are not rigid and overlap between spectral regions is possible. In the previous section we defined several characteristic properties of electromagnetic radiation, including its energy, velocity, amplitude, frequency, phase angle, polarization, and direction of propagation. A spectroscopic measurement is possible only if the photon’s interaction with the sample leads to a change in one or more of these characteristic properties. We will divide spectroscopy into two broad classes of techniques. In one class of techniques there is a transfer of energy between the photon and the sample. Table 10.1.1
provides a list of several representative examples. electron spin resonance nuclear magnetic resonance atomic fluorescence spectroscopy In absorption spectroscopy a photon is absorbed by an atom or molecule, which undergoes a transition from a lower-energy state to a higher-energy, or excited state (Figure 10.1.4
). The type of transition depends on the photon’s energy. The electromagnetic spectrum in Figure 10.1.3
, for example, shows that absorbing a photon of visible light promotes one of the atom’s or molecule’s valence electrons to a higher-energy level. When an molecule absorbs infrared radiation, on the other hand, one of its chemical bonds experiences a change in vibrational energy. When it absorbs electromagnetic radiation the number of photons passing through a sample decreases. The measurement of this decrease in photons, which we call , is a useful analytical signal. Note that each energy level in Figure 10.1.4
has a well-defined value because each is quantized. Absorption occurs only when the photon’s energy, \(h \nu\), matches the difference in energy, \(\Delta E\), between two energy levels. A plot of absorbance as a function of the photon’s energy is called an . Figure 10.1.5
, for example, shows the absorbance spectrum of cranberry juice. When an atom or molecule in an excited state returns to a lower energy state, the excess energy often is released as a photon, a process we call ( ). There are several ways in which an atom or a molecule may end up in an excited state, including thermal energy, absorption of a photon, or as the result of a chemical reaction. Emission following the absorption of a photon is also called , and that following a chemical reaction is called . A typical emission spectrum is shown in Figure 10.1.6
. Molecules also can release energy in the form of heat. We will return to this point later in the chapter. In the second broad class of spectroscopic techniques, the electromagnetic radiation undergoes a change in amplitude, phase angle, polarization, or direction of propagation as a result of its refraction, reflection, scattering, diffraction, or dispersion by the sample. Several representative spectroscopic techniques are listed in Table 10.1.2
. The spectroscopic techniques in and use instruments that share several common basic components, including a source of energy, a means for isolating a narrow range of wavelengths, a detector for measuring the signal, and a signal processor that displays the signal in a form convenient for the analyst. In this section we introduce these basic components. Specific instrument designs are considered in later sections. You will find a more detailed treatment of these components in the for this chapter. All forms of spectroscopy require a source of energy. In absorption and scattering spectroscopy this energy is supplied by photons. Emission and photoluminescence spectroscopy use thermal, radiant (photon), or chemical energy to promote the analyte to a suitable excited state. . A source of electromagnetic radiation must provide an output that is both intense and stable. Sources of electromagnetic radiation are classified as either continuum or line sources. A emits radiation over a broad range of wavelengths, with a relatively smooth variation in intensity (Figure 10.1.7
). A , on the other hand, emits radiation at selected wavelengths (Figure 10.1.8
). Table 10.1.3
provides a list of the most common sources of electromagnetic radiation. . The most common sources of thermal energy are flames and plasmas. A flame source uses a combustion of a fuel and an oxidant to achieve temperatures of 2000–3400 K. Plasmas, which are hot, ionized gases, provide temperatures of 6000–10000 K. . Exothermic reactions also may serve as a source of energy. In chemiluminescence the analyte is raised to a higher-energy state by means of a chemical reaction, emitting characteristic radiation when it returns to a lower-energy state. When the chemical reaction results from a biological or enzymatic reaction, the emission of radiation is called bioluminescence. Commercially available “light sticks” and the flash of light from a firefly are examples of chemiluminescence and bioluminescence. In Nessler’s original colorimetric method for ammonia, which was described at the beginning of the chapter, the sample and several standard solutions of ammonia are placed in separate tall, flat-bottomed tubes. As shown in Figure 10.1.9
, after adding the reagents and allowing the color to develop, the analyst evaluates the color by passing ambient light through the bottom of the tubes and looking down through the solutions. By matching the sample’s color to that of a standard, the analyst is able to determine the concentration of ammonia in the sample. In Figure 10.1.9
every wavelength of light from the source passes through the sample. This is not a problem if there is only one absorbing species in the sample. If the sample contains two components, then a quantitative analysis using Nessler’s original method is impossible unless the standards contains the second component at the same concentration it has in the sample. To overcome this problem, we want to select a wavelength that only the analyte absorbs. Unfortunately, we can not isolate a single wavelength of radiation from a continuum source, although we can narrow the range of wavelengths that reach the sample. As seen in Figure 10.1.10
, a wavelength selector always passes a narrow band of radiation characterized by a , an , and a maximum throughput of radiation. The effective bandwidth is defined as the width of the radiation at half of its maximum throughput. The ideal wavelength selector has a high throughput of radiation and a narrow effective bandwidth. A high throughput is desirable because the more photons that pass through the wavelength selector, the stronger the signal and the smaller the background noise. A narrow effective bandwidth provides a higher , with spectral features separated by more than twice the effective bandwidth being resolved. As shown in Figure 10.1.11
, these two features of a wavelength selector often are in opposition. A larger effective bandwidth favors a higher throughput of radiation, but provide less resolution. Decreasing the effective bandwidth improves resolution, but at the cost of a noisier signal [Jiang, S.; Parker, G. A. , October, 38–43]. For a qualitative analysis, resolution usually is more important than noise and a smaller effective bandwidth is desirable; however, in a quantitative analysis less noise usually is desirable. . The simplest method for isolating a narrow band of radiation is to use an absorption or interference . Absorption filters work by selectively absorbing radiation from a narrow region of the electromagnetic spectrum. Interference filters use constructive and destructive interference to isolate a narrow range of wavelengths. A simple example of an absorption filter is a piece of colored glass. A purple filter, for example, removes the complementary color green from 500–560 nm. Commercially available absorption filters provide effective bandwidths of 30–250 nm, although the throughput at the low end of this range often is only 10% of the source’s emission intensity. Interference filters are more expensive than absorption filters, but have narrower effective bandwidths, typically 10–20 nm, with maximum throughputs of at least 40%. . A filter has one significant limitation—because a filter has a fixed nominal wavelength, if we need to make measurements at two different wavelengths, then we must use two different filters. A is an alternative method for selecting a narrow band of radiation that also allows us to continuously adjust the band’s nominal wavelength. The construction of a typical monochromator is shown in Figure 10.1.12
. Radiation from the source enters the monochromator through an entrance slit. The radiation is collected by a collimating mirror, which reflects a parallel beam of radiation to a diffraction grating. The diffraction grating is an optically reflecting surface with a large number of parallel grooves (see insert to Figure 10.1.12
). The diffraction grating disperses the radiation and a second mirror focuses the radiation onto a planar surface that contains an exit slit. In some monochromators a prism is used in place of the diffraction grating. Radiation exits the monochromator and passes to the detector. As shown in Figure 10.1.12
, a monochromator converts a source of radiation at the entrance slit to a source of finite effective bandwidth at the exit slit. The choice of which wavelength exits the monochromator is determined by rotating the diffraction grating. A narrower exit slit provides a smaller effective bandwidth and better resolution than does a wider exit slit, but at the cost of a smaller throughput of radiation. Polychromatic means many colored. Polychromatic radiation contains many different wavelengths of light. Monochromatic means one color, or one wavelength. Although the light exiting a monochromator is not strictly of a single wavelength, its narrow effective bandwidth allows us to think of it as monochromatic. Monochromators are classified as either fixed-wavelength or scanning. In a fixed-wavelength monochromator we manually select the wavelength by rotating the grating. Normally a fixed-wavelength monochromator is used for a quantitative analysis where measurements are made at one or two wavelengths. A scanning monochromator includes a drive mechanism that continuously rotates the grating, which allows successive wavelengths of light to exit from the monochromator. A scanning monochromator is used to acquire spectra, and, when operated in a fixed-wavelength mode, for a quantitative analysis. . An provides an alternative approach for wavelength selection. Instead of filtering or dispersing the electromagnetic radiation, an interferometer allows source radiation of all wavelengths to reach the detector simultaneously (Figure 10.1.13
). Radiation from the source is focused on a beam splitter that reflects half of the radiation to a fixed mirror and transmits the other half to a moving mirror. The radiation recombines at the beam splitter, where constructive and destructive interference determines, for each wavelength, the intensity of light that reaches the detector. As the moving mirror changes position, the wavelength of light that experiences maximum constructive interference and maximum destructive interference also changes. The signal at the detector shows intensity as a function of the moving mirror’s position, expressed in units of distance or time. The result is called an or a time domain spectrum. The time domain spectrum is converted mathematically, by a process called a Fourier transform, to a spectrum (a frequency domain spectrum) that shows intensity as a function of the radiation’s energy. The mathematical details of the Fourier transform are beyond the level of this textbook. You can consult the chapter’s additional resources for additional information. In comparison to a monochromator, an interferometer has two significant advantages. The first advantage, which is termed , is the greater throughput of source radiation. Because an interferometer does not use slits and has fewer optical components from which radiation is scattered and lost, the throughput of radiation reaching the detector is \(80-200 \times\) greater than that for a monochromator. The result is less noise. The second advantage, which is called , is a savings in the time needed to obtain a spectrum. Because the detector monitors all frequencies simultaneously, a spectrum takes approximately one second to record, as compared to 10–15 minutes when using a scanning monochromator. In Nessler’s original method for determining ammonia ( ) the analyst’s eye serves as the detector, matching the sample’s color to that of a standard. The human eye, of course, has a poor range—it responds only to visible light—and it is not particularly sensitive or accurate. Modern detectors use a sensitive to convert a signal consisting of photons into an easily measured electrical signal. Ideally the detector’s signal, , is a linear function of the electromagnetic radiation’s power, , \[S=k P+D \nonumber\] where is the detector’s sensitivity, and is the detector’s , or the background current when we prevent the source’s radiation from reaching the detector. There are two broad classes of spectroscopic transducers: thermal transducers and photon transducers. Table 10.1.4
provides several representative examples of each class of transducers. Transducer is a general term that refers to any device that converts a chemical or a physical property into an easily measured electrical signal. The retina in your eye, for example, is a transducer that converts photons into an electrical nerve impulse; your eardrum is a transducer that converts sound waves into a different electrical nerve impulse. . Phototubes and photomultipliers use a photosensitive surface that absorbs radiation in the ultraviolet, visible, or near IR to produce an electrical current that is proportional to the number of photons reaching the transducer (Figure 10.1.14
). Other photon detectors use a semiconductor as the photosensitive surface. When the semiconductor absorbs photons, valence electrons move to the semiconductor’s conduction band, producing a measurable current. One advantage of the Si photodiode is that it is easy to miniaturize. Groups of photodiodes are gathered together in a linear array that contains 64–4096 individual photodiodes. With a width of 25 μm per diode, a linear array of 2048 photodiodes requires only 51.2 mm of linear space. By placing a along the monochromator’s focal plane, it is possible to monitor simultaneously an entire range of wavelengths. . Infrared photons do not have enough energy to produce a measurable current with a photon transducer. A thermal transducer, therefore, is used for infrared spectroscopy. The absorption of infrared photons increases a thermal transducer’s temperature, changing one or more of its characteristic properties. A pneumatic transducer, for example, is a small tube of xenon gas with an IR transparent window at one end and a flexible membrane at the other end. Photons enter the tube and are absorbed by a blackened surface, increasing the temperature of the gas. As the temperature inside the tube fluctuates, the gas expands and contracts and the flexible membrane moves in and out. Monitoring the membrane’s displacement produces an electrical signal. A transducer’s electrical signal is sent to a where it is displayed in a form that is more convenient for the analyst. Examples of signal processors include analog or digital meters, recorders, and computers equipped with digital acquisition boards. A signal processor also is used to calibrate the detector’s response, to amplify the transducer’s signal, to remove noise by filtering, or to mathematically transform the signal. If the retina in your eye and the eardrum in your ear are transducers, then your brain is the signal processor. | 23,471 | 3,769 |
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The following experiments provide useful introductions to the statistical analysis of data in the analytical chemistry laboratory. A more comprehensive discussion of the analysis of data, which includes all topics considered in this chapter as well as additional material, are found in many textbook on statistics or data analysis; several such texts are listed here. The importance of defining statistical terms is covered in the following papers. The detection of outliers, particularly when working with a small number of samples, is discussed in the following papers. The following papers provide additional information on error and uncertainty, including the propagation of uncertainty. Consult the following resources for a further discussion of detection limits. The following articles provide thoughts on the limitations of statistical analysis based on significance testing. The following resources provide additional information on using Excel, including reports of errors in its handling of some statistical procedures. To learn more about using R, consult the following resources. The following papers provide insight into visualizing data. | 1,164 | 3,770 |
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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 Learning Objective of this module is to determine the contribution of each component gas to the total pressure of a mixture of gases. 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: \[P=n \bigg(\dfrac{RT}{V}\bigg) = n \times \rm const. \tag{10.6.1}\] 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 . We can write it mathematically as where \(P_{tot}\) is the total pressure and the other terms are the partial pressures of the individual gases (up to \(n\) component 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 \(n\) component gases, the total pressure is given by Equation 10.6.4 restates Equation 10.6.3 in a more general form and makes it explicitly clear that, at constant temperature and volume, the pressure exerted by a gas depends on only 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 Equation 10.6.4 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 the example below: 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 \(O_2\) 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 The number of moles of \(He\) is \[n_{\rm He}=\rm\dfrac{326.4\;g}{4.003\;g/mol}=81.54\;mol\] The number of moles of \(O_2\) is \[n_{\rm O_2}=\rm \dfrac{51.2\;g}{32.00\;g/mol}=1.60\;mol\] We can now use the ideal gas law to calculate the partial pressure of each: \[P_{\rm He}=\dfrac{n_{\rm He}RT}{V}=\rm\dfrac{81.54\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=196.2\;atm\] \[P_{\rm O_2}=\dfrac{n_{\rm O_2}RT}{V}=\rm\dfrac{1.60\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=3.85\;atm\] The total pressure is the sum of the two partial pressures: \[P_{\rm tot}=P_{\rm He}+P_{\rm O_2}=\rm(196.2+3.85)\;atm=200.1\;atm\] 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. Answer: \(P_{CH_4}=137 \; atm\); \(P_{C_2H_6}=13.4\; atm\); \(P_{tot}=151\; atm\) The composition of a gas mixture can be described by the mole fractions of the gases present. The mole fraction (\(X\)) 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 (\(n_{tot}\)): \[x_A=\dfrac{\text{moles of A}}{\text{total moles}}= \dfrac{n_A}{n_{tot}} =\dfrac{n_A}{n_A+n_B+\cdots}\tag{10.6.5}\] The mole fraction is a dimensionless quantity between 0 and 1. If \(x_A = 1.0\), then the sample is pure \(A\), not a mixture. If \(x_A = 0\), 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: \(P_A = n_ART/V\) and \(P_{tot} = n_tRT/V\). The ratio of the two is thus \[\dfrac{P_A}{P_{tot}}=\dfrac{n_ART/V}{n_{tot}RT/V} = \dfrac{n_A}{n_{tot}}=x_A \tag{10.6.6}\] Rearranging this equation gives \[P_A = x_AP_{tot} \tag{10.6.7}\] 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. By volume, Earth’s atmosphere is about 78% \(N_2\), 21% \(O_2\), and 0.9% \(Ar\), with trace amounts of gases such as \(CO_2\), \(H_2O\), and others. This means that 78% of the particles present in the atmosphere are \(N_2\); hence the mole fraction of \(N_2\) is 78%/100% = 0.78. Similarly, the mole fractions of \(O_2\) and \(Ar\) are 0.21 and 0.009, respectively. Using Equation 10.6.7, 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 \(O_2\) 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 Equation 10.6.7. The mole fraction of any gas \(A\) is given by \[x_A=\dfrac{P_A}{P_{tot}}\] where \(P_A\) is the partial pressure of \(A\) and \(P_{tot}\) is the total pressure. For example, the mole fraction of \(CO_2\) is given as: \[x_{\rm CO_2}=\rm\dfrac{48\;mmHg}{767\;mmHg}=0.063\] The following table gives the values of \(x_A\) for the gases in the exhaled air. 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. | 8,480 | 3,771 |
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Demonstration of Charles' Law The volume of a 1 L Florence flask and rubber hose is 1125 mL (fill with water, get volume, then empty). The flask is connected to a 15 mL Pipette used as a water manometer, which in turn is connected to a leveling bulb. The flask is immersed in a 3 L low form beaker of room temperature (25 C) water, and the manometer connected, and the levelling jar adjusted so the water levels are equal and P = P . Read the volume. About 30 g of ice is added to the water to lower the temperature a degree or two. The water levels are adjusted, and the new volume determined. Calculate ΔT/ΔV = ~0.267 C/mL To reach V = 0, must decrease 1125 mL * 0.267 C/mL = 300 C Absolute zero = 25 C - 300 = -275 C | 735 | 3,772 |
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. In addition to these individual basis; please contact The human eye is sensitive to what fraction of the electromagnetic spectrum, assuming a typical spectral range of 10 to 10 Hz? If we came from the planet Krypton and had x-ray vision (i.e., if our eyes were sensitive to x-rays in addition to visible light), how would this fraction be changed? What is the frequency in megahertz corresponding to each wavelength? What is the frequency in megahertz corresponding to each wavelength? Line spectra are also observed for molecular species. Given the following characteristic wavelengths for each species, identify the spectral region (ultraviolet, visible, etc.) in which the following line spectra will occur. Given 1.00 mol of each compound and the wavelength of absorbed or emitted light, how much energy does this correspond to? What is the speed of a wave in meters per second that has a wavelength of 1250 m and a frequency of 2.36 × 10 s ? A wave travels at 3.70 m/s with a frequency of 4.599 × 10 Hz and an amplitude of 1.0 m. What is its wavelength in nanometers? An AM radio station broadcasts with a wavelength of 248.0 m. What is the broadcast frequency of the station in kilohertz? An AM station has a broadcast range of 92.6 MHz. What is the corresponding wavelength range in meters for this reception? An FM radio station broadcasts with a wavelength of 3.21 m. What is the broadcast frequency of the station in megahertz? An FM radio typically has a broadcast range of 82–112 MHz. What is the corresponding wavelength range in meters for this reception? A microwave oven operates at a frequency of approximately 2450 MHz. What is the corresponding wavelength? Water, with its polar molecules, absorbs electromagnetic radiation primarily in the infrared portion of the spectrum. Given this fact, why are microwave ovens used for cooking food? The energy of a photon is directly proportional to the frequency of the electromagnetic radiation. Quantized: harp, tire size, speedboat waves; continuous: human voice, colors of light, car speed. What is the energy of a photon of light with each wavelength? To which region of the electromagnetic spectrum does each wavelength belong? How much energy is contained in each of the following? To which region of the electromagnetic spectrum does each wavelength belong? A 6.023 x 10 photons are found to have an energy of 225 kJ. What is the wavelength of the radiation? Use the data in Table 2.1.1 to calculate how much more energetic a single gamma-ray photon is than a radio-wave photon. How many photons from a radio source operating at a frequency of 8 × 10 Hz would be required to provide the same amount of energy as a single gamma-ray photon with a frequency of 3 × 10 Hz? Use the data in Table 2.1.1 to calculate how much more energetic a single x-ray photon is than a photon of ultraviolet light. A radio station has a transmitter that broadcasts at a frequency of 100.7 MHz with a power output of 50 kW. Given that 1 W = 1 J/s, how many photons are emitted by the transmitter each second? 532 nm Is the spectrum of the light emitted by isolated atoms of an element discrete or continuous? How do these spectra differ from those obtained by heating a bulk sample of a solid element? Explain your answers. Explain why each element has a characteristic emission and absorption spectra. If spectral emissions had been found to be continuous rather than discrete, what would have been the implications for Bohr’s model of the atom? Explain the differences between a ground state and an excited state. Describe what happens in the spectrum of a species when an electron moves from a ground state to an excited state. What happens in the spectrum when the electron falls from an excited state to a ground state? What phenomenon causes a neon sign to have a characteristic color? If the emission spectrum of an element is constant, why do some neon signs have more than one color? How is light from a laser different from the light emitted by a light source such as a light bulb? Describe how a laser produces light. Using a Bohr model and the transition from = 2 to = 3 in an atom with a single electron, describe the mathematical relationship between an emission spectrum and an absorption spectrum. What is the energy of this transition? What does the sign of the energy value represent in this case? What range of light is associated with this transition? If a hydrogen atom is excited from an = 1 state to an = 3 state, how much energy does this correspond to? Is this an absorption or an emission? What is the wavelength of the photon involved in this process? To what region of the electromagnetic spectrum does this correspond? The hydrogen atom emits a photon with a 486 nm wavelength, corresponding to an electron decaying from the = 4 level to which level? What is the color of the emission? An electron in a hydrogen atom can decay from the = 3 level to = 2 level. What is the color of the emitted light? What is the energy of this transition? Calculate the wavelength and energy of the photon that gives rise to the third line in order of increasing energy in the Lyman series in the emission spectrum of hydrogen. In what region of the spectrum does this wavelength occur? Describe qualitatively what the absorption spectrum looks like. The wavelength of one of the lines in the Lyman series of hydrogen is 121 nm. In what region of the spectrum does this occur? To which electronic transition does this correspond? The emission spectrum of helium is shown. Estimate what change in energy (Δ ) gives rise to each line? Removing an electron from solid potassium requires 222 kJ/mol. Would you expect to observe a photoelectric effect for potassium using a photon of blue light (λ = 485 nm)? What is the longest wavelength of energy capable of ejecting an electron from potassium? What is the corresponding color of light of this wavelength? The binding energy of an electron is the energy needed to remove an electron from its lowest energy state. According to Bohr’s postulates, calculate the binding energy of an electron in a hydrogen atom. There are 6.02 x 10 atoms in 1g of hydrogen atoms What wavelength in nanometers is required to remove such an electron from one hydrogen atom? As a radio astronomer, you have observed spectral lines for hydrogen corresponding to a state with = 320, and you would like to produce these lines in the laboratory. Is this feasible? Why or why not? 656 nm; red light = 2, blue-green light 97.2 nm, 2.04 × 10 J/photon, ultraviolet light, absorption spectrum is a single dark line at a wavelength of 97.2 nm Violet: 390 nm, 307 kJ/mol photons; Blue-purple: 440 nm, 272 kJ/mol photons; Blue-green: 500 nm, 239 kJ/mol photons; Orange: 580 nm, 206 kJ/mol photons; Red: 650 nm, 184 kJ/mol photons 1313 kJ/mol, λ ≤ 91.1 nm Explain what is meant by each term and illustrate with a sketch: How does Einstein’s theory of relativity illustrate the wave–particle duality of light? What properties of light can be explained by a wave model? What properties can be explained by a particle model? In the modern theory of the electronic structure of the atom, which of de Broglie’s ideas have been retained? Which proved to be incorrect? According to Bohr, what is the relationship between an atomic orbit and the energy of an electron in that orbit? Is Bohr’s model of the atom consistent with Heisenberg’s uncertainty principle? Explain your answer. The development of ideas frequently builds on the work of predecessors. Complete the following chart by filling in the names of those responsible for each theory shown. How much heat is generated by shining a carbon dioxide laser with a wavelength of 1.065 μm on a 68.95 kg sample of water if 1.000 mol of photons is absorbed and converted to heat? Is this enough heat to raise the temperature of the water by 4°C? Show the mathematical relationship between energy and mass and between wavelength and mass. What is the effect of doubling the What is the de Broglie wavelength of a 39 g bullet traveling at 1020 m/s ± 10 m/s? What is the minimum uncertainty in the bullet’s position? What is the de Broglie wavelength of a 6800 tn aircraft carrier traveling at 18 ± 0.1 knots (1 knot = 1.15 mi/h)? What is the minimum uncertainty in its position? Calculate the mass of a particle if it is traveling at 2.2 × 10 m/s and has a frequency of 6.67 × 10 Hz. If the uncertainty in the velocity is known to be 0.1%, what is the minimum uncertainty in the position of the particle? Determine the wavelength of a 2800 lb automobile traveling at 80 mi/h ± 3%. How does this compare with the diameter of the nucleus of an atom? You are standing 3 in. from the edge of the highway. What is the minimum uncertainty in the position of the automobile in inches? = 112.3 kJ, Δ = 0.3893°C, over ten times more light is needed for a 4.0°C increase in temperature 1.7 × 10 m, uncertainty in position is ≥ 1.4 × 10 m 9.1 × 10 kg, uncertainty in position ≥ 2.6 m Why does an electron in an orbital with = 1 in a hydrogen atom have lower energy than a free electron ( = ∞)? What four variables are required to fully describe the position of any object in space? In quantum mechanics, one of these variables is not explicitly considered. Which one and why? Chemists generally refer to the square of the wave function rather than to the wave function itself. Why? Orbital energies of species with only one electron are defined by only one quantum number. Which one? In such a species, is the energy of an orbital with = 2 greater than, less than, or equal to the energy of an orbital with = 4? Justify your answer. In each pair of subshells for a hydrogen atom, which has the higher energy? Give the principal and the azimuthal quantum number for each pair. What is the relationship between the energy of an orbital and its average radius? If an electron made a transition from an orbital with an average radius of 846.4 pm to an orbital with an average radius of 476.1 pm, would an emission spectrum or an absorption spectrum be produced? Why? In making a transition from an orbital with a principal quantum number of 4 to an orbital with a principal quantum number of 7, does the electron of a hydrogen atom emit or absorb a photon of energy? What would be the energy of the photon? To what region of the electromagnetic spectrum does this energy correspond? What quantum number defines each of the following? In an attempt to explain the properties of the elements, Niels Bohr initially proposed electronic structures for several elements with orbits holding a certain number of electrons, some of which are in the following table: How many subshells are possible for = 3? What are they? How many subshells are possible for = 5? What are they? What value of corresponds to a subshell? How many orbitals are in this subshell? What value of corresponds to an subshell? How many orbitals are in this subshell? State the number of orbitals and electrons that can occupy each subshell. State the number of orbitals and electrons that can occupy each subshell. How many orbitals and subshells are found within the principal shell = 6? How do these orbital energies compare with those for = 4? How many nodes would you expect a 4 orbital to have? A 5 orbital? A orbital is found to have one node in addition to the nodal plane that bisects the lobes. What would you predict to be the value of ? If an orbital has two nodes, what is the value of ? Three subshells, with = 0 ( ), = 1 ( ), and = 2 ( ). A subshell has = 2 and contains 5 orbitals. A principal shell with = 6 contains six subshells, with = 0, 1, 2, 3, 4, and 5, respectively. These subshells contain 1, 3, 5, 7, 9, and 11 orbitals, respectively, for a total of 36 orbitals. The energies of the orbitals with = 6 are higher than those of the corresponding orbitals with the same value of for = 4. A set of four quantum numbers specifies each wave function. What information is given by each quantum number? What does the specified wave function describe? List two pieces of evidence to support the statement that electrons have a spin. The periodic table is divided into blocks. Identify each block and explain the principle behind the divisions. Which quantum number distinguishes the horizontal rows? Identify the element with each ground state electron configuration. Identify the element with each ground state electron configuration. Propose an explanation as to why the noble gases are inert. How many magnetic quantum numbers are possible for a 4 subshell? A 3 subshell? How many orbitals are in these subshells? How many magnetic quantum numbers are possible for a 6 subshell? A 4 subshell? How many orbitals does each subshell contain? If = 2 and = 2, give all the allowed combinations of the four quantum numbers ( , , , ) for electrons in the corresponding 3 subshell. Give all the allowed combinations of the four quantum numbers ( , , , ) for electrons in a 4 subshell. How many electrons can the 4 orbital accommodate? How would this differ from a situation in which there were only three quantum numbers ( , , )? Given the following sets of quantum numbers ( , , , ), identify each principal shell and subshell. Is each set of quantum numbers allowed? Explain your answers. List the set of quantum numbers for each electron in the valence shell of each element. List the set of quantum numbers for each electron in the valence shell of each element. Sketch the shape of the periodic table if there were three possible values of for each electron (+½, −½, and 0); assume that the Pauli principle is still valid. Predict the shape of the periodic table if eight electrons could occupy the subshell. If the electron could only have spin +½, what would the periodic table look like? If three electrons could occupy each orbital, what would be the electron configuration of each species? If Hund’s rule were not followed and maximum pairing occurred, how many unpaired electrons would each species have? How do these numbers compare with the number found using Hund’s rule? Write the electron configuration for each element in the ground state. Write the electron configuration for each element in the ground state. Give the complete electron configuration for each element. Give the complete electron configuration for each element. Write the valence electron configuration for each element: Using the Pauli exclusion principle and Hund’s rule, draw valence orbital diagrams for each element. Using the Pauli exclusion principle and Hund’s rule, draw valence orbital diagrams for each element. How many unpaired electrons does each species contain? How many unpaired electrons does each species contain? For each element, give the complete electron configuration, draw the valence electron configuration, and give the number of unpaired electrons present. Use an orbital diagram to illustrate the aufbau principle, the Pauli exclusion principle, and Hund’s rule for each element. For a 4 subshell, = 4 and = 1. The allowed values of the magnetic quantum number, , are therefore +1, 0, −1, corresponding to three 4 orbitals. For a 3 subshell, = 3 and = 2. The allowed values of the magnetic quantum number, , are therefore +2, +1, 0, −1, −2, corresponding to five 3 orbitals. | 15,428 | 3,775 |
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In potentiometry we measure the potential of an electrochemical cell under static conditions. Because no current—or only a negligible current—flows through the electrochemical cell, its composition remains unchanged. For this reason, potentiometry is a useful quantitative method of analysis. The first quantitative potentiometric applications appeared soon after the formulation, in 1889, of the Nernst equation, which relates an electrochemical cell’s potential to the concentration of electroactive species in the cell [Stork, J. T. , , 344A–351A]. For an on-line introduction to much of the material in this section, see Analytical Electrochemistry: Potentiometry by Erin Gross, Richard S. Kelly, and Donald M. Cannon, Jr., a resource that is part of the Analytical Sciences Digital Library. Potentiometry initially was restricted to redox equilibria at metallic electrodes, which limited its application to a few ions. In 1906, Cremer discovered that the potential difference across a thin glass membrane is a function of pH when opposite sides of the membrane are in contact with solutions that have different concentrations of H O . This discovery led to the development of the glass pH electrode in 1909. Other types of membranes also yield useful potentials. For example, in 1937 Kolthoff and Sanders showed that a pellet of AgCl can be used to determine the concentration of Ag . Electrodes based on membrane potentials are called ion-selective electrodes, and their continued development extends potentiometry to a diverse array of analytes. As shown in , we use a potentiometer to determine the difference between the potential of two electrodes. The potential of one electrode—the working or indicator electrode—responds to the analyte’s activity and the other electrode—the counter or reference electrode—has a known, fixed potential. In this section we introduce the conventions for describing potentiometric electrochemical cells, and the relationship between the measured potential and the analyte’s activity. In we noted that a chemical reaction’s equilibrium position is a function of the activities of the reactants and products, not their concentrations. To be correct, we should write the Nernst equation in terms of activities. So why didn’t we use activities in when we calculated redox titration curves? There are two reasons for that choice. First, concentrations are always easier to calculate than activities. Second, in a redox titration we determine the analyte’s concentration from the titration’s end point, not from the potential at the end point. The only reasons for calculating a titration curve is to evaluate its feasibility and to help us select a useful indicator. In most cases, the error we introduce by assuming that concentration and activity are identical is too small to be a significant concern. In potentiometry we cannot ignore the difference between activity and concentration. Later in this section we will consider how we can design a potentiometric method so that we can ignore the difference between activity and concentration. See to review our earlier discussion of activity and concentration. A schematic diagram of a typical potentiometric electrochemical cell is shown in Figure 11.2.1
. The electrochemical cell consists of two half-cells, each of which contains an electrode immersed in a solution of ions whose activities determine the electrode’s potential. A that contains an inert electrolyte, such as KCl, connects the two half-cells. The ends of the salt bridge are fixed with porous frits, which allow the electrolyte’s ions to move freely between the half-cells and the salt bridge. This movement of ions in the salt bridge completes the electrical circuit. By convention, we identify the electrode on the left as the and assign to it the oxidation reaction; thus \[\mathrm{Zn}(s) \rightleftharpoons \text{ Zn}^{2+}(a q)+2 e^{-} \nonumber\] The electrode on the right is the , where the reduction reaction occurs. \[\mathrm{Ag}^{+}(a q)+e^{-} \rightleftharpoons \mathrm{Ag}(s) \nonumber\] The potential of the electrochemical cell in Figure 11.2.1
is for the reaction \[\mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \rightleftharpoons 2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(\mathrm{aq}) \nonumber\] We also define potentiometric electrochemical cells such that the cathode is the indicator electrode and the anode is the reference electrode. The reason for separating the electrodes is to prevent the oxidation reaction and the reduction reaction from occurring at one of the electrodes. For example, if we place a strip of Zn metal in a solution of AgNO , the reduction of Ag to Ag occurs on the surface of the Zn at the same time as a potion of the Zn metal oxidizes to Zn . Because the transfer of electrons from Zn to Ag occurs at the electrode’s surface, we can not pass them through the potentiometer. Although Figure 11.2.1
provides a useful picture of an electrochemical cell, it is not a convenient way to represent it (Imagine having to draw a picture of each electrochemical cell you are using!). A more useful way to describe an electrochemical cell is a shorthand notation that uses symbols to identify different phases and that lists the composition of each phase. We use a vertical slash (|) to identify a boundary between two phases where a potential develops, and a comma (,) to separate species in the same phase or to identify a boundary between two phases where no potential develops. Shorthand cell notations begin with the anode and continue to the cathode. For example, we describe the electrochemical cell in Figure 11.2.1
using the following shorthand notation. \[\text{Zn}(s) | \text{ZnCl}_2(aq, a_{\text{Zn}^{2+}} = 0.0167) || \text{AgNO}_3(aq, a_{\text{Ag}^+} = 0.100) | \text{Ag} (s) \nonumber\] The double vertical slash (||) represents the salt bridge, the contents of which we usually do not list. Note that a double vertical slash implies that there is a potential difference between the salt bridge and each half-cell. What are the anodic, the cathodic, and the overall reactions responsible for the potential of the electrochemical cell in Figure 11.2.2
? Write the shorthand notation for the electrochemical cell. The oxidation of Ag to Ag occurs at the anode, which is the left half-cell. Because the solution contains a source of Cl , the anodic reaction is \[\mathrm{Ag}(s)+\mathrm{Cl}^{-}(aq) \rightleftharpoons\text{ AgCl}(s)+e^{-} \nonumber\] The cathodic reaction, which is the right half-cell, is the reduction of Fe to Fe . \[\mathrm{Fe}^{3+}(a q)+e^{-}\rightleftharpoons \text{ Fe}^{2+}(a q) \nonumber\] The overall cell reaction, therefore, is \[\mathrm{Ag}(s)+\text{ Fe}^{3+}(a q)+\text{ Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s)+\text{ Fe}^{2+}(a q) \nonumber\] The electrochemical cell’s shorthand notation is \[\text{Ag}(s) | \text{HCl} (aq, a_{\text{Cl}^{-}} = 0.100), \text{AgCl} (\text{sat’d}) || \text{FeCl}_2(aq, a_{\text{Fe}^{2+}} = 0.0100), \text{ Fe}^{3+}(aq,a_{\text{Fe}^{3+}} = 0.0500) | \text{Pt} (s) \nonumber\] Note that the Pt cathode is an inert electrode that carries electrons to the reduction half-reaction. The electrode itself does not undergo reduction. Write the reactions occurring at the anode and the cathode for the potentiometric electrochemical cell with the following shorthand notation. Pt( ) | H ( ), H ( ) || Cu ( ) | Cu( ) The oxidation of H to H occurs at the anode \[\mathrm{H}_{2}(g)\rightleftharpoons2 \mathrm{H}^{+}(a q)+2 e^{-} \nonumber\] and the reduction of Cu to Cu occurs at the cathode. \[\mathrm{Cu}^{2+}(a q)+2 e^{-}\rightleftharpoons\mathrm{Cu}(s) \nonumber\] The overall cell reaction, therefore, is \[\mathrm{Cu}^{2+}(a q)+\text{ H}_{2}(g)\rightleftharpoons2 \mathrm{H}^{+}(a q)+\mathrm{Cu}(s) \nonumber\] The potential of a potentiometric electrochemical cell is \[E_{\text {cell }}=E_{\text {cathode }}-E_{\text {anode }} \label{11.1}\] where and are reduction potentials for the redox reactions at the cathode and the anode, respectively. Each reduction potential is given by the Nernst equation \[E=E^{\circ}-\frac{R T}{n F} \ln Q \nonumber\] where is the standard-state reduction potential, is the gas constant, is the temperature in Kelvins, is the number of electrons in the redox reaction, is Faraday’s constant, and is the reaction quotient. At a temperature of 298 K (25 C) the Nernst equation is \[E=E^{\circ}-\frac{0.05916}{n} \log Q \label{11.2}\] where is in volts. Using Equation \ref{11.2}, the potential of the anode and cathode in Figure 11.2.1
are \[E_\text{anode} = E_{\text{Zn}^{2+}/\text{Zn}}^{\circ} - \frac {0.05916} {2} \log \frac{1} {a_{\text{Zn}^{2+}}} \nonumber\] \[E_\text{anode} = E_{\text{Ag}^{+}/\text{Ag}}^{\circ} - \frac {0.05916} {1} \log \frac{1} {a_{\text{Ag}^{+}}} \nonumber\] Even though an oxidation reaction is taking place at the anode, we define the anode's potential in terms of the corresponding reduction reaction and the standard-state reduction potential. See for a review of using the Nernst equation in calculations. Substituting and into Equation \ref{11.1}, along with the activities of Zn and Ag and the standard-state reduction potentials, gives as \[E_\text{cell} = \left( E_{\text{Ag}^{+}/\text{Ag}}^{\circ} - \frac {0.05916} {1} \log \frac{1} {a_{\text{Ag}^{+}}} \right) - \left( E_{\text{Zn}^{2+}/\text{Zn}}^{\circ} - \frac {0.05916} {2} \log \frac{1} {a_{\text{Zn}^{2+}}} \right) \nonumber\] \[E_\text{cell} = \left( 0.7996 - \frac {0.05916} {1} \log \frac{1} {0.100} \right) - \left( -0.7618 - \frac {0.05916} {2} \log \frac{1} {0.0167} \right) = 1.555 \text{ V} \nonumber\] You will find values for the standard-state reduction potentials in . What is the potential of the electrochemical cell shown in ? Substituting and into Equation \ref{11.1}, along with the concentrations of Fe , Fe , and Cl and the standard-state reduction potentials gives \[E_\text{cell} = \left( E_{\text{Fe}^{3+}/\text{Fe}^{2+}}^{\circ} - \frac {0.05916} {1} \log \frac{a_{\text{Fe}^{2+}}} {a_{\text{Fe}^{3+}}} \right) - \left( E_{\text{AgCl/Ag}}^{\circ} - \frac {0.05916} {1} \log a_{\text{Cl}^-} \right) \nonumber\] \[E_\text{cell} = \left( 0.771 - \frac {0.05916} {1} \log \frac{0.0100} {0.0500} \right) - \left( 0.2223 - \frac {0.05916} {1} \log (0.100) \right) = 0.531 \text{ V} \nonumber\] What is the potential for the electrochemical cell in if the activity of H in the anodic half-cell is 0.100, the fugacity of H in the anodic half-cell is 0.500, and the activity of Cu in the cathodic half-cell is 0.0500? Fugacity, \(f\), is the equivalent term for the activity of a gas. Making appropriate substitutions into Equation \ref{11.1} and solving for gives its value as \[E_\text{cell} = \left( E_{\text{Cu}^{2+}/\text{Cu}}^{\circ} - \frac {0.05916} {2} \log \frac{1} {a_{\text{Cu}^{2+}}} \right) - \left( E_{\text{H}^{+}/\text{H}_2}^{\circ} - \frac {0.05916} {2} \log \frac{f_{\text{H}_2}} {a_{\text{H}^+}^2} \right) \nonumber\] \[E_\text{cell} = \left( 0.3419 - \frac {0.05916} {2} \log \frac{1} {0.0500} \right) - \left( 0.0000 - \frac {0.05916} {2} \log \frac{0.500} {(0.100)^2} \right) = 0.3537 \text{ V} \nonumber\] In potentiometry, we assign the reference electrode to the anodic half-cell and assign the indicator electrode to the cathodic half-cell. Thus, if the potential of the cell in is +1.50 V and the activity of Zn is 0.0167, then we can solve the following equation for \[1.50 \text{ V} = \left( 0.7996 - \frac {0.05916} {1} \log \frac{1} {a_{\text{Ag}^+}} \right) - \left( -0.7618 - \frac {0.05916} {2} \log \frac{1} {0.0167} \right) \nonumber\] obtaining an activity of 0.0118. What is the activity of Fe in an electrochemical cell similar to that in if the activity of Cl in the left-hand cell is 1.0, the activity of Fe in the right-hand cell is 0.015, and is +0.546 V? Making appropriate substitutions into Equation \ref{11.1} \[0.546 = \left( 0.771 - \frac {0.05916} {1} \log \frac{0.0100} {a_{\text{Fe}^{3+}}} \right) - \left( 0.2223 - \frac {0.05916} {1} \log (1.0) \right) \nonumber\] and solving for gives its activity as 0.0135. What is the activity of Cu in the electrochemical cell in if the activity of H in the anodic half-cell is 1.00 with a fugacity of 1.00 for H , and an of +0.257 V? Making appropriate substitutions into Equation \ref{11.1} \[0.257 \text{ V} = \left( 0.3419 - \frac {0.05916} {2} \log \frac{1} {a_{\text{Cu}^{2+}}} \right) - \left( 0.0000 - \frac {0.05916} {2} \log \frac{1.00} {(1.00)^2} \right) \nonumber\] and solving for gives its activity as \(1.35 \times 10^{-3}\). Despite the apparent ease of determining an analyte’s activity using the Nernst equation, there are several problems with this approach. One problem is that standard-state potentials are temperature-dependent and the values in reference tables usually are for a temperature of 25 C. We can overcome this problem by maintaining the electrochemical cell at 25 C or by measuring the standard-state potential at the desired temperature. Another problem is that a standard-sate reduction potential may have a significant matrix effect. For example, the standard-state reduction potential for the Fe /Fe redox couple is +0.735 V in 1 M HClO , +0.70 V in 1 M HCl, and +0.53 V in 10 M HCl. The difference in potential for equimolar solutions of HCl and HClO is the result of a difference in the activity coefficients for Fe and Fe in these two media. The shift toward a more negative potential with an increase in the concentration of HCl is the result of chloride’s ability to form a stronger complex with Fe than with Fe . We can minimize this problem by replacing the standard-state potential with a matrix-dependent formal potential. Most tables of standard-state potentials, including those in , include selected formal potentials. Finally, a more serious problem is the presence of additional potentials in the electrochemical cell not included in Equation \ref{11.1}. In writing the shorthand notation for an electrochemical cell we use a double slash (||) to indicate the salt bridge, suggesting a potential exists at the interface between each end of the salt bridge and the solution in which it is immersed. The origin of this potential is discussed in the following section. A develops at the interface between two ionic solution if there is a difference in the concentration and mobility of the ions. Consider, for example, a porous membrane that separations a solution of 0.1 M HCl from a solution of 0.01 M HCl (Figure 11.2.3
a). Because the concentration of HCl on the membrane’s left side is greater than that on the right side of the membrane, H and Cl will diffuse in the direction of the arrows. The mobility of H , however, is greater than that for Cl , as shown by the difference in the lengths of their respective arrows. Because of this difference in mobility, the solution on the right side of the membrane develops an excess concentration of H and a positive charge (Figure 11.2.3
b). Simultaneously, the solution on the membrane’s left side develops a negative charge because there is an excess concentration of Cl . We call this difference in potential across the membrane a junction potential and represent it as . The magnitude of the junction potential depends upon the difference in the concentration of ions on the two sides of the interface, and may be as large as 30–40 mV. For example, a junction potential of 33.09 mV has been measured at the interface between solutions of 0.1 M HCl and 0.1 M NaCl [Sawyer, D. T.; Roberts, J. L., Jr. , Wiley-Interscience: New York, 1974, p. 22]. A salt bridge’s junction potential is minimized by using a salt, such as KCl, for which the mobilities of the cation and anion are approximately equal. We also can minimize the junction potential by incorporating a high concentration of the salt in the salt bridge. For this reason salt bridges frequently are constructed using solutions that are saturated with KCl. Nevertheless, a small junction potential, generally of unknown magnitude, is always present. When we measure the potential of an electrochemical cell, the junction potential also contributes to ; thus, we rewrite Equation \ref{11.1} as \[E_{\text {cell }}=E_{\text {cathode }}-E_{\text {anode }}+E_{j} \nonumber\] to include its contribution. If we do not know the junction potential’s actual value—which is the usual situation—then we cannot directly calculate the analyte’s concentration using the Nernst equation. Quantitative analytical work is possible, however, if we use one of the standardization methods—external standards, the method of standard additions, or internal standards—discussed in . In a potentiometric electrochemical cell one of the two half-cells provides a fixed reference potential and the potential of the other half-cell responds the analyte’s concentration. By convention, the reference electrode is the anode; thus, the short hand notation for a potentiometric electrochemical cell is reference electrode || indicator electrode and the cell potential is \[E_{\mathrm{cell}}=E_{\mathrm{ind}}-E_{\mathrm{ref}}+E_{j} \nonumber\] The ideal reference electrode provides a stable, known potential so that we can attribute any change in to the analyte’s effect on the indicator electrode’s potential. In addition, it should be easy to make and to use the reference electrode. Three common reference electrodes are discussed in this section. Although we rarely use the (SHE) for routine analytical work, it is the reference electrode used to establish standard-state potentials for other half-reactions. The SHE consists of a Pt electrode immersed in a solution in which the activity of hydrogen ion is 1.00 and in which the fugacity of H is 1.00 (Figure 11.2.4
). A conventional salt bridge connects the SHE to the indicator half-cell. The short hand notation for the standard hydrogen electrode is \[\text{Pt}(s), \text{ H}_{2}\left(g, f_{\mathrm{H}_{2}}=1.00\right) | \text{ H}^{+}\left(a q, a_{\mathrm{H}^{+}}=1.00\right) \| \nonumber\] and the standard-state potential for the reaction \[\mathrm{H}^{+}(a q)+e^{-}=\frac{1}{2} \mathrm{H}_{2}(g) \nonumber\] is, by definition, 0.00 V at all temperatures. Despite its importance as the fundamental reference electrode against which we measure all other potentials, the SHE is rarely used because it is difficult to prepare and inconvenient to use. A calomel reference electrode is based on the following redox couple between Hg Cl and Hg (calomel is the common name for Hg Cl ) \[\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+2 e^{-}\rightleftharpoons2 \mathrm{Hg}(l)+2 \mathrm{Cl}^{-}(a q) \nonumber\] for which the potential is \[E=E_{\mathrm{Hg}_{2} \mathrm{Cl}_{2} / \mathrm{Hg}}^{\mathrm{o}}-\frac{0.05916}{2} \log \left(a_{\text{Cl}^-}\right)^{2}=+0.2682 \mathrm{V}-\frac{0.05916}{2} \log \left(a_{\text{Cl}^-}\right)^{2} \nonumber\] The potential of a calomel electrode, therefore, depends on the activity of Cl in equilibrium with Hg and Hg Cl . As shown in Figure 11.2.5
, in a (SCE) the concentration of Cl is determined by the solubility of KCl. The electrode consists of an inner tube packed with a paste of Hg, Hg Cl , and KCl, situated within a second tube that contains a saturated solution of KCl. A small hole connects the two tubes and a porous wick serves as a salt bridge to the solution in which the SCE is immersed. A stopper in the outer tube provides an opening for adding addition saturated KCl. The short hand notation for this cell is \[\mathrm{Hg}(l) | \mathrm{Hg}_{2} \mathrm{Cl}_{2}(s), \mathrm{KCl}(a q, \text { sat'd }) \| \nonumber\] Because the concentration of Cl is fixed by the solubility of KCl, the potential of an SCE remains constant even if we lose some of the inner solution to evaporation. A significant disadvantage of the SCE is that the solubility of KCl is sensitive to a change in temperature. At higher temperatures the solubility of KCl increases and the electrode’s potential decreases. For example, the potential of the SCE is +0.2444 V at 25 C and +0.2376 V at 35 C. The potential of a calomel electrode that contains an unsaturated solution of KCl is less dependent on the temperature, but its potential changes if the concentration, and thus the activity of Cl , increases due to evaporation. For example, the potential of a calomel electrode is +0.280 V when the concentration of KCl is 1.00 M and +0.336 V when the concentration of KCl is 0.100 M. If the activity of Cl is 1.00, the potential is +0.2682 V. Another common reference electrode is the , which is based on the reduction of AgCl to Ag. \[\operatorname{AgCl}(s)+e^{-} \rightleftharpoons \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) \nonumber\] As is the case for the calomel electrode, the activity of Cl determines the potential of the Ag/AgCl electrode; thus \[E = E_\text{AgCl/Ag}^{\circ}-0.05916 \log a_{\text{Cl}^-} = 0.2223 \text{ V} - 0.05916 \log a_{\text{Cl}^-} \nonumber\] When prepared using a saturated solution of KCl, the electrode's potential is +0.197 V at 25 C. Another common Ag/AgCl electrode uses a solution of 3.5 M KCl and has a potential of +0.205 V at 25 C. As you might expect, the potential of a Ag/AgCl electrode using a saturated solution of KCl is more sensitive to a change in temperature than an electrode that uses an unsaturated solution of KCl. A typical Ag/AgCl electrode is shown in Figure 11.2.6
and consists of a silver wire, the end of which is coated with a thin film of AgCl, immersed in a solution that contains the desired concentration of KCl. A porous plug serves as the salt bridge. The electrode’s short hand notation is \[\operatorname{Ag}(s) | \operatorname{Ag} \mathrm{Cl}(s), \mathrm{KCl}\left(a q, a_{\mathrm{Cl}^{-}}=x\right) \| \nonumber\] The standard state reduction potentials in most tables are reported relative to the standard hydrogen electrode’s potential of +0.00 V. Because we rarely use the SHE as a reference electrode, we need to convert an indicator electrode’s potential to its equivalent value when using a different reference electrode. As shown in the following example, this is easy to do. The potential for an Fe /Fe half-cell is +0.750 V relative to the standard hydrogen electrode. What is its potential if we use a saturated calomel electrode or a saturated silver/silver chloride electrode? When we use a standard hydrogen electrode the potential of the electrochemical cell is \[E_\text{cell} = E_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E_\text{SHE} = 0.750 \text{ V} -0.000 \text{ V} = 0.750 \text{ V} \nonumber\] We can use the same equation to calculate the potential if we use a saturated calomel electrode \[E_\text{cell} = E_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E_\text{SHE} = 0.750 \text{ V} -0.2444 \text{ V} = 0.506 \text{ V} \nonumber\] or a saturated silver/silver chloride electrode \[E_\text{cell} = E_{\text{Fe}^{3+}/\text{Fe}^{2+}} - E_\text{SHE} = 0.750 \text{ V} -0.197 \text{ V} = 0.553 \text{ V} \nonumber\] Figure 11.2.7
provides a pictorial representation of the relationship between these different potentials. The potential of a \(\text{UO}_2^+\)/U half-cell is –0.0190 V relative to a saturated calomel electrode. What is its potential when using a saturated silver/silver chloride electrode or a standard hydrogen electrode? When using a saturated calomel electrode, the potential of the electro- chemical cell is \[E_\text{cell} = E_{\text{UO}_2^+/\text{U}^{4+}} - E_\text{SCE} \nonumber\] Substituting in known values \[-0.0190 \text{ V} = E_{\text{UO}_2^+/\text{U}^{4+}} - 0.2444 \text{ V} \nonumber\] and solving for \(E_{\text{UO}_2^+/\text{U}^{4+}}\) gives its value as +0.2254 V. The potential relative to the Ag/AgCl electrode is \[E_\text{cell} = E_{\text{UO}_2^+/\text{U}^{4+}} - E_\text{Ag/AgCl} = 0.2254 \text{ V} - 0.197 \text{ V} = 0.028 \text{ V} \nonumber\] and the potential relative to the standard hydrogen electrode is \[E_\text{cell} = E_{\text{UO}_2^+/\text{U}^{4+}} - E_\text{SHE} = 0.2254 \text{ V} - 0.000 \text{ V} = 0.2254 \text{ V} \nonumber\] In potentiometry, the potential of the indicator electrode is proportional to the analyte’s activity. Two classes of indicator electrodes are used to make potentiometric measurements: metallic electrodes, which are the subject of this section, and ion-selective electrodes, which are covered in the next section. If we place a copper electrode in a solution that contains Cu , the electrode’s potential due to the reaction \[\mathrm{Cu}^{2+}(a q)+2 e^{-} \rightleftharpoons \mathrm{Cu}(s) \nonumber\] is determined by the activity of Cu . \[E=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\mathrm{o}}-\frac{0.05916}{2} \log \frac{1}{a_{\mathrm{Cu}^{2+}}}=+0.3419 \mathrm{V}-\frac{0.05916}{2} \log \frac{1}{a_{\mathrm{Cu}^{2+}}} \nonumber\] If copper is the indicator electrode in a potentiometric electrochemical cell that also includes a saturated calomel reference electrode \[\mathrm{SCE} \| \mathrm{Cu}^{2+}\left(a q, a_{\mathrm{Cu^{2+}}}=x\right) | \text{Cu}(s) \nonumber\] then we can use the cell potential to determine an unknown activity of Cu in the indicator electrode’s half-cell \[E_{\text{cell}}= E_{\text { ind }}-E_{\text {SCE }}+E_{j}= +0.3419 \mathrm{V}-\frac{0.05916}{2} \log \frac{1}{a_{\mathrm{Cu}^{2+}}}-0.2224 \mathrm{V}+E_{j} \nonumber\] An indicator electrode in which the metal is in contact with a solution containing its ion is called an . In general, if a metal, , is in a solution of , the cell potential is \[E_{\mathrm{call}}=K-\frac{0.05916}{n} \log \frac{1}{a_{M^{n+}}}=K+\frac{0.05916}{n} \log a_{M^{n+}} \nonumber\] where is a constant that includes the standard-state potential for the / redox couple, the potential of the reference electrode, and the junction potential. Note that including in the constant means we do not need to know the junction potential’s actual value; however, the junction potential must remain constant if is to maintain a constant value. For a variety of reasons—including the slow kinetics of electron transfer at the metal–solution interface, the formation of metal oxides on the electrode’s surface, and interfering reactions—electrodes of the first kind are limited to the following metals: Ag, Bi, Cd, Cu, Hg, Pb, Sn, Tl, and Zn. Many of these electrodes, such as Zn, cannot be used in acidic solutions because they are easily oxidized by H . \[\mathrm{Zn}(s)+2 \mathrm{H}^{+}(a q)\rightleftharpoons \text{ H}_{2}(g)+\mathrm{Zn}^{2+}(a q) \nonumber\] The potential of an electrode of the first kind responds to the activity of . We also can use this electrode to determine the activity of another species if it is in equilibrium with . For example, the potential of a Ag electrode in a solution of Ag is \[E=0.7996 \mathrm{V}+0.05916 \log a_{\mathrm{Ag}^{+}} \label{11.3}\] If we saturate the indicator electrode’s half-cell with AgI, the solubility reaction \[\operatorname{Agl}(s)\rightleftharpoons\operatorname{Ag}^{+}(a q)+\mathrm{I}^{-}(a q) \nonumber\] determines the concentration of Ag ; thus \[a_{\mathrm{Ag}^{+}}=\frac{K_{\mathrm{sp}, \mathrm{Agl}}}{a_{\text{I}^-}} \label{11.4}\] where , is the solubility product for AgI. Substituting Equation \ref{11.4} into Equation \ref{11.3} \[E=0.7996 \text{ V}+0.05916 \log \frac{K_{\text{sp, Agl}}}{a_{\text{I}^-}} \nonumber\] shows that the potential of the silver electrode is a function of the activity of I . If we incorporate this electrode into a potentiometric electrochemical cell with a saturated calomel electrode \[\mathrm{SCE} \| \mathrm{AgI}(s), \text{ I}^-\left(a q, a_{\text{I}^-}=x\right) | \mathrm{Ag}(\mathrm{s}) \nonumber\] then the cell potential is \[E_{\mathrm{cell}}=K-0.05916 \log a_{\text{I}^-} \nonumber\] where is a constant that includes the standard-state potential for the Ag /Ag redox couple, the solubility product for AgI, the reference electrode’s potential, and the junction potential. If an electrode of the first kind responds to the activity of an ion in equilibrium with , we call it an . Two common electrodes of the second kind are the calomel and the silver/silver chloride reference electrodes. In an electrode of the second kind we link together a redox reaction and another reaction, such as a solubility reaction. You might wonder if we can link together more than two reactions. The short answer is yes. An electrode of the third kind, for example, links together a redox reaction and two other reactions. Such electrodes are less common and we will not consider them in this text. An electrode of the first kind or second kind develops a potential as the result of a redox reaction that involves the metallic electrode. An electrode also can serve as a source of electrons or as a sink for electrons in an unrelated redox reaction, in which case we call it a The Pt cathode in and is a redox electrode because its potential is determined by the activity of Fe and Fe in the indicator half-cell. Note that a redox electrode’s potential often responds to the activity of more than one ion, which limits its usefulness for direct potentiometry. If metals were the only useful materials for constructing indicator electrodes, then there would be few useful applications of potentiometry. In 1906, Cremer discovered that the potential difference across a thin glass membrane is a function of pH when opposite sides of the membrane are in contact with solutions that have different concentrations of H O . The existence of this led to the development of a whole new class of indicator electrodes, which we call (ISEs). In addition to the glass pH electrode, ion-selective electrodes are available for a wide range of ions. It also is possible to construct a membrane electrode for a neutral analyte by using a chemical reaction to generate an ion that is monitored with an ion-selective electrode. The development of new membrane electrodes continues to be an active area of research. Figure 11.2.8
shows a typical potentiometric electrochemical cell equipped with an ion-selective electrode. The short hand notation for this cell is \[\text { ref (sample) }\left\|[\mathrm{A}]_{\text { samp }}\left(a q, a_{\mathrm{A}}=x\right) |[\mathrm{A}]_{\text { int }}\left(a q, a_{\mathrm{A}}=y\right)\right\| \text { ref (internal) } \nonumber\] where the ion-selective membrane is represented by the vertical slash that separates the two solutions that contain analyte: the sample solution and the ion-selective electrode’s internal solution. The potential of this electrochemical cell includes the potential of each reference electrode, a junction potential, and the membrane’s potential \[E_\text{cell} = E_\text{ref(int)} - E_\text{ref(samp)} + E_\text{mem} + E_j \label{11.5}\] where is the potential across the membrane and The notations ref(sample) and ref(internal) represent a reference electrode immersed in the sample and a reference electrode immersed in the ISE’s internal solution. Because the junction potential and the potential of the two reference electrodes are constant, any change in reflects a change in the membrane’s potential. The analyte’s interaction with the membrane generates a membrane potential if there is a difference in its activity on the membrane’s two sides. Current is carried through the membrane by the movement of either the analyte or an ion already present in the membrane’s matrix. The membrane potential is given by the following Nernst-like equation \[E_{\mathrm{mem}}=E_{\mathrm{asym}}-\frac{R T}{z F} \ln \frac{\left(a_{A}\right)_{\mathrm{int}}}{\left(a_{A}\right)_{\mathrm{samp}}} \label{11.6}\] where ( ) is the analyte’s activity in the sample, ( ) is the analyte’s activity in the ion-selective electrode’s internal solution, and is the analyte’s charge. Ideally, is zero when ( ) = ( ) . The term , which is an , accounts for the fact that usually is not zero under these conditions. For now we simply note that a difference in the analyte’s activity results in a membrane potential. As we consider different types of ion-selective electrodes, we will explore more specifically the source of the membrane potential. Substituting Equation \ref{11.6} into Equation \ref{11.5}, assuming a temperature of 25 C, and rearranging gives \[E_{\mathrm{cell}}=K+\frac{0.05916}{z} \log \left(a_{A}\right)_{\mathrm{samp}} \label{11.7}\] where is a constant that includes the potentials of the two reference electrodes, the junction potentials, the asymmetry potential, and the analyte's activity in the internal solution. Equation \ref{11.7} is a general equation and applies to all types of ion-selective electrodes. A membrane potential results from a chemical interaction between the analyte and active sites on the membrane’s surface. Because the signal depends on a chemical process, most membranes are not selective toward a single analyte. Instead, the membrane potential is proportional to the concentration of each ion that interacts with the membrane’s active sites. We can rewrite Equation \ref{11.7} to include the contribution to the potential of an interferent, \[E_\text{cell} = K + \frac {0.05916} {z_A} \log \left\{ a_A + K_{A,I}(a_I)^{z_A/z_I} \right\} \nonumber\] where and are the charges of the analyte and the interferent, and is a selectivity coefficient that accounts for the relative response of the interferent. The selectivity coefficient is defined as \[K_{A,I} = \frac {(a_A)_e} {(a_I)_e^{z_A/z_I}} \label{11.8}\] where ( ) and ( ) are the activities of analyte and the interferent that yield identical cell potentials. When the selectivity coefficient is 1.00, the membrane responds equally to the analyte and the interferent. A membrane shows good selectivity for the analyte when is significantly less than 1.00. Selectivity coefficients for most commercially available ion-selective electrodes are provided by the manufacturer. If the selectivity coefficient is not known, it is easy to determine its value experimentally by preparing a series of solutions, each of which contains the same activity of interferent, ( ) , but a different activity of analyte. As shown in Figure 11.2.9
, a plot of cell potential versus the log of the analyte’s activity has two distinct linear regions. When the analyte’s activity is significantly larger than \(\times\) ( ) , the potential is a linear function of log( ), as given by Equation \ref{11.7}. If \(\times\) ( ) is significantly larger than the analyte’s activity, however, the cell’s potential remains constant. The activity of analyte and interferent at the intersection of these two linear regions is used to calculate . Sokalski and co-workers described a method for preparing ion-selective electrodes with significantly improved selectivities [Sokalski, T.; Ceresa, A.; Zwicki, T.; Pretsch, E. . , , 11347–11348]. For example, a conventional Pb ISE has a \(\log K_{\text{Pb}^{2+}/\text{Mg}^{2+}}\) of –3.6. If the potential for a solution in which the activity of Pb is \(4.1 \times 10^{-12}\) is identical to that for a solution in which the activity of Mg is 0.01025, what is the value of \(\log K_{\text{Pb}^{2+}/\text{Mg}^{2+}}\) for their ISE? Making appropriate substitutions into Equation \ref{11.8}, we find that \[K_{\text{Pb}^{2+}/\text{Mg}^{2+}} = \frac {(a_{\text{Pb}^{2+}})_e} {(a_{\text{Mg}^{2+}})_e^{z_{\text{Pb}^{2+}}/z_{\text{Mg}^{2+}}}} = \frac {4.1 \times 10^{-12}} {(0.01025)^{+2/+2}} = 4.0 \times 10^{-10} \nonumber\] The value of \(\log K_{\text{Pb}^{2+}/\text{Mg}^{2+}}\), therefore, is –9.40. A ion-selective electrode for \(\text{NO}_2^-\) has log values of –3.1 for F , –4.1 for \(\text{SO}_4^{2-}\), –1.2 for I , and –3.3 for \(\text{NO}_3^-\). Which ion is the most serious interferent and for what activity of this interferent is the potential equivalent to a solution in which the activity of \(\text{NO}_2^-\) is \(2.75 \times 10^{-4}\)? The larger the value of the more serious the interference. Larger values for correspond to more positive (less negative) values for log ; thus, I , with a of \(6.3 \times 10^{-2}\), is the most serious of these interferents. To find the activity of I that gives a potential equivalent to a \(\text{NO}_2^-\) activity of \(2.75 \times 10^{-4}\), we note that \[a_{\text{NO}_2^-}=K_{A, I} \times a_{\text{I}^-} \nonumber\] Making appropriate substitutions \[2.75 \times 10^{-4}=\left(6.3 \times 10^{-2}\right) \times a_{\mathrm{I}^-} \nonumber\] and solving for \(a_{\text{I}^-}\) gives its activity as \(4.4 \times 10^{-3}\). The first commercial were manufactured using Corning 015, a glass with a composition that is approximately 22% Na O, 6% CaO, and 72% SiO . When immersed in an aqueous solution for several hours, the outer approximately 10 nm of the membrane’s surface becomes hydrated, resulting in the formation of negatively charged sites, —SiO . Sodium ions, Na , serve as counter ions. Because H binds more strongly to —SiO than does Na , they displace the sodium ions \[\mathrm{H}^{+}+-\mathrm{SiO}^{-} \mathrm{Na}^{+}\rightleftharpoons-\mathrm{SiO}^{-} \mathrm{H}^{+}+\mathrm{Na}^{+} \nonumber\] explaining the membrane’s selectivity for H . The transport of charge across the membrane is carried by the Na ions. The potential of a glass electrode using Corning 015 obeys the equation \[E_{\mathrm{cell}}=K+0.05916 \log a_{\mathrm{H}^{+}} \label{11.9}\] over a pH range of approximately 0.5 to 9. At more basic pH values the glass membrane is more responsive to other cations, such as Na and K For a Corning 015 glass membrane, the selectivity coefficient is \(\approx 10^{-11}\). What is the expected error if we measure the pH of a solution in which the activity of H is \(2 \times 10^{-13}\) and the activity of Na is 0.05? A solution in which the actual activity of H , ( ) , is \(2 \times 10^{-13}\) has a pH of 12.7. Because the electrode responds to both H and Na , the apparent activity of H , ( ) , is \[(a_{\text{H}^+})_\text{app} = (a_{\text{H}^+})_\text{act} + (K_{\text{H}^+ / \text{Na}^+} \times a_{\text{Na}^+}) = 2 \times 10^{-13} + (10^{-11} \times 0.05) = 7 \times 10^{-13} \nonumber\] The apparent activity of H is equivalent to a pH of 12.2, an error of –0.5 pH units. Replacing Na O and CaO with Li O and BaO extends the useful pH range of glass membrane electrodes to pH levels greater than 12. Glass membrane pH electrodes often are available in a combination form that includes both the indicator electrode and the reference electrode. The use of a single electrode greatly simplifies the measurement of pH. An example of a typical combination electrode is shown in Figure 11.2.10
. The observation that the Corning 015 glass membrane responds to ions other than H (see Example 11.2.6
) led to the development of glass membranes with a greater selectivity for other cations. For example, a glass membrane with a composition of 11% Na O, 18% Al O , and 71% SiO is used as an ion-selective electrode for Na . Other glass ion-selective electrodes have been developed for the analysis of Li , K , Rb , Cs , \(\text{NH}_4^+\), Ag , and Tl . Table 11.2.1
provides several examples. \(K_{\mathrm{Na}^{+} / \mathrm{H}^{+}}=1000\) \(K_{\mathrm{Na}^{+} / \mathrm{K}^{+}}=0.001\) \(K_{\mathrm{Na}^{+} / \mathrm{Li}^{+}}=0.001\) \(K_{\mathrm{Li}^{+} / \mathrm{Na}^{+}}=0.3\) \(K_{\mathrm{Li}^{+} / \mathrm{K}^{+}}=0.001\) \(K_{\mathrm{K}^{+} / \mathrm{Na}^{+}}=0.05\) Selectivity coefficients are approximate; values found experimentally may vary substantially from the listed values. See Cammann, K. , Springer-Verlag: Berlin, 1977. Because an ion-selective electrode’s glass membrane is very thin—it is only about 50 μm thick—they must be handled with care to avoid cracks or breakage. Glass electrodes usually are stored in a storage buffer recommended by the manufacturer, which ensures that the membrane’s outer surface remains hydrated. If a glass electrode dries out, it is reconditioned by soaking for several hours in a solution that contains the analyte. The composition of a glass membrane will change over time, which affects the electrode’s performance. The average lifetime for a typical glass electrode is several years. A s has a membrane that consists of either a polycrystalline inorganic salt or a single crystal of an inorganic salt. We can fashion a polycrystalline solid-state ion-selective electrode by sealing a 1–2 mm thick pellet of AgS—or a mixture of AgS and a second silver salt or another metal sulfide—into the end of a nonconducting plastic cylinder, filling the cylinder with an internal solution that contains the analyte, and placing a reference electrode into the internal solution. Figure 11.2.11
shows a typical design. The NaCl in a salt shaker is an example of polycrystalline material because it consists of many small crystals of sodium chloride. The NaCl salt plates used in IR spectroscopy (see ), on the other hand, are an example of a single crystal of sodium chloride. The membrane potential for a Ag S pellet develops as the result of a difference in the extent of the solubility reaction \[\mathrm{Ag}_{2} \mathrm{S}(s)\rightleftharpoons2 \mathrm{Ag}^{+}(a q)+\mathrm{S}^{2-}(a q) \nonumber\] on the membrane’s two sides, with charge carried across the membrane by Ag ions. When we use the electrode to monitor the activity of Ag , the cell potential is \[E_{\text {cell }}=K+0.05916 \log a_{\mathrm{Ag}^{+}} \nonumber\] The membrane also responds to the activity of \(\text{S}^{2-}\), with a cell potential of \[E_{\mathrm{cell}}=K-\frac{0.05916}{2} \log a_{\text{S}^{2-}} \nonumber\] If we combine an insoluble silver salt, such as AgCl, with the Ag S, then the membrane potential also responds to the concentration of Cl with a cell potential of \[E_{\text {cell }}=K-0.05916 \log a_{\mathrm{Cl}^{-}} \nonumber\] By mixing Ag S with CdS, CuS, or PbS, we can make an ion-selective electrode that responds to the activity of Cd , Cu , or Pb . In this case the cell potential is \[E_{\mathrm{cell}}=K+\frac{0.05916}{2} \ln a_{M^{2+}} \nonumber\] where is the activity of the metal ion. Table 11.2.2
provides examples of polycrystalline, Ag S-based solid-state ion-selective electrodes. The selectivity of these ion-selective electrodes depends on the relative solubility of the compounds. A Cl ISE using a Ag S/AgCl membrane is more selective for Br ( = 10 ) and for I ( = 10 ) because AgBr and AgI are less soluble than AgCl. If the activity of Br is sufficiently high, AgCl at the membrane/solution interface is replaced by AgBr and the electrode’s response to Cl decreases substantially. Most of the polycrystalline ion-selective electrodes listed in Table 11.2.2
operate over an extended range of pH levels. The equilibrium between S and HS limits the analysis for S to a pH range of 13–14. \(K_{\text{Ag}^+/\text{Cu}^{2+}} = 10^{-6}\) \(K_{\text{Ag}^+/\text{Pb}^{2+}} = 10^{-10}\) Hg interferes \(K_{\text{Cd}^{2+}/\text{Fe}^{2+}} = 200\) \(K_{\text{Cd}^{2+}/\text{Pb}^{2+}} = 6\) Ag , Hg , and Cu must be absent \(K_{\text{Cu}^{2+}/\text{Fe}^{3+}} = 10\) \(K_{\text{Cu}^{2+}/\text{Cu}^{+}} = 10^{-6}\) Ag and Hg must be absent \(K_{\text{Pb}^{2+}/\text{Fe}^{3+}} = 1\) \(K_{\text{Pb}^{2+}/\text{Cd}^{2+}} = 1\) Ag , Hg , and Cu must be absent \(K_{\text{Br}^-/\text{I}^{-}} = 5000\) \(K_{\text{Br}^-/\text{Cl}^{-}} = 0.005\) \(K_{\text{Br}^-/\text{OH}^{-}} = 10^{-5}\) S must be absent \(K_{\text{Cl}^-/\text{I}^{-}} = 10^{6}\) \(K_{\text{Cl}^-/\text{Br}^{-}} = 100\) \(K_{\text{Cl}^-/\text{OH}^{-}} = 0.01\) S must be absent \(K_{\text{I}^-/\text{S}^{2-}} = 30\) \(K_{\text{I}^-/\text{Br}^{-}} = 10^{-4}\) \(K_{\text{I}^-/\text{Cl}^{-}} = 10^{-6}\) \(K_{\text{I}^-/\text{OH}^{-}} = 10^{-7}\) \(K_{\text{SCN}^-/\text{I}^{-}} = 10^{3}\) \(K_{\text{SCN}^-/\text{Br}^{-}} = 100\) \(K_{\text{SCN}^-/\text{Cl}^{-}} = 0.1\)\(K_{\text{SCN}^-/\text{OH}^{-}} = 0.01\) S must be absent Selectivity coefficients are approximate; values found experimentally may vary substantially from the listed values. See Cammann, K. , Springer-Verlag: Berlin, 1977. The membrane of a F ion-selective electrode is fashioned from a single crystal of LaF , which usually is doped with a small amount of EuF to enhance the membrane’s conductivity. Because EuF provides only two F ions—compared to the three F ions in LaF —each EuF produces a vacancy in the crystal’s lattice. Fluoride ions pass through the membrane by moving into adjacent vacancies. As shown in , the LaF membrane is sealed into the end of a non-conducting plastic cylinder, which contains a standard solution of F , typically 0.1 M NaF, and a Ag/AgCl reference electrode. The membrane potential for a F ISE results from a difference in the solubility of LaF on opposite sides of the membrane, with the potential given by \[E_{\mathrm{cell}}=K-0.05916 \log a_{\mathrm{F}^-} \nonumber\] One advantage of the F ion-selective electrode is its freedom from interference. The only significant exception is OH ( = 0.1), which imposes a maximum pH limit for a successful analysis. Below a pH of 4 the predominate form of fluoride in solution is HF, which does not contribute to the membrane potential. For this reason, an analysis for fluoride is carried out at a pH greater than 4. What is the maximum pH that we can tolerate if we need to analyze a solution in which the activity of F is \(1 \times 10^{-5}\) with an error of less than 1%? In the presence of OH the cell potential is \[E_{\mathrm{cell}}=K-0.05916\left\{a_{\mathrm{F}^-}+K_{\mathrm{F}^- / \mathrm{OH}^{-}} \times a_{\mathrm{OH}^-}\right\} \nonumber\] To achieve an error of less than 1%, the term \(K_{\mathrm{F}^- / \mathrm{OH}^{-}} \times a_{\mathrm{OH}^-}\) must be less than 1% of ; thus \[K_{\mathrm{F}^- / \mathrm{OH}^-} \times a_{\mathrm{OH}^{-}} \leq 0.01 \times a_{\mathrm{F}^-} \nonumber\] \[0.10 \times a_{\mathrm{OH}^{-}} \leq 0.01 \times\left(1.0 \times 10^{-5}\right) \nonumber\] Solving for gives the maximum allowable activity for OH as \(1 \times 10^{-6}\), which corresponds to a pH of less than 8. Suppose you wish to use the nitrite-selective electrode in to measure the activity of \(\text{NO}_2^-\). If the activity of \(\text{NO}_2^-\) is \(2.2 \times 10^{-4}\), what is the maximum pH you can tolerate if the error due to OH must be less than 10%? The selectivity coefficient for OH , \(K_{\text{NO}_2^-/\text{OH}^-}\), is 630. Do you expect the electrode to have a lower pH limit? Clearly explain your answer. In the presence of OH the cell potential is \[E_{\mathrm{cell}}=K-0.05916 \log \left\{a_{\mathrm{NO}_{2}^-}+K_{\mathrm{NO}_{2}^- / \mathrm{OH}^{-}} \times a_{\mathrm{OH}^{-}}\right\} \nonumber\] To achieve an error of less than 10%, the term \(K_{\mathrm{NO}_{2}^- / \mathrm{OH}^{-}} \times a_{\mathrm{OH}^{-}}\) must be less than 1% of \(a_{\text{NO}_2^-}\); thus \[K_{\mathrm{NO}_{2}^- / \mathrm{OH}^{-}} \times a_{\mathrm{OH}^-} \leq 0.10 \times a_{\mathrm{NO}_{2}^-} \nonumber\] \[630 \times a_{\mathrm{OH}^{-}} \leq 0.10 \times\left(2.2 \times 10^{-4}\right) \nonumber\] Solving for gives its maximum allowable activity as \(3.5 \times 10^{-8}\), which corresponds to a pH of less than 6.54. The electrode does have a lower pH limit. Nitrite is the conjugate weak base of HNO , a species to which the ISE does not respond. As shown by the ladder diagram below, at a pH of 4.15 approximately 10% of nitrite is present as HNO . A minimum pH of 4.5 is the usual recommendation when using a nitrite ISE. This corresponds to a \(\text{NO}_2^- / \text{HNO}_2\) ratio of \[\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log \frac{\left[\mathrm{NO}_{2}^{-}\right]}{\left[\mathrm{HNO}_{2}\right]} \nonumber\] \[4.5=3.15+\log \frac{\left[\mathrm{NO}_{2}^{-}\right]}{\left[\mathrm{HNO}_{2}\right]} \nonumber\] \[\frac{\left[\mathrm{NO}_{2}^{-}\right]}{\left[\mathrm{HNO}_{2}\right]} \approx 22 \nonumber\] Thus, at a pH of 4.5 approximately 96% of nitrite is present as \(\text{NO}_2^-\). Unlike a glass membrane ion-selective electrode, a solid-state ISE does not need to be conditioned before it is used, and it may be stored dry. The surface of the electrode is subject to poisoning, as described above for a Cl ISE in contact with an excessive concentration of Br . If an electrode is poisoned, it can be returned to its original condition by sanding and polishing the crystalline membrane. Poisoning simply means that the surface has been chemically modified, such as AgBr forming on the surface of a AgCl membrane. Another class of ion-selective electrodes uses a hydrophobic membrane that contains a liquid organic complexing agent that reacts selectively with the analyte. Three types of organic complexing agents have been used: cation exchangers, anion exchangers, and neutral ionophores. A membrane potential exists if the analyte’s activity is different on the two sides of the membrane. Current is carried through the membrane by the analyte. An is a ligand whose exterior is hydrophobic and whose interior is hydrophilic. The crown ether shown here is one example of a neutral ionophore. One example of a is that for Ca , which uses a porous plastic membrane saturated with the cation exchanger di-( -decyl) phosphate. As shown in Figure 11.2.12
, the membrane is placed at the end of a non-conducting cylindrical tube and is in contact with two reservoirs. The outer reservoir contains di-( -decyl) phosphate in di- -octylphenylphosphonate, which soaks into the porous membrane. The inner reservoir contains a standard aqueous solution of Ca and a Ag/AgCl reference electrode. Calcium ion-selective electrodes also are available in which the di-( -decyl) phosphate is immobilized in a polyvinyl chloride (PVC) membrane that eliminates the need for the outer reservoir. The membrane potential for the Ca ISE develops as the result of a difference in the extent of the complexation reaction \[\mathrm{Ca}^{2+}(a q)+2\left(\mathrm{C}_{10} \mathrm{H}_{21} \mathrm{O}\right)_{2} \mathrm{PO}_{2}^{-}(mem) \rightleftharpoons \mathrm{Ca}\left[\left(\mathrm{C}_{10} \mathrm{H}_{21} \mathrm{O}\right)_{2} \mathrm{PO}_{2}\right]_2 (mem) \nonumber\] on the two sides of the membrane, where ( ) indicates a species that is present in the membrane. The cell potential for the Ca ion-selective electrode is \[E_{\mathrm{cell}}=K+\frac{0.05916}{2} \log a_{\mathrm{ca}^{2+}} \nonumber\] The selectivity of this electrode for Ca is very good, with only Zn showing greater selectivity. Table 11.2.3
lists the properties of several liquid-based ion-selective electrodes. An electrode using a liquid reservoir can be stored in a dilute solution of analyte and needs no additional conditioning before use. The lifetime of an electrode with a PVC membrane, however, is proportional to its exposure to aqueous solutions. For this reason these electrodes are best stored by covering the membrane with a cap along with a small amount of wetted gauze to maintain a humid environment. Before using the electrode it is conditioned in a solution of analyte for 30–60 minutes. \(K_{\text{Ca}^{2+}/\text{Zn}^{2+}} = 1-5\) \(K_{\text{Ca}^{2+}/\text{Al}^{3+}} = 0.90\) \(K_{\text{Ca}^{2+}/\text{Mn}^{2+}} = 0.38\) \(K_{\text{Ca}^{2+}/\text{Cu}^{2+}} = 0.070\) \(K_{\text{Ca}^{2+}/\text{Mg}^{2+}} = 0.032\) \(K_{\text{K}^{+}/\text{Rb}^{+}} = 1.9\) \(K_{\text{K}^{+}/\text{Cs}^{+}} = 0.38\) \(K_{\text{K}^{+}/\text{Li}^{+}} = 10^{-4}\) \(K_{\text{Li}^{+}/\text{H}^{+}} = 1\) \(K_{\text{Li}^{+}/\text{Na}^{+}} = 0.03\) \(K_{\text{Li}^{+}/\text{K}^{+}} = 0.007\) \(K_{\text{NH}_4^{+}/\text{K}^{+}} = 0.12\) \(K_{\text{NH}_4^{+}/\text{H}^{+}} = 0.016\) \(K_{\text{NH}_4^{+}/\text{Li}^{+}} = 0.0042\) \(K_{\text{NH}_4^{+}/\text{Na}^{+}} = 0.002\) \(K_{\text{ClO}_4^{-}/\text{OH}^{-}} = 1\) \(K_{\text{ClO}_4^{-}/\text{I}^{-}} = 0.012\) \(K_{\text{ClO}_4^{-}/\text{NO}_3^{-}} = 0.0015\) \(K_{\text{ClO}_4^{-}/\text{Br}^{-}} = 5.6 \times 10^{-4}\) \(K_{\text{ClO}_4^{-}/\text{Cl}^{-}} = 2.2 \times 10^{-4}\) \(K_{\text{NO}_3^{-}/\text{Cl}^{-}} = 0.006\) \(K_{\text{NO}_3^{-}/\text{F}^{-}} = 9 \times 10^{-4}\) Selectivity coefficients are approximate; values found experimentally may vary substantially from the listed values. See Cammann, K. , Springer-Verlag: Berlin, 1977. A number of membrane electrodes respond to the concentration of a dissolved gas. The basic design of a , as shown in Figure 11.2.13
, consists of a thin membrane that separates the sample from an inner solution that contains an ion-selective electrode. The membrane is permeable to the gaseous analyte, but impermeable to nonvolatile components in the sample’s matrix. The gaseous analyte passes through the membrane where it reacts with the inner solution, producing a species whose concentration is monitored by the ion-selective electrode. For example, in a CO electrode, CO diffuses across the membrane where it reacts in the inner solution to produce H O . \[\mathrm{CO}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons\text{ HCO}_{3}^{-}(a q)+\text{ H}_{3} \mathrm{O}^{+}(a q) \label{11.10}\] The change in the activity of H O in the inner solution is monitored with a pH electrode, for which the cell potential is given by Equation \ref{11.9}. To find the relationship between the activity of H O in the inner solution and the activity of CO in the inner solution we rearrange the equilibrium constant expression for reaction \ref{11.10}; thus \[a_{\mathrm{H}_{3} \mathrm{O}^{+}}=K_{\mathrm{a}} \times \frac{a_{\mathrm{CO}_{2}}}{a_{\mathrm{HCO}_{3}^{-}}} \label{11.11}\] where is the equilibrium constant. If the activity of \(\text{HCO}_3^-\) in the internal solution is sufficiently large, then its activity is not affected by the small amount of CO that passes through the membrane. Substituting Equation \ref{11.11} into Equation \ref{11.9} gives \[E_{\mathrm{cell}}=K^{\prime}+0.05916 \log a_{\mathrm{co}_{2}} \nonumber\] where K′ is a constant that includes the constant for the pH electrode, the equilibrium constant for reaction \ref{11.10} and the activity of \(\text{HCO}_3^-\) in the inner solution. Table 11.2.4
lists the properties of several gas-sensing electrodes. The composition of the inner solution changes with use, and both the inner solution and the membrane must be replaced periodically. Gas-sensing electrodes are stored in a solution similar to the internal solution to minimize their exposure to atmospheric gases. 10 mM NaHCO 10 mM NaCl 10 mM NH Cl 0.1 M KNO 20 mM NaNO 0.1 M KNO 1 mM NaHSO pH 5 Cammann, K. , Springer-Verlag: Berlin, 1977. The approach for developing gas-sensing electrodes can be modified to create potentiometric electrodes that respond to a biochemically important species. The most common class of potentiometric biosensors are , in which we trap or immobilize an enzyme at the surface of a potentiometric electrode. The analyte’s reaction with the enzyme produces a product whose concentration is monitored by the potentiometric electrode. Potentiometric biosensors also have been designed around other biologically active species, including antibodies, bacterial particles, tissues, and hormone receptors. One example of an enzyme electrode is the urea electrode, which is based on the catalytic hydrolysis of urea by urease \[\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons 2 \mathrm{NH}_{4}^{+}(a q)+\text{ CO}_{3}^{-}(a q) \nonumber\] Figure 11.2.14
shows one version of the urea electrode, which modifies a gas-sensing NH electrode by adding a dialysis membrane that traps a pH 7.0 buffered solution of urease between the dialysis membrane and the gas permeable membrane [(a) Papastathopoulos, D. S.; Rechnitz, G. A. , , 17–26; (b) Riechel, T. L. . , , 640–642]. An NH electrode, as shown in Table 11.2.4
, uses a gas-permeable membrane and a glass pH electrode. The NH diffuses across the membrane where it changes the pH of the internal solution. When immersed in the sample, urea diffuses through the dialysis membrane where it reacts with the enzyme urease to form the ammonium ion, \(\text{NH}_4^+\), which is in equilibrium with NH . \[\mathrm{NH}_{4}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(l ) \rightleftharpoons \text{ H}_{3} \mathrm{O}^{+}(a q)+\text{ NH}_{3}(a q) \nonumber\] The NH , in turn, diffuses through the gas permeable membrane where a pH electrode measures the resulting change in pH. The electrode’s response to the concentration of urea is \[E_{\text {cell }}=K-0.05916 \log a_{\text {urea }} \label{11.12}\] Another version of the urea electrode (Figure 11.2.15
) immobilizes the enzyme urease in a polymer membrane formed directly on the tip of a glass pH electrode [Tor, R.; Freeman, A. , , 1042–1046]. In this case the response of the electrode is \[\mathrm{pH}=K a_{\mathrm{urea}} \label{11.13}\] Few potentiometric biosensors are available commercially. As shown in Figure 11.2.14
and Figure 11.2.15
, however, it is possible to convert an ion-selective electrode or a gas-sensing electrode into a biosensor. Several representative examples are described in Table 11.2.5
, and additional examples can be found in this chapter’s additional resources. Complied from Cammann, K. , Springer-Verlag: Berlin, 1977 and Lunte, C. E.; Heineman, W. R. “Electrochemical techniques in Bioanalysis,” in Steckham, E. ed. , Vol. 143, Springer-Verlag: Berlin, 1988, p.8. Abbreviations for biologically active phase: E = enzyme; B = bacterial particle; T = tissue. The potentiometric determination of an analyte’s concentration is one of the most common quantitative analytical techniques. Perhaps the most frequent analytical measurement is the determination of a solution’s pH, a measurement we will consider in more detail later in this section. Other areas where potentiometry is important are clinical chemistry, environmental chemistry, and potentiometric titrations. Before we consider representative applications, however, we need to examine more closely the relationship between cell potential and the analyte’s concentration and methods for standardizing potentiometric measurements. The Nernst equation relates the cell potential to the analyte’s activity. For example, the Nernst equation for a metallic electrode of the first kind is \[E_{\mathrm{cll}}=K+\frac{0.05916}{n} \log a_{M^{n+}} \label{11.14}\] where is the metal ion’s activity. When we use a potentiometric electrode, however, our goal is to determine the analyte’s concentration. As we learned in an ion’s activity is the product of its concentration, [ ], and a matrix-dependent activity coefficient, \(\gamma_{Mn^{n+}}\). \[a_{M^{n+}}=\left[M^{n+}\right] \gamma_{M^{n+}} \label{11.15}\] Substituting Equation \ref{11.15} into Equation \ref{11.14} and rearranging, gives \[E_{\mathrm{cell}}=K+\frac{0.05916}{n} \log \gamma_{M^{n+}}+\frac{0.05916}{n} \log \left[M^{n+}\right] \label{11.16}\] We can solve Equation \ref{11.16} for the metal ion’s concentration if we know the value for its activity coefficient. Unfortunately, if we do not know the exact ionic composition of the sample’s matrix—which is the usual situation—then we cannot calculate the value of \(\gamma_{Mn^{n+}}\). There is a solution to this dilemma. If we design our system so that the standards and the samples have an identical matrix, then the value of \(\gamma_{Mn^{n+}}\) remains constant and Equation \ref{11.16} simplifies to \[E_{\mathrm{cell}}=K^{\prime}+\frac{0.05916}{n} \log \left[M^{n+}\right] \nonumber\] where \(K^{\prime}\) includes the activity coefficient. Before we can determine the concentration of analyte in a sample, we must standardize the electrode. If the electrode’s response obeys the Nernst equation, then we can determine the constant using a single external standard. Because a small deviation from the ideal slope of ± / or ± / is not unexpected, we usually use two or more external standards. To review the use of external standards, see . In the absence of interferents, a calibration curve of versus log , where is the analyte, is a straight-line. A plot of versus log[ ], however, may show curvature at higher concentrations of analyte as a result of a matrix-dependent change in the analyte’s activity coefficient. To maintain a consistent matrix we add a high concentration of an inert electrolyte to all samples and standards. If the concentration of added electrolyte is sufficient, then the difference between the sample’s matrix and the matrix of the standards will not affect the ionic strength and the activity coefficient essentially remains constant. The inert electrolyte added to the sample and the standards is called a (TISAB). The concentration of Ca in a water sample is determined using the method of external standards. The ionic strength of the samples and the standards is maintained at a nearly constant level by making each solution 0.5 M in KNO . The measured cell potentials for the external standards are shown in the following table. \(1.00 \times 10^{-5}\) What is the concentration of Ca in a water sample if its cell potential is found to be –0.084 V? Linear regression gives the calibration curve in Figure 11.2.16
, with an equation of \[E_{\mathrm{cell}}=0.027+0.0303 \log \left[\mathrm{Ca}^{2+}\right] \nonumber\] Substituting the sample’s cell potential gives the concentration of Ca as \(2.17 \times 10^{-4}\) M. Note that the slope of the calibration curve, which is 0.0303, is slightly larger than its ideal value of 0.05916/2 = 0.02958; this is not unusual and is one reason for using multiple standards. One reason that it is not unusual to find that the experimental slope deviates from its ideal value of 0.05916/ is that this ideal value assumes that the temperature is 25°C. Another approach to calibrating a potentiometric electrode is the method of standard additions. First, we transfer a sample with a volume of and an analyte concentration of into a beaker and measure the potential, ( ) . Next, we make a standard addition by adding to the sample a small volume, , of a standard that contains a known concentration of analyte, , and measure the potential, ( )std. If is significantly smaller than , then we can safely ignore the change in the sample’s matrix and assume that the analyte’s activity coefficient is constant. Example 11.2.9
demonstrates how we can use a one-point standard addition to determine the concentration of analyte in a sample. To review the method of standard additions, see . The concentration of Ca in a sample of sea water is determined using a Ca ion-selective electrode and a one-point standard addition. A 10.00-mL sample is transferred to a 100-mL volumetric flask and diluted to volume. A 50.00-mL aliquot of the sample is placed in a beaker with the Ca ISE and a reference electrode, and the potential is measured as –0.05290 V. After adding a 1.00-mL aliquot of a \(5.00 \times 10^{-2}\) M standard solution of Ca the potential is –0.04417 V. What is the concentration of Ca in the sample of sea water? To begin, we write the Nernst equation before and after adding the standard addition. The cell potential for the sample is \[\left(E_{\mathrm{cell}}\right)_{\mathrm{samp}}=K+\frac{0.05916}{2} \log C_{\mathrm{samp}} \nonumber\] and that following the standard addition is \[\left(E_{\mathrm{cell}}\right)_{\mathrm{std}}=K+\frac{0.05916}{2} \log \left\{ \frac {V_\text{samp}} {V_\text{tot}}C_\text{samp} + \frac {V_\text{std}} {V_\text{tot}}C_\text{std} \right\} \nonumber\] where is the total volume ( + ) after the standard addition. Subtracting the first equation from the second equation gives \[\Delta E = \left(E_{\mathrm{cell}}\right)_{\mathrm{std}} - \left(E_{\mathrm{cell}}\right)_{\mathrm{samp}} = \frac{0.05916}{2} \log \left\{ \frac {V_\text{samp}} {V_\text{tot}}C_\text{samp} + \frac {V_\text{std}} {V_\text{tot}}C_\text{std} \right\} - \frac{0.05916}{2}\log C_\text{samp} \nonumber\] Rearranging this equation leaves us with \[\frac{2 \Delta E_{cell}}{0.05916} = \log \left\{ \frac {V_\text{samp}} {V_\text{tot}} + \frac {V_\text{std}C_\text{std}} {V_\text{tot}C_\text{samp}} \right\} \nonumber\] Substituting known values for \(\Delta E\), , , and , \[\begin{array}{l}{\frac{2 \times\{-0.04417-(-0.05290)\}}{0.05916}=} \\ {\log \left\{\frac{50.00 \text{ mL}}{51.00 \text{ mL}}+\frac{(1.00 \text{ mL})\left(5.00 \times 10^{-2} \mathrm{M}\right)}{(51.00 \text{ mL}) C_{\mathrm{samp}}}\right\}} \\ {0.2951=\log \left\{0.9804+\frac{9.804 \times 10^{-4}}{C_{\mathrm{samp}}}\right\}}\end{array} \nonumber\] and taking the inverse log of both sides gives \[1.973=0.9804+\frac{9.804 \times 10^{-4}}{C_{\text {samp }}} \nonumber\] Finally, solving for gives the concentration of Ca as \(9.88 \times 10^{-4}\) M. Because we diluted the original sample of seawater by a factor of 10, the concentration of Ca in the seawater sample is \(9.88 \times 10^{-3}\) M. Most potentiometric electrodes are selective toward the free, uncomplexed form of the analyte, and do not respond to any of the analyte’s complexed forms. This selectivity provides potentiometric electrodes with a significant advantage over other quantitative methods of analysis if we need to determine the concentration of free ions. For example, calcium is present in urine both as free Ca ions and as protein-bound Ca ions. If we analyze a urine sample using atomic absorption spectroscopy, the signal is propor- tional to the total concentration of Ca because both free and bound calcium are atomized. Analyzing urine with a Ca ISE, however, gives a signal that is a function of only free Ca ions because the protein-bound Ca can not interact with the electrode’s membrane. 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 F in toothpaste provides an instructive example of a typical procedure. The description here is based on Kennedy, J. H. , Harcourt Brace Jaovanovich: San Diego, 1984, p. 117–118. The concentration of fluoride in toothpastes that contains soluble F is determined with a F ion-selective electrode using a calibration curve prepared with external standards. Although the F ISE is very selective (only OH with a of 0.1 is a significant interferent), Fe and Al interfere with the analysis because they form soluble fluoride complexes that do not interact with the ion-selective electrode’s membrane. This interference is minimized by reacting any Fe and Al with a suitable complexing agent. Prepare 1 L of a standard solution of 1.00% w/v SnF and transfer it to a plastic bottle for storage. Using this solution, prepare 100 mL each of standards that contain 0.32%, 0.36%, 0.40%, 0.44% and 0.48% w/v SnF , adding 400 mg of malic acid to each solution as a stabilizer. Transfer the standards to plastic bottles for storage. Prepare a total ionic strength adjustment buffer (TISAB) by mixing 500 mL of water, 57 mL of glacial acetic acid, 58 g of NaCl, and 4 g of disodium DCTA ( -1,2-cyclohexanetetraacetic acid) in a 1-L beaker, stirring until dissolved. Cool the beaker in a water bath and add 5 M NaOH until the pH is between 5–5.5. Transfer the contents of the beaker to a 1-L volumetric flask and dilute to volume. Prepare each external standard by placing approximately 1 g of a fluoride-free toothpaste, 30 mL of distilled water, and 1.00 mL of standard into a 50-mL plastic beaker and mix vigorously for two min with a stir bar. Quantitatively transfer the resulting suspension to a 100-mL volumetric flask along with 50 mL of TISAB and dilute to volume with distilled water. Store the entire external standard in a 250-mL plastic beaker until you are ready to measure the potential. Prepare toothpaste samples by obtaining an approximately 1-g portion and treating in the same manner as the standards. Measure the cell potential for the external standards and the samples using a F ion-selective electrode and an appropriate reference electrode. When measuring the potential, stir the solution and allow two to three minutes to reach a stable potential. Report the concentration of F in the toothpaste %w/w SnF . 1. The total ionic strength adjustment buffer serves several purposes in this procedure. Identify these purposes. The composition of the TISAB has three purposes: (a) The high concentration of NaCl (the final solutions are approximately 1 M NaCl) ensures that the ionic strength of each external standard and each sample is essentially identical. Because the activity coefficient for fluoride is the same in all solutions, we can write the Nernst equation in terms of fluoride’s concentration instead of its activity. (b) The combination of glacial acetic acid and NaOH creates an acetic acid/acetate buffer of pH 5–5.5. As shown in Figure 11.2.17
, the pH of this buffer is high enough to ensure that the predominate form of fluoride is F instead of HF. This pH also is sufficiently acidic that it avoids an interference from OH (see ). (c) DCTA is added as a complexing agent for Fe or Al , preventing the formation of \(\text{FeF}_6^{3-}\) or \(\text{AlF}_6^{3-}\). 2. Why is a fluoride-free toothpaste added to the standard solutions? Adding a fluoride-free toothpaste protects against any unaccounted for matrix effects that might influence the ion-selective electrode’s response. This assumes, of course, that the matrices of the two toothpastes are otherwise similar. 3. The procedure specifies that the standards and the sample should be stored in plastic containers. Why is it a bad idea to store the solutions in glass containers? The fluoride ion is capable of reacting with glass to form SiF . 4. Suppose your calibration curve has a slope of –57.98 mV for each 10-fold change in the concentration of F . The ideal slope from the Nernst equation is –59.16 mV per 10-fold change in concentration. What effect does this have on the quantitative analysis for fluoride in toothpaste? No effect at all! This is why we prepare a calibration curve using multiple standards. With the availability of inexpensive glass pH electrodes and pH meters, the determination of pH is one of the most common quantitative analytical measurements. The potentiometric determination of pH, however, is not without complications, several of which we discuss in this section. One complication is confusion over the meaning of pH [Kristensen, H. B.; Saloman, A.; Kokholm, G. , , 885A–891A]. The conventional definition of pH in most general chemistry textbooks is \[\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \label{11.17}\] As we now know, pH actually is a measure of the activity of H . \[\mathrm{pH}=-\log a_{\mathrm{H}^{+}} \label{11.18}\] Try this experiment—find several general chemistry textbooks and look up in each textbook’s index. Turn to the appropriate pages and see how it is defined. Next, look up or in each textbook’s index and see if these terms are indexed. Equation \ref{11.17} only approximates the true pH. If we calculate the pH of 0.1 M HCl using Equation \ref{11.17}, we obtain a value of 1.00; the solution’s actual pH, as defined by Equation \ref{11.18}, is 1.1 [Hawkes, S. J. , , 747–749]. The activity and the concentration of H are not the same in 0.1 M HCl because the activity coefficient for H is not 1.00 in this matrix. Figure 11.2.18
shows a more colorful demonstration of the difference between activity and concentration. A second complication in measuring pH is the uncertainty in the relationship between potential and activity. For a glass membrane electrode, the cell potential, ( )samp, for a sample of unknown pH is \[(E_{\text{cell}})_\text {samp} = K-\frac{R T}{F} \ln \frac{1}{a_{\mathrm{H}^{+}}}=K-\frac{2.303 R T}{F} \mathrm{pH}_{\mathrm{samp}} \label{11.19}\] where includes the potential of the reference electrode, the asymmetry potential of the glass membrane, and any junction potentials in the electrochemical cell. All the contributions to are subject to uncertainty, and may change from day-to-day, as well as from electrode-to-electrode. For this reason, before using a pH electrode we calibrate it using a standard buffer of known pH. The cell potential for the standard, ( ) , is \[\left(E_{\text {ccll}}\right)_{\text {std}}=K-\frac{2.303 R T}{F} \mathrm{p} \mathrm{H}_{\mathrm{std}} \label{11.20}\] where pH is the standard’s pH. Subtracting Equation \ref{11.20} from Equation \ref{11.19} and solving for pH gives \[\text{pH}_\text{samp} = \text{pH}_\text{std} - \frac{\left\{\left(E_{\text {cell}}\right)_{\text {samp}}-\left(E_{\text {cell}}\right)_{\text {std}}\right\} F}{2.303 R T} \label{11.21}\] which is the operational definition of pH adopted by the International Union of Pure and Applied Chemistry [Covington, A. K.; Bates, R. B.; Durst, R. A. , , 531–542]. Calibrating a pH electrode presents a third complication because we need a standard with an accurately known activity for H . Table 11.2.6
provides pH values for several primary standard buffer solutions accepted by the National Institute of Standards and Technology. saturated (at 25 C) KHC H4O (tartrate) 0.05 m KH C H O (citrate) — Values taken from Bates, R. G. , 2nd ed. Wiley: New York, 1973. See also Buck, R. P., et. al.“Measurement of pH. Definition, Standards, and Procedures,” , , 2169–2200. All concentrations are molal (m). To standardize a pH electrode using two buffers, choose one near a pH of 7 and one that is more acidic or basic depending on your sample’s expected pH. Rinse your pH electrode in deionized water, blot it dry with a laboratory wipe, and place it in the buffer with the pH closest to 7. Swirl the pH electrode and allow it to equilibrate until you obtain a stable reading. Adjust the “Standardize” or “Calibrate” knob until the meter displays the correct pH. Rinse and dry the electrode, and place it in the second buffer. After the electrode equilibrates, adjust the “Slope” or “Temperature” knob until the meter displays the correct pH. Some pH meters can compensate for a change in temperature. To use this feature, place a temperature probe in the sample and connect it to the pH meter. Adjust the “Temperature” knob to the solution’s temperature and calibrate the pH meter using the “Calibrate” and “Slope” controls. As you are using the pH electrode, the pH meter compensates for any change in the sample’s temperature by adjusting the slope of the calibration curve using a Nernstian response of 2.303 / . Because of their selectivity for analytes in complex matricies, ion-selective electrodes are important sensors for clinical samples. The most common analytes are electrolytes, such as Na , K , Ca , H , and Cl , and dissolved gases such as CO . For extracellular fluids, such as blood and urine, the analysis can be made . An analysis, however, requires a much smaller electrode that we can insert directly into a cell. Liquid-based membrane microelectrodes with tip diameters smaller than 1 μm are constructed by heating and drawing out a hard-glass capillary tube with an initial diameter of approximately 1–2 mm (Figure 11.2.19
). The microelectrode’s tip is made hydrophobic by dipping into a solution of dichlorodimethyl silane, and an inner solution appropriate for the analyte and a Ag/AgCl wire reference electrode are placed within the microelectrode. The microelectrode is dipped into a solution of the liquid complexing agent, which through capillary action draws a small volume of the liquid complexing agent into the tip. Potentiometric microelectrodes have been developed for a number of clinically important analytes, including H , K , Na , Ca , Cl , and I [Bakker, E.; Pretsch, E. , , 612–618]. Although ion-selective electrodes are used in environmental analysis, their application is not as widespread as in clinical analysis. Although standard potentiometric methods are available for the analysis of CN , F , NH , and \(\text{NO}_3^-\) in water and wastewater, other analytical methods generally provide better detection limits. One potential advantage of an ion-selective electrode is the ability to incorporate it into a flow cell for the continuous monitoring of wastewater streams. One method for determining the equivalence point of an acid–base titration is to use a pH electrode to monitor the change in pH during the titration. A potentiometric determination of the equivalence point is possible for acid–base, complexation, redox, and precipitation titrations, as well as for titrations in aqueous and nonaqueous solvents. Acid–base, complexation, and precipitation potentiometric titrations usually are monitored with an ion-selective electrode that responds the analyte, although an electrode that responds to the titrant or a reaction product also can be used. A redox electrode, such as a Pt wire, and a reference electrode are used for potentiometric redox titrations. More details about potentiometric titrations are found in . The working range for most ion-selective electrodes is from a maximum concentration of 0.1–1 M to a minimum concentration of \(10^{-5}-10^{-11}\) M [(a) Bakker, E.; Pretsch, E. , , 420A–426A; (b) Bakker, E.; Pretsch, E. , , 199–207]. This broad working range extends from major analytes to ultratrace analytes, and is significantly greater than many other analytical techniques. To use a conventional ion-selective electrode we need a minimum sample volume of several mL (a macro sample). Microelectrodes, such as the one shown in , are used with an ultramicro sample, although care is needed to ensure that the sample is representative of the original sample. The accuracy of a potentiometric analysis is limited by the error in measuring . Several factors contribute to this measurement error, including the contribution to the potential from interfering ions, the finite current that passes through the cell while we measure the potential, differences between the analyte’s activity coefficient in the samples and the standard solutions, and junction potentials. We can limit the effect of an interfering ion by including a separation step before the potentiometric analysis. Modern high impedance potentiometers minimize the amount of current that passes through the electrochemical cell. Finally, we can minimize the errors due to activity coefficients and junction potentials by matching the matrix of the standards to that of the sample. Even in the best circumstances, however, a difference of approximately ±1 mV for samples with equal concentrations of analyte is not unusual. We can evaluate the effect of uncertainty on the accuracy of a potentiometric measurement by using a propagation of uncertainty. For a membrane ion-selective electrode the general expression for potential is \[E_{\mathrm{cell}}=K+\frac{R T}{z F} \ln \left[ A\right] \nonumber\] where is the analyte’s, , charge. From in Chapter 4, the uncertainty in the cell potential, \(\Delta E_\text{cell}\) is \[\triangle E_{\text {cell}}=\frac{R T}{z F} \times \frac{\Delta [A]}{[A]} \nonumber\] Rearranging and multiplying through by 100 gives the percent relative error in concentration as \[\% \text { relative error }=\frac{\Delta[A]}{[A]} \times 100=\frac{\triangle E_{\mathrm{cell}}}{R T / z F} \times 100 \label{11.22}\] The relative error in concentration, therefore, is a function of the measurement error for the electrode’s potential, \(\Delta E_\text{cell}\), and the analyte’s charge. Table 11.2.7
provides representative values for ions with charges of ±1 and ±2 at a temperature of 25 C. Accuracies of 1–5% for monovalent ions and 2–10% for divalent ions are typical. Although Equation \ref{11.22} applies to membrane electrodes, we can use if for a metallic electrode by replacing with . Precision in potentiometry is limited by variations in temperature and the sensitivity of the potentiometer. Under most conditions—and when using a simple, general-purpose potentiometer—we can measure the potential with a repeatability of ±0.1 mV. Using Table 11.2.7
, this corresponds to an uncertainty of ±0.4% for monovalent analytes and ±0.8% for divalent analytes. The reproducibility of potentiometric measurements is about a factor of ten poorer. The sensitivity of a potentiometric analysis is determined by the term / or / in the Nernst equation. Sensitivity is best for smaller values of or . As described earlier, most ion-selective electrodes respond to more than one analyte; the selectivity for the analyte, however, often is significantly greater than the sensitivity for the interfering ions. The manufacturer of an ion-selective usually provides an ISE’s selectivity coefficients, which allows us to determine whether a potentiometric analysis is feasible for a given sample. In comparison to other techniques, potentiometry provides a rapid, relatively low-cost means for analyzing samples. The limiting factor when analyzing a large number of samples is the need to rinse the electrode between samples. The use of inexpensive, disposable ion-selective electrodes can increase a lab’s sample throughput. Figure 11.2.20
shows one example of a disposable ISE for Ag [Tymecki, L.; Zwierkowska, E.; Głąb, S.; Koncki, R. , , 482–488]. Commercial instruments for measuring pH or potential are available in a variety of price ranges, and includes portable models for use in the field. | 81,352 | 3,777 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Geochemistry_(Lower)/02%3A_The_Hydrosphere/2.03%3A_Chemistry_and_geochemistry_of_the_oceans |
The composition of the ocean has attracted the attention of some of the more famous names in science, including Robert Boyle, Antoine Lavoisier and Edmund Halley. Their early investigations tended to be difficult to reproduce, owing to the different conditions under which they crystallized the various salts. As many as 54 salts, double salts and hydrated salts can be obtained by evaporating seawater to dryness. At least 73 elements are now known to be present in seawater. These values, expressed in parts per thousand, are for seawater of 35% salinity. The best way of characterizing seawater is in terms of its ionic content, shown above. The remarkable thing about seawater is the constancy of its relative ionic composition. The overall salt content, known as the salinity (grams of salts contained in 1 kg of seawater), varies slightly within the range of 32-37.5%, corresponding to a solution of about 0.7% salt content. The ratios of the concentrations of the different ions, however, are quite constant, so that a measurement of Cl concentration is sufficient to determine the overall composition and total salinity. Although most elements are found in seawater only at trace levels, marine organisms may selectively absorb them and make them more detectable. Iodine, for example, was discovered in marine algae (seaweeds) 14 years before it was found in seawater. Other elements that were not detected in seawater until after they were found in marine organisms include barium, cobalt, copper, lead, nickel, silver and zinc. Si , presumably deriving from cosmic ray bombardment of Ar, has been discovered in marine sponges. Reflecting this constant ionic composition is the pH, which is usually maintained in the narrow range of 7.8-8.2, compared with 1.5 to 11 for fresh water. The major buffering action derives from the carbonate system, although ion exchange between Na+ in the water and H in clay sediments has recently been recognized to be a significant factor. The major ionic constituents whose concentrations can be determined from the salinity are known as conservative substances. Their constant relative concentrations are due to the large amounts of these species in the oceans in comparison to their small inputs from river flow. This is another way of saying that their residence times are very large. A number of other species, mostly connected with biological activity, are subject to wide variations in concentration. These include the nutrients NO , NO , NH , and HPO , which may become depleted near the surface in regions of warmth and light. As was explained in the preceding subsection on coastal upwelling, offshore prevailing winds tend to drive Western coastal surface waters out to sea, causing deeper and more nutrient-rich water to be drawn to the surface. This upwelled water can support a large population of phytoplankton and thus of zooplankton and fish. The best-known example of this is the anchovy fishery off the coast of Peru, but the phenomenon occurs to some extent on the West coasts of most continents, including our own. Other non-conservative components include Ca and SiO . These ions are incorporated into the solid parts of marine organisms, which sink to greater depths after the organisms die. The silica gradually dissolves, since the water is everywhere undersaturated in this substance. Calcium carbonate dissolves at intermediate depths, but may reprecipitate in deep waters owing to the higher pressure. Thus the concentrations of Ca and of SiO tend to vary with depth. The gases O and CO , being intimately involved with biological activity, are also non-conservative, as are N O and CO. Most of the organic carbon in seawater is present as dissolved material, with only about 1-2% in particulates. The total organic carbon content ranges between 0.5 mg/L in deep water to 1.5 mg/L near the surface. There is still considerable disagreement about the composition of the dissolved organic matter; much of it appears to be of high molecular weight, and may be polymeric. Substances qualitatively similar to the humic acids found in soils can be isolated. The greenish color that is often associated with coastal waters is due to a mixture of fluorescent, high molecular weight substances of undetermined composition known as . It is likely that the significance of the organic fraction of seawater may be much greater than its low abundance would suggest. For one thing, many of these substances are lipid-like and tend to adsorb onto surfaces. It has been shown that any particle entering the ocean is quickly coated with an organic surface film that may influence the rate and extent of its dissolution or decomposition. Certain inorganic ions may be strongly complexed by humic-like substances. The surface of the ocean is mostly covered with an organic film, only a few molecular layers thick. This is believed to consist of hydrocarbons, lipids, and the like, but glycoproteins and proteoglycans have been reported. If this film is carefully removed from a container of seawater, it will quickly be reconstituted. How significant this film is in its effects on gas exchange with the atmosphere is not known. The salinity of the ocean appears to have been about the same for at least the last 200 million years. There have been changes in the relative amounts of some species, however; the ratio of Na/K has increased from about 1:1 in ancient ocean sediments to its present value of 28:1. Incorporation of calcium into sediments by the action of marine organisms has depleted the Ca/Mg ratio from 1:1 to 1:3. input to ocean dissolved
in seawater in dead
organisms loss to sediments residence
time, y If the composition of the ocean has remained relatively unchanged with time, the continual addition of new mineral substances by the rivers and other sources must be exactly balanced by their removal as sediment, possibly passing through one or more biological systems in the process. In 1715 Edmund Halley suggested that the age of the ocean (and thus presumably of the world) might be estimated from the rate of salt transport by rivers. When this measurement was actually carried out in 1899, it gave an age of only 90 million years. This is somewhat better than the calculation made in 1654 by James Ussher, the Anglican Archbishop of Armagh, Ireland, based on his interpretation of the Biblical book of Genesis, that the world was created at 9 A.M. on October 23, 4004 BC, but it is still far too recent, being about when the dinosaurs became extinct. What Halley actually described was the residence time, which is about right for Na but much to long for some of the minor elements of seawater. The commonly stated view that the salt content of the oceans derives from surface runoff that contains the products of weathering and soil leaching is not consistent with the known compositions of the major river waters (See Table). The halide ions are particularly over-represented in seawater, compared to fresh water. These were once referred to as “excess volatiles”, and were attributed to volcanic emissions. With the discovery of plate tectonics, it became apparent that the locations of seafloor spreading at which fresh basalt flows up into the ocean from the mantle are also sources of mineral-laden water. Some of this may be seawater that has cycled through a hot porous region and has been able to dissolve some of the mineral material owing to the high temperature. Much of the water, however, is “juvenile” water that was previously incorporated into the mantle material and has never before been in the liquid phase. The substances introduced by this means (and by volcanic activity) are just the elements that are “missing” from river waters. Estimates of what fraction of the total volume of the oceans is due to juvenile water (most of it added in the early stages of mantle differentiation that began a billion years ago) range from 30 to 90%. The oceans can be regarded as a product of a giant acid-base titration in which the carbonic acid present in rain reacts with the basic materials of the lithosphere. The juvenile water introduced at locations of ocean-floor spreading is also acidic, and is partly neutralized by the basic components of the basalt with which it reacts. Surface rocks mostly contain aluminum, silicon and oxygen combined with alkali and alkaline-earth metals, mainly potassium, sodium and calcium. The CO and volcanic gases in rainwater react with this material to form a solution of the metal ion and HCO , in which is suspended some hydrated SiO . The solid material left behind is a clay such as kaolinite, Al Si O (OH) . This first forms as a friable coating on the surface of the weathered rock; later it becomes a soil material, then an alluvial deposit, and finally it may reach the sea as a suspended sediment. Here it may undergo a number of poorly-understood transformations to other clay sediments such as illites. Sea floor spreading eventually transports these sediments to a subduction region under a continental block, where the high temperatures and pressures permit reactions that transform it into hard rock such as granite, thus completing the geochemical cycle. Deep-sea hydrothermal vents are now recognized to be another significant route for both the addition and removal of ionic substances from seawater. Although the relative concentrations of most of the elements in seawater are constant throughout the oceans, there are certain elements that tend to have highly uneven distributions vertically, and to a lesser extent horizontally. Neglecting the highly localized effects of undersea springs and volcanic vents, these variations are direct results of the removal of these elements from seawater by organisms; if the sea were sterile, its chemical composition would be almost uniform. Plant life can exist only in the upper part of the ocean where there is sufficient light available to drive photosynthesis. These plants, together with the animals that consume them, extract nutrients from the water, reducing the concentrations of certain elements in the upper part of the sea. When these organisms die, they fall toward the lower depths of the ocean as particulate material. On the way down, some of the softer particles, deriving from tissue, may be consumed by other animals and recycled. Eventually, however, the nutrient elements that were incorporated into organisms in the upper part of the ocean will end up in the colder, dark, and essentially lifeless lower part. Mixing between the upper and lower reservoirs of the ocean is quite slow, owing to the higher density of the colder water; the average residence time of a water molecule in the lower reservoir is about 1600 years. Since the volume of the upper reservoir is only about 1/20 of that of the lower, a water molecule stays in the upper reservoir for only about 80 years. Except for dissolved oxygen, all elements required by living organisms are depleted in the upper part of the ocean with respect to the lower part. In the case of the major nutrients P, N and Si, the degree of depletion is sufficiently complete (around 95%) to limit the growth of organisms at the surface. These three elements are said to be biolimiting. A few other biointermediate elements show partial depletion in surface waters: Ca (1%), C (15%), Ba (75%). The organic component of plants and animals has the average composition C N P . It is remarkable that the ratio of N:P in seawater (both surface and deep) is also 15:1; this raises the interesting question of to what extent the ocean and life have co-evolved. In the deep part of the ocean the elemental ratio corresponds to C N P, but of course with much larger absolute amounts of these elements. Eventually some of this deeper water returns to the surface where the N and P are quickly taken up by plants. But since plants can only utilize 80 out of every 800 carbon atoms, 90 percent of the carbon will remain in dissolved form, mostly as HCO . To work out the balance of Ca and Si used in the hard parts of organisms, we add these elements to the average composition of the lower reservior to get Ca Si C N P. Particulate carbon falls into the deep ocean in the ratio of about two atoms in organic tissue to one atom in the form of calcite. This makes the overall composition of detrital material something like C N P; i.e., 80 organic C’s and 40 in CaCO . Accompanying these 40 calcite units will be 40 Ca atoms, but this represents a minor depletion of the 3200 Ca atoms that eventually return to the surface, so this element is only slightly depleted in the upper waters. Silicon, being far less abundant, is depleted to a much greater extent. A continual rain of particulate material from dead organisms falls through the ocean. This shower is comprised of three major kinds of material: calcite (CaCO ), silica (SiO ), and organic matter. The first two come from the hard parts of both plants and animals (mainly microscopic animals such as foraminifera and radiolarians). The organic matter is derived mainly from the soft tissues of organisms, and from animal fecal material. Some of this solid material dissolves before it reaches the ocean floor, but not usually before it enters the deep ocean where it will remain for about 1600 years. The remainder of this material settles onto the floor of the sea, where it forms one component of a layer of sediments that provide important information about the evolution of the sea and of the earth. Over a short time scale of months to years, these sediments are in quasi-equilibrium with the seawater. On a scale of millions of years, the sediments are merely way-stations in the geochemical cycling of material between the earth’s surface and its interior. The oceanic sediments have three main origins: Our main interest lies with the silica and calcium carbonate, since these substances form a crucial part of the biogeological cycle. Also, their distributions in the ocean are not uniform- a fact that must tell us something. The skeletons of diatoms and radiolarians are the principal sources of silica sediments. Since the ocean is everywhere undersaturated with respect to silica, only the most resistant parts of these skeletons reach the bottom of the deep ocean and get incorporated into sediments. Silica sediments are less common in the Atlantic ocean, owing to the lower content of dissolved silica. The parts of the ocean where these sediments are increasing most rapidly correspond to regions of upwelling, where deep water that is rich in dissolved silica rises to the surface where the silica is rapidly fixed by organisms. Where upwelling is absent, the growth of the organisms is limited, and little silica is precipitated. Since deep waters tend to flow from the Atlantic into the Pacific ocean where most of the upwelling occurs, Atlantic waters are depleted in silica, and silica sediments are not commonly found in this ocean. For calcium carbonate, the situation is quite different. In the first place, surface waters are everywhere supersaturated with respect to both calcite and aragonite, the two common crystal forms of CaCO . Secondly, Ca and HCO are never limiting factors in the growth of the coccoliths (plants) and forams (animals) that precipitate CaCO ; their production depends on the availability of phosphate and nitrogen. Because these elements are efficiently recycled before they fall into the deep ocean, their supply does not depend on upwelling, and so the production of solid is more uniformly distributed over the world’s oceans. More importantly, however, the chances that a piece of carbonate skeleton will end up as sediment will be highly dependent on both the local CO concentration and the depth of the ocean floor. These factors give rise to small-scale variations in the production of carbonate sediments that can be quite wide-ranging. New crust is being generated and moving away from the crests of the mid-ocean ridges at a rate of a few centimetres per year. Although the crests of these ridges are relatively high points, projecting to within about 3000 m of the surface, the continual injection of new material prevents sediments from accumulating in these areas. Farther from the crests, carbonate sediments do build up, eventually reaching a depth of about 500 m, but by this time the elevation has dropped off below the saturation horizon, so from this point on the carbonate sediments are overlaid by red clay. If we drill a hole down through a part of the ocean floor that is presently below the saturation horizon, the top part of the drill core will consist of clay, followed by CaCO at greater depths. The core may also contain regions in which silica predominates. Since silica production is very high in equatorial regions, the appearance of such a layer suggests that this particular region of the oceanic crust has moved across the equator. Page last modified: 21.01.2008 For information about this Web site or to contact the author,
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The average amount of salt to use in dough is about 1.75% to 2.25% based on the flour used. Some authorities recommend that the amount of salt used should be based on the actual quantity of water used in making the dough, namely about 30 g per L (1 oz. per qt.) of water. During the hot summer months, many bakers find it advantageous to use slightly more salt than in the winter as a safeguard against the development of any undesirable changes in the dough fermentation. Salt should never be dissolved in the same water in which yeast is dissolved. It is an antiseptic and dehydrates yeast cells and can even kill part of them, which means that less power is in the dough and a longer fermentation is needed. In bread made by the sponge dough method and in liquid fermentation systems, a small amount of salt included in the first stage strengthens the gluten. | 874 | 3,780 |
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Electroplating is the process of plating one metal onto another by hydrolysis, most commonly for decorative purposes or to prevent corrosion of a metal. There are also specific types of electroplating such as copper plating, silver plating, and chromium plating. The Purposes of Electroplating: The cathode would be the piece to be plated and the anode would be either a sacrificial anode or an inert anode, normally either platinum or carbon (graphite form). Sometimes plating occurs on racks or barrels for efficiency when plating many products. Please refer to electrolysis for more information. In the figure below, the Ag ions are being drawn to the surface of the spoon and it eventually becomes plated. The process is undergone using silver as the anode, and a screw as the cathode. The electrons are transferred from the anode to the cathode and is underwent in a solution containing silver. Electroplating was first discovered by Luigi Brugnatelli in 1805 through using the electrodeposition process for the electroplating of gold. However his discovery was not noted as he was disregarded by the French Academy of Science as well as Napolean Bonaparte. But a couple of decades later, John Wright managed to use potassium cyanide as an electrolyte for gold and silver. He discovered that potassium cyanide was in fact an efficient electrolyte. The Elkington cousins later in 1840 used potassium cyanide as their electrolyte and managed to create a feasible electroplating method for gold and silver. They attained a patent for electroplating and this method became widely spread throughout the world from England. Electroplating method has gradually become more efficient and advanced through the use of more eco-friendly formulas and by using direct current power supplies. There are many different metals that can be used in plating and so determining the right electrolyte is important for the quality of plating. Some electrolytes are acids, bases, metal salts or molten salts. When choosing the type of electrolyte some things to keep in mind are corrosion, resistance, brightness or reflectivity, hardness, mechanical strength, ductility, and wear resistance. The purpose of preparing the surface before beginning to plate another metal onto it is to ensure that it is clean and free of contaminants, which may interfere with the bonding. Contamination often prevents deposition and lack of adhesion. Normally this is done in three steps: cleaning, treatment and rinsing. Cleaning usually consists of using certain solvents such as alkaline cleaners, water, or acid cleaners in order to remove layers of oil on the surface. Treatment includes surface modification which is the hardening of the parts and applying metal layers. Rinsing leads to the final product and is the final touch to electroplating.Two certain methods of preparing the surface are physical cleaning and chemical cleaning. Chemical cleaning consists of using solvents that are either surface-active chemicals or chemicals which react with the metal/surface. In physical cleaning there is mechanical energy being applied in order to remove contaminants. Physical cleaning includes brush abrasion and ultrasonic agitation. There are different processes by which people can electroplate metals such as by mass plating (also barrel plating), rack plating, continuous plating, and line plating. Each process has its own set of procedures which allow for the ideal plating. Most electroplating coatings can be separated into these categories: | 3,541 | 3,783 |
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. In addition to these individual basis; please contact 1. When a reaction is described as "having reached equilibrium" this means that the forward reaction rate is now equal to the reverse reaction rate. In regards to the amounts or concentrations of the reactants and the products, there is no change due to the forward reaction rate being equal to the reverse reaction rate. 2. It is not correct to say that the reaction has "stopped" when it has reached equilibrium because it is not necessarily a static process where it can be assumed that the reaction rates cancel each other out to equal zero or be "stopped" but rather a dynamic process in which reactants are converted to products at the same rate products are converted to reactants. For example, a soda has carbon dioxide dissolved in the liquid and carbon dioxide between the liquid and the cap that is constantly being exchanged with each other. The system is in equilibrium and the reaction taking place is: \(CO_{2}\,(g)+2\,H_{2}O\,(l)\rightleftharpoons H_{2}CO_3\,(aq)\). 3. Chemical equilibrium is described as a dynamic process because there is a movement in which the forward and reverse reactions occur at the same rate to reach a point where the amounts or concentrations of the reactants and products are unchanging with time. Chemical equilibrium can be described in a saturated solution of \(NaCl\) as on the microscopic level \(Na^+\) and \(Cl^−\) ions continuously leave the surface of an \(NaCl\) crystal to enter the solution, while at the same time \(Na^+\) and \(Cl^−\) ions in solution precipitate on the surface of the crystal. At the macroscopic level, the salt can be seen to dissolve or not dissolve depending whether chemical equilibrium was established. 4. a. Exists in a state of equilibrium as the chemical reaction that occurs in the body is: \(Hb\,(aq)+4\,H_{2}O\,(l)\rightleftharpoons Hb(O_{2})_{4}\,(aq)\). \(H_{2}\,(g)+I_{2}\,(g) \rightleftharpoons 2\,HI\,(g)\) \(2\,NO\,(g)+O_{2}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)\) \(\dfrac{1}{2}\,H_2\,(g)+12\,I_{2}\,(g) \rightleftharpoons HI\,(g)\) \(cis-stilbene\,(soln) \rightleftharpoons trans-silbene\,(soln)\) Give the overall reaction equation and show that \(K = K_1 \times K_2 \times K_3\). 1. By reversing the reactants and products for an equilibrium reaction, the equilibrium constant becomes: \(K′ = \dfrac{1}{K}\). 2. a. This equilibrium is homogenous as all substances are in the same state. b. This equilibrium is heterogeneous as not all substances are in the same state. c. This equilibrium is homogeneous as all substances are in the same state. d. This equilibrium is heterogeneous as not all substances are in the same state. 3. a. This equilibrium is heterogeneous as not all substances are in the same state. b. This equilibrium is heterogeneous as not all substances are in the same state. c. This equilibrium is heterogeneous as not all substances are in the same state. d. This equilibrium is heterogeneous as not all substances are in the same state. 4. According to Le Chatelier’s principle, equilibrium will shift in the direction to counteract the effect of a constraint (such as concentration of a reactant, pressure, and temperature). Thus, in an endothermic reaction, the equilibrium shifts to the right-hand side when the temperature is increased which increases the equilibrium constant and the equilibrium shifts to the left-hand side when the temperature is decreased which decreases the equilibrium constant. 5. After sufficient industrial production of \(NO\) by the reaction of \(N_{2}\,(g)+O_{2}\,(g) \rightleftharpoons 2\,NO_\,(g)\) at elevated temperatures to drive the reaction toward the formation of the product, the reaction mixture is cooled quickly because it quenches the reaction and prevents the system from reverting to the low-temperature equilibrium composition that favors the reactants. 6. To differentiate between a system that has reached equilibrium and one that is reacting slowly that changes in concentrations are difficult to observe we can use Le Chatelier’s principle to observe any shifts in the reaction upon addition of a constraint (such as concentration, pressure, or temperature). 7. The relationship between the equilibrium constant, the concentration of each component of a system, and the rate constants for the forward and reverse reactions considering a reaction of a general form: \(a\,A+b\,B \rightleftharpoons c\,C+d\,D\) is \(K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}=\dfrac{k_f}{k_r}\) 8. a. \(K=\dfrac{[CO_2,H_2]}{[CO,H_2O]}\) \(K_p=\dfrac{(P_{CO_2})(P_{H_2})}{(P_{CO})(P_{H_2O})}\) b. \(K=\dfrac{[PCl_5]}{[PCl_3,Cl_2]}\) \(K_p=\dfrac{(P_{PCl_5})}{(P_{Cl_3})(P_{Cl_2})}\) c. \(K=\dfrac{[O_2]^3}{[O_3]^2}\) \(K_p=\dfrac{(P_{O_2})^3}{(P_{O_3})^2}\) 9. a. \(K=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\) \(K_p=\dfrac{(P_{NO_{2}})^2}{(P_{NO})^2(P_{O_2})}\) b. \(K=\dfrac{[HI]}{[H_2]^{\dfrac{1}{2}}[O_2]}\) \(K_p=\dfrac{(P_{HI})}{(P_{H_{2}})^{\frac{1}{2}}(P_{O_{2}})}\) c. \(K=\dfrac{trans-stilbene}{cis-stilbene}\) \(K_p=\dfrac{(P_{trans-stilbene})}{(P_{cis-stilbene})}\) 10. It is incorrect to state that pure liquids, pure solids, and solvents are not part of an equilibrium constant expression because they are not reactive enough or cause a change in the concentrations of the ions or the species that exist in the gas phase. 11. \(K=\dfrac{[SO_3]^2}{[O_2]^3}\) \(K_p=\dfrac{(P_{SO_3})^2}{(P_{O_2})^3}\) b. \(K=\dfrac{[CO]^2}{[CO_2]}\) \(K_p=\dfrac{(P_{CO})^2}{(P_{CO_2})}\) c. \(K=\dfrac{[SO_2]^2}{[O_2]^3}\) \(K_p=\dfrac{(P_{SO_2})^2}{(P_{O_2})^3}\) 12. a. \(K=[O_2]\) \(K_p=(P_{O_{2}})\) b. \(K=\dfrac{[HI]^2}{[H_2]}\) \(K_p=\dfrac{(P_{HI})^2}{(P_{H_2})}\) c. \(K=[NH_3]^2[CO_2]\) \(K_p=(P_{NH_3})^2(P_{CO_2})\) 13. \(K=\dfrac{[B]}{[A]^2} \rightarrow 1=\frac{[B]}{[A]^2} \rightarrow [A]^2=[B] \rightarrow [A]=\sqrt{B}\) \(K\) and \(K_p\) vary by \(RT\), but it largely depends on \(T\) as \(R\) is constant. A raise or decrease in temperature would cause a difference. \(K_p=K(RT)^{Δn}=K(RT)^{-1}=\dfrac{K}{RT}\) \(Δn=(total\,moles\,of\,gas\,on\,the\,product\,side)-(total\,of\,moles\,on\,the\,reactant\,side)=1-2=−1\) 14. \(K=\dfrac{K_1}{K_2}={\dfrac{\dfrac{[OH^−,HCO_3^−]}{[CO_3^{2−}]}}{\dfrac{[OH^−,H_2CO_3]}{[HCO_3^−]}}=\dfrac{[HCO_3^−]^2}{[CO_3^{2-},H_2CO_3]}}\) \(CO_3^{2-}\,(g)+H_2CO_{3}\,(g) \rightleftharpoons 2\,HCO_3^{-}\,(g)\) 15. \(K = K_1 \times K_2 \times K_3=\frac{[ES]}{[E,S]}\times\frac{[ESO_2]}{[ES,O_2]}\times\frac{[E,P]}{[ESO_2]}=\frac{[P]}{[S,O_2]}\) \(S+O_2 \rightleftharpoons P\) How are these two expressions mathematically related to the equilibrium constant expression for \[N_{2}\,(g)+3\,H_{2}\,(g) \rightleftharpoons 2\,NH_{3}\,(g) ?\] a. \(2\,NO\,(g) + O_{2}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)\) b. \(\frac{1}{2}H_{2}\,(g)+\frac{1}{2}I_{2}\,(g) \rightleftharpoons HI\,(g)\) c. \(CaCO_{3}\,(s) + 2\,HOCl\,(aq) \rightleftharpoons Ca^{2+}\,(aq) + 2\,OCl^−\,(aq) + H_2O\,(l) + CO_{2}\,(g)\) 6. Calculate \(K\) and \(K_p\) for each reaction. 7. Calculate \(K\) and \(K_p\) for each reaction. 8. Determine \(K\) and \(K_p\) (where applicable) for each reaction. 9. Determine \(K\) and \(K_p\) for each reaction. 10. The equilibrium constant expression for a reaction is \(\dfrac{[CO_2]^2}{[SO_2]^2[O_2]}\). What is the balanced chemical equation for the overall reaction if one of the reactants is \(Na_2CO_{3}\,(s)\)? 11. The equilibrium constant expression for a reaction is \(\dfrac{[NO,H_{2}O]^{\dfrac{3}{2}}}{[NH_3,O_2]^{\dfrac{5}{4}}}\). What is the balanced chemical equation for the overall reaction? 12. Given \(K =\dfrac{k_f}{k_r}\), what happens to the magnitude of the equilibrium constant if the reaction rate of the forward reaction is doubled? What happens if the reaction rate of the reverse reaction for the overall reaction is decreased by a factor of 3? 13. The value of the equilibrium constant for \[2\,H_{2}\,(g)+S_{2}\,(g) \rightleftharpoons 2\,H_2S\,(g)\] is \(1.08 \times 10^7\) at 700°C. What is the value of the equilibrium constant for the following related reactions 1. In the given equilibrium reaction where \(K = 0.892\approx1\) has a concentration of the reactants that is approximately equal to the concentration of the products so neither formation of the reactants or products is favored. In the given equilibrium reaction where \(K = 3.25 \times 10^8>1\) has a concentration of the products that is relatively small compared to the concentration of the reactants so the formation of the products is favored. In the given equilibrium reaction where \(K = 5.26 \times 10^{−11}<1\) has a concentration of products that is relatively large compared to that of the concentration of the reactants so the formation of the reactants is favored. 2. a. \(K=\dfrac{[NO_2]^{2}}{[N_2O_4]}\) b. \(K=\dfrac{[NO_2]}{[N_2O_4]^{\dfrac{1}{2}}}\) Although the equilibrium constant expressions have a 2:1 ratio of concentration for the products to the concentration of the reactants for the same species involved to get the \(K\) value for a. We would need to square it to get the \(K\) value for b. 3. \(K’=\dfrac{[NH_3]}{[N_2]^{\dfrac{1}{2}}[H_2]^{\dfrac{3}{2}}}\) \(K’’=\dfrac{[NH_3]^{\dfrac{2}{3}}}{[N_2]^{\dfrac{1}{2}}[H_2]}\) \(K=\dfrac{[NH_3]^2}{[N_2,H_2]^3}\) \(K'=K^{\dfrac{1}{2}}\) \(K''=K^{\dfrac{1}{3}}\) 4. a. \(K=\dfrac{[H_2]^2[CO_2]}{[H_{2}O]^2}\) b. \(K=\dfrac{[SbCl_5]}{[SbCl_3,Cl_2]}\) c. \(K=\dfrac{[O_2]^3}{[O_3]^2}\) 5. 6. a. \(K=\dfrac{[NO]^2[Br]}{[NOBr]^2}=\frac{[1.29\,M]^2[10.52\,M]}{[0.423\,M]^2}=97.8\) \(K_p=K(RT)^{Δn}=(97.8)((0.08206\frac{L\cdot atm}{mol\cdot K})((727+273.15)K))^{3-2}=8.03x10^{4}\) b. \(K_p=K(RT)^{Δn}=K(RT)^{2-1}=K(RT) \rightarrow K=\dfrac{K_p}{RT}=\frac{63.1}{(0.08206\frac{L\cdot atm}{mol\cdot K})(1,200\,K)}=6.41\) \(K=\dfrac{(P_{CO})^2}{P_{CO_2}}=\frac{(76.8\,atm)^2}{93.5\,atm}=63.1\) 7. \(K=\dfrac{[NO_{2}]^2}{[N_{2}O_{4}]}=\frac{[0.001368\,M]}{[0.149316\,M]}=1.25x10^{-5}\) \([NO_2]=(2)(0.150\,M)(0.00456)=0.001368\,M\) \([N_2O_4]=(0.150\,M)(1-0.00456)=0.149316\,M\) \(K_p=K(RT)^{Δn}=(1.25x10^{-5})((0.08206\frac{L\cdot atm}{mol\cdot K})((-40+273.15)K))^{2-1}=2.39x10^{-4}\) b. \(K=\dfrac{[CH_{3}OH]}{[CO,H_{2}]^2}=\frac{[14.5\,M]}{[1.05\,M,1.21\,M]^2}=9.47\) \([CH_{3}OH]=7,193\,g\,CH_{3}OH \times \frac{1\,mol\,CH_{3}OH}{32.04\,g\,CH_{3}OH} \times \frac{1}{15.5\,L}=14.5\,M\) \([CO]=457.7\,g\,CO \times \frac{1\,mol\,CO}{28.01\,g\,CO} \times \frac{1}{15.5\,L}=1.05\,M\) \([H_{2}]=37.8\,g\,H_{2} \times \frac{1\,mol\,H_{2}}{2.02\,g\,H_{2}} \times \frac{1}{15.5\,L}=1.21\,M\) \(K_p=K(RT)^{Δn}=(9.47)((0.08206\frac{L\cdot atm}{mol\cdot K})((227+273.15)K))^{1-3}=5.62x10^{-2}\) 8. a. \(K=\dfrac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}=\frac{[1.00 \times 10^{-3}\,M]^2[1.20 \times 10^{-3}\,M]}{[3.32 \times\ 10^{-3}\,M]^2}=1.09 \times 10^{-4}\) \(K_p=K(RT)^{Δn}=(1.09 \times 10^{-4})((0.08206\frac{L\cdot atm}{mol\cdot K})((1065+273.15)K))^{3-2}=1.20 \times 10^{-2}\) b. \(K=[OH^{-}]^2[Ba^{2+}]=[0.2136\,M]^2[0.1068\,M]=4.87 \times 10^{-3}\) \([OH^{-}]=0.0534\,mol\,OH^{-} \times \frac{1}{0.25\,L}=0.2136\,M\) \([Ba^{2+}]=0.0267\,mol\,Ba^{2+} \times \frac{1}{0.25\,L}=0.1068\,M\) \(K_p=K(RT)^{Δn}=(4.87 \times 10^{-3})((0.08206\frac{L\cdot atm}{mol\cdot K})((25+273.15)K))^{3-1}=2.92\) 9. a. \(K=\dfrac{[NO]^2[Cl_{2}]}{[NOCl]^2}=4.59 \times 10^{-4}\) \(K_p=K(RT)^{Δn}=(4.59 \times 10^{-4})((0.08206\frac{L\cdot atm}{mol\cdot K})(500K))^{3-2}=1.88 \times 10^{-2}\) b. \(K=\dfrac{[PCl_5]}{[Cl_{2},PCl_{3}]}=\frac{[9.80 \times 10^{-2}\,M]}{[4.77 \times 10^{-1}\,M,7.28 \times 10^{-3}\,M]}=28.2\) \([PCl_{5}]=10.2\,g\,PCl_{5} \times \frac{1\,mol\,PCl_{5}}{208.2388\,g\,PCl_{5}} \times \frac{1}{0.5\,L}=9.80 \times 10^{-2}\,M\) \([Cl_{2}]=16.9\,g\,Cl_{2} \times \frac{1\,mol\,Cl_{2}}{70.9\,g\,Cl_{2}} \times \frac{1}{0.5\,L}=4.77 \times 10^{-1}\,M\) \([PCl_{3}]=0.500\,g\,PCl_{3} \times \frac{1\,mol\,PCl_{3}}{137.33\,g\,PCl_{3}} \times \frac{1}{0.5\,L}=7.28 \times 10^{-3}\,M\) \(K_p=K(RT)^{Δn}=(28.2)((0.08206\frac{L\cdot atm}{mol\cdot K})(250+273.15)K)^{1-2}=6.57 \times 10^{-1}\) 10. \(2\,SO_{2}\,(g)+O_{2}\,(g)+2\,Na_{2}CO_{3}\,(s) \rightleftharpoons 2\,CO_{2}\,(g)+2\,Na_{2}SO_{4}\,(s)\) 11. \(NH_{3}\,(g) + \frac{5}{4}\,O_{2}\,(g)⇌NO \,(g)+\frac{3}{2}\,H_{2}O\,(g)\) a. \(K= \dfrac{[H_{2}S]}{[H_{2},S_{2}]^\frac{1}{2}}=K’^{\frac{1}{2}}=(1.08 \times 10^{7})^{\frac{1}{2}}=3.29 \times 10^{3}\) b. \(K= \dfrac{[H_{2}S]^4}{[H_{2}]^4[S_{2}]^2}=K’^{2}=(1.08 \times 10^{7})^{2}=1.17 \times 10^{14}\) c. \(K= \dfrac{[H_{2},S_2]^\frac{1}{2}}{[H_{2}S]}=K’^{-\frac{1}{2}}= (1.08 \times 10^{7})^{-\frac{1}{2}}=3.04 \times 10^{-4}\) 1. The magnitude of the equilibrium constant for a reaction depends on the form in which the chemical reaction is written. For example, writing a chemical reaction in different but chemically equivalent forms causes the magnitude of the equilibrium constant to be different but can be related by comparing their respective magnitudes. 2. a. When \(K\) is very large the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion. \(K= \dfrac{[C]}{[A,B]^2}=\frac{[C]}{very\,small}=\frac{1}{0}= \infty \rightarrow [C]= \infty\) b. When \(K\) is very small the reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of the reactants. \(K=\dfrac{[C]}{[A,B]^2}=\frac{very\,small}{[A,B]^2}=\frac{0}{1}=0 \rightarrow [C]=0\) Simplifying assumptions should not be used if the equilibrium constant is not known to be very large or very small. 1. In both cases, the equilibrium constant will remain the same as it does not depend on the concentrations. 2. No, the data is not consistent with what I would expect to occur because enthalpy is positive indicating that the reaction is endothermic thus heat is on the left side of the reaction. As the temperature is raised \(P_{O_2}\) would be expected to increase to counteract the constraint. 3. If the initial pressure of \(N_2O_4\) was doubled then \(K_p\) is one half of the original value. \(K_p=\dfrac{(P_{NO_{2}})^2}{(P_{N_{2}O_{4}})} \rightarrow 1.7 \times 10^{-1} = \frac{(2x)^2}{2.6 \times 10^2 -x} \rightarrow 44.2-0.17x=4x^2 \rightarrow x \approx 3.303\,atm\) \(P_{N_{2}O_{4}}=2.6 \times 10^2-x=260-3.303=2.6 \times 10^2\,atm\) \(P_{NO_{2}}=2x=(2)(3.303)=6.6\,atm\) 4. \(K=\frac{[HI]^2}{[H_2,I_2]}=\frac{0.345\,M}{(0.047\,M)(0.047\,M)}=157\) \(K_p=K(RT)^{Δn}=(157)((0.08206\frac{L\cdot atm}{mol\cdot K})(430+273.15)K)^{2-2}=157\) 5. \(Maximum\;Percent\;Yield=\frac{Actual}{Theoretical} \times 100\%=\frac{212.593}{376.127} \times 100\%=56.52\% \approx 57\%\) \(PV=nRT \rightarrow P=\frac{(1.999\;mol)(0.08206\frac{L\cdot atm}{mol\cdot K})(300+273.15)K}{0.250\;L}=376.127\;atm\) \([CO]=56.0\;g\;CO \times \frac{1\;mol\;CO}{28.01\,g\,CO}=1.999\,mol\,CO\) \(K_p=\frac{P_{CH_{3}OH}}{(P_{CO})(P_{H_2})^{2}} \rightarrow 1.3 \times 10^{-4}= \frac{x}{(376.02-x)(100^2)} \rightarrow 1.3= \frac{x}{376.07-x} \rightarrow 488.965-1.3x=x \rightarrow 488.965=2.3x \rightarrow x=212.593\,atm\) \(K_p=\frac{(P_{CH_{3}OH})}{(P_{CO})(P_{H_2})^2} \rightarrow 1.3 \times 10^{-4}= \frac{357.320}{(376.127-357.320)(P_{H_{2}})^2} \rightarrow 1.3 \times 10^{-4}= \frac{357.320}{(18.80635)(P_{H_{2}})^2} \rightarrow 0.002444(P_{H_{2}})^2=357.320 \rightarrow (P_{H_{2}})=382.300 \approx 3.8 \times 10^2\,atm\) \(Minimum\,Percent\,Yield=\frac{Actual}{Theoretical} \times 100 \% \rightarrow 95\%=\frac{x}{376.127} \times 100\% \rightarrow x=357.320\,atm\) 6. \(K_p=\frac{(P_B)^2(P_C)}{P_A}=\frac{[2x]^2[x]}{[0.969-x]}=\frac{4x^3}{0.969-x}\) 7. \(P_{CO_{2}}=P_{tot}-P_{NH_{3}}=P_{tot}-0.242\,atm\) \(P_{tot}=P_{NH_3}+P_{CO_2}=0.242\,atm+P_{CO_2}\) \(K_p=(P_{NH_3})^2(P_{CO_2})=(0.242\,atm)^2(P_{CO_2})\) 8. a. \(At\,375\,K:K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{2.9 \times 10^{-2}}{((0.08206\frac{L\cdot atm}{mol\cdot K})(375\,K))^{2-1}}=7.80 \times 10^{-2}\) \(At\,303\,K:K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{2.9 \times 10^{-2}}{((0.08206\frac{L\cdot atm}{mol\cdot K})(303\,K))^{2-1}}=1.17 \times 10^{-3}\) b. \(K=\frac{[SO_{2},Cl_{2}]}{[SO_{2}Cl_{2}]} \rightarrow [SO_{2}Cl_{2}]=\frac{[0.200\,M,0.100\,M]}{7.80 \times 10^{-2}}=2.56 \times 10^{-1}\,M\) c. If the sample given in part b is cooled to 303 \(K\), the pressure inside the bulb would decrease. 9. \(K_p=\frac{(P_B)^b}{(P_A)^a}=\frac{((\frac{n_B}{V})(RT))^b}{((\frac{n_A}{V})(RT))^a}=\frac{[B]^{b}(RT)^b}{[A]^a(RT)^a}=K(RT)^{b-a}=K(RT)^{Δn}\) \(PV=nRT \rightarrow P=\frac{n}{V}RT\) \(K=\frac{[B]^{b}}{[A]^{a}}\) \(Δn=b-a\) 10. \(P_T=P_I+P_{I_2}=\sqrt{K_pP_{I_2}}+P_{I_2}\) \(K_p=\frac{(P_I)^2}{P_{I_2}} \rightarrow (P_I)^2=K_p(P_{I_2}) \rightarrow P_I=\sqrt{K_pP_{I_2}}\) 11. The graph should be a positive linear correlation. \([Br_2\,(l)]=1.0\,g\,Br_2 \times \frac{1\,mol\,Br_2}{159.808\,g\,Br_2} \times \frac{1}{0.1\,L}=6.26 \times 10^{-2}\) \(K=\frac{[Br_2\,(aq)]}{[Br_2\,(l)]}=\frac{[Br_2\,(aq)]}{1}=[Br_2\,(aq)]\) 12. \(K=\frac{[isobutane]}{[n-butane]}=\frac{x}{1-x}\) 13. \(P_{NH_3}=2x=2(0.0387)=7.73 \times 10^{-2}\,atm\) \(P_{CO_2}=x=3.87 \times 10^{-2}\,atm\) \(K_p=(P_{NH_3})^2(P_{CO_2})=(2x)^2(x)=4x^3=4(0.0387)^3=2.32 \times 10^{-4}\) \(P_{tot}=P_{NH_3}+P_{CO_2} \rightarrow 0.116=2x+x \rightarrow 0.116=3x \rightarrow x=0.0387\) If the concentration of \(CO_{2}\) is doubled and then equilibrates to its initial equilibrium partial pressure +x atm, the concentration of \(NH_{3}\) should also be doubled for the system to restore equilibrium. 14. \(P_{COCl_{2}}=9.34 \times 10^{-2}-x=9.34 \times 10^{-2}-9.34 \times 10^{-22}=9.34 \times 10^{-2}\,atm\) \(P_{CO}=x=9.34 \times 10^{-22}\,atm\) \(P_{Cl_{2}}=x=9.34 \times 10^{-22}\,atm\) Assume that the equilibrium mainly lies on the reactants side because the \(K_p\) value is less than 1. \(K_p=\frac{(P_{CO})(P_{Cl_{2}})}{(P_{COCl_{2}})} \rightarrow 2.2 \times 10^{-10} =\frac{x^{2}}{9.34 \times 10^{-2}-x} \rightarrow 2.0548 \times 10^{-11}-2.2 \times 10^{-10}x=x^{2} \rightarrow x^{2}+2.2 \times 10^{-10}x-2.0548 \times 10^{-11}=0 \rightarrow x=9.34 \times 10^{-22}\) \(PV=nRT \rightarrow P=\frac{nRT}{V}=(3.05 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})(100+273.15)K=9.34 \times 10^{-2}\) 15. \([H_{4}IO_{6}^{-}]=x=1.6 \times 10^{-3}\,mol\) \(K=\frac{[H_{4}IO_{6}^{-}]}{[IO_{4}^{-}]} \rightarrow 3.5 \times 10^{-2} =\frac{x}{(0.0448-x)} \rightarrow x=1.568 \times 10^{-3}\) \(IO_{4}^{-}:50\,mL\,IO_{4}^{-} \times \frac{1\,L\,IO_{4}^{-}}{1,000\,mL\,IO_{4}^{-}} \times \frac{0.896\,mol\,IO_{4}^{-}}{1\,L\,IO_{4}^{-}}=0.0448\,mol\) 16. \(PV=nRT \rightarrow \frac{P}{RT}=\frac{n}{V} \rightarrow \frac{12.9468}{(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K}=3.5 \times 10^{-1}\,M\) \(K_p=\frac{(P_{IBr})^2}{(P_{I_{2}})(P_{Br_{2}})} \rightarrow 1.2 \times 10^{-2} = \frac{2x}{(1.096-x)(1.479-x)}=\frac{2x}{x^2-2.575x+1.62098} \rightarrow 0.012x^2-0.0309x+0.0194518=2x \rightarrow 0.012x^2-2.0309+0.0194518=0 \rightarrow x=12.9468\) \(PV=nRT \rightarrow P=\frac{nRT}{V}=(2.92 \times 10^{-2}\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K=1.096\,atm\) \([I_{2}]=7.4\,g\,I_{2} \times \frac{1\,mol\,I_{2}}{253\,g\,I_{2}} \times \frac{1}{1.00\,L}=2.92 \times 10^{-2}\,M\) \(PV=nRT \rightarrow P={nRT}{V}=(3.94 \times 10^{-2}\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K=1.479\,atm\) \([Br_{2}]=6.3\,g\,Br_{2} \times \frac{1\,mol\,Br_{2}}{159.808\,g\,Br_{2}}=3.94 \times 10^{-2}\,M\) 17. \([N_{2}]=0.5-12x=0.5-12(0.000471330)=0.494\,M\) \([CH_{3}NH_{2}]=2x=2(0.0004713300=9.43 \times 10^{-4}\,M\) \(If\,the\,concentration\,of\,H_{2}\,is\,doubled\,,then\,K=\frac{[CH_{3}NH_{2}]^{2}}{[N_2,H_2]^5}=\frac{(9.43 \times 10^{-4})^{2}}{(0.494)(1.998)^5}=5.65 \times 10^{-8}\) \(2 \times [H_{2}]=2(1.0-2.5x)=2(1.0-2.5(0.000471330))=1.998\,M\) \(K=\frac{[CH_{3}NH_{2}]^2}{[N_2,H_2]^5} \rightarrow 1.8 \times 10^{-6}=\frac{(2x)^2}{(0.5-x)(1.0-5x)^5} \rightarrow x=0.000471330\,M\) \(H_2\) \([N_2]=1.00\,mol\,N_2 \times \frac{1}{2.00\,L}=0.5\,M\) \([H_2]=2.00\,mol\,H_2 \times \frac{1}{2.00\,L}=1.0\,M\) 1. During a set of experiments, graphs were drawn of [reactants] versus [products] at equilibrium. Using Figure 15.8 and Figure 15.9 as your guides, sketch the shape of each graph using appropriate labels. 2. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium? 3. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium? 1. a. According to Figure 15.8, we could obtain a graph with the x-axis labeled \([H_2O]\,(l)\,(M)\) and y-axis labeled \([H_2O]\,(g)\,(M)\). The graph should have a positive linear correlation. For any equilibrium concentration of \(H_2O\,(g)\), there is only one equilibrium \(H_2O\;(l)\). Because the magnitudes of the two concentrations are directly proportional, a large \([H_2O]\,(g)\) at equilibrium requires a large \([H_2O]\,(l)\) and vice versa. In this case, the slope of the line is equal to \(K\). b. According to Figure 15.9, we could obtain a graph with the x-axis labeled \([O2]\,(M)\) and y-axis labeled \([MgO]\,(M)\). Because \(O_2\,(g)\) is the only one in gaseous form, the graph would depend on the concentration of \(O_2\). c. According to Figure 15.8, we could obtain a graph with the x-axis labeled \([O_3]\,(M)\) and y-axis labeled \([O2]\,(M)\). The graph should have a positive linear correlation. For every \(3\,O_2\,(g)\) there is \(2\,O_3\,(g)\). Because the magnitudes of the two concentrations are directly proportional, a large \([O3]\,(g)\) at equilibrium requires a large \([O_2]\,(g)\) and vice versa. In this case, the slope of the line is equal to \(K\). d. According to figure 15.8, we could obtain a graph with the x-axis labeled \([O2]\,(M)\) and y-axis labeled \([SO2]\,(M)\). The graph should have a positive linear correlation. For every \(3\,O_{2}\,(g)\) there is \(2\,SO_2\,(g)\). Because the magnitudes of the two concentrations are directly proportional, a large \(O_2\,(g)\) at equilibrium requires a large \(SO_2\,(g)\) and vice versa. In this case, the slope of the line is equal to \(K\). 2. a. \(K=[Na_{2}CO_{3},CO_2,H_{2}O]\) If \([CO_2]\) is doubled, \([H_2O]\) should be halved if the system is to maintain equilibrium. b. \(K=\frac{[NF_2]^2}{[N_{2}F_{4}]}\) If \([NF_2]\) is decreased by a factor of 2, then \([N_{2}F_{4}]\) must also be decreased by a factor of 2 if the system is to maintain equilibrium. c. \(K=\frac{[HI]^2}{[H_2,I_{2}]}\) If \([I_{2}]\) is doubled then \([HI]\) must also be doubled if the system is to maintain equilibrium. 3. \(K=\dfrac{[CH_4,H_2S]^2}{[CS_2,H_2]^4}\) If \([CS_2]\) is doubled then \([H_2]\) must be decreased by a factor of 2√4≅ 1.189 if the system is to maintain equilibrium. b. \(K=\dfrac{[PCl_3]}{[Cl_2,PCl_5]}\) If \([Cl_2]\) is halved then \([PCl_5]\) must also be halved if the system is to maintain equilibrium. c. \(K=\dfrac{[NO]^4[H_2O]^6}{[NH_3,O_2]^5}\) If \([NO]\) is doubled then \([H_2O]\) must also be multiplied by 22/3≅1.587 if the system is to maintain equilibrium. atm atm atm \[CoO\,(s)+H_{2}\,(g) \rightleftharpoons Co\,(s)+H_2O\,(g) \text{ with } K=67\] \[CoO\,(s)+CO\,(g) \rightleftharpoons Co\,(s)+CO_{2}\,(g) \text{ with } K=490\] The system is not at equilibrium at each of these higher pressures. To reach equilibrium, the reaction will proceed to the right to decrease the pressure because the equilibrium partial pressure is less than the total pressure. \(K_p=\frac{[NH_{3}]^2}{[N_{2},H_{2}]^3}=\frac{[15.20]^2}{[19.17,65.13]^3}=4.4 \times 10^{-5}\) \(K_p=\frac{[NH_{3}]^{2}}{[N_{2},H_{2}]^3}=\frac{[321.6]^2}{[56.74,220.8]^3}= 1.7 \times 10^{-4}\) 2. 1. \(K=\frac{[B,C]}{[A]}=\frac{[2.50,2.50]}{[2.50]}=2.50\) 2. \(K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{19.0}{((0.08206\frac{L\cdot atm}{mol\cdot K})(200+273.15)K)^{2-1})}=0.49\) \(K_p=\frac{(P_B)(P_C)}{(P_A)}=\frac{(1.75)(14.15)}{1.30}=19.0\) 3. \(K=\frac{[B,C]}{[A]^{2}}=\frac{(18.72)(6.51)}{12.61}=9.7\) Experiment 1 is about the same as the given \(K\) value and thus considered to be about equilibrium. The second experiment has a \(K\) value that is about 1 so neither the formation of the reactants or products is favored. The third experiment has a \(K\) value that is larger than 1 so the formation of the products is favored. 3. a. \(K=\frac{[H_{2}O]}{[H_{2}]}\) \(K=\frac{[CO_{2}]}{[Co]}\) b. \([H_2]=[CO]=0.316\,mol\,H_{2} \times \frac{1}{1.00\,L}=0.316\,M\) \([CoO]=0.5\,mol\,CoO \times \frac{1}{1.00\,L}=0.5\,M\) Reaction 1: \(PV=nRT \rightarrow P=\frac{nRT}{V}=(4.65 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=0.314\) \([H_{2}]=0.316-x=0.316-0.311=4.65x10^{-3}\) \(PV=nRT \rightarrow P=\frac{nRT}{V}=(0.311)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=21.0\) \([H_{2}O]=x=0.311\) \(K=\frac{[H_{2}O]}{[H_{2}]} \rightarrow 67=\frac{x}{0.316-x} \rightarrow x=0.311\) Reaction 2: \(PV=nRT \rightarrow P=\frac{nRT}{V}=(0.001)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=6.75 \times 10^{-2}\,atm\) \([CO]=0.316-x=0.316-0.315=0.001\,M\) \(PV=nRT \rightarrow P=\frac{nRT}{V}=(0.315)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=21.3\,atm\) \([CO_{2}]=x=0.315\,M\) \(K=\frac{[CO_{2}]}{[Co]} \rightarrow 490=\frac{x}{0.316-x} \rightarrow x=0.315\) c. \(H_{2}\,(g)+CO_{2}\,(g) \rightleftharpoons CO\,(g)+H_{2}O\,(g)\) \(K_p=\frac{(P_{CO})(P_{H_{2}O})}{(P_{H_{2}})(P_{CO_{2}})}=\frac{(6.75 \times 10^{-2})(21)}{(0.314)(21.3)}=0.21\) d. The shape of the graphs [reactants] versus [products] does not change as the amount of \(CoO\) changes because it is a solid. 4. \(PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{n}{V}=\frac{0.101798}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.5 \times 10^{-4}\,M\,HI\) \([HI]=2x=2(0.050899)=0.101798\,atm\) \(K_p=\frac{(P_{HI})^2}{(P_{H_{2}})(P_{I_{2}})} \rightarrow 54.5=\frac{(2x)^2}{(3.2 \times 10^{-1}-x)(5.16 \times 10^{-2}-x)} \rightarrow x=0.050899\,atm\) \(PV=nRT \rightarrow P=\frac{nRT}{V}=(5.6 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=3.2 \times 10^{-1}\,atm\) \([H_{2}]=1.12 \times 10^{-2}\,mol\,H_{2} \times \frac{1}{2.0\,L}=5.6 \times 10^{-3}\,M\) \(PV=nRT \rightarrow P=\frac{nRT}{V}=(9.0 \times 10^{-4})(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=5.16 \times 10^{-2}\,atm\) \([I_2]= 1.8 \times 10^{-3}\,mol\,I_{2} \times \frac{1}{2.0\,L}=9.0 \times 10^{-4}\,M\) For excess hydrogen: \(Q=\frac{[HI]}{[H_{2},I_{2}]}=\frac{1.8 \times 10^{-3}}{(594.410)(1.09 \times 10^{-3})}=2.8 \times 10^{-3}\) The reaction will proceed to the right to reach equilibrium. \(PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{0.103162}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.8 \times 10^{-3}\,M\) \([HI]=2x=2(0.051581)=0.103162\,atm\) \(PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{10.375}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=594.410\,M\) \([H_{2}]=10.427-x=10.427-0.051581=10.375\,atm\) \(PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{1.9 \times 10^{-5}}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.09 \times 10^{-3}\,M\) \([I_{2}]=5.16 \times 10^{-2}-x=5.16 \times 10^{-2}-0.051581=1.9 \times 10^{-5}\,atm\) \(K_p=\frac{(P_{HI})^2}{(P_{H_{2}})(P_{I_{2}})} \rightarrow 54.5=\frac{(2x)^2}{(10.427-x)(5.16 \times 10^{-2}-x)} \rightarrow x=0.051581\,atm\) \(PV=nRT \rightarrow P=\frac{nRT}{V}=(0.182\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=10.427\,atm\) \([H_{2}]=3.64 \times 10^{-1}\,mol\,H_{2} \times \frac{1}{2.0\,L}=0.182\,M\) | 27,654 | 3,785 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Key_Elements_of_Green_Chemistry_(Lucia)/02%3A_Life-Cycle_Analysis/2.02%3A_LCA_LCIA_Concepts |
Assumptions inherent in an LCA study are apt to change the results and conclusions derived from analysis. In addition, many different types of studies require various levels data collection and analysis. The goal and scope of a LCA defines its intent, targeted audience, and use. The intended use informs further decisions for scope, functional unit of comparison, and data collection. For example, if a LCA study is used internally, a full review panel of LCA experts is not required; however, when providing public environmental claims about a competing product, a review is required. A life cycle inventory (LCI) is the most laborious step of a LCA: data is collected and organized. It often involves contacting companies, accumulating literature sources, and building models using life cycle assessment software. Materials flows, types of
materials, product life time, and product energy requirements are collected in the LCI phase. A life cycle impact assessment (LCIA) part of the analysis process collects life cycle inventory data and delivers environmental impacts values. This process greatly reduces the complexity of the data set from hundreds of inputs to 10 or fewer impact categories for decision-making. There are many different methods for LCIA based on location, goals, and scope. The interpretation step derives from what was found in the other steps for the generation of new information. It is not the last step but iterative. When it is done, the study assumptions, goals, scope, and methods are refined to suit the needs of the study. The first step is defining the goal to give the aim and what it encompasses. There are two types of LCA objectives: (1) descriptive and (2) change-oriented. The descriptive types look at broader aspects of an issues, e.g., how much of the world’s carbon dioxide emissions are derived from commuters (light duty vehicles). These broader environmental questions fall within the domain of descriptive LCAs. The second type of LCA is change-oriented, in which two options for fulfilling a function are compared. Typical examples of change-oriented LCAs are paper vs. plastic, flying vs. driving, and gas vs. electric heating. These types of studies can guide the choice of methods to reduce environmental impacts. The intended audience is another part of the goal and scope. The audience may include interest groups such as policy makers, company marketing groups, or product development teams. Additionally, interest groups should be identified. These include companies, funding sources, target audiences, and expert reviewers. It is noted that the intended use of the LCA may be different from the end use because the information may be relevant to other decisions and analyses beyond the original intent. One specific LCA to compare two products is a “comparative assertion disclosed to the public”. In this type of study, “environmental claims regarding the superiority or equivalence of one product vs a competing product which performs the same function” are communicated. These types of studies must follow ISO 14044 standards with the nine steps for a “comparative assertion”. The scope definition serves the purpose of communicating to the audience what is included and what is excluded. Depending on the goal, there are several types of scopes including cradle-to-gate, cradle-to-grave, and gate-to-gate. There are other words commonly used to describe these scopes: The scope must be carefully selected in consideration of the potential implications not including product stages or phases in the scope of the work. For example, a product may have lower product emissions, but have a shorter lifetime than an alternative product that would not be communicated in a process stage diagram as seen in Figure \(\Page {1}\) These types of diagrams list the major unit steps considered and clearly show what is not included. Temporal boundaries are also established in the scope. Assumptions relating to time can have a large influence on the results. A study timeframe should be picked which will best capture the impacts of the product or processes. A 100-year window is a common temporal boundary, for example, in global warming. In a 100-year temporal window, impacts occurring after 100 years are not part of the overall analysis. Other aspects to be included in scope are technology and geographical regions. Many studies are spatially dependent so LCA results are not broadly applicable to other regions. Products or services from older technologies often have different impacts than current technologies. Thus, it is important to communicate the type and stage of the technology. In addition, allocation procedure impact assessment methods and should be reported. A functional unit is the primary measure of a product or service. ISO states that “the functional unit defines the quantification of the identified functions (performance characteristics) of the product. The primary purpose of a functional unit is to provide a reference to which the inputs and outputs are related. This reference is necessary to ensure comparability of LCA results. The functional unit can be a service, mass of material, or an amount of energy. Selecting appropriate functional units is critical to creating an unbiased analysis. For example, when comparing trains to cars for transportation, the comparison may suffer from the inability to correlate energy inputs and outputs. The real purpose of the train would be to deliver a larger number of people to a specific centralized location. For this example, a better functional unit may be impacts of a train delivering a specific number of people over a specified distance. The results will then be normalized to distance for more reasonable correlations and assessments. Data collection for an LCA is the most time-intensive and laborious step. Cut-off criteria are used to expedite the process. Cut-off criteria define a level of product content or other parameter to which the study will not consider. One example: materials contents less than 1% of the total product mass are not considered. This allows the LCA practitioner to focus on data from the main flows of the system while systematically eliminating flows which may not influence the results. In addition to data collection, the life cycle inventory (LCI) step is a very laborious aspect of life cycle assessment. The data collected for the product, production process, and product life cycle are used in the impact assessment to determine the environmental impacts. Collecting consistent, transparent, and accurate LCI data is critical to the success of an overall LCA. Example data collected: Tracking the material flows into and out of the defined system is the first step of LCI. After the materials flows have been determined through interviews, literature searches, and measurement, LCA software can be used to track the material process’s elementary flows to and from the environment. Elementary flows originate in the environment and are mined or retrieved for use in a process or flows that are released from processes to the environment and are not used by other processes. These elementary flows are the actual materials used and materials released to the environment as a result of the studied product system. In Figure \(\Page {2}\), the two types of LCI data can be seen. On the top half of the figure, process flows such as products, services and other goods are listed. The lower half lists elementary flows such as chemicals released to soil or air. Life cycle impact assessment (LCIA) is among the last steps of LCA. The purpose of a LCIA “is to provide additional information to assess life cycle inventory (LCI) results and help users better understand the environmental significance of natural resource use and environmental releases”. The LCIA helps provide significance and results for easier decision making; however, it is important to understand it does not directly measure the impacts of chemical releases to the environment as an environmental risk assessment does. The third step of LCIA follows sequentially after the LCI using the many flows to and from the environment developed in the LCI. These LCI flows, without an impact assessment step, are not easily interpreted and understanding the significance of emissions is impossible. The LCIA is different from a risk assessment measuring absolute values of environmental impacts in that the LCIA helps determine the significance of emissions and impacts in relation to the study scope. The absolute value of the impacts cannot be determined by the LCIA due to (Margni and Curran 2012): Even though the LCIA has limitations, it is useful in determining what impacts matter, what unit processes are contributing the most through hot spot analysis and identify best scenario options when environmental tradeoffs occur. According to ISO there are three mandatory processes of a LCIA including Selection of impact categories, Classification, and Characterization, Figure \ . | 9,031 | 3,786 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity_of_Alpha_Hydrogens/Alpha_Alkylation |
Enolates can act as a nucleophile in S 2 type reactions. The alpha alkylation reaction involves an α hydrogen being replaced with an alkyl group. This reaction is one of the more important for enolates because a carbon-carbon bond is formed. These alkylations are affected by the same limitations as S 2 reactions previously discussed. Good leaving groups like chloride, bromide, iodide, tosylate, should be used. Also, secondary and tertiary leaving groups should not be used because of poor reactivity and possible competition with elimination reactions. Lastly, it is important to use a strong base, such as LDA or sodium amide, for this reaction. Using a weaker base such as hydroxide or an alkoxide leaves the possibility of multiple alkylation’s occurring. Stpe 1: Enolate formation Step 2: S 2 attack Please write the structure of the product for the following reactions. | 896 | 3,787 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Averill_et_al./09.E%3A_Molecular_Geometry_(Exercises) |
" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact ♦ Sulfur hexafluoride (SF ) is a very stable gas that is used in a wide range of applications because it is nontoxic, nonflammable, and noncorrosive. Unfortunately, it is also a very powerful “greenhouse gas” that is about 22,000 times more effective at causing global warming than the same mass of CO . ♦ The elevated concentrations of chlorine monoxide (ClO) that accompany ozone depletions in Earth’s atmosphere can be explained by a sequence of reactions. In the first step, chlorine gas is split into chlorine atoms by sunlight. Each chlorine atom then catalyzes the decomposition of ozone through a chlorine monoxide intermediate. ♦ Saccharin is an artificial sweetener that was discovered in 1879. For several decades, it was used by people who had to limit their intake of sugar for medical reasons. Because it was implicated as a carcinogen in 1977, however, warning labels are now required on foods and beverages containing saccharin. The structure of this sweetener is as follows: ♦ Pheromones are chemical signals used for communication between members of the same species. For example, the bark beetle uses an aggregation pheromone to signal other bark beetles to congregate at a particular site in a tree. Bark beetle infestations can cause severe damage because the beetles carry a fungal infection that spreads rapidly and can kill the tree. One of the components of this aggregation pheromone has the following structure: Carbon monoxide is highly poisonous because it binds more strongly than O to the iron in red blood cells, which transport oxygen in the blood. Consequently, a victim of CO poisoning suffocates from a lack of oxygen. Draw a molecular orbital energy-level diagram for CO. What is the highest occupied molecular orbital? Are any of the molecular orbitals degenerate? If so, which ones? There are six electron groups, the molecular geometry is octahedral, and the hybridization of S is . | 2,069 | 3,788 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/10%3A_Fundamentals_of_Acids_and_Bases/10.02%3A_Aqueous_Solutions-_pH_and_Titrations |
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic. As you will see in the lesson that follows this one, plays an essential role in acid-base chemistry as we ordinarily know it. To even those who know very little about chemistry, the term is recognized as a measure of "acidity", so the major portion of this unit is devoted to the definition of pH and of the pH scale. But since these topics are intimately dependant on the properties of water and its ability do dissociate into hydrogen and hydroxyl ions, we begin our discussion with this topic. We end this lesson with a brief discussion of acid-base titration— probably the most frequently carried-out chemistry laboratory operation in the world. The ability of acids to react with bases depends on the tendency of hydrogen ions to combine with hydroxide ions to form water: H + OH → H O ( ) This tendency happens to be very great, so the reaction is practically complete— but not "completely" complete; a few stray H and OH ions will always be present. What's more, this is true even if you start with the purest water attainable. This means that in pure water, the reaction, the "dissociation" of water H O → H + OH Liquids that contain ions are able to conduct an electric current. Pure water is practically an insulator, but careful experiments show that even the most highly purified water exhibits a very slight conductivity that corresponds to a concentration of both the H ion and OH ions of almost exactly 1.00 × 10 mol L at 25°C. All chemical reactions that take place in a single phase (such as in a solution) are theoretically "incomplete" and are said to be reversible. What of water molecules in a liter of water are dissociated 1 L of water has a mass of 1000 g. The number of moles in 1000 g of H O is
(1000 g)/(18 g mol ) = 55.5 mol. This corresponds to (55.5 mol)(6.02E23 mol-1) = 3.34E25 H O molecules. An average of 10 mole, or (10 )(6.02E23) = 6.0E16 H O molecules will be dissociated at any time. The fraction of dissociated water molecules is therefore (6.0E16)/(3.3E25) = 1.8E–9. Thus we can say that only about two out of every billion (10 ) water molecules will be dissociated. The degree of dissociation of water is so small that you might wonder why it is even mentioned here. The reason stems from an important relationship that governs the concentrations of H and OH ions in aqueous solutions: in which the square brackets [ ] refer to the concentrations (in moles per litre) of the substances they enclose. The quantity 1.00 x 10 is commonly denoted by . Its value varies slightly with temperature, pressure, and the presence of other ions in the solution. This expression is known as the , and it applies to aqueous solutions, not just to pure water. The consequences of this are far-reaching, because it implies that if the concentration of H is large, that of OH will be small, and vice versa. This leads to the following , which you must know: Take special note of the following definition: The values of these concentrations are constrained by Eq. . Thus, in a , both the hydrogen- and hydroxide ion concentrations are 1.00 × 10 mol L : [H ,OH ] = [1.00 × 10 ,1.00 × 10 ] =1.00 × 10 Hydrochloric acid is a typical strong acid that is totally dissociated in solution: HCl → H + Cl A 1.0 solution of HCl in water therefore does not really contain any significant concentration of HCl molecules at all; it is a solution in of H and Cl in which the concentrations of both ions are 1.0 mol L . The concentration of hydroxide ion in such a solution, according to Eq , is [OH ] = ( )/[H ] = (1.00 x 10 ) / (1 mol L ) = 1.00 x 10 mol L . Similarly, the concentration of hydrogen ion in a solution made by dissolving 1.0 mol of sodium hydroxide in water will be 1.00 x 10 mol L . When dealing with a range of values (such as the variety of hydrogen ion concentrations encountered in chemistry) that spans many powers of ten, it is convenient to represent them on a more compressed logarithmic scale. By convention, we use the to denote hydrogen ion concentrations: or conversely, . This notation was devised by the Danish chemist Soren Sorenson (1868-1939) in 1909. There are several accounts of why he chose "pH"; a likely one is that the letters stand for the French term , meaning "power of hydrogen"— "power" in the sense of an exponent. It has since become common to represent other small quantities in "p-notation". Two that you need to know in this course are the following: pOH = – log [OH ] p = – log (= 14 when = 1.00 × 10 ) Note that pH and pOH are expressed as numbers without any units, since logarithms must be dimensionless. Recall from Eq that [H ,OH ] = 1.00 × 10 ; if we write this in "p-notation" it becomes In a neutral solution at 25°C, pH = pOH = 7.0. As pH increases, pOH diminishes; a higher pH corresponds to an alkaline solution, a lower pH to an acidic solution. In a solution with [H ] = 1 M , the pH would be 0; in a 0.00010 solution of H , it would be 4.0. Similarly, a 0.00010 M solution of NaOH would have a pOH of 4.0, and thus a pH of 10.0. It is very important that you thoroughly understand the pH scale, and be able to convert between [H ] or [OH ] and pH in both directions. The pH of blood must be held very close to 7.40. Find the hydroxide ion concentration that corresponds to this pH. The pOH will be (14.0 – 7.40) = 6.60.
[OH ] = 10 = 10 = 2.51 x 10 The range of possible pH values runs from about 0 to 14. The word "about" in the above statement reflects the fact that at very high concentrations (10 M hydrochloric acid or sodium hydroxide, for example,) a significant fraction of the ions will be associated into neutral pairs such as H ·Cl , thus reducing the concentration of “available” ions to a smaller value which we will call the . It is the of H and OH that determines the pH and pOH. For solutions in which ion concentrations don't exceed 0.1 M, the formulas pH = –log [H ] and pOH = –log[OH ] are generally reliable, but don't expect a 10.0 M solution of a strong acid to have a pH of exactly –1.00! The table shown here will help give you a general feeling for where common substances fall on the pH scale. Notice especially that The colors of many dye-like compounds depend on the pH, and can serve as useful to determine whether the pH of a solution is above or below a certain value. The best known of these is of course , which has served as a means of distinguishing between acidic and alkaline substances since the early 18th century. is a popular make-it-yourself indicator. Most indicator dyes show only one color change, and thus are only able to determine whether the pH of a solution is greater or less than the value that is characteristic of a particular indicator. By combining a variety of dyes whose color changes occur at different pHs, a "universal" indicator can be made. Commercially-prepared pH test papers of this kind are available for both wide and narrow pH ranges. Since acids and bases readily react with each other, it is experimentally quite easy to find the amount of acid in a solution by determining how many moles of base are required to neutralize it. This operation is called , and you should already be familiar with it from your work in the Laboratory. We can titrate an acid with a base, or a base with an acid. The substance whose concentration we are determining (the ) is the substance being titrated; the substance we are adding in measured amounts is the . The idea is to add titrant until the titrant has reacted with all of the analyte; at this point, the number of moles of titrant added tells us the concentration of base (or acid) in the solution being titrated. 36.00 ml of a solution of HCl was titrated with 0.44 M KOH. The volume of KOH solution required to neutralize the acid solution was 27.00 ml. What was the concentration of the HCl? Solution: The number of moles of titrant added was
(.027 L)(.44 mol L ) = .0119 mol. Because one mole of KOH reacts with one mole of HCl, this is also the number of moles of HCl; its concentration is therefore
(.0119 mol) ÷ (.036 L) = 0.33 . The course of a titration can be followed by plotting the pH of the solution as a function of the quantity of titrant added. The figure shows two such curves, one for a strong acid (HCl) and the other for a weak acid, acetic acid, denoted by HAc. Looking first at the HCl curve, notice how the pH changes very slightly until the acid is almost neutralized. At that point, which corresponds to the vertical part of the plot, just one additional drop of NaOH solution will cause the pH to jump to a very high value— almost as high as that of the pure NaOH solution. Compare the curve for HCl with that of HAc. For a weak acid, the pH jump near the neutralization point is less steep. Notice also that the pH of the solution at the neutralization point is greater than 7. These two characteristics of the titration curve for a weak acid are very important for you to know. If the acid or base is polyprotic, there will be a jump in pH for each proton that is titrated. In the example shown here, a solution of carbonic acid H CO is titrated with sodium hydroxide. The first equivalence point (at which the H CO has been converted entirely into bicarbonate ion HCO ) occurs at pH 8.3. The solution is now identical to one prepared by dissolving an identical amount of sodium bicarbonate in water. Addition of another mole equivalent of hydroxide ion converts the bicarbonate into carbonate ion and is complete at pH 10.3; an identical solution could be prepared by dissolving the appropriate amount of sodium carbonate in water. When enough base has been added to react completely with the hydrogens of a monoprotic acid, the equivalence point has been reached. If a strong acid and strong base are titrated, the pH of the solution will be 7.0 at the equivalence point. However, if the acid is a weak one, the pH will be greater than 7; the “neutralized” solution will not be “neutral” in terms of pH. For a polyprotic acid, there will be an equivalence point for each titratable hydrogen in the acid. These typically occur at pH values that are 4-5 units apart, but they are occasionally closer, in which case they may not be readily apparent in the titration curve. The key to a successful titration is knowing when the equivalence point has been reached. The easiest way of finding the equivalence point is to use an dye; this is a substance whose color is sensitive to the pH. One such indicator that is commonly encountered in the laboratory is ; it is colorless in acidic solution, but turns intensely red when the solution becomes alkaline. If an acid is to be titrated, you add a few drops of phenolphthalein to the solution before beginning the titration. As the titrant is added, a local red color appears, but quickly dissipates as the solution is shaken or stirred. Gradually, as the equivalence point is approached, the color dissipates more slowly; the trick is to stop the addition of base after a single drop results in a permanently pink solution. Different indicators change color at different pH values. Since the pH of the equivalence point varies with the strength of the acid being titrated, one tries to fit the indicator to the particular acid. One can titrate polyprotic acids by using a suitable combination of several indicators, as is illustrated above for carbonic acid. | 11,664 | 3,789 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Electrolytic_Cells/Electrolysis_I |
In this chapter, we have described various galvanic cells in which a spontaneous chemical reaction is used to generate electrical energy. In an electrolytic cell, however, the opposite process, called , occurs: an external voltage is applied to drive a nonspontaneous reaction. In this section, we look at how electrolytic cells are constructed and explore some of their many commercial applications. If we construct an electrochemical cell in which one electrode is copper metal immersed in a 1 M Cu solution and the other electrode is cadmium metal immersed in a \(\,1\; M\, Cd^{2+}\) solution and then close the circuit, the potential difference between the two compartments will be 0.74 V. The cadmium electrode will begin to dissolve (Cd is oxidized to Cd ) and is the anode, while metallic copper will be deposited on the copper electrode (Cu is reduced to Cu), which is the cathode (Figure \(\Page {1a}\)). The overall reaction is as follows: \[ \ce{Cd (s) + Cu^{2+} (aq) \rightarrow Cd^{2+} (aq) + Cu (s)} \nonumber \] with \(E°_{cell} = 0.74\; V\) This reaction is thermodynamically spontaneous as written (\(ΔG^o < 0\)): \[ \begin{align*} \Delta G^\circ &=-nFE^\circ_\textrm{cell} \\[4pt] &=-(\textrm{2 mol e}^-)[\mathrm{96,485\;J/(V\cdot mol)}](\mathrm{0.74\;V}) \\[4pt] &=-\textrm{140 kJ (per mole Cd)} \end{align*} \nonumber \] In this direction, the system is acting as a galvanic cell. In an electrolytic cell, an external voltage is applied to drive a reaction. The reverse reaction, the reduction of Cd by Cu, is thermodynamically nonspontaneous and will occur only with an input of 140 kJ. We can force the reaction to proceed in the reverse direction by applying an electrical potential greater than 0.74 V from an external power supply. The applied voltage forces electrons through the circuit in the reverse direction, converting a galvanic cell to an electrolytic cell. Thus the copper electrode is now the anode (Cu is oxidized), and the cadmium electrode is now the cathode (Cd is reduced) (Figure \(\Page {1b}\)). The signs of the cathode and the anode have switched to reflect the flow of electrons in the circuit. The half-reactions that occur at the cathode and the anode are as follows: \[\ce{Cd^{2+}(aq) + 2e^{−} \rightarrow Cd(s)}\label{20.9.3} \] with \(E^°_{cathode} = −0.40 \, V\) \[\ce{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{−}} \label{20.9.4} \] with \(E^°_{anode} = 0.34 \, V \) \[\ce{Cd^{2+}(aq) + Cu(s) \rightarrow Cd(s) + Cu^{2+}(aq) } \label{20.9.5} \] with \(E^°_{cell} = −0.74 \: V\) Because \(E^°_{cell} < 0\), the overall reaction—the reduction of \(Cd^{2+}\) by \(Cu\)—clearly occur spontaneously and proceeds only when sufficient electrical energy is applied. The differences between galvanic and electrolytic cells are summarized in Table \(\Page {1}\). At sufficiently high temperatures, ionic solids melt to form liquids that conduct electricity extremely well due to the high concentrations of ions. If two inert electrodes are inserted into molten \(\ce{NaCl}\), for example, and an electrical potential is applied, \(\ce{Cl^{-}}\) is oxidized at the anode, and \(\ce{Na^{+}}\) is reduced at the cathode. The overall reaction is as follows: \[\ce{ 2NaCl (l) \rightarrow 2Na(l) + Cl2(g)} \label{20.9.6} \] This is the reverse of the formation of \(\ce{NaCl}\) from its elements. The product of the reduction reaction is liquid sodium because the melting point of sodium metal is 97.8°C, well below that of \(\ce{NaCl}\) (801°C). Approximately 20,000 tons of sodium metal are produced commercially in the United States each year by the electrolysis of molten \(\ce{NaCl}\) in a Downs cell (Figure \(\Page {2}\)). In this specialized cell, \(\ce{CaCl2}\) (melting point = 772°C) is first added to the \(\ce{NaCl}\) to lower the melting point of the mixture to about 600°C, thereby lowering operating costs. Similarly, in the Hall–Heroult process used to produce aluminum commercially, a molten mixture of about 5% aluminum oxide (Al O ; melting point = 2054°C) and 95% cryolite (Na AlF ; melting point = 1012°C) is electrolyzed at about 1000°C, producing molten aluminum at the cathode and CO gas at the carbon anode. The overall reaction is as follows: \[\ce{2Al2O3(l) + 3C(s) -> 4Al(l) + 3CO2(g)} \label{20.9.7} \] Oxide ions react with oxidized carbon at the anode, producing CO (g). There are two important points to make about these two commercial processes and about the electrolysis of molten salts in general. In the Hall–Heroult process, C is oxidized instead of O or F because oxygen and fluorine are more electronegative than carbon, which means that C is a weaker oxidant than either O or F . Similarly, in the Downs cell, we might expect electrolysis of a NaCl/CaCl mixture to produce calcium rather than sodium because Na is slightly less electronegative than Ca (χ = 0.93 versus 1.00, respectively), making Na easier to oxidize and, conversely, Na more difficult to reduce. In fact, the reduction of Na to Na is the observed reaction. In cases where the electronegativities of two species are similar, other factors, such as the formation of complex ions, become important and may determine the outcome. If a molten mixture of MgCl and KBr is electrolyzed, what products will form at the cathode and the anode, respectively? identity of salts electrolysis products The possible reduction products are Mg and K, and the possible oxidation products are Cl and Br . Because Mg is more electronegative than K (χ = 1.31 versus 0.82), it is likely that Mg will be reduced rather than K. Because Cl is more electronegative than Br (3.16 versus 2.96), Cl is a stronger oxidant than Br . Electrolysis will therefore produce Br at the anode and Mg at the cathode. Predict the products if a molten mixture of AlBr and LiF is electrolyzed. Br and Al Electrolysis can also be used to drive the thermodynamically nonspontaneous decomposition of water into its constituent elements: H and O . However, because pure water is a very poor electrical conductor, a small amount of an ionic solute (such as H SO or Na SO ) must first be added to increase its electrical conductivity. Inserting inert electrodes into the solution and applying a voltage between them will result in the rapid evolution of bubbles of H and O (Figure \(\Page {3}\)). The reactions that occur are as follows: For a system that contains an electrolyte such as Na SO , which has a negligible effect on the ionization equilibrium of liquid water, the pH of the solution will be 7.00 and [H ] = [OH ] = 1.0 × 10 . Assuming that \(P_\mathrm{O_2}\) = \(P_\mathrm{H_2}\) = 1 atm, we can use the standard potentials to calculate E for the overall reaction: \[\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log(P_\mathrm{O_2}P^2_\mathrm{H_2}) \\ &=-\textrm{1.23 V}-\left(\dfrac{\textrm{0.0591 V}}{4}\right)\log(1)=-\textrm{1.23 V}\end{align} \label{20.9.11} \] Thus E is −1.23 V, which is the value of E° if the reaction is carried out in the presence of 1 M H rather than at pH 7.0. In practice, a voltage about 0.4–0.6 V greater than the calculated value is needed to electrolyze water. This added voltage, called an , represents the additional driving force required to overcome barriers such as the large activation energy for the formation of a gas at a metal surface. Overvoltages are needed in all electrolytic processes, which explain why, for example, approximately 14 V must be applied to recharge the 12 V battery in your car. In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. The p-block metals and most of the transition metals are in this category, but metals in high oxidation states, which form oxoanions, cannot be reduced to the metal by simple electrolysis. Active metals, such as aluminum and those of groups 1 and 2, react so readily with water that they can be prepared only by the electrolysis of molten salts. Similarly, any nonmetallic element that does not readily oxidize water to O can be prepared by the electrolytic oxidation of an aqueous solution that contains an appropriate anion. In practice, among the nonmetals, only F cannot be prepared using this method. Oxoanions of nonmetals in their highest oxidation states, such as NO , SO , PO , are usually difficult to reduce electrochemically and usually behave like spectator ions that remain in solution during electrolysis. In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. In a process called , a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis. Electroplating is used to enhance the appearance of metal objects and protect them from corrosion. Examples of electroplating include the chromium layer found on many bathroom fixtures or (in earlier days) on the bumpers and hubcaps of cars, as well as the thin layer of precious metal that coats silver-plated dinnerware or jewelry. In all cases, the basic concept is the same. A schematic view of an apparatus for electroplating silverware and a photograph of a commercial electroplating cell are shown in Figure \(\Page {4}\). The half-reactions in electroplating a fork, for example, with silver are as follows: The overall reaction is the transfer of silver metal from one electrode (a silver bar acting as the anode) to another (a fork acting as the cathode). Because \(E^o_{cell} = 0\, V\), it takes only a small applied voltage to drive the electroplating process. In practice, various other substances may be added to the plating solution to control its electrical conductivity and regulate the concentration of free metal ions, thus ensuring a smooth, even coating. If we know the stoichiometry of an electrolysis reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction. Conversely, we can use stoichiometry to determine the combination of current and time needed to produce a given amount of material. The quantity of material that is oxidized or reduced at an electrode during an electrochemical reaction is determined by the stoichiometry of the reaction and the amount of charge that is transferred. For example, in the reaction \[\ce{Ag^{+}(aq) + e^{−} → Ag(s)} \nonumber \] 1 mol of electrons reduces 1 mol of \(\ce{Ag^{+}}\) to \(\ce{Ag}\) metal. In contrast, in the reaction \[\ce{Cu^{2+}(aq) + 2e^{−} → Cu(s)} \nonumber \] 1 mol of electrons reduces only 0.5 mol of \(\ce{Cu^{2+}}\) to \(\ce{Cu}\) metal. Recall that the charge on 1 mol of electrons is 1 faraday (1 F), which is equal to 96,485 C. We can therefore calculate the number of moles of electrons transferred when a known current is passed through a cell for a given period of time. The total charge (\(q\) in coulombs) transferred is the product of the current (\(I\) in amperes) and the time (\(t\), in seconds): \[ q = I \times t \label{20.9.14} \] The stoichiometry of the reaction and the total charge transferred enable us to calculate the amount of product formed during an electrolysis reaction or the amount of metal deposited in an electroplating process. For example, if a current of 0.60 A passes through an aqueous solution of \(\ce{CuSO4}\) for 6.0 min, the total number of coulombs of charge that passes through the cell is as follows: \[\begin{align*} q &= \textrm{(0.60 A)(6.0 min)(60 s/min)} \\[4pt] &=\mathrm{220\;A\cdot s} \\[4pt] &=\textrm{220 C} \end{align*} \nonumber \] The number of moles of electrons transferred to \(\ce{Cu^{2+}}\) is therefore \[\begin{align*} \textrm{moles e}^- &=\dfrac{\textrm{220 C}}{\textrm{96,485 C/mol}} \\[4pt] &=2.3\times10^{-3}\textrm{ mol e}^- \end{align*} \nonumber \] Because two electrons are required to reduce a single Cu ion, the total number of moles of Cu produced is half the number of moles of electrons transferred, or 1.2 × 10 mol. This corresponds to 76 mg of Cu. In commercial electrorefining processes, much higher currents (greater than or equal to 50,000 A) are used, corresponding to approximately 0.5 F/s, and reaction times are on the order of 3–4 weeks. A silver-plated spoon typically contains about 2.00 g of Ag. If 12.0 h are required to achieve the desired thickness of the Ag coating, what is the average current per spoon that must flow during the electroplating process, assuming an efficiency of 100%? mass of metal, time, and efficiency current required We must first determine the number of moles of Ag corresponding to 2.00 g of Ag: \(\textrm{moles Ag}=\dfrac{\textrm{2.00 g}}{\textrm{107.868 g/mol}}=1.85\times10^{-2}\textrm{ mol Ag}\) The reduction reaction is Ag (aq) + e → Ag(s), so 1 mol of electrons produces 1 mol of silver. Using the definition of the faraday, The current in amperes needed to deliver this amount of charge in 12.0 h is therefore \[\begin{align*}\textrm{amperes} &=\dfrac{1.78\times10^3\textrm{ C}}{(\textrm{12.0 h})(\textrm{60 min/h})(\textrm{60 s/min})}\\
& =4.12\times10^{-2}\textrm{ C/s}=4.12\times10^{-2}\textrm{ A}\end{align*} \nonumber \] Because the electroplating process is usually much less than 100% efficient (typical values are closer to 30%), the actual current necessary is greater than 0.1 A. A typical aluminum soft-drink can weighs about 29 g. How much time is needed to produce this amount of Al(s) in the Hall–Heroult process, using a current of 15 A to reduce a molten Al O /Na AlF mixture? 5.8 h Electroplating: In electrolysis, an external voltage is applied to drive a reaction. The quantity of material oxidized or reduced can be calculated from the stoichiometry of the reaction and the amount of charge transferred. Relationship of charge, current and time: \[ q = I \times t \nonumber \] In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. Electrolysis can also be used to produce H and O from water. In practice, an additional voltage, called an overvoltage, must be applied to overcome factors such as a large activation energy and a junction potential. Electroplating is the process by which a second metal is deposited on a metal surface, thereby enhancing an object’s appearance or providing protection from corrosion. The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction. | 14,641 | 3,790 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Reactivity_of_Alkanes/Alkane_Heats_of_Combustion |
The combustion of carbon compounds, especially hydrocarbons, has been the most important source of heat energy for human civilizations throughout recorded history. The practical importance of this reaction cannot be denied, but the massive and uncontrolled chemical changes that take place in combustion make it difficult to deduce mechanistic paths. Using the combustion of propane as an example, we see from the following equation that every covalent bond in the reactants has been broken and an entirely new set of covalent bonds have formed in the products. No other common reaction involves such a profound and pervasive change, and the mechanism of combustion is so complex that chemists are just beginning to explore and understand some of its elementary features. \[\ce{CH3CH2CH3} + 5 \ce{O2} \rightarrow 3 \ce{CO2} + 4\ce{H2O} + \text{heat} \label{1}\] Two points concerning this reaction are important: \[\ce{CH3CH2CH3} + 4 \ce{O2} \rightarrow \ce{CO2} + 2 \ce{CO} + 4\ce{H2O} + \text{heat} \label{2}\] From the previous discussion, we might expect isomers to have identical heats of combustion. However, a few simple measurements will disabuse this belief. Thus, the heat of combustion of pentane is –782 kcal/mole, but that of its 2,2-dimethylpropane (neopentane) isomer is –777 kcal/mole. Differences such as this reflect subtle structural variations, including the greater of 1º-C–H versus 2º-C–H bonds and steric crowding of neighboring groups. In small-ring cyclic compounds ring strain can be a major contributor to thermodynamic stability and chemical reactivity. The following table lists heat of combustion data for some simple cycloalkanes and compares these with the increase per CH unit for long chain alkanes. The chief source of ring strain in smaller rings is and . As noted elsewhere, cyclopropane and cyclobutane have large contributions of both strains, with angle strain being especially severe. Changes in chemical reactivity as a consequence of angle strain are dramatic in the case of cyclopropane, and are also evident for cyclobutane. Some examples are shown in the following diagram. The cyclopropane reactions are additions, many of which are initiated by electrophilic attack. The pyrolytic conversion of β-pinene to myrcene probably takes place by an initial rupture of the 1:6 bond, giving an allylic 3º-diradical, followed immediately by breaking of the 5:7 bond. | 2,422 | 3,791 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/02%3A_Gases/2.02%3A_The_Ideal_Gas_Law |
The combines the empirical laws into a single expression. It also predicts the existence of a single, universal gas constant, which turns out to be one of the most important fundamental constants in science. \[pV = nRT \nonumber \] The ideal gas law constant is of fundamental importance and can be expressed in a number of different sets of units. The ideal gas law, as derived here, is based entirely on empirical data. It represents “limiting ideal behavior.” As such, deviations from the behavior suggested by the ideal gas law can be understood in terms of what conditions are required for ideal behavior to be followed (or at least approached.) As such, it would be nice if there was a theory of gases that would suggest the form of the ideal gas law and also the value of the gas law constant. As it turns out, the kinetic molecular theory of gases does just that! | 885 | 3,792 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Molecular_Geometry/Limitations_of_VSEPR |
Athough the model is useful in predicting molecular geometry, it fails to predict the shapes of isoelectronic species and transition metal compounds. This model does not take relative sizes of substituents and stereochemically inactive lone pairs into account. As a result, VSEPR cannot be applied to heavy d-block species that experience the stereochemical . The VSEPR model is a powerful tool used by chemists to predict the shapes of molecules; yet like many other theories, it has exceptions and limitations. This article discusses in detail the various limitations of VSEPR and gives explanations for these exceptions. Isoelectronic species are elements, ions and molecules that share the same number of electrons. According to the VSEPR model, chemists determine the shape of the molecules based on numbers of valence electrons (i.e. bond pairs and lone pairs). However, two isoelectronic species, can differ in geometry despite the fact that they have the same numbers of valence electrons. For example, both IF and [TeF ] have 56 valence electrons; VSEPR theory predicts both to be pentagonal bipyramidal. However, electron diffraction and X-ray diffraction data indicate that the equatorial F atoms of Te-F VSEPR TeF7 The VSEPR model also fails to predict the structure of certain compounds because it does not take relative sizes of the substituents and stereochemically inactive lone pairs into account. Elements in the d-block have relatively high atomic masses and they tend to have stereochemically inactive electron pairs. In other words, valence shell s-electrons in VSEPR SeCl TeCl BrF bipyramidal VSEPR stereochemical stereochemically | 1,677 | 3,793 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Concept_Development_Studies_in_Chemistry_(Hutchinson)/12%3A__The_Kinetic_Molecular_Theory |
We assume an understanding of the atomic molecular theory postulates, including that all matter is composed of discrete particles. The elements consist of identical atoms, and compounds consist of identical molecules, which are particles containing small whole number ratios of atoms. We also assume that we have determined a complete set of relative atomic weights, allowing us to determine the molecular formula for any compound. Finally, we assume a knowledge of the , and the observations from which it is derived. Our continuing goal is to relate the properties of the atoms and molecules to the properties of the materials which they comprise. As simple examples, we compare the substances water, carbon dioxide, and nitrogen. Each of these is composed of molecules with few (two or three) atoms and low molecular weight. However, the physical properties of these substances are very different. Carbon dioxide and nitrogen are gases at room temperature, but it is well known that water is a liquid up to \(100^\text{o} \text{C}\). To liquefy nitrogen, we must cool it to \(-196^\text{o} \text{C}\), so the boiling temperatures of water and nitrogen differ by about \(300^\text{o} \text{C}\). Water is a liquid for a rather large temperature range, freezing at \(0^\text{o} \text{C}\). In contrast, nitrogen is a liquid for a very narrow range of temperatures, freezing at \(-210^\text{o} \text{C}\). Carbon dioxide poses yet another very different set of properties. At atmospheric pressure, carbon dioxide gas cannot be liquefied at all: cooling the gas to \(-60^\text{o} \text{C}\) converts it directly to solid "dry ice". As is commonly observed, warming dry ice does not produce any liquid, as the solid sublimes directly to gas. Why should these materials, whose molecules do not seem all that different, behave so differently? What are the important characteristics of these molecules which produce these physical properties? It is important to keep in mind that these are properties of the bulk materials. At this point, it is not even clear that the concept of a molecule is useful in answering these questions about melting or boiling. There are at least two principal questions that arise about the . First, it is interesting to ask whether this law always holds true, or whether there are conditions under which the pressure of the gas cannot be calculated from \(\frac{nRT}{V}\). We thus begin by considering the limitations of the validity of the . We shall find that the ideal gas law is only approximately accurate and that there are variations which do depend upon the nature of the gas. Second, then, it is interesting to ask why the ideal gas law should ever hold true. In other words, why are the variations not the rule rather than the exception? To answer these questions, we need a model which will allow us to relate the properties of bulk materials to the characteristics of individual molecules. We seek to know what happens to a gas when it is compressed into a smaller volume, and why it generates a greater resisting pressure when compressed. Perhaps most fundamentally of all, we seek to know what happens to a substance when it is heated. What property of a gas is measured by the temperature? To design a systematic test for the validity of the , we note that the value of \(\frac{PV}{nRT}\), calculated, from the observed values of \(P\), \(V\), \(n\), and \(T\), should always be equal to 1, exactly. Deviation of \(\frac{PV}{nRT}\) from 1 indicates a violation of the . We thus measure the pressure for several different gases under a variety of conditions by varying \(n\), \(V\), and \(T\), and we calculate the ratio \(\frac{PV}{nRT}\) for these conditions. In Figure 12.1, the value of this ratio is plotted for several gases as a function of the "particle density" of the gas in moles, \(\frac{n}{V}\). To make the analysis of this plot more convenient, the particle density is given in terms of the particle density of an ideal gas at room temperature and atmospheric pressure (i.e. the density of air), which is \(0.04087 \: \frac{\text{mol}}{\text{L}}\). In Figure 12.1, a particle density of 10 means that the particle density of the gas is 10 times the particle density of air at room temperature. The x-axis in the figure is thus unitless. Note that \(\frac{PV}{nRT}\) on the y-axis is also unitless and has value exactly 1 for an ideal gas. We observe in the data in this figure that \(\frac{PV}{nRT}\) is extremely close to 1 for particle densities which are close to that of normal air. Therefore, deviations from the are not expected under "normal" conditions. This is not surprising, since , , and the were all observed under normal conditions. Figure 12.1 also shows that, as the particle density increases above the normal range, the value of \(\frac{PV}{nRT}\) starts to vary from 1, and the variation depends on the type of gas we are analyzing. However, even for particle densities 10 times greater than that of air at atmospheric pressure, the is accurate to a few percent. Thus, to observe any significant deviations from \(PV = nRT\), we need to push the gas conditions to somewhat more extreme values. The results for such extreme conditions are shown in Figure 12.2. Note that the densities considered are large numbers corresponding to very high pressures. Under these conditions, we find substantial deviations from the . In addition, we see that the pressure of the gas (and thus \(\frac{PV}{nRT}\)) does depend strongly on which type of gas we are examining. Finally, this figure shows that deviations from the can generate pressures either greater than or less than that predicted by the . For low densities for which the is valid, the pressure of a gas is independent of the nature of the gas, and is therefore independent of the characteristics of the particles of that gas. We can build on this observation by considering the significance of a low particle density. Even at the high particle densities considered in Figure 12.2, all gases have low density in comparison to the densities of liquids. To illustrate, we note that 1 gram of liquid water at its boiling point has a volume very close to 1 milliliter. In comparison, this same 1 gram of water, once evaporated into steam, has a volume of over 1700 milliliters. How does this expansion by a factor of 1700 occur? It is not credible that the individual water molecules suddenly increase in size by this factor. The only plausible conclusion is that the distance between gas molecules has increased dramatically. Therefore, it is a characteristic of a gas that the molecules are far apart from one another. In addition, the lower the density of the gas the farther apart the molecules must be, since the same number of molecules occupies a larger volume at lower density. We reinforce this conclusion by noting that liquids and solids are virtually incompressible, whereas gases are easily compressed. This is easily understood if the molecules in a gas are very far apart from one another, in contrast to the liquid and solid where the molecules are so close as to be in contact with one another. We add this conclusion to the observations in Figures 12.1 and 12.2 that the pressure exerted by a gas depends only on the number of particles in the gas and is independent of the type of particles in the gas, provided that the density is low enough. This requires that the gas particles be far enough apart. We conclude that the holds true because there is sufficient distance between the gas particles that the identity of the gas particles becomes irrelevant. Why should this large distance be required? If gas particle A were far enough away from gas particle B that they experience no electrical or magnetic interaction, then it would not matter what types of particles A and B were. Nor would it matter what the sizes of particles A and B were. Finally, then, we conclude from this reasoning that the validity of the ideal gas law rests on the presumption that there are no interactions of any type between gas particles. We recall at this point our purpose in these observations. Our primary concern in this study is attempting to relate the properties of individual atoms or molecules to the properties of mass quantities of the materials composed of these atoms or molecules. We now have extensive quantitative observations on some specific properties of gases, and we proceed with the task of relating these to the particles of these gases. By taking an atomic molecule view of a gas, we can postulate that the pressure observed is a consequence of the collisions of the individual particles of the gas with the walls of the container. This presumes that the gas particles are in constant motion. The pressure is, by definition, the force applied per area, and there can be no other origin for a force on the walls of the container than that provided by the particles themselves. Furthermore, we observe easily that the pressure exerted by the gas is the same in all directions. Therefore, the gas particles must be moving equally in all directions, implying quite plausibly that the motions of the particles are random. To calculate the force generated by these collisions, we must know something about the motions of the gas particles so that we know, for example, each particle's velocity upon impact with the wall. This is too much to ask: there are perhaps \(10^{20}\) particles or more, and following the path of each particle is out of the question. Therefore, we seek a model which permits calculation of the pressure without this information. Based on our observations and deductions, we take as the postulates of our model: This model is the . We now look to see where this model leads. To calculate the pressure generated by a gas of \(N\) particles contained in a volume \(V\), we must calculate the force \(F\) generated per area \(A\) by collisions against the walls. To do so, we begin by determining the number of collisions of particles with the walls. The number of collisions we observe depends on how long we wait. Let's measure the pressure for a period of time \(\Delta t\) and calculate how many collisions occur in that time period. For a particle to collide with the wall within the time \(\Delta t\), it must start close enough to the wall to impact it in that period of time. If the particle is traveling with speed \(v\), then the particle must be within a distance \(v \Delta t\) of the wall to hit it. Also, if we are measuring the force exerted on the area \(A\), the particle must hit that area to contribute to our pressure measurement. For simplicity, we can view the situation pictorially in Figure 12.3. We assume that the particles are moving perpendicularly to the walls. (This is clearly not true. However, very importantly, this assumption is only made to simplify the mathematics of our derivation. It is not necessary to make this assumption, and the result is not affected by the assumption.) In order for a particle to hit the area \(A\) marked on the wall, it must lie within the cylinder shown, which is of length \(v \Delta t\) and cross-sectional area \(A\). The volume of this cylinder is \(A v \Delta t\), so the number of particles contained in the cylinder is \(\left( A v \Delta t \right) \times \frac{N}{V}\). Not all of these particles collide with the wall during \(\Delta t\), though, since most of them are not traveling in the correct direction. There are six directions for a particle to go, corresponding to plus or minus direction in x, y, or z. Therefore, on average, the fraction of particles moving in the correct direction should be \(\frac{1}{6}\), assuming as we have that the motions are all random. Therefore, the number of particles which impact the wall in time \(\Delta t\) is \(\left( A v \Delta t \right) \times \frac{N}{6V}\). The force generated by these collisions is calculated from Newton's equation, \(F = ma\), where \(a\) is the acceleration due to the collisions. Consider first a single particle moving directly perpendicular to a wall with velocity \(v\) as in Figure 12.3. We note that, when the particle collides with the wall, the wall does not move, so the collision must generally conserve the energy of the particle. Then the particle's velocity after the collision must be \(-v\), since it is now traveling in the opposite direction. Thus, the change in velocity of the particle by the time \(\Delta t\), we find that the total acceleration (change in velocity per time) is \(\frac{2A N v^2}{6V}\), and the force imparted on the wall due to collisions is found by multiplying by the mass of the particles: \[F = \frac{2ANmv^2}{6V}\] To calculate the pressure, we divide by the area \(A\), to find that \[P = \frac{Nmv^2}{3V}\] or, rearranged for comparison to , \[PV = \frac{Nmv^2}{3}\] Since we have assumed that the particles travel with constant speed \(v\), then the right side of this equation is a constant. Therefore the product of pressure times volume, \(PV\), is a constant, in agreement with . Furthermore, the product \(PV\) is proportional to the number of particles, also in agreement with the . Therefore, the model we have developed to describe an ideal gas is consistent with our experimental observations. We can draw two very important conclusions from this derivation. First, the inverse relationship observed between pressure and volume and the independence of this relationship on the type of gas analyzed are both due to the lack of interactions between gas particles. Second, the lack of interactions is in turn due to the great distances between gas particles, a fact which will be true provided that the density of the gas is low. The absence of temperature in the above derivation is notable. The other gas properties have all been incorporated, yet we have derived an equation which omits temperature all together. The problem is that, as we discussed at length above, the temperature was somewhat arbitrarily defined. In fact, it is not precisely clear what has been measured by the temperature. We defined the temperature of a gas in terms of the volume of mercury in a glass tube in contact with the gas. It is perhaps then no wonder that such a quantity does not show up in a mechanical derivation of the gas properties. On the other hand, the temperature does appear prominently in the . Therefore, there must be a greater significance (and less arbitrariness) to the temperature than might have been expected. To discern this significance, we rewrite the last equation above in the form \[PV = \frac{2}{3} N \left( \frac{1}{2} mv^2 \right)\] The last quantity in parentheses can be recognized as the kinetic energy of an individual gas particle, and \(N \left( \frac{1}{2} mv^2 \right)\) must be the total kinetic energy (KE) of the gas. Therefore \[PV = \frac{2}{3} KE\] Now we insert the for \(PV\) to find that \[KE = \frac{3}{2} nRT\] This is an extremely important conclusion, for it reveals the answer to the question of what property is measured by the temperature. We see now that the temperature is a measure of the total kinetic energy of the gas. Thus, when we heat a gas, elevating its temperature, we are increasing the average kinetic energy of the gas particles, causing them to move, on average, more rapidly. We are at last in a position to understand the observations of deviations from the . The most important assumption of our model of the behavior of an ideal gas is that the gas molecules do not interact. This allowed us to calculate the force imparted on the wall of the container due to a single particle collision without worrying about where the other particles were. In order for a gas to disobey the , the conditions must be such that this assumption is violated. What do the deviations from ideality tell us about the gas particles? Starting with very low density and increasing the density as in Figure 12.1, we find that, for many gases, the value of \(\frac{PV}{nRT}\) falls below 1. One way to state this result is that, for a given value of \(V\), \(n\), and \(T\), the pressure of the gas is less than it would have been for an ideal gas. This must be the result of the interactions of the gas particles. In order for the pressure to be reduced, the force of the collisions of the particles with the walls must be less than is predicted by our model of an ideal gas. Therefore, the effect of the interactions is to slow the particles as they approach the walls of the container. This means that an individual particle approaching a wall must experience a force acting to pull it back into the body of the gas. Hence, the gas particles are confined in closer proximity to one another. At this closer range, the attractions of individual particles become significant. It should not be surprising that these attractive forces depend on what the particles are. We note in Figure 12.1 that deviation from the is greater for ammonia than for nitrogen, and greater for nitrogen than for helium. Therefore, the attractive interactions of ammonia molecules are greater than those of nitrogen molecules, which are in turn greater than those of helium atoms. We analyze this conclusion in more detail below. Continuing to increase the density of the gas, we find in Figure 12.2 that the value of \(\frac{PV}{nRT}\) begins to rise, eventually exceeding 1 and continuing to increase. Under these conditions, therefore, the pressure of the gas is greater than we would have expected from our model of non-interacting particles. What does this tell us? The gas particles are interacting in such a way as to increase the force of the collisions of the particles with the walls. This requires that the gas particles repel one another. As we move to higher density, the particles are forced into closer and closer proximity. We can conclude that gas particles at very close range experience strong repulsive forces away from one another. Our model of the behavior of gases can be summarized as follows: at low density, the gas particles are sufficiently far apart that there are no interactions between them. In this case, the pressure of the gas is independent of the nature of the gas and agrees with the . At somewhat higher densities, the particles are closer together and the interaction forces between the particles are attractive. The pressure of the gas now depends on the strength of these interactions and is lower than the value predicted by the . At still higher densities, the particles are excessively close together, resulting in repulsive interaction forces. The pressure of the gas under these conditions is higher than the value predicted by the . The postulates of the provide us a way to understand the relationship between molecular properties and the physical properties of bulk amounts of substance. As a distinct example of such an application, we now examine the boiling points of various compounds, focusing on hydrides of sixteen elements in the main group (Groups IV through VII). These are given in Table 12.1. In tabular form, there are no obvious trends here, and therefore no obvious connection to the structure or bonding in the molecules. The data in the table are displayed in a suggestive form, however, in Figure 12.4; the boiling point of each hydride is plotted according to which period (row) of the periodic table the main group element belongs. For example, the Period 2 hydrides (\(\ce{CH_4}\), \(\ce{NH_3}\), \(\ce{H_2O}\), and \(\ce{HF}\)) are grouped in a column to the left of the figure, followed by a column for the Period 3 hydrides (\(\ce{SiH_4}\), \(\ce{PH_3}\), \(\ce{H_2S}\), \(\ce{HCl}\)), etc. Now a few trends are more apparent. First, the lowest boiling points in each period are associated with the Group IV hydrides (\(\ce{CH_4}\), \(\ce{SiH_4}\), \(\ce{GeH_4}\), \(\ce{SnH_4}\)), and the highest boiling points in each period belong to the Group VI hydrides (\(\ce{H_2O}\), \(\ce{H_2S}\), \(\ce{H_2Se}\), \(\ce{H_2Te}\)). For this reason, the hydrides belonging to a single group have been connected in Figure 12.4. Second, we notice that, with the exceptions of \(\ce{NH_3}\), \(\ce{H_2O}\), and \(\ce{HF}\), the boiling points of the hydrides always increase in a single group as we go down the periodic table: for example, in Group IV, the boiling points increase in the order \(\ce{CH_4} < \ce{SiH_4} < \ce{GeH_4} < \ce{SnH_4}\). Third, we can also say that the hydrides from Period 2 appear to have unusually high boiling points for \(\ce{CH_4}\), which as noted has the lowest boiling point of all. We begin our analysis of these trends by assuming that there is a relationship between the boiling points of these compounds and the structure and bonding in their molecules. Recalling our kinetic molecular model of gases and liquids, we recognize that a primary difference between these two phases is that the strength of the interaction between the molecules in the liquid is much greater than that in the gas, due to the proximity of the molecules in the liquid. In order for a molecule to leave the liquid phase and enter into the gas phase, it must possess sufficient energy to overcome the interactions it has with other molecules in the liquid. Also recalling the kinetic molecular description, we recognize that, on average, the energies of molecules increase with increasing temperature. We can conclude from these two statements that a high boiling point implies that significant energy is required to overcome intermolecular interactions. Conversely, a substance with a low boiling point must have weak intermolecular interactions, surmountable even at low temperature. In light of these conclusions, we can now look at Figure 12.4 as directly (though qualitatively) revealing the comparative strengths of intermolecular interactions of the various hydrides. For example, we can conclude that, amongst the hydrides considered here, the intermolecular interactions are greatest between \(\ce{H_2O}\) molecules and weakest between \(\ce{CH_4}\) molecules. We examine the three trends in this figure, described above, in light of the strength of intermolecular forces. First, the most dominant trend in the boiling points is that, within a single group, the boiling points of the hydrides increase as we move the periodic table. This is true in all four groups in Figure 12.4; the only exceptions to this trend are \(\ce{NH_3}\), \(\ce{H_2O}\), and \(\ce{HF}\). We can conclude that, with notable exceptions, intermolecular interactions increase with increasing atomic number of the central atom in the molecule. This is true whether the molecules of the group considered have dipole moments (as in Groups V, VI, and VII) or not (as in Group IV). We can infer that the large intermolecular attractions for molecules with large central atoms arises from the large number of charged particles in these molecules. This type of interaction arises from forces referred to as or . These forces are believed to arise from the instantaneous interactions of the charged particles from one molecule with the charged particles in an adjacent molecule. Although these molecules may not be polar individually, the nuclei in one molecule may attract the electrons in a second molecule, thus inducing an instantaneous dipole in the second molecule. In turn, the second molecule induces a dipole in the first. Thus, two non-polar molecules can interact as if there were dipole-dipole attractions between them, with positive and negative charges interacting and attracting. The tendency of a molecule to have an induced dipole is called the of the molecule. The more charged particles there are in a molecule, the more a molecule is and the greater the attractions arising from dispersion forces will be. Second, we note that, without exception, the Group IV hydrides must have the weakest intermolecular interactions in each period. As noted above, these are the only hydrides that have no dipole moment. Consequently, in general, molecules without dipole moments have weaker interactions than molecules which are polar. We must qualify this carefully, however, by noting that the nonpolar \(\ce{SnH_4}\) has a higher boiling point than the polar \(\ce{PH_3}\) and \(\ce{HCl}\). We can conclude from these comparisons that the increased polarizability of molecules with heavier atoms can offset the lack of a molecular dipole. Third, and most importantly, we note that the intermolecular attractions involving \(\ce{NH_3}\), \(\ce{H_2O}\), and \(\ce{HF}\) must be uniquely and unexpectedly large, since their boiling points are markedly out of line with those of the rest of their groups. The common feature of these molecules is that they contain small atomic number atoms which are strongly electronegative, which have lone pairs, and which are bonded to hydrogen atoms. Molecules without these features do not have unexpectedly high boiling points. We can deduce from these observations that the hydrogen atoms in each molecule are unusually strongly attracted to the lone pair electrons on the strongly electronegative atoms with the same properties in other molecules. This intermolecular attraction of a hydrogen atom to an electronegative atom is referred to as . It is clear from our boiling point data that hydrogen bonding interactions are much stronger than either dispersion forces or dipole-dipole attractions. Explain the significance to the development of the kinetic molecular model of the observation that the ideal gas law works well only at low pressure. Explain the significance to the development of the kinetic molecular model of the observation that the pressure predicted by the ideal gas law is independent of the type of gas. Sketch the value of \(\frac{PV}{nRT}\) as a function of density for two gases, one with strong intermolecular attractions and one with weak intermolecular attractions but strong repulsions. Give a brief molecular explanation for the observation that the pressure of a gas at fixed temperature increases proportionally with the density of the gas. Give a brief molecular explanation for the observation that the pressure of a gas confined to a fixed volume increases proportionally with the temperature of the gas. Give a brief molecular explanation for the observation that the volume of a balloon increases roughly proportionally with the temperature of the gas inside the balloon. Explain why there is a correlation between high boiling point and strong deviation from the . Referring to Figure 12.4, explain why the hydride of the Group IV element always has the lowest boiling point in each period. Explain why the Period 2 hydrides except \(\ce{CH_4}\) all have high boiling points, and explain why \(\ce{CH_4}\) is an exception. ; Chemistry) | 26,810 | 3,794 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Medicinal_Chemistry/Basic_Aspects_of_Drug_Activity |
While there are several types of exceptions, the effects of most drugs result from their interaction with functional macromolecular components of the organism. Such interaction alters the function of the pertinent cellular component and thereby initiates the series of biochemical and physiological changes that are characteristic of the response to the drug. The term receptor is used to denote the component of the organism with which the chemical agent interacts. By virtue of interactions with such receptors, drugs do not create effects but merely modulate ongoing function. Thus, drugs cannot impart a new function to a cell. The affinity of a drug for a specific macromolecular component of the cell and its intrinsic activity are intimately related to its chemical structure. The relationship is frequently quite stringent, and relatively minor modifications in the drug molecule, particularly such subtle changes as stereochemistry, may result in major changes in pharmacological properties. Exploitation of structure-activity relationships has on many occasions led to the synthesis of valuable therapeutic agents. Since changes in molecular configuration need to alter all actions and effects of a drug equally, it is sometimes possible to develop a congener with a more favorable ratio of therapeutic to toxic effects, enhanced selectivity among different cells or tissues, or more acceptable secondary characteristics than those of the parent drug. In addition, effective therapeutic agents have been fashioned by developing structurally related competitive antagonists of other drugs or of endogenous substances known to be important in biochemical or physiological function. Minor modifications of structure can also have profound effects on the pharmacokinetic properties of drugs. Enantemerism can be produced by sp hybridized carbon atoms. Because free rotation about the chiral carbon is not possible, two stable forms of the molecule can exist. A molecule with two nonidentical asymmetric centers can exist as (2 = 4) four stereo isomers. Interaction with biological receptors can differ greatly between two enantomers, even to the point of no binding. There are numerous examples among drug molecules where only one isomer exhibits the desired pharmacology. Some isomers may even cause side effects or entirely different effects than its mirror image. Different isomers can be used in different cases depending on the desired effect. Clinically, D(-) ephedrine is used to a large extent as an anti-asthmatic and, formerly, as a presser amine to restore low blood pressure as a result of trauma. L(+) pseudo-ephedrine is used primarily as a nasal decongestant. If the biological receptor has at least three binding sites, the receptor easily can differentiate enantomers (see figures below). The R(-)isomer has three points of interaction and is held in the conformation shown to maximize binding energy, whereas, the S(+)isomer can have only two sites of interaction. It should be noted that the structure of alpha and beta adrenergic receptors are not entirely known. Also we should not forget that there is also enantioselectivity with respect to pharmacokinetics, such as, absorption, distribution, metabolism, and excretion. Cell signaling is the method in which cells communicate between each other in order to coordinate their activities and react to changes in their environments. Cell signaling involves a signal molecule (an agonist) and a specific signal receptor. Agonistic drugs are those which mimick natural signaling molecules and couse similar effects, while antagonists compete or inhibit agonists and hamper their effects. Agonist drugs mimic cell signaling molecules by activating the same receptor sites and causing similar effects. Both are described quantitatively by the same methods. If one assumes that an agonist drug interacts reversibly with its receptor and that the resultant effect is proportional to the number of receptors occupied, the following reaction can be written: \[ \text{Drug (D)} + \text{Receptor (R)} \ce{ <=>[k_1,k_2]} DR \rightarrow \text{Effect}\] This reaction sequence is analogous to the interaction of substrate with enzyme, and the magnitude of effect can be analyzed in a manner similar to that for enzymatic product formation. The application equation is identical in form with the : \[ \text{Effect} = \dfrac{(\text{Maximal Effect}) [D]}{K_d+[D]} \label{2}\] where[D] is the concentration of free drug and K (equal to K / K ) is the dissociation constant for the drug-receptor complex. This equation describes a simple rectangular hyperbola . There is no effect at [D] = 0; the effect is half-maximal when [D] = K that is when half of the receptors are occupied; the maximal effect is approached asymptotically as [D] increases above K (Figure A below). Therefore, doubling the dose does not double the drug effect, but creates a less than two-fold consequence. It is frequently convenient to plot the magnitude of effect versus log [D], since a wide range of drug concentrations is easily displayed and a portion of the curve is more linear. In this case, the result is the familiar sigmoidal log dose-effect curve (Figure B below). Equation \ref{2} can be rearranged by taking the reciprocal of both sides. The graph of the resulting equation is called the : \[ \dfrac{1}{\text{Effect}} = \dfrac{K_D}{(\text{Max Effect} [D])} + \dfrac{1}{\text{Max Effect}} \label{3}\] A plot of 1/Effect versus 1/[D] yields a straight line that intersects the Y-axis at 1/(Max. Effect) and that has a slope equal to K /(Max. Effect). Extrapolation of this line to the X-axis yields the value of -1/K (Figure C below). Thus, values of \(K_D\) and Max. Effect can be readily calculated from such a plot. This representation is especially useful for analyzing drug antagonism. . Certain drugs, called antagonists, interact with receptor sites to inhibit the actions of an agonists, while initiating no effect themselves. In this type of interaction, called competitive inhibition, the 1/Effect intercept of the plot of 1/[D] versus 1/[S] (Figure C) is the same in the presence and absence of an inhibitor, although the slope is different. This reflects the fact that the Max. Effect is not altered by a competitive inhibitor. The identification of competitive inhibition is that it can be overcome by a sufficiently high concentration of substrate. A noncompetitive antagonist prevents the agonist from producing any effect at a given receptor site. This could result from irreversible interaction of the antagonist at any site to prevent binding of agonist. It could also follow reversible interaction with any component of the system so as to decrease the effect binding of agonist. These results may be conceptualized as removal of receptor from the system. The Max. Effect is reduced but the agonist's ability to still act normally at the receptor site is unaltered. The affinity of the agonist for the receptor and its potency are thus unaltered. On a Lineweaver-Burk plot (Fig. C), the intercept on the 1/Effect axis is increased and the new slope, is steeper. In contrast with the Max. Effect, K is not affected by this kind of inhibition and so the x-intercept is unaltered. Noncompetitive inhibition cannot be overcome by increasing the agonist concentration. Antagonists may thus be classified as acting reversibly or irreversibly. If the antagonist binds at the active site for the agonist, reversible antagonists will be competitive and irreversible antagonists will be noncompetitive. If binding is elsewhere, however, these simple rules do not hold, and any combination is possible. Down load the spreadsheet used to make Figures A, B, and C and change K and Max Effect to see the resulting changes in the graphs. (This spreadsheet can be also used in Microsoft Excel). There are two primary factors contributing to drug interactions. First, many drugs are bound to plasma proteins and this binding serves as a reservoir of inactive drugs. If a second drug displaces a drug already bound to the protein (by competing for the same protein), more of the previously bound drug will be able to pass out of the bloodstream and be available to the receptor and thus a more intense effect may be produced: the displacing drug would increase the effects or the toxicity of the displaced drug. Second, a drug that is metabolized by the liver may induce new enzymes, which can then metabolize any of a variety of new drugs. Thus, an enzyme-inducing drug such as pentobarbital will decrease the activity of other drugs metabolized by the liver by increasing their rates of metabolism. Therefore, a wide variety of factors influence a drug's action in the body, making the use of more than one drug at a time risky, whether they are used separately or mixed in a concoction. Pharmacokinetics deals with the absorption, distribution, biotransformation, and excretion of drugs. These factors, coupled with dosage, determine the concentration of a drug at its sites of action and, hence, the intensity of its effects as a function of time. Many basic principles of biochemistry and enzymology and the physical and chemical principles that govern the active and passive transfer and the distribution of substances across biological membranes are readily applied to the understanding of this important aspect of medicinal chemistry. Drugs are eliminated from the body either unchanged or as metabolites. Excretory organs, the lung excluded, eliminate polar compounds more efficiently than substances with high lipid solubility. Lipid-soluble drugs are thus not readily eliminated until they are metabolized to more polar compounds. The kidney is the most important organ for elimination of drugs and their metabolites. Substances excreted in the feces are mainly unabsorbed orally ingested drugs or metabolites excreted in the bile and not reabsorbed from the intestinal tract. Excretion of drugs in milk is important not because of the amounts eliminated but because the excreted drugs are potential sources of unwanted pharmacological effects in the nursing infant. Pulmonary excretion is important mainly for the elimination of anesthetic gases and vapors: occasionally, small quantities of other drugs of metabolites are excreted by this route. Drug elimination follows first-order kinetics. To illustrate first order kinetics we might consider what would happen if we were instantly inject (with an IV) a person with a drug, collect blood samples at various times and measure the plasma concentrations Cp of the drug. We might see a steady decrease in concentration as the drug is eliminated, as shown in the figure below. If we measure the slope of this curve at a number of times we are actually measuring the rate of change of concentration at each time point. This can be written mathematically as Equation \ref{4}: \[\dfrac{dC_p}{dt} = - k_{el}C_p \label{4}\] where \(k_{el}\) is an elimination constant. Can you make a plot from this equation to calculate Kel? How can this equation be rearranged to determine Kel? If we integrate, we find that the plasma concentration at a given time is Equation \ref{5}: \[ C_p = C_P^o e^{-k_{el}t} \label{5}\] where Cp is the initial plasma concentration. In the process of deriving this equation we can calculate the half life to be Equation 6: \[t_{1/2} = \dfrac{0.693}{k_{el}}\label{6}\] So far we have considered elimination by excretion into urine only. Usually drugs are eliminated by excretion AND metabolism. Schematically this can be represented as: where ke is the excretion rate constant and km is the metabolism rate constant. Here we have two parallel pathways for elimination although there could be more pathways, ie. excretion by exhalation, in sweat, or as is commonly the case, more than one metabolite. The differential equations for the two components shown in this diagram could be written just as we did above. We can also use the equations above to calculate the plasma concentration at any time when we know kel and Cp . However, usually we don't know Cp ahead of time, but we do know the dose. To calculate Cp we need to know the volume that the drug is distributed into. That is, the apparent volume of the mixing container, the body. This apparent volume of distribution is not a physiological volume. It won't be lower than blood or plasma volume but it can may be much larger than body volume for some drugs. It is a mathematical 'fudge' factor relating the amount of drug in the body and the concentration of drug in the measured compartment, usually plasma. This can be expressed as Equation 7: or Immediately after an intravenous dose is administered, the amount of drug in the body is the dose. Thus we get equation 8: or Some example values for apparent volume of distribution are listed in the table below. The last figure, for digoxin, is much larger than body volume. The drug must be extensively distributed into tissue, leaving low concentrations in the plasma, thus the body as a whole appears to have a large volume, of distribution. Remember, this is not a physiological volume. From Equations 8 and 5 we can produce Equation 9: Here are some * to try. Clearance can be defined as the volume of plasma which is completely cleared of drug per unit time. The symbol is CL and the units are ml/min, L/hr, i.e. volume per time. Another way of looking at Clearance is to consider the drug being eliminated from the body ONLY via the kidneys. (If we were to also assume that all of the drug that reaches the kidneys is removed from the plasma then we have a situation where the clearance of the drug is equal to the plasma flow rate to the kidneys. All of the plasma reaching the kidneys would be cleared of drug.) The amount cleared by the body per unit time is dU/dt, the rate of elimination (also the rate of excretion in this example). To calculate the volume which contains that amount we can divide by Cp. That is the volume = amount/concentration. Thus equation 11: equation12 is
As we have defined the term here it is the total body clearance. We have considered that the drug is cleared totally by excretion in urine. Below we will see that the total body clearance can be divided into a clearance due to renal excretion and that due to metabolism. Clearance is a useful term when talking of drug elimination since it can be related to the efficiency of the organs of elimination and blood flow to the organ of elimination. It is useful in investigating mechanisms of elimination and renal or hepatic function in cases of reduced clearance of test substances. Also the units of clearance, volume/time (e.g. ml/min) are easier to visualize, compared with elimination rate constant (units 1/time, e.g. 1/hr). Total body clearance, CL, can be separated into clearance due to renal elimination, CLr and clearance due to metabolism, CLm. CLr = ke * V (renal clearance) and CLm = km * V (metabolic clearance) where these are combined to give equation 13: CL = CLr + CLm Equation 11 can be rearranged to get: thus a plot of dU/dt versus Cp will give a straight line through the origin with a slope equal to the clearance, CL. Most commonly the absorption process of oral administration follows first order kinetics. Even though many oral dosage forms are solids which must dissolve before being absorbed the overall absorption process can often be considered to be a single first order process. At least that's the assumption we will use for now. Schematically this model can be represented as:
The differential equation for Xg is Equation 14: This is similar to the equation for after an IV injection. Integrating this we get Equation 15: Xg = Xg * e = F * Dose * e where F is the fraction of the dose which is absorbed, the bioavailability. For the amount of drug in the body Xp ( = V * Cp), the differential equation is Equation 16: The first term of this equation is ka * Xg which is absorption, and the second term is kel * V * Cp which is elimination. Even without integrating this equation we can get an idea of the plasma concentration time curve. At the start Xg >> V * Cp therefore the value of is positive, the slope will be positive and Cp will increase. With increasing time Xg will decrease, while initially Cp is increasing, therefore there will be a time when ka * Xg = kel * V * Cp. At this time will be zero and there will be a peak in the plasma concentration. At even later times Xg 0, and will become negative and Cp will decrease. Now we will integrate the equation. Starting with the differential equation we can substitute Xg = Xg0 *e . If we use F * DOSE for Xg where F is the fraction of the dose absorbed, the integrated equation for Cp versus time is Equation 17: Notice that the right hand side of this equation (Equation 17) is a constant multiplied by the difference of two exponential terms. A biexponential equation. We can plot Cp as a constant times the difference between two exponential curves. If we plot each exponential separately we get the following: Notice that the difference starts at zero, increases, and finally decreases again. By adding the two lines in the second plot we get the actual plasma concentration. Plotting this difference we get:. We can calculate the plasma concentration at anytime if we know the values of all the parameters of Equation 17. We can also calculate the time of peak concentration using the equation: As an example we could calculate the peak plasma concentration given that F = 0.9, DOSE = 600 mg, ka = 1.0 hr , kel = 0.15 hr , and V = 30 liter. = 2.23 hour = 21.18 x [ 0.7157 - 0.1075] = 12.9 mg/L As another example we could consider what would happen with ka = 0.2 hr instead of 1.0 hr = 5.75 hour = 72 x (0.4221 - 0.3166) = 7.6 mg/L lower and slower than before. Using Equation 17 above we can create a spreadsheet that will describe the effects of multiple doses given at different time intervals (see plot below). Using a graph such as this, we can coordinate drug doses with proper time intervals in order to will keep drug concentrations at there optimum levels.. Plot of plasma concentration due to multiple oral doses of a hypothetical drug. The doses are 200 mg every 8 hours; the fraction absorbed was 1.00; the elimination constant is 0.2/hr; the absorption constant is 1.0/hr; and the volume concentration is 30 L. | 18,413 | 3,795 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/08%3A_Solutions/8.10%3A_Ions_and_Electrolytes/8.10.9D%3A_8.10.9D%3A_Ionic_migration |
The conductance of an electrolytic solution results from the movement of the ions it contains as they migrate toward the appropriate electrodes. But the picture we tend to have in our minds of these ions moving in a orderly, direct march toward an electrode is wildly mistaken. The thermally-induced random motions of molecules is known as . The term refers specifically to the movement of ions due to an externally-applied electrostatic field. The average thermal energy at temperatures within water's liquid range (given by ) is sufficiently large to dominate the movement of ions even in the presence of an applied electric field. This means that the ions, together with the water molecules surrounding them, are engaged in a wild dance as they are buffeted about by thermal motions (which include Brownian motion). If we now apply an external electric field to the solution, the chaotic motion of each ion is supplemented by an occasional jump in the direction dictated by the interaction between the ionic charge and the field. But this is really a surprisingly tiny effect: It can be shown that in a typical electric field of 1 volt/cm, a given ion will experience only about one field-directed (non-random) jump for every 10 random jumps. This translates into an average migration velocity of roughly 10 m sec (10 mm sec ). Given that the radius of the H O molecule is close to 10 m, it follows that about 1000 such jumps are required to advance beyond a single solvent molecule! All ionic solutions contain at least two kinds of ions (a cation and an anion), but may contain others as well. In the late 1870's, the physicist Friedrich Kohlrausch noticed that the limiting equivalent conductivities of salts that share a common ion exhibit constant differences. 34.9 These differences represent the differences in the conductivities of the ions that are shared between the two salts. The fact that these differences are identical for two pairs of salts such as KCl/LiCl and KNO /LiNO tells us that the mobilities of the non-common ions K and LI are not affected by the accompanying anions. This principle is known as , which states that , Each ionic species makes a contribution to the conductivity of the solution that depends only on the nature of that particular ion, and is independent of the other ions present. Kohlrausch's law can be expressed as Λ = Σ λ + Σ λ This means that we can assign a λ to each kind of ion: Limiting ionic equivalent conductivities at 25°C, S cm eq Just as a compact table of thermodynamic data enables us to predict the chemical properties of a very large number of compounds, this compilation of equivalent conductivities of twenty different species yields reliable estimates of the of Λ values for five times that number of salts. One useful application of Kohlrausch's law is to estimate the limiting equivalent conductivities of weak electrolytes which, as we observed above, cannot be found by extrapolation. Thus for acetic acid CH COOH ("HAc"), we combine the λ values for H O and CH COO given in the above table: Λ = λ + λ Movement of a migrating ion through the solution is brought about by a force exerted by the applied electric field. This force is proportional to the field strength and to the ionic charge. Calculations of the frictional drag are based on the premise that the ions are spherical (not always true) and the medium is continuous (never true) as opposed to being composed of discrete molecules. Nevertheless, the results generally seem to be realistic enough to be useful. According to Newton's law, a constant force exerted on a particle will it, causing it to move faster and faster unless it is restrained by an opposing force. In the case of electrolytic conductance, the opposing force is frictional drag as the ion makes its way through the medium. The magnitude of this force depends on the radius of the ion and its primary hydration shell, and on the viscosity of the solution. Eventually these two forces come into balance and the ion assumes a constant average velocity which is reflected in the values of λ tabulated in the table above. The relation between λ and the velocity (known as the μ ) is easily derived, but we will skip the details here, and simply present the results: Anions are conventionally assigned negative μ values because they move in opposite directions to the cations; the values shown here are absolute values |μ |. Note also that the units are cm/sec per volt/cm, hence the cm term. Absolute limiting mobilities of ions at 25°C, (cm volt sec ) × 100 As with the limiting conductivities, the trends in the mobilities can be roughly correlated with the charge and size of the ion. (Recall that negative ions tend to be larger than positive ions.) In electrolytic conduction, ions having different charge signs move in opposite directions. Conductivity measurements give only the sum of the positive and negative ionic conductivities according to Kohlrausch's law, but they do not reveal how much of the charge is carried by each kind of ion. Unless their mobilities are the same, cations and anions do not contribute equally to the total electric current flowing through the cell. Recall that an electric current is defined as a flow of electric charges; the current in amperes is the number of coulombs of charge moving through the cell per second. Because ionic solutions contain equal quantities of positive and negative charges, it follows that the current passing through the cell consists of positive charges moving toward the cathode, and negative charges moving toward the anode. But owing to mobility differences, cations and ions do not usually carry identical fractions of the charge. Transference numbers are often referred to as ; either term is acceptable in the context of electrochemistry. The fraction of charge carried by a given kind of ion is known as the \(t_{\pm}\). For a solution of a simple binary salt, \[ t_+ = \dfrac{\lambda_+}}{\lambda_+ + \lambda_-}\] and \[ t_- = \dfrac{\lambda_-}}{\lambda_+ + \lambda_-}\] By definition, \[t_+ + t_– = 1.\] To help you visualize the effects of non-identical transference numbers, consider a solution of M X in which = 0.75 and = 0.25. Let the cell be divided into three [imaginary] sections as we examine the distribution of cations and anions at three different stages of current flow. Transference numbers can be determined experimentally by observing the movement of the boundary between electrolyte solutions having an ion in common, such as LiCl and KCl: In this example, K has a higher transference number than Li , but don't try to understand why the KCl boundary move to the left; the details of how this works are rather complicated and not important for the purposes of this this course. You may have noticed from the tables above that the hydrogen- and hydroxide ions have extraordinarily high equivalent conductivities and mobilities. This is a consequence of the fact that unlike other ions which need to bump and nudge their way through the network of hydrogen-bonded water molecules, these ions are in this network. By simply changing the H O partners they hydrogen-bond with, they can migrate "virtually". In effect, what migrates is the hydrogen-bonds, rather than the physical masses of the ions themselves. This process is known as the . The shifting of the hydrogen bonds occurs when the rapid thermal motions of adjacent molecules brings a particular pair into a more favorable configuration for hydrogen bonding within the local molecular network. Bear in mind that what we refer to as "hydrogen ions" H are really H O . It has been proposed that the larger aggregates H O and H O are important intermediates in this process. It is remarkable that this virtual migration process was proposed by Theodor Grotthuss in 1805 — just five years after the discovery of electrolysis, and he didn't even know the correct formula for water; he thought its structure was H–O–O–H. These two diagrams will help you visualize the process. The successive downward rows show the first few "hops" made by the virtual H and OH ions as they move in opposite directions toward the appropriate electrodes. (Of course, the same mechanism is operative in the absence of an external electric field, in which case of the hops will be in random directions.) Covalent bonds are represented by black lines, and hydrogen bonds by gray lines. | 8,476 | 3,796 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Exemplars/Corrosion/Galvanization |
Galvanization is a metal coating process in which a ferrous part is coated with a thin layer of zinc. The zinc coating seals the surface of the part from the environment, preventing oxidation and weathering from occurring. The primary method of galvanization is “hot dip galvanization”, which has been in use for over 150 years. While the idea of coating a part in molten zinc was first proposed by chemist Paul Jacques Malouin in 1742, the process was not put into practice until patented by chemist Stanislas Sorel in 1836. Sorel’s process has changed little since then, and still involves coating a part in molten zinc after cleaning it with an acid solution and coating the part in flux. Galvanization helps to extend the life of steel parts by providing a barrier between the steel and the atmosphere, preventing iron oxide from forming on the surface of the steel. Galvanization also provides superior corrosion resistance to parts exposed to the environment. Galvanization provides a cost-effective solution for coating steel parts, specifically those that will receive significant environmental exposure over their lifetime. | 1,152 | 3,797 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Free_Radical_Reactions/Chlorination_of_Methane |
If a mixture of methane and chlorine is exposed to a flame, it explodes - producing carbon and hydrogen chloride. This is not a very useful reaction! The reaction we are going to explore is a more gentle one between methane and chlorine in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction - a reaction brought about by light. \[ \ce{CH4 + Cl_2 -> CH_3Cl + HCl} \nonumber\] The organic product is chloromethane. One of the hydrogen atoms in the methane has been replaced by a chlorine atom, so this is a substitution reaction. However, the reaction does not stop there, and all the hydrogens in the methane can in turn be replaced by chlorine atoms. Multiple substitution is dealt with on a separate page, and you will find a link to that at the bottom of this page. The mechanism involves a chain reaction. During a chain reaction, for every reactive species you start off with, a new one is generated at the end - and this keeps the process going. The over-all process is known as , or as a . Cl \(\rightarrow\) 2Cl CH + Cl \(\rightarrow\)CH + HCl CH + Cl \(\rightarrow\)CH Cl + Cl 2Cl \(\rightarrow\)Cl CH + Cl \(\rightarrow\) CH C l CH + CH \(\rightarrow\)CH CH | 1,243 | 3,798 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Properties_of_Matter |
We are all surrounded by matter on a daily basis. Anything that we use, touch, eat, etc. is an example of matter. Matter can be defined or described as anything that takes up space, and it is composed of miniscule particles called atoms. It must display the two properties of mass and volume. The different types of matter can be distinguished through two components: composition and properties. The composition of matter refers to the different components of matter along with their relative proportions. The properties of matter refer to the qualities/attributes that distinguish one sample of matter from another. These properties are generally grouped into two categories: physical or chemical. Physical properties can be observed or measured without changing the composition of matter. Physical properties are used to observe and describe matter. Physical properties of materials and systems are often described as intensive and extensive properties. This classification relates to the dependency of the properties upon the size or extent of the system or object in question. An intensive property is a bulk property, meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system. Examples of intensive properties include temperature, refractive index, density, and hardness of an object. When a diamond is cut, the pieces maintain their intrinsic hardness (until their size reaches a few atoms thick). In contrast, an extensive property is additive for independent, non-interacting subsystems. The property is proportional to the amount of material in the system. : A physical property that will be the same regardless of the amount of matter. : A physical property that will change if the amount of matter changes. Change in which the matter's physical appearance is altered, but composition remains unchanged. A takes place without any changes in molecular composition. The same element or compound is present before and after the change. The same molecule is present through out the changes. Physical changes are related to physical properties since some measurements require that changes be made. The are: Solid, Liquid, Gas When liquid water (\(H_2O\)) freezes into a solid state (ice), it appears changed; However, this change is only physical as the the composition of the constituent molecules is the same: 11.19% hydrogen and 88.81% oxygen by mass. Figure \(\Page {2}\) Chemical properties of matter describes its "potential" to undergo some chemical change or reaction by virtue of its composition. What elements, electrons, and bonding are present to give the potential for chemical change. It is quite difficult to define a chemical property without using the word "change". Eventually you should be able to look at the formula of a compound and state some chemical property. At this time this is very difficult to do and you are not expected to be able to do it. For example hydrogen has the potential to ignite and explode given the right conditions. This is a chemical property. Metals in general have they chemical property of reacting with an acid. Zinc reacts with hydrochloric acid to produce hydrogen gas. This is a chemical property. Chemical change results in one or more substances of entirely different composition from the original substances. The elements and/or compounds at the start of the reaction are rearranged into new product compounds or elements. A alters the composition of the original matter. Different elements or compounds are present at the end of the chemical change. The atoms in compounds are rearranged to make new and different compounds. Corrosion is the unwanted oxidation of metals resulting in metal oxides. \[2 Mg + O_2 \rightarrow 2 MgO\] The following questions are multiple choice. 1. Milk turns sour. This is a ________________ 2. HCl being a strong acid is a __________, Wood sawed in two is ___________ 3. CuSO is dissolved in water 4. Aluminum Phosphate has a density of 2.566 g/cm3 5. Which of the following are examples of matter? 6. The formation of gas bubbles is a sign of what type of change? 7. True or False: Bread rising is a physical property. 8. True or False: Dicing potatoes is a physical change. 9. Is sunlight matter? 10. The mass of lead is a _____________property. | 4,340 | 3,799 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.14%3A_The_Free_Energy |
In the previous section, we were careful to differentiate between the entropy change occurring in the reaction system Δ , on the one hand, and the entropy change occurring in the surroundings, Δ , given by –Δ / , on the other. By doing this we were able to get a real insight into what controls the direction of a reaction and why. In terms of calculations, though, it is a nuisance having to look up both entropy and enthalpy data in order to determine the direction of a reaction. For reasons of convenience, therefore, chemists usually combine the entropy and the enthalpy into a new function called the , or more simply the free energy, which is given the symbol . If free-energy tables are available, they are all that is needed to predict the direction of a reaction at the temperature for which the tables apply. In order to introduce free energy, let us start with the inequality \[-\frac{\Delta H}{T}+ {\Delta S}_{sys} > 0 \nonumber \] This inequality must be true if a reaction occurring at constant pressure in surroundings at constant temperature is to be spontaneous. It is convenient to multiply this inequality by ; it then becomes \[-\Delta H + T \Delta S > 0 \nonumber \] (From now on we will abandon the subscript "sys".) If –Δ + Δ is greater than zero, it follows that multiplying it by –1 produces a quantity which is than zero, that is, \[\Delta H - T \Delta S < 0 \label{3} \] This latest inequality can be expressed very neatly in terms of the free energy , which is defined by the equation \[G=H-TS \nonumber \] When a chemical reaction occurs at constant temperature, the free energy will change from an initial value of , given by \[G_{1}=H_{1}=TS_{1} \nonumber \] to a final value \[G_{2}=H_{2}-TS_{2} \nonumber \] The change in free energy ΔG will thus be \[\Delta G = G_{2} - G_{1} = H_{2} - H_{1} - T(S_{2}-S_{1}) \nonumber \] or \[\Delta G = \Delta H - T\Delta S \nonumber \] Feeding this result back into inequality \(\ref{3}\) gives the result \[\Delta G = \Delta H - T \Delta S <0 \nonumber \] \[\Delta G < 0 \label{10} \] This very important and useful result tells us that when a spontaneous chemical reaction occurs (at constant temperature and pressure), the . In other words a . If we have available the necessary free-energy data in the form of tables, it is now quite easy to determine whether a reaction is spontaneous or not. We merely calculate Δ for the reaction using the tables. If Δ turns out to be positive, the reaction is nonspontaneous, but if it turns out to be negative, then by virtue of Eq. \(\ref{10}\) we can conclude that it is spontaneous. Data on free energy are usually presented in the form of a table of values of . The standard free energy of formation of a substance is defined as the free-energy change which results when 1 mol of substance is prepared from its elements at the standard pressure of 1 atm and a given temperature, usually 298 K. It is given the symbol Δ °. A table of values of Δ ° (298 K) for a limited number of substances is given in the following table. This table is used in exactly the same way as a table of . This type of table enables us to find Δ values for any reaction occurring at 298 K and 1 atm pressure, provided only that all the substances involved in the reaction appear in the table. The two following examples illustrate such usage. Determine whether the following reaction is spontaneous or not: \[\ce{4NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(l)}\qquad 1 \text{ atm, 298K} \nonumber \] Following exactly the same rules used for standard enthalpies of formation, we have \(\Delta G_{m}^{\circ} = \Sigma \Delta G_{f}^{\circ}\text{ (products)} - \Sigma \Delta G_{f}^{\circ} \text{ (reactants)}\) \(\qquad = \ce{4 \Delta G_{f}^{\circ}(NO) + 6 \Delta G_{f}^{\circ}(H2O) - 4\Delta G_{f}^{\circ}(NH3) - 5\Delta G_{f}^{\circ}(O2)}\) \(\Delta G_{m}^{\circ} = [4 \times 86.7 + 6 \times (-273.3) - 4 \times (-16.7) - 5 \times 0.0]\frac{ \text{kJ }}{\text{mol}}\) \(\qquad = -1010\frac{\text{ kJ}}{\text{ mol}}\) Since \(\Delta G_{m}^{\circ} \) is very negative, we conclude that this reaction is spontaneous. The reaction of NH with O is very slow, so that when NH is released into the air, no noticeable reaction occurs. In the presence of a catalyst, though, NH burns with a yellowish flame in O . This reaction is very important industrially, since the NO produced from it can be reacted further with O and H O to form HNO : \[\ce{2NO + \frac{3}{2}O2 + H2O→2HNO3} \nonumber \] Nitric acid, HNO is used mainly in the manufacture of nitrate fertilizers but also in the manufacture of explosives. Determine whether the following reaction is spontaneous or not: \[\ce{2NO(g) + 2CO(g) → 2CO2(g) + N2 (g)}\qquad 1 \text{ atm, 298K} \nonumber \] Following previous procedure we have \((\Delta G_{m}^{\circ} = (-2\times 394.4 + 0.0 - 2 \times 86.7 + 2 \times 137.3)\frac{\text{ kJ}}{\text{ mol}}\) \(\qquad = -687.6\frac{ \text{ kJ}}{\text{ mol}}\) The reaction is thus spontaneous. This example is an excellent illustration of how useful thermodynamics can be. Since both NO and CO are air pollutants produced by the internal-combustion engine, this reaction provides a possible way of eliminating both of them in one reaction, killing two birds with one stone. A thee-way catalytic converter is able to perform the equivalent of this reaction. The reduction step coverts NO to O and N . Then, in the oxidation step, CO and O are converted to CO . If Δ ° had turned out be +695 kJ mol , the reaction would be nonspontaneous and there would be no point at all in developing such a device. We quite often encounter situations in which we need to know the value of Δ ° for a reaction at a temperature other than 298 K. Although extensive thermodynamic tables covering a large range of temperatures are available, we can also obtain approximate values for Δ from the relationship \[\Delta G_{m}^{\circ}=\Delta H_{m}^{\circ}-T\Delta S_{m}^{\circ} \nonumber \] If we assume, as we did previously, that neither Δ ° nor Δ ° varies much as the temperature changes from 298 K to the temperature in question, we can then use the values of Δ °(298 K) obtained from the and Δ °(298 K) obtained from the to calculate Δ ° for the temperature in question. Using the and the , calculate Δ ° and Δ ° for the reaction \[\ce{CH4(g) + H2O(g) → 3 H2(g) + CO(g)} \qquad 1 \text{ atm} \nonumber \] From the tables we find \(\Delta H_{m}^{\circ}(298 \text{ K})= 3\Delta H_{f}^{\circ}(\text{H}_{2})+\Delta H_{f}^{\circ}\text{(CO)} - \Delta H_{f}^{º}( \text{CH}_{4}) - \Delta H_{f}^{\circ}(\text{H}_{2}\text{O})\) \(\qquad = (3 \times 0.0 - 110.6 + 74.8 + 241.8) \frac{\text{ kJ}}{\text{ mol}} = +206.1\frac{\text{kJ}}{\text{ mol}}\) \(\Delta S_{m}^{\circ}(298K) = (3 \times 130.6 + 197.6 - 187.9 - 188.7)\frac{\text{ J}}{\text{ mol K}} = +212.8\frac{\text{ J}}{\text{ mol K}}\) \(\Delta G_{m}^{\circ}=\Delta H^{\circ}(298 \text{ K}) - T\Delta S^{\circ}(298 \text{ K})\) \(\qquad =206.1\frac{ \text{ kJ}}{\text{ mol}} - 600 \times 212.8\frac{\text{ J}}{\text{ mol}}\) \(\qquad = (206.1 - 127.7)\frac{\text{kJ}}{\text{ mol}} = +78.4\frac{\text{ kJ}}{\text{ mol}}\) At 1200 K by contrast \(\Delta G_{m}^{\circ}=206.1\frac{\text{ kJ}}{\text{ mol}} - 1200 \times 212.8\frac{\text{ J}}{\text{ mol}}\) \(\qquad = (206.1 - 255.4)\frac{\text{ kJ}}{\text{ mol}} = -49.3\frac{\text{ kJ}}{\text{ mol}}\) From more extensive tables we find that accurate values of the free-energy change are \(\Delta G_{m}^{\circ}(600 \text{ K}) = +72.6\frac{\text{ kJ}}{\text{ mol}}\) and \(\Delta G_{m}^{\circ}(1200 \text{ K})=-77.7\frac{\text{ kJ} }{\text{ mol}}\). Our approximate value at 1200 K is thus about 50 percent in error. Nevertheless it predicts the right for \(\Delta G\), a result which is adequate for most purposes. | 7,812 | 3,800 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/10%3A_Liquids_and_Solids/10.6%3A_Lattice_Structures_in_Crystalline_Solids |
Over 90% of naturally occurring and man-made solids are crystalline. Most solids form with a regular arrangement of their particles because the overall attractive interactions between particles are maximized, and the total intermolecular energy is minimized, when the particles pack in the most efficient manner. The regular arrangement at an atomic level is often reflected at a macroscopic level. In this module, we will explore some of the details about the structures of metallic and ionic crystalline solids, and learn how these structures are determined experimentally. We will begin our discussion of crystalline solids by considering elemental metals, which are relatively simple because each contains only one type of atom. A pure metal is a crystalline solid with metal atoms packed closely together in a repeating pattern. Some of the properties of metals in general, such as their malleability and ductility, are largely due to having identical atoms arranged in a regular pattern. The different properties of one metal compared to another partially depend on the sizes of their atoms and the specifics of their spatial arrangements. We will explore the similarities and differences of four of the most common metal crystal geometries in the sections that follow. The structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its . The unit cell consists of lattice points that represent the locations of atoms or ions. The entire structure then consists of this unit cell repeating in three dimensions, as illustrated in Figure \(\Page {1}\). Let us begin our investigation of crystal lattice structure and unit cells with the most straightforward structure and the most basic unit cell. To visualize this, imagine taking a large number of identical spheres, such as tennis balls, and arranging them uniformly in a container. The simplest way to do this would be to make layers in which the spheres in one layer are directly above those in the layer below, as illustrated in Figure \(\Page {2}\). This arrangement is called , and the unit cell is called the or primitive cubic unit cell. In a simple cubic structure, the spheres are not packed as closely as they could be, and they only “fill” about 52% of the volume of the container. This is a relatively inefficient arrangement, and only one metal (polonium, Po) crystallizes in a simple cubic structure. As shown in Figure \(\Page {3}\), a solid with this type of arrangement consists of planes (or layers) in which each atom contacts only the four nearest neighbors in its layer; one atom directly above it in the layer above; and one atom directly below it in the layer below. The number of other particles that each particle in a crystalline solid contacts is known as its . For a polonium atom in a simple cubic array, the coordination number is, therefore, six. In a simple cubic lattice, the unit cell that repeats in all directions is a cube defined by the centers of eight atoms, as shown in Figure \(\Page {4}\). Atoms at adjacent corners of this unit cell contact each other, so the edge length of this cell is equal to two atomic radii, or one atomic diameter. A cubic unit cell contains only the parts of these atoms that are within it. Since an atom at a corner of a simple cubic unit cell is contained by a total of eight unit cells, only one-eighth of that atom is within a specific unit cell. And since each simple cubic unit cell has one atom at each of its eight “corners,” there is \(8×\dfrac{1}{8}=1\) atom within one simple cubic unit cell. The edge length of the unit cell of alpha polonium is 336 pm. Alpha polonium crystallizes in a simple cubic unit cell: (a) Two adjacent Po atoms contact each other, so the edge length of this cell is equal to two Po atomic radii: \(l = 2r\). Therefore, the radius of Po is \[r=\mathrm{\dfrac{l}{2}=\dfrac{336\: pm}{2}=168\: pm}\nonumber \] (b) Density is given by \[\mathrm{density=\dfrac{mass}{volume}}.\nonumber \] The density of polonium can be found by determining the density of its unit cell (the mass contained within a unit cell divided by the volume of the unit cell). Since a Po unit cell contains one-eighth of a Po atom at each of its eight corners, a unit cell contains one Po atom. The mass of a Po unit cell can be found by: \[\mathrm{1\: Po\: unit\: cell×\dfrac{1\: Po\: atom}{1\: Po\: unit\: cell}×\dfrac{1\: mol\: Po}{6.022\times 10^{23}\:Po\: atoms}×\dfrac{208.998\:g}{1\: mol\: Po}=3.47\times 10^{−22}\:g}\nonumber \] The volume of a Po unit cell can be found by: \[V=l^3=\mathrm{(336\times 10^{−10}\:cm)^3=3.79\times 10^{−23}\:cm^3}\nonumber \] (Note that the edge length was converted from pm to cm to get the usual volume units for density.) Therefore, the density of \[\mathrm{Po=\dfrac{3.471\times 10^{−22}\:g}{3.79\times 10^{−23}\:cm^3}=9.16\: g/cm^3}\nonumber \] The edge length of the unit cell for nickel is 0.3524 nm. The density of Ni is 8.90 g/cm . Does nickel crystallize in a simple cubic structure? Explain. No. If Ni were simple cubic, its density would be given by: \[\mathrm{1\: Ni\: atom×\dfrac{1\: mol\: Ni}{6.022\times 10^{23}\:Ni\: atoms}×\dfrac{58.693\:g}{1\: mol\: Ni}=9.746\times 10^{−23}\:g}\nonumber \] \[V=l^3=\mathrm{(3.524\times 10^{−8}\:cm)^3=4.376\times 10^{−23}\:cm^3}\nonumber \] Then the density of Ni would be \[(\mathrm{=\dfrac{9.746\times 10^{−23}\:g}{4.376\times 10^{−23}\:cm^3}=2.23\: g/cm^3}\nonumber \] Since the actual density of Ni is not close to this, Ni does not form a simple cubic structure. Most metal crystals are one of the four major types of unit cells. For now, we will focus on the three cubic unit cells: simple cubic (which we have already seen), , and —all of which are illustrated in Figure \(\Page {5}\). (Note that there are actually seven different lattice systems, some of which have more than one type of lattice, for a total of 14 different types of unit cells. We leave the more complicated geometries for later in this module.) Some metals crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and an atom in the center, as shown in Figure \(\Page {6}\). This is called a . Atoms in the corners of a BCC unit cell do not contact each other but contact the atom in the center. A BCC unit cell contains two atoms: one-eighth of an atom at each of the eight corners (\(8×\dfrac{1}{8}=1\) atom from the corners) plus one atom from the center. Any atom in this structure touches four atoms in the layer above it and four atoms in the layer below it. Thus, an atom in a BCC structure has a coordination number of eight. Atoms in BCC arrangements are much more efficiently packed than in a simple cubic structure, occupying about 68% of the total volume. Isomorphous metals with a BCC structure include K, Ba, Cr, Mo, W, and Fe at room temperature. (Elements or compounds that crystallize with the same structure are said to be .) Many other metals, such as aluminum, copper, and lead, crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and at the centers of each face, as illustrated in Figure \(\Page {7}\). This arrangement is called a . A FCC unit cell contains four atoms: one-eighth of an atom at each of the eight corners (\(8×\dfrac{1}{8}=1\) atom from the corners) and one-half of an atom on each of the six faces (\(6×\dfrac{1}{2}=3\) atoms from the faces). The atoms at the corners touch the atoms in the centers of the adjacent faces along the face diagonals of the cube. Because the atoms are on identical lattice points, they have identical environments. Atoms in an FCC arrangement are packed as closely together as possible, with atoms occupying 74% of the volume. This structure is also called . In , there are three repeating layers of hexagonally arranged atoms. Each atom contacts six atoms in its own layer, three in the layer above, and three in the layer below. In this arrangement, each atom touches 12 near neighbors, and therefore has a coordination number of 12. The fact that FCC and CCP arrangements are equivalent may not be immediately obvious, but why they are actually the same structure is illustrated in Figure \(\Page {8}\). Because closer packing maximizes the overall attractions between atoms and minimizes the total intermolecular energy, the atoms in most metals pack in this manner. We find two types of closest packing in simple metallic crystalline structures: CCP, which we have already encountered, and shown in Figure \(\Page {9}\). Both consist of repeating layers of hexagonally arranged atoms. In both types, a second layer (B) is placed on the first layer (A) so that each atom in the second layer is in contact with three atoms in the first layer. The third layer is positioned in one of two ways. In , atoms in the third layer are directly above atoms in the first layer (i.e., the third layer is also type A), and the stacking consists of alternating type A and type B close-packed layers (i.e., ABABAB⋯). In CCP, atoms in the third layer are not above atoms in either of the first two layers (i.e., the third layer is type C), and the stacking consists of alternating type A, type B, and type C close-packed layers (i.e., ABCABCABC⋯). About two–thirds of all metals crystallize in closest-packed arrays with coordination numbers of 12. Metals that crystallize in an HCP structure include Cd, Co, Li, Mg, Na, and Zn, and metals that crystallize in a CCP structure include Ag, Al, Ca, Cu, Ni, Pb, and Pt. Calcium crystallizes in a face-centered cubic structure. The edge length of its unit cell is 558.8 pm. In an FCC structure, Ca atoms contact each other across the diagonal of the face, so the length of the diagonal is equal to four Ca atomic radii (d = 4 ). Two adjacent edges and the diagonal of the face form a right triangle, with the length of each side equal to 558.8 pm and the length of the hypotenuse equal to four Ca atomic radii: \[\begin{align*} a^2+a^2 &=d^2 \\[4pt] \mathrm{(558.8\:pm)^2+(558.5\:pm)^2} &=(4r)^2 \end{align*} \nonumber \] Solving this gives \[r=\mathrm{\sqrt{\dfrac{(558.8\:pm)^2+(558.5\:pm)^2}{16}}}=\textrm{197.6 pmg for a Ca radius}. \nonumber \] Density is given by \(\mathrm{density=\dfrac{mass}{volume}}\). The density of calcium can be found by determining the density of its unit cell: for example, the mass contained within a unit cell divided by the volume of the unit cell. A face-centered Ca unit cell has one-eighth of an atom at each of the eight corners (\(8 \times \dfrac{1}{8}=1\)atom) and one-half of an atom on each of the six faces \(6×\dfrac{1}{2}=3\) atoms), for a total of four atoms in the unit cell. The mass of the unit cell can be found by: \[\mathrm{1\: Ca\: unit\: cell×\dfrac{4\: Ca\: atoms}{1\: Ca\: unit\: cell}×\dfrac{1\: mol\: Ca}{6.022\times 10^{23}\:Ca\: atoms}×\dfrac{40.078\:g}{1\: mol\: Ca}=2.662\times 10^{−22}\:g} \nonumber \] The volume of a Ca unit cell can be found by: \[V=a^3=\mathrm{(558.8\times 10^{−10}\:cm)^3=1.745\times 10^{−22}\:cm^3} \nonumber \] (Note that the edge length was converted from pm to cm to get the usual volume units for density.) Then, the density of polonium: \[\mathrm{Po=\dfrac{2.662\times 10^{−22}\:g}{1.745\times 10^{−22}\:cm^3}=1.53\: g/cm^3} \nonumber \] Silver crystallizes in an FCC structure. The edge length of its unit cell is 409 pm. 144 pm 10.5 g/cm In general, a unit cell is defined by the lengths of three axes ( , , and ) and the angles ( , , and ) between them, as illustrated in Figure \(\Page {10}\). The axes are defined as being the lengths between points in the space lattice. Consequently, unit cell axes join points with identical environments. There are seven different lattice systems, some of which have more than one type of lattice, for a total of fourteen different unit cells, which have the shapes shown in Figure \(\Page {11}\). Ionic crystals consist of two or more different kinds of ions that usually have different sizes. The packing of these ions into a crystal structure is more complex than the packing of metal atoms that are the same size. Most monatomic ions behave as charged spheres, and their attraction for ions of opposite charge is the same in every direction. Consequently, stable structures for ionic compounds result (1) when ions of one charge are surrounded by as many ions as possible of the opposite charge and (2) when the cations and anions are in contact with each other. Structures are determined by two principal factors: the relative sizes of the ions and the ratio of the numbers of positive and negative ions in the compound. In simple ionic structures, we usually find the anions, which are normally larger than the cations, arranged in a closest-packed array. (As seen previously, additional electrons attracted to the same nucleus make anions larger and fewer electrons attracted to the same nucleus make cations smaller when compared to the atoms from which they are formed.) The smaller cations commonly occupy one of two types of (or interstices) remaining between the anions. The smaller of the holes is found between three anions in one plane and one anion in an adjacent plane. The four anions surrounding this hole are arranged at the corners of a tetrahedron, so the hole is called a . The larger type of hole is found at the center of six anions (three in one layer and three in an adjacent layer) located at the corners of an octahedron; this is called an . Figure \(\Page {12}\) illustrates both of these types of holes. Depending on the relative sizes of the cations and anions, the cations of an ionic compound may occupy tetrahedral or octahedral holes, as illustrated in Figure \(\Page {13}\). Relatively small cations occupy tetrahedral holes, and larger cations occupy octahedral holes. If the cations are too large to fit into the octahedral holes, the anions may adopt a more open structure, such as a simple cubic array. The larger cations can then occupy the larger cubic holes made possible by the more open spacing. There are two tetrahedral holes for each anion in either an HCP or CCP array of anions. A compound that crystallizes in a closest-packed array of anions with cations in the tetrahedral holes can have a maximum cation:anion ratio of 2:1; all of the tetrahedral holes are filled at this ratio. Examples include Li O, Na O, Li S, and Na S. Compounds with a ratio of less than 2:1 may also crystallize in a closest-packed array of anions with cations in the tetrahedral holes, if the ionic sizes fit. In these compounds, however, some of the tetrahedral holes remain vacant. Zinc sulfide is an important industrial source of zinc and is also used as a white pigment in paint. Zinc sulfide crystallizes with zinc ions occupying one-half of the tetrahedral holes in a closest-packed array of sulfide ions. What is the formula of zinc sulfide? Because there are two tetrahedral holes per anion (sulfide ion) and one-half of these holes are occupied by zinc ions, there must be \(\dfrac{1}{2}×2\), or 1, zinc ion per sulfide ion. Thus, the formula is ZnS. Lithium selenide can be described as a closest-packed array of selenide ions with lithium ions in all of the tetrahedral holes. What it the formula of lithium selenide? \(\ce{Li2Se}\) The ratio of octahedral holes to anions in either an HCP or CCP structure is 1:1. Thus, compounds with cations in octahedral holes in a closest-packed array of anions can have a maximum cation:anion ratio of 1:1. In NiO, MnS, NaCl, and , for example, all of the octahedral holes are filled. Ratios of less than 1:1 are observed when some of the octahedral holes remain empty. Aluminum oxide crystallizes with aluminum ions in two-thirds of the octahedral holes in a closest-packed array of oxide ions. What is the formula of aluminum oxide? Because there is one octahedral hole per anion (oxide ion) and only two-thirds of these holes are occupied, the ratio of aluminum to oxygen must be \(\dfrac{2}{3}\):1, which would give \(\mathrm{Al_{2/3}O}\). The simplest whole number ratio is 2:3, so the formula is Al O . The white pigment titanium oxide crystallizes with titanium ions in one-half of the octahedral holes in a closest-packed array of oxide ions. What is the formula of titanium oxide? \(\ce{TiO2}\) In a simple cubic array of anions, there is one cubic hole that can be occupied by a cation for each anion in the array. In CsCl, and in other compounds with the same structure, all of the cubic holes are occupied. Half of the cubic holes are occupied in SrH , UO , SrCl , and CaF . Different types of ionic compounds often crystallize in the same structure when the relative sizes of their ions and their stoichiometries (the two principal features that determine structure) are similar. Many ionic compounds crystallize with cubic unit cells, and we will use these compounds to describe the general features of ionic structures. When an ionic compound is composed of cations and anions of similar size in a 1:1 ratio, it typically forms a simple cubic structure. Cesium chloride, CsCl, (Figure \(\Page {14}\)) is an example of this, with Cs and Cl having radii of 174 pm and 181 pm, respectively. We can think of this as chloride ions forming a simple cubic unit cell, with a cesium ion in the center; or as cesium ions forming a unit cell with a chloride ion in the center; or as simple cubic unit cells formed by Cs ions overlapping unit cells formed by Cl ions. Cesium ions and chloride ions touch along the body diagonals of the unit cells. One cesium ion and one chloride ion are present per unit cell, giving the l:l stoichiometry required by the formula for cesium chloride. Note that there is no lattice point in the center of the cell, and CsCl is not a BCC structure because a cesium ion is not identical to a chloride ion. We have said that the location of lattice points is arbitrary. This is illustrated by an alternate description of the CsCl structure in which the lattice points are located in the centers of the cesium ions. In this description, the cesium ions are located on the lattice points at the corners of the cell, and the chloride ion is located at the center of the cell. The two unit cells are different, but they describe identical structures. When an ionic compound is composed of a 1:1 ratio of cations and anions that differ significantly in size, it typically crystallizes with an FCC unit cell, like that shown in Figure \(\Page {15}\). Sodium chloride, NaCl, is an example of this, with Na and Cl having radii of 102 pm and 181 pm, respectively. We can think of this as chloride ions forming an FCC cell, with sodium ions located in the octahedral holes in the middle of the cell edges and in the center of the cell. The sodium and chloride ions touch each other along the cell edges. The unit cell contains four sodium ions and four chloride ions, giving the 1:1 stoichiometry required by the formula, NaCl. The cubic form of zinc sulfide, zinc blende, also crystallizes in an FCC unit cell, as illustrated in Figure \(\Page {16}\). This structure contains sulfide ions on the lattice points of an FCC lattice. (The arrangement of sulfide ions is identical to the arrangement of chloride ions in sodium chloride.) The radius of a zinc ion is only about 40% of the radius of a sulfide ion, so these small Zn ions are located in alternating tetrahedral holes, that is, in one half of the tetrahedral holes. There are four zinc ions and four sulfide ions in the unit cell, giving the empirical formula ZnS. A calcium fluoride unit cell, like that shown in Figure \(\Page {17}\), is also an FCC unit cell, but in this case, the cations are located on the lattice points; equivalent calcium ions are located on the lattice points of an FCC lattice. All of the tetrahedral sites in the FCC array of calcium ions are occupied by fluoride ions. There are four calcium ions and eight fluoride ions in a unit cell, giving a calcium:fluorine ratio of 1:2, as required by the chemical formula, CaF . Close examination of Figure \(\Page {17}\) will reveal a simple cubic array of fluoride ions with calcium ions in one half of the cubic holes. The structure cannot be described in terms of a of points on the fluoride ions because the fluoride ions do not all have identical environments. The orientation of the four calcium ions about the fluoride ions differs. If we know the edge length of a unit cell of an ionic compound and the position of the ions in the cell, we can calculate ionic radii for the ions in the compound if we make assumptions about individual ionic shapes and contacts. The edge length of the unit cell of LiCl (NaCl-like structure, FCC) is 0.514 nm or 5.14 Å. Assuming that the lithium ion is small enough so that the chloride ions are in contact, calculate the ionic radius for the chloride ion. Note: The length unit angstrom, Å, is often used to represent atomic-scale dimensions and is equivalent to 10 m. On the face of a LiCl unit cell, chloride ions contact each other across the diagonal of the face: Drawing a right triangle on the face of the unit cell, we see that the length of the diagonal is equal to four chloride radii (one radius from each corner chloride and one diameter—which equals two radii—from the chloride ion in the center of the face), so \(d = 4r\). From the , we have: \[a^2+a^2=d^2 \nonumber \] which yields: \[\mathrm{(0.514\:nm)^2+(0.514\:nm)^2}=(4r)^2=16r^2 \nonumber \] Solving this gives: \[r=\mathrm{\sqrt{\dfrac{(0.514\:nm)^2+(0.514\:nm)^2}{16}}=0.182\: nm\:(1.82\: Å)\:for\: a\: Cl^−\: radius.} \nonumber \] The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion. The radius of the chloride ion is 1.82 Å. The radius of the potassium ion is 1.33 Å. It is important to realize that values for ionic radii calculated from the edge lengths of unit cells depend on numerous assumptions, such as a perfect spherical shape for ions, which are approximations at best. Hence, such calculated values are themselves approximate and comparisons cannot be pushed too far. Nevertheless, this method has proved useful for calculating ionic radii from experimental measurements such as X-ray crystallographic determinations. The size of the unit cell and the arrangement of atoms in a crystal may be determined from measurements of the of X-rays by the crystal, termed . is the change in the direction of travel experienced by an electromagnetic wave when it encounters a physical barrier whose dimensions are comparable to those of the wavelength of the light. X-rays are electromagnetic radiation with wavelengths about as long as the distance between neighboring atoms in crystals (on the order of a few Å). When a beam of monochromatic X-rays strikes a crystal, its rays are scattered in all directions by the atoms within the crystal. When scattered waves traveling in the same direction encounter one another, they undergo , a process by which the waves combine to yield either an increase or a decrease in amplitude (intensity) depending upon the extent to which the combining waves’ maxima are separated (Figure \(\Page {18}\)). When X-rays of a certain wavelength, , are scattered by atoms in adjacent crystal planes separated by a distance, , they may undergo constructive interference when the difference between the distances traveled by the two waves prior to their combination is an integer factor, , of the wavelength. This condition is satisfied when the angle of the diffracted beam, , is related to the wavelength and interatomic distance by the equation: \[nλ=2d\sin \theta \label{Eq1} \] This relation is known as the in honor of W. H. , the English physicist who first explained this phenomenon. Figure \(\Page {18}\) illustrates two examples of diffracted waves from the same two crystal planes. The figure on the left depicts waves diffracted at the Bragg angle, resulting in constructive interference, while that on the right shows diffraction and a different angle that does not satisfy the Bragg condition, resulting in destructive interference. An X-ray diffractometer, such as the one illustrated in Figure \(\Page {20}\), may be used to measure the angles at which X-rays are diffracted when interacting with a crystal as described earlier. From such measurements, the Bragg equation may be used to compute distances between atoms as demonstrated in the following example exercise. In a diffractometer, X-rays with a wavelength of 0.1315 nm were used to produce a diffraction pattern for copper. The first order diffraction ( = 1) occurred at an angle = 25.25°. Determine the spacing between the diffracting planes in copper. The distance between the planes is found by solving the Bragg equation (Equation \(\ref{Eq1}\)) for . This gives \[d=\dfrac{nλ}{2\sinθ}=\mathrm{\dfrac{1(0.1315\:nm)}{2\sin(25.25°)}=0.154\: nm}\nonumber \] A crystal with spacing between planes equal to 0.394 nm diffracts X-rays with a wavelength of 0.147 nm. What is the angle for the first order diffraction? 21.9° The discovery of the structure of in 1953 by Francis and James is one of the great achievements in the history of science. They were awarded the 1962 Nobel Prize in Physiology or Medicine, along with Maurice , who provided experimental proof of DNA’s structure. British chemist Rosalind made invaluable contributions to this monumental achievement through her work in measuring X-ray diffraction images of DNA. Early in her career, Franklin’s research on the structure of coals proved helpful to the British war effort. After shifting her focus to biological systems in the early 1950s, Franklin and doctoral student Raymond Gosling discovered that DNA consists of two forms: a long, thin fiber formed when wet (type “B”) and a short, wide fiber formed when dried (type “A”). Her X-ray diffraction images of DNA provided the crucial information that allowed Watson and Crick to confirm that DNA forms a double helix, and to determine details of its size and structure. Franklin also conducted pioneering research on viruses and the that contains their genetic information, uncovering new information that radically changed the body of knowledge in the field. After developing ovarian cancer, Franklin continued to work until her death in 1958 at age 37. Among many posthumous recognitions of her work, the Chicago Medical School of Finch University of Health Sciences changed its name to the Rosalind Franklin University of Medicine and Science in 2004, and adopted an image of her famous X-ray diffraction image of DNA as its official university logo. The structures of crystalline metals and simple ionic compounds can be described in terms of packing of spheres. Metal atoms can pack in hexagonal closest-packed structures, cubic closest-packed structures, body-centered structures, and simple cubic structures. The anions in simple ionic structures commonly adopt one of these structures, and the cations occupy the spaces remaining between the anions. Small cations usually occupy tetrahedral holes in a closest-packed array of anions. Larger cations usually occupy octahedral holes. Still larger cations can occupy cubic holes in a simple cubic array of anions. The structure of a solid can be described by indicating the size and shape of a unit cell and the contents of the cell. The type of structure and dimensions of the unit cell can be determined by X-ray diffraction measurements. ). | 27,565 | 3,801 |
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Because of the coulombic force of attraction or repulsion between them, . Suppose we have charges of +1 and –1 μC separated by 1 cm, for example. The charges could be separated by hand, and by the time the length of a football field lay between them, their attractive force would be negligible. Expenditure of muscle energy (0.898 J, to be exact) will be necessary to carry out such a separation. That is, because the charges attract each other, we must do work to pull them apart. According to , the muscle energy expanded to pull the opposite charges apart cannot be destroyed. We say that the 0.898 J is gained by the two charges and stored as potential energy. (symbol ) is the energy which one or more bodies have because of their . We can always regain this energy by reversing the process during which it was stored. If the two opposite charges are returned to their original separation of 1 cm, their potential energy will by 0.898 J. The energy released will appear as kinetic energy, as heat, or in some other form, but it cannot be destroyed. If we had taken two particles both of which had a charge of +1 μC for our example of potential energy, work would have been required to push them together against their repulsive force. Their potential energy would increase as they were brought together from the ends of a football field, and 0.898 J would be required to move them to a distance of 1 cm apart. Because the potential energy of like-charged particles increases as they are brought closer together, while that of opposite-charged particles decreases, it is convenient to assign a value of zero potential energy to two charged particles which are a long distance apart. Bringing a pair of positive charges (or a pair of negative charges) closer together increases to a positive value. Bringing one positive and one negative particle together decreases , giving a negative value. | 1,918 | 3,802 |
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We have examined the theoretical implications and practical applications of John Dalton’s ideas about atoms in our discussion on , and . Clearly the atomic theory is a powerful tool which aids our thinking about how much of one substance can combine with (or be produced from) a given quantity of another. The theory is much less helpful, however, when we try to speculate about what holds the atoms together in molecules such as Br , HgBr and Hg Br . As you have seen, techniques are available for determination of the formula of a new compound, but Dalton’s theory is of little value in formulas. Neither does it tell us which elements are likely to combine with which, nor indicate what chemical and physical properties are to be expected of the compounds which form. The ability to make predictions about chemical reactivity and properties is very important because it guides chemists’ efforts to synthesize new substances which are of value to society at large. Medicines, metals, transistors, plastics, textiles, fertilizers, and many other things that we take for granted today have been made possible by detailed knowledge of chemical and physical properties. Such knowledge also permits greater understanding of how the natural world works and what changes (favorable or detrimental) may be brought about by human activities. Knowledge of chemical reactivity and properties may be approached on both the macroscopic and microscopic levels. Macroscopically this involves what is called . The person who first carries out a chemical reaction describes what happened, usually in terms of a balanced equation, and lists properties of any new substances. This enables other scientists to repeat the experiment if they wish. Even if the work is not carried out again, the descriptive report allows prediction of what would happen if it were repeated. The microscopic approach uses theory to predict which substances will react with which. During the past century Dalton’s atomic theory has been modified so that it can help us to remember the properties of elements and compounds. We now attribute structure to each kind of atom and expect atoms having similar structures to undergo similar reactions. Such work has led to the classification of groups of elements, for instance the , , , and many more. The additional complication of learning about atomic structure is repaid many fold by the increased ability of our microscopic model to predict macroscopic properties. In the following sections, you will see that a number of quite different kinds of experiments contributed to the extension of Dalton’s atomic theory to and atomic structure. The periodic variation of and the successful correlation of macroscopic properties indicate that atoms must have certain specific ways of connecting to other atoms. It is reasonable to assume that valence depends on some underlying atomic structure. Atoms which are similar in structure should exhibit the same valence and have similar chemical and physical properties. While the periodicity was initially based upon atomic weight, based upon weight implied some other property led to periodicity. The property on which periodicity is based is the of atoms. Our model for electronic structure is both scientifically and philosophically interesting, because it is based on a wave model for electrons. To learn more about our current electronic structure for atoms and the discoveries that led to this understanding, watch the video below. The discovery of implied that one kind of atom could change into another. This can he explained if atoms have structure. A change in that structure may produce a new kind of atom. Experiments with cathode-ray tubes indicated that , which are very light and carry a negative charge, are present in all atoms. Rutherford’s interpretation of the Geiger-Marsden experiment suggested that electrons occupy most of the volume of the atom while most of the mass is concentrated in a small positively charged . Moseley’s x-ray spectra and the existence of made it quite clear that Dalton’s emphasis on the importance of atomic weight would have to he dropped. The chemical behavior of an atom is determined by how many protons are in the nucleus. Changing the number of neutrons changes the atomic mass but has very little effect on chemistry. The identity of an element depends on its atomic number, not on its atomic weight. If the periodic law is restated as “When the elements are listed in order of increasing atomic , their properties vary periodically,” there are no exceptions. | 4,601 | 3,803 |
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In a , the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \label{14.4.5}\] If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s ). The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows: \[[A] = [A]_0e^{−kt} \label{14.4.6}\] where [A] is the initial concentration of reactant A at = 0; is the rate constant; and is the base of the natural logarithms, which has the value 2.718 to three decimal places. Recall that an integrated rate law gives the relationship between reactant concentration and time. predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of A and : \[\ln[A] = \ln[A]_0 − kt \label{14.4.7}\] Because has the form of the algebraic equation for a straight line, = + , with = \ln[A] and = \ln[A] , a plot of \ln[A] versus for a first-order reaction should give a straight line with a slope of − and an intercept of \ln[A] . Either the differential rate law ( ) or the integrated rate law ( ) can be used to determine whether a particular reaction is first order. First-order reactions are very common. We have already encountered two examples of first-order reactions: the hydrolysis of aspirin and the reaction of -butyl bromide with water to give -butanol. Another reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin. Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare, but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows: Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in \(\Page {1}\) is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in \(\Page {1}\) have been studied extensively to find ways of maximizing the concentration of the active species. If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order. The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in ble \(\Page {1}\). The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin. Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in \(\Page {1}\) shows that the reaction rate doubles [(1.8 × 10 M/min) ÷ (9.0 × 10 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10 M/min) ÷ (9.0 × 10 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = [cisplatin] . Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of Table \(\Page {1}\). For example, substituting the values for Experiment 3 into , 3.6 × 10 M/min = (0.024 M) 1.5 × 10 min = Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug. At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction: \[\mathrm{CH_3CH_2Cl(g)}\xrightarrow{\Delta}\mathrm{HCl(g)}+\mathrm{C_2H_4(g)} \nonumber\] Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction. balanced chemical equation, initial concentrations of reactant, and initial rates of reaction reaction order and rate constant Use measured concentrations and rate data from any of the experiments to find the rate constant. The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate. Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH CH Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH CH Cl]. This behavior is characteristic of a first-order reaction, for which the rate law is rate = [CH CH Cl]. We can calculate the rate constant ( ) using any row in the table. Selecting Experiment 1 gives the following: 1.60 × 10 M/s = (0.010 M) 1.6 × 10 s = Sulfuryl chloride (SO Cl ) decomposes to SO and Cl by the following reaction: SO Cl (g) → SO (g) + Cl (g) Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction. first order; = 2.2 × 10 s We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. Part (a) in shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C. The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in . The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in for = 100 min ([cisplatin] = 0.0086 M) and = 1000 min ([cisplatin] = 0.0022 M), The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min because the times plotted on the horizontal axes in parts (a) and (b) in are in minutes rather than seconds. The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions. The First-Order Integrated Rate Law Equation: If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M ( = 1.6 × 10 s ) ? initial concentration, rate constant, and time interval concentration at specified time and time required to obtain particular concentration The exponential form of the integrated rate law for a first-order reaction ( ) is [A] = [A] . Having been given the initial concentration of ethyl chloride ([A] ) and having the rate constant of = 1.6 × 10 s , we can use the rate law to calculate the concentration of the reactant at a given time . Substituting the known values into the integrated rate law, We could also have used the logarithmic form of the integrated rate law ( ): To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for . gives the following: In the exercise above, you found that the decomposition of sulfuryl chloride (SO Cl ) is first order, and you calculated the rate constant at 320°C. Use the form(s) of the integrated rate law to find the amount of SO Cl that remains after 20 h if a sample with an original concentration of 0.123 M is heated at 320°C. How long would it take for 90% of the SO Cl to decompose? 0.0252 M; 29 h Example Using the First-Order Integrated Rate Law Equation: | 9,489 | 3,804 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Named_Reactions/Hell-Volhard-Zelinsky_reaction |
An aldehyde or ketone in possession of an α hydrogen will be in equilibrium with its enol tautomer. This aspect of aldehydes and ketones allows electrophilic addition to occur at the α hydrogen. Carboxylic acids, however, don't generally form stable enols, so alpha addition is more difficult to achieve with carboxylic acids than aldehydes and ketones. The Hell Volhard Zelinsky reaction demonstrates a method for alpha addition with a carboxylic acid. The gist of the method is to convert the carboxylic acid into a derivative that does undergo tautomerization and then to carry out alpha addition upon that form. In the Hell Volhard Zelinsky reaction PBr3 is used to replace the carboxylic OH with a bromide, resulting in a carboxylic acid bromide. The acyl bromide can then tautomerize to an enol. This enol is then made to react with Br2 at the α position forming an α-bromo acyl bromide. Reaction of the α-bromo acyl bromide with the original carboxylic acid yields the α-bromo carboxylic acid product and regenerates the acyl bromide intermediate. | 1,072 | 3,805 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Fused_Benzene_Ring_Compounds |
Benzene rings may be joined together (fused) to give larger polycyclic aromatic compounds. A few examples are drawn below, together with the approved numbering scheme for substituted derivatives. The peripheral carbon atoms (numbered in all but the last three examples) are all bonded to hydrogen atoms. Unlike benzene, all the C-C bond lengths in these fused ring aromatics are not the same, and there is some localization of the pi-electrons. The six benzene rings in coronene are fused in a planar ring; whereas the six rings in hexahelicene are not joined in a larger ring, but assume a helical turn, due to the crowding together of the terminal ring atoms. This helical configuration renders the hexahelicene molecule chiral, and it has been resolved into stable enantiomers. As these extended aromatic compounds become larger, the ratio of hydrogen to carbon decreases. For example, the symmetrical hexacyclic compound coronene has a H/C ratio =1/2, compared with 1 for benzene. If we were to imagine fused ring systems of this kind to be further extended in space, the H/C ratio would approach zero, and the resulting compound would be a form of carbon. Such a carbon allotrope exists and is called graphite. Another well-characterized carbon allotrope is diamond. The structures for these two forms of carbon are very different, and are displayed below. Diamond is an extended array of sp hybridized carbon atoms; whereas, graphite consists of overlapping sheets of sp hybridized carbon atoms arranged in a hexagonal pattern. You may examine models of partial diamond and graphite structures by clicking on the appropriate structure below. A comparison of the coronene and corannulene models discloses an interesting difference in their shapes. Coronene is absolutely flat and, aside from the peripheral hydrogens, resembles a layer of graphite. Its very high melting point reflects this resemblance. Corannulene, on the other hand, is slightly curved, resulting in a bowl-like shape. If we extend the structure of corannulene by adding similar cycles of five benzene rings, the curvature of the resulting molecule should increase, and eventually close into a sphere of carbon atoms. The archetypical compound of this kind (C ) has been named buckminsterfullerene because of its resemblance to the geodesic structures created by Buckminster Fuller. It is a member of a family of similar carbon structures that are called fullerenes. These materials represent a third class of carbon allotropes. Alternating views of the C fullerene structure are shown on the right, together with a soccer ball-like representation of the 12 five and 20 six-membered rings composing its surface. Precise measurement by Atomic Force Microscopy (AFM) has shown that the C-C bond lengths of the six-membered rings are not all equal, and depend on whether the ring is fused to a five or six-membered beighbor. Interest in the fullerenes has led to the discovery of a related group of carbon structures referred to as nanotubes. As shown in the following illustration, nanotubes may be viewed as rolled up segments of graphite. The chief structural components are six-membered rings, but changes in tube diameter, branching into side tubes and the capping of tube ends is accomplished by fusion with five and seven-membered rings. Many interesting applications of these unusual structures have been proposed. A model of a nanotube will be displayed by clicking on the diagram ), | 3,481 | 3,806 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/05%3A_Gases/5.02%3A_Pressure-_The_Result_of_Particle_Collisions |
The earth’s atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes—for example, when your ears “pop” during take-off and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure \(\Page {1}\)). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container. Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant. Pressure is defined as the force exerted on a given area: \[P=\dfrac{F}{A} \label{9.2.1} \] Since pressure is directly proportional to force and inversely proportional to area (Equation \ref{9.2.1}), pressure can be increased either by either the amount of force or by the area over which it is applied. Correspondingly, pressure can be decreased by either the force or the area. Let’s apply the definition of pressure (Equation \ref{9.2.1}) to determine which would be more likely to fall through thin ice in Figure \(\Page {2}\).—the elephant or the figure skater? A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft (footprint area of 250 in ), so the pressure exerted by each foot is about 14 lb/in : \[\mathrm{pressure\: per\: elephant\: foot=14,000\dfrac{lb}{elephant}×\dfrac{1\: elephant}{4\: feet}×\dfrac{1\: foot}{250\:in^2}=14\:lb/in^2} \label{9.2.2} \] The figure skater weighs about 120 lbs, supported on two skate blades, each with an area of about 2 in , so the pressure exerted by each blade is about 30 lb/in : \[\mathrm{pressure\: per\: skate\: blade=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: blades}×\dfrac{1\: blade}{2\:in^2}=30\:lb/in^2} \label{9.2.3} \] Even though the elephant is more than one hundred times heavier than the skater, it exerts less than one-half of the pressure and would therefore be less likely to fall through thin ice. On the other hand, if the skater removes her skates and stands with bare feet (or regular footwear) on the ice, the larger area over which her weight is applied greatly reduces the pressure exerted: \[\mathrm{pressure\: per\: human\: foot=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: feet}×\dfrac{1\: foot}{30\:in^2}=2\:lb/in^2} \label{9.2.4} \] The SI unit of pressure is the , with 1 Pa = 1 N/m , where N is the newton, a unit of force defined as 1 kg m/s . One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal (1 kPa = 1000 Pa) or (1 bar = 100,000 Pa). In the United States, pressure is often measured in pounds of force on an area of one square inch— —for example, in car tires. Pressure can also be measured using the unit , which originally represented the average sea level air pressure at the approximate latitude of Paris (45°). Table \(\Page {1}\) provides some information on these and a few other common units for pressure measurements The United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in. Hg into: This is a unit conversion problem. The relationships between the various pressure units are given in Table 9.2.1. A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar? 0.974 atm; 740 mm Hg; 98.7 kPa; 0.987 bar We can measure atmospheric pressure, the force exerted by the atmosphere on the earth’s surface, with a (Figure \(\Page {3}\)). A barometer is a glass tube that is closed at one end, filled with a nonvolatile liquid such as mercury, and then inverted and immersed in a container of that liquid. The atmosphere exerts pressure on the liquid outside the tube, the column of liquid exerts pressure inside the tube, and the pressure at the liquid surface is the same inside and outside the tube. The height of the liquid in the tube is therefore proportional to the pressure exerted by the atmosphere. If the liquid is water, normal atmospheric pressure will support a column of water over 10 meters high, which is rather inconvenient for making (and reading) a barometer. Because mercury (Hg) is about 13.6-times denser than water, a mercury barometer only needs to be \(\dfrac{1}{13.6}\) as tall as a water barometer—a more suitable size. Standard atmospheric pressure of 1 atm at sea level (101,325 Pa) corresponds to a column of mercury that is about 760 mm (29.92 in.) high. The was originally intended to be a unit equal to one millimeter of mercury, but it no longer corresponds exactly. The pressure exerted by a fluid due to gravity is known as , : \[p=hρg \label{9.2.5} \] where Show the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury = \(13.6 \,g/cm^3\). The hydrostatic pressure is given by Equation \ref{9.2.5}, with \(h = 760 \,mm\), \(ρ = 13.6\, g/cm^3\), and \(g = 9.81 \,m/s^2\). Plugging these values into the Equation \ref{9.2.5} and doing the necessary unit conversions will give us the value we seek. (Note: We are expecting to find a pressure of ~101,325 Pa:) \[\mathrm{101,325\:\mathit{N}/m^2=101,325\:\dfrac{kg·m/s^2}{m^2}=101,325\:\dfrac{kg}{m·s^2}} \nonumber \] \[\begin {align*}
p&\mathrm{=\left(760\: mm×\dfrac{1\: m}{1000\: mm}\right)×\left(\dfrac{13.6\: g}{1\:cm^3}×\dfrac{1\: kg}{1000\: g}×\dfrac{( 100\: cm )^3}{( 1\: m )^3}\right)×\left(\dfrac{9.81\: m}{1\:s^2}\right)}\\[4pt]
&\mathrm{=(0.760\: m)(13,600\:kg/m^3)(9.81\:m/s^2)=1.01 \times 10^5\:kg/ms^2=1.01×10^5\mathit{N}/m^2} \\[4pt] & \mathrm{=1.01×10^5\:Pa} \end {align*} \nonumber \] Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/cm . 10.3 m A manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm, one arm that connects to the gas to be measured, and a nonvolatile liquid (usually mercury) in between. As with a barometer, the distance between the liquid levels in the two arms of the tube ( in the diagram) is proportional to the pressure of the gas in the container. An open-end manometer (Figure \(\Page {3}\)) is the same as a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and the atmosphere. The pressure of a sample of gas is measured at sea level with an open-end Hg (mercury) manometer, as shown below. Determine the pressure of the gas in: The pressure of the gas equals the hydrostatic pressure due to a column of mercury of height 13.7 cm plus the pressure of the atmosphere at sea level. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 13.7 cm of Hg plus atmospheric pressure.) The pressure of a sample of gas is measured at sea level with an open-end Hg manometer, as shown below Determine the pressure of the gas in: Blood pressure is measured using a device called a sphygmomanometer (Greek = “pulse”). It consists of an inflatable cuff to restrict blood flow, a manometer to measure the pressure, and a method of determining when blood flow begins and when it becomes impeded (Figure \(\Page {5}\)). Since its invention in 1881, it has been an essential medical device. There are many types of sphygmomanometers: manual ones that require a stethoscope and are used by medical professionals; mercury ones, used when the most accuracy is required; less accurate mechanical ones; and digital ones that can be used with little training but that have limitations. When using a sphygmomanometer, the cuff is placed around the upper arm and inflated until blood flow is completely blocked, then slowly released. As the heart beats, blood forced through the arteries causes a rise in pressure. This rise in pressure at which blood flow begins is the the peak pressure in the cardiac cycle. When the cuff’s pressure equals the arterial systolic pressure, blood flows past the cuff, creating audible sounds that can be heard using a stethoscope. This is followed by a decrease in pressure as the heart’s ventricles prepare for another beat. As cuff pressure continues to decrease, eventually sound is no longer heard; this is the the lowest pressure (resting phase) in the cardiac cycle. Blood pressure units from a sphygmomanometer are in terms of millimeters of mercury (mm Hg). Throughout the ages, people have observed clouds, winds, and precipitation, trying to discern patterns and make predictions: when it is best to plant and harvest; whether it is safe to set out on a sea voyage; and much more. We now face complex weather and atmosphere-related challenges that will have a major impact on our civilization and the ecosystem. Several different scientific disciplines use chemical principles to help us better understand weather, the atmosphere, and climate. These are meteorology, climatology, and atmospheric science. is the study of the atmosphere, atmospheric phenomena, and atmospheric effects on earth’s weather. Meteorologists seek to understand and predict the weather in the short term, which can save lives and benefit the economy. Weather forecasts (Figure \(\Page {5}\)) are the result of thousands of measurements of air pressure, temperature, and the like, which are compiled, modeled, and analyzed in weather centers worldwide. In terms of weather, low-pressure systems occur when the earth’s surface atmospheric pressure is lower than the surrounding environment: Moist air rises and condenses, producing clouds. Movement of moisture and air within various weather fronts instigates most weather events. The atmosphere is the gaseous layer that surrounds a planet. Earth’s atmosphere, which is roughly 100–125 km thick, consists of roughly 78.1% nitrogen and 21.0% oxygen, and can be subdivided further into the regions shown in Figure \(\Page {7}\): the exosphere (furthest from earth, > 700 km above sea level), the thermosphere (80–700 km), the mesosphere (50–80 km), the stratosphere (second lowest level of our atmosphere, 12–50 km above sea level), and the troposphere (up to 12 km above sea level, roughly 80% of the earth’s atmosphere by mass and the layer where most weather events originate). As you go higher in the troposphere, air density and temperature both decrease. Climatology is the study of the climate, averaged weather conditions over long time periods, using atmospheric data. However, climatologists study patterns and effects that occur over decades, centuries, and millennia, rather than shorter time frames of hours, days, and weeks like meteorologists. Atmospheric science is an even broader field, combining meteorology, climatology, and other scientific disciplines that study the atmosphere. Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers. | 12,164 | 3,807 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.01%3A_Prelude_to_Electrochemistry |
It is also possible to produce a flow of electricity as a result of a spontaneous chemical reaction. A chemical system which can cause a current to flow in this way is called a or a . An example of a galvanic cell with which you are almost certainly familiar is a flashlight battery. Since an electrical current is a flow of electrons or other charged particles, it should come as no surprise that both electrolytic and galvanic cells involve . In an electrolytic cell electric energy supplied from an outside source causes a nonspontaneous reaction to occur. A galvanic (or voltaic) cell, on the other hand. harnesses a spontaneous reaction to produce electric current. In either kind of cell the electrode at which oxidation occurs is called the anode and the electrode at which reduction occurs is the cathode. Electrolytic cells have numerous commercial applications. Chlorine, sodium hydroxide, hydrogen, aluminum, magnesium, sodium, calcium, and high-purity copper are some of the more important chemicals produced by electrolysis. of metals such as chromium, silver, nickel, zinc, and tin is also quite important. In any electrolysis reaction can be related to the electric charge which passes through the cell by means of the Faraday constant , which equals 9.649 × 10 C mol . A galvanic cell may be such as \[ \text{Zn}│\text{Zn}^{2+} (1\; M)║ \text{Ag}^{2+}(1\; M)│\text{Ag} \label{1} \] When a cell is written this way, \( \dfrac{1}{2} \)it is always assumed that the left-hand electrode is an anode and an oxidation half-equation occurs there. The right-hand electrode must then be taken as the cathode and a reduction half-equation is assumed to occur there. The cell reaction is the sum of these two half-equations. If it is spontaneous, our assumptions about anode on the left and cathode on the right were correct. Electrons will be forced into an external circuit on the left and the cell emf is taken to be positive. If the cell reaction written according to the above convention turns out to be nonspontaneous, then its reverse will be spontaneous. Our assumptions about which electrode is the anode and which the cathode must also be reversed, and the cell emf is given a negative sign. Equation \(\ref{1}\) is lots of fun. Because cell emf values indicate whether a process is spontaneous, they are quite useful. They are additive and are conventionally reported as standard electrode potentials. These refer to the emf of a cell with a hydrogen-gas electrode on the left and the electrode whose potential is reported on the right. The standard electrode potential is directly related to the standard free energy change for a reaction, thus allowing direct determination of Δ °. Many galvanic cells are of commercial importance. These include dry cells, mercury cells, rechargeable Ni-Cd batteries, and lead storage cells. Fuel cells, in which a continuous supply of both oxidizing and reducing agent is supplied, may eventually become important because of their high efficiencies. | 3,027 | 3,808 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.2%3A_Hybrid_Atomic_Orbitals |
Thinking in terms of overlapping atomic orbitals is one way for us to explain how chemical bonds form in diatomic molecules. However, to understand how molecules with more than two atoms form stable bonds, we require a more detailed model. As an example, let us consider the water molecule, in which we have one oxygen atom bonding to two hydrogen atoms. Oxygen has the electron configuration 1 2 2 , with two unpaired electrons (one in each of the two 2 orbitals). Valence bond theory would predict that the two O–H bonds form from the overlap of these two 2 orbitals with the 1 orbitals of the hydrogen atoms. If this were the case, the bond angle would be 90°, as shown in Figure \(\Page {1}\), because orbitals are perpendicular to each other. Experimental evidence shows that the bond angle is 104.5°, not 90°. The prediction of the valence bond theory model does not match the real-world observations of a water molecule; a different model is needed. Quantum-mechanical calculations suggest why the observed bond angles in H O differ from those predicted by the overlap of the 1 orbital of the hydrogen atoms with the 2 orbitals of the oxygen atom. The mathematical expression known as the wave function, , contains information about each orbital and the wavelike properties of electrons in an isolated atom. When atoms are bound together in a molecule, the wave functions combine to produce new mathematical descriptions that have different shapes. This process of combining the wave functions for atomic orbitals is called and is mathematically accomplished by the , , (a technique that we will encounter again later). The new orbitals that result are called . The valence orbitals in an oxygen atom are a 2 orbital and three 2 orbitals. The valence orbitals in an oxygen atom in a water molecule differ; they consist of four equivalent hybrid orbitals that point approximately toward the corners of a tetrahedron (Figure \(\Page {2}\)). Consequently, the overlap of the O and H orbitals should result in a tetrahedral bond angle (109.5°). The observed angle of 104.5° is experimental evidence for which quantum-mechanical calculations give a useful explanation: Valence bond theory must include a hybridization component to give accurate predictions. The following ideas are important in understanding hybridization: In the following sections, we shall discuss the common types of hybrid orbitals. The beryllium atom in a gaseous BeCl molecule is an example of a central atom with no lone pairs of electrons in a linear arrangement of three atoms. There are two regions of valence electron density in the BeCl molecule that correspond to the two covalent Be–Cl bonds. To accommodate these two electron domains, two of the Be atom’s four valence orbitals will mix to yield two hybrid orbitals. This hybridization process involves mixing of the valence orbital with one of the valence orbitals to yield two equivalent that are oriented in a linear geometry (Figure \(\Page {3}\)). In this figure, the set of orbitals appears similar in shape to the original orbital, but there is an important difference. The number of atomic orbitals combined always equals the number of hybrid orbitals formed. The orbital is one orbital that can hold up to two electrons. The set is two equivalent orbitals that point 180° from each other. The two electrons that were originally in the orbital are now distributed to the two orbitals, which are half filled. In gaseous BeCl , these half-filled hybrid orbitals will overlap with orbitals from the chlorine atoms to form two identical σ bonds. We illustrate the electronic differences in an isolated Be atom and in the bonded Be atom in the orbital energy-level diagram in Figure \(\Page {4}\). These diagrams represent each orbital by a horizontal line (indicating its energy) and each electron by an arrow. Energy increases toward the top of the diagram. We use one upward arrow to indicate one electron in an orbital and two arrows (up and down) to indicate two electrons of opposite spin. When atomic orbitals hybridize, the valence electrons occupy the newly created orbitals. The Be atom had two valence electrons, so each of the orbitals gets one of these electrons. Each of these electrons pairs up with the unpaired electron on a chlorine atom when a hybrid orbital and a chlorine orbital overlap during the formation of the Be–Cl bonds. Any central atom surrounded by just two regions of valence electron density in a molecule will exhibit hybridization. Other examples include the mercury atom in the linear HgCl molecule, the zinc atom in Zn(CH ) , which contains a linear C–Zn–C arrangement, and the carbon atoms in and CO . The valence orbitals of a central atom surrounded by three regions of electron density consist of a set of three and one unhybridized orbital. This arrangement results from hybridization, the mixing of one orbital and two orbitals to produce three identical hybrid orbitals oriented in a trigonal planar geometry (Figure \(\Page {5}\)). Although quantum mechanics yields the “plump” orbital lobes as depicted in Figure \(\Page {5}\), sometimes for clarity these orbitals are drawn thinner and without the minor lobes, as in Figure \(\Page {6}\), to avoid obscuring other features of a given illustration. We will use these “thinner” representations whenever the true view is too crowded to easily visualize. The observed structure of the borane molecule, BH suggests hybridization for boron in this compound. The molecule is trigonal planar, and the boron atom is involved in three bonds to hydrogen atoms ( Figure \(\Page {7}\)). We can illustrate the comparison of orbitals and electron distribution in an isolated boron atom and in the bonded atom in BH as shown in the orbital energy level diagram in Figure \(\Page {8}\). We redistribute the three valence electrons of the boron atom in the three hybrid orbitals, and each boron electron pairs with a hydrogen electron when B–H bonds form. Any central atom surrounded by three regions of electron density will exhibit hybridization. This includes molecules with a lone pair on the central atom, such as ClNO (Figure \(\Page {9}\)), or molecules with two single bonds and a double bond connected to the central atom, as in formaldehyde, CH O, and ethene, H CCH . The valence orbitals of an atom surrounded by a tetrahedral arrangement of bonding pairs and lone pairs consist of a set of four . The hybrids result from the mixing of one orbital and all three orbitals that produces four identical hybrid orbitals (Figure \(\Page {10}\)). Each of these hybrid orbitals points toward a different corner of a tetrahedron. A molecule of methane, CH , consists of a carbon atom surrounded by four hydrogen atoms at the corners of a tetrahedron. The carbon atom in methane exhibits hybridization. We illustrate the orbitals and electron distribution in an isolated carbon atom and in the bonded atom in CH in Figure \(\Page {11}\). The four valence electrons of the carbon atom are distributed equally in the hybrid orbitals, and each carbon electron pairs with a hydrogen electron when the C–H bonds form. In a methane molecule, the 1 orbital of each of the four hydrogen atoms overlaps with one of the four orbitals of the carbon atom to form a sigma (σ) bond. This results in the formation of four strong, equivalent covalent bonds between the carbon atom and each of the hydrogen atoms to produce the methane molecule, CH . The structure of ethane, C H is similar to that of methane in that each carbon in ethane has four neighboring atoms arranged at the corners of a tetrahedron—three hydrogen atoms and one carbon atom (Figure \(\Page {10}\)). However, in ethane an orbital of one carbon atom overlaps end to end with an orbital of a second carbon atom to form a σ bond between the two carbon atoms. Each of the remaining hybrid orbitals overlaps with an orbital of a hydrogen atom to form carbon–hydrogen σ bonds. The structure and overall outline of the bonding orbitals of ethane are shown in Figure \(\Page {12}\). The orientation of the two CH groups is not fixed relative to each other. Experimental evidence shows that rotation around σ bonds occurs easily. An hybrid orbital can also hold a lone pair of electrons. For example, the nitrogen atom in ammonia is surrounded by three bonding pairs and a lone pair of electrons directed to the four corners of a tetrahedron. The nitrogen atom is hybridized with one hybrid orbital occupied by the lone pair. The molecular structure of water is consistent with a tetrahedral arrangement of two lone pairs and two bonding pairs of electrons. Thus we say that the oxygen atom is hybridized, with two of the hybrid orbitals occupied by lone pairs and two by bonding pairs. Since lone pairs occupy more space than bonding pairs, structures that contain lone pairs have bond angles slightly distorted from the ideal. Perfect tetrahedra have angles of 109.5°, but the observed angles in ammonia (107.3°) and water (104.5°) are slightly smaller. Other examples of hybridization include CCl , PCl , and NCl . To describe the five bonding orbitals in a trigonal bipyramidal arrangement, we must use five of the valence shell atomic orbitals (the orbital, the three orbitals, and one of the orbitals), which gives five . With an octahedral arrangement of six hybrid orbitals, we must use six valence shell atomic orbitals (the orbital, the three orbitals, and two of the orbitals in its valence shell), which gives six . These hybridizations are only possible for atoms that have orbitals in their valence subshells (that is, not those in the first or second period). In a molecule of phosphorus pentachloride, PCl , there are five P–Cl bonds (thus five pairs of valence electrons around the phosphorus atom) directed toward the corners of a trigonal bipyramid. We use the 3 orbital, the three 3 orbitals, and one of the 3 orbitals to form the set of five hybrid orbitals (Figure \(\Page {13}\)) that are involved in the P–Cl bonds. Other atoms that exhibit hybridization include the sulfur atom in SF and the chlorine atoms in ClF and in \(\ce{ClF4+}\). (The electrons on fluorine atoms are omitted for clarity.) The sulfur atom in sulfur hexafluoride, SF , exhibits hybridization. A molecule of sulfur hexafluoride has six bonding pairs of electrons connecting six fluorine atoms to a single sulfur atom. There are no lone pairs of electrons on the central atom. To bond six fluorine atoms, the 3 orbital, the three 3 orbitals, and two of the 3 orbitals form six equivalent hybrid orbitals, each directed toward a different corner of an octahedron. Other atoms that exhibit hybridization include the phosphorus atom in \(\ce{PCl6-}\), the iodine atom in the interhalogens \(\ce{IF6+}\), IF , \(\ce{ICl4-}\), \(\ce{IF4-}\) and the xenon atom in XeF . The hybridization of an atom is determined based on the number of regions of electron density that surround it. The geometrical arrangements characteristic of the various sets of hybrid orbitals are shown in Figure \(\Page {16}\). These arrangements are identical to those of the electron-pair geometries predicted by VSEPR theory. VSEPR theory predicts the shapes of molecules, and hybrid orbital theory provides an explanation for how those shapes are formed. To find the hybridization of a central atom, we can use the following guidelines: It is important to remember that hybridization was devised to rationalize experimentally observed molecular geometries, not the other way around. The model works well for molecules containing small central atoms, in which the valence electron pairs are close together in space. However, for larger central atoms, the valence-shell electron pairs are farther from the nucleus, and there are fewer repulsions. Their compounds exhibit structures that are often not consistent with VSEPR theory, and hybridized orbitals are not necessary to explain the observed data. For example, we have discussed the H–O–H bond angle in H O, 104.5°, which is more consistent with hybrid orbitals (109.5°) on the central atom than with 2 orbitals (90°). Sulfur is in the same group as oxygen, and H S has a similar Lewis structure. However, it has a much smaller bond angle (92.1°), which indicates much less hybridization on sulfur than oxygen. Continuing down the group, tellurium is even larger than sulfur, and for H Te, the observed bond angle (90°) is consistent with overlap of the 5 orbitals, without invoking hybridization. We invoke hybridization where it is necessary to explain the observed structures. Ammonium sulfate is important as a fertilizer. What is the hybridization of the sulfur atom in the sulfate ion, \(\ce{SO4^2-}\)? The Lewis structure of sulfate shows there are four regions of electron density. What is the hybridization of the selenium atom in SeF ? The selenium atom is hybridized. Urea, NH C(O)NH , is sometimes used as a source of nitrogen in fertilizers. What is the hybridization of each nitrogen and carbon atom in urea? The Lewis structure of urea is The carbon atom is surrounded by three regions of electron density, positioned in a trigonal planar arrangement. The hybridization in a trigonal planar electron pair geometry is (Figure \(\Page {16}\)), which is the hybridization of the carbon atom in urea. Acetic acid, H CC(O)OH, is the molecule that gives vinegar its odor and sour taste. What is the hybridization of the two carbon atoms in acetic acid? H , ; (O)OH, We can use hybrid orbitals, which are mathematical combinations of some or all of the valence atomic orbitals, to describe the electron density around covalently bonded atoms. These hybrid orbitals either form sigma (σ) bonds directed toward other atoms of the molecule or contain lone pairs of electrons. We can determine the type of hybridization around a central atom from the geometry of the regions of electron density about it. Two such regions imply hybridization; three, hybridization; four, hybridization; five, hybridization; and six, hybridization. Pi (π) bonds are formed from unhybridized atomic orbitals ( or orbitals). | 14,274 | 3,809 |
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The three common phases (or states) of matter are gases, liquids, and solids. Gases have the lowest density of the three, are highly compressible, and completely fill any container in which they are placed. Gases behave this way because their intermolecular forces are relatively weak, so their molecules are constantly moving independently of the other molecules present. Solids, in contrast, are relatively dense, rigid, and incompressible because their intermolecular forces are so strong that the molecules are essentially locked in place. Liquids are relatively dense and incompressible, like solids, but they flow readily to adapt to the shape of their containers, like gases. We can therefore conclude that the sum of the intermolecular forces in liquids are between those of gases and solids. Figure \(\Page {1}\) compares the three states of matter and illustrates the differences at the molecular level. The state of a given substance depends strongly on conditions. For example, H O is commonly found in all three states: solid ice, liquid water, and water vapor (its gaseous form). Under most conditions, we encounter water as the liquid that is essential for life; we drink it, cook with it, and bathe in it. When the temperature is cold enough to transform the liquid to ice, we can ski or skate on it, pack it into a snowball or snow cone, and even build dwellings with it. Water vapor is a component of the air we breathe, and it is produced whenever we heat water for cooking food or making coffee or tea. Water vapor at temperatures greater than 100°C is called steam. Steam is used to drive large machinery, including turbines that generate electricity. The properties of the three states of water are summarized in Table \(\Page {1}\). The geometric structure and the physical and chemical properties of atoms, ions, and molecules usually do depend on their physical state; the individual water molecules in ice, liquid water, and steam, for example, are all identical. In contrast, the macroscopic properties of a substance depend strongly on its physical state, which is determined by intermolecular forces and conditions such as temperature and pressure. Figure \(\Page {2}\) shows the locations in the periodic table of those elements that are commonly found in the gaseous, liquid, and solid states. Except for hydrogen, the elements that occur naturally as gases are on the right side of the periodic table. Of these, all the noble gases (group 18) are monatomic gases, whereas the other gaseous elements are diatomic molecules ( ). Oxygen can also form a second allotrope, the highly reactive triatomic molecule ozone ( ), which is also a gas. In contrast, bromine (as Br ) and mercury (Hg) are liquids under normal conditions (25°C and 1.0 atm, commonly referred to as “room temperature and pressure”). Gallium (Ga), which melts at only 29.76°C, can be converted to a liquid simply by holding a container of it in your hand or keeping it in a non-air-conditioned room on a hot summer day. The rest of the elements are all solids under normal conditions. All of the gaseous elements (other than the monatomic noble gases) are molecules. W All gaseous substances are characterized by weak interactions between the constituent molecules or atoms. Defining Gas Pressure: Bulk matter can exist in three states: gas, liquid, and solid. Gases have the lowest density of the three, are highly compressible, and fill their containers completely. Elements that exist as gases at room temperature and pressure are clustered on the right side of the periodic table; they occur as either monatomic gases (the noble gases) or diatomic molecules (some halogens, N , O ). | 3,707 | 3,810 |
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The laws that describe the behavior of gases were well established long before anyone had developed a coherent model of the properties of gases. In this section, we introduce a theory that describes why gases behave the way they do. The theory we introduce can also be used to derive laws such as the ideal gas law from fundamental principles and the properties of individual particles. The kinetic molecular theory of gases explains the laws that describe the behavior of gases. Developed during the mid-19th century by several physicists, including the Austrian Ludwig Boltzmann (1844–1906), the German Rudolf Clausius (1822–1888), and the Englishman James Clerk Maxwell (1831–1879), who is also known for his contributions to electricity and magnetism, this theory is based on the properties of individual particles as defined for an ideal gas and the fundamental concepts of physics. Thus the kinetic molecular theory of gases provides a molecular explanation for observations that led to the development of the ideal gas law. The kinetic molecular theory of gases is based on the following five postulates: Although the molecules of real gases have nonzero volumes and exert both attractive and repulsive forces on one another, for the moment we will focus on how the kinetic molecular theory of gases relates to the properties of gases we have been discussing. In the following sections, we explain how this theory must be modified to account for the behavior of real gases. Postulates 1 and 4 state that gas molecules are in constant motion and collide frequently with the walls of their containers. The collision of molecules with their container walls results in a (impulse) from molecules to the walls (Figure \(\Page {2}\)). The to the wall perpendicular to \(x\) axis as a molecule with an initial velocity \(u_x\) in \(x\) direction hits is expressed as: \[\Delta p_x=2mu_x \label{10.7.1} \] The , a number of collisions of the molecules to the wall per unit area and per second, increases with the molecular speed and the number of molecules per unit volume. \[f\propto (u_x) \times \Big(\dfrac{N}{V}\Big) \label{10.7.2} \] The pressure the gas exerts on the wall is expressed as the product of impulse and the collision frequency. \[P\propto (2mu_x)\times(u_x)\times\Big(\dfrac{N}{V}\Big)\propto \Big(\dfrac{N}{V}\Big)mu_x^2 \label{10.7.3} \] At any instant, however, the molecules in a gas sample are traveling at different speed. Therefore, we must replace \(u_x^2\) in the expression above with the average value of \(u_x^2\), which is denoted by \(\overline{u_x^2}\). The overbar designates the average value over all molecules. The exact expression for pressure is given as : \[P=\dfrac{N}{V}m\overline{u_x^2} \label{10.7.4} \] Finally, we must consider that there is nothing special about \(x\) direction. We should expect that \[\overline{u_x^2}= \overline{u_y^2}=\overline{u_z^2}=\dfrac{1}{3}\overline{u^2}. \nonumber \] Here the quantity \(\overline{u^2}\) is called the defined as the average value of square-speed (\(u^2\)) over all molecules. Since \[u^2=u_x^2+u_y^2+u_z^2 \nonumber \] for each molecule, then \[\overline{u^2}=\overline{u_x^2}+\overline{u_y^2}+\overline{u_z^2}. \nonumber \] By substituting \(\dfrac{1}{3}\overline{u^2}\) for \(\overline{u_x^2}\) in the expression above, we can get the final expression for the pressure: \[P=\dfrac{1}{3}\dfrac{N}{V}m\overline{u^2} \label{10.7.5} \] Because volumes and intermolecular interactions are negligible, postulates 2 and 3 state that all gaseous particles behave identically, regardless of the chemical nature of their component molecules. This is the essence of the ideal gas law, which treats all gases as collections of particles that are identical in all respects except mass. Postulate 2 also explains why it is relatively easy to compress a gas; you simply decrease the distance between the gas molecules. Postulate 5 provides a molecular explanation for the temperature of a gas. Postulate 5 refers to the kinetic energy of the molecules of a gas \((\overline{e_K})\), which can be represented as \((T)\) \[\overline{e_K}=\dfrac{1}{2}m\overline{u^2}=\dfrac{3}{2}\dfrac{R}{N_A}T \label{10.7.6} \] where \(N_A\) is the Avogadro's constant. The total translational kinetic energy of 1 mole of molecules can be obtained by multiplying the equation by \(N_A\): \[N_A\overline{e_K}=\dfrac{1}{2}M\overline{u^2}=\dfrac{3}{2}RT \label{10.7.7} \] where \(M\) is the molar mass of the gas molecules and is related to the molecular mass by \(M=N_Am\). By rearranging the equation, we can get the relationship between the root-mean square speed (\(u_{\rm rms}\)) and the temperature. The rms speed (\(u_{\rm rms}\)) is the square root of the sum of the squared speeds divided by the number of particles: \[u_{\rm rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \label{10.7.8} \] where \(N\) is the number of particles and \(u_i\) is the speed of particle \(i\). The relationship between \(u_{\rm rms}\) and the temperature is given by: \[u_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \label{10.7.9} \] In Equation \(\ref{10.7.9}\), \(u_{\rm rms}\) has units of meters per second; consequently, the units of molar mass \(M\) are kilograms per mole, temperature \(T\) is expressed in kelvins, and the ideal gas constant \(R\) has the value 8.3145 J/(K•mol). Equation \(\ref{10.7.9}\) shows that \(u_{\rm rms}\) of a gas is proportional to the square root of its Kelvin temperature and inversely proportional to the square root of its molar mass. The root mean-square speed of a gas increase with increasing temperature. At a given temperature, heavier gas molecules have slower speeds than do lighter ones. The rms speed and the average speed do not differ greatly (typically by less than 10%). The distinction is important, however, because the rms speed is the speed of a gas particle that has average kinetic energy. Particles of different gases at the same temperature have the same average kinetic energy, not the same average speed. In contrast, the most probable speed (vp) is the speed at which the greatest number of particles is moving. If the average kinetic energy of the particles of a gas increases linearly with increasing temperature, then Equation \(\ref{10.7.8}\) tells us that the rms speed must also increase with temperature because the mass of the particles is constant. At higher temperatures, therefore, the molecules of a gas move more rapidly than at lower temperatures, and vp increases. At a given temperature, all gaseous particles have the same average kinetic energy but not the same average speed. The speeds of eight particles were found to be 1.0, 4.0, 4.0, 6.0, 6.0, 6.0, 8.0, and 10.0 m/s. Calculate their average speed (\(v_{\rm av}\)) root mean square speed (\(v_{\rm rms}\)), and most probable speed (\(v_{\rm m}\)). particle speeds average speed (\(v_{\rm av}\)), root mean square speed (\(v_{\rm rms}\)), and most probable speed (\(v_{\rm m}\)) Use Equation \(\ref{10.7.6}\) to calculate the average speed and Equation \(\ref{10.7.8}\) to calculate the rms speed. Find the most probable speed by determining the speed at which the greatest number of particles is moving. The average speed is the sum of the speeds divided by the number of particles: \[v_{\rm av}=\rm\dfrac{(1.0+4.0+4.0+6.0+6.0+6.0+8.0+10.0)\;m/s}{8}=5.6\;m/s \nonumber \] The rms speed is the square root of the sum of the squared speeds divided by the number of particles: \[v_{\rm rms}=\rm\sqrt{\dfrac{(1.0^2+4.0^2+4.0^2+6.0^2+6.0^2+6.0^2+8.0^2+10.0^2)\;m^2/s^2}{8}}=6.2\;m/s \nonumber \] The most probable speed is the speed at which the greatest number of particles is moving. Of the eight particles, three have speeds of 6.0 m/s, two have speeds of 4.0 m/s, and the other three particles have different speeds. Hence \(v_{\rm m}=6.0\) m/s. The \(v_{\rm rms}\) of the particles, which is related to the average kinetic energy, is greater than their average speed. At any given time, what fraction of the molecules in a particular sample has a given speed? Some of the molecules will be moving more slowly than average, and some will be moving faster than average, but how many in each situation? Answers to questions such as these can have a substantial effect on the amount of product formed during a chemical reaction. This problem was solved mathematically by Maxwell in 1866; he used statistical analysis to obtain an equation that describes the distribution of molecular speeds at a given temperature. Typical curves showing the distributions of speeds of molecules at several temperatures are displayed in Figure \(\Page {3}\). Increasing the temperature has two effects. First, the peak of the curve moves to the right because the most probable speed increases. Second, the curve becomes broader because of the increased spread of the speeds. Thus increased temperature increases the of the most probable speed but decreases the relative number of molecules that have that speed. Although the mathematics behind curves such as those in Figure \(\Page {3}\) were first worked out by Maxwell, the curves are almost universally referred to as Boltzmann distributions, after one of the other major figures responsible for the kinetic molecular theory of gases. We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously. The temperature of a 4.75 L container of N gas is increased from 0°C to 117°C. What is the qualitative effect of this change on the temperatures and volume effect of increase in temperature Use the relationships among pressure, volume, and temperature to predict the qualitative effect of an increase in the temperature of the gas. A sample of helium gas is confined in a cylinder with a gas-tight sliding piston. The initial volume is 1.34 L, and the temperature is 22°C. The piston is moved to allow the gas to expand to 2.12 L at constant temperature. What is the qualitative effect of this change on the no change no change no change no change decreases decreases decreases The behavior of ideal gases is explained by the . Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same . The actual values of speed and kinetic energy are not the same for all particles of a gas but are given by a , in which some molecules have higher or lower speeds (and kinetic energies) than average. | 10,663 | 3,811 |
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Kinetic methods of analysis use the rate of a chemical or s physical process to determine an analyte’s concentration. Three types of kinetic methods are discussed in this chapter: chemical kinetic methods, radiochemical methods, and flow injection methods. Chemical kinetic methods use the rate of a chemical reaction and either its integrated or its differential rate law. For an integral method, we determine the concentration of analyte—or the concentration of a reactant or product that is related stoichiometrically to the analyte—at one or more points in time following the reaction’s initiation. The initial concentration of analyte is then determined using the integrated form of the reaction’s rate law. Alternatively, we can measure the time required to effect a given change in concentration. In a differential kinetic method we measure the rate of the reaction at a time , and use the differential form of the rate law to determine the analyte’s concentration. Chemical kinetic methods are particularly useful for reactions that are too slow for other analytical methods. For reactions with fast kinetics, automation allows for sampling rates of more than 100 samples/h. Another important application of chemical kinetic methods is the quantitative analysis of enzymes and their substrates, and the characterization of enzyme catalysis. Radiochemical methods of analysis take advantage of the decay of radioactive isotopes. A direct measurement of the rate at which a radioactive isotope decays is used to determine its concentration. For an analyte that is not naturally radioactive, neutron activation can be used to induce radio- activity. Isotope dilution, in which we spike a radioactively-labeled form of analyte into the sample, is used as an internal standard for quantitative work. In flow injection analysis we inject the sample into a flowing carrier stream that usually merges with additional streams of reagents. As the sample moves with the carrier stream it both reacts with the contents of the carrier stream and with any additional reagent streams, and undergoes dispersion. The resulting fiagram of signal versus time bears some resemblance to a chromatogram. Unlike chromatography, however, flow injection analysis is not a separation technique. Because all components in a sample move with the carrier stream’s flow rate, it is possible to introduce a second sample before the first sample reaches the detector. As a result, flow injection analysis is ideally suited for the rapid throughput of samples. alpha particle competitive inhibitor equilibrium method gamma ray inhibitor intermediate rate kinetic method Michaelis constant noncompetitive inhibitor positron rate constant scintillation counter substrate uncompetitive inhibitor beta particle curve-fitting method fiagram Geiger counter initial rate isotope Lineweaver-Burk plot negatron one-point fixed-time integral method quench rate law steady-state approximation tracer variable time integral method centrifugal analyzer enzyme flow injection analysis half-life integrated rate law isotope dilution manifold
neutron activation peristaltic pump rate rate method stopped-flow analyzer two-point fixed-time integral method | 3,239 | 3,814 |
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. In addition to these individual basis; please contact 1. a. \(\underset{\text{acid}}{HSO^−_{4}\,(aq)} + \underset{\text{base}}{H_2O\,(l)} \rightleftharpoons \underset{\text{conjugate base}}{SO^{2−}_{4}\,(aq)} + \underset{\text{conjugate acid}}{H_3O^+\,(aq)}\) b. \(\underset{\text{base}}{C_3H_7NO_{2}\,(aq)} + \underset{\text{acid}}{H_3O^+\,(aq)} \rightleftharpoons \underset{\text{conjugate acid}}{C_3H_8NO^+_{2}\,(aq)} + \underset{\text{conjugate base}}{H_2O\,(l)}\) c. \(\underset{\text{acid}}{HOAc\,(aq)} + \underset{\text{base}}{NH_{3}\,(aq)} \rightleftharpoons \underset{\text{conjugate base}}{CH_3CO^−_{2}\,(aq)} + \underset{\text{conjugate acid}}{NH^+_{4}\,(aq)}\) d. \(\underset{\text{acid}}{SbF_{5}\,(aq)} + \underset{\text{base}}{2\,HF\,(aq)} \rightleftharpoons \underset{\text{conjugate acid}}{H_2F^+\,(aq)} + \underset{\text{conjugate base}}{SbF_6^−(aq)}\) 2. a. \(\underset{\text{acid}}{HF\,(aq)} + \underset{\text{base}}{H_2O\,(l)} \rightleftharpoons \underset{\text{conjugate acid}}{H_{3}O^{+}\,(aq)} + \underset{\text{conjugate base}}{F^{-}\,(aq)}\) b. \(\underset{\text{base}}{CH_{3}CH_{2}NH_{2}\,(aq)} + \underset{\text{acid}}{H_{2}O\,(l)} \rightleftharpoons \underset{\text{conjugate acid}}{CH_{3}CH_{2}NH_{3}^{+}\,(aq)} + \underset{\text{conjugate base}}{OH^{-}\,(aq)}\) c. \(\underset{\text{acid}}{C_{3}H_{7}NO_{2}\,(aq)} + \underset{\text{base}}{OH^{-}\,(aq)} \rightleftharpoons \underset{\text{conjugate base}}{C_{3}H_{6}NO_{2}^{-}\,(aq)} + \underset{\text{conjugate acid}}{H_{2}O\,(l)}\) d. \(\underset{\text{base}}{CH_{3}CO_{2}H\,(aq)} + \underset{\text{acid}}{2\,HF\,(aq)} \rightleftharpoons \underset{\text{conjugate acid}}{CH_{3}C(OH)_{2}^{+}\,(aq)} + \underset{\text{conjugate base}}{HF_{2}^{-}(aq)}\) 3. \(\underset{\text{base}}{NaH\,(s)} + \underset{\text{acid}}{H_{2}O\,(l)} \rightleftharpoons \underset{\text{conjugate acid}}{H_{2}\,(g)} + \underset{\text{conjugate base}}{NaOH\,(aq)}\) 4. a. \(K_a=\frac{[CO_{3}^{2-},H_{3}O^{+}]}{[HCO_{3}^{-}]}\) b. \(K_a=\frac{[formate,H_{3}O^{+}]}{[formic\,acid]}\) c. \(K_a=\frac{[H_{2}PO_{4}^{-},H_{3}O^{+}]}{[H_{3}PO_{4}]}\) 5. a. \(K_b=\frac{[CO_{3}^{2-},H_{3}O^{+}]}{[HCO_{3}^{-}]}\) b. \(K_b=\frac{[NH_{3},OH^{-}]}{[NH_{2}^{-}]}\) c. \(K_b=\frac{[HS^{-},OH^{-}]}{[S^{2-}]}\) 6. Strong acids have the smaller \(pK_a\). a. Equilibrium lies primarily to the right because \(HBr\) (\(pK_a=-8.7\)) is a stronger acid than \(H_{3}O^{+}\) (\(pK_a=-1.7\)) and \(H_{2}O\) (\(pK_a=14\)) is a stronger base than \(Br^-\) ( \(pK_a=-8.7\)). b. Equilibrium lies primarily to the left because \(H_{2}\) (\(pK_a=36\)) is a stronger acid than \(NH_{3}\) (\(pK_a=38\)) and (\(NaNH_2\)) (\(pK_a=38\)) is a stronger base than \(NaH\) (\(pK_a=35\)). c. Equilibrium lies primarily to the left because \(CH_{3}OH\) (\(pK_a=17\)) is a stronger acid than \(NH_{3}\) (\(pK_a=38\)) and \(NH_{2}^{-}\) (\(pK_a=38\)) is a stronger base than \(OCH_{3}^{-}\) (\(pK_a=25\)). d. Equilibrium lies to the right because \(HCl\) (\(pK_a=-7\)) is a stronger acid than \(NH_{4}^{+}\) (\(pK_a=9.3\)) and \(NH_{3}\) is a stronger base than \(Cl^{-}\) (\(pK_a=-7\)). 7. To identify the strongest base we can determine their weakest conjugate acid. The conjugate acids of \(CH_{3}^{-}\), \(NH_{2}^{-}\), and \(S_{2}^{-}\) are \(CH_{4}\), \(NH_{3}\), and \(HS^{-}\), respectively. Next, we consider that acidity increases with positive charge on the molecule, thus ruling out that \(S_{2}^{-}\) is the weakest base. Finally, we consider that acidity increases with electronegativity, therefore \(NH_{3}\) is the second most basic and \(CH_{4}\) is the most basic. To distinguish between the strength of the acids \(HIO_3\), \(H_{2}SO_{4}\), and \(HClO_4\) we can consider that the higher electronegativity and oxidation state of the central nonmetal is the more acidic, therefore the order of acidity is: \(HIO_3\)<\(H_{2}SO_{4}\)<\(HClO_4\) because electronegativity and oxidation state increases as follows: \(I(+5)<S(+6)<Cl(+7)\). 8. It is not accurate to say that a 2.0 M solution of \(H_2SO_4\), which contains two acidic protons per molecule, is 4.0 M in \(H^+\) because a 2.0 M solution of \(H_2SO_4\) is equivalent to 4.0 N in \(H^+\). \(\frac{2.0\,mol\,H_{2}SO_{4}}{1\,L} \times \frac{2\,eq\,H^{+}}{1\,mol\,H_{2}SO_{4}}=\frac{4\,eq\,H^{+}}{L}=4\,N\,H^{+}\) 9. Alkalinity is a measure of acid neutralizing capability. The basicity of the soil is defined this way because bases such as \(HCO_{3}^{-}\) and \(CO_{3}^{2-}\) can neutralize acids in soil. Because most soil has a pH between 6 and 8, alkalinity can be estimated by its carbonate species alone. At a near neutral pH, most carbonate species are bicarbonate. 10. Aqueous solutions of salts such as \(CaCl_{2}\) are neutral because it is created from hydrochloric acid (a strong acid) and calcium hydroxide (a strong base). An aqueous solution of \(NaNH_2\) is basic because it can deprotonate alkynes, alcohols, and a host of other functional groups with acidic protons such as esters and ketones. 11. a. \(Li_3N\) is a base because the lone pair on the nitrogen can accept a proton. b. \(NaH\) is a base because the hydrogen has a negative charge. c. \(KBr\) is neutral because it is formed from \(HBr\) (a strong acid) and \(KOH\) (a strong base). d. \(C_2H_5NH_3Cl\) is acidic because it can donate a proton. 12. a. The pH is expected to increase. \(\underset{\text{acid}}{LiCH_{3}\,(aq)} + \underset{\text{base}}{H_2O\,(l)} \rightleftharpoons \underset{\text{conjugate base}}{LiOH\,(aq)} + \underset{\text{conjugate acid}}{CH_{4}\,(aq)}\) b. The pH is expected to increase. \(\underset{\text{acid}}{MgCl_{2}\,(aq)} + \underset{\text{base}}{H_2O\,(l)} \rightleftharpoons \underset{\text{conjugate acid}}{2\,HCl\,(aq)} + \underset{\text{conjugate base}}{MgO\,(aq)}\) c. The pH is expected to remain the same. \(K_{2}O\,(aq)+H_2O\,(l) \rightleftharpoons 2\,KOH\,(aq)\) d. The pH is expected to increase. \(\underset{\text{acid}}{(CH_3)_2NH_2^+Br^−\,(aq)} + \underset{\text{base}}{H_2O\,(l)} \rightleftharpoons \underset{\text{conjugate acid}}{H_3O^{+}\,(aq)} + \underset{\text{conjugate base}}{(CH_{3})_{2}NH\,(aq)}\) 13. \(Sn(H_2O)_4^{2+}\) is expected to be more acidic than \(Pb(H_2O)_4^{2+}\) because \(Sn\) is more electronegative than \(Pb\). 14. \(Sn(H_2O)_6^{4+}\) is expected to be more acidic than \(Sn(H_2O)_4^{2+}\) because the charge on \(Sn\) is greater (\(4^+>2^+\)). 15. Yes, it is possible the order of increasing base strength is: \(LiH<NaH<RbH<CsH\) because increasing base strength is dependent on decreasing electronegativity. Given solutions with the same initial concentration of each acid, which would have the highest percent ionization? Given solutions with the same initial concentration of each base, which would have the highest percent ionization? 1. Acids in order of increasing strength: \(acid\,B<acid\,C<acid\,A\). Given the same initial concentration of each acid, the highest percent of ionization is acid A because it is the strongest acid. 2. Bases in order of increasing strength: \(base\,A<base\,C<base\,B\). Given the solutions with the same initial concentration of each base, the higher percent of ionization is base A because it is the weakest base. 3. a. \(pK_a+pK_b=14 \rightarrow pK_a=14-pK_b=14-3.80=10.2\) \(K_a=10^{-pK_a}=10^{-10.2}=6.31 \times 10^{-11}\) b. \(pK_a+pK_b=14 \rightarrow pK_a=14-pK_b=14-7.90=6.10\) \(K_a=10^{-pK_a}=10^{-6.10}=7.94 \times 10^{-7}\) c. \(pK_a+pK_b=14 \rightarrow pK_a=14-pK_b=14-7.90=3.000 \times 10^{-1}\) \(K_a=10^{-pK_a}=10^{-3.000 \times 10^{-1}}=-5.012 \times 10^{-1}\) d. \(pK_a+pK_b=14 \rightarrow pK_a=14-pK_b=14-1.40=12.6\) \(K_a=10^{-pK_a}=10^{-12.6}=2.51 \times 10^{-13}\) e. \(pK_a+pK_b=14 \rightarrow pK_a=14-pK_b=14-7.90=16.5\) \(K_a=10^{-pK_a}=10^{-16.5}=3.16 \times 10^{-17}\) 4. \(pK_a+pK_b=14 \rightarrow pK_b=14-pK_a=14-4.20=9.80\) \(K_b=10^{-pK_b}=10^{-9.80}=1.58 \times 10^{-10}\) 5. \(pK_a+pK_b=14 \rightarrow pK_a=14-pK_b=14-4.80=9.20\) \(K_a=10^{-pK_a}=10^{-9.20}=6.31 \times 10^{-10}\) \[HMnO_{4}\,(aq) \rightleftharpoons H^+\,(aq) + MnO^−_{4}\,(aq)\] \[MnO^−_{4}\,(aq)+H_2O\,(l) \rightarrow HMnO_{4}\,(aq) + OH^−\,(aq)\] 1. \[K_{auto} = \dfrac{[H_3O^+,OH^−]}{[H_2O]^2}\] \[K_w = [H_3O^+,OH^−] = K_{auto}[H_2O]^2\] 2. This will affect \(K_w\) as it is dependent on temperature. As the temperature increases, an endothermic process occurs (energy must be absorbed to break the bonds). Consequently, according to Le Chatelier, an increase in temperature favors the forward reaction thus the position of equilibrium shifts toward the right-hand side and \(K_w\) becomes larger. a. \(HNO_3\,(aq)+H_2O\,(l) \rightleftharpoons H_3O^{+}\,(aq)+ HNO_{2}^{-}\,(aq)\) b. \(KOH\,(s)+H_2O\,(l) \rightleftharpoons K^{-}\,(aq)+OH^{-}\,(aq)\) c. \(Ca(OH)_{2}\,(s)+H_2O\,(l) \rightleftharpoons Ca^{2+}\,(aq)+2\,OH^{-}\,(aq)\) d. \(H_2SO_4\, (aq)+H_2O\,(l) \rightleftharpoons HSO_4^{-}\,(aq)+H^{+}\,(aq)\) 5. \(H_{2}O\,(l) \rightleftharpoons H^{+}\,(aq)+OH^{-}\,(aq)\) \(K_w=[H^{+},OH^{-}]\) 1. \[K_{H_2SO_4}=[H_3SO_4^+,HSO_4^−]=K[H_2SO_4]_2\] \[[H_3SO_4^+] = 0.3\,M\] So the fraction ionized is 0.02. 2. The solution is basic because the \(pH=-log([H_3O^{+}])=-log(2.48 \times 10^{−8})=7.61>7\). 3. \(pH+pOH=14 \rightarrow pOH=14-pH=14-5.63=8.37\) \([OH^{-}]=10^{-pOH}=-4.27 \times 10^{-9}\) The \(pH=5.63<7\), therefore the solution is acidic. 4. a. The solution is acidic. \(pH=-log([H_3O^{+}])=-log(8.6 \times 10^{−3})=2.1<7\) b. The solution is basic. \(pH=-log([H_3O^{+}])=-log(3.7 \times 10^{−9})=8.4>7\) c. The solution is acidic. \(pH=-log([H_3O^{+}])=-log(2.1 \times 10^{−7})=6.7<7\) d. The solution is acidic. \(pH=-log([H_3O^{+}])=-log(1.4 \times 10^{−6})=5.9<7\) 5. a. \(pH=-log([H_3O^{+}])=-log(0.15)=0.82\) \(pH+pOH=14 \rightarrow pOH=14-pH=14-0.82=13\) b. \(pOH=-log([OH^{-}])=-log(0.03)=2\) \(pH+pOH=14 \rightarrow pH=14-pOH=14-2=10\) c. \(pH=-log([H_3O^{+}])=-log(2.3 \times 10^{−3})=2.6\) \(pH+pOH=14 \rightarrow pOH=14-pH=14-2.6=11\) d. \(pOH=-log([OH^{-}])=-log(9.78 \times 10^{−2})=1.01\) \(pH+pOH=14 \rightarrow pH=14-pOH=14-1.01=13.0\) e. \(pH=-log([H_3O^{+}])=-log(0.00017)=3.8\) \(pH+pOH=14 \rightarrow pOH=14-pH=14-3.8=10\) f. \(pH=-log([H_3O^{+}])=-log(5.78)=-0.762\) \(pH+pOH=14 \rightarrow pOH=14-pH=14-(-0.762)=14.8\) 6. a. \(25.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{2.3 \times 10^{-2}\,mol}{1\,L} \times \frac{1}{100\,mL \times \frac{1\,L}{1,000\,mL}}=0.060\,M\,HCl\) \(pH=-log([H_3O^{+}])=-log(0.060)=1.22\) \(pH+pOH=14 \rightarrow pOH=14-pH=14-1.22=12.78\) b. \(5.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{1.87\,mol}{1\,L} \times \frac{1}{125\,mL \times \frac{1\,L}{1,000\,mL}}=7.5 \times 10^{-2}\,M\,NaOH\) \(pOH=-log([OH^{-}])=-log(7.5 \times 10^{-2})=1.1\) \(pH+pOH=14 \rightarrow pH=14-pOH=14-1.1=12.9\) c. \(5.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{5.98\,mol}{1\,L} \times \frac{1}{100\,mL \times \frac{1\,L}{1,000\,mL}}=0.20\,M\,HCl\) \(pH=-log([H_3O^{+}])=-log(0.20)=0.70\) \(pH+pOH=14 \rightarrow pOH=14-pH=14-0.70=13.3\) d. \(25.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{3.7\,mol}{1\,L} \times \frac{1}{250\,mL \times \frac{1\,L}{1,000\,mL}}=0.370\,M\,HNO_3\) \(pH=-log([H_3O^{+}])=-log(0.370)=0.432\) \(pH+pOH=14 \rightarrow pOH=14-pH=14-0.432=13.568\) e. \(35.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.046\,mol}{1\,L} \times \frac{1}{500\,mL \times \frac{1\,L}{1,000\,mL}}=3 \times 10^{-3}\,M\,HI\) \(pH=-log([H_3O^{+}])=-log(3 \times 10^{-3})=2.52\) \(pH+pOH=14 \rightarrow pOH=14-pH=14-2.52=11.48\) f. \(15.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.0087\,mol}{1\,L} \times \frac{1}{125\,mL \times \frac{1\,L}{1,000\,mL}}=5.20 \times 10^{-4}\,M\,KOH\) \(pOH=-log([OH^{-}])=-log(5.20 \times 10^{-4})=3.28\) \(pH+pOH=14 \rightarrow pH=14-pOH=14-3.28=10.72\) 7. \([H^+]=10^{-pH}=10^{-1.5}=3.2 \times 10^{-2}\,M\) 8. a. \([H^{+}]=10^{-11.4}=3.98 \times 10^{-12}\,M\) b. \([H^{+}]=10^{-6.5}=3.2 \times 10^{-7}\,M\) c. \([H^{+}]=10^{-3.5}=3.2 \times 10^{-4}\,M\) d. \([H^{+}]=10^{-8.5}=3.2 \times 10^{-9}\,M\) e. \([H^{+}]=10^{-4.2}=6.3 \times 10^{-5}\,M\) 9. 2.9 mg \(HCl\) \([H^{+}]=10^{-pH}=10^{-3.50}=3.1622 \times 10^{-4}\,M\) \(x\,mg\,HCl \times \frac{1\,g\,HCl}{1,000\,mg\,HCl} \times \frac{1\,mol\,HCl}{36.46\,g\,HCl} \times \frac{1}{250\,mL\,H_2O \times \frac{1\,L}{1,000\,mL\,H_2O}}=3.1622 \times 10^{-4}\,M\,HCl \rightarrow \frac{x\,mol\,HCl}{9115\,L\,HCl}=3.1622 \times 10^{-4}\,M\,HCl \rightarrow x\,mol\,HCl=3.1622\times 10^{-4}\,M \times 9115\,L\,HCl \rightarrow x\,mol\,HCl=2.9\,mol\,HCl \rightarrow x=2.9\) 10. To prepare the stock solution, \(2.11 \times 10^{-2}\,L\) of \(0.500\,M\,NaOH\) solution is required. \(\frac{0.03162\,mol\,NaOH}{1\,L\,NaOH} \times \frac{1\,L\,NaOH}{0.5\,mol\,NaOH} \times 0.333\,L=2.11 \times 10^{-2}\,g\,NaOH\). \([OH^{-}]=10^{-1.5}=0.03162\,M\) \(pH+pOH=14 \rightarrow pOH=14-12.50=1.5\) 1. a. The most important factor in determining the stronger acid is considering the inductive effect. Chlorine is an electron-withdrawing group. It pulls electron density away from the compound by means of the inductive effect through the sigma bond. In considering the conjugate base of \(CH_3CCl_2CH_2CO_2H\), Chlorine absorbs some of the electron density or excess negative charge on the oxygen atom. This causes the C bonded to the attached Chlorine atoms to be partially positive. The conjugate base of \(CH_3CCl_2CH_2CO_2H\) is more stable, thus more acidic than the conjugate base of \(CH_3CH_2CH_2CO_2H\). b. The most important factor in determining the stronger acid is knowing the \(pK_a\) values for functional groups. The \(pK_a\) of alcohol is about 16 while the \(pK_a\) of a carboxylic acid is about 5. Therefore, \(CH_3CO_2H\) is more acidic than \(CH_3CH_2OH\). c. The most important factor in determining the stronger acid is electronegativity. The chlorine atom is more electronegative than the bromine atom, therefore \(HClO\) is more acidic than \(HBrO\). d. The most important factor in determining the stronger acid is considering resonance. The \(\ce{CH_3C(=O)NH_2}\) has a resonance which increases the stability of the conjugate base (therefore increasing acidity) because the negative charge can be delocalized. Thus, \(\ce{CH_3C(=O)NH_2}\) is more acidic than \(CH_3CH_2NH_2\). e. The most important factor in determining the stronger acid is considering oxidation states on the central nonmetal. \(H_3AsO_4\) has an oxidation state of +5 which is larger and thus more acidic than \(H_3AsO_3\) which has an oxidation state of +3. \(CF_3S^− < CH_3S^− < OH^−\) (strongest base) \(NH_3\); \(Cl\) atoms withdraw electron density from \(N\) in \(Cl_2NH\). 2. It is expected that the stronger acid is \(C_6H_5NH_3^+\) because in considering the conjugate base the lone pair of electrons on nitrogen is involved in resonance, hence the molecule is stable. 3. \(H_2Se\) is a weaker acid than HBr because \(Br\) is more electronegative than \(Se\) thus more stable. 4. \(HClO_3>CH_3PO_3H_2>H_3PO_4\) This is because \(H_3PO_4\) is a polyprotic acid which contains more than one ionizable proton, and the protons are lost in a stepwise manner. The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion. Thus, acid strength decreases with the loss of subsequent protons, and, correspondingly, the \(pK_a\) increases which indicate it is the most basic. The conjugate base of \(ClO^{3-}\) has a much smaller charge to volume ratio, thus most stable and acidic. 5. \(CF_{3}S^{-}<CH_{3}S^{-}<OH^{-}\) \(CF_{3}S^{-}\) is the most acidic because of the three electronegative Fluorines. \(OH^{-}\) is the strongest base in water. Thus, \(CH_{3}S^{-}\) is in between these aqueous solutions. 6. \(HNO_2<HClO_2<HNO_3\) \(HNO_3\) has resonance stabilization, therefore it is the most acidic. Between \(HClO_2\) and \(HNO_2\), \(Cl\) is the most electronegative therefore \(HClO_2\) is more acidic than \(HNO_2\). 7. I expect \(H_2SO_3\) to be the stronger acid because it is more electronegative or has greater attraction which means it’ll be less inclined to share their electrons with a proton. 8. \(CF_3OH\) is a stronger acid than \(CH_3OH\) in aqueous solution because \(F\) is more electronegative than \(H\). 9. It would be expected that \(NH_3\) be a stronger base than \(Cl_2NH\) because it is electronegative due to the two \(Cl\) atoms. 2. Explain why \(SiF_4\) can act as a Lewis Acid. 4. What is the product of the reaction of \(CO_2 + OH^- \rightarrow \) ? 5. In the reactions below, which is the Lewis Acid and/or which is the Lewis Base? \([PtCl_6]^{2-}\) w of \(AgCl\) + \(NH_3\) pr 1. a. \(NH_3\) is a lewis base because nitrogen has a lone pair of electrons to "donate." b. \(Ag^{+}\) is a lewis acid because it has an unfilled octet and thus is able to accept a pair of electrons. c. \(Ni^{+}\) is a lewis acid because it has an unfilled octet and thus is able to accept a pair of electrons. d. \(Pt^{4+}\) is a lewis acid because it has an unfilled octet and thus is able to accept a pair of electrons. e. \(H_2O\) is a lewis base because oxygen has two lone pairs of electrons to "donate." f. \(SO_2\) is a lewis acid because sulfur has an unfilled octet and thus is able to accept a pair of electrons. 2. SiF 3. 4. This reaction forms a bicarbonate ion. \(CO_2 + OH^- \rightarrow O--COH=O\). 5. a. Lewis Acid: \(H^+\), Lewis Base: \(NH_3\) b. Lewis Acid: \(H^+\), Lewis Base: \(H_2O\) 6. \(Pt^{4+}\) is the Lewis acid and \(Cl^-\) is the Lewis base. 7. \[AgCl + 2\,NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-\] Construct a table comparing how OH , NH , H O, and BCl are classified according to the Arrhenius, the Brønsted–Lowry, and the Lewis definitions of acids and bases Describe how the proton \((H^{+})\) can simultaneously behave as an Arrhenius acid, a Brønsted–Lowry acid, and a Lewis acid. Would you expect aluminum to form compounds with covalent bonds or coordinate covalent bonds? Explain your answer. Classify each compound as a Lewis acid or a Lewis base and justify your choice. a. \(AlCl_{3}\) b. \(CH_{3}N\) c. \(IO_{3}^{-}\) Explain how a carboxylate ion \((RCO_{2}^{-})\) can act as both a Brønsted–Lowry base and a Lewis base. 1. Arrhenius Acid Arrhenius Base An Arrhenius acid is a molecule that when dissolved in water it will donate an \(H^{+}\) in solution. An Arrhenius base is a molecule that when dissolved in water it will donate an \(OH^{-}\) in solution. A Brønsted–Lowry acid is a molecule that when dissolved in a solution it will donate an \(H^{+}\) in solution. A Brønsted–Lowry base is a molecule that when dissolved in a solution it will donate an atom or ion capable of accepting or bonding to a free proton in solution. A Lewis acid is an atom or molecule that accepts an electron pair. A Lewis base is an atom or molecule that donates an electron pair. 2. The proton \((H^{+})\) can simultaneously behave as an Arrhenius acid because when it is dissolved in water it will donate itself. \(H^{+}+H_{2}O \rightleftharpoons H_{3}O^{+}\) The proton \((H^{+})\) can simultaneously behave as a Brønsted–Lowry acid because when it is dissolved in solution it will donate itself. \(H^{+}+B^{-} \rightleftharpoons HB\) The proton \((H^{+})\) can simultaneously behave as a Lewis acid as it can accept an electron pair. \(H^{+}+B^{-} \rightleftharpoons HB\) 3. It is expected that Aluminum forms a coordinate covalent bond as it can participate in a Lewis acid and a Lewis base interaction. For example \(Al^{3+}+H_{2}O \rightleftharpoons [Al(OH_2)_{6}]^{3+}\) 4. a. \(AlCl_{3}\) is a Lewis acid as \(Al\) can accept an electron pair. b. \(CH_{3}N\) is a Lewis base as \(N\) can donate an electron pair. c. \(IO_{3}^{-}\) is a Lewis base as \(I\) can donate an electron pair. 5. The carboxylate ion \((RCO_{2}^{-})\) can act as Brønsted–Lowry base because when dissolved in a solution the electron rich \(O\) is capable of accepting a proton. The carboxylate ion \((RCO_{2}^{-})\) can act as a Lewis base because the electron rich \(O\) can donate an electron pair. 1. In each reaction, identify the Lewis acid and the Lewis base and complete the reaction by writing the products(s). a. (CH ) O + AlCl b. SnCl + 2 Cl 2. Use Lewis dot symbols to depict the reaction of BCl with dimethyl ether [(CH ) O]. How is this reaction similar to that in which a proton is added to ammonia? 1. a. The Lewis acid is \(AlCl_{3}\) and the Lewis base is \((CH_{3})_{2}O\). \((CH_{3})_{2}O+AlCl_{3} \rightleftharpoons AlCl_{3} \cdot O(CH_{3})_{2}\) b. The Lewis acid is \(SnCl_{4}\) and the Lewis base is \(Cl^{-}\). \(SnCl_{4}+2\,Cl^{-} \rightleftharpoons SnCl_{6}^{2-}\) 2. \(BCl_{3}+(CH_{3})_{2}O \rightleftharpoons BCl_{3}\cdot O(CH_{3})_{2}\) This reaction is similar to that in which a proton is added to ammonia as it also involves a Lewis acid and a Lewis base interaction. | 20,944 | 3,815 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Voltaic_Cells/The_Cell_Potential |
The batteries in your remote and the engine in your car are only a couple of examples of how chemical reactions create power through the flow of electrons. The cell potential is the way in which we can measure how much voltage exists between the two half cells of a battery. We will explain how this is done and what components allow us to find the voltage that exists in an electrochemical cell. The cell potential, \(E_{cell}\), is the measure of the potential difference between two half cells in an electrochemical cell. The potential difference is caused by the ability of electrons to flow from one half cell to the other. Electrons are able to move between electrodes because the chemical reaction is a redox reaction. A redox reaction occurs when a certain substance is oxidized, while another is reduced. During oxidation, the substance loses one or more electrons, and thus becomes positively charged. Conversely, during reduction, the substance gains electrons and becomes negatively charged. This relates to the measurement of the cell potential because the difference between the potential for the reducing agent to become oxidized and the oxidizing agent to become reduced will determine the cell potential. The cell potential (E ) is measured in voltage (V), which allows us to give a certain value to the cell potential. An electrochemical cell is comprised of two half cells. In one half cell, the oxidation of a metal electrode occurs, and in the other half cell, the reduction of metal ions in solution occurs. The half cell essentially consists of a metal electrode of a certain metal submerged in an aqueous solution of the same metal ions. The electrode is connected to the other half cell, which contains an electrode of some metal submerged in an aqueous solution of subsequent metal ions. The first half cell, in this case, will be marked as the anode. In this half cell, the metal in atoms in the electrode become oxidized and join the other metal ions in the aqueous solution. An example of this would be a copper electrode, in which the Cu atoms in the electrode loses two electrons and becomes Cu . The Cu ions would then join the aqueous solution that already has a certain molarity of Cu ions. The electrons lost by the Cu atoms in the electrode are then transferred to the second half cell, which will be the cathode. In this example, we will assume that the second half cell consists of a silver electrode in an aqueous solution of silver ions. As the electrons are passed to the Ag electrode, the Ag ions in solution will become reduced and become an Ag atom on the Ag electrode. In order to balance the charge on both sides of the cell, the half cells are connected by a salt bridge. As the anode half cell becomes overwhelmed with Cu ions, the negative anion of the salt will enter the solution and stabilized the charge. Similarly, in the cathode half cell, as the solution becomes more negatively charged, cations from the salt bridge will stabilize the charge. For electrons to be transferred from the anode to the cathode, there must be some sort of energy potential that makes this phenomenon favorable. The potential energy that drives the redox reactions involved in electrochemical cells is the potential for the anode to become oxidized and the potential for the cathode to become reduced. The electrons involved in these cells will from the anode, which has a higher potential to become oxidized to the cathode, which has a lower potential to become oxidized. This is analogous to a rock falling from a cliff in which the rock will fall from a higher potential energy to a lower potential energy. The difference between the anode's potential to become reduced and the cathode's potential to become reduced is the cell potential. \[E^o_{Cell}= E^o_{Red,Cathode} - E^o_{Red,Anode}\] Note: Here is the list of the all the components: All of these components create the Electrochemical Cell. The image above is an electrochemical cell. The voltmeter at the very top in the gold color is what measures the cell voltage, or the amount of energy being produced by the electrodes. This reading from the voltmeter is called the voltage of the electrochemical cell. This can also be called the potential difference between the half cells, E . Volts are the amount of energy for each electrical charge; 1V=1J/C: V= voltage, J=joules, C=coulomb. The voltage is basically what propels the electrons to move. If there is a high voltage, that means there is high movement of electrons. The voltmeter reads the transfer of electrons from the anode to the cathode in Joules per Coulomb. The image above is called the cell diagram. The cell diagram is a representation of the overall reaction in the electrochemical cell. The chemicals involved are what are actually reacting during the reduction and oxidation reactions. (The spectator ions are left out). In the cell diagram, the anode half cell is always written on the left side of the diagram, and in the cathode half cell is always written on the right side of the diagram. Both the anode and cathode are seperated by two vertical lines (ll) as seen in the blue cloud above. The electrodes (yellow circles) of both the anode and cathode solutions are seperated by a single vertical line (l). When there are more chemicals involved in the aqueous solution, they are added to the diagram by adding a comma and then the chemical. For example, in the image above, if copper wasn't being oxidized alone, and another chemical like K was involved, you would denote it as (Cu, K) in the diagram. The cell diagram makes it easier to see what is being oxidized and what is being reduced. These are the reactions that create the cell potential. The standard cell potential (\(E^o_{cell}\)) is the difference of the two electrodes, which forms the voltage of that cell. To find the difference of the two half cells, the following equation is used: \[E^o_{Cell}= E^o_{Red,Cathode} - E^o_{Red,Anode} \tag{1a}\] with The units of the potentials are typically measured in volts (V). Note that this equation can also be written as a sum rather than a difference \[E^o_{Cell}= E^o_{Red,Cathode} + E^o_{Ox,Anode} \tag{1b}\] where we have switched our strategy from taking the difference between two reduction potentials (which are traditionally what one finds in reference tables) to taking the sum of the oxidation potential and the reduction potential (which are the reactions that actually occur). Since E^o_{Red}=-E^o_{Ox}, the two approaches are equivalent. The example will be using the picture of the Copper and Silver cell diagram. The oxidation half cell of the redox equation is: Cu(s) → Cu (aq) + 2e E = -0.340 V where we have negated the reduction potential E = 0.340 V, which is the quantity we found from a list of standard reduction potentials, to find the oxidation potential E . The reduction half cell is: ( Ag + e → Ag(s) ) x2 E = 0.800 V where we have multiplied the reduction chemical equation by two in order to balance the electron count . Voltage is energy per charge, not energy per reaction, so it does not need to account for the number of reactions required to produce or consume the quantity of charge you are using to balance the equation. The chemical equations can be summed to find: Cu(s) + 2Ag + 2e → Cu (aq) + 2Ag(s) + 2e and simplified to find the overall reaction: Cu(s) + 2Ag → Cu (aq) + 2Ag(s) where the potentials of the half-cell reactions can be summed E = E +E E = 0.800 V + ( 0.340 V) E = 0.460V to find that the standard cell potential of this cell is 0.460 V. Note that since E^o_{Red}=-E^o_{Ox} we could have accomplished the same thing by taking the difference of the reduction potentials, where the absent or doubled negation accounts for the fact that the of the reduction reaction is what actually occurs. E = E -E E = 0.800V - 0.340V E = 0.460V The table below is a list of important standard electrode potentials in the reduction state. To determine oxidation electrodes, the reduction equation can simply be flipped and its potential changed from positive to negative (and vice versa). When using the half cells below, instead of changing the potential the equation below can be used without changing any of the potentials from positive to negative (and vice versa): E = E - E E = 2.71V= +0.401V - E {Al(OH) ] (aq)/Al(s)} E {[Al(OH) ] (aq)/Al(s)} = 0.401V - 2.71V = -2.31V Confirm this on the table of standard reduction potentials | 8,500 | 3,817 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Metabolism/Catabolism/Electron_Transport_Chain_II |
The majority of the energy conserved during catabolism reactions occurs near the end of the metabolic series of reactions in the electron transport chain. The gets its name from the fact electrons are transported to meet up with oxygen from respiration at the end of the chain. The overall electron chain transport reaction is: 2 H + 2 e + 1/2 O → H O + energy Notice that 2 hydrogen ions, 2 electrons, and an oxygen molecule react to form as a product water with energy released is an exothermic reaction. This relatively straight forward reaction actually requires or more steps. The energy released is coupled with the formation of three ATP molecules per every use of the electron transport chain. The electron transport chain is initiated by the reaction of an organic (intermediate in metabolic reactions) with the coenzyme NAD (nicotinamide adenine dinucleotide). This is an oxidation reaction where 2 hydrogen atoms (or 2 hydrogen ions and 2 electrons) are removed from the organic metabolite. (The organic metabolites are usually from the citric acid cycle and the oxidation of fatty acids--details in following pages.) The reaction can be represented simply where M = any metabolite. MH + NAD → NADH + H + M: + energy One hydrogen is removed with 2 electrons as a hydride ion (H ) while the other is removed as the positive ion (H ). Usually the metabolite is some type of alcohol which is oxidized to a ketone. is a coenzyme containing the B-vitamin, nicotinamide, shown on a previous page. The purpose of the other seven steps in the electron transport chain is threefold: Once the NADH has been made from a metabolite in the citric acid cycle inside of the mitochondria, it interacts with the first complex 1 enzyme, known as NADH reductase. This complex 1 contains a coenzyme flavin mononucleotide (FMN) which is similar to FAD. The sequence of events is that the NADH, plus another hydrogen ion enter the enzyme complex and pass along the 2 hydrogen ions, ultimately to an interspace in the mitochondria. These hydrogen ions, acting as a pump, are utilized by ATP synthetase to produce an ATP for every two hydrogen ions produced. Three complexes (1, 3, 4) act in this manner to produce 2 hydrogen ions each, and thus will produce 3 ATP for every use of the complete electron transport chain. In addition, NADH passes along 2 electrons to first FMN, then to an iron-sulfur protein (FeS), and finally to coenzyme Q. The net effect of these reactions are to regenerate coenzyme NAD . This regeneration of reactants occurs in many of the reactions so that a cycling effect occurs. The NAD is ready to react further with metabolites in the citric acid cycle. Coenzyme Q, which also picks up an additional 2 hydrogen ions to make CoQH , is soluble in the lipid membrane and can move through the membrane to come into contact with enzyme complex 3. In summary, the very first enzyme complex in the electron transport chain is coupled with the formation of ATP. The coupled reaction may be written as: MH + NAD → NADH + H + M + energy ADP + P + energy → ATP + H O Coenzyme QH carrying an extra 2 electrons and 2 hydrogen ions now starts a cascade of events through enzyme complex 3, also known as cytochrome reductase bc. Cytochromes are very similar to the structure of myoglobin or hemoglobin. The significant feature is the heme structure containing the iron ions, initially in the +3 state and changed to the +2 state by the addition of an electron. The CoQH (yellow)passes along the 2 electrons first to cytochrome (blue) b1 heme (magenta), then b2 heme , then to an iron-sulfur protein (green), then to cytochrome c1 (red with black heme), and finally to cytochrome c (not shown). Co Q is represented by the inhibitor antimycin (yellow) in the graphic. In the meantime the 2 hydrogen ions are channeled to the interspace of the mitochondria for ultimate conversion into ATP. Refer to the middle graphic: Cytochrome c is a small molecule which is also able to move in the lipid membrane layer and diffuses toward cytochrome a complex 4. At this time it continues the transport of the electrons, and provides the third and final time that 2 hydrogen ions are channeled to the interspace of the mitochondria for ultimate conversion into ATP. ATP synthetase is also found at numerous locations in the bilayer membrane of the mitochondria. by the pumping action of the re-entry of the hydrogen ions through the ATP synthetase. Finally, oxygen has diffused into the cell and the mitochondria for the finally reaction of metabolism. Oxygen atom reacts with the 2 electrons and 2 hydrogens to . | 4,644 | 3,819 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Acid_Rain/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 3,820 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Polymerization_of_Alkenes |
This page looks at the polymerisation of alkenes to produce polymers like poly(ethene) (usually known as polythene, and sometimes as polyethylene), poly(propene) (old name: polypropylene), PVC and PTFE. It also looks briefly at how the structure of the polymers affects their properties and uses. In common with everything else on this page, this is an example of addition polymerisation. An addition reaction is one in which two or more molecules join together to give a single product. During the polymerisation of ethene, thousands of ethene molecules join together to make poly(ethene) - commonly called polythene. The number of molecules joining up is very variable, but is in the region of 2000 to 20000. Conditions: : Poly(propene) is manufactured using Ziegler-Natta and other modern catalysts. There are three variants on the structure of poly(propene) which you may need to know about, but we'll start from the beginning with a general structure which fits all of them. Poly(chloroethene) is commonly known by the initials of its old name, PVC. Poly(chloroethene) is made by polymerising chloroethene, CH =CHCl. Working out its structure is no different from working out the structure of poly(propene) (see above). As long as you draw the chloroethene molecule in the right way, the structure is pretty obvious. The equation is usually written: It doesn't matter which carbon you attach the chlorine to in the original molecule. Just be consistent on both sides of the equation. The polymerisation process produces mainly atactic polymer molecules - with the chlorines orientated randomly along the chain. The structure is no different from atactic poly(propene) - just replace the CH groups by chlorine atoms. Because of the way the chlorine atoms stick out from the chain at random, and because ot their large size, it is difficult for the chains to lie close together. Poly(chloroethene) is mainly amorphous with only small areas of crystallinity. You normally expect amorphous polymers to be more flexible than crystalline ones because the forces of attraction between the chains tend to be weaker. However, pure poly(chloroethene) tends to be rather hard and rigid. This is because of the presence of additional dipole-dipole interactions due to the polarity of the carbon-chlorine bonds. Chlorine is more electronegative than carbon, and so attracts the electrons in the bond towards itself. That makes the chlorine atoms slightly negative and the carbons slightly positive. These permanent dipoles add to the attractions due to the temporary dipoles which produce the dispersion forces. You may have come across this under the brand names of Teflon or Fluon. Structurally, PTFE is just like poly(ethene) except that each hydrogen in the structure is replaced by a fluorine atom. The PTFE chains tend to pack well and PTFE is fairly crystalline. Because of the fluorine atoms, the chains also contain more electrons (for an equal length) than a corresponding poly(ethene) chain. Taken together (the good packing and the extra electrons) that means that the van der Waals dispersion forces will be stronger than in even high density poly(ethene). Jim Clark ( ) | 3,188 | 3,821 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Diels-Alder_Cycloaddition |
The unique character of conjugated dienes manifests itself dramatically in the . A cycloaddition reaction is the concerted bonding together of two independent pi-electron systems to form a new ring of atoms. When this occurs, two pi-bonds are converted to two sigma-bonds, the simplest example being the hypothetical combination of two ethene molecules to give cyclobutane. This does not occur under normal conditions, but the cycloaddition of 1,3-butadiene to cyanoethene (acrylonitrile) does, and this is an example of the Diels-Alder reaction. The following diagram illustrates two cycloadditions, and introduces several terms that are useful in discussing reactions of this kind. In the hypothetical ethylene dimerization on the left, each reactant molecule has a pi-bond (colored orange) occupied by two electrons. The cycloaddition converts these pi-bonds into new sigma-bonds (colored green), and this transformation is then designated a [2+2] cycloaddition, to enumerate the reactant pi-electrons that change their bonding location. The Diels-Alder reaction is an important and widely used method for making six-membered rings, as shown on the right. The reactants used in such reactions are a conjugated diene, simply referred to as the , and a double or triple bond coreactant called the , because it combines with (has an affinity for) the diene. The Diels-Alder cycloaddition is classified as a [4+2] process because the diene has four pi-electrons that shift position in the reaction and the dienophile has two. The Diels-Alder reaction is a single step process, so the diene component must adopt a cis-like conformation in order for the end carbon atoms (#1 & #4) to bond simultaneously to the dienophile. Such conformations are called , the referring to the single bond connecting the two double bonds. The s-cis and s-trans conformers of 1,3-butadiene are shown in the preceding diagram. For many acyclic dienes the s-trans conformer is more stable than the s-cis conformer (due to steric crowding of the end groups), but the two are generally in rapid equilibrium, permitting the use of all but the most hindered dienes as reactants in Diels-Alder reactions. In its usual form, the diene component is electron rich, and the best dienophiles are electron poor due to electron withdrawing substituents such as CN, C=O & NO . The initial bonding interaction reflects this electron imbalance, with the two new sigma-bonds being formed simultaneously, but not necessarily at equal rates. We noted earlier that addition reactions of alkenes often exhibited stereoselectivity, in that the reagent elements in some cases added syn and in other cases anti to the the plane of the double bond. Both reactants in the Diels-Alder reaction may demonstrate stereoisomerism, and when they do it is found that the relative configurations of the reactants are preserved in the product (the adduct). The following drawing illustrates this fact for the reaction of 1,3-butadiene with (E)-dicyanoethene. The trans relationship of the cyano groups in the dienophile is preserved in the six-membered ring of the adduct. Likewise, if the terminal carbons of the diene bear substituents, their relative configuration will be retained in the adduct. Using the earlier terminology, we could say that bonding to both the diene and the dienophile is syn. An alternative description, however, refers to the planar nature of both reactants and terms the bonding in each case to be (i.e. to or from the same face of each plane). This stereospecificity also confirms the synchronous nature of the 1,4-bonding that takes place. These features are illustrated by the following eight examples, one of which does not give a Diels-Alder cycloaddition. Try to predict the course of each reaction before checking the answers. The formation of a new six-membered ring should be apparent in every case where reaction occurs. There is no reaction in example because this diene cannot adopt a s-cis orientation. In examples at least one of the reactants is cyclic so that the product has more than one ring, but the newly formed ring is always six-membered. In example the the same cyclic compound acts as both the diene colored blue) and the dienophile (colored red). The adduct has three rings, two of which are the five-membered rings present in the reactant, and the third is the new six-membered ring (shaded light yellow). Example has an alkyne as a dienophile (colored red), so the adduct retains a double bond at that location. This double bond could still serve as a dienophile, but in the present case the diene is sufficiently hindered to retard a second cycloaddition. The quinone dienophile in reaction has two dienophilic double bonds. However, the double bond with two methyl substituents is less reactive than the unsubstituted dienophile due in part to the electron donating properties of the methyl groups and in part to steric hindrance. The stereospecificity of the Diels-Alder reaction is demonstrated by examples . In the stereogenic centers lie on the dienophile, whereas in these centers are on the diene. In all cases the configuration of the reactant is preserved in the adduct. Cyclic dienes, such as those in examples , give bridged bicyclic adducts for which an additional configurational feature must be designated. As shown in the following diagram, there are two possible configurations for compounds of this kind. If a substituent (colored magenta here) is oriented cis to the longest or more unsaturated bridge (colored blue here), it is said to be . When directed trans to the bridge it is . When the Diels-Alder reaction forms bridged bicyclic adducts and an unsaturated substituent is located on this bicyclic structure (as in ), the chief product is normally the endo isomer " ". Example does not merit such a nomenclature, since stereoisomeric orientations of the substituent are not possible. ), | 5,953 | 3,822 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Decomposition_Reactions |
reactions are very common, and this word is used so much that many chemists just say "decomp". When people use this word in the lab, they might just mean that something didn't work, or that a chemical reacted in an unwanted way, especially while it was sitting in a bottle for a long time. The official meaning of decomposition is a little bit more specific, and means a reaction in which one chemical splits into two or more chemicals, like this: \[A \rightarrow B + C\] Decomposition reactions are often undesirable, but not always. For instance, many explosions are decompositions, and explosives are very important for many purposes other than weapons. Decomposition reactions might be hard to predict at first. The decomposition reactions in intro chemistry classes often result from heating a substance. For instance, when heated or struck, a salt of a complex anion (chlorate, carbonate, azide) may lose a gas (oxygen, carbon dioxide or nitrogen) leaving behind a simpler salt or metal. This could happen explosively, depending on the compound. Or, when heated, a metal hydroxide loses water to form the metal oxide (the reverse of the ). Here are some examples: \[CaCO_{3}(s) \rightarrow CaO(s) + CO_{2}(g)\] \[2NaN_{3}(s) \rightarrow 2Na(s) + 3N_{2}g)\] \[2KClO_{3}(s) \rightarrow 2KCl(s) + 3O_{2}(g)\] \[Cu(OH)_{2}(s) \rightarrow CuO(s) + H_{2}O(g\; or\; l)\] Students often get confused into thinking that combination reactions, the opposite of decomposition reactions, are called composition reactions. Actually, composition is not a type of reaction, rather, it has a different meaning. means the ratio of elements in a compound, such as 75% C and 25% H. | 1,691 | 3,823 |
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