url
stringlengths 90
342
| html
stringlengths 602
98.8k
| text_length
int64 602
98.8k
| __index_level_0__
int64 0
5.02k
|
|---|---|---|---|
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/11%3A_Solutions/11.5%3A_Phase_Equilibrium_in_Solutions_-_Nonvolatile_Solutes |
Many of the physical properties of solutions differ significantly from those of the pure substances discussed in earlier chapters, and these differences have important consequences. For example, the limited temperature range of liquid water (0°C–100°C) severely limits its use. Aqueous solutions have both a lower freezing point and a higher boiling point than pure water. Probably one of the most familiar applications of this phenomenon is the addition of ethylene glycol (“antifreeze”) to the water in an automobile radiator. This solute lowers the freezing point of the water, preventing the engine from cracking in very cold weather from the expansion of pure water on freezing. Antifreeze also enables the cooling system to operate at temperatures greater than 100°C without generating enough pressure to explode. Changes in the freezing point and boiling point of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are called colligative properties (from the Latin , meaning “bound together” as in a quantity). As we will see, the vapor pressure and osmotic pressure of solutions are also colligative properties. When we determine the number of particles in a solution, it is important to remember that not all solutions with the same molarity contain the same concentration of solute particles. Consider, for example, 0.01 M aqueous solutions of sucrose, \(\ce{NaCl}\), and \(\ce{CaCl_2}\). Because sucrose dissolves to give a solution of neutral molecules, the concentration of solute particles in a 0.01 M sucrose solution is 0.01 M. In contrast, both \(\ce{NaCl}\) and \(\ce{CaCl_2}\) are ionic compounds that dissociate in water to yield solvated ions. As a result, a 0.01 M aqueous solution of \(\ce{NaCl}\) contains 0.01 M \(\ce{Na^{+}}\) ions and 0.01 M \(\ce{Cl^{−}}\) ions, for a total particle concentration of 0.02 M. Similarly, the \(\ce{CaCl_2}\) solution contains 0.01 M \(\ce{Ca^{2+}}\) ions and 0.02 M \(\ce{Cl^{−}}\) ions, for a total particle concentration of 0.03 M. These values are correct for dilute solutions, where the dissociation of the compounds to form separately solvated ions is complete. At (typically >1 M), especially with salts of small, highly charged ions (such as \(\ce{Mg^{2+}}\) or \(\ce{Al^{3+}}\)), or in solutions with less polar solvents, dissociation to give separate ions is often incomplete. The sum of the concentrations of the dissolved solute particles dictates the physical properties of a solution. In the following discussion, we must therefore keep the chemical nature of the solute firmly in mind. A greater discussion of this is below. Adding a nonvolatile solute, one whose vapor pressure is too low to measure readily, to a volatile solvent decreases the vapor pressure of the solvent. We can understand this phenomenon qualitatively by examining Figure \(\Page {1}\), which is a schematic diagram of the surface of a solution of glucose in water. In an aqueous solution of glucose, a portion of the surface area is occupied by nonvolatile glucose molecules rather than by volatile water molecules. As a result, fewer water molecules can enter the vapor phase per unit time, even though the surface water molecules have the same kinetic energy distribution as they would in pure water. At the same time, the rate at which water molecules in the vapor phase collide with the surface and reenter the solution is unaffected. The net effect is to shift the dynamic equilibrium between water in the vapor and the liquid phases, decreasing the vapor pressure of the solution compared with the vapor pressure of the pure solvent. Figure \(\Page {2}\) shows two beakers, one containing pure water and one containing an aqueous glucose solution, in a sealed chamber. We can view the system as having two competing equilibria: water vapor will condense in both beakers at the same rate, but water molecules will evaporate more slowly from the glucose solution because fewer water molecules are at the surface. Eventually all of the water will evaporate from the beaker containing the liquid with the higher vapor pressure (pure water) and condense in the beaker containing the liquid with the lower vapor pressure (the glucose solution). If the system consisted of only a beaker of water inside a sealed container, equilibrium between the liquid and vapor would be achieved rather rapidly, and the amount of liquid water in the beaker would remain constant. If the particles of a solute are essentially the same size as those of the solvent and both solute and solvent have roughly equal probabilities of being at the surface of the solution, then the effect of a solute on the vapor pressure of the solvent is proportional to the number of sites occupied by solute particles at the surface of the solution. Doubling the concentration of a given solute causes twice as many surface sites to be occupied by solute molecules, resulting in twice the decrease in vapor pressure. The relationship between solution composition and vapor pressure is therefore \[P_A=\chi_AP^0_A \label{13.5.1}\] where \(P_A\) is the vapor pressure of component A of the solution (in this case the solvent), \(\chi_A\) is the mole fraction of \(A\) in solution, and \(P^0_A\) is the vapor pressure of pure \(A\). Equation \(\ref{13.5.1}\) is known as , after the French chemist who developed it. If the solution contains only a single nonvolatile solute (\(B\)), then we can relate the mole fraction of the solvent to the solute \[\chi_A + \chi_B = 1 \nonumber\] and we can substitute \(\chi_A = 1 − \chi_B\) into Equation \ref{13.5.1} to obtain \[\begin{align*} P_A &=(1−\chi_B)P^0_A \\[4pt] &=P^0_A−\chi_BP^0_A \end{align*}\] Rearranging and defining \(ΔP_A=P^0_A−P_A\), we obtain a relationship between the decrease in vapor pressure and the mole fraction of nonvolatile solute: \[ \begin{align} P^0_A−P_A &=ΔP_A \\[4pt] &=\chi_BP^0_A \label{13.5.3} \end{align}\] We can solve vapor pressure problems in either of two ways: by using Equation \(\ref{13.5.1}\) to calculate the actual vapor pressure above a solution of a nonvolatile solute, or by using Equation \(\ref{13.5.3}\) to calculate the decrease in vapor pressure caused by a specified amount of a nonvolatile solute. Ethylene glycol (\(\ce{HOCH_2CH_2OH}\)), the major ingredient in commercial automotive antifreeze, increases the boiling point of radiator fluid by lowering its vapor pressure. At 100°C, the vapor pressure of pure water is 760 mmHg. Calculate the vapor pressure of an aqueous solution containing 30.2% ethylene glycol by mass, a concentration commonly used in climates that do not get extremely cold in winter. : identity of solute, percentage by mass, and vapor pressure of pure solvent : vapor pressure of solution : : A 30.2% solution of ethylene glycol contains 302 g of ethylene glycol per kilogram of solution; the remainder (698 g) is water. To use Raoult’s law to calculate the vapor pressure of the solution, we must know the mole fraction of water. Thus we must first calculate the number of moles of both ethylene glycol (EG) and water present: \[moles \; EG=(302 \;\cancel{g}) \left( \dfrac{1\; mol}{62.07\; \cancel{g}} \right)=4.87\; mol\; EG \nonumber\] \[moles \; \ce{H2O} =(698 \;\cancel{g}) \left( \dfrac{1\; mol}{18.02\; \cancel{g}} \right)=38.7\; mol\; H_2O \nonumber\] The mole fraction of water is thus \[\chi_{H_2O}=\dfrac{38.7\; \cancel{mol} \; H_2O}{38.7\; \cancel{mol}\; H_2O +4.87 \cancel{mol}\; EG} =0.888 \nonumber\] From Raoult’s law (Equation \(\ref{13.5.1}\)), the vapor pressure of the solution is \[ \begin{align*} P_{\ce{H2O}} &=(\chi_{\ce{H2O}})(P^0_{\ce{H2O}}) \\[4pt] &=(0.888)(760\; mmHg) =675 \;mmHg \end{align*}\] Alternatively, we could solve this problem by calculating the mole fraction of ethylene glycol and then using Equation \(\ref{13.5.3}\) to calculate the resulting decrease in vapor pressure: \[\begin{align*} \chi_{EG} &=\dfrac{4.87\; mol\; EG}{4.87\; mol\; EG+38.7\; mol\; H_2O} \\[4pt] &=0.112 \end{align*}\] \[\begin{align*} ΔP_{\ce{H2O}} &=(\chi_{EG})(P^0_{\ce{H2O}}) \\[4pt] &=(0.112)(760\; mmHg)=85.1\; mmHg \end{align*}\] \[P_{\ce{H2O}}=P^0_{\ce{H2O}}−ΔP_{\ce{H2O}}=760\; mmHg−85.1\; mmHg=675\; mmHg \nonumber\] The same result is obtained using either method. Seawater is an approximately 3.0% aqueous solution of \(\ce{NaCl}\) by mass with about 0.5% of other salts by mass. Calculate the decrease in the vapor pressure of water at 25°C caused by this concentration of \(\ce{NaCl}\), remembering that 1 mol of \(\ce{NaCl}\) produces 2 mol of solute particles. The vapor pressure of pure water at 25°C is 23.8 mmHg. 0.45 mmHg. This may seem like a small amount, but it constitutes about a 2% decrease in the vapor pressure of water and accounts in part for the higher humidity in the north-central United States near the Great Lakes, which are freshwater lakes. The decrease therefore has important implications for climate modeling. Recall that the normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm. If a nonvolatile solute lowers the vapor pressure of a solvent, it must also affect the boiling point. Because the vapor pressure of the solution at a given temperature is less than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent. Thus the boiling point of a solution is always greater than that of the pure solvent. We can see why this must be true by comparing the phase diagram for an aqueous solution with the phase diagram for pure water (Figure \(\Page {4}\)). The vapor pressure of the solution is less than that of pure water at all temperatures. Consequently, the liquid–vapor curve for the solution crosses the horizontal line corresponding to \(P = 1\, atm\) at a higher temperature than does the curve for pure water. The boiling point of a solution with a nonvolatile solute is greater than the boiling point of the pure solvent. The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure. As we have just discussed, the decrease in the vapor pressure is proportional to the concentration of the solute in the solution. Hence the magnitude of the increase in the boiling point must also be proportional to the concentration of the solute (Figure \(\Page {5}\)). We can define the boiling point elevation (\(ΔT_b\)) as the difference between the boiling points of the solution and the pure solvent: \[ΔT_b=T_b−T^0_b \label{13.5.8}\] where \(T_b\) is the boiling point of the solution and \(T^0_b\) is the boiling point of the pure solvent. We can express the relationship between \(ΔT_b\) and concentration as follows \[ΔT_b = mK_b \label{13.5.9}\] where m is the concentration of the solute expressed in molality, and \(K_b\) is the of the solvent, which has units of °C/m. Table \(\Page {1}\) lists characteristic \(K_b\) values for several commonly used solvents. For relatively dilute solutions, the magnitude of both properties is proportional to the solute concentration. The concentration of the solute is typically expressed as molality rather than mole fraction or molarity for two reasons. First, because the density of a solution changes with temperature, the value of molarity also varies with temperature. If the boiling point depends on the solute concentration, then by definition the system is not maintained at a constant temperature. Second, molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value (a mole fraction can be only between zero and one). Using molality allows us to eliminate nonsignificant zeros. According to Table \(\Page {1}\), the molal boiling point elevation constant for water is 0.51°C/m. Thus a 1.00 m aqueous solution of a nonvolatile molecular solute such as glucose or sucrose will have an increase in boiling point of 0.51°C, to give a boiling point of 100.51°C at 1.00 atm. The increase in the boiling point of a 1.00 m aqueous \(\ce{NaCl}\) solution will be approximately twice as large as that of the glucose or sucrose solution because 1 mol of \(\ce{NaCl}\) produces 2 mol of dissolved ions. Hence a 1.00 m \(\ce{NaCl}\) solution will have a boiling point of about 101.02°C. In Example \(\Page {1}\), we calculated that the vapor pressure of a 30.2% aqueous solution of ethylene glycol at 100°C is 85.1 mmHg less than the vapor pressure of pure water. We stated (without offering proof) that this should result in a higher boiling point for the solution compared with pure water. Now that we have seen why this assertion is correct, calculate the boiling point of the aqueous ethylene glycol solution. : composition of solution : boiling point : Calculate the molality of ethylene glycol in the 30.2% solution. Then use Equation \(\ref{13.5.9}\) to calculate the increase in boiling point. : From Example \(\Page {1}\), we know that a 30.2% solution of ethylene glycol in water contains 302 g of ethylene glycol (4.87 mol) per 698 g of water. The molality of the solution is thus \[\begin{align*} \text{molality of ethylene glycol} &= \left(\dfrac{4.87 \;mol}{698 \; \cancel{g} \;H_2O} \right) \left(\dfrac{1000\; \cancel{g}}{1 \;kg} \right) \\[4pt] &=6.98\, m \end{align*}\] From Equation \(\ref{13.5.9}\), the increase in boiling point is therefore \[\begin{align*} ΔT_b &= m K_b \\[4pt] &=(6.98 \cancel{m})(0.51°C/\cancel{m}) \\[4pt] &=3.6°C \end{align*}\] The boiling point of the solution is thus predicted to be 104°C. With a solute concentration of almost 7 m, however, the assumption of a dilute solution used to obtain Equation \(\ref{13.5.9}\) may not be valid. Assume that a tablespoon (5.00 g) of \(\ce{NaCl}\) is added to 2.00 L of water at 20.0°C, which is then brought to a boil to cook spaghetti. At what temperature will the water boil? 100.04°C, or 100°C to three significant figures. (Recall that 1 mol of \(\ce{NaCl}\) produces 2 mol of dissolved particles. The small increase in temperature means that adding salt to the water used to cook pasta has essentially no effect on the cooking time.) The phase diagram in Figure \(\Page {4}\) shows that dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point. The solid–liquid curve for the solution crosses the line corresponding to \(P = 1\,atm\) at a lower temperature than the curve for pure water. We can understand this result by imagining that we have a sample of water at the normal freezing point temperature, where there is a dynamic equilibrium between solid and liquid. Water molecules are continuously colliding with the ice surface and entering the solid phase at the same rate that water molecules are leaving the surface of the ice and entering the liquid phase. If we dissolve a nonvolatile solute such as glucose in the liquid, the dissolved glucose molecules will reduce the number of collisions per unit time between water molecules and the ice surface because some of the molecules colliding with the ice will be glucose. Glucose, though, has a very different structure than water, and it cannot fit into the ice lattice. Consequently, the presence of glucose molecules in the solution can only decrease the rate at which water molecules in the liquid collide with the ice surface and solidify. Meanwhile, the rate at which the water molecules leave the surface of the ice and enter the liquid phase is unchanged. The net effect is to cause the ice to melt. The only way to reestablish a dynamic equilibrium between solid and liquid water is to lower the temperature of the system, which decreases the rate at which water molecules leave the surface of the ice crystals until it equals the rate at which water molecules in the solution collide with the ice. By analogy to our treatment of boiling point elevation,the freezing point depression (\(ΔT_f\)) is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution: \[ ΔT_f=T^0_f−T_f \label{13.5.10}\] where \(T^0_f\) is the freezing point of the pure solvent and \(T_f\) is the freezing point of the solution. The order of the terms is reversed compared with Equation \(\ref{13.5.8}\) to express the freezing point depression as a positive number. The relationship between \(ΔT_f\) and the solute concentration is given by an equation analogous to Equation \(\ref{13.5.9}\): \[ΔT_f = mK_f \label{13.5.11}\] where \(m\) is the molality of the solution and \(K_f\) is the molal freezing point depression constant for the solvent (in units of °C/m). Like \(K_b\), each solvent has a characteristic value of \(K_f\) (see Table \(\Page {1}\)). Freezing point depression depends on the total number of dissolved nonvolatile solute particles, just as with boiling point elevation. Thus an aqueous \(\ce{NaCl}\) solution has twice as large a freezing point depression as a glucose solution of the same molality. People who live in cold climates use freezing point depression to their advantage in many ways. For example, salt is used to melt ice and snow on roads and sidewalks, ethylene glycol is added to engine coolant water to prevent an automobile engine from being destroyed, and methanol is added to windshield washer fluid to prevent the fluid from freezing. The decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus a pure liquid all depend on the total number of dissolved nonvolatile solute particles. In colder regions of the United States, \(\ce{NaCl}\) or \(\ce{CaCl_2}\) is often sprinkled on icy roads in winter to melt the ice and make driving safer. Use the data in the Figure below to estimate the concentrations of two saturated solutions at 0°C, one of \(\ce{NaCl}\) and one of \(\ce{CaCl_2}\), and calculate the freezing points of both solutions to see which salt is likely to be more effective at melting ice. : solubilities of two compounds : concentrations and freezing points : : A. From Figure above, we can estimate the solubilities of \(\ce{NaCl}\) and \(\ce{CaCl_2}\) to be about 36 g and 60 g, respectively, per 100 g of water at 0°C. The corresponding concentrations in molality are \[m_{NaCl}=\left(\dfrac{36 \; \cancel{g \;NaCl}}{100 \;\cancel{g} \;H_2O}\right)\left(\dfrac{1\; mol\; NaCl}{58.44\; \cancel{ g\; NaCl}}\right)\left(\dfrac{1000\; \cancel{g}}{1\; kg}\right)=6.2\; m \nonumber\] \[m_{CaCl_2}=\left(\dfrac{60\; \cancel{g\; CaCl_2}}{100\;\cancel{g}\; H_2O}\right)\left(\dfrac{1\; mol\; CaCl_2}{110.98\; \cancel{g\; CaCl_2}}\right)\left(\dfrac{1000 \;\cancel{g}}{1 kg}\right)=5.4\; m \nonumber \] The lower formula mass of \(\ce{NaCl}\) more than compensates for its lower solubility, resulting in a saturated solution that has a slightly higher concentration than \(\ce{CaCl2}\). B. Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of \(\ce{NaCl}\) and \(\ce{CaCl2}\), respectively, the actual concentrations of the dissolved species in the two saturated solutions are 2 × 6.2 m = 12 m for \(\ce{NaCl}\) and 3 × 5.4 m = 16 m for \(\ce{CaCl2}\). The resulting freezing point depressions can be calculated using Equation \(\ref{13.5.11}\): \[\ce{NaCl}: ΔT_f=mK_f=(12\; \cancel{m})(1.86°C/\cancel{m})=22°C \nonumber\] \[\ce{CaCl2}: ΔT_f=mK_f=(16\;\cancel{m})(1.86°C/\cancel{m})=30°C \nonumber\] Because the freezing point of pure water is 0°C, the actual freezing points of the solutions are −22°C and −30°C, respectively. Note that \(\ce{CaCl2}\) is substantially more effective at lowering the freezing point of water because its solutions contain three ions per formula unit. In fact, \(\ce{CaCl2}\) is the salt usually sold for home use, and it is also often used on highways. Because the solubilities of both salts decrease with decreasing temperature, the freezing point can be depressed by only a certain amount, regardless of how much salt is spread on an icy road. If the temperature is significantly below the minimum temperature at which one of these salts will cause ice to melt (say −35°C), there is no point in using salt until it gets warmer Calculate the freezing point of the 30.2% solution of ethylene glycol in water whose vapor pressure and boiling point we calculated in Examples \(\Page {1}\) and \(\Page {3}\). −13.0°C Arrange these aqueous solutions in order of decreasing freezing points: 0.1 m \(\ce{KCl}\), 0.1 m glucose, 0.1 m \(\ce{SrCl2}\), 0.1 m ethylene glycol, 0.1 m benzoic acid, and 0.1 m \(\ce{HCl}\). : molalities of six solutions relative freezing points : : Because the molal concentrations of all six solutions are the same, we must focus on which of the substances are strong electrolytes, which are weak electrolytes, and which are nonelectrolytes to determine the actual numbers of particles in solution. \(\ce{KCl}\), \(\ce{SrCl_2}\), and \(\ce{HCl}\) are , producing two, three, and two ions per formula unit, respectively. Benzoic acid is a weak electrolyte (approximately one particle per molecule), and glucose and ethylene glycol are both nonelectrolytes (one particle per molecule). The molalities of the solutions in terms of the total particles of solute are: \(\ce{KCl}\) and \(\ce{HCl}\), 0.2 m; \(SrCl_2\), 0.3 m; glucose and ethylene glycol, 0.1 m; and benzoic acid, 0.1–0.2 m. Because the magnitude of the decrease in freezing point is proportional to the concentration of dissolved particles, the order of freezing points of the solutions is: glucose and ethylene glycol (highest freezing point, smallest freezing point depression) > benzoic acid > \(\ce{HCl}\) = \(\ce{KCl}\) > \(\ce{SrCl_2}\). Arrange these aqueous solutions in order of increasing freezing points: 0.2 m \(\ce{NaCl}\), 0.3 m acetic acid, 0.1 m \(\ce{CaCl_2}\), and 0.2 m sucrose. 0.2 m \(\ce{NaCl}\) (lowest freezing point) < 0.3 m acetic acid ≈ 0.1 m \(\ce{CaCl_2}\) < 0.2 m sucrose (highest freezing point) Colligative properties can also be used to determine the molar mass of an unknown compound. One method that can be carried out in the laboratory with minimal equipment is to measure the freezing point of a solution with a known mass of solute. This method is accurate for dilute solutions (≤1% by mass) because changes in the freezing point are usually large enough to measure accurately and precisely. By comparing \(K_b\) and \(K_f\) values in Table \(\Page {1}\), we see that changes in the boiling point are smaller than changes in the freezing point for a given solvent. Boiling point elevations are thus more difficult to measure precisely. For this reason, freezing point depression is more commonly used to determine molar mass than is boiling point elevation. Because of its very large value of \(K_f\) (37.8°C/m), d-(+)-camphor (Table \(\Page {1}\)) is often used to determine the molar mass of organic compounds by this method. A 7.08 g sample of elemental sulfur is dissolved in 75.0 g of \(CS_2\) to create a solution whose freezing point is −113.5°C. Use these data to calculate the molar mass of elemental sulfur and thus the formula of the dissolved \(\ce{S_n}\) molecules (i.e., what is the value of \(n\)?). : masses of solute and solvent and freezing point : molar mass and number of \(\ce{S}\) atoms per molecule : : A The first step is to calculate the freezing point depression using Equation \(\ref{13.5.10}\): \[ΔT_f=T^0_f−T_f=−112.1°C−(−113.5°C)=1.4°C \nonumber\] Then Equation \(\ref{13.5.11}\) gives \[m=\dfrac{ΔT_f}{K_f}=\dfrac{1.4° \cancel{C}}{3.74° \cancel{C}/m}=0.37\;m \nonumber\] B The total number of moles of solute present in the solution is \[\text{moles solute}=\left(\dfrac{0.37 mol}{\cancel{kg}}\right) (75.0\; g) \left(\dfrac{1 kg}{1000\; g}\right)=0.028 \;mol \nonumber\] C We now know that 0.708 g of elemental sulfur corresponds to 0.028 mol of solute. The molar mass of dissolved sulfur is thus \[\text{molar mass}=\dfrac{7.08\; g}{0.028\; mol}=260\; g/mol \nonumber\] The molar mass of atomic sulfur is 32 g/mol, so there must be 260/32 = 8.1 sulfur atoms per mole, corresponding to a formula of \(\ce{S_8}\). One of the byproducts formed during the synthesis of \(C_{60}\) is a deep red solid containing only carbon. A solution of 205 mg of this compound in 10.0 g of \(CCl_4\) has a freezing point of −23.38°C. What are the molar mass and most probable formula of the substance? 847 g/mol; \(\ce{C_{70}}\) Osmotic pressure is a colligative property of solutions that is observed using a semipermeable membrane, a barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions. The net flow of solvent through a semipermeable membrane is called osmosis (from the Greek osmós, meaning “push”). The direction of net solvent flow is always from the side with the lower concentration of solute to the side with the higher concentration. Osmosis can be demonstrated using a U-tube like the one shown in Figure \(\Page {6}\), which contains pure water in the left arm and a dilute aqueous solution of glucose in the right arm. A net flow of water through the membrane occurs until the levels in the arms eventually stop changing, which indicates that equilibrium has been reached. The osmotic pressure (\(\Pi\)) of the glucose solution is the difference in the pressure between the two sides, in this case the heights of the two columns. Although the semipermeable membrane allows water molecules to flow through in either direction, the rate of flow is not the same in both directions because the concentration of water is not the same in the two arms. The net flow of water through the membrane can be prevented by applying a pressure to the right arm that is equal to the osmotic pressure of the glucose solution. Just as with any other colligative property, the osmotic pressure of a solution depends on the concentration of dissolved solute particles. Osmotic pressure obeys a law that resembles the ideal gas equation: \[\Pi=\dfrac{nRT}{V}=MRT \label{13.5.12}\] where \(M\) is the number of moles of solute per unit volume of solution (i.e., the molarity of the solution), \(R\) is the ideal gas constant, and \(T\) is the absolute temperature. As shown in Example \(\Page {7}\), osmotic pressures tend to be quite high, even for rather dilute solutions. When placed in a concentrated salt solution, certain yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding medium. Suppose that the yeast cells are placed in an aqueous solution containing 4.0% \(\ce{NaCl}\) by mass; the solution density is 1.02 g/mL at 25°C. : concentration, density, and temperature of \(\ce{NaCl}\) solution; internal osmotic pressure of cell : osmotic pressure of \(\ce{NaCl}\) solution and concentration of glycerol needed : : A The solution contains 4.0 g of \(\ce{NaCl}\) per 100 g of solution. Using the formula mass of \(\ce{NaCl}\) (58.44 g/mol) and the density of the solution (1.02 g/mL), we can calculate the molarity: \[ \begin{align*} M_{NaCl} &=\dfrac{moles\; NaCl}{\text{liter solution}} \\[4pt] &=\left(\dfrac{4.0 \; \cancel{g} \;NaCl}{58.44\; \cancel{g}/mol\; NaCl}\right)\left(\dfrac{1}{100\; \cancel{g \;solution}}\right)\left(\dfrac{1.02\; \cancel{g\; solution}}{1.00\; \cancel{mL}\; solution}\right)\left(\dfrac{1000\; \cancel{mL}}{1\; L}\right) \\[4pt] &= 0.70\; M\; \ce{NaCl} \end{align*}\] Because 1 mol of \(\ce{NaCl}\) produces 2 mol of particles in solution, the total concentration of dissolved particles in the solution is (2)(0.70 M) = 1.4 M. B Now we can use Equation \ref{13.5.12} to calculate the osmotic pressure of the solution: \[ \begin{align*} \Pi &=MRT \\[4pt] &=(1.4 \;mol/L)\left[ 0.0821\; (L⋅atm)/(K⋅mol) \right ] (298\; K)\\[4pt] &=34 \;atm\end{align*}\] C If the yeast cells are to exactly balance the external osmotic pressure, they must produce enough glycerol to give an additional internal pressure of (34 atm − 7.3 atm) = 27 atm. Glycerol is a nonelectrolyte, so we can solve Equation \ref{13.5.12} for the molarity corresponding to this osmotic pressure: \[ \begin{align*} M&=\dfrac{\Pi}{RT}\\[4pt] &=\dfrac{27\; \cancel{atm}}{[0.0821(L⋅\cancel{atm})/(\cancel{K}⋅mol)] (298 \;\cancel{K})}\\[4pt] &=1.1 \;M \;\text{glycerol} \end{align*}\] In solving this problem, we could also have recognized that the only way the osmotic pressures can be the same inside the cells and in the solution is if the concentrations of dissolved particles are the same. We are given that the normal concentration of dissolved particles in the cells is 0.3 M, and we have calculated that the \(\ce{NaCl}\) solution is effectively 1.4 M in dissolved particles. The yeast cells must therefore synthesize enough glycerol to increase the internal concentration of dissolved particles from 0.3 M to 1.4 M—that is, an additional 1.1 M concentration of glycerol. Assume that the fluids inside a sausage are approximately 0.80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100°C to learn why experienced cooks pierce the semipermeable skin of sausages before boiling them. 24 atm Because of the large magnitude of osmotic pressures, osmosis is extraordinarily important in biochemistry, biology, and medicine. Virtually every barrier that separates an organism or cell from its environment acts like a semipermeable membrane, permitting the flow of water but not solutes. The same is true of the compartments inside an organism or cell. Some specialized barriers, such as those in your kidneys, are slightly more permeable and use a related process called dialysis, which permits both water and small molecules to pass through but not large molecules such as proteins. The same principle has long been used to preserve fruits and their essential vitamins over the long winter. High concentrations of sugar are used in jams and jellies not for sweetness alone but because they greatly increase the osmotic pressure. Thus any bacteria not killed in the cooking process are dehydrated, which keeps them from multiplying in an otherwise rich medium for bacterial growth. A similar process using salt prevents bacteria from growing in ham, bacon, salt pork, salt cod, and other preserved meats. The effect of osmotic pressure is dramatically illustrated in Figure \(\Page {7}\), which shows what happens when red blood cells are placed in a solution whose osmotic pressure is much lower or much higher than the internal pressure of the cells. In addition to capillary action, trees use osmotic pressure to transport water and other nutrients from the roots to the upper branches. Evaporation of water from the leaves results in a local increase in the salt concentration, which generates an osmotic pressure that pulls water up the trunk of the tree to the leaves. Finally, a process called reverse osmosis can be used to produce pure water from seawater. As shown schematically in Figure \(\Page {8}\), applying high pressure to seawater forces water molecules to flow through a semipermeable membrane that separates pure water from the solution, leaving the dissolved salt behind. Large-scale desalinization plants that can produce hundreds of thousands of gallons of freshwater per day are common in the desert lands of the Middle East, where they supply a large proportion of the freshwater needed by the population. Similar facilities are now being used to supply freshwater in southern California. Small, hand-operated reverse osmosis units can produce approximately 5 L of freshwater per hour, enough to keep 25 people alive, and are now standard equipment on US Navy lifeboats. Thus far we have assumed that we could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the actual concentration of dissolved particles in an electrolyte solution. We have used this simple model to predict such properties as freezing points, melting points, vapor pressure, and osmotic pressure. If this model were perfectly correct, we would expect the freezing point depression of a 0.10 m solution of sodium chloride, with 2 mol of ions per mole of \(\ce{NaCl}\) in solution, to be exactly twice that of a 0.10 m solution of glucose, with only 1 mol of molecules per mole of glucose in solution. In reality, this is not always the case. Instead, the observed change in freezing points for 0.10 m aqueous solutions of \(\ce{NaCl}\) and \(\ce{KCl}\) are significantly less than expected (−0.348°C and −0.344°C, respectively, rather than −0.372°C), which suggests that fewer particles than we expected are present in solution. The relationship between the actual number of moles of solute added to form a solution and the apparent number as determined by colligative properties is called the van ’t Hoff factor (\(i\)) and is defined as follows: \[i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \label{13.5.13}\] Named for Jacobus Hendricus van ’t Hoff (1852–1911), a Dutch chemistry professor at the University of Amsterdam who won the first Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions. As the solute concentration increases, the van ’t Hoff factor decreases. The van ’t Hoff factor is therefore . The lower the van ’t Hoff factor, the greater the deviation. As the data in Table \(\Page {2}\) show, the van ’t Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions apparently contain fewer particles than predicted by the number of ions per formula unit. As the concentration of the solute increases, the van ’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution. Instead, some of the ions exist as ion pairs, a cation and an anion that for a brief time are associated with each other without an intervening shell of water molecules (Figure \(\Page {9}\)). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as \(Mg^{2+}\), \(Al^{3+}\), \(SO_4^{2−}\), and \(PO_4^{3−}\) have a greater tendency to form ion pairs because of their strong electrostatic interactions. The actual number of solvated ions present in a solution can be determined by measuring a colligative property at several solute concentrations. A 0.0500 M aqueous solution of \(\ce{FeCl3}\) has an osmotic pressure of 4.15 atm at 25°C. Calculate the van ’t Hoff factor \(i\) for the solution. : solute concentration, osmotic pressure, and temperature : van ’t Hoff factor : : A If \(\ce{FeCl_3}\) dissociated completely in aqueous solution, it would produce four ions per formula unit (one \(\ce{Fe^{3+}(aq)}\) and three \(\ce{Cl^{-}(aq)} ions) for an effective concentration of dissolved particles of 4 × 0.0500 M = 0.200 M. The osmotic pressure would be \[\begin{align*} \Pi &=MRT \\[4pt] &=(0.200 \;mol/L) \left[0.0821\;(L⋅atm)/(K⋅mol) \right] (298\; K)=4.89\; atm \end{align*}\] B The observed osmotic pressure is only 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is 4.15 atm/4.89 atm = 0.849, which indicates that the solution contains (0.849)(4) = 3.40 moles of particles per mole of \(\ce{FeCl_3}\) dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm: \[4.15\; atm=M \left( \dfrac{0.0821 \;(L⋅atm)}{(K⋅mol)}\right) (298 \;K) \nonumber\] or after rearranging \[M = 0.170 mol \nonumber\] The ratio of this value to the expected value of 0.200 M is 0.170 M/0.200 M = 0.850, which again gives us (0.850)(4) = 3.40 mole of particles per mole of \(\ce{FeCl_3}\) dissolved. From Equation \(\ref{13.5.13}\), the van ’t Hoff factor for the solution is \[i=\dfrac{\text{3.40 particles observed}}{\text{1 formula unit}\; \ce{FeCl_3}}=3.40 \nonumber\] Calculate the van ’t Hoff factor for a 0.050 m aqueous solution of \(\ce{MgCl2}\) that has a measured freezing point of −0.25°C. 2.7 (versus an ideal value of 3). The colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as . The boiling point elevation (\(ΔT_b\)) and freezing point depression (\(ΔT_f\)) of a solution are defined as the differences between the boiling and freezing points, respectively, of the solution and the pure solvent. Both are proportional to the molality of the solute. When a solution and a pure solvent are separated by a semipermeable membrane, a barrier that allows solvent molecules but not solute molecules to pass through, the flow of solvent in opposing directions is unequal and produces an osmotic pressure, which is the difference in pressure between the two sides of the membrane. Osmosis is the net flow of solvent through such a membrane due to different solute concentrations. Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through. In more concentrated solutions, or in solutions of salts with highly charged ions, the cations and anions can associate to form ion pairs, which decreases their effect on the colligative properties of the solution. The extent of ion pair formation is given by the van ’t Hoff factor (i), the ratio of the apparent number of particles in solution to the number predicted by the stoichiometry of the salt. | 37,922 | 1,173 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/02._Fundamental_Concepts_of_Quantum_Mechanics/Photons |
A photon is a tiny particle that comprises waves of . As shown by Maxwell, photons are just electric fields traveling through space. Photons have no charge, no resting mass, and travel at the speed of light. Photons are emitted by the action of , although they can be emitted by other methods including radioactive decay. Since they are extremely small particles, the contribution of wavelike characteristics to the behavior of photons is significant. In diagrams, individual photons are represented by a squiggly arrow. Photons are often described as energy packets. This is a very fitting analogy, as a photon contains energy that cannot be divided. This energy is stored as an oscillating electric field. These fields may oscillate at almost any frequency. Although they have never been observed, the longest theoretical wavelength of light is the size of the universe, and some theories predict the shortest possible at the Planck length. These packets of energy can be transmitted over vast distances with no decay in energy or speed. Photons travel at the speed of light, 2.997x10 m/s in empty space. The speed of a photon through space can be directly derived from the speed of an electric field through free space. Maxwell unveiled this proof in 1864. Even though photons have no mass, they have an observable momentum which follows the equation. The momentum of photons leads to interesting practical applications such as optical tweezers. Generally speaking, photons have similar properties to electromagnetic waves. Each photon has a wavelength and a frequency. The wavelength is defined as the distance between two peaks of the electric field with the same vector. The frequency of a photon is defined as how many wavelengths a photon propagates each second. Unlike an electromagnetic wave, a photon cannot actually be of a color. Instead, a photon will correspond to light of a given color. As color is defined by the capabilities of the human eye, a single photon cannot have color because it cannot be detected by the human eye. In order for the retina to detect and register light of a given color, several photons must act on it. Only when many photons act in unison on the retina, as an electromagnetic wave, can color be perceived. The most accurate descriptions we have about the nature of photons are given by Maxwell's equations. Maxwell's equations mathematically predict how photons move through space. Fundamentally, an electric field undergoing flux will create an orthogonal magnetic field. The flux of the magnetic field then recreates the electric field. The creation and destruction of each corresponding wave allows the wave pair to move through space at the speed of light. Maxwell's equations correctly describe the nature of individual photons within the framework of quantum dynamics. Photons can be generated in many different ways. This section will discuss some of the ways photons may be emitted. As photons are electric field propagating through space, the emission of photons requires the movement of charged particles. As a substance is heated, the atoms within it vibrate at higher energies. These vibrations rapidly change the shape and energies of electron orbitals. As the energy of the electrons changes, photons emitted and absorbed at energies corresponding to the energy of the change. is what causes light bulbs to glow, and the heat of an object to be felt from a great distance. The simplification of objects as blackbodies allows indirect temperature calculation of distant objects. Astronomers and kitchen infrared thermometers use this principle every day. Photons may be when electons fall from an excited state to a lower energy state(usually the ground state). The technical term for this drop in energy is a relaxation. Electrons undergoing this type of emission will produce a very distinctive set of photons based on the available energy levels of their environment. This set of possible photons is the basis for an emission spectrum. Florescence is special case of spontaneous emission. In florescence, the energy of a photon emitted does not match the energy used to excite the electron. An electron will fluoresce when it loses a considerable amount of energy to its surroundings before undergoing a relaxation. Generally florescence is employed in a laboratory setting to visualize the presence of target molecules. UV light is used to excite electrons, which then emit light at visible wavelengths that researchers can see. An excited electron can be artificially caused to relax to a lower energy state by a photon matching the difference between these energy states. The electric field's phase and orientation of the resultant photon, as well as its energy and direction will be identical to that of the incident photon. The light produced by stimulated emission is said to be coherent as it is similar in every way to the photon that caused it. Lasers produce coherent electromagnetic radiation by stimulated emission. Electrons with extremely high kinetic energy, such as those in particle accelerators, will produce high energy photons when their path is altered. This alteration is accomplished by a strong magnetic field. All free electrons will emit light in this manner, but synchrotron radiation has special practical implications. Synchrotron radiation is currently the best technology available for producing directional x-ray radiation at precise frequencies. Synchrotrons, such as the Advanced Light Source (ALS) at Lawrence Berkeley Labs and Stanford Synchrotron Radiation Light Source (SSRL) are hotbeds of x-ray spectroscopy due to the excellent quality of x-rays produced. Certain types of can involve the release of high energy photons. One such type of decay is a nuclear isomerization. In an isomerization, a nucleus rearranges itself to a more stable configuration and emits a gamma ray. While it is only theorized to occur, proton decay will also emit extremely high energy photons. Light incident on a metal plate may cause electrons to break loose from the plate surface (Fig. 1). This interaction between light and electrons is called the photoelectric effect. The photoelectric effect provided the first conclusive evidence that beams of light was made of quantized particles. The energy required to eject an electron from the surface of the metal is usually on the same order of magnitude as the ionization energy. As metals generally have ionization energies of several electron-volts, the photoelectric effect is generally observed using visible light or light of even higher energy. At the time this phenomenon was studied, light was thought to travel in waves. Contrary to what the wave model of light predicted, an increase in the intensity of light resulted in an increase in current, not an increase in the kinetic energy of the emitted electron. Einstein later explained this difference by showing that light was comprised of quantized packets of energy called photons. His work on the photoelectric effect earned him the Nobel Prize. The photoelectric effect has many practical applications, as current may be generated from a light source. Generally, the photoelectric effect is used as a component in switches that respond to light. Some examples are nightlights and photomultipliers. Usually the current is so small that it must be amplified in order to be an effective switch The energy of a photon is a discrete quantity determined by its frequency. This result can be determined experimentally by studying the photoelectric effect. The kinetic energy of an emitted electron varies directly with the frequency of the incident light. If the experimental values of these energies are fitted to a line, the slope of that line is Planck's constant. The point at which electrons begin to be emitted from the surface is called the threshold frequency, and is denoted by \(\nu_0\). The principle of conservation of energy dictates that the energy of a photon must all go somewhere. Assuming that the energy \(h\nu_0\) is the initial energy requirement to pry an electron from its orbital, the kinetic energy of a photon is equal to the kinetic energy of the emitted electron plus the ionization energy. Therefore the energy of a free photon becomes \(E = h\nu\) where nu is the frequency of the photon and h is Planck's constant. Fig. 2, Photoelectric effect results The results from a photoelectric experiment are shown in Figure 2. \(\nu_0\) is the minimum frequency at which electrons start to be detected. The solid lines represent the actual observed kinetic energies of released electrons. The dotted red line shows how a linear result can be obtained by tracing back to the y axis. Electrons cannot actually have negative kinetic energies. Whereas the double slit experiment initially indicated that a beam of light was a wave, more advanced experiments confirm the electron as a particle with wavelike properties. The diffraction of a beam of light though a double slit is observed to diffract producing constructive and destructive interference. Modern technology allows the emission and detection of single photons. In an experiment conducted by Philippe Grangier, a single photon is passed through a double slit. The photon then is detected on the other side of the slits. Across a large sample size, a trend in the final position of the photons can be determined. Under the wave model of light, an interference pattern will be observed as the photon splits over and over to produce a pattern. However, the results disagree with the wave model of light. Each photon emitted corresponds with a single detection on the other side of the slits(Fig. 3). With a certain probability, each photon is be detected at 100% strength. Over a series of measurements, photons produce the same interference pattern expected of a beam of photons. When one slit is closed, no interference pattern is observed and each photon travels in a linear path through the open slit. This interference has a profound implication which is that photons do not necessarily interact with each other to produce an interference pattern. Instead, they interact and interfere with . Furthermore, this shows that the electron does not pass through one slit or the other, but rather passes through both slits simultaneously. Richard Feynman's theory of quantum electrodynamics explains this phenomenon by asserting that a photon will travel not in a single path, but all possible paths in the universe. The interference between these paths will give the probability of the photon taking any given path, as the majority of the paths cancel with each other. He has used this theory to explain the nature of wide ranges of the actions of photons, such as reflection and refraction, with absolute precision. 1) The peak wavelength of a light bulb is 500 nm. Calculate the energy of a single photon at this wavelength. \(E = h\nu\) \(\nu = \dfrac{c}{\lambda}\) \(E = h*\dfrac{c}{\lambda}\) \( = 6.626x10^{-34}m^2kg/s^2*\dfrac{3.00x10^{8}m/s}{500x10^{-9}m}\) \( = 3.97x10^{-19}J\) 2) The work function of an metal surface is 9.4eV. What is the frequency of a photon which ejects an electron from this surface at 420km/s? \(h\nu_0 = 9.4eV x 1.6x10^{-19}J/eV = 1.51x10^{-18} J\) \(KE = \dfrac{1}{2}mv^2 = h\nu-h\nu_0\) \(\dfrac{1}{2} ( 9.11x10^{-31}kg)/{(420,000km/s)}^2 = 6.626x10^{-34}m^2kg/s*\nu - 1.51x10^{-18}J\) \(\nu = 2.28x10^{15}hz\) 3) A single photon passes through a double slit 20nm apart. A photomultiplier detects at least one particle in the 20 nm directly behind the slit. What fraction of the photon is detected here? The entire photon is detected. Protons are quantized particles. Although they can pass through both slits, it is still a single particle and will be detected accordingly. 4) A photon removes an electron from an atom. The kinetic energy of the exiting electron is found to be less than that of the photon that removed it. Why isn't the energy the same? Recall the photoelectric equation : \(KE = h\nu-h\nu_0\). This equation relates the energies of photons and electrons from an ejection. The second term of the equation, \(-h\nu_0\) is the amount of energy required to remove an electron from its orbital. The extra energy goes into breaking the association of an electron with a nucleus. Keep in mind that for a metal this is not the ionization energy due to the delocalization of electrons involved in metallic bonding. 5) Keeping in mind the relationship between the energy and frequency of light, design an experiment to determine if photons lose energy as they travel through space. One possible experiment utilizes the photoelectric effect. A light source is shone on a piece of metal, and the kinetic energy of ejected electrons is calculated. By shining the light at different distances from the metal plate, individual photons may be shown to undergo lossless transmission. The experiment will show that while the number of electrons ejected may decrease as a function of distance, their kinetic energy will remain the same. | 13,065 | 1,174 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.10%3A_Enthalpy_of_Fusion_and_Enthalpy_of_Vaporization/10.10.01%3A_Astronomy-_Water_on_Mars |
Mars, with an atmosphere of 95.32% CO and 0.03% water and temperatures (measured at 1.5 meters above the surface) ranging from + 1° F, ( -17.2° C) to -178° F (-107° C), is not at all Earth-like, but the two planets have a weather phenomenon in common: A " " is a rain or snow shower that evaporates before it reaches the surface, and it appears that virga snowstorms occur on Mars . We can use thermodynamic parameters developed on Earth to understand weather phenomena on Mars, and conditions there are right for snow to evaporate directly into gaseous water (a process called " "). There are also dry ice storms on Mars that are analogous to thunderstorms on Earth, where the precipitation is solid dry ice. Virgas on Earth can be disastrous to aviation and cause strong damaging winds on the ground because of their heat effects. When a rain virga occurs, the raindrops absorb heat from the air, creating a "microburst" of cold air that descends rapidly, and the heat removed by sublimation of snowshowers is much greater. Although there is some putative evidence for liquid water on Mars, most water exists as ice, mixed in with dry ice (frozen carbon dioxide). Solar radiation may raise equatorial soil temperature to +81°F (27° C) briefly, but the polar cap surface temperatures are always much colder. But even at the highest surface temperatues, most ice appears to sublime. This is because the mean surface level atmospheric pressure is only 600 Pa (0.6 kPa, 0.6% of Earth's 101.3 kPa). The solid carbon dioxide (dry ice) on the surface of Mars sublimes, just as dry ice sublimes on earth. We can understand these effects in the Earth and Martian atmospheres in terms of the and of water, described in more familiar terms: When heat energy is supplied to a solid, like ice, at a steady rate we find that the temperature climbs steadily until the melting point is reached and the first signs of liquid formation become evident. Thereafter, even though we are still supplying heat energy to the system, the temperature remains constant as long as both liquid and solid are present. Only when the last vestiges of the solid have disappeared does the temperature start to climb again. A large quantity of energy must be supplied to a solid in order to melt it. On a microscopic level melting involves separating molecules which attract each other. This requires an increase in the potential energy of the molecules, and the necessary energy is supplied by the heating coil. The kinetic energy of the molecules (rotation, vibration, and limited translation) remains constant during phase changes, because the temperature does not change. The heat energy which a solid absorbs when it melts is called the or heat of fusion and is usually quoted on a molar basis. (The word means the same thing as “melting.”) When 1 mol of ice, for example, is melted, we find from experiment that 6.01 kJ are needed. The molar enthalpy of fusion of ice is thus +6.01 kJ mol , and we can write \[\ce{H2O (s) -> H2O (l)} \nonumber \]
(0°C) ΔH = 6.01 kJ mol If we continue adding heat after the solid melts, the temperature again gradually increases until the liquid begins to boil. At this point, the temperature remains constant until the has been supplied. Once all the liquid has been converted to vapor, the temperature again rises. In the case of water the molar enthalpy of vaporization is 40.67 kJ mol . In other words \[\ce{H2O (l) -> H2O (g)} \nonumber \]
(100°C) ΔH = 40.67 kJ mol The enthalpies of fusion and vaporization both depend on temperature and pressure, as you can see in the table of selected molar enthalpies below. Solids like ice which have strong intermolecular forces have much higher values than those like CO with weak ones. Even on Earth Mars, CO sublimes rather than melting, then evaporating. If snow in a virga evaporates at, say 0°C, in removes heat from the surroundings equal to the sum of the heat of fusion and vaporization, which at 0°C is 45.051 + 6.007= 51.058 kJ/mol. \[\ce{H2O (s) -> H2O (g)} \nonumber \]
(0°C) ΔH = 51.058 kJ mol or 2.83 kJ/g While the temperature of the water does not change during sublimation, the surrounding atmosphere is cooled dramatically. This can be noted as snow on the ground disappears without melting as it sublimes, or as frozen clothes on a clothesline dry without melting. The enthalpy of sublimation for CO at atmospheric pressure, and −78.5 °C (−109.3°F). is \[\ce{CO2 (s) -> CO2 (g)} \nonumber \] ΔH = 25.2 kJ mol or 0.57 kJ/g. Note that much less heat is required to sublime CO ( ) than H O( ). \(\Page {1}\) Molar Enthalpies of Fusion and Vaporization of Selected Substances. *www1.lsbu.ac.uk/water/data.html Heat energy is absorbed when a liquid boils because molecules which are held together by mutual attraction in the liquid are jostled free of each other as the gas is formed. Such a separation requires energy. In general the energy needed differs from one liquid to another depending on the magnitude of the intermolecular forces. We can thus expect liquids with strong intermolecular forces to have larger enthalpies of vaporization. The list of enthalpies of vaporization given in the table bears this out. Two other features of the table deserve mention. One is the fact that the enthalpy of vaporization of a substance is always higher than its enthalpy of fusion. When a solid melts, the molecules are not separated from each other to nearly the same extent as when a liquid boils. Second, there is a close correlation between the enthalpy of vaporization and the boiling point measured on the thermodynamic scale of temperature. Periodic trends in boiling point closely follow periodic trends in heat of vaporiation. If we divide the one by the other, we find that the result is often in the range of 75 to 90 J K mol . To a first approximation therefore the . This interesting result is called . An equivalent rule does not hold for fusion. The energy required to melt a solid and the temperature at which this occurs depend on the structure of the crystal as well as on the magnitude of the intermolecular forces. From ChemPRIME: 10.9: Enthalpy of Fusion and Enthalpy of Vaporization | 6,209 | 1,175 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Vibrational_Spectroscopy/Raman_Spectroscopy/Raman%3A_Theory |
The phenomenon of Raman scattering of light was first postulated by Smekai in 1923 and first observed experimentally in 1928 by Raman and Krishnan. Raman scattering is most easily seen as the change in frequency for a small percentage of the intensity in a monochromatic beam as the result of coupling between the incident radiation and vibrational energy levels of molecules. A vibrational mode will be Raman active only when it changes the polariazbility of the molecule. \[ \Delta E=hc\tilde{\nu} _{M} \] \[ E_2-E_1=hc\tilde{\nu}_M \] According to the classical theory of electromagnetic radiation, electric and magnetic fields oscillating at a given frequency are able to give out electromagnetic radiation of the same frequency. One could use electromagnetic radiation theory to explain light scattering phenomena. For a majority of systems, only an induced electric dipole moment is taken into consideration. This dipole moment which is induced by the electric field could be expressed by the power series \[ \mathbf{\mu} =\mathbf{\mu}^{\left (1 \right )}+\mathbf{\mu}^{\left (2 \right )}+\mathbf{\mu}^{\left (3 \right )}+\cdots\] where \[ \mathbf{\mu}^{\left (1 \right )}=\mathbf{\alpha} \cdot \mathit{\mathbf{E}}\] \( \mathbf{\mu}^{\left (2 \right )}=\frac{1}{2}\mathbf{\beta} \cdot \mathit{\mathbf{E}}\mathit{\mathbf{E}}\) \( \mathbf{\mu}^{\left (3 \right )}=\frac{1}{6}\mathbf{\gamma } \cdot\mathit{\mathbf{E}}\mathit{\mathbf{E}}\mathit{\mathbf{E}}\) is termed the polarizability tensor. It is a second-rank tensor with all the components in the unit of CV m . Typically, orders of magnitude for components in , , and are as follows, , 10 CV m ; , 10 CV m ; and , 10 CV m . According to the values, the contributions of and are quite small unless electric field is very high. Since Rayleigh and Raman scattering are observed quite readily with very much lower electric field intensities, one may expect to explain Rayleigh and Raman scattering in terms of only. We shall now consider the interaction of a molecular system with the harmonically oscillating electric field in the frequency . To make the explanation easily, we shall ignore the rotation but just consider the vibration part. It is to be expected that the polarizability will be a function of the nuclear coordinates. The variation of components in polarizability tensor with vibrational coordinates is expressed in a Taylor series \( \alpha_{ij} =\left (\alpha_{ij} \right )_{0} +\sum_{k} \left (\frac{\partial \alpha_{ij} }{\partial Q_{k}} \right )_{0}Q_{k}+\frac{1}{2}\sum_{k,l}\left (\frac{\partial^2 \alpha_{ij}}{\partial Q_{k}\partial Q_{l}} \right )_{0}Q_{k}Q_{l}+\cdots \) where (α ) is the α value at the equilibrium configuration, , are normal coordinates of vibration at frequencies , . We shall make a harmonic approximation to neglect the terms which involve powers of higher than first. After initially fixing our attention on one normal mode, , we could get \( \alpha_{ij} =\left (\alpha_{ij} \right )_{0} +\left (\frac{\partial \alpha_{ij} }{\partial Q_{k}} \right )_{0}Q_{k}\) As for a harmonic vibration, \[ Q_{k}=Q_{k0} \cos \left ( \omega_k t + \delta _k \right )\] Then we could get the expression of tensor resulting from k-th vibration, \( \mathbf{\alpha}_{k}=\mathbf{\alpha }_{0}+\left (\frac{\partial \mathbf{\alpha}_{k}}{\partial Q_{k}} \right )_{0}Q_{k0}\cos\left (\omega_{k} t+\delta _{k} \right )\) Now, under the influence of electromagnetic radiation at frequency , the induced electric dipole moment is expressed, \[ \mathbf{\mu} ^{\left (1 \right )}=\mathbf{\alpha} _{k}\cdot \mathbf{E}_{0}\cos\omega_{0} t=\mathbf{\alpha} _{0}\cdot \mathbf{E}_{0}\cos\omega_{0} t+\left (\frac{\partial \mathbf{\alpha} _{k}}{\partial Q_{k}} \right )_{0}\cdot \mathbf{E}_{0}Q_{k0}\cos\omega_{0} t\cos\left (\omega_{k} t+\delta _{k} \right )t =\] \[\mathbf{\alpha}_{0}\cdot \mathbf{E}_{0}\cos\omega_{0} t+\dfrac{1}{2}\left (\frac{\partial \mathbf{\alpha}_{k}}{\partial Q_{k}} \right )_{0}\cdot \mathbf{E}_{0}Q_{k0}\cos\left (\omega_{0} t+\omega_{k} t+\delta_k \right )+\dfrac{1}{2}\left (\dfrac{\partial \mathbf{\alpha}_{k}}{\partial Q_{k}} \right )_{0}\cdot \mathbf{E}_0 Q_{k0}\cos\left (\omega_{0}t-\omega_{k} t-\delta_{k} \right )\] We see that the linear induced dipole moment has three components with different frequencies, \(\mathbf{\alpha} _{0}\cdot \mathbf{E}_{0}\cos\omega_{0} t\) which gives rise to radiation at and accounts for the Rayleigh scattering; \[\frac{1}{2}\left(\frac{\partial \mathbf{\alpha}_{k}}{\partial Q_{k}}\right)_{0}\cdot\mathbf{E}_{0}Q_{k0}\cos\left(\omega_{0}t\omega_{k}t\delta_{k}\right)\] which gives rise to radiation at - and accounts for the Stokes Raman scattering. From these mathematical manipulations, there emerges a useful qualitative picture of the mechanisms of Rayleigh and Raman scattering in terms of classical radiation theory. Rayleigh scattering comes from the dipole oscillating at induced in the molecule by the electric field of the incident radiation at frequency . Raman scattering arises from the dipole moment oscillating at ± produced by the modulation of dipole oscillating at with molecular vibration at frequency . In other words, the frequencies we observe in Raman scattering are beat frequencies of the radiation frequency and the molecular vibrational frequency . According to the quantum theory, radiation is emitted or absorbed as a result of a system making a downward or upward transition between two discrete energy levels. A quantum theory of spectroscopic processes should, therefore, treat the radiation and molecule together as a complete system, and explore how energy is transferred between the radiation and the molecule as a result of their interaction. A transition between energy levels of the molecular systems takes place with the emission or absorption of radiation, provided a transition moment associated with the initial and final molecular states is non-zero. The transition moment could be defined as \( \mathbf{M}_{fi}=\left \langle \Psi _{f}\mid \mathbf{\mu} \mid \Psi _{i}\right \rangle\) in the Dirac bracket notation, where and are the wave function of initial and final states, respectively, and is the dipole moment operator. As we have discussed in the classical part, the linear induced dipole moment could be expressed by \[ \mathbf{\mu}^{\left (1 \right )}=\mathbf{\alpha} \cdot \mathit{\mathbf{E}}\] Therefore, in quantum mechanical treatment, if a transition from an initial state to a final state is induced by incident radiation at frequency , the transition moment is given by \[ \mathbf{\mu} _{fi}^{\left (1 \right )}=\left \langle \Psi _{f}\mid \mathbf{\alpha } \mid \Psi _{i}\right \rangle\cdot \mathbf{E}\] Now we will examine in more detail the nature of a typical matrix element of the polarizability tensor, like [ ] , for Raman scattering. Just like in the classical theory, we will ignore the rotational wave function and consider the vibrational part only. \[ \left [\alpha _{xy} \right ]_{fi}=\left \langle \Phi _{f}\mid \alpha_{xy} \mid \Phi _{i}\right \rangle\] where is the vibrational wave function. In classical theory, \[ \alpha_{xy} =\left (\alpha_{xy} \right )_{0} +\sum_{k} \left (\frac{\partial \alpha_{xy} }{\partial Q_{k}} \right )_{0}Q_{k}+\frac{1}{2}\sum_{k,l}\left (\frac{\partial^2 \alpha_{ij}}{\partial Q_{k}\partial Q_{l}} \right )_{0}Q_{k}Q_{l}+\cdots \] Introducing the quantum part, we obtain (just consider the first order term) \( \left [\alpha _{xy} \right ]_{fi}=\left (\alpha _{xy} \right )_{0}\left \langle \Phi _{f}\mid \Phi _{i}\right \rangle+\sum_{k}\left (\frac{\partial \alpha _{xy}}{\partial Q_{k}} \right )_{0}\left \langle \Phi _{f}\mid Q_{k}\mid \Phi _{i}\right \rangle\) In the harmonic oscillator model, the total vibrational wave function is product of the harmonic oscillator wave functions for each of the normal modes of vibration. Thus, for , \[ \Phi _{i}=\prod_{k}\Phi _{v_{k}^{i}}\left (Q_{k} \right )\] where and \[ \left \langle \Phi _{v_{k}^{f}}\left (Q_{k} \right )\mid Q_{k}\mid \Phi _{v_{k}^{i}}\left (Q_{k} \right )\right \rangle=\left\{\begin{matrix} 0 \; \; for \; v_{k}^{f}= v_{k}^{i}\\ \left (v_{k}^{i}+1 \right )^{\frac{1}{2}}b_{v_{k}}\; \; for \; v_{k}^{f}= v_{k}^{i}+1 \\ \left (v_{k}^{i} \right )^{\frac{1}{2}}b_{v_{k}}\; \; for \; v_{k}^{f}= v_{k}^{i}-1 \end{matrix}\right.\] where \[ b_{v_{k}}=\sqrt{\frac{h}{8\pi^{2} v_{k}}}\] We are now able to find out the conditions which have to be satisfied if the transition moment is non-zero. We will consider the zero order term which accounts for the Rayleigh scattering part first. This term is none-zero only if \(v_{k}^{f}=v_{k}^{i}\) which means none of the vibrational quantum numbers change during this transition from initial state to final state . Thus, for Rayleigh scattering, the quantum mechanical treatment and the classical theory give the same results. Then we could go to the first order term \(\sum_{k}\left (\frac{\partial \alpha _{xy}}{\partial Q_{k}} \right )_{0}\left \langle \Phi _{f}\mid Q_{k}\mid \Phi _{i}\right \rangle\) which accounts for the Raman scattering part. For the -th summand, it is zero unless every term in the product is non-zero, and to achieve this the following conditions must be satisfied, for all modes except the -th: the vibrational quantum numbers must be the same during the transition, i.e. \(v_{j}^{f}=v_{j}^{i}\) where j≠k; and, for the -th mode, and the vibrational quantum number must change by one unit, i.e. \(v_{k}^{f}=v_{k}^{i}\pm 1\). The transition moment is associated with Stokes Raman scattering for Δ =1, and with anti-Stokes Raman scattering for Δ =-1. These conditions are a result of the properties of harmonic oscillator wave functions. It follows from these arguments that, in the harmonic approximation, only vibrational fundamentals, i.e., transitions with only one vibrational quantum number changes by one unit, can be observed in the Raman scattering. However, this Δ =±1 restriction is only a necessary but not a sufficient condition for the occurrence of Raman scattering at the -th vibrational mode. The vibrational mode should be Raman active, i.e., at least one of the elements of the derived polarizability tensor should be non-zero. It can be rigorously established by group theory that the elements of the derived polarizability will be non-zero only if they have the same symmetry with the second order terms, i.e., x , y , z , xy, yz, xz. In other words, the irreducible representation of a certain vibrational mode should have a basis in x , y , z , xy, yz or xz. Finally, we could establish a much better basis for determining selection rules for vibrational transitions in the Raman effect, if we consider the properties of the vibrational transition polarizability components, rather than the derived prolarizability tensor components. For fundamental vibrational transitions, where in the initial state all vibrational quantum numbers are zero and in the final state only the -th vibrational quantum number has changed to unity, \( \left [\alpha _{xy} \right ]_{fi}=\left \langle \Phi _{1}\mid \alpha_{xy} \mid \Phi _{0}\right \rangle\) According to the group theory, this integral will be non-zero, only if and ( ) belong to the same symmetry species, which implies that, under each symmetry operation of the molecule in question, and ( ) transform in the same way. This constitutes a general selection rule for the Raman activity of a fundamental transition. In its most general way, covering all types of transitions, the selection rule is as follows: a transition between two states, and , is Raman forbidden unless at least one of the triple products of the type belongs to a representation whose structure contains the totally symmetric species. | 11,844 | 1,177 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Competition_between_substitution_and_elimination |
Having discussed the many factors that influence nucleophilic substitution and elimination reactions of alkyl halides, we must now consider the practical problem of predicting the most likely outcome when a given alkyl halide is reacted with a given nucleophile. As we noted earlier, several variables must be considered, .In general, in order for an SN1 or E1 reaction to occur, the relevant carbocation intermediate must be relatively stable. Strong nucleophile favor substitution, and strong bases, especially strong hindered bases (such as tert-butoxide) favor elimination. The nature of the halogen substituent on the alkyl halide is usually not very significant if it is Cl, Br or I. In cases where both S 2 and E2 reactions compete, chlorides generally give more elimination than do iodides, since the greater electronegativity of chlorine increases the acidity of beta-hydrogens. Indeed, although alkyl fluorides are relatively unreactive, when reactions with basic nucleophiles are forced, elimination occurs (note the high electronegativity of fluorine). The following table summarizes the expected outcome of alkyl halide reactions with nucleophiles. It is assumed that the alkyl halides have one or more beta-hydrogens, making elimination possible; and that low dielectric solvents (e.g. acetone, ethanol, tetrahydrofuran & ethyl acetate) are used. When a high dielectric solvent would significantly influence the reaction this is noted in red. . Nucleophile
( Weak Bases: I , Br , SCN , N ,
CH CO , RS , CN etc. )
( Strong Bases: HO , RO )
( H O, ROH, RSH, R N ) Alkyl Group ), | 1,668 | 1,178 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/21%3A_Spectra_and_Structure_of_Atoms_and_Molecules/21.06%3A_The_Visible_and_Ultraviolet_Spectra_of_Molecules-_Molecular_Orbitals |
When molecules absorb or emit in the ultraviolet and visible regions of the spectrum, this almost always corresponds to the transition of an from a low-energy to a high-energy orbital, or vice versa. One might expect the spectra of molecules to be like the atomic line spectra shown in Figure \(\Page {1}\), but in fact molecular spectra are very different. Consider, for example, the absorption spectrum of the rather beautiful purple-violet gas I . This molecule strongly absorbs photons whose wavelengths are between 440 and 600 nm, and much of the orange, yellow, and green components of white light are removed. The light which passes through a sample of I is mainly blue and red. When analyzed with an average quality spectroscope, this light gives the spectrum shown in Figure \(\Page {2}\) . Instead of the few discrete lines typical of atoms, we now have a broad, apparently continuous, absorption band. This is typical of molecules. Why is there this difference between atomic and molecular spectra? An answer begins to appear if we use a somewhat more expensive spectroscope. Figure \(\Page {2b}\) shows a tracing of the I spectrum made with such an instrument. What originally appeared to he a continuous band is now shown to consist of a very large number of very narrow, closely spaced lines. Thus the broad absorption band of I is actually made up of discrete lines. The reason molecules give rise to such an enormous number of lines is that molecules can vibrate and rotate in a very large number of ways while atoms cannot. Furthermore both rotational levels and vibrational motion are quantized. When a molecule absorbs a photon of light and an electron is excited to a higher orbital, the molecule will not be stationary either before or after the absorption of the photon. Because of the large number of energy possibilities both before and after the transition, a very large number of lines of slightly different wavelengths is obtained. A careful analysis of these lines yields much valuable information about the way in which the molecule rotates and vibrates. In particular, very accurate values of and bond lengths can be obtained from a study of the fine structure of an absorption band like that shown in Figure \(\Page {2b}\). | 2,278 | 1,179 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/10%3A_Solids_Liquids_and_Phase_Transitions/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 | 1,180 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Molecular_Orbital_Theory/Molecular_Orbitals_of_Li_to_F |
The molecular orbital theory (MO) has been introduced for the diatomic hydrogen molecules. The same method can be applied to other diatomic molecules, but involving more than the 1 atomic orbitals. For the second period elements, the 2 and 2 orbitals are important for MO considerations. A linear combination of properly oriented atomic orbitals for the formation of sigma and pi bonds. The formation of bonds from the linear combination of atomic orbitals is the same as that of the valence bond theory. For simplicity, we are not going into the details of the theory, but simply show you how to construct the MO energy level diagram. In the discussion of electronic configurations of many-electron atoms, the variation of energy levels of the atomic orbitals was given. All the corresponding levels become more negative as the atomic number increases. The energy levels and of the second period are given in the table on the right. The energy level ranges from -521 to -4680 kJ mol for these elements. The energy levels also become more negative, but the decrease (because they are negative) is not as rapid as that of the levels. Thus, the differences - increase as the atomic numbers increase. A qualitative diagram showing the changes of energy levels of atomic orbitals is given below: The 2 and 2 energy levels of \(\ce{O}\) and \(\ce{F}\) are very far apart. The combination of the 2 orbitals from the two atoms form a sigma bonding and sigma antibonding orbitals in a way very similar to the case of the hydrogen molecules, because the 2 orbitals have little to do with the 2 orbitals. On the other hand, the three 2 orbitals of each \(\ce{O}\) (or \(\ce{F}\)) atom can form one sigma and two pi bonds and their corresponding antibonding molecular orbitals. The interaction of the 2 orbitals for the sigma bond is stronger, and the levels of sigma and anti sigma bonds are farther apart than those of pi and anti pi bonds. Thus, the relative energy level diagram of \(\ce{O2}\) and \(\ce{F2}\) has the following arrangement: The electronic configuration for \(\ce{O2}\) is: s s* s p p* This electronic configuration indicates a bond order of 2, and the bond can be represented by \(\ce{O=O}\). There is no net bonding from the orbitals, because the number of bonding electrons equals the number of antibonding electrons. The two electons in * cancel two of the 6 bonding electrons ( ). Therefore, there are 4 total bonding electrons. The two electrons in the * orbitals have the same spin, and they are responsible for the paramagnetism of oxygen. As an exercise, as well as the conclusion regarding the \(\ce{F2}\) molecule. The electronic configuration for \(\ce{F2}\) is: s s* s p p* This electronic configuration shows a single \(\ce{F-F}\) bond in the molecule for the reasons given for the \(\ce{O2}\) molecule. The bond lengths and bond energies of \(\ce{O2}\) and \(\ce{F2}\) (shown on the right) correspond to \(\ce{O=O}\) and \(\ce{F-F}\) respectively. The bond energy is higher for \(\ce{O=O}\) than for \(\ce{F-F}\) due to the double \(\ce{O=O}\) bond, and its \(\ce{O=O}\) bond length is shorter than that of \(\ce{F-F}\). Recently, the study of the energies of electrons in molecules revealed that the relative energy levels of molecular orbitals of \(\ce{Li2}\) to \(\ce{N2}\) are different from those of \(\ce{O2}\) and \(\ce{F2}\). The explanation for the difference comes from the consideration of hybrid atomic orbitals. Because the 2 energy levels and 2 energy levels for \(\ce{Li}\) to \(\ce{N}\) are relatively close, the 2 orbitals are influenced by the 2 orbitals. This influence makes the bonding orbitals stronger than, and the antibonding orbitals weaker than, those formed by pure 2 orbitals. This process is called . Due to mixing, the orbital is weakened, and the * is also affected. These effects cause the relative order to change, and the typical relative energy levels for \(\ce{Li2}\), \(\ce{Be2}\), \(\ce{B2}\), \(\ce{C2}\) and \(\ce{N2}\) to have the following diagram: The electronic configurations agree with the experimental bond lengths and bond energies of homonuclear diatomic molecules of second-period elements. They are given in a table below. The argument regarding bond lengths, bond orders, and bond energies given for \(\ce{O2}\) and \(\ce{F2}\) above applies to all these molecules. Note also that \(\ce{B2}\) and \(\ce{O2}\) are paramagnetic due to the unpaired electrons in the molecular orbitals. Other molecules in this group are diamagnetic. | 4,580 | 1,181 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/20%3A_Carbohydrates/20.11%3A_The_Generation_of_Energy_from_Carbohydrate_Metabolism |
We will consider here the reverse process of photosynthesis, namely how carbohydrates, especially glucose, are converted to energy by being broken down into carbon dioxide and water. A general summary of the several stages involved is shown in Figure 20-8. Initially, the storage fuels or foodstuffs (fats, carbohydrates, and proteins) are hydrolyzed into smaller components (fatty acids and glycerol, glucose and other simple sugars, and amino acids). In the next stage, these simple fuels are degraded further to two-carbon fragments that are delivered as the \(\ce{CH_3C=O}\) group (ethanoyl, or acetyl) in the form of the thioester of coenzyme A, \(\ce{CH_3COS}\) . The structure of this compound and the manner in which fatty acids are degraded has been considered in , and amino acid metabolism is discussed briefly in . This section is concerned mainly with the pathway by which glucose is metabolized by the process known as . In the conversion of glucose to \(\ce{CH_3COS}\) , two carbons are oxidized to carbon dioxide with consumption of two oxygen molecules: \[\ce{C_6H_{12}O_6} + 2 \textbf{CoA} \ce{SH} + \left[ 2 \ce{O_2} \right] \rightarrow 2 \ce{CH_3COS} \textbf{CoA} + 4 \ce{H_2O} + 2 \ce{CO_2} \tag{20-5}\] For further oxidation to occur, the \(\ce{CH_3COS}\) must enter the next stage of metabolism, whereby the \(\ce{CH_3C=O}\) group is converted to \(\ce{CO_2}\) and \(\ce{H_2O}\). This stage is known variously as the citric acid cycle, the tricarboxylic acid cycle, or the Krebs cycle, in honor of H. A. Krebs (Nobel Prize, 1953), who first recognized its cyclic nature in 1937. We can write an equation for the process as if it involved oxygen: \[2 \ce{CH_3COS} \textbf{CoA} + \left[ 4 \ce{O_2} \right] \rightarrow 4 \ce{CO_2} + 2 \ce{H_2O} + 2 \textbf{CoA} \ce{SH} \tag{20-6}\] Notice that combination of the reactions of Equations 20-5 and 20-6, glycolysis plus the citric acid cycle, oxidizes glucose completely to \(\ce{CO_2}\) and \(\ce{H_2O}\): \[\ce{C_6H_{12}O_6} + \left[ 6 \ce{O_2} \right] \rightarrow 6 \ce{CO_2} + 6 \ce{H_2O} \tag{20-7}\] But, as you will see, none of the steps uses molecular oxygen directly. Hence there must be a stage in metabolism whereby molecular oxygen is linked to production of oxidizing agents that are consumed in glycolysis and in the citric acid cycle. The coupling of oxygen into the metabolism of carbohydrates is an extremely complex process involving transport of the oxygen to the cells by an oxygen carrier such as hemoglobin, myoglobin, or hemocyanin. This is followed by a series of reactions, among which \(\ce{NADH}\) is converted to \(\ce{NAD}^\oplus\) with associated formation of three moles of ATP from three moles of ADP and inorganic phosphate. Another electron-carrier is flavin adenine dinucleotide (\(\ce{FAD}\); ), which is reduced to \(\ce{FADH_2}\) with an associated production of two moles of ATP from two moles of ADP. These processes are known as and can be expressed by the equations: Oxidative phosphorylation resembles photophosphorylation, discussed in , in that electron transport in photosynthesis also is coupled with ATP formation. By suitably juggling Equations 20-7 through 20-9, we find that the metabolic oxidation of one mole of glucose is achieved by ten moles of \(\ce{NAD}^\oplus\) and two moles of \(\ce{FAD}\): The overall result is production of 36 moles of ATP from ADP and phosphate per mole of glucose oxidized to \(\ce{CO_2}\) and \(\ce{H_2O}\). Of these, 34 ATPs are produced according to Equation 20-10 and, as we shall see, two more come from glycolysis. Glycolysis is the sequence of steps that converts glucose into two \(\ce{C_3}\) fragments with the production of ATP. The \(\ce{C_3}\) product of interest here is 2-oxopropanoate (pyruvate): There are features in this conversion that closely resemble the dark reactions of photosynthesis, which build a \(\ce{C_6}\) chain (fructose) from \(\ce{C_3}\) chains ( ). For example, the reactants are either phosphate esters or mixed anhydrides, and the phosphorylating agent is ATP: \[\ce{ROH} + \text{ATP} \rightarrow \ce{RO-PO_3^2-} + \text{ADP} + \ce{H}^\oplus\] Furthermore, rearrangements occur that interconvert an aldose and ketose, and the cleavage of a \(\ce{C_6}\) chain into two \(\ce{C_3}\) chains is achieved by a reverse aldol condensation: Also, oxidation of an aldehyde to an acid is accomplished with \(\ce{NAD}^\oplus\). There is a related reaction in photosynthesis ( ) that accomplishes the reduction of an acid to an aldehyde and is specific for \(\ce{NADPH}\), not \(\ce{NADH}\): First, glucose is phosphorylated to glucose 6-phosphate with ATP. Then an aldose \(\rightleftharpoons\) ketose rearrangement converts glucose 6-phosphate into fructose 6-phosphate. A second phosphorylation with ATP gives fructose 1,6-diphosphate: At this stage the enzyme aldolase catalyzes the aldol cleavage of fructose 1,6-diphosphate. One product is glyceraldehyde 3-phosphate and the other is 1,3-dihydroxypropanone phosphate. Another ketose \(\rightleftharpoons\) aldose equilibrium converts the propanone into the glyceraldehyde derivative: The next step oxidizes glyceraldehyde 3-phosphate with \(\ce{NAD}^\oplus\) in the presence of phosphate with the formation of 1,3-diphosphoglycerate: The mixed anhydride of phosphoric acid and glyceric acid then is used to convert ADP to ATP and form 3-phosphoglycerate. Thereafter the sequence differs from that in photosynthesis. The next few steps accomplish the formation of pyruvate by transfer of the phosphoryl group from \(\ce{C_3}\) to \(\ce{C_2}\) followed by dehydration to phosphoenolpyruvate. Phosphoenolpyruvate is an effective phosphorylating agent that converts ADP to ATP and forms pyruvate: The net reaction at this point produces more ATP than is consumed in the phosphorylation of glucose and fructose. What happens thereafter depends on the organism. With yeast and certain other microorganisms, pyruvate is decarboxylated and reduced to ethanol. The end result of glycolysis in this instance is . In higher organisms, pyruvate can be stored temporarily as a reduction product (lactate) or it can be oxidized further to give \(\ce{CH_3COS}\) and \(\ce{CO_2}\). The \(\ce{CH_3COS}\) then enters the citric acid cycle to be oxidized to \(\ce{CO_2}\) and \(\ce{H_2O}\), as discussed in the next section: Glycolysis to the pyruvate or lactate stage liberates heat, which can help keep the organism warm and produce ATP from ADP for future conversion into energy. However, glycolysis does not directly involve oxygen and does not liberate \(\ce{CO_2}\), as we might expect from the overall process of the metabolic conversion of glucose to carbon dioxide and water (Equation 20-10). The liberation of \(\ce{CO_2}\) occurs subsequent to pyruvate formation in a process called variously, the citric acid cycle, the Krebs cycle, or the tricarboxylic acid (TCA) cycle. The initial step, which is not really part of the cycle, is conversion of pyruvate to ethanoyl (acetyl ): To achieve the oxidation of acetyl on a continuing basis, intermediates consumed in certain steps must be regenerated in others. Thus we have a situation similar to that in the Calvin cycle ( ), whereby the first stage of the cycle achieved the desired reaction (\(\ce{CO_2}\) formation) and the second stage is designed to regenerate intermediates necessary to perpetuate the cycle. The entry point is the reaction between acetyl and a four-carbon unit, 2-oxobutanedioic acid. An aldol-type addition of the \(\ce{CH_3CO}\) group to this \(\ce{C_4}\) keto acid extends the chain to a branched \(\ce{C_6}\) acid (as citric acid): Dehydration-rehydration of citrate converts it to isocitrate: From here, oxidation of the hydroxyl function with \(\ce{NAD}^\oplus\) gives a keto acid, which loses \(\ce{CO_2}\) readily ( ) and affords 2-oxopentanedioate: We now have a \(\ce{C_5}\) keto acid that can be oxidized in the same way as the \(\ce{C_3}\) keto acid, pyruvic acid, to give a butanedioyl : Two molecules of \(\ce{CO_2}\) now have been produced and the remaining part of the citric acid cycle is concerned with regeneration of the for forming acetyl from 2-oxopropanoate, and also with regenerating the 2-oxobutanedioate, which is the precursor of citrate. The steps involved are The hydrolysis of the acyl in the first step is used for energy storage by conversion of guanosine diphosphate (GDP) to guanosine triphosphate (GTP): The hydration of the -butenedioate ( ) and the final oxidation reaction ( ) have been discussed previously. There is an alternative route, called the , by which glucose enters the glycolytic sequence to pyruvate. This route achieves the oxidative decarboxylation of glucose to give ribose, as the 5-phosphate ester. The essential steps are The net result is that three pentoses are converted into two molecules of fructose and one of glyceraldehyde \(\left( 3 \ce{C_5} \rightarrow 2 \ce{C_6} + \ce{C_3} \right)\). The relationship of the pentose-phosphate pathway to glycolysis is shown in Figure 20-11. The steps involved in the pentose shunt are readily reversible, but there are several steps in glycolysis that are not. These are the phosphorylation steps (Figure 20-9). Yet, there has to be a return route from pyruvate to glucose. This route is called and, in animals, takes place in the liver. We shall not discuss the steps in gluconeogenesis except to indicate again that they are not all the reverse of glycolysis. For comparison, the steps that differ are indicated in Figure 20-9 by dashed lines. Why is lactate formed from pyruvate in the metabolism of glucose? Pyruvate \( + \: \ce{NADH} + \ce{H}^\oplus \rightarrow\) lactate \(+ \: \ce{NAD}^\oplus\) is a dead-end path, but it does furnish the \(\ce{NAD}^\oplus\) needed for glycolysis in active muscle. This route for forming \(\ce{NAD}^\oplus\) is important, because in circumstances of physical exertion, the rate of production of \(\ce{NAD}^\oplus\) from oxidative phosphorylation may be slower than the demand for \(\ce{NAD}^\oplus\), in which case a temporary supply is available from the pyruvate \(\rightarrow\) lactate reduction. The lactate so formed builds up in muscle tissue under conditions of physical exertion and is apt to cause muscles to "cramp". The excess lactate so formed ultimately is removed by being converted back to pyruvate by oxidation with \(\ce{NAD}^\oplus\). The beauty of the metabolic cycle through pyruvate, shown in summary in Figure 20-11, is the way it can be tapped at various points according to whether the organism requires ATP (from glycolysis), \(\ce{NADH}\) (from pentose shunt), or \(\ce{NAD}^\oplus\) (from the lactate siding). and (1977) | 10,736 | 1,184 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Examples/Examples_of_Catalysis/5._Examples_of_Other_Catalytic_Reactions_in_Organic_Chemistry |
This page looks a few odds and ends of examples of catalysts used in organic chemistry. It includes the formation of epoxyethane from ethene, and several reactions from benzene chemistry - Friedel-Crafts reactions and halogenation. Epoxyethane is manufactured by reacting ethene with a limited amount of oxygen in the presence of a silver catalyst at a temperature of about 250 - 300°C and a pressure of less than 15 atmospheres. Because the solid silver is catalysing a gas reaction, this is an example of heterogeneous catalysis. The reaction is exothermic and the temperature has to be carefully controlled to prevent further oxidation of the ethene to carbon dioxide and water. Benzene reacts with chlorine or bromine in the presence of a catalyst. The catalyst is either aluminium chloride (or aluminium bromide if you are reacting benzene with bromine) or iron. Strictly speaking iron isn't a catalyst, because it gets permanently changed during the reaction. It reacts with some of the chlorine or bromine to form iron(III) chloride, FeCl , or iron(III) bromide, FeBr . These compounds act as the catalyst and behave exactly like aluminium chloride in these reactions. The reaction between benzene and chlorine in the presence of either aluminium chloride or iron gives chlorobenzene. or: The reaction between benzene and bromine in the presence of either aluminium bromide or iron gives bromobenzene. Iron is usually used because it is cheaper and more readily available. or: Alkylation involves replacing a hydrogen atom on a benzene ring by an alkyl group like methyl or ethyl. This is another example of the use of aluminium chloride as a catalyst. Benzene is treated with a chloroalkane (for example, chloromethane or chloroethane) in the presence of aluminium chloride as a catalyst. The equation shows the reaction using a methyl group, but any other alkyl group could be used in the same way. Substituting a methyl group gives methylbenzene - once known as toluene. or: An acyl group is an alkyl group attached to a carbon-oxygen double bond. Acylation means substituting an acyl group into something - in this case, into a benzene ring. The most commonly used acyl group is CH CO-. This is called the ethanoyl group. In the example which follows we are substituting a CH CO- group into the ring, but you could equally well use any other alkyl group instead of the CH . The most reactive substance containing an acyl group is an acyl chloride (also known as an acid chloride). Benzene is treated with a mixture of ethanoyl chloride, CH COCl, and aluminium chloride as the catalyst. A ketone called phenylethanone is formed. or: | 2,654 | 1,185 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Statistical_Mechanics/Boltzmann_Average/Proof_that_%CE%B2_%3D_1%2F%2FkT |
The pressure in the state \(j\) is given by \(p_j = - (\partial E_j/\partial V)\). The average energy is \[\bar{E}=\frac{\displaystyle \sum_{J} E_{j}(N, V) e^{-\beta E_{f}(N, n)}}{\displaystyle \sum_{j} e^{-\beta E / N, n}} \label{I}\] The average pressure is \[\bar{p}=\frac{\displaystyle \sum p_{j}(N, V) e^{-\beta E_{f}(N, n)}}{\displaystyle \sum_{j} e^{-\beta E_{f}(N, n)}} \label{II}\] According the Gibbs postulate the average energy, average pressure and other average mechanical properties calculated using the partition function are equal to their thermodynamic counterparts. Note that some authors use \(\bar{E}\) and \(\bar{p}\) bar for the average quantities and elsewhere the angle bracket notation is used. These are equivalent notations. If we differentiate the expression for the average energy we can treat the denominator, \(Q\) as a function of \(V\) as well since it represents a sum over \(\exp(-\beta E_j(N,V)\). Since \(E_j\) appears both in the exponent and as a function multiplying the exponent we have \[\left(\frac{\partial E}{\partial V}\right)_{N, \beta}=\frac{\displaystyle \sum\left(\frac{\partial E_{j}}{\partial V}\right) e^{-\beta E / N, n}}{Q}-\frac{\displaystyle \sum-\beta\left(\frac{\partial E_{j}}{\partial V}\right) E_{f} e^{-\beta E / N, n}}{Q} -\frac{\displaystyle \sum_{j} E_{j}^{-\beta E_{j} / N, n} \displaystyle \sum_{j}-\beta\left(\frac{\partial E_{j}}{\partial V}\right) e^{-\beta E_{f}(N, n)}}{Q^{2}}\] Here we used the quotient rule to take the derivative. This can written compactly as \[\left(\frac{\partial E}{\partial V}\right)_{N, \beta}=-\bar{p}+\beta \overline{E p}-\beta \bar{E} \bar{p}\] We can differentiate Equation \ref{II} with respect to \(\beta\) to obtain \[\left(\frac{\partial \bar{p}}{\partial \beta}\right)_{N, V}=\bar{E} \bar{p}-\overline{E p}\] The two derivative expressions can be combined to give \[\left(\frac{\partial E}{\partial V}\right)_{N, \beta}+\beta\left(\frac{\partial \bar{p}}{\partial \beta}\right)_{N, V}=-p\] This can be compared to the thermodynamic equation of state \[\left(\frac{\partial E}{\partial V}\right)_{T}-T\left(\frac{\partial p}{\partial T}\right)_{V}=-p \label{IV}\] This can be derived as follows from \[dE = TdS – PdV.\] First take the derivative of both sides with respect to \(V\) at constant \(T\). Now, note that \[\left(\frac{\partial p}{\partial T}\right)_{V}=\left(\frac{\partial S}{\partial V}\right)_{T}\] This is known as a . It is obtained from \[dA = SdT + PdV \label{III}\] From the fact that \(A\) is a state function (the ) we know that the second cross derivatives must be equal. That is: \[\left(\frac{\partial A}{\partial V \partial T}\right)=\left(\frac{\partial A}{\partial T \partial V}\right)\] And from inspection of Equation \ref{III} we see that \[\left(\frac{\partial A}{\partial T}\right)_{V}=S\] and \[\left(\frac{\partial A}{\partial V}\right)_{T}=P\] Finally, using the relation \[T\left(\frac{\partial}{\partial T}\right)=-\frac{1}{T}\left(\frac{\partial}{\partial(1 / T)}\right)\] Showing that this is true is a little tricky. For example, we can define \(F = 1/T\). Then \[\dfrac{\partial F}{\partial T} = \dfrac{-1}{T^2}\] and \[\dfrac{\partial F}{\partial F} = 1\] So we can write \[\dfrac{\partial F}{\partial T} = \dfrac{-1}{T^2} \left(\dfrac{\partial F}{\partial F}\right)\] or \[\dfrac{\partial }{\partial T} = \dfrac{-1}{T^2} \left(\dfrac{\partial }{\partial F}\right)\] which gives Equation \ref{IV} when both sides are multiplied by \(T\). \[\left(\frac{\partial E}{\partial V}\right)_{T}+\frac{1}{T}\left(\frac{\partial p}{\partial(1 / T)}\right)_{V}=-p\] The comparison with the above equation shows that \(\beta \propto 1/T\). This proves that \(\beta = constant/T\). The constant turns out to be \(k_B\) or Boltzmann’s constant by comparison with expressions for the average energy or average pressure with known thermodynamic equations. | 3,904 | 1,187 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/09%3A_Chemical_Kinetics/9.05%3A_The_Effect_of_Temperature_on_Reaction_Rates |
When molecules collide, the kinetic energy of the molecules can be used to stretch, bend, and ultimately break bonds, leading to chemical reactions. If molecules move too slowly with little kinetic energy, or collide with improper orientation, they do not react and simply bounce off each other. However, if the molecules are moving fast enough with a proper collision orientation, such that the kinetic energy upon collision is greater than the minimum energy barrier, then a reaction occurs. The minimum energy requirement that must be met for a chemical reaction to occur is called the activation energy, \(E_a\). The reaction pathway is similar to what happens in Figure \(\Page {1}\). To get to the other end of the road, an object must roll with enough speed to completely roll over the hill of a certain height. The faster the object moves, the more kinetic energy it has. If the object moves too slowly, it does not have enough kinetic energy necessary to overcome the barrier; as a result, it eventually rolls back down. In the same way, there is a minimum amount of energy needed in order for molecules to break existing bonds during a chemical reaction. If the kinetic energy of the molecules upon collision is greater than this minimum energy, then bond breaking and forming occur, forming a new product (provided that the molecules collide with the proper orientation). The activation energy (\(E_a\)), labeled \(\Delta{G^{\ddagger}}\) in Figure \(\Page {2}\), is the energy difference between the reactants and the activated complex, also known as transition state. In a chemical reaction, the transition state is defined as the highest-energy state of the system. If the molecules in the reactants collide with enough kinetic energy and this energy is higher than the transition state energy, then the reaction occurs and products form. In other words, the higher the activation energy, the harder it is for a reaction to occur and vice versa. Overcoming the energy barrier from thermal energy involves addressing the fraction of the molecules that possess enough kinetic energy to react at a given temperature. According to , a population of molecules at a given temperature is distributed over a variety of kinetic energies that is described by the . The two distribution plots shown here are for a lower temperature and a higher temperature . The area under each curve represents the total number of molecules whose energies fall within particular range. The shaded regions indicate the number of molecules which are sufficiently energetic to meet the requirements dictated by the two values of that are shown. It is clear from these plots that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) proceed more rapidly at higher temperatures. By 1890 it was common knowledge that higher temperatures speed up reactions, often doubling the rate for a 10-degree rise, but the reasons for this were not clear. Finally, in 1899, the Swedish chemist Svante Arrhenius (1859-1927) combined the concepts of activation energy and the Boltzmann distribution law into one of the most important relationships in physical chemistry: Take a moment to focus on the meaning of this equation, neglecting the factor for the time being. First, note that this is another form of the exponential decay law discussed in the previous section of this series. What is "decaying" here is not the concentration of a reactant as a function of time, but the magnitude of the rate constant as a function of the exponent . And what is the significance of this quantity? Recalling that is the , it becomes apparent that the exponent is just the ratio of the activation energy to the average kinetic energy. The larger this ratio, the smaller the rate (hence the negative sign). This means that high temperature and low activation energy favor larger rate constants, and thus speed up the reaction. Because these terms occur in an exponent, their effects on the rate are quite substantial. Svante August Arrhenius (19 February 1859 – 2 October 1927) was a Swedish scientist, originally a physicist, but often referred to as a chemist, and one of the founders of the science of physical chemistry. He received the Nobel Prize for Chemistry in 1903, becoming the first Swedish Nobel laureate, and in 1905 became director of the Nobel Institute where he remained until his death. The Arrhenius equation, Arrhenius definition of an acid, lunar crater Arrhenius, the mountain of Arrheniusfjellet and the Arrhenius Labs at Stockholm University are named after him. Today, Arrhenius is best known for his study published in 1896, on the greenhouse effect. The two plots in Figure \(\Page {4}\) show the effects of the activation energy (denoted here by ) on the rate constant. Even a modest activation energy of 50 kJ/mol reduces the rate by a factor of 10 . Looking at the role of temperature, a similar effect is observed. (If the -axis were in "kilodegrees" the slopes would be more comparable in magnitude with those of the kilojoule plot at the above right.) The Arrhenius equation \[k = A e^{-E_a/RT} \label{1}\] can be written in a non-exponential form that is often more convenient to use and to interpret graphically (Figure \(\Page {4}\)). Taking the logarithms of both sides and separating the exponential and pre-exponential terms yields \[ \ln k = \ln \left(Ae^{-E_a/RT} \right) = \ln A + \ln \left(e^{-E_a/RT}\right) \label{2}\] \[\ln k = \ln A + \dfrac{-E_a}{RT} = \left(\dfrac{-E_a}{R}\right) \left(\dfrac{1}{T}\right) + \ln A \label{3}\] which is the equation of a straight line whose slope is \(–E_a /R\). This affords a simple way of determining the activation energy from values of observed at different temperatures, by plotting \(\ln k\) as a function of \(1/T\). For the isomerization of cyclopropane to propene, the following data were obtained (calculated values shaded in pink): From the calculated slope, we have – ( / ) = –3.27 × 10 K =– (8.314 J mol K ) (–3.27 × 10 K) = 273 kJ mol This activation energy is high, which is not surprising because a carbon-carbon bond must be broken in order to open the cyclopropane ring. (C–C bond energies are typically around 350 kJ/mol.) This is why the reaction must be carried out at high temperature. Because the ln vs.-1/ plot yields a straight line, it is often convenient to estimate the activation energy from experiments at only two temperatures. The ln term is eliminated by subtracting the expressions for the two ln- terms via \[ \ln k_{1}=\ln A - \dfrac{E_{a}}{k_{B}T_1} \] at \(T_1\) and \[ \ln k_{2}=\ln A - \dfrac{E_{a}}{k_{B}T_2} \] at \(T_2\). By \[ \ln A = \ln k_{2} + \dfrac{E_{a}}{k_{B}T_2} \] and substitute for \(\ln A\) into the first equation: \[ \ln k_{1}= \ln k_{2} + \dfrac{E_{a}}{k_{B}T_2} - \dfrac{E_{a}}{k_{B}T_1} \] This simplifies to: \[ \ln k_{1} - \ln k_{2} = -\dfrac{E_{a}}{k_{B}T_1} + \dfrac{E_{a}}{k_{B}T_2} \] \[ \ln \dfrac{k_{1}}{k_{2}} = -\dfrac{E_{a}}{k_{B}} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )\] A widely used rule-of-thumb for the temperature dependence of a reaction rate is that a rise in the temperature approximately doubles the rate. This is not generally true, especially when a strong covalent bond must be broken. For a reaction that does show this behavior, what would the activation energy be? Center the ten degree interval at 300 K. Substituting into the above expression yields \[E_a = \dfrac{(8.314)(\ln 2/1)}{\dfrac{1}{295} – \dfrac{1}{305}} = \dfrac{(8.314)(0.693)}{0.00339 K^{-1} – 0.00328 \; K^{-1}} \] = (5.76 J mol K ) / (0.00011 K ) = 52400 J mol = It takes about 3.0 minutes to cook a hard-boiled egg in Los Angeles, but at the higher altitude of Denver, where water boils at 92°C, the cooking time is 4.5 minutes. Use this information to estimate the activation energy for the coagulation of egg albumin protein. The ratio of the rate constants at the elevations of Los Angeles and Denver is 4.5/3.0 = 1.5, and the respective temperatures are \(373 \; \rm{K }\) and \(365\; \rm{K}\). With the subscripts 2 and 1 referring to Los Angeles and Denver respectively: \[E_a = \dfrac{(8.314)(\ln 1.5)}{\dfrac{1}{365\; \rm{K}} – \dfrac{1}{373 \; \rm{K}}} = \dfrac{(8.314)(0.405)}{0.00274 \; \rm{K^{-1}} – 0.00268 \; \rm{K^{-1}}}\] \[\\ = \dfrac{(3.37\; \rm{J\; mol^{–1} K^{–1}})}{5.87 \times 10^{-5}\; \rm{K^{–1}}} = 5740\; \rm{ J\; mol^{–1}} = 5.73 \; \rm{kJ \;mol^{–1}}\] : This low value seems reasonable because thermal denaturation of proteins primarily involves the disruption of relatively weak hydrogen bonds; no covalent bonds are broken (although can interfere with this interpretation). Up to this point, the pre-exponential term, in the Arrhenius equation, has been ignored because it is not directly involved in relating temperature and activation energy, which is the main practical use of the equation. However, because multiplies the exponential term, its value clearly contributes to the value of the rate constant and thus of the rate. Recall that the exponential part of the Arrhenius equation expresses the fraction of reactant molecules that possess enough kinetic energy to react, as governed by the . This fraction can run from zero to nearly unity, depending on the magnitudes of \(E_a\) and of the temperature. If this fraction were 0, the Arrhenius law would reduce to \[k = A\] In other words, \(A\) is the fraction of molecules that would react if either the activation energy were zero, or if the kinetic energy of all molecules exceeded \(E_a\) — admittedly, an uncommon scenario (although barrierless reactions have been characterized). What would limit the rate constant if there were no activation energy requirements? The most obvious factor would be the rate at which reactant molecules come into contact. This can be calculated from kinetic molecular theory and is known as the or r, \(Z\). In some reactions, the relative orientation of the molecules at the point of collision is important, so a geometrical or (commonly denoted by \(\rho\) (Greek lower case ) can be defined. In general, we can express \(A\) as the product of these two factors: \[A=Z\rho\] Values of ρ are generally very difficult to assess; they are sometime estimated by comparing the observed rate constant with the one in which \(A\) is assumed to be the same as \(Z\). Usually, the more complex the reactant molecules, the lower the steric factors. The deviation from unity can have different causes: the molecules are not spherical, so different geometries are possible; not all the kinetic energy is delivered into the right spot; the presence of a solvent (when applied to solutions) and other factors (Figure \(\Page {4}\)). ) | 10,901 | 1,188 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/10%3A_Enzyme_Kinetics/10.08%3A_The_Effect_of_Temperature_on_Enzyme_Kinetics |
Enzymes are generally globular proteins, acting alone or in larger complexes. Like all proteins, enzymes are linear chains of amino acids that fold to produce a three-dimensional structure. The sequence of the amino acids specifies the structure which in turn determines the catalytic activity of the enzyme. Although structure determines function, a novel enzyme's activity yet be predicted from its structure alone. Enzyme structures unfold (denature) when heated or exposed to chemical denaturants and this disruption to the structure typically causes a loss of activity. Protein folding is key to whether a globular protein or a membrane protein can do its job correctly. It must be folded into the right shape to function. But hydrogen bonds, which play a big part in folding, are rather weak, and it does not take much heat, acidity, or other stress to break some and form others, denaturing the protein. This is one reason why tight homeostasis is physiologically necessary in many life forms. Denaturation is a process in which proteins or nucleic acids lose the quaternary structure, tertiary structure and secondary structure which is present in their native state, by application of some external stress or compound such as a strong acid or base, a concentrated inorganic salt, an organic solvent (e.g., alcohol or chloroform), radiation or heat. If proteins in a living cell are denatured, this results in disruption of cell activity and possibly cell death. Denatured proteins can exhibit a wide range of characteristics, from conformational change and loss of solubility to aggregation due to the exposure of hydrophobic groups. Enzyme denaturation is normally linked to temperatures above a species' normal level; as a result, enzymes from bacteria living in volcanic environments such as hot springs are prized by industrial users for their ability to function at high temperatures, allowing enzyme-catalyzed reactions to be operated at a very high rate. | 1,989 | 1,189 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Free_Radical_Reactions_of_Alkenes/Radical_Additions%3A_Anti-Markovnikov_Product_Formation |
Anti-Markovnikov rule describes the regiochemistry where the substituent is bonded to a less substituted carbon, rather than the more substitued carbon. This process is quite unusual, as carboncations which are commonly formed during alkene, or alkyne reactions tend to favor the more substitued carbon. This is because substituted carbocation allow more hyperconjugation and indution to happen, making the carbocation more stable. This process was first explained by Morris Selig Karasch in his paper: 'The Addition of Hydrogen Bromide to Allyl Bromide' in 1933. Examples of Anti-Markovnikov includes Hydroboration-Oxidation and Radical Addition of HBr. A free radical is any chemical substance with unpaired electron. The more substituents the carbon is connected to, the more substituted is that carbon. For example: Tertiary carbon (most substituted), Secondary carbon (medium substituted), primary carbon (least substituted) Anti-Markovnikov Radical Addition of can happen to HBr and there be presence of Hydrogen Peroxide (H O ). Hydrogen Peroxide is essential for this process, as it is the chemical which starts off the chain reaction in the initiation step. be used in radical reactions, because in their radical reaction one of the radical reaction steps: Initiation is Endothermic, as recalled from Chem 118A, this means the reaction is unfavorable. To demonstrate the anti-Markovnikov regiochemistry, I will use 2-Methylprop-1-ene as an example below: Hydrogen Peroxide is an unstable molecule, if we heat it, or shine it with sunlight, two free radicals of OH will be formed. These OH radicals will go on and attack HBr, which will take the Hydrogen and create a Bromine radical. Hydrogen radical do not form as they tend to be extremely unstable with only one electron, thus bromine radical which is more stable will be readily formed. The Bromine Radical will go on and attack the carbon of the alkene. This is because after the bromine radical attacked the alkene a carbon radical will be formed. A carbon radical is more stable when it is at a more substituted carbon due to induction and hyperconjugation. Thus, the radical will be formed at the more substituted carbon, while the bromine is bonded to the less substituted carbon. After a carbon radical is formed, it will go on and attack the hydrogen of a HBr, which a bromine radical will be formed again. There are also Termination Steps, but we do not concern about the termination steps as they are just the radicals combining to create waste products. For example two bromine radical combined to give bromine. This radical addition of bromine to alkene by radical addition reaction will go on until all the alkene turns into bromoalkane, and this process will take some time to finish. Please give the product(s) of the reactions below: | 2,839 | 1,191 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/25%3A_Solutions_II_-_Nonvolatile_Solutes/25.04%3A_Osmotic_Pressure_can_Determine_Molecular_Masses |
Some membrane materials are permeable for some molecules, but not for others. This is often a matter of the of the molecules, but it can also be a question of of the molecule in the barrier material. Many biological membranes have semipermeable properties and osmosis is therefore an important biological process. Figure 25.4.1
shows a simple osmotic cell. Both compartments contain water, but the one on the right also contains a solute whose molecules (represented by green circles) are too large to pass through the membrane. Many artificial and natural substances are capable of acting as semi-permeable membranes. For example, the walls of most plant and animal cells fall into this category. If solvent molecules can pass through the membrane, but solute molecules (or ions) cannot, solvent molecules will spontaneously migrate across the membrane to increase the solution's volume and thus reduce its concentration. If the solution is ideal, this process is in many ways analogous to the spontaneous increase in volume of a gas allowed to expand against vacuum. Of course the volume of the 'solute-gas' is limited by the availability of solvent and, if done under gravity in a U-shaped tube, by the build up of hydrostatic pressure. This pressure is known as the \(\Pi\). At equilibrium we can write: \[μ^*(T,P) = μ^{sln}(T,P+ Π,a_1) \nonumber \] \[μ^*(T,P) = μ^*(T,P+ Π)+ RT\ln a_1 \nonumber \] From Gibbs energy (\(dG\)) in its natural variables (\(P,T\)) we know that: \[ \left(\dfrac{∂G}{∂P} \right)_{T,x} = V \nonumber \] Taking the partial versus \(x_1\) we get: \[ \left( \dfrac{∂μ^*}{∂P} \right)_{T,x_{j}} = \bar{V}^*_{1} \nonumber \] This means we can integrate over the molar volume to convert \(μ^*(T,P+ Π)\) to a different pressure: \[μ^*(T,P+\Pi) = μ^*(T,P) + \int_P^{P+ \Pi} \bar{V}^*_{1} dP \nonumber \] Thus we get: \[μ^*(T,P) = μ^*(T,P+ \Pi)+ RT\ln a_1 \nonumber \] \[μ^*(T,P) = μ^*(T,P)+ \Pi\bar{V}^*_{1}+ RT\ln a_1 \nonumber \] Once again using the ideal approximation: \[\ln a_1 \ln x_1 ≈ -x_2 \nonumber \] we get: \[RT x_2 = \Pi\bar{V}^*_{1} \nonumber \] \[x_2 = \dfrac{n_2}{n_1+n_2}≈\dfrac{n_2}{n_1} \nonumber \] The combination gives an expression involving the molarity: \[\Pi=RTc \nonumber \] Where \(c\) is the molar concentration. Osmosis can be used in reverse, if we apply about 30 bar to sea water we can obtain fresh water on the other side of a suitable membrane. This process is used in some places, but better membranes would be desirable and they easily get clogged. The resulting water is not completely salt-free and this means that if used for agriculture the salt may accumulate on the field over time. Both melting point depression and boiling point elevation only facilitate the determination of relatively molar weights. The need for such measurements is no longer felt because we now have good techniques to determine the structure of most small to medium size molecules. For polymers this is a different matter. They usually have a molecular weight (mass) and determining it is an important topic of polymer science. Osmometry is still of some practical usefulness. It is also colligative and able to measure up to about 8000 daltons. Many polymers are much bigger than that. Their mass distribution is usually determined by different means. The polymers is dissolved and led over a chromatographic column usually based on size-exclusion. The effluent is then probed as function of the elution time by a combination of techniques: The latter two provide information on the molar mass distribution but they give a different moment of that distribution. The combination of techniques gives an idea not only of how much material there is of a given molar mass but also of the linearity or degree of branching of the chains. Nevertheless melting point depression is still used in a somewhat different application. When a slightly impure solid is melted its melting point in depressed. Also the melting process is not sudden but takes place over the whole trajectory from typically a lower eutectic temperature up to the depressed melting point (the liquid line in the phase diagram). In organic synthesis the melting behavior is often used as a first convenient indication of purity. In a differential scanning calorimetry (DSC) experiment the melting peak becomes progressively skewed towards lower temperatures at higher impurity levels. The shape of the curve can be modeled with a modified version of the melting point depression expression. This yields a value for the total impurity level in the solid. This technique is used in the pharmaceutical industry for quality control purposes. | 4,668 | 1,192 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Ideal_Systems/Thermodynamics_of_Mixing |
When solids, liquids or gases are combined, the thermodynamic quantities of the system experience a change as a result of the mixing. This module will discuss the effect that mixing has on a solution’s , , and , with a specific focus on the mixing of two gases. A solution is created when two or more components mix homogeneously to form a single phase. Studying solutions is important because most chemical and biological life processes occur in systems with multiple components. Understanding the thermodynamic behavior of mixtures is integral to the study of any system involving either ideal or non-ideal solutions because it provides valuable information on the molecular properties of the system. Most real gases behave like ideal gases at standard temperature and pressure. This allows us to combine our knowledge of ideal systems and solutions with standard state thermodynamics in order to derive a set of equations that quantitatively describe the effect that mixing has on a given gas-phase solution’s thermodynamic quantities. Unlike the extensive properties of a one-component system, which rely only on the amount of the system present, the extensive properties of a solution depend on its temperature, pressure and composition. This means that a mixture must be described in terms of the partial molar quantities of its components. The total Gibbs free energy of a two-component solution is given by the expression \[ G=n_1\overline{G}_1+n_2\overline{G} _2 \label{1}\] where The molar Gibbs energy of an ideal gas can be found using the equation \[\overline{G}=\overline{G}^\circ+RT\ln \frac{P}{1 bar} \label{2}\] where \( \overline{G}^\circ\) is the standard molar Gibbs energy of the gas at 1 bar, and P is the pressure of the system. In a mixture of ideal gases, we find that the system’s partial molar Gibbs energy is equivalent to its chemical potential, or that \[ \overline{G}_i=\mu_i \label{3} \] This means that for a solution of ideal gases, Equation \(\ref{2}\) can become \[\overline{G}_i=\mu_i=\mu^\circ_i+RT \ln \frac{P_i}{1 bar} \label{4}\] where Now pretend we have two gases at the same temperature and pressure, gas 1 and gas 2. The Gibbs energy of the system before the gases are mixed is given by Equation \(\ref{1}\), which can be combined with Equation \(\ref{4}\) to give the expression \[G_{initial}=n_1(\mu^\circ_1+RT \ln P)+n_2(\mu^\circ_2+RT \ln P) \label{5}\] If gas 1 and gas 2 are then mixed together, they will each exert a partial pressure on the total system, \(P_1\) and \(P_2\), so that \(P_1+ P_2= P\). This means that the final Gibbs energy of the final solution can be found using the equation \[G_{final}=n_1(\mu^\circ_1+RT \ln P_1)+n_2(\mu^\circ_2+RT \ln P_2) \label{6}\] The Gibbs energy of mixing, \(Δ_{mix}G\), can then be found by subtracting \(G_{initial}\) from \(G_{final}\). \[ \begin{align} Δ_{mix}G &= G_{final} - G_{initial}\\[4pt] &=n_1RT \ln \frac{P_1}{P}+n_2RT \ln \frac{P_2}{P} \\[4pt] &=n_1 RT \ln \chi_1+n_2 RT \ln \chi_2 \label{7} \end{align} \] where \[P_i = \chi_iP\] and \(\chi_i\) is the mole fraction of gas \(i\). This equation can be simplified further by knowing that the mole fraction of a component is equal to the number of moles of that component over the total moles of the system, or \[\chi_i = \dfrac{n_i}{n}.\] Equation \ref{7} then becomes \[\Delta_{mix} G=nRT(\chi_1 \ln \chi_1 + \chi_2 \ln \chi_2) \label{8}\] This expression gives us the effect that mixing has on the Gibbs free energy of a solution. Since \(\chi_1\) and \(\chi_2\) are mole fractions that range from 0 to 1, we can conclude that \(Δ_{mix}G \) will be a negative number. This is consistent with the idea that gases mix spontaneously at constant pressure and temperature. Figure \(\Page {1}\) shows that when two gases mix, it can really be seen as two gases expanding into twice their original volume. This greatly increases the number of available microstates, and so we would therefore expect the entropy of the system to increase as well. Thermodynamic studies of an ideal gas’s dependence of Gibbs free energy of temperature have shown that \[ \left( \dfrac {d G} {d T} \right )_P=-S \label{9}\] This means that differentiating Equation \(\ref{8}\) at constant pressure with respect to temperature will give an expression for the effect that mixing has on the entropy of a solution. We see that \[ \begin{align} \left( \dfrac {d G_{mix}} {d T} \right)_P &=nR(x_1 \ln x_1+x_2 \ln x_2) \\[4pt] &=-\Delta_{mix} S \end{align}\] \[\Delta_{mix} S=-nR(x_1 \ln x_1+x_2 \ln x_2) \label{10}\] Since the mole fractions again lead to negative values for ln x and ln x , the negative sign in front of the equation makes Δ S positive, as expected. This agrees with the idea that mixing is a spontaneous process. We know that in an ideal system \(\Delta G= \Delta H-T \Delta S\), but this equation can also be applied to the thermodynamics of mixing and solved for the enthalpy of mixing so that it reads \[\Delta_{mix} H=\Delta_{mix} G+T\Delta_{mix} S \label{11}\] Plugging in our expressions for \(Δ_{mix}G\) (Equation \(\ref{8}\)) and \(Δ_{mix}S\) (Equation \(\ref{10}\)) , we get \[\Delta_{mix} H=nRT(x_1 \ln x_1+x_2 \ln x_2)+T \left[-nR(x_1 \ln x_1+x_2 \ln x_2) \right] = 0\] This result makes sense when considering the system. The molecules of ideal gas are spread out enough that they do not interact with one another when mixed, which implies that no heat is absorbed or produced and results in a \(Δ_{mix}H\) of zero. Figure \(\Page {2}\) illustrates how \(TΔ_{mix}S\) and \(Δ_{mix}G\) change as a function of the mole fraction so that \(Δ_{mix}H\) of a solution will always be equal to zero (this is for the mixing of two ideal gasses). | 5,718 | 1,193 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/17%3A_Solubility_and_Complexation_Equilibria/17.03%3A_Factors_That_Affect_Solubility |
The solubility product of an ionic compound describes the concentrations of in equilibrium with a solid, but what happens if some of the cations become associated with anions rather than being completely surrounded by solvent? Then predictions of the total solubility of the compound based on the assumption that the solute exists solely as discrete ions would differ substantially from the actual solubility, as would predictions of ionic concentrations. In general, four situations explain why the solubility of a compound may be other than expected: ion pair formation, the incomplete dissociation of molecular solutes, the formation of complex ions, and changes in pH. The first two situations are described in this section, the formation of complex ions is discussed in , and changes in pH are discussed in . An ion pair consists of a cation and an anion that are in intimate contact in solution, rather than separated by solvent ( ). The ions in an ion pair are held together by the same attractive electrostatic forces that we discussed in for ionic solids. As a result, the ions in an ion pair migrate as a single unit, whose net charge is the sum of the charges on the ions. In many ways, we can view an ion pair as a species intermediate between the ionic solid (in which each ion participates in many cation–anion interactions that hold the ions in a rigid array) and the completely dissociated ions in solution (where each is fully surrounded by water molecules and free to migrate independently). As illustrated for calcium sulfate in the following equation, a second equilibrium must be included to describe the solubility of salts that form ion pairs:
\[\mathrm{CaSO_4(s)}\rightleftharpoons\mathrm{Ca^{2+}}\cdot\underset{\textrm{ion pair}}{\mathrm{SO_4^{2-}(aq)}}\rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{SO_4^{2-}(aq)} \tag{17.2.1}\] The ion pair is represented by the symbols of the individual ions separated by a dot, which indicates that they are associated in solution. The formation of an ion pair is a dynamic process, just like any other equilibrium, so a particular ion pair may exist only briefly before dissociating into the free ions, each of which may later associate briefly with other ions. Ion-pair formation can have a major effect on the measured solubility of a salt. For example, the measured for calcium sulfate is 4.93 × 10 at 25°C. The solubility of CaSO should be 7.02 × 10 M if the only equilibrium involved were as follows:
In fact, the experimentally measured solubility of calcium sulfate at 25°C is 1.6 × 10 M, almost twice the value predicted from its . The reason for the discrepancy is that the concentration of ion pairs in a saturated CaSO solution is almost as high as the concentration of the hydrated ions. Recall that the magnitude of attractive electrostatic interactions is greatest for small, highly charged ions. Hence ion pair formation is most important for salts that contain M and M ions, such as Ca and La , and is relatively unimportant for salts that contain monopositive cations, except for the smallest, Li . We therefore expect a saturated solution of CaSO to contain a high concentration of ion pairs and its solubility to be greater than predicted from its . The formation of ion pairs increases the solubility of a salt. A molecular solute may also be more soluble than predicted by the measured concentrations of ions in solution due to . This is particularly common with weak organic acids. (For more information about weak organic acids, see ). Although strong acids (HA) dissociate completely into their constituent ions (H and A ) in water, weak acids such as carboxylic acids do not ( = 1.5 × 10 ). However, the molecular (undissociated) form of a weak acid (HA) is often quite soluble in water; for example, acetic acid (CH CO H) is completely miscible with water. Many carboxylic acids, however, have only limited solubility in water, such as benzoic acid (C H CO H), with = 6.25 × 10 . Just as with calcium sulfate, we need to include an additional equilibrium to describe the solubility of benzoic acid:
\[ C_6H_5CO_2H_{(s)} \rightleftharpoons C_6H_5CO_2H_{(aq)} \rightleftharpoons C_6H_5CO^−_{2(aq)} + H^+_{(aq)} \tag{17.2.3}\] In a case like this, measuring only the concentration of the ions grossly underestimates the total concentration of the organic acid in solution. In the case of benzoic acid, for example, the pH of a saturated solution at 25°C is 2.85, corresponding to [H ] = [C H CO ] = 1.4 × 10 M. The total concentration of benzoic acid in the solution, however, is 2.8 × 10 M. Thus approximately 95% of the benzoic acid in solution is in the form of hydrated neutral molecules—C H CO H(aq)—and only about 5% is present as the dissociated ions ( ). Incomplete dissociation of a molecular solute that is miscible with water can increase the solubility of the solute. Although ion pairs, such as Ca ·SO , and undissociated electrolytes, such as C H CO H, are both electrically neutral, there is a major difference in the forces responsible for their formation. Simple electrostatic attractive forces between the cation and the anion hold the ion pair together, whereas a polar covalent O−H bond holds together the undissociated electrolyte. There are four explanations why the solubility of a compound can differ from the solubility indicated by the concentrations of ions: (1) formation, in which an anion and a cation are in intimate contact in solution and not separated by solvent, (2) the incomplete dissociation of molecular solutes, (3) the formation of complex ions, and (4) changes in pH. An ion pair is held together by electrostatic attractive forces between the cation and the anion, whereas incomplete dissociation results from intramolecular forces, such as polar covalent O–H bonds. Do you expect the actual molar solubility of LaPO to be greater than, the same as, or less than the value calculated from its ? Explain your reasoning. Do you expect the difference between the calculated molar solubility and the actual molar solubility of Ca (PO ) to be greater than or less than the difference in the solubilities of Mg (PO ) ? Why? Write chemical equations to describe the interactions in a solution that contains Mg(OH) , which forms ion pairs, and in one that contains propanoic acid (CH CH CO H), which forms a hydrated neutral molecule. Draw representations of Ca(IO ) in solution Ferric phosphate has a molar solubility of 5.44 × 10 in 1.82 M Na PO . Predict its . The actual is 1.3 × 10 . Explain this discrepancy. 9.90 × 10 the solubility is much higher than predicted by due to the formation of ion pairs (and/or phosphate complexes) in the sodium phosphate solution. | 6,753 | 1,194 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.15%3A_Writing_Lewis_Structures_for_Molecules/6.15.01%3A_An_Excess_of_Bonds |
There are a number of cases where the normal valences of the atoms involved do not predict the correct skeleton structure. For example, thionyl chloride, SOCl , is found to have both chlorines and the oxygen atom bonded to sulfur: This exceeds the usual valence of 2 for sulfur, while oxygen has one less bond than we might have expected. Molecules which deviate in this way from the usual valence rules often contain at least one atom (such as S) from the third row or below in the periodic table. One or more oxygen atoms are usually bonded just to that third-row atom instead of linking a pair of other atoms. Usually the atom which occupies the central position in the skeleton is written first in the molecular formula, although sometimes H (which forms only one bond and cannot be the central atom) precedes it. Some examples (with the central atom in italics) are: OCl , OCl , H O , O Br , and O. (In the last case, one of the two nitrogens occupies the central position.) In such molecules the deviation from the normal valence occurs because at least one electron-pair bond contains two electrons which were originally associated with the atom. Such a bond is called a coordinate or a . An example is the bond between sulfur and oxygen in SOCl : Both electrons in the S―O bond were originally valence electrons of sulfur. Therefore the sharing of this electron pair adds nothing to the valence shell of sulfur, and sulfur can form one more bond than would be predicted by its normal valence. Neither electron in the coordinate covalent bond was originally associated with oxygen, and a single bond (both electrons) is sufficient to provide an octet when added to oxygen’s six valence electrons. Hence oxygen forms one less bond than expected. It should be noted that there is nothing to distinguish a coordinate covalent bond from any other covalent bond once a structural formula has been drawn. A pair of electrons is still a pair of electrons no matter where it came from. The distinction is merely one we make when trying to fit electron pairs into octets around each atom. Structural formulas for the other examples of unusual valence we have mentioned are shown below with coordinate covalent bonds indicated in color: It is good practice to draw out the complete Lewis diagram for each of these molecules, differentiating electrons from different nuclei with different symbols such as and ●, and satisfy yourself that they obey the octet rule. | 2,488 | 1,195 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/06%3A_Bonding_in_Organic_Molecules/6.04%3A_Electron_Repulsion_and_Bond_Angles._Orbital_Hybridization |
In predicting bond angles in small molecules, we find we can do a great deal with the simple idea that charges produce forces while charges produce forces. We will have electron-nuclear attractions, electron-electron repulsions, and nucleus-nucleus repulsions. Let us first consider the case of a molecule with just electron-pair bonds, as might be expected to be formed by combination of beryllium and hydrogen to give beryllium hydride, \(H:Be:H\). The problem will be how to formulate the bonds and how to predict what the \(H-Be-H\) angle, \(\theta\), will be: We might formulate a second \(\sigma\) bond involving the \(2p\) orbital, but a new problem arises as to where the hydrogen should be located relative to the beryllium orbital. Is it as in \(2\), \(3\), or some other way? The \(Be\) and \(H\) nuclei will be farther apart in \(2\) than they will be in \(3\) or any other similar arrangement, so there will be less with \(2\). We therefore expect the hydrogen to locate along a line going through the greatest extension of the \(2p\) orbital. According to this simple picture, beryllium hydride should have two different types of \(H-Be\) bonds - one as in \(1\) and the other as in \(2\). This is intuitively unreasonable for such a simple compound. Furthermore, the \(H-Be-H\) bond angle is unspecified by this picture because the \(2s\) \(Be\) orbital is spherically symmetrical and could form bonds equally well in any direction. However, if we forget about the orbitals and only consider the possible repulsions between the electron pairs, and between the hydrogen nuclei, we can see that these repulsions will be minimized when the \(H-Be-H\) bond angle is \(180^\text{o}\). Thus arrangement \(5\) should be more favorable than \(4\), with a \(H-Be-H\) angle less than \(180^\text{o}\): Unfortunately, we cannot check this particular bond angle by experiment because \(BeH_2\) is unstable and reacts with itself to give a high-molecular-weight solid. However, a number of other compounds, such as \(\left( CH_3 \right)_2 Be\), \(BeCl_2\), \(\left( CH_3 \right)_2 Hg\), \(HgF_2\), and \(\left( CH_3 \right)_2 Zn\), are known to have \(\sigma\) bonds involving \(\left( s \right)^1 \left( p \right)^1\) valence states. Measurements of the bond angles at the metal of these substances in the vapor state has shown them to be uniformly \(180^\text{o}\). How are the \(s\) and \(p\) orbitals deployed in this kind of bonding? It turns out that . The degree of overlap will depend on the sizes of the orbital and, particularly, on how far out they extend from the nucleus. Figure 6-7 shows how far \(2s\) and \(2p\) orbitals extend relative to one another. Bonding with these orbitals as in \(1\) and \(2\) does not utilize the overlapping power of the orbitals to the fullest extent. With \(1\) we have overlap that uses only part of the \(2s\) orbital, and with \(2\), only a part of the \(2p\) orbital. Molecules such as \(BeH_2\) can be formulated with and with the aid of the concept of . This concept, published independently by L. Pauling and J. C. Slater in 1931, involves determining which (if any) combinations of \(s\) and \(p\) orbitals may overlap better and make more effective bonds than do the individual \(s\) and \(p\) orbitals. The mathematical procedure for orbital hybridization predicts that an \(s\) and a \(p\) orbital of one atom can form two stronger covalent bonds if they combine to form two new orbitals called (Figure 6-8). Each \(sp\)-hybrid orbital has an overlapping power of 1.93, compared to the pure \(s\) orbital taken as unity and a pure \(p\) orbital as 1.73. Bond angles of \(180^\text{o}\) are expected for bonds to an atom using \(sp\)-hybrid orbitals and, of course, this also is the angle we expect on the basis of our consideration of minimum electron-pair and internuclear repulsions. Henceforth, we will proceed on the basis that molecules of the type \(X:M:X\) may form \(sp\)-hybrid bonds. On the basis of repulsion between electron pairs and between nuclei, molecules such as \(BH_3\), \(B \left( CH_3 \right)_3\), \(BF_3\), and \(AlCl_3\), in which the central atom forms covalent bonds using the valence-state electronic configuration \(\left( s \right)^1 \left( p_x \right)^1 \left( p_y \right)^1\), are expected to be planar with bond angles of \(120^\text{o}\). For example, With atoms such as carbon and silicon, the valence-state electronic configuration to form covalent bonds has to be \(\left( s \right)^1 \left( p_x \right)^1 \left( p_y \right)^1 \left( p_z \right)^1\). Repulsion between the electron pairs and between the attached nuclei will be minimized by formation of a tetrahedral arrangement of the bonds. the same geometry is predicted from hybridization one one \(s\) and three \(p\) orbitals, which gives four directed at angles of \(109.5^\text{o}\) to each other. The predicted relative overlapping power of \(sp^3\)-hybrid orbitals is 2.00 (Figure 6-10). and (1977) | 5,000 | 1,196 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/20%3A_Entropy_and_The_Second_Law_of_Thermodynamics/20.01%3A_Energy_Does_not_Determine_Spontaneity |
There are many spontaneous events in nature. If you open the valve in both cases a spontaneous event occurs. In the first case the gas fills the evacuated chamber, in the second the gases will mix. The state functions \(U\) and \(H\) do not give us a clue what will happen. You might think that only those events are spontaneous that produce heat. Not so: Clearly the first law is not enough to describe nature. The development of the new state function entropy has brought us much closer to a complete understanding of how heat and work are related: Two problems remain: Let us start with the latter. Yes we can use S to explain the odd paradox between w and q both being forms energy on the one hand, but the conversion being easier in one direction than the other, but we have introduced the concept purely as a phenomenon on iis own. Scientifically there is nothing wrong with such a phenomenological theory except that experience tells us that if you can understand the phenomenon itself better your theory becomes more powerful. To understand entropy better we need to leave the macroscopic world and look at what happens on a molecular level and do statistics over many molecules. First, let us do a bit more statistics of the kind we will need. If we have n distinguishable objects, say playing cards we can arrange them in a large number of ways. For the first object in our series we have \(n\) choices, for number two we have \(n-1\) choices (the first one being spoken for) etc. This means that in total we have \[W=n(n-1)(n-2)….4.3.2.1 = n!\;\text{ choices.} \nonumber \] The quantity \(W\) is usually called the number of realizations in thermodynamics. The above is true if the objects are all distinguishable. If they fall in groups within which they are not distinguishable we have to correct for all the swaps within these groups that do not produce a distinguishably new arrangement. This means that \(W\) becomes \(\frac{n!}{a!b!c!…z!}\) where a,b, c to z stands for the size of the groups. (Obviously \(a+b+c+..+z = n\)) In thermodynamincs our 'group of objects' is typically an ensemble of systems, think of size N and so the factorial become horribly large. This makes it necessary to work with logarithms. Fortunately there is a good approximation (by Stirling) for a logarithmic factorial: \[\ln N! \approx N \ln N-N \nonumber \] In Europe nosey little children who are curious to find out where their newborn little brother or sister came from, often get told that the stork brought it during the night. When you look at the number of breeding pairs of this beautiful bird in e.g. Germany since 1960 you see a long decline to about 1980 when the bird almost got extinct. After that the numbers go up again due to breeding programs mostly. The human birth rate in the country follows a pretty much identical curve and the correlation between the two is very high (>0.98 or so). (Dr H. Sies in Nature (volume 332 page 495; 1988). Does this prove that storks indeed bring babies? No, it does show causality, just a correlation due a common underlying factor. In this case that is the choices made by the German people. First they concentrated on working real hard and having few children to get themselves out of the poverty WWII had left behind and neglected the environment, then they turned to protect the environment and opened the doors to immigration of people, mostly from Muslim countries like Turkey or Morocco, that usually have larger families. The lesson from this is that you can only conclude causality you are sure that there are no other intervening factors. The expansion of an ideal gas against vacuum is really a wonderful , because but a spontaneous expansion and a change in entropy. No energy change, no heat, no work, no change in mass, no interactions, nothing. In fact, it does not even matter whether we consider it an isolated process or not. We might as well do so. Physicists and Physical Chemists love to find such experiments that allows them to retrace . All this means that if we look at what happens at the atomic level, we should be able to retrace the of the entropy change. As we have seen before, the available energy states of particles in a box depend on the size of the box. \[ E_{kin} = \dfrac{h^2}{8ma^2} \left[ n_1^2 + n_2^2 + n_2^2 \right] \nonumber \] Clearly if the side (\(a\) and therefore the volume of the box changes, the energy spacing between the states will become smaller. Therefore during our expansion against vacuum, the energy states inside the box are changing. Because \(U\) does not change the energy \(\langle E \rangle\) is constant. Of course this average is taken over a great number of molecules ( ) in the gas (the ), but let's look at just two of them and for simplicity let us assume that the energy of the states are equidistant (rather than quadratic in the quantum numbers n). As you can see there is more than one way to skin a cat, or in this case to the same average \(\langle E \rangle\) of the complete ensemble (of only two particles admittedly). Before expansion I have shown three realizations W , W , W that add up to the same \(\langle E \rangle\). After expansion however, there are more energy states available and the schematic figure shows twice as many realizations W in the same energy interval. Boltzmann was the first to postulate that this is what is at the root of the entropy function, not so much the (total) energy itself (that stays the same!), but the number of ways the energy can be distributed in the ensemble. Note that because the ensemble average (or total) energy is identical, we could also say that the various realizations \(W\) represent the degree of \(Ω\) of the ensemble. Boltzmann considered a much larger ( ) ensemble consisting of a great number of identical (e.g. molecules, but it could also be planets or so). If each of our systems already has a large number of energy states the systems can all have the same (total) energy but distributed in rather different ways. This means that two systems within the ensemble can either have the same distribution or a different one. Thus we can divide the ensemble A in subgroups a having the same energy distribution and calculate the number of ways to distribute energy in the ensemble \(A\) as \[W= \dfrac{A!}{a1!a2!…}. \nonumber \] Boltzmann postulated that entropy was directly related to the number of realizations W, that is the number of ways the same energy can be distributed in the ensemble. This leads immediately to the concept of order versus disorder, e.g, if the number of realizations is \(W=1\), all systems must be in the same state (W=A! / A!0!0!0!…) which is a very orderly arrangement of energies. If we were to add two ensembles to each other the total number of possible arrangements W becomes the product W W but the entropies should be additive. As logarithms transform products into additions Boltzmann assumed that the relation between W and S should be logarithmic and wrote: \[S= k \ln W \nonumber \] Again, if we consider a very ordered state, e.g. where all systems are in the ground state the number of realizations A!/A!= 1 so that the entropy is zero. If we have a very messy system where the number of ways to distribute energy over the many many different states is very large S becomes very large. Thus entropy is very large. This immediately gives us the driving force for the expansion of a gas into vacuum or the mixing of two gases. We simply get more energy states to play with, this increases W. This means an increase in S. This leads to a spontaneous process. | 7,634 | 1,197 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/19%3A_Nuclear_Chemistry/19.2%3A_Nuclear_Decay_Processes |
The relative abundances of the elements in the known universe vary by more than 12 orders of magnitude. For the most part, these differences in abundance cannot be explained by differences in nuclear stability. Although the Fe nucleus is the most stable nucleus known, the most abundant element in the known universe is not iron, but hydrogen ( H), which accounts for about 90% of all atoms. In fact, H is the raw material from which all other elements are formed. In this section, we explain why H and He together account for at least 99% of all the atoms in the known universe. We also describe the nuclear reactions that take place in stars, which transform one nucleus into another and create all the naturally occurring elements. The relative abundances of the elements in the known universe and on Earth relative to silicon are shown in . The data are estimates based on the characteristic emission spectra of the elements in stars, the absorption spectra of matter in clouds of interstellar dust, and the approximate composition of Earth as measured by geologists. The data in illustrate two important points. First, except for hydrogen, the most abundant elements have even atomic numbers. Not only is this consistent with the known trends in nuclear stability, but it also suggests that heavier elements are formed by combining helium nuclei (Z = 2). Second, the relative abundances of the elements in the known universe and on Earth are often very different, as indicated by the data in Table \(\Page {1}\) for some common elements. Some of these differences are easily explained. For example, nonmetals such as H, He, C, N, O, Ne, and Kr are much less abundant relative to silicon on Earth than they are in the rest of the universe. These elements are either noble gases (He, Ne, and Kr) or elements that form volatile hydrides, such as NH , CH , and H O. Because Earth’s gravity is not strong enough to hold such light substances in the atmosphere, these elements have been slowly diffusing into outer space ever since our planet was formed. Argon is an exception; it is relatively abundant on Earth compared with the other noble gases because it is continuously produced in rocks by the radioactive decay of isotopes such as K. In contrast, many metals, such as Al, Na, Fe, Ca, Mg, K, and Ti, are relatively abundant on Earth because they form nonvolatile compounds, such as oxides, that cannot escape into space. Other metals, however, are much less abundant on Earth than in the universe; some examples are Ru and Ir. This section explains some of the reasons for the great differences in abundances of the metallic elements. All the elements originally present on Earth (and on other planets) were synthesized from hydrogen and helium nuclei in the interiors of stars that have long since exploded and disappeared. Six of the most abundant elements in the universe (C, O, Ne, Mg, Si, and Fe) have nuclei that are integral multiples of the helium-4 nucleus, which suggests that helium-4 is the primary building block for heavier nuclei. Elements are synthesized in discrete stages during the lifetime of a star, and some steps occur only in the most massive stars known ( ). Initially, all stars are formed by the aggregation of interstellar “dust,” which is mostly hydrogen. As the cloud of dust slowly contracts due to gravitational attraction, its density eventually reaches about 100 g/cm , and the temperature increases to about 1.5 × 10 K, forming a dense plasma of ionized hydrogen nuclei. At this point, self-sustaining nuclear reactions begin, and the star “ignites,” creating a yellow star like our sun. In the first stage of its life, the star is powered by a series of nuclear fusion reactions that convert hydrogen to helium: The overall reaction is the conversion of four hydrogen nuclei to a helium-4 nucleus, which is accompanied by the release of two positrons, two \(\gamma\) rays, and a great deal of energy: \[4_1^1\textrm H\rightarrow\,_2^4\textrm{He}+2_{+1}^0\beta+2_0^0\gamma\label{Eq2} \] These reactions are responsible for most of the enormous amount of energy that is released as sunlight and solar heat. It takes several billion years, depending on the size of the star, to convert about 10% of the hydrogen to helium. Once large amounts of helium-4 have been formed, they become concentrated in the core of the star, which slowly becomes denser and hotter. At a temperature of about 2 × 10 K, the helium-4 nuclei begin to fuse, producing beryllium-8: \[2_2^4\textrm{He}\rightarrow\,_4^8\textrm{Be}\label{Eq3} \] Although beryllium-8 has both an even mass number and an even atomic number, its also has a low neutron-to-proton ratio (and other factors beyond the scope of this text) that makes it unstable; it decomposes in only about 10 s. Nonetheless, this is long enough for it to react with a third helium-4 nucleus to form carbon-12, which is very stable. Sequential reactions of carbon-12 with helium-4 produce the elements with even numbers of protons and neutrons up to magnesium-24: \[_4^8\textrm{Be}\xrightarrow{_2^4\textrm{He}}\,_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\label{Eq4} \] So much energy is released by these reactions that it causes the surrounding mass of hydrogen to expand, producing a red giant that is about 100 times larger than the original yellow star. As the star expands, heavier nuclei accumulate in its core, which contracts further to a density of about 50,000 g/cm3, so the core becomes even hotter. At a temperature of about 7 × 10 K, carbon and oxygen nuclei undergo nuclear fusion reactions to produce sodium and silicon nuclei: \[_6^{12}\textrm C+\,_6^{12}\textrm C\rightarrow \,_{11}^{23}\textrm{Na}+\,_1^1\textrm H\label{Eq5} \] \[_6^{12}\textrm C+\,_8^{16}\textrm O\rightarrow \,_{14}^{28}\textrm{Si}+\,_0^0\gamma\label{Eq6} \] At these temperatures, carbon-12 reacts with helium-4 to initiate a series of reactions that produce more oxygen-16, neon-20, magnesium-24, and silicon-28, as well as heavier nuclides such as sulfur-32, argon-36, and calcium-40: \[_6^{12}\textrm C\xrightarrow{_2^4\textrm{He}}\,_8^{16}\textrm O\xrightarrow{_2^4\textrm{He}}\,_{10}^{20}\textrm{Ne}\xrightarrow{_2^4\textrm{He}}\,_{12}^{24}\textrm{Mg}\xrightarrow{_2^4\textrm{He}}\,_{14}^{28}\textrm{Si}\xrightarrow{_2^4\textrm{He}}\,_{16}^{32}\textrm S\xrightarrow{_2^4\textrm{He}}\,_{18}^{36}\textrm{Ar}\xrightarrow{_2^4\textrm{He}}\,_{20}^{40}\textrm{Ca}\label{Eq7} \] The energy released by these reactions causes a further expansion of the star to form a red supergiant, and the core temperature increases steadily. At a temperature of about 3 × 10 K, the nuclei that have been formed exchange protons and neutrons freely. This equilibration process forms heavier elements up to iron-56 and nickel-58, which have the most stable nuclei known. None of the processes described so far produces nuclei with Z > 28. All naturally occurring elements heavier than nickel are formed in the rare but spectacular cataclysmic explosions called ( ). When the fuel in the core of a very massive star has been consumed, its gravity causes it to collapse in about 1 s. As the core is compressed, the iron and nickel nuclei within it disintegrate to protons and neutrons, and many of the protons capture electrons to form neutrons. The resulting neutron star is a dark object that is so dense that atoms no longer exist. Simultaneously, the energy released by the collapse of the core causes the supernova to explode in what is arguably the single most violent event in the universe. The force of the explosion blows most of the star’s matter into space, creating a gigantic and rapidly expanding dust cloud, or nebula ( ). During the extraordinarily short duration of this event, the concentration of neutrons is so great that multiple neutron-capture events occur, leading to the production of the heaviest elements and many of the less stable nuclides. Under these conditions, for example, an iron-56 nucleus can absorb as many as 64 neutrons, briefly forming an extraordinarily unstable iron isotope that can then undergo multiple rapid β-decay processes to produce tin-120: \[_{26}^{56}\textrm{Fe}+64_0^1\textrm n\rightarrow \,_{26}^{120}\textrm{Fe}\rightarrow\,_{50}^{120}\textrm{Sn}+24_{-1}^0\beta\label{Eq8} \] Although a supernova occurs only every few hundred years in a galaxy such as the Milky Way, these rare explosions provide the only conditions under which elements heavier than nickel can be formed. The force of the explosions distributes these elements throughout the galaxy surrounding the supernova, and eventually they are captured in the dust that condenses to form new stars. Based on its elemental composition, our sun is thought to be a second- or third-generation star. It contains a considerable amount of cosmic debris from the explosion of supernovas in the remote past. The reaction of two carbon-12 nuclei in a carbon-burning star can produce elements other than sodium. Write a balanced nuclear equation for the formation of reactant and product nuclides balanced nuclear equation Use conservation of mass and charge to determine the type of nuclear reaction that will convert the reactant to the indicated product. Write the balanced nuclear equation for the reaction. How many neutrons must an iron-56 nucleus absorb during a supernova explosion to produce an arsenic-75 nucleus? Write a balanced nuclear equation for the reaction. 19 neutrons; \(_{26}^{56}\textrm{Fe}+19_0^1\textrm n \rightarrow \,_{26}^{75}\textrm{Fe}\rightarrow \,_{33}^{75}\textrm{As}+7_{-1}^0\beta\) Hydrogen and helium are the most abundant elements in the universe. Heavier elements are formed in the interior of stars via multiple neutron-capture events. By far the most abundant element in the universe is hydrogen. The fusion of hydrogen nuclei to form helium nuclei is the major process that fuels young stars such as the sun. Elements heavier than helium are formed from hydrogen and helium in the interiors of stars. Successive fusion reactions of helium nuclei at higher temperatures create elements with even numbers of protons and neutrons up to magnesium and then up to calcium. Eventually, the elements up to iron-56 and nickel-58 are formed by exchange processes at even higher temperatures. Heavier elements can only be made by a process that involves multiple neutron-capture events, which can occur only during the explosion of a supernova. | 10,576 | 1,199 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Exercises%3A_Physical_and_Theoretical_Chemistry/Data-Driven_Exercises/Osmotic_Pressure_and_Polymer_Molecular_Weight_Determination |
: The molecular weight of a polymer is determined by analyzing osmotic pressure data. : An introductory knowledge of ideal versus non-ideal solution behavior and colligative properties. : This exercise should be carried out within a software environment that can generate a best-fit line for an x-y data set. The software should also report uncertainties in the determined slope and y-intercept. You will also be graphing the data along with the fitted function. Suppose you are given an unknown biomolecular substance, such as a protein, RNA strand, or polysaccharide, and you are then asked to determine the molecular weight of the substance. What experimental methods are available for determining the molecular weight of a large molecule (or perhaps a synthetic polymer or biopolymer)?
One of the more precise techniques involves the measurement of osmotic pressure. The instrument for carrying out these measurements is called an osmometer. The device consists of a semipermeable membrane that separates two solution compartments. The semipermeable membrane is made of a material that permeates the solvent (not the solute). If the membrane separates pure solvent from a solution, an osmotic pressure exists across the membrane which, in turn, drives the flow of solvent through the membrane from the pure solvent compartment to the solution compartment. The flow of solvent that occurs due to a concentration gradient across the membrane is called osmosis.
Osmotic pressure is a colligative property, which means that it is proportional to the concentration of solute. The van ’t Hoff equation is often presented in introductory chemistry for calculating osmotic pressure (\(Π\)) from the moles of solute (\(n_{solute}\)) that occupy a given volume (\(V\)) and the absolute temperature (\(T\)) of the solution; \[ \Pi = \dfrac{n_{solute}RT}{V} \label{1}\] Note the similarity between Equation \ref{1} and the ideal gas equation (\(P=nRT/V\)). But just like the ideal gas equation, the van’t Hoff equation is only valid for an ideal system. In this case, Equation \ref{1} is only valid for an ideal solution, which is a hypothetical solution in which the solute-solvent, solvent-solvent, and solute-solute interactions are all equivalent. Since all non-volatile, non-electrolytic solutions approach , Equation \ref{1} is actually a limiting law, and should be written in the form \[ \lim_{n_{solute} \rightarrow 0} \Pi = \dfrac{n_{solute}RT}{V} \label{2}\] When using an osmometer, it is more convenient to express concentration in terms of the 'grams' of solute per liter, \[c =\dfrac{g}{L}\] whereby we can make the following substitution in Equation \ref{2}: \[ \dfrac{n_{solute}}{V} = \dfrac{c}{M},\] where \(M\) is the molecular weight of the solute. In this fashion, and with minor rearrangement, Equation \ref{2} can be written as \[ \lim_{c \rightarrow 0} \dfrac{\Pi}{c} = \dfrac{RT}{M} \label{3}\] According to Equation \ref{3}, the molecular weight of a solute can be obtained by plotting osmotic pressure divided by \(c\) versus concentration and extrapolating the data back to \(c = 0\). Since Equation \ref{3} is only exact in the dilute limit, we can recognize this relationship as the first term in a more general power series expansion in \(c\), \[\dfrac{\Pi}{c} = RT \left( \dfrac{1}{M} + A_2c + A_3c^2 + \ldots \right) \label{4}\] where \(A_2\) and \(A_3\) are called the second and third virial coefficients, respectively. These coefficients are empirically determined constants for a given solute-solvent system, and also depend on temperature. According to statistical mechanical solution theory, \(A_2\) represents the interaction of a single solute particle with the solvent, and higher order virial coefficients are associated with correspondingly larger number solute particle cluster interactions with the solvent.
The value of \(A_2\) is of practical importance to those who are interested in working with a particular solute-solvent system. A negative value of \(A_2\) is indicative of a ‘good solvent’ (i.e. the solute will be highly soluble in the solvent due to favorable solute-solvent intermolecular interactions). A positive value of \(A_2\) indicates the solute is insoluble in the solvent. Since \(A_2\) is temperature dependent, there will exist a special temperature for each solute-solvent system where \(A_2= 0\). This temperature is called the , and represents the theoretical temperature at which an infinite molecular weight solute just precipitates from solution. When a solvent has a Flory Θ-temperature that lies somewhere between ambient to slightly elevated temperature conditions, then that solvent is often referred to as a 'Θ-solvent' for that solute. Among other things, saying that a solvent is a 'Θ-solvent' implies that it is a great solvent for recrystallizing the solute (for purification purposes).
In this exercise, you will be given the osmotic pressure of a polymer solution, cellulose tricaproate in dimethylformamide, over a range of concentrations and at three different temperatures. You will determine the molecular weight (M) of the polymer, the second virial coefficient (\(A_2\)) at each temperature, and the Flory Θ-temperature for this system. Cellulose is a natural polymer that makes up nearly 80% of all the biomass on earth (cotton is nearly 100% cellulose). Structurally, cellulose is a linear chain of glucose monomers, linked together at the 1,4 positions of the glucose ring. These glucose chains can be hundreds of units long, and the corresponding molecular weights of these polymers can be several hundred thousand grams per mole.
The following table contains osmotic pressure data at three different temperatures for a cellulose derivative called 'cellulose tricaproate' dissolved in dimethylformamide at a number of concentrations. The data was obtained from Krigbaum, W.R. and Sperling, L.H. Journal of Physical Chemistry, 64, 99 (1960). | 5,961 | 1,201 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/18%3A_Partition_Functions_and_Ideal_Gases/18.E%3A_Partition_Functions_and_Ideal_Gases_(Exercises) |
These are homework exercises to accompany of McQuarrie and Simon's "Physical Chemistry: A Molecular Approach" Textmap. 1) If the nucleus has a spin of \(s_n\), then its spin degeneracy \(g_n = 2 s_n + 1\). The diatomic molecule formed from such a nucleus will have g n 2 spin functions which have to be combined to form symmetric and antisymmetric functions. Carry out an analysis similar to that of H 2 for D 2 where the deuterium nucleus has a spin of 1. 2) Derive the thermodynamic functions from the polyatomic rotational partition function. 3) Carry out the integration for the rotational partition function of the symmetric top. 4) Calculate the total partition function and the thermodynamic functions of water at 1000K. The three moments of inertia of water are 1.02, 1.91 and 2.92 in 10 - 47 kg m 2 . The symmetry number is 2. The vibrational data in given in Fig. 3.5. Assume a non- degenerate electronic ground state. 5) Verify that the symmetry numbers for methane, benzene and SF 6 are 12, 12 and 24 respectively. 6) The ground state of Na is a doublet (two states with the same energy). Assuming this to be the zero of energy and assuming that the next energy level to be 2 eV higher than the ground state, calculate q el . 7) The bond length \(r_{eq}\) of \(O_2\) is 1.2 Å. The moment of inertia \(I\) is \(mr_{eq}^2 / 2\) where m of O is \(16 X 1.66 times 10^{-27}\; kg\). Calculate \(\tilde{B}\) and the rotational partition function of \(O_2\) at 300 K. 8) The vibrational frequency ν of ICl is 384 cm -1 . What is its vibrational partition function at 300 K? What is the fraction of molecules in the ground state (n = 0) and the first excited state n = 1? 9) Calculate the translational partition function of N 2 at 300 K. For volume, use the molar volume at 300 K. 10) An isotope exchange reaction between isotopes of bromine is 79 79 81 81 79 81 2 Br Br Br Br Br Br + The fundamental vibrational frequency of 79 81 Br Br 323.33 cm -1 . All the molecules can be assumed to have the same bond length and have a singlet ground electronic state. Calculate the equilibrium constant at 300K and 1000K. 11) For the reaction I 2 ↔ 2I, calculate the equilibrium constant at 1000K. The relevant data are as follows. The ground electronic state of I is 2 3/ 2 P whose degeneracy is 4. The rotational and vibrational frequencies of I 2 are 0.0373 cm -1 and 214.36 cm -1 respectively. The dissociation energy of I 2 is 1.5422 eV. 12) The representative molecular data for a few molecules is given in table 3.1. Using the relevant data, calculate the equilibrium constant for the reaction H 2 + Cl 2 I HCl at 1000K. What is the value of the equilibrium constant as T → ∞? 13) Eq. (3.50) is related to the Giauque function. Estimate the total molar Giauque function for molecules that behave as harmonic oscillators-rigid rotors. 14) The energy of a molecule in the rigid rotor – harmonic oscillator approximation is E vib, rot is (n +1/2) hν + B J (J+1) Real molecules deviate from this behaviour due to the existence of anharmonicity (anharmonicity constant x e ), centrifugal distortion (centrifugal distortion constant D ) and the interaction between vibration and rotation (α is the coupling constant between the vibrational and rotational modes). The expression for the energy when these affects are included is 2 2 2 1 1 1 2 2 2 , ( ) ( 1) ( ) ( 1) ( ) ( 1) vib rot e E n h BJ J x n h DJ J n J J ν ν α = + + + − + + + − + + Here, the third term is due to anharmonicity, the fourth term is due to centrifugal distortion and the last term is due to the interaction between vibration and rotation. Calculate the q vib, rot which includes the effects of these distortions. Using the data in Table 8.6, calculate the fraction of sodium atoms in the first excited state at temperatures 300 K, 1000 K, and 2000 K. Using Equation 18.10, we can calculate the fraction of sodium atoms in the first excited state, with \( g_{e1} = 2 , g_{e2} = 2 , g_{e3} = 4 , g_{e4} = 2 \) : \[ f_2 = \dfrac{2e^{-\beta \epsilon_{e2}}} {2+2e^{-\beta \epsilon_{e2}} + 4e^{-\beta \epsilon_{e3}} + 2e^{-\beta \epsilon_{e4}} + ... } \] Using the data in Table 8.6, the numerator of this fraction becomes \[ 2 exp [ - \dfrac{16956.183 cm^{-1}}{(0.6950 cm^{-1} K^{-1} ) T} ] \] and the denominator becomes \( 2+ 2 exp [ - \dfrac{16956.183 cm^{-1}}{(0.6950 cm^{-1} K^{-1} ) T} ] + 4 exp [ - \dfrac{16973.379 cm^{-1}}{(0.6950 cm^{-1} K^{-1} ) T} ] + 2 exp [ - \dfrac{25739.86 cm^{-1}}{(0.6950 cm^{-1} K^{-1} ) T} ] + ...\) Using these values, we can find the values of \( f_2 \) at the different temperatures. \( f_2 ( T = 300 K) = 4.8 x 10^{-36} \) \( f_2 ( T = 1000 K) = 2.5 x 10^{-11} \) \( f_2 ( T = 2000 K) = 5.0 x 10^{-6} \) Using the data in the table, calculate the fraction of hydrogen atoms in the first excited state at \(400\ K\), \(1800\ K\), and \(2100\ K\). Use the equation: \[f_{2} = \dfrac{g_{e2}e^{- \beta \varepsilon _{e2} }}{g_{e1}+g_{e2}e^{- \beta \varepsilon _{e2} }+g_{e3}e^{- \beta \varepsilon _{e3} }+ \cdots}\] and \[\beta \approx \dfrac{1}{T\ 0.6950\ cm^{-1}K^{-1}}\] to get: \[f_{2} = \dfrac{2e^{- \dfrac{ 82\ 258.907\ cm}{T\ 0.6950\ cm^{-1}K^{-1}}}}{2+2e^{- \dfrac{ 82\ 258.907\ cm}{T\ 0.6950\ cm^{-1}K^{-1}} }+2e^{- \dfrac{ 82\ 258.942\ cm}{T\ 0.6950\ cm^{-1}K^{-1}} }+ 4e^{- \dfrac{ 82\ 259.272\ cm}{T\ 0.6950\ cm^{-1}K^{-1}}}}\] \(f_{2}(400\ K) = 0.2498\) \(f_{2}(900\ K) = 0.2499\) \(f_{2}(2100\ K) = 0.2500\) | 5,450 | 1,202 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Free_Energy/Gibbs_(Free)_Energy |
Gibbs free energy, denoted \(G\), combines and into a single value. The change in free energy, \(\Delta G\), is equal to the sum of the enthalpy plus the product of the temperature and entropy of the system. \( \Delta G\) can predict the direction of the chemical reaction under two conditions: If \(ΔG\) is positive, then the reaction is nonspontaneous (i.e., an the input of external energy is necessary for the reaction to occur) and if it is negative, then it is spontaneous (occurs without external energy input). This quantity is the energy associated with a chemical reaction that can be used to do work, and is the sum of its enthalpy (H) and the product of the temperature and the entropy (S) of the system. This quantity is defined as follows: \[ G= H-TS \label{1.1} \] or more completely as \[ G= U+PV-TS \label{1.2} \] where Spontaneous - is a reaction that is consider to be natural because it is a reaction that occurs by itself without any external action towards it. Non spontaneous - needs constant external energy applied to it in order for the process to continue and once you stop the external action the process will cease. When solving for the equation, if change of G is negative, then it's spontaneous. If change of G if positive, then it's non spontaneous. The symbol that is commonly used for FREE ENERGY is G. can be more properly consider as "standard free energy change" In chemical reactions involving the changes in thermodynamic quantities, a variation on this equation is often encountered: \[ \underset{\text {change in free energy} }{\Delta G } = \underset{ \text {change in enthalpy}}{ \Delta H } - \underset{\text {(temperature) change in entropy}}{T \Delta S} \label{1.3} \] Calculate ∆G at 290 K for the following reaction: \[\ce{2NO(g) + O2(g) \rightarrow 2NO2(g)} \nonumber \] Given now all you have to do is plug in all the given numbers into Equation 3 above. Remember to divide \(\Delta S\) by 1000 \(J/kJ\) so that after you multiply by temperature, \(T\), it will have the same units, \(kJ\), as \(\Delta H\). \[\Delta S = -150 \cancel{J}/K \left( \dfrac{1\; kJ}{1000\;\cancel{J}} \right) = -0.15\; kJ/K \nonumber \] and substituting into Equation 3: \[\begin{align*} ∆G &= -120\; kJ - (290 \;\cancel{K})(-0.150\; kJ/\cancel{K}) \\[4pt] &= -120 \;kJ + 43 \;kJ \\[4pt] &= -77\; kJ \end{align*} \] What is the \(\Delta G\) for this formation of ammonia from nitrogen and hydrogen gas. \[\ce{N_2 + 3H_2 \rightleftharpoons 2NH_3} \nonumber \] The Standard free energy formations: NH =-16.45 H =0 N =0 \[\Delta G=-32.90\;kJ \;mol^{-1} \nonumber \] Since the changes of entropy of chemical reaction are not measured readily, thus, entropy is not typically used as a criterion. To obviate this difficulty, we can use \(G\). The sign of ΔG indicates the direction of a chemical reaction and determine if a reaction is spontaneous or not. The factors affect of a reaction (assume and are independent of temperature): Note: The standard Gibbs energy change (at which reactants are converted to products at 1 bar) for: \[ aA + bB \rightarrow cC + dD \label{1.4} \] \[ \Delta r G^o = c \Delta _fG^o (C) + d \Delta _fG^o (D) - a \Delta _fG^o (A) - b \Delta _fG^o (B) \label{1.5} \] \[\Delta _fG^0 = \sum v \Delta _f G^0 (\text {products}) - \sum v \Delta _f G^0 (\text {reactants}) \label{1.6} \] The standard-state free energy of reaction ( \(\Delta G^o\)) is defined as the free energy of reaction at standard state conditions: \[ \Delta G^o = \Delta H^o - T \Delta S^o \label{1.7} \] The standard-state free energy of formation is the change in free energy that occurs when a compound is formed from its elements in their most thermodynamically stable states at standard-state conditions. In other words, it is the difference between the free energy of a substance and the free energies of its constituent elements at standard-state conditions: \[ \Delta G^o = \sum \Delta G^o_{f_{products}} - \sum \Delta G^o_{f_{reactants}} \label{1.8} \] Used the below information to determine if \(NH_4NO_{3(s)}\) will dissolve in water at room temperature. This question is essentially asking if the following reaction is spontaneous at room temperature. \[\ce{NH4NO3(s) \overset{H_2O} \longrightarrow NH4(aq)^{+} + NO3(aq)^{-}} \nonumber \] This would normally only require calculating \(\Delta{G^o}\) and evaluating its sign. However, the \(\Delta{G^o}\) values are not tabulated, so they must be calculated manually from calculated \(\Delta{H^o}\) and \(\Delta{S^o}\) values for the reaction. \[ \Delta H^o = \sum n\Delta H^o_{f_{products}} - \sum m\Delta H^o_{f_{reactants}} \nonumber \] \[ \Delta H^o= \left[ \left( 1\; mol\; NH_3\right)\left(-132.51\;\dfrac{kJ}{mol} \right) + \left( 1\; mol\; NO_3^- \right) \left(-205.0\;\dfrac{kJ}{mol}\right) \right] \nonumber \] \[- \left[ \left(1\; mol\; NH_4NO_3 \right)\left(-365.56 \;\dfrac{kJ}{mol}\right) \right] \nonumber \] \[ \Delta H^o = -337.51 \;kJ + 365.56 \; kJ= 28.05 \;kJ \nonumber \] \[ \Delta S^o = \sum n\Delta S^o_{f_{products}} - \sum S\Delta H^o_{f_{reactants}} \nonumber \] \[ \Delta S^o= \left[ \left( 1\; mol\; NH_3\right)\left(113.4 \;\dfrac{J}{mol\;K} \right) + \left( 1\; mol\; NO_3^- \right) \left(146.6\;\dfrac{J}{mol\;K}\right) \right] \nonumber \] \[- \left[ \left(1\; mol\; NH_4NO_3 \right)\left(151.08 \;\dfrac{J}{mol\;K}\right) \right] \nonumber \] \[ \Delta S^o = 259.8 \;J/K - 151.08 \; J/K= 108.7 \;J/K \nonumber \] These values can be substituted into the free energy equation \[T_K = 25\;^oC + 273.15K = 298.15\;K \nonumber \] \[\Delta{S^o} = 108.7\; \cancel{J}/K \left(\dfrac{1\; kJ}{1000\;\cancel{J}} \right) = 0.1087 \; kJ/K \nonumber \] \[\Delta{H^o} = 28.05\;kJ \nonumber \] Plug in \(\Delta H^o\), \(\Delta S^o\) and \(T\) into Equation 1.7 \[\Delta G^o = \Delta H^o - T \Delta S^o \nonumber \] \[\Delta G^o = 28.05\;kJ - (298.15\; \cancel{K})(0.1087\;kJ/ \cancel{K}) \nonumber \] \[\Delta G^o= 28.05\;kJ - 32.41\; kJ \nonumber \] \[\Delta G^o = -4.4 \;kJ \nonumber \] This reaction is spontaneous at room temperature since \(\Delta G^o\) is negative. Therefore \(NH_4NO_{3(s)}\) will dissolve in water at room temperature. Calculate \(\Delta{G}\) for the following reaction at \(25\; ^oC\). Will the reaction occur spontaneously? \[NH_{3(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)} \nonumber \] given for the reaction calculate \(\Delta{G}\) from the formula \[\Delta{G} = \Delta{H} - T\Delta{S} \nonumber \] but first we need to convert the units for \(\Delta{S}\) into kJ/K (or convert \(\Delta{H}\) into J) and temperature into Kelvin The definition of Gibbs energy can then be used directly \[\Delta{G} = \Delta{H} - T\Delta{S} \nonumber \] \[\Delta{G} = -176.0 \;kJ - (298 \cancel{K}) (-0.284.8\; kJ/\cancel{K}) \nonumber \] \[\Delta{G} = -176.0 \;kJ - (-84.9\; kJ) \nonumber \] \[\Delta{G} = -91.1 \;kJ \nonumber \] Yes, this reaction is spontaneous at room temperature since \(\Delta{G}\) is negative. Let's consider the following reversible reaction: \[ A + B \leftrightharpoons C + D \label{1.9} \] The following equation relates the standard-state free energy of reaction with the free energy at any point in a given reaction (not necessarily at standard-state conditions): \[ \Delta G = \Delta G^o + RT \ln Q \label{1.10} \] At equilibrium, ΔG = 0 and Q=K. Thus the equation can be arranged into: \[\Delta{G} = \Delta{G}^o + RT \ln \dfrac{[C,D]}{[A,B]} \label{1.11} \] with The Gibbs free energy depends primarily on the reactants' nature and concentrations (expressed in the term and the logarithmic term of Equation 1.11, respectively). At equilibrium, \(\Delta{G} = 0\): no driving force remains The equilibrium constant is defined as \[K_{eq} = \dfrac{[C,D]}{[A,B]} \label{1.14} \] When \(K_{eq}\) is large, almost all reactants are converted to products. Substituting \(K_{eq}\) into Equation 1.14, we have: Rearrange, \[K_{eq} = 10^{-\Delta{G}^{o}/(2.303RT)} \label{1.17} \] This equation is particularly interesting as it relates the free energy difference under standard conditions to the properties of a system at equilibrium (which is rarely at standard conditions). What is for isomerization of dihydroxyacetone phosphate to glyceraldehyde 3-phosphate? If at equilibrium, we have \(K_{eq} = 0.0475\) at 298 K and pH 7. We can calculate: \[\Delta{G}^{o} = -2.303\;RT log_{10} K_{eq}= (-2.303) * (1.98 * 10^{-3}) * 298 * (log_{10} 0.0475) = 1.8 \;kcal/mol \nonumber \] Given: From equation 2: = 1.8 kcal/mol + 2.303 RT log (3*10 M/2*10 M) = -0.7 kcal/mol Under non-standard conditions (which is essential all reactions), the spontaneity of reaction is determined by , not . The relates the standard-state cell potential with the cell potential of the cell at any moment in time: \[ E = E^o - \dfrac {RT}{nF} \ln Q \label{1.18} \] with By rearranging this equation we obtain: \[ E = E^o - \dfrac {RT}{nF} \ln Q \label{1.19} \] multiply the entire equation by \(nF\) \[ nFE = nFE^o - RT \ln Q \label{1.20} \] which is similar to: \[ \Delta G = \Delta G^o + RT \ln Q \label{1.21} \] By juxtaposing these two equations: \[ nFE = nFE^o - RT \ln Q \label{1.22} \] \[ \Delta G = \Delta G^o + RT \ln Q \label{1.23} \] it can be concluded that: \[ \Delta G = -nFE \label{1.24} \] Therefore, \[ \Delta G^o = -nFE^o \label{1.25} \] | 9,290 | 1,203 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/03%3A_Rate_Laws/3.03%3A_The_Rate_Law/3.3.03%3A_Reaction_Order |
The reaction order is the relationship between the concentrations of species and the rate of a reaction. The of a rate law is the sum of the exponents of its concentration terms. Once the of a reaction has been determined, that same law can be used to understand more fully the composition of the reaction mixture. More specifically, the reaction order is the exponent to which the concentration of that species is raised, and it indicates to what extent the concentration of a species affects the rate of a reaction, as well as which species has the greatest effect. For the decomposition with a rate law of k[ ], this exponent is 1 (and thus is not explicitly shown); this reaction is therefore a first order reaction. It can also be said that the reaction is "first order in ". For more complicated rate laws, the overall reaction order and the orders with respect to each component are used. As an example, consider the following reaction, \[ A + 3B + 2C \rightarrow \text{products} \nonumber \] whose experimental rate law is given by: \[\text{rate} = k[A,B]^2 \nonumber \] This reaction is third-order overall, first-order in A, second-order in B, and zero-order in C. means that the rate is independent of the concentration of a particular reactant. Of course, enough C must be present to allow the equilibrium mixture to form. For the reaction: \[ aA + bB \longrightarrow P \nonumber \] The rate law is as follows: \[ rate=k[A]^x[B]^y \nonumber \] The order of a reaction is not necessarily an integer. The following orders are possible: The rate of oxidation of bromide ions by bromate in an acidic aqueous solution, \[6H^+ + BrO_3^– + 5Br^– \rightarrow 3 Br_2 + 3 H_2O \nonumber \] is found to follow the following rate law: \[\text{rate} = k[Br^-,BrO_3^-,H^+]^2 \nonumber \] What happens to the rate if, in separate experiments, [ ] is doubled; the pH is increased by one unit; the solution is diluted to twice its volume, with the pH held constant using a buffer? For chemical reactions that require only one elementary step, the values of and are equal to the stoichiometric coefficients of each reactant. For chemical reactions that require more than one elementary step, this is not always the case. However, there are many simple ways of determining the order of a reaction. One very popular method is known as the differential method. The differential method, also known as the initial rates method, uses an experimental data table to determine the order of a reaction with respect to the reactants used. Below is an example of a table corresponding with the following chemical reaction: \[ A + B \longrightarrow P \nonumber \] When looking at the experiments in the table above, it is important to note factors that change between experiments. In order to determine the reaction order with respect to A, one must note in which experiment A is changing; that is, between experiments 1 and 2. Write a rate law equation based on the chemical reaction above. This is the rate law: \[\text{rate} = k[A]^x[B]^y \nonumber \] Next, the rate law equation from experiment 2 must be divided by the rate law equation for experiment 1. Notice that the [B] term cancels out, leaving "x" as the unknown variable. Simple algebra reveals that x = 0. The same steps must be taken for determining the reaction order with respect to B. However, in this case experiments 1 and 3 are used. After working through the problem and canceling out [A] from the equation, y = 1. Finding the reaction order for the whole process is the easy addition of x and y: n = 0 + 1. Therefore, n = 1 After finding the reaction order, several pieces of information can be obtained, such as . Other methods that can be used to solve for reaction order include the , the half-life method, and the . 1. Define "reaction order." Use the following information to solve questions 2 and 3: Given the rate law equation: \[\text{rate} = k[A]^1[B]^2 \nonumber \] 2. Determine: a) the reaction order with respect to A, b) the reaction order with respect to B, and c) the total reaction order for the equation. 3. Assuming the reaction occurs in one elementary step, propose a chemical equation using P as the symbol for your product. Use the data table below to answer questions 4 and 5: 4. Use the differential method to determine the reaction order with respect to A (x) and B (y). What is the total reaction order (n)? 5. What is the rate constant, k? | 4,466 | 1,204 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acids_and_Bases_in_Aqueous_Solutions/The_Hydronium_Ion |
Owing to the overwhelming excess of \(\ce{H2O}\) molecules in aqueous solutions, a bare hydrogen ion has no chance of surviving in water. The hydrogen ion in aqueous solution is no more than a proton, a bare nucleus. Although it carries only a single unit of positive charge, this charge is concentrated into a volume of space that is only about a hundred-millionth as large as the volume occupied by the smallest atom. (Think of a pebble sitting in the middle of a sports stadium!) The resulting extraordinarily high of the proton strongly attracts it to any part of a nearby atom or molecule in which there is an excess of negative charge. In the case of water, this will be the lone pair (unshared) electrons of the oxygen atom; the tiny proton will be buried within the lone pair and will form a shared-electron (coordinate) bond with it, creating a , \(\ce{H3O^{+}}\). In a sense, \(\ce{H2O}\) is acting as a base here, and the product \(\ce{H3O^{+}}\) is the conjugate acid of water: Although other kinds of dissolved ions have water molecules bound to them more or less tightly, the interaction between H and \(\ce{H2O}\) is so strong that writing “\(\ce{H^{+}(aq)}\)” hardly does it justice, although it is formally correct. The formula \(\ce{H3O^{+}}\) more adequately conveys the sense that it is both a molecule in its own right, and is also the conjugate acid of water. The equation \[\ce{HA → H^{+} + A^{–}} \nonumber\] is so much easier to write that chemists still use it to represent acid-base reactions in contexts in which the proton donor-acceptor mechanism does not need to be emphasized. Thus, it is permissible to talk about “hydrogen ions” and use the formula \(\ce{H^{+}}\) in writing chemical equations as long as you remember that they are not to be taken literally in the context of aqueous solutions. Interestingly, experiments indicate that the proton does not stick to a single \(\ce{H2O}\) molecule, but changes partners many times per second. This molecular promiscuity, a consequence of the uniquely small size and mass the proton, allows it to move through the solution by rapidly hopping from one \(\ce{H2O}\) molecule to the next, creating a new \(\ce{H3O^{+}}\) ion as it goes. The overall effect is the same as if the \(\ce{H3O^{+}}\) ion itself were moving. Similarly, a hydroxide ion, which can be considered to be a “proton hole” in the water, serves as a landing point for a proton from another \(\ce{H2O}\) molecule, so that the OH ion hops about in the same way. The ion is an important factor when dealing with that occur in aqueous solutions. Its concentration relative to hydroxide is a direct measure of the of a solution. It can be formed when an acid is present in water or simply in pure water. It's chemical formula is \(\ce{H3O^{+}}\). It can also be formed by the combination of a H ion with an \(\ce{H2O}\) molecule. The hydronium ion has a trigonal pyramidal and is composed of three hydrogen atoms and one oxygen atom. There is a lone pair of electrons on the oxygen giving it this shape. The bond angle between the atoms is 113 degrees. \[H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H^+_{(aq)}\] As H ions are formed, they bond with \(\ce{H2O}\) molecules in the solution to form \(\ce{H3O^{+}}\) (the hydronium ion). This is because hydrogen ions do not exist in aqueous solutions, but take the form of the hydronium ion, \(\ce{H3O^{+}}\). A reversible reaction is one in which the reaction goes both ways. In other words, the water molecules dissociate while the OH ions combine with the H ions to form water. Water has the ability to attract H ions because it is a molecule. This means that it has a partial charge, in this case the charge is negative. The partial charge is caused by the fact that oxygen is more than hydrogen. This means that in the bond between hydrogen and oxygen, oxygen "pulls" harder on the shared electrons thus causing a partial negative charge on the molecule and causing it to be attracted to the positive charge of H to form . Another way to describe why the water molecule is considered polar is through the concept of . The electron geometry of water is and the molecular geometry is . This bent geometry is asymmetrical, which causes the molecule to be polar and have a dipole moment, resulting in a partial charge. An overall reaction for the dissociation of water to form can be seen here: \[2 H_2O_{(l)} \rightleftharpoons OH^-_{(aq)} + H_3O^+_{(aq)}\] Hydronium not only forms as a result of the dissociation of water, but also forms when water is in the presence of an . As the acid dissociates, the H ions bond with water molecules to form as seen here when hydrochloric acid is in the presence of water: \[HCl (aq) + H_2O \rightarrow H_3O^+ (aq) + Cl^-(aq)\] In a sample of pure water, the hydronium concentration is \(1 \times 10^{-7}\) moles per liter (0.0000001 M) at room temperature. The equation to find the pH of a solution using its hydronium concentration is: \[pH = -\log {(H_3O^+)}\] Using this equation, we find the pH of pure water to be 7. This is considered to be neutral on the . The pH can either go up or down depending on the change in concentration. If the hydronium concentration increases, the pH decreases, causing the solution to become more acidic. This happens when an acid is introduced. As H ions dissociate from the acid and bond with water, they form hydronium ions, thus increasing the hydronium concentration of the solution. If the hydronium concentration decreases, the pH increases, resulting in a solution that is less acidic and more basic. This is caused by the \(OH^-\) ions that dissociate from bases. These ions bond with H ions from the dissociation of water to form \(\ce{H2O}\) rather than hydronium ions. A variation of the equation can be used to calculate the hydronium concentration when a pH is given to us: \[H_3O^+ = 10^{-pH} \] When the pH of 7 is plugged into this equation, we get a concentration of 0.0000001 M as we should. Learning to use mathematical formulas to calculate the acidity and basicity of solutions can be difficult. Here is a video tutorial on the subject of calculating hydronium ion concentrations: It is believed that, on average, every hydronium ion is attracted to six water molecules that are not attracted to any other hydronium ions. This topic is still currently under debate and no real answer has been found. 1. Remembering the equation: pH = -log[H O] Plug in what is given: \(pH = -\log[2.6 \times 10^{-4}\;M]\) When entered into a calculator: pH = 3.6 2. Remembering the equation: [H O] = 10 Plug in what is given: [H O] = 10 When entered into a calculator: 1.995x10 M 3. Determine pH the same way we did in question one: pH = -log[3.6x10 ] pH = 7.4 Because this pH is above 7 it is considered to be basic. 4. First write out the balanced equation of the reaction: \[HCl_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + Cl^-_{(aq)}\] Notice that the amount of HCl is equal to the amount of \(\ce{H3O^{+}}\) produced due to the fact that all of the are one. So if we can figure out concentration of HCl we can figure out concentration of hydronium. Notice that the amount of HCl given to us is provided in grams. This needs to be changed to moles in order to find concentration: \[12.2\;g\; HCl \times \dfrac{1 \;mol\; HCl}{36.457\; g} = 0.335\; mol\; HCl\] Concentration is defined as moles per liter so we convert the 500mL of water to liters and get .5 liters. \[\dfrac{0.335\; mol\; HCl}{0.5\; L} = 0.67\;M\] Using this concentration we can obtain pH: pH = -log[.67M] \[pH = 0.17\] 5. Acids cause burns because they dehydrate the cells they are exposed to. This is caused by the dissociation that occurs in acids where H ions are formed. These H ions bond with water in the cell and thus dehydrate them to cause cell damage and burns. | 7,879 | 1,205 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Exercises%3A_Physical_and_Theoretical_Chemistry/Data-Driven_Exercises/Determination_of_Virial_Coefficients_for_Argon_Gas_at_323_K |
The pressure (\(P\)), volume (\(V\)), and temperature (\(T\)) of a given amount of gas can be used to calculate the compression factor (\(Z\)), \[ Z = \dfrac{PV_m}{RT} \label{1}\] where \(V_m\) is the molar volume of the gas. \(Z\) is exactly equal to 1 for an ideal gas at all pressures. Deviations away from \(Z=1\) arise from intermolecular interactions. A plot of \(Z\) versus pressure for an arbitrary gas is shown in Figure \ref{1}. Three regimes can be identified in this graph; The variation in the compression factor as a function of pressure can be expressed as a power series according to the expression \[ Z = 1 +B'p + C'p^2 + D'p^3 ... \label{2}\] where \(B’\), \(C’\), and \(D’\) are the second, third, and fourth virial coefficients, respectively. These coefficients have a characteristic value for each gas and are temperature dependent. When applying Equation \ref{2} to a gas from low to moderate pressures, one generally neglects higher order terms in the series. Equation \ref{2} is called a virial equation, and is one of the most accurate equations of state for a gas, provided one retains a sufficient number of terms in the expansion. For some problems, it is more convenient to write the virial equation as a power series in reciprocal molar volume \[ Z = 1 + B \dfrac{1}{V_m} +C \dfrac{1}{V_m^2} +D \dfrac{1}{V_m^3} \label{3}\] where by convention the primes (') are normally dropped from the virial coefficients to distinguish them from the coefficients that are present in Equation \ref{2}.
This exercise will involve downloading a set of pressure-volume data for Argon gas at 323.15 K and analyzing the data to determine the virial coefficients \(B\), \(C\), and \(D\) in Equation \ref{3}. on the following link and save the data to an appropriate folder. The data was obtained from Blancett, A.L. and Hall, K.R., Physica, 47, 75-91 (1970). The first column of numbers is the pressure of argon gas in units of atmospheres and the second column of numbers is the corresponding molar volume of the gas in units of Liters/mol. The temperature of the gas was held constant throughout data collection at 323.15 Kelvin. P-V Data for Argon at 323.15 K | 2,192 | 1,207 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/05%3A_Orbital_Picture_of_Bonding-_Orbital_Combinations_Hybridization_Theory_and_Molecular_Orbitals/5.04%3A_Hybridization_of_Carbon |
We will now reproduce the hybridization process for carbon, but instead of taking The process is shown below. As shown, the three resulting orbitals are equivalent in energy, but the remaining orbital has not been affected. It still retains its original energy and shape. Again, according to VSEPR theory, equivalent orbitals will arrange themselves in 3-D space to be as far apart from each other as possible. Therefore, the three equivalent orbitals will arrange themselves in a configuration. That is to say, the carbon nucleus will be at the center of an equilateral triangle, and the three orbitals will point to the corners of that triangle. A of this arrangement is shown below. In this top view, the cannot be seen because it also arranges itself to be as far apart from the orbitals as possible. That is to say, , with one lobe coming out of the plane of the page and the other going behind the page. To see this arrangement clearly, we must switch to a of the orbital system. When two hybridized carbon atoms approach each other to bond, two orbitals approach each other head to head, and The bond formed by the orbitals is a sigma bond, and The process is shown below. The illustration above tries to convey a basic feature of the bond as compared to the sigma bond. The sigma bond is short and strong. As a rule, The bond, on the other hand, is relatively long and diffuse. This has some implications in the properties and chemical reactivity of sigma and bonds. The electrons in the sigma bond (or sigma electrons) are more tightly bound to the nucleus and don’t move too much. In other words, they are more The electrons in the bond (or electrons) are less tightly bound by the nucleus, and therefore . Under certain conditions, they have the capability to become , that is to say, they can move in the molecular skeleton from one atom to another, or even become spread over several atoms, according to principles we’ll study later. At the same time, in chemical reactions where electrons are to be traded, . It is relatively easy to break a bond compared to the sigma bond. The principles of all this chemistry will be discussed later in the course. . By this definition, the simplest possible alkene must contain two carbon atoms. It is called . Below is a Lewis and a line-angle representation of ethene, which is sometimes informally called . Notice that although C–H bonds are not usually shown in line-angle formulas, sometimes they are included for enhanced clarity. In this case a pure line-angle formula for ethene would look awkward because it would resemble an equal sign (=). Notice that It simply shows the two together as two equal dashes. The orbital picture better represents the actual nature of the two types of bonds. Additional examples are shown below. For a full discussion of the structure of alkenes refer to chapter 7 of the Wade textbook. Observe that where is the total number of carbon atoms. The process for understanding the hybridization process for carbon is basically an extension of the other two types ( and ). You should try to work out this scheme on your own and see if your predictions agree with those presented in the textbook. sp hybridization gives rise to the formation of hydrocarbons known as alkynes. , and have around the carbons comprising the triple bond. Therefore, the ideal angle between the sp hybrid orbitals is Some examples of alkynes are shown below. For additional information refer to chapter 9 of the Wade textbook. Observe that the where is the total number of carbon atoms. | 3,618 | 1,208 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Period/Period_3_Elements/Acid-base_Behavior_of_the_Oxides |
This page discusses the reactions of the oxides of Period 3 elements (sodium to chlorine) with water, and with acids or bases where relevant (as before, argon is omitted because it does not form an oxide). The oxides of interest are given below: The trend in acid-base behavior can be summarized as follows: Acidity increases from left to right, ranging from strongly basic oxides on the left to strongly acidic ones on the right, with an amphoteric oxide (aluminum oxide) in the middle. An amphoteric oxide is one which shows both acidic and basic properties. This trend applies only to the highest oxides of the individual elements (see the top row of the table), in the highest oxidation states for those elements. The pattern is less clear for other oxides. Non-metal oxide acidity is defined in terms of the acidic solutions formed in reactions with water—for example, sulfur trioxide reacts with water to forms sulfuric acid. They will all, however, react with bases such as sodium hydroxide to form salts such as sodium sulfate as explored in detail below. Sodium oxide is a simple strongly basic oxide. It is basic because it contains the oxide ion, O , which is a very strong base with a high tendency to combine with hydrogen ions. : Sodium oxide reacts exothermically with cold water to produce sodium hydroxide solution. A concentrated solution of sodium oxide in water will have pH 14. \[ Na_2O + H_2O \rightarrow 2NaOH\] As a strong base, sodium oxide also reacts with acids. For example, it reacts with dilute hydrochloric acid to produce sodium chloride solution. \[Na_2O + 2HCl \rightarrow 2NaCl + H_2O\] Magnesium oxide is another simple basic oxide, which also contains oxide ions. However, it is not as strongly basic as sodium oxide because the oxide ions are not as weakly-bound. In the sodium oxide, the solid is held together by attractions between 1+ and 2- ions. In magnesium oxide, the attractions are between 2+ and 2- ions. Because of the higher charge on the metal, more energy is required to break this association. Even considering other factors (such as the energy released from ion-dipole interactions between the cations and water), the net effect is that reactions involving magnesium oxide will always be less exothermic than those of sodium oxide. At first glance, magnesium oxide powder does not appear to react with water. However, the pH of the resulting solution is about 9, indicating that hydroxide ions have been produced. In fact, s \[MgO + H_2O \rightarrow Mg(OH)_2\] Magnesium oxide reacts with acids as predicted for a simple metal oxide. For example, it reacts with warm dilute hydrochloric acid to give magnesium chloride solution. \[MgO + 2HCl \rightarrow MgCl_2+H_2O\] Describing the properties of aluminum oxide can be confusing because it exists in a number of different forms. One of those forms is very unreactive (known chemically as alpha-Al O ) and is produced at high temperatures. Aluminium amphoteric Aluminum oxide is insoluble in water and does not react like sodium oxide and magnesium oxide. The oxide ions are held too strongly in the solid lattice to react with the water. Aluminum oxide contains oxide ions, and thus reacts with acids in the same way sodium or magnesium oxides do. Aluminum oxide reacts with hot dilute hydrochloric acid to give aluminum chloride solution. \[Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O\] This reaction and others display the amphoteric nature of aluminum oxide. Aluminum oxide also displays acidic properties, as shown in its reactions with bases such as sodium hydroxide. Various aluminates (compounds in which the aluminum is a component in a negative ion) exist, which is possible because aluminum can form covalent bonds with oxygen. This is possible because the electronegativity difference between aluminum and oxygen is small, unlike the difference between sodium and oxygen, for example (electronegativity increases across a period) tetrahydroxoaluminate: \[Al_2O_3 + 2NaOH +3H_2O \rightarrow 2NaAl(OH)_4\] Silicon is too similar in electronegativity to oxygen to form ionic bonds. Therefore, because silicon dioxide does not contain oxide ions, it has no basic properties. In fact, it is very weakly acidic, reacting with strong bases. Silicon dioxide does not react with water, due to the thermodynamic difficulty of breaking up its network covalent structure. : Silicon dioxide reacts with hot, concentrated sodium hydroxide solution, forming a colorless solution of sodium silicate: \[SiO_2 + 2NaOH \rightarrow Na_2SiO_3 + H2O\] In another \[SiO_2 + CaO \rightarrow CaSiO_3\] Two phosphorus oxides, phosphorus(III) oxide, P O , and phosphorus(V) oxide, P O , are considered here. Phosphorus(III) oxide reacts with cold water to produce a solution of the weak acid, H PO —known as phosphorous acid, orthophosphorous acid or phosphonic acid: \[P_4O_6 + 6H_2O \rightarrow 4H_3PO_3\] The fully-protonated acid structure is shown below: The protons remain associated until water is added; even then, because phosphorous acid is a weak acid, few acid molecules are deprotonated. Phosphorous acid has a pK of 2.00, which is more acidic than common organic acids like ethanoic acid (pK = 4.76). Phosphorus(III) oxide is unlikely to be reacted directly with a base. In phosphorous acid, the two hydrogen atoms in the -OH groups are acidic, but the third hydrogen atom is not. Therefore, there are two possible reactions with a base like sodium hydroxide, depending on the amount of base added: \[ NaOH + H_3PO_3 \rightarrow NaH_2PO_3 + H_2O\] \[ 2NaOH + H_3PO_3 \rightarrow Na_2HPO_3 + 2H_2O\] In the first reaction, only one of the protons reacts with the hydroxide ions from the base. In the second case (using twice as much sodium hydroxide), both protons react. If instead phosphorus(III) oxide is reacted directly with sodium hydroxide solution, the same salts are possible: \[4NaOH + P_4O_6 + 2H_2O \rightarrow 4NaH_2PO_3\] \[9NaOH + P_4O_6 \rightarrow 4Na_2HPO_3 + 2H_2O\] Phosphorus(V) oxide reacts violently with water to give a solution containing a mixture of acids, the nature of which depends on the reaction conditions. Only one acid is commonly considered, phosphoric(V) acid, H PO (also known as phosphoric acid or as orthophosphoric acid). \[P_4O_{10} + 6H_2O \rightarrow 4H_3PO_4\] This time the fully protonated acid has the following structure: Phosphoric(V) acid is another weak acid with a pK of 2.15, marginally weaker than phosphorous acid. Solutions of each of these acids with concentrations around 1 mol dm have a pH of about 1. \[ NaOH + H_3PO_4 \rightarrow NaH_2PO_4 + H_2O\] \[ 2NaOH + H_3PO_4 \rightarrow Na_2HPO_4 + 2H_2O\] \[ 3NaOH + H_3PO_4 \rightarrow Na_3PO_4 + 3H_2O\] Similar to phosphorus (III) oxide, if phosphorus(V) oxide reacts directly with sodium hydroxide solution, the same possible salt as in the third step (and only this salt) is formed: \[12NaOH + P_4O_{10} \rightarrow 4Na_3PO_4 + 6H_2O\] Two oxides are considered: sulfur dioxide, SO , and sulfur trioxide, SO . Sulfur dioxide is fairly soluble in water, reacting to give a solution of sulfurous acid (also known as sulfuric(IV) acid), H SO , as shown in the reaction below. This species only exists in solution, and any attempt to isolate it gives off sulfur dioxide. \[ SO_2 + H_2O \rightarrow H_2SO_3\] The protonated acid has the following structure: Sulfurous acid is also a relatively weak acid, with a pK of around 1.8, but slightly stronger than the two phosphorus-containing acids above. A reasonably concentrated solution of sulfurous acid has a pH of about 1. Sulfur dioxide also reacts directly with bases such as sodium hydroxide solution. Bubbling sulfur dioxide through sodium hydroxide solution first forms sodium sulfite solution, followed by sodium hydrogen sulfite solution if the sulfur dioxide is in excess. \[ SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O\] \[Na_2SO_3 + H_2O \rightarrow 2NaHSO_3\] Another important reaction of sulfur dioxide is with the base calcium oxide to form calcium sulfite (also known as calcium sulfate(IV)). This is of the important methods of removing sulfur dioxide from flue gases in power stations. \[CaO + SO_2 \rightarrow CaSO_3\] Sulfur trioxide reacts violently with water to produce a fog of concentrated sulfuric acid droplets. \[ SO_3 + H_2O \rightarrow H_2SO_4\] Pure, fully-protonated sulfuric acid has the structure: Sulfuric acid is a strong acid, and solutions will typically have a pH around 0. The acid reacts with water to give a hydronium ion (a hydrogen ion in solution) and a hydrogen sulfate ion. This reaction runs essentially to completion: \[ H_2SO_4 (aq) + H_2O (l) \rightarrow H_3P^+ + HSO_4^- (aq)\] The second proton is more difficult to remove. In fact, the hydrogen sulfate ion is a relatively weak acid, similar in strength to the acids discussed above. This reaction is more appropriately described as an equilibrium: \[ HSO_4^- (aq) + H_2O \rightleftharpoons H_3O^+ (aq) + SO_4^{2-} (aq)\] It is useful if you understand the reason that sulfuric acid is a stronger acid than sulfurous acid. You can apply the same reasoning to other acids that you find on this page as well. Sulfuric acid is stronger than sulfurous acid because when a hydrogen ion is lost from one of the -OH groups on sulfuric acid, the negative charge left on the oxygen is spread out (delocalized) over the ion by interacting with the doubly-bonded oxygen atoms. It follows that more double bonded oxygen atoms in the ion make more delocalization possible; more delocalization leads to greater stability, making the ion less likely to recombine with a hydrogen ion and revert to the non-ionized acid. Sulfurous acid only has one double bonded oxygen, whereas sulfuric acid has two; the extra double bond provides much more effective delocalization, a much more stable ion, and a stronger acid. Sulfuric acid displays all the reactions characteristic of a strong acid. For example, a reaction with sodium hydroxide forms sodium sulfate; in this reaction, both of the acidic protons react with hydroxide ions as shown: \[2NaOH +H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O\] In principle, sodium hydrogen sulfate can be formed by using half as much sodium hydroxide; in this case, only one of the acidic hydrogen atoms is removed. Sulfur trioxide itself also reacts directly with bases such as calcium oxide, forming calcium sulfate: \[ CaO + SO_3 \rightarrow CaSO_4\] Chlorine forms several oxides, but only two (chlorine(VII) oxide, Cl O , and chlorine(I)oxide, Cl O) are considered here. Chlorine(VII) oxide is also known as dichlorine heptoxide, and chlorine(I) oxide as dichlorine monoxide. Chlorine(VII) oxide is the highest oxide of chlorine—the chlorine atom is in its maximum oxidation state of +7. It continues the trend of the highest oxides of the Period 3 elements towards being stronger acids. Chlorine(VII) oxide reacts with water to give the very strong acid, chloric(VII) acid, also known as perchloric acid. \[ Cl_2O_7 + H_2O \rightarrow 2HClO_4\] As in sulfuric acid, the pH of typical solutions of perchloric acid are around 0. When the chlorate(VII) ion (perchlorate ion) forms by loss of a proton (in a reaction with water, for example), the charge is delocalized over every oxygen atom in the ion. That makes the ion very stable, making chloric(VII) acid very strong. Chloric(VII) acid reacts with sodium hydroxide solution to form a solution of sodium chlorate(VII): \[ NaOH + HClO_4 \rightarrow NaClO_4 + H2O\] Chlorine(VII) oxide itself also reacts directly with sodium hydroxide solution to give the same product: \[ 2NaOH + Cl_2O_7 \rightarrow 2NaClO_4 + H_2O\] Chlorine(I) oxide is far less acidic than chlorine(VII) oxide. It reacts with water to some extent to give chloric(I) acid, \(HOCl^-\) also known as hypochlorous acid. \[ Cl_2O + H_2O \rightleftharpoons 2HOCl\] The structure of chloric(I) acid is exactly as shown by its formula, HOCl. It has no doubly-bonded oxygens, and no way of delocalizing the charge over the negative ion formed by loss of the hydrogen. Therefore, the negative ion formed not very stable, and readily reclaims its proton to revert to the acid. Chloric(I) acid is very weak (pK = 7.43) and reacts with sodium hydroxide solution to give a solution of sodium chlorate(I) (sodium hypochlorite): \[ NaOH + HOCl \rightarrow NaOCl + H_2O\] Chlorine(I) oxide also reacts directly with sodium hydroxide to give the same product: \[2NaOH + Cl_2O \rightarrow 2NaOCl + H_2O\] Jim Clark ( ) | 12,528 | 1,210 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/19%3A_The_Distribution_of_Outcomes_for_Multiple_Trials/19.05%3A_Problems |
1. Leland got a train set for Christmas. It came with seven rail cars. (We say that all seven cars are “distinguishable.”) Four of the rail cars are box cars and three are tank cars. If we distinguish between permutations in which the box cars are coupled (lined up) differently but not between permutations in which tank cars are coupled differently, how many ways can the seven cars be coupled so that all of the tank cars are together? What are they? What formula can we use to compute this number? (Hint: We can represent one of the possibilities as \(b_1b_2b_3b_4T\). This is one of the possibilities in which the first four cars behind the engine are all box cars. There are \(4!\) such possibilities; that is, there are \(4!\) possible permutations for placing the four box cars.) 2. If we don’t care about the order in which the box cars are coupled, and we don’t care about the order in which the tank cars are coupled, how many ways can the rail cars in problem 1 be coupled so that all of the tank cars are together? What are they? What formula can we use to compute this number? 3. If we distinguish between permutations in which either the box cars or the tank cars in problem 1 are ordered differently, how many ways can the rail cars be coupled so that all of the tank cars are together? What formula can we use to compute this number? 4. How many ways can all seven rail cars in problem 1 be coupled if the tank cars need not be together? 5. If, as in the previous problem, we distinguish between permutations in which any of the rail cars are ordered differently, how many ways can the rail cars be coupled so that not all of the tank cars are together? 6. If we distinguish between box cars and tank cars, but we do not distinguish one box car from another box car, and we do not distinguish one tank car from another tank car, how many ways can the rail cars in problem 1 be coupled? 7. If Leland gets five flat cars for his birthday, he will have four box cars, three tank cars and five flat cars. How many ways will Leland be able to couple (permute) these twelve rail cars? 8. If we distinguish between box cars and tank cars, between box cars and flat cars, and between tank cars and flat cars, but we do not distinguish one box car from another box car, and we do not distinguish one tank car from another tank car, and we do not distinguish one flat car from another flat car, how many ways can the rail cars in problem seven be coupled? What formula can we use to compute this number? 9. We are given four distinguishable marbles, labeled \(A--D\), and two cups, labeled \(1\) and \(2\). We want to explore the number of ways we can put two marbles in cup \(1\) and two marbles in cup \(2\). This is the number of combinations, \(C\left(2,2\right)\), for the population set \(N_1=2\), \(N_2=2\). (a) One combination is \({\left[AB\right]}_1{\left[CD\right]}_2\). Find the remaining combinations. What is \(C\left(2,2\right)\)? (b) There are four permutations for the combination given in (a):\(\ {\left[AB\right]}_1{\left[CD\right]}_2\); \({\left[BA\right]}_1{\left[CD\right]}_2\); \({\left[AB\right]}_1{\left[DC\right]}_2\); \({\left[BA\right]}_1{\left[DC\right]}_2\). Find all of the permutations for each of the remaining combinations. (c) How many permutations are there for each combination? (d) Write down all of the possible permutations of marbles \(A--D\). Show that there is a one-to-one correspondence with the permutations in (b). (e) Show that the total number of permutations is equal to the number of combinations times the number of permutations possible for each combination. 10. We are given seven distinguishable marbles, labeled \(A--G\), and two cups, labeled \(1\) and \(2\). We want to find the number of ways we can put three marbles in cup \(1\) and four marbles in cup\(\ 2\). That is, we seek \(C\left(3,4\right)\), the number of combinations in which \(N_1=3\) and \(N_2=4\). \({\left[ABC\right]}_1{\left[DEFG\right]}_2\) is one such combination. (a) How many different ways can these marbles be placed in different orders without exchanging any marbles between cup \(1\) and cup \(2\)? (This is the number of permutations associated with this combination.) (b) Find a different combination with \(N_1=3\) and \(N_2=4\). (c) How many permutations are possible for the marbles in (b)? How many permutations are possible for any combination with \(N_1=3\) and \(N_2=4\)? (d) If \(C\left(3,4\right)\) is the number of combinations in which \(N_1=3\) and \(N_2=4\), and if \(P\) is the number of permutations for each such combination, what is the total number of permutations possible for 7 marbles? (e) How else can one express the number of permutations possible for 7 marbles? (f) Equate your conclusions in (d) and (e). Find \(C\left(3,4\right)\). 11. (a) Calculate the probabilities of 0, 1, 2, 3, and 4 heads in a series of four tosses of an unbiased coin. The event of 2 heads is\(\ 20\%\) of these five events. Note particularly the probability of the event: 2 heads in 4 tosses. (b) Calculate the probabilities of 0, 1, 2, 3, , 8, and 9 heads in a series of nine tosses of an unbiased coin. The events of 4 heads and 5 heads comprise \(20\%\) of these ten cases. Calculate the probability of 4 heads or 5 heads; i.e., the probability of being in the middle \(20\%\) of the possible events. (c) Calculate the probabilities of 0, 1, 2, 3, , 13, and 14 heads in a series of fourteen tosses of an unbiased coin. The events of 6 heads, 7 heads, and 8 heads comprise 20% of these fifteen cases. Calculate the probability of 6, 7, or 8 heads; i.e., the probability of being in the middle \(20\%\) of the possible events. (d) What happens to the probabilities for the middle \(20\%\) of possible events as the number of tosses becomes very large? How does this relate to the fraction heads in a series of tosses when the total number of tosses becomes very large? 12. Let the value of the outcome heads be one and the value of the outcome tails be zero. Let the “score” from a particular simultaneous toss of \(n\) coins be \[\mathrm{score}=1\times \left(\frac{number\ of\ heads}{number\ of\ coins}\right)\ +0\times \left(\frac{number\ of\ tails}{number\ of\ coins}\right)\] Let us refer to the distribution of scores from tosses of \(n\) coins as the “\(S_n\) distribution.” (a) The \(S_1\) distribution comprises two outcomes: \(\mathrm{\{}\)1 head, 0 tail\(\mathrm{\}}\) and \(\mathrm{\{}\)0 head, 1 tail\(\mathrm{\}}\). What is the mean of the \(S_1\) distribution? (b) What is the variance of the \(S_1\) distribution? (c) What is the mean of the \(S_n\) distribution? (d) What is the variance of the \(S_n\) distribution? 13. Fifty unbiased coins are tossed simultaneously. (a) Calculate the probability of 25 heads and 25 tails. (b) Calculate the probability of 23 heads and 27 tails. (c) Calculate the probability of 3 heads and 47 tails. (d) Calculate the ratio of your results for parts (a) and (b). (e) Calculate the ratio of your results for parts (a) and (c). 14. For \(N=3,\ 6\) and \(10\), calculate\(\) (a) The exact value of \(N!\) (b) The value of \(N!\) according to the approximation \[N!\approx N^N \left(2\pi N\right)^{1/2}\mathrm{exp}\left(-N\right)\mathrm{exp}\left(\frac{1}{12N}\right)\] (c) The value of N! according to the approximation \[N!\approx N^N \left(2\pi N\right)^{1/2}\mathrm{exp}\left(-N\right)\] (d) The value of N! according to the approximation \[N!\approx N^N\mathrm{exp}\left(-N\right)\] (e) The ratio of the value in (b) to the corresponding value in (a). (f) The ratio of the value in (c) to the corresponding value in (a). (g) The ratio of the value in (d) to the corresponding value in (a). (h) Comment. 15. Find , \(d ~ \ln N! /dN\) using each of the approximations \[N!\approx N^N \left(2\pi N\right)^{1/2} \mathrm{exp}\left(-N\right)\mathrm{exp}\left(\frac{1}{12N}\right)\approx N^N \left(2\pi N\right)^{1/2} \mathrm{exp}\left(-N\right)\approx N^N\mathrm{exp}\left(-N\right)\] How do the resulting approximations for \(d ~ \ln N! /dN\) compare to one another as \(N\) becomes very large? 16. There are three energy levels available to any one molecule in a crystal of the substance. Consider a crystal containing \(1000\) molecules. These molecules are distinguishable because each occupies a unique site in the crystalline lattice. How many combinations (microstates) are associated with the population set \(N_1=800\), \(N_2=150\), \(N_3=50\)? | 8,459 | 1,211 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/10%3A_Multi-electron_Atoms/Quantum_Numbers_for_Atoms |
A total of four quantum numbers are used to describe completely the movement and trajectories of each electron within an atom. The combination of all quantum numbers of all electrons in an atom is described by a wave function that complies with the Schrödinger equation. Each electron in an atom has a unique set of quantum numbers; according to the , no two electrons can share the same combination of four quantum numbers. Quantum numbers are important because they can be used to determine the electron configuration of an atom and the probable location of the atom's electrons. Quantum numbers are also used to understand other characteristics of atoms, such as ionization energy and the atomic radius. In atoms, there are a total of four quantum numbers: the principal quantum number ( ), the orbital angular momentum quantum number ( ), the magnetic quantum number ( ), and the electron spin quantum number ( ). The principal quantum number, \(n\), describes the energy of an electron and the most probable distance of the electron from the nucleus. In other words, it refers to the size of the orbital and the energy level an electron is placed in. The number of subshells, or \(l\), describes the shape of the orbital. It can also be used to determine the number of angular nodes. The magnetic quantum number, , describes the energy levels in a subshell, and refers to the spin on the electron, which can either be up or down. The principal quantum number, \(n\), designates the principal electron shell. Because describes the most probable distance of the electrons from the nucleus, the larger the number is, the farther the electron is from the nucleus, the larger the size of the orbital, and the larger the atom is. can be any positive integer starting at 1, as \(n=1\) designates the first principal shell (the innermost shell). The first principal shell is also called the ground state, or lowest energy state. This explains why \(n\) can not be 0 or any negative integer, because there exists no atoms with zero or a negative amount of energy levels/principal shells. When an electron is in an excited state or it gains energy, it may jump to the second principle shell, where \(n=2\). This is called absorption because the electron is "absorbing" photons, or energy. Known as emission, electrons can also "emit" energy as they jump to lower principle shells, where n decreases by whole numbers. As the energy of the electron increases, so does the principal quantum number, e.g., = 3 indicates the third principal shell, = 4 indicates the fourth principal shell, and so on. \[n=1,2,3,4…\] If = 7, what is the principal electron shell? If an electron jumped from energy level = 5 to energy level = 3, did absorption or emission of a photon occur? Emission, because energy is lost by release of a photon. The orbital angular momentum quantum number \(l\) determines the shape of an orbital, and therefore the angular distribution. The number of angular nodes is equal to the value of the angular momentum quantum number \(l\). (For more information about angular nodes, see .) Each value of \(l\) indicates a specific s, p, d, f subshell (each unique in shape.) The value of \(l\) is dependent on the principal quantum number \(n\). Unlike \(n\), the value of \(l\) can be zero. It can also be a positive integer, but it cannot be larger than one less than the principal quantum number (\(n-1\)): \[l=0, 1, 2, 3, 4…, (n-1)\] If \(n = 7\), what are the possible values of \(l\)? Since \(l\) can be zero or a positive integer less than (\(n-1\)), it can have a value of 0, 1, 2, 3, 4, 5 or 6. If \(l = 4\), how many angular nodes does the atom have? The number of angular nodes is equal to the value of , so the number of nodes is also 4. The magnetic quantum number \(m_l\) determines the number of orbitals and their orientation within a subshell. Consequently, its value depends on the orbital angular momentum quantum number \(l\). Given a certain \(l\), \(m_l\) is an interval ranging from \(–l\) to \(+l\), so it can be zero, a negative integer, or a positive integer. \[m_l= -l, (-l +1),( -l +2),…, -2, -1, 0, 1, 2, … (l – 1), (l – 2), +l\] Example: If \(n=3\), and \(l=2\), then what are the possible values of \(m_l\)? Since \(m_l\) must range from \(–l\) to \(+l\), then \(m_l\) can be: -2, -1, 0, 1, or 2. Unlike \(n\), \(l\), and \(m_l\), the electron spin quantum number \(m_s\) does not depend on another quantum number. It designates the direction of the electron spin and may have a spin of +1/2, represented by↑, or –1/2, represented by ↓. This means that when \(m_s\) is positive the electron has an upward spin, which can be referred to as "spin up." When it is negative, the electron has a downward spin, so it is "spin down." The significance of the electron spin quantum number is its determination of an atom's ability to generate a magnetic field or not. ( .) \[m_s= \pm \dfrac{1}{2}\] List the possible combinations of all four quantum numbers when \(n=2\), \(l=1\), and \(m_l=0\). The fourth quantum number is independent of the first three, allowing the first three quantum numbers of two electrons to be the same. Since the spin can be +1/2 or =1/2, there are two combinations: Can an electron with \(m_s=1/2\) have a downward spin? No, if the value of \(m_s\) is positive, the electron is "spin up." The value of the principal quantum number n is the level of the principal electronic shell (principal level). All orbitals that have the same n value are in the same principal level. For example, all orbitals on the second principal level have a principal quantum number of n=2. When the value of n is higher, the number of principal electronic shells is greater. This causes a greater distance between the farthest electron and the nucleus. As a result, the size of the atom and its increases. Because the atomic radius increases, the electrons are farther from the nucleus. Thus it is easier for the atom to expel an electron because the nucleus does not have as strong a pull on it, and the decreases. Which orbital has a higher ionization energy, one with \(n=3\) or \(n=2\)? The orbital with n=2, because the closer the electron is to the nucleus or the smaller the atomic radius, the more energy it takes to expel an electron. The number of values of the orbital angular number l can also be used to identify the number of subshells in a principal electron shell: After looking at the examples above, we see that the value of n is equal to the number of subshells in a principal electronic shell: To identify what type of possible subshells n has, these subshells have been assigned letter names. The value of l determines the name of the subshell: Therefore: We can designate a principal quantum number, n, and a certain subshell by combining the value of n and the name of the subshell (which can be found using l). For example, 3p refers to the third principal quantum number (n=3) and the p subshell (l=1). What is the name of the orbital with quantum numbers n=4 and l=1? Knowing that the principal quantum number n is 4 and using the table above, we can conclude that it is 4p. What is the name of the oribital(s) with quantum number n=3? 3s, 3p, and 3d. Because n=3, the possible values of l = 0, 1, 2, which indicates the shapes of each subshell. The number of in a subshell is equivalent to the number of values the magnetic quantum number ml takes on. A helpful equation to determine the number of orbitals in a subshell is 2l +1. This equation will not give you the value of ml, but the number of possible values that ml can take on in a particular orbital. For example, if l=1 and ml can have values -1, 0, or +1, the value of 2l+1 will be three and there will be three different orbitals. The names of the orbitals are named after the subshells they are found in: In the figure below, we see examples of two orbitals: the p orbital (blue) and the s orbital (red). The red s orbital is a 1s orbital. To picture a 2s orbital, imagine a layer similar to a cross section of a jawbreaker around the circle. The layers are depicting the atoms angular nodes. To picture a 3s orbital, imagine another layer around the circle, and so on and so on. The p orbital is similar to the shape of a dumbbell, with its orientation within a subshell depending on m . The shape and orientation of an orbital depends on l and m . To visualize and organize the first three quantum numbers, we can think of them as constituents of a house. In the following image, the roof represents the principal quantum number n, each level represents a subshell l, and each room represents the different orbitals ml in each subshell. The s orbital, because the value of ml can only be 0, can only exist in one plane. The p orbital, however, has three possible values of ml and so it has three possible orientations of the orbitals, shown by Px, Py, and Pz. The pattern continues, with the d orbital containing 5 possible orbital orientations, and f has 7: Another helpful visual in looking at the possible orbitals and subshells with a set of quantum numbers would be the electron orbital diagram. (For more electron orbital diagrams, see .) The characteristics of each quantum number are depicted in different areas of this diagram. Dr. (Japan Fine Ceramics Center) | 9,339 | 1,212 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/20%3A_Periodic_Trends_and_the_s-Block_Elements/20.04%3A_The_Alkali_Metals_(Group_1) |
The alkali metals are so reactive that they are never found in nature in elemental form. Although some of their ores are abundant, isolating them from their ores is somewhat difficult. For these reasons, the group 1 elements were unknown until the early 19th century, when Sir Humphry Davy first prepared sodium (Na) and potassium (K) by passing an electric current through molten alkalis. (The ashes produced by the combustion of wood are largely composed of potassium and sodium carbonate.) Lithium (Li) was discovered 10 years later when the Swedish chemist Johan Arfwedson was studying the composition of a new Brazilian mineral. Cesium (Cs) and rubidium (Rb) were not discovered until the 1860s, when Robert Bunsen conducted a systematic search for new elements. Known to chemistry students as the inventor of the Bunsen burner, Bunsen’s spectroscopic studies of ores showed sky blue and deep red emission lines that he attributed to two new elements, Cs and Rb, respectively. Francium (Fr) is found in only trace amounts in nature, so our knowledge of its chemistry is limited. All the isotopes of Fr have very short half-lives, in contrast to the other elements in group 1. Davy was born in Penzance, Cornwall, England. He was a bit of a wild man in the laboratory, often smelling and tasting the products of his experiments, which almost certainly shortened his life. He discovered the physiological effects that cause nitrous oxide to be called “laughing gas” (and became addicted to it!), and he almost lost his eyesight in an explosion of nitrogen trichloride (NCl ), which he was the first to prepare. Davy was one of the first to recognize the utility of Alessandro Volta’s “electric piles” (batteries). By connecting several “piles” in series and inserting electrodes into molten salts of the alkali metals and alkaline earth metals, he was able to isolate six previously unknown elements as pure metals: sodium, potassium, calcium, strontium, barium, and magnesium. He also discovered boron and was the first to prepare phosphine (PH ) and hydrogen telluride (H Te), both of which are highly toxic. Bunsen was born and educated in Göttingen, Germany. His early work dealt with organic arsenic compounds, whose highly toxic nature and explosive tendencies almost killed him and did cost him an eye. He designed the Bunsen burner, a reliable gas burner, and used it and emission spectra to discover cesium (named for its blue line) and rubidium (named for its red line). Because the alkali metals are among the most potent reductants known, obtaining them in pure form requires a considerable input of energy. Pure lithium and sodium for example, are typically prepared by the electrolytic reduction of molten chlorides: \[\mathrm{LiCl(l)}\rightarrow\mathrm{Li(l)}+\frac{1}{2}\mathrm{Cl_2(g)} \label{21.15}\] In practice, CaCl is mixed with LiCl to lower the melting point of the lithium salt. The electrolysis is carried out in an argon atmosphere rather than the nitrogen atmosphere typically used for substances that are highly reactive with O and water because Li reacts with nitrogen gas to form lithium nitride (Li N). Metallic sodium is produced by the electrolysis of a molten mixture of NaCl and CaCl . In contrast, potassium is produced commercially from the reduction of KCl by Na, followed by the fractional distillation of K(g). Although rubidium and cesium can also be produced by electrolysis, they are usually obtained by reacting their hydroxide salts with a reductant such as Mg: \[2RbOH_{(s)} + Mg_{(s)} \rightarrow 2Rb_{(l)} + Mg(OH)_{2(s)} \label{21.6}\] Massive deposits of essentially pure NaCl and KCl are found in nature and are the major sources of sodium and potassium. The other alkali metals are found in low concentrations in a wide variety of minerals, but ores that contain high concentrations of these elements are relatively rare. No concentrated sources of rubidium are known, for example, even though it is the 16th most abundant element on Earth. Rubidium is obtained commercially by isolating the 2%–4% of Rb present as an impurity in micas, minerals that are composed of sheets of complex hydrated potassium–aluminum silicates. Alkali metals are recovered from silicate ores in a multistep process that takes advantage of the pH-dependent solubility of selected salts of each metal ion. The steps in this process are leaching, which uses sulfuric acid to dissolve the desired alkali metal ion and Al from the ore; basic precipitation to remove Al from the mixture as Al(OH) ; selective precipitation of the insoluble alkali metal carbonate; dissolution of the salt again in hydrochloric acid; and isolation of the metal by evaporation and electrolysis. Figure \(\Page {1}\) illustrates the isolation of liquid lithium from a lithium silicate ore by this process. Various properties of the group 1 elements are summarized in Table \(\Page {1}\). In keeping with overall periodic trends, the atomic and ionic radii increase smoothly from Li to Cs, and the first ionization energies decrease as the atoms become larger. As a result of their low first ionization energies, the alkali metals have an overwhelming tendency to form ionic compounds where they have a +1 charge. All the alkali metals have relatively high electron affinities because the addition of an electron produces an anion (M−) with an ns electron configuration. The densities of the elements generally increase from Li to Cs, reflecting another common trend: because the atomic masses of the elements increase more rapidly than the atomic volumes as you go down a group, the densest elements are near the bottom of the periodic table. An unusual trend in the group 1 elements is the smooth decrease in the melting and boiling points from Li to Cs. As a result, Cs (melting point = 28.5°C) is one of only three metals (the others are Ga and Hg) that are liquids at body temperature (37°C). The standard reduction potentials (E°) of the alkali metals do not follow the trend based on ionization energies. Unexpectedly, lithium is the strongest reductant, and sodium is the weakest (Table \(\Page {1}\)). Because Li is much smaller than the other alkali metal cations, its hydration energy is the highest. The high hydration energy of Li more than compensates for its higher ionization energy, making lithium metal the strongest reductant in aqueous solution. This apparent anomaly is an example of how the physical or the chemical behaviors of the elements in a group are often determined by the subtle interplay of opposing periodic trends. All alkali metals are electropositive elements with an ns valence electron configuration, forming the monocation (M ) by losing the single valence electron. Because removing a second electron would require breaking into the (n − 1) closed shell, which is energetically prohibitive, the chemistry of the alkali metals is largely that of ionic compounds that contain M ions. However, as we discuss later, the lighter group 1 elements also form a series of organometallic compounds that contain polar covalent M–C bonds. All the alkali metals react vigorously with the halogens (group 17) to form the corresponding ionic halides, where \(X\) is a halogen: \[2M_{(s)} + X_{2(s, l, g)} \rightarrow 2M^+X^−_{(s)} \label{21.7}\] Similarly, the alkali metals react with the heavier chalcogens (sulfur, selenium, and tellurium in group 16) to produce metal chalcogenides, where Y is S, Se, or Te: \[2M_{(s)} + Y_{(s)} \rightarrow M_2Y_{(s)} \label{21.8}\] When excess chalcogen is used, however, a variety of products can be obtained that contain chains of chalcogen atoms, such as the sodium polysulfides (Na S , where n = 2–6). For example, Na S contains the S ion, which is V shaped with an S–S–S angle of about 103°. The one-electron oxidation product of the trisulfide ion (S ) is responsible for the intense blue color of the gemstones lapis lazuli and blue ultramarine (Figure \(\Page {2}\)). Reacting the alkali metals with oxygen, the lightest element in group 16, is more complex, and the stoichiometry of the product depends on both the metal:oxygen ratio and the size of the metal atom. For instance, when alkali metals burn in air, the observed products are Li O (white), Na O (pale yellow), KO (orange), RbO (brown), and CsO (orange). Only Li O has the stoichiometry expected for a substance that contains two M cations and one O ion. In contrast, Na O contains the O (peroxide) anion plus two Na cations. The other three salts, with stoichiometry MO , contain the M cation and the O (superoxide) ion. Because O is the smallest of the three oxygen anions, it forms a stable ionic lattice with the smallest alkali metal cation (Li ). In contrast, the larger alkali metals—potassium, rubidium, and cesium—react with oxygen in air to give the metal superoxides. Because the Na cation is intermediate in size, sodium reacts with oxygen to form a compound with an intermediate stoichiometry: sodium peroxide. Under specific reaction conditions, however, it is possible to prepare the oxide, peroxide, and superoxide salts of all five alkali metals, except for lithium superoxide (LiO ). The chemistry of the alkali metals is largely that of ionic compounds containing the M ions. The alkali metal peroxides and superoxides are potent oxidants that react, often vigorously, with a wide variety of reducing agents, such as charcoal or aluminum metal. For example, Na O is used industrially for bleaching paper, wood pulp, and fabrics such as linen and cotton. In submarines, Na O and KO are used to purify and regenerate the air by removing the CO produced by respiration and replacing it with O . Both compounds react with CO in a redox reaction in which O or O is simultaneously oxidized and reduced, producing the metal carbonate and O : \[2Na_2O_{2(s)} + 2CO_{2(g)} \rightarrow 2Na_2CO_{3(s)} + O_{2(g)} \label{21.9}\] \[4KO_{2(s)} + 2CO_{2(g)} \rightarrow 2K_2CO_{3(s)} + 3O_{2(g)} \label{21.10}\] The presence of water vapor, the other product of respiration, makes KO even more effective at removing CO because potassium bicarbonate, rather than potassium carbonate, is formed: \[4KO_{2(s)} + 4CO_{2(g)} + 2H_2O_{(g)} \rightarrow 4KHCO_{3(s)} + 3O_{2(g)} \label{21.11}\] Notice that 4 mol of CO are removed in this reaction, rather than 2 mol in Equation 21.10. Lithium, the lightest alkali metal, is the only one that reacts with atmospheric nitrogen, forming lithium nitride (Li N). Lattice energies again explain why the larger alkali metals such as potassium do not form nitrides: packing three large K cations around a single relatively small anion is energetically unfavorable. In contrast, all the alkali metals react with the larger group 15 elements phosphorus and arsenic to form metal phosphides and arsenides (where Z is P or As): \[12M_{(s)} + Z_{4(s)} \rightarrow 4M_3Z_{(s)} \label{21.12}\] Because of lattice energies, only lithium forms a stable oxide and nitride. The alkali metals react with all group 14 elements, but the compositions and properties of the products vary significantly. For example, reaction with the heavier group 14 elements gives materials that contain polyatomic anions and three-dimensional cage structures, such as K Si whose structure is shown here. In contrast, lithium and sodium are oxidized by carbon to produce a compound with the stoichiometry M C (where M is Li or Na): \[ 2M_{(s)} + 2C_{(s)} \rightarrow M_2C_{2(s)} \label{21.13}\] The same compounds can be obtained by reacting the metal with acetylene (C H ). In this reaction, the metal is again oxidized, and hydrogen is reduced: \[ 2M_{(s)} + C_2H_{2(g)} \rightarrow M_2C_{2(s)} + H_{2(g)} \label{21.14}\] The acetylide ion (C ), formally derived from acetylene by the loss of both hydrogens as protons, is a very strong base. Reacting acetylide salts with water produces acetylene and MOH(aq). The heavier alkali metals (K, Rb, and Cs) also react with carbon in the form of graphite. Instead of disrupting the hexagonal sheets of carbon atoms, however, the metals insert themselves between the sheets of carbon atoms to give new substances called (part (a) in Figure \(\Page {3}\)). The stoichiometries of these compounds include MC and MC , which are black/gray; MC and MC , which are blue; and MC , which is bronze (part (b) in Figure \(\Page {3}\)). The remarkably high electrical conductivity of these compounds (about 200 times greater than graphite) is attributed to a net transfer of the valence electron of the alkali metal to the graphite layers to produce, for example, K C . All the alkali metals react directly with gaseous hydrogen at elevated temperatures to produce ionic hydrides (M H ): \[2M_{(s)} + H_{2(g)} \rightarrow 2MH_{(s)} \label{21.15a}\] All are also capable of reducing water to produce hydrogen gas: \[\mathrm{M(s)}+\mathrm{H_2O(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{MOH(aq)} \label{21.16}\] Alkali metal cations are found in a wide variety of ionic compounds. In general, any alkali metal salt can be prepared by reacting the alkali metal hydroxide with an acid and then evaporating the water: \[2MOH_{(aq)} + H_2SO_{4(aq)} \rightarrow M_2SO_{4(aq)} + 2H_2O_{(l)} \label{21.17}\] \[MOH_{(aq)} + HNO_{3(aq)} \rightarrow MNO_{3(aq)} + H_2O_{(l)} \label{21.18}\] Hydroxides of alkali metals also can react with organic compounds that contain an acidic hydrogen to produce a salt. An example is the preparation of sodium acetate (CH CO Na) by reacting sodium hydroxide and acetic acid: \[CH_3CO_2H_{(aq)} + NaOH_{(s)} \rightarrow CH_3CO_2Na_{(aq)} + H_2O_{(l)} \label{21.19}\] Soap is a mixture of the sodium and potassium salts of naturally occurring carboxylic acids, such as palmitic acid [CH (CH ) CO H] and stearic acid [CH (CH ) CO H]. Lithium salts, such as lithium stearate [CH (CH ) CO Li], are used as additives in motor oils and greases. Because of their low positive charge (+1) and relatively large ionic radii, alkali metal cations have only a weak tendency to react with simple Lewis bases to form metal complexes. Complex formation is most significant for the smallest cation (Li ) and decreases with increasing radius. In aqueous solution, for example, Li forms the tetrahedral [Li(H O) ] complex. In contrast, the larger alkali metal cations form octahedral [M(H O) ] complexes. Complex formation is primarily due to the electrostatic interaction of the metal cation with polar water molecules. Because of their high affinity for water, anhydrous salts that contain Li and Na ions (such as Na SO ) are often used as drying agents. These compounds absorb trace amounts of water from nonaqueous solutions to form hydrated salts, which are then easily removed from the solution by filtration. Because of their low positive charge (+1) and relatively large ionic radii, alkali metal cations have only a weak tendency to form complexes with simple Lewis bases. Electrostatic interactions also allow alkali metal ions to form complexes with certain cyclic polyethers and related compounds, such as crown ethers and cryptands. As discussed in Chapter 13, are cyclic polyethers that contain four or more oxygen atoms separated by two or three carbon atoms. All crown ethers have a central cavity that can accommodate a metal ion coordinated to the ring of oxygen atoms, and crown ethers with rings of different sizes prefer to bind metal ions that fit into the cavity. For example, 14-crown-4, with the smallest cavity that can accommodate a metal ion, has the highest affinity for Li , whereas 18-crown-6 forms the strongest complexes with K . are more nearly spherical analogues of crown ethers and are even more powerful and selective complexing agents. Cryptands consist of three chains containing oxygen that are connected by two nitrogen atoms (part (b) in Figure 13.7). They can completely surround (encapsulate) a metal ion of the appropriate size, coordinating to the metal by a lone pair of electrons on each O atom and the two N atoms. Like crown ethers, cryptands with different cavity sizes are highly selective for metal ions of particular sizes. Crown ethers and cryptands are often used to dissolve simple inorganic salts such as KMnO in nonpolar organic solvents. A remarkable feature of the alkali metals is their ability to dissolve reversibly in liquid ammonia. Just as in their reactions with water, reacting alkali metals with liquid ammonia eventually produces hydrogen gas and the metal salt of the conjugate base of the solvent—in this case, the amide ion (NH ) rather than hydroxide: \[\mathrm{M(s)}+\mathrm{NH_3(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{M^+(am)}+\mathrm{NH_2^-(am)} \label{21.20}\] where the (am) designation refers to an ammonia solution, analogous to (aq) used to indicate aqueous solutions. Without a catalyst, the reaction in Equation 21.20 tends to be rather slow. In many cases, the alkali metal amide salt (MNH ) is not very soluble in liquid ammonia and precipitates, but when dissolved, very concentrated solutions of the alkali metal are produced. One mole of Cs metal, for example, will dissolve in as little as 53 mL (40 g) of liquid ammonia. The pure metal is easily recovered when the ammonia evaporates. Solutions of alkali metals in liquid ammonia are intensely colored and good conductors of electricity due to the presence of solvated electrons (e , NH ), which are not attached to single atoms. A solvated electron is loosely associated with a cavity in the ammonia solvent that is stabilized by hydrogen bonds. Alkali metal–liquid ammonia solutions of about 3 M or less are deep blue (Figure \(\Page {5}\)) and conduct electricity about 10 times better than an aqueous NaCl solution because of the high mobility of the solvated electrons. As the concentration of the metal increases above 3 M, the color changes to metallic bronze or gold, and the conductivity increases to a value comparable with that of the pure liquid metals. In addition to solvated electrons, solutions of alkali metals in liquid ammonia contain the metal cation (M ), the neutral metal atom (M), metal dimers (M ), and the metal anion (M ). The anion is formed by adding an electron to the singly occupied ns valence orbital of the metal atom. Even in the absence of a catalyst, these solutions are not very stable and eventually decompose to the thermodynamically favored products: M NH and hydrogen gas (Equation 21.20). Nonetheless, the solvated electron is a potent reductant that is often used in synthetic chemistry. Compounds that contain a metal covalently bonded to a carbon atom of an organic species are called . The properties and reactivities of organometallic compounds differ greatly from those of either the metallic or organic components. Because of its small size, lithium, for example, forms an extensive series of covalent organolithium compounds, such as methyllithium (LiCH ), which are by far the most stable and best-known group 1 organometallic compounds. These volatile, low-melting-point solids or liquids can be sublimed or distilled at relatively low temperatures and are soluble in nonpolar solvents. Like organic compounds, the molten solids do not conduct electricity to any significant degree. Organolithium compounds have a tendency to form oligomers with the formula (RLi) , where R represents the organic component. For example, in both the solid state and solution, methyllithium exists as a tetramer with the structure shown in Figure \(\Page {6}\), where each triangular face of the Li tetrahedron is bridged by the carbon atom of a methyl group. Effectively, the carbon atom of each CH group is using a single pair of electrons in an sp hybrid lobe to bridge three lithium atoms, making this an example of two-electron, four-center bonding. Clearly, such a structure, in which each carbon atom is apparently bonded to six other atoms, cannot be explained using any of the electron-pair bonding schemes. Molecular orbital theory can explain the bonding in methyllithium, but the description is beyond the scope of this text. Because sodium remains liquid over a wide temperature range (97.8–883°C), it is used as a coolant in specialized high-temperature applications, such as nuclear reactors and the exhaust valves in high-performance sports car engines. Cesium, because of its low ionization energy, is used in photosensors in automatic doors, toilets, burglar alarms, and other electronic devices. In these devices, cesium is ionized by a beam of visible light, thereby producing a small electric current; blocking the light interrupts the electric current and triggers a response. Compounds of sodium and potassium are produced on a huge scale in industry. Each year, the top 50 industrial compounds include NaOH, used in a wide variety of industrial processes; Na CO , used in the manufacture of glass; K O, used in porcelain glazes; and Na SiO , used in detergents. Several other alkali metal compounds are also important. For example, Li CO is one of the most effective treatments available for manic depression or bipolar disorder. It appears to modulate or dampen the effect on the brain of changes in the level of neurotransmitters, which are biochemical substances responsible for transmitting nerve impulses between neurons. Consequently, patients who take “lithium” do not exhibit the extreme mood swings that characterize this disorder. For each application, choose the more appropriate substance based on the properties and reactivities of the alkali metals and their compounds. Explain your choice in each case. application and selected alkali metals appropriate metal for each application Use the properties and reactivities discussed in this section to determine which alkali metal is most suitable for the indicated application. Indicate which of the alternative alkali metals or their compounds given is more appropriate for each application. Predict the products of each reaction and then balance each chemical equation. reactants products and balanced chemical equation Determine whether one of the reactants is an oxidant or a reductant or a strong acid or a strong base. If so, a redox reaction or an acid–base reaction is likely to occur. Identify the products of the reaction. If a reaction is predicted to occur, balance the chemical equation. The balanced chemical equation is 2Na(s) + O (g) → Na O (s). The balanced chemical equation is Li O(s) + H O(l) → 2LiOH(aq). The balanced chemical equation is as follows: \(\mathrm{K(s)}+\mathrm{CH_3OH(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{CH_3OK(soln)}\). Two moles of lithium are required to balance the equation: 2Li(s) + CH Cl(l) → LiCl(s) + CH Li(soln). We conclude that the two substances will not react with each other. Predict the products of each reaction and balance each chemical equation. The alkali metals are potent reductants whose chemistry is largely that of ionic compounds containing the M ion. Alkali metals have only a weak tendency to form complexes with simple Lewis bases. The first alkali metals to be isolated (Na and K) were obtained by passing an electric current through molten potassium and sodium carbonates. The alkali metals are among the most potent reductants known; most can be isolated by electrolysis of their molten salts or, in the case of rubidium and cesium, by reacting their hydroxide salts with a reductant. They can also be recovered from their silicate ores using a multistep process. Lithium, the strongest reductant, and sodium, the weakest, are examples of the physical and chemical effects of opposing periodic trends. The alkali metals react with halogens (group 17) to form ionic halides; the heavier chalcogens (group 16) to produce metal chalcogenides; and oxygen to form compounds, whose stoichiometry depends on the size of the metal atom. The peroxides and superoxides are potent oxidants. The only alkali metal to react with atmospheric nitrogen is lithium. Heavier alkali metals react with graphite to form graphite intercalation compounds, substances in which metal atoms are inserted between the sheets of carbon atoms. With heavier group 14 elements, alkali metals react to give polyatomic anions with three-dimensional cage structures. All alkali metals react with hydrogen at high temperatures to produce the corresponding hydrides, and all reduce water to produce hydrogen gas. Alkali metal salts are prepared by reacting a metal hydroxide with an acid, followed by evaporation of the water. Both Li and Na salts are used as drying agents, compounds that are used to absorb water. Complexing agents such as crown ethers and cryptands can accommodate alkali metal ions of the appropriate size. Alkali metals can also react with liquid ammonia to form solutions that slowly decompose to give hydrogen gas and the metal salt of the amide ion (NH ). These solutions, which contain unstable solvated electrons loosely associated with a cavity in the solvent, are intensely colored, good conductors of electricity, and excellent reductants. Alkali metals can react with organic compounds that contain an acidic proton to produce salts. They can also form organometallic compounds, which have properties that differ from those of their metallic and organic components. | 25,289 | 1,213 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/13%3A_Intermolecular_Forces/13.05%3A_The_Structure_and_Properties_of_Water |
With 70% of our earth being ocean water and 65% of our bodies being water, it is hard to not be aware of how important it is in our lives. There are 3 different forms of water, or H O: (ice), (water), and (steam). Because water seems so ubiquitous, many people are unaware of the unusual and unique properties of water, including: If you look at the periodic table and locate tellurium (atomic number: 52), you find that the boiling points of hydrides decrease as molecule size decreases. So the hydride for tellurium: has a boiling point of . Moving up, the next hydride would be with a boiling point of . One more up and you find that has a boiling point at . The next hydride would be . And we all know that the boiling point of water is . So despite its molecular weight, water has an incredibly boiling point. This is because water requires more energy to break its hydrogen bonds before it can then begin to boil. The same concept is applied to freezing point as well, as seen in the table below. The boiling and freezing points of water enable the molecules to be very slow to boil or freeze, this is important to the ecosystems living in water. If water was very easy to freeze or boil, drastic changes in the environment and so in oceans or lakes would cause all the organisms living in water to die. This is also why sweat is able to cool our bodies. Besides mercury, water has the highest for all liquids. Water's high surface tension is due to the hydrogen bonding in water molecules. Water also has an exceptionally high . Vaporization occurs when a liquid changes to a gas, which makes it an endothermic reaction. Water's heat of vaporization is 41 kJ/mol. is inversely related to intermolecular forces, so those with stronger intermolecular forces have a lower vapor pressure. Water has very strong intermolecular forces, hence the low vapor pressure, but it's even lower compared to larger molecules with low vapor pressures. All substances, including water, become less dense when they are heated and more dense when they are cooled. So if water is cooled, it becomes more dense and forms ice. Water is one of the few substances whose solid state can float on its liquid state! Why? Water continues to become more dense until it reaches 4°C. After it reaches 4°C, it becomes dense. When freezing, molecules within water begin to move around more slowly, making it easier for them to form hydrogen bonds and eventually arrange themselves into an open crystalline, hexagonal structure. Because of this open structure as the water molecules are being held further apart, the volume of water about 9%. So molecules are more tightly packed in water's liquid state than its solid state. This is why a can of soda can explode in the freezer. It is very rare to find a compound that lacks carbon to be a liquid at standard temperatures and pressures. So it is unusual for water to be a liquid at room temperature! Water is liquid at room temperature so it's able to move around quicker than it is as solid, enabling the molecules to form fewer hydrogen bonds resulting in the molecules being packed more closely together. Each water molecule links to four others creating a tetrahedral arrangement, however they are able to move freely and slide past each other, while ice forms a solid, larger hexagonal structure. As water boils, its hydrogen bonds are broken. Steam particles move very far apart and fast, so barely any hydrogen bonds have the time to form. So, less and less hydrogen bonds are present as the particles reach the critical point above steam. The lack of hydrogen bonds explains why steam causes much worse burns that water. Steam contains all the energy used to break the hydrogen bonds in water, so when steam hits your face you first absorb the energy the steam has taken up from breaking the hydrogen bonds it its liquid state. Then, in an exothermic reaction, steam is converted into liquid water and heat is released. This heat adds to the heat of boiling water as the steam condenses on your skin. Because of water's polarity, it is able to dissolve or dissociate many particles. Oxygen has a slightly negative charge, while the two hydrogens have a slightly positive charge. The slightly negative particles of a compound will be attracted to water's hydrogen atoms, while the slightly positive particles will be attracted to water's oxygen molecule; this causes the compound to dissociate. Besides the explanations above, we can look to some attributes of a water molecule to provide some more reasons of water's uniqueness: The properties of water make it suitable for organisms to survive in during differing weather conditions. Ice freezes as it expands, which explains why ice is able to float on liquid water. During the winter when lakes begin to freeze, the surface of the water freezes and then moves down toward deeper water; this explains why people can ice skate on or fall through a frozen lake. If ice was not able to float, the lake would freeze from the bottom up killing all ecosystems living in the lake. However ice floats, so the fish are able to survive under the surface of the ice during the winter. The surface of ice above a lake also shields lakes from the cold temperature outside and insulates the water beneath it, allowing the lake under the frozen ice to stay liquid and maintain a temperature adequate for the ecosystems living in the lake to survive. | 5,465 | 1,214 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/05%3A_Experimental_Methods/5.01%3A_Determining_Reaction_Order |
Kinetics are used to study of the rate of the reaction and how factors such as temperature, concentration of the reactants, and catalysts affect it. The reaction order shows how the concentration of reactants affects the reaction rate. Determining the reaction rate also helps determine the reaction mechanism. For the reaction \(\ce{A \rightarrow B}\) the rate is given by \[\text{rate}=[\text{A}]^x[\text{B}]^y \nonumber \] The overall rate is the sum of these superscripts. Although there is an infinite variety of rates laws possible, three simple reaction orders commonly taught: zeroth, first, and second order. In zero order reactions, the disappearance of reactants is \[\dfrac{-d[\text{A}]}{dt}= k[\text{A}]^0= k \nonumber \] Its integrated form is \[[\text{A}]=-kt+[\text{A}]_0 \nonumber \] In first order reactions, the disappearance of reactants is \[\dfrac{-d[\text{A}]}{dt}=k[\text{A}]^1 \nonumber \] The integrated form is \[[\text{A}]=[\text{A}]_0 e^{-kt} \label{1st} \] In second order reactions with two reactant species, the rate of disappearance of \(\text{A}\) is \[\dfrac{-d[\text{A}]}{dt}= k[\text{A},\text{B}] \nonumber \] The integrated form is \[\dfrac{1}{[\text{B}]_0-[\text{A}]_0}\ln \dfrac{[\text{B},\text{A}]_0}{[\text{B}]_0[\text{A}]}=kt \label{2nd} \] when \([\text{B}]_0>[\text{A}]_0\). When \([\text{B}]_0>>[\text{A}]_0\), then \([\text{B}] \approx [\text{B}]_0\) and Equation \(\ref{2nd}\) becomes \[\dfrac{1}{[\text{B}]_0-[\text{A}]_0}\ln \dfrac{[\text{B},\text{A}]_0}{[\text{B}]_0[\text{A}]} \approx \dfrac{1}{[\text{B}]_0}\ln \dfrac{[\text{A}]_0}{[\text{A}]}=kt \nonumber \] or \[ [\text{A}] = [\text{A}]_0 e^{-[\text{B}]kt} \nonumber \] This functional form of the decay kinetics is similar ot the first order kinetics in Equation \(\ref{1st}\) and the system is said to operate under pseudo-first order kinetics. In second order reactions with one reactant, the disappearance of reactants is \[\dfrac{-d[\text{A}]}{dt}= k[\text{A}]^2 \nonumber \] The integrated form is \[\dfrac{1}{[\text{A}]}= kt+\dfrac{1}{[\text{A}]_0} \nonumber \] The graphical method makes use of the concentrations of reactants. It is most useful when one reactant is isolated by having the others in large excess. A series of standards is used to make a Beer's Law plot, the best-fit equation of which is used to determine the concentration of the isolated reactant. To test for the order of reaction with regard to that reactant, three plots are made. The first is concentration of the isolated reactant versus time. The second is of inverse concentration versus time, while the third is of the natural log of concentration versus time. These graphs, respectively, show zero, first, and second order dependence on the specific reactant. The graph that is most linear shows the order of the reaction with regard to that reactant. \[\ce{ CH_3COCH_3 + Br_2 + H^+ \rightarrow CH_3COCH_2Br + HBr} \nonumber \] For example, in the acid-catalyzed bromination of acetone, the concentration of bromine is isolated and measured, while the concentrations of the hydrogen ion and acetone are not. The three graphs from a run are below. The graph of concentration versus time is the one that is most linear, so the order of the bromination reaction with respect to bromine is zero. The slope of the best-fit line gives \(-k_{obs}\). In theory, the other reactants could be isolated like bromine was, but the data from bromine can be used to determine the reaction order with regard to them as well. It is found by comparing two reactions where only the concentration of the reactant in question is changed. If is the reaction order with regard to acetone, \[ p = \dfrac{ \log \dfrac{k_{obsII}}{k_{obsI}}}{ \log u } \nonumber \] The observed rate constant at the higher concentration of acetone is \(k_{obsII}\), while \(k_{obsI}\) is the observed rate constant at the lower concentration of acetone. The ratio of the higher concentration to the lower concentration is given by . The process is repeated for the hydrogen ion. The mathematical method is useful when the means to graph are not available. It is essentially determining the slope of the plot that "linearizes the data". This requires plotting concentration versus time data. As in the graphical method, the inverse and natural log of the concentration must be calculated. A linear plot has a constant slope, so the slopes calculated from two pairs of adjacent points should be the same.Take three consecutive points from the concentration versus time data. Calculate \( \dfrac{\Delta y}{\Delta x}\) for the first and second points. The concentration is the \(y\) value, while time is the \(x\) value. Do the same for the second and third point. If the reaction is zero order with regard to the reactant, the numbers will be the same. If not, then calculate the slope for the inverse concentration versus time data or natural log of the concentration versus time data. The slope of the first two points is \[m= \dfrac{0.00344\,\text{M} - 0.00349\,\text{M}}{4.42\, \text{min}-4.25\, \text{min}} \nonumber \] or \[m= -2.9 \times 10^{-4}\dfrac{\text{M}}{\text{min}} \nonumber \] The slope of the second two points is \[\dfrac{0.00476\, \text{M}-0.00479\,\text{M}}{4.58 \,\text{min}-4.42\, \text{min}} \nonumber \] or \[-3.1 \times 10^{-4}\dfrac{\text{M}}{\text{min}} \nonumber \] These are approximately the same, so the bromine depletion follows zero order kinetics. Another method uses half lives. For a zero-order reaction, \[t_{1/2}=\dfrac{[\text{A}]_0}{2k} \nonumber \] For a first-order reaction, \[t_{1/2}=\dfrac{\ln 2}{k} \nonumber \] For a second-order reaction, \[t_{1/2}=\dfrac{1}{[\text{A}]_0k} \nonumber \] If an increase in reactant increases the half life, the reaction has zero-order kinetics. If it has no effect, it has first-order kinetics. If the increase in reactant decreases the half life, the reaction has second-order kinetics. | 5,937 | 1,216 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/17%3A_Solubility_and_Complexation_Equilibria/17.04%3A_The_Formation_of_Complex_Ions |
Previously, you learned that metal ions in aqueous solution are hydrated—that is, surrounded by a shell of usually four or six water molecules. A hydrated ion is one kind of a (or, simply, complex), a species formed between a central metal ion and one or more surrounding , molecules or ions that contain at least one lone pair of electrons, such as the [Al(H O) ] ion. A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions, such as Cu or Ru , have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions. As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu ion {[Cu(H O) ] }. Because it is a stronger base than H O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH ) (H O) ] ion. Formation of the [Cu(NH ) (H O) ] complex is accompanied by a dramatic color change, as shown in . The solution changes from the light blue of [Cu(H O) ] to the blue-violet characteristic of the [Cu(NH ) (H O) ] ion. The replacement of water molecules from [Cu(H O) ] by ammonia occurs in sequential steps. Omitting the water molecules bound to Cu for simplicity, we can write the equilibrium reactions as follows: The sum of the stepwise reactions is the overall equation for the formation of the complex ion: The hydrated Cu ion contains six H O ligands, but the complex ion that is produced contains only four \(NH_3\) ligands, not six. \[Cu^{2+}_{(aq)} + 4NH_{3(aq)} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)} \label{17.3.2}\] The equilibrium constant for the formation of the complex ion from the hydrated ion is called the . The equilibrium constant expression for has the same general form as any other equilibrium constant expression. In this case, the expression is as follows: \[K_\textrm f=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}},\mathrm{NH_3}]^4}=2.1\times10^{13}=K_1K_2K_3K_4\label{17.3.3}\] Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu (aq) ion is represented as Cu for simplicity. As for any equilibrium, the larger the value of the equilibrium constant (in this case, ), the more stable the product. With = 2.1 × 10 , the [Cu(NH ) (H O) ] complex ion is very stable. The formation constants for some common complex ions are listed in . Data from , 15th ed. (1999).
Example \(\Page {1}\) If 12.5 g of Cu(NO ) •6H O is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of Cu (aq)? mass of Cu salt and volume and concentration of ammonia solution equilibrium concentration of Cu (aq) Adding an ionic compound that contains Cu to an aqueous ammonia solution will result in the formation of [Cu(NH ) ] (aq), as shown in . We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible. The initial concentration of Cu from the amount of added copper nitrate prior to any reaction is as follows: Because the stoichiometry of the reaction is four NH to one Cu , the amount of NH required to react completely with the Cu is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 10 ), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH ) ] in the first step is complete and allow some of it to dissociate into Cu and NH until equilibrium has been reached. If we define as the amount of Cu produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH ) ] is − , and the change in the concentration of ammonia is +4 , as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations. Substituting the final concentrations into the expression for the formation constant ( ) and assuming that << 0.0846, which allows us to remove from the sum and difference, The value of indicates that our assumption was justified. The equilibrium concentration of Cu (aq) in a 1.00 M ammonia solution is therefore 2.1 × 10 M. The ferrocyanide ion {[Fe(CN) ] } is very stable, with a of 1 × 10 . Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of K [Fe(CN) ]. 1.2 × 10 M What happens to the solubility of a sparingly soluble salt if a ligand that forms a stable complex ion is added to the solution? One such example occurs in conventional black-and-white photography. Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. AgBr is a sparingly soluble salt, with a of 5.35 × 10 at 25°C. When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted AgBr on the film is removed using a complexation reaction to dissolve the sparingly soluble salt. The reaction for the dissolution of silver bromide is as follows: \[AgBr_{(s)} \rightleftharpoons Ag^+_{(aq)} + Br^{−}_{(aq)} \label{17.3.4a}\] with \[K_{sp} = 5.35 \times 10^{−13} \text{ at 25°C} \label{17.3.4b}\] The equilibrium lies far to the left, and the equilibrium concentrations of Ag and Br ions are very low (7.31 × 10 M). As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. Le Chatelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S O ). In photographic processing, excess AgBr is dissolved using a concentrated solution of sodium thiosulfate. The reaction of Ag with thiosulfate is as follows: \[Ag^+_{(aq)} + 2S_2O^{2−}_{3(aq)} \rightleftharpoons [Ag(S_2O_3)_2]^{3−}_{(aq)} \label{17.3.5a}\] with \[K_f = 2.9 \times 10^{13} \label{17.3.5b}\] The magnitude of the equilibrium constant indicates that almost all Ag ions in solution will be immediately complexed by thiosulfate to form [Ag(S O ) ] . We can see the effect of thiosulfate on the solubility of AgBr by writing the appropriate reactions and adding them together: Comparing with shows that the formation of the complex ion increases the solubility of AgBr by approximately 3 × 10 . The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled. If a complex ion has a large , the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts. Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl ). Calculate the solubility of AgCl in each situation: At 25°C, = 1.77 × 10 for AgCl and = 1.1 × 10 for AgCl . of AgCl, of AgCl , and KCl concentration solubility of AgCl in water and in KCl solution with and without the formation of complex ions Thus the solubility of AgCl in pure water at 25°C is 1.33 × 10 M. If the common ion effect were the only important factor, we would predict that AgCl is approximately five orders of magnitude less soluble in a 1.0 M KCl solution than in water. If we let equal the solubility of AgCl in the KCl solution, then at equilibrium [AgCl ] = and [Cl ] = 1.0 − . Substituting these quantities into the equilibrium constant expression for the net reaction and assuming that << 1.0, That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10 M solution of the AgCl complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 10 times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.) Calculate the solubility of mercury(II) iodide (HgI ) in each situation: = 2.9 × 10 for HgI and = 6.8 × 10 for [HgI ] . Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca and Mg , which are present in high concentrations in “hard” water. The precipitation of these salts produces a bathtub ring and gives a gray tinge to clothing. Adding a complexing agent such as pyrophosphate (O POPO , or P O ) or triphosphate (P O ) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large: with However, phosphates can cause environmental damage by promoting eutrophication, the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. The development of phosphate substitutes is an area of intense research. Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Water treated in this way has a saltier taste due to the presence of Na , but it contains fewer dissolved minerals. Another application of complexing agents is found in medicine. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. MRI is based on the magnetic properties of the H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately ( ). One of the most important metal ions for this application is Gd , which with seven unpaired electrons is highly paramagnetic. Because Gd (aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. The complexing agents used for gadolinium are ligands such as DTPA (diethylene triamine pentaacetic acid), whose fully protonated form is shown here. Figure \(\Page {2}\): An MRI Image of the Heart, Arteries, and Veins. When a patient is injected with a paramagnetic metal cation in the form of a stable complex known as an MRI contrast agent, the magnetic properties of water in cells are altered. Because the different environments in different types of cells respond differently, a physician can obtain detailed images of soft tissues. The formation of complex ions can substantially increase the solubility of sparingly soluble salts if the complex ion has a large . A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant ( ). The formation of a complex ion by adding a complexing agent increases the solubility of a compound. | 13,352 | 1,217 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/10%3A_Multi-electron_Atoms/Quantum_Mechanical_H_Atom |
A hydrogen atom consists of a single proton orbited by a single electron. When compared to the electron, the proton has such a large mass that it may be considered stationary while the electron circles around it. It is known that this model is acceptable when the reduced mass of the system is used. Reduced mass is defined below for two masses 1 and 2. \[ \mu = \dfrac{m_1m_2}{m_1+m_2}\] Using the reduced mass effectively converts the two-body problem (two moving and interacting bodies in space) into a one-body problem (a single electron moving about a fixed point). The basic Sch dinger equation is \[\hat{H}\Psi=E\Psi\] where \(\hat{H}\) is the Hamiltonian operator, E is the energy of the particle and \(\Psi\) is the particle's wavefunction that describes its spatial probability. The Hamiltonian operator is the sum of the kinetic energy operator and potential energy operator. The kinetic energy operator is the same for all models but the potential energy changes and is the defining parameter. The hydrogen atom's electron wavefunctions can be described using a variation of the - (RRHO) model. Rather than following a parabolic potential energy surface, the electron associated with the hydrogen experiences an exponential coulombic interaction. This RRHO Hamiltonian combines the kinetic energy elements of both previous models as well as an associated potential energy (as that in the harmonic oscillator scenario). The Schr dinger Equation for the RRHO model involves the Hamiltonian operator acting on a wavefunction that similarly reflects both the rigid rotor and harmonic oscillator models. In order for the Hamiltonian to fit the model, all component Hamiltonian elements need only be added together. \[\hat{H}_{RRHO}=\hat{H}_{RR}(\theta{,}\phi{})+\hat{H}_{HO}(r)\] \[\hat{H}_{RRHO}=\frac{1}{2mr_0^2}\hat{L}^2(\theta{,}\phi{})+\frac{-\hbar{^2}}{2\mu{}}\frac{d^2}{dr^2}+V(r)\] As shown below, the solution wavefunction will be a multiplicative combination of the two model solutions. \[\psi{_{RRHO}}=\psi{}_{RR}*\psi{}_{OH}\] \((\hat{H}_{RR}+\hat{H}_{HO})(\psi{_{RR}}*\psi{_{HO}})=\psi{_{HO}}\hat{H}_{RR}\psi{_{RR}}+\psi{_{RR}}\hat{H}_{HO}\psi{_{HO}}\) \((\hat{H}_{RR}+\hat{H}_{HO})(\psi{_{RR}}*\psi{_{HO}})=\psi{_{HO}}*E_{RR}\psi{_{RR}}+\psi{_{RR}}*E_{HO}\psi{_{HO}}\) \[(] (\hat{H}_{RR}+\hat{H}_{HO})(\psi_{RR}*\psi_{HO})=E_{RR}(\psi{_{RR}}*\psi{_{HO}})+E_{HO}(\psi_{RR}*\psi_{HO}) \] \(\hat{H}_{RRHO}\psi{_{RRHO}}=E_{RR}\psi{_{RRHO}}+E_{HO}\psi{_{RRHO}}") }} Thus, \( E_{RRHO}=E_{RR}+E_{HO}\) After further refinement the Hamiltonian operator for the hydrogen atom is found to be \[\hat{H}=-\dfrac{\hbar{^2}}{2m_e}\bigtriangledown{^2}-\dfrac{e^2}{4\pi{}\epsilon{r}}\] where the Laplacian operator is defined as \[ \bigtriangledown{^2}=\dfrac{\partial{^2}}{\partial{x^2}}+\dfrac{\partial{^2}}{\partial{y^2}}+\dfrac{\partial{^2}}{\partial{z^2}}\] To solve the Schrödinger Equation for the hydrogen atom, it is simplest to perform the quantum mechanical calculations using spherical coordinates (based on the three variables r, \(\theta\) and \(\phi\)). After appropriate adjustments are made to compensate for the change of variables, the Schr dinger equation becomes: \[-\hbar{^2}\dfrac{\partial{}}{\partial{r}}\left(r^2\dfrac{\partial{}\psi{}}{\partial{r}}\right)+\hat{L}^2\psi{}+2m_er^2[V(r)-E]\psi{(}r,\theta{,}\phi{)}=0\] where \( V(r)=\dfrac{e^2}{4\pi{}\epsilon{_0}r}\) is the Coulombic (electrostatic) potential between the nucleus and electron and \(\hat{L}^2\) is the angular momentum operator also found from the quantum mechanical rigid rotor model. Additionally, the assumption must be made that the wavefunction is of a form such that it can be arranged as the product of two functions using different variables (separation of variables). It is found that the wavefunction can be described as \[ \psi{(}r,\theta{,}\phi{)}=R_{nl}(r)Y_l^m(\theta{,}\phi{)}\] where \(R_{nl}(r)\) is the and is defined as \(R_{nl}(r)=N_{nl}e^{r/na_0}r^lL_{n+l}^{2l+1}\left(\frac{2r}{na_0}\right)\) Y is a spherical harmonic function, identical to the set of solutions to the rigid rotor quantum mechanical model. The function \( L_{n+l}^{2l+1}\left(\frac{2r}{na_0}\right) \) is an associated Laguerre polynomial, determined by the two quantum numbers \(l")}} and n. However, simply keeping the notation of R(r) will suffice for the following mathematics. If this separated wavefunction is used in the Schrödinger Equation and the substitution \(\hat{L}^2Y_l^{m_l}(\theta{,}\phi{)}=\hbar{^2}l(l+1)Y_l^{m_l}(\theta{,}\phi{)})\) is carried out, it is found that all of the angular elements of the equation cancel out to yield an equation containing solely radial functions. \(-\hbar{^2}\dfrac{\partial{}}{\partial{r}}\left(r^2\dfrac{\partial{}}{\partial{r}}\left(R(r)Y_l^{m_l}(\theta{,}\phi{})\right)\right)+R(r)\hbar{^2}l(l+1)Y_l^{m_l}(\theta{},\phi{})+2m_er^2[V(r)-E]R(r)Y_l^{m_l}(\theta{},\phi{})\) \(-\hbar{^2}\dfrac{d}{dr}\left(r^2\dfrac{\partial{R}}{\partial{r}}\right)Y_l^{m_l}(\theta{},\phi{})+\hbar{^2}l(l+1)R(r)Y_l^{m_l}(\theta{},\phi{})+2m_er^2[V(r)-E]R(r)Y_l^{m_l}(\theta{},\phi{})=0\) \[ -\hbar{^2}\dfrac{d}{dr}\left(r^2\dfrac{\partial{R}}{\partial{r}}\right)+\hbar{^2}l(l+1)R(r)+2m_er^2[V(r)-E]R(r)=0\] \[ \dfrac{-\hbar{^2}}{2m_er^2}\dfrac{d}{dr}\left(r^2\dfrac{dR}{dr}\right)+\left[\dfrac{\hbar{^2}l(l+1)}{2m_er^2}+V(r)-E\right]R(r)=0\] To solve this Schrödinger equation the power series method, the same method used to solve the , has to be used. 1. Show that the energy is dependent only on the radial portion of the wavefunction. | 5,581 | 1,218 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/21%3A_Resonance_and_Molecular_Orbital_Methods/21.05%3A_Application_of_the_MO_Method_to_13-Butadiene |
To treat the \(\pi\)-electron system of 1,3-butadiene by simple MO theory, we combine the four \(p\) carbon orbitals of an atomic-orbital model, such as \(17\), to obtain four molecular orbitals: We can estimate a stabilization energy for butadiene from heats of hydrogenation, and it is useful to compare the values obtained with the calculated delocalization energy. Thus the heat of hydrogenation of 1,3-butadiene is \(57.1 \: \text{kcal}\), whereas that of ethene is \(32.8 \: \text{kcal}\) and of propene \(30.1 \: \text{kcal}\). If ethene is used as the model alkene, the stabilization energy of 1,3-butadiene is \(\left( 2 \times 32.8 - 57.1 \right) = 8.5 \: \text{kcal}\), whereas with propene as the model, it would be \(\left( 2 \times 30.1 - 57.1 \right) = 3.1 \: \text{kcal}\). The bond energies (Table 4-3) in combination with the heat of formation at \(25^\text{o}\) \(\left( 26.33 \: \text{kcal} \right)\) give a stabilization energy of \(5.0 \: \text{kcal}\). and (1977) | 1,006 | 1,219 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/20%3A_Periodic_Trends_and_the_s-Block_Elements/20.02%3A_Overview_of_Periodic_Trends |
As we begin our summary of periodic trends, recall that the single most important unifying principle in understanding the chemistry of the elements is the systematic increase in atomic number, accompanied by the orderly filling of atomic orbitals by electrons, which leads to periodicity in such properties as atomic and ionic size, ionization energy, electronegativity, and electron affinity. The same factors also lead to periodicity in valence electron configurations, which for each group results in similarities in oxidation states and the formation of compounds with common stoichiometries. The most important periodic trends in atomic properties are summarized in Figure \(\Page {1}\). Recall that these trends are based on periodic variations in a single fundamental property, the (Z ), which increases from left to right and from top to bottom in the periodic table. The diagonal line in Figure \(\Page {1}\) separates the metals (to the left of the line) from the nonmetals (to the right of the line). Because metals have relatively low electronegativities, they tend to lose electrons in chemical reactions to elements that have relatively high electronegativities, forming compounds in which they have positive oxidation states. Conversely, nonmetals have high electronegativities, and they therefore tend to gain electrons in chemical reactions to form compounds in which they have negative oxidation states. The semimetals lie along the diagonal line dividing metals and nonmetals. It is not surprising that they tend to exhibit properties and reactivities intermediate between those of metals and nonmetals. Because the elements of groups 13, 14, and 15 span the diagonal line separating metals and nonmetals, their chemistry is more complex than predicted based solely on their valence electron configurations. The chemistry of the second-period element of each group (n = 2: Li, Be, B, C, N, O, and F) differs in many important respects from that of the heavier members, or congeners, of the group. Consequently, the elements of the third period (n = 3: Na, Mg, Al, Si, P, S, and Cl) are generally more representative of the group to which they belong. The anomalous chemistry of second-period elements results from three important characteristics: small radii, energetically unavailable d orbitals, and a tendency to form pi (π) bonds with other atoms. In contrast to the chemistry of the second-period elements, the chemistry of the third-period elements is more representative of the chemistry of the respective group. Due to their small radii, second-period elements have electron affinities that are less negative than would be predicted from general periodic trends. When an electron is added to such a small atom, increased electron–electron repulsions tend to destabilize the anion. Moreover, the small sizes of these elements prevent them from forming compounds in which they have more than four nearest neighbors. Thus BF forms only the four-coordinate, tetrahedral BF ion, whereas under the same conditions AlF forms the six-coordinate, octahedral AlF ion. Because of the smaller atomic size, simple binary ionic compounds of second-period elements also have more covalent character than the corresponding compounds formed from their heavier congeners. The very small cations derived from second-period elements have a high charge-to-radius ratio and can therefore polarize the filled valence shell of an anion. As such, the bonding in such compounds has a significant covalent component, giving the compounds properties that can differ significantly from those expected for simple ionic compounds. As an example, LiCl, which is partially covalent in character, is much more soluble than NaCl in solvents with a relatively low dielectric constant, such as ethanol (ε = 25.3 versus 80.1 for H O). Because d orbitals are never occupied for principal quantum numbers less than 3, the valence electrons of second-period elements occupy 2s and 2p orbitals only. The energy of the 3d orbitals far exceeds the energy of the 2s and 2p orbitals, so using them in bonding is energetically prohibitive. Consequently, electron configurations with more than four electron pairs around a central, second-period element are simply not observed. You may recall that the role of d orbitals in bonding in main group compounds with coordination numbers of 5 or higher remains somewhat controversial. In fact, theoretical descriptions of the bonding in molecules such as SF have been published without mentioning the participation of d orbitals on sulfur. Arguments based on d-orbital availability and on the small size of the central atom, however, predict that coordination numbers greater than 4 are unusual for the elements of the second period, which is in agreement with experimental results. One of the most dramatic differences between the lightest main group elements and their heavier congeners is the tendency of the second-period elements to form species that contain multiple bonds. For example, N is just above P in group 15: N contains an N≡N bond, but each phosphorus atom in tetrahedral P forms three P–P bonds. This difference in behavior reflects the fact that within the same group of the periodic table, the relative energies of the π bond and the sigma (σ) bond differ. A C=C bond, for example, is approximately 80% stronger than a C–C bond. In contrast, an Si=Si bond, with less p-orbital overlap between the valence orbitals of the bonded atoms because of the larger atomic size, is only about 40% stronger than an Si–Si bond. Consequently, compounds that contain both multiple and single C to C bonds are common for carbon, but compounds that contain only sigma Si–Si bonds are more energetically favorable for silicon and the other third-period elements. Another important trend to note in main group chemistry is the chemical similarity between the lightest element of one group and the element immediately below and to the right of it in the next group, a phenomenon known as (Figure \(\Page {2}\)) There are, for example, significant similarities between the chemistry of Li and Mg, Be and Al, and B and Si. Both BeCl and AlCl have substantial covalent character, so they are somewhat soluble in nonpolar organic solvents. In contrast, although Mg and Be are in the same group, MgCl behaves like a typical ionic halide due to the lower electronegativity and larger size of magnesium. The refers to the empirical observation that the heavier elements of groups 13–17 often have oxidation states that are lower by 2 than the maximum predicted for their group. For example, although an oxidation state of +3 is common for group 13 elements, the heaviest element in group 13, thallium (Tl), is more likely to form compounds in which it has a +1 oxidation state. There appear to be two major reasons for the inert-pair effect: increasing ionization energies and decreasing bond strengths. In moving down a group in the p-block, increasing ionization energies and decreasing bond strengths result in an inert-pair effect. The ionization energies increase because filled (n − 1)d or (n − 2)f subshells are relatively poor at shielding electrons in ns orbitals. Thus the two electrons in the ns subshell experience an unusually high effective nuclear charge, so they are strongly attracted to the nucleus, reducing their participation in bonding. It is therefore substantially more difficult than expected to remove these ns electrons, as shown in Table \(\Page {1}\) by the difference between the first ionization energies of thallium and aluminum. Because Tl is less likely than Al to lose its two ns electrons, its most common oxidation state is +1 rather than +3. Source of data: John A. Dean, Lange’s Handbook of Chemistry, 15th ed. (New York: McGraw-Hill, 1999). Going down a group, the atoms generally became larger, and the overlap between the valence orbitals of the bonded atoms decreases. Consequently, bond strengths tend to decrease down a column. As shown by the M–Cl bond energies listed in Table \(\Page {1}\), the strength of the bond between a group 13 atom and a chlorine atom decreases by more than 30% from B to Tl. Similar decreases are observed for the atoms of groups 14 and 15. The net effect of these two factors—increasing ionization energies and decreasing bond strengths—is that in going down a group in the p-block, the additional energy released by forming two additional bonds eventually is not great enough to compensate for the additional energy required to remove the two ns electrons. Based on the positions of the group 13 elements in the periodic table and the general trends outlined in this section, positions of elements in the periodic table classification, oxidation-state stability, and chemical reactivity From the position of the diagonal line in the periodic table separating metals and nonmetals, classify the group 13 elements. Then use the trends discussed in this section to compare their relative stabilities and chemical reactivities. Based on the positions of the group 14 elements C, Si, Ge, Sn, and Pb in the periodic table and the general trends outlined in this section, The chemistry of the third-period element in a group is most representative of the chemistry of the group because the chemistry of the second-period elements is dominated by their small radii, energetically unavailable d orbitals, and tendency to form π bonds with other atoms. The most important unifying principle in describing the chemistry of the elements is that the systematic increase in atomic number and the orderly filling of atomic orbitals lead to periodic trends in atomic properties. The most fundamental property leading to periodic variations is the effective nuclear charge (Z ). Because of the position of the diagonal line separating metals and nonmetals in the periodic table, the chemistry of groups 13, 14, and 15 is relatively complex. The second-period elements (n = 2) in each group exhibit unique chemistry compared with their heavier congeners because of their smaller radii, energetically unavailable d orbitals, and greater ability to form π bonds with other atoms. Increasing ionization energies and decreasing bond strengths lead to the inert-pair effect, which causes the heaviest elements of groups 13–17 to have a stable oxidation state that is lower by 2 than the maximum predicted for their respective groups. | 10,435 | 1,220 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Nucleophilic_Addition_Reactions/The_Addition_of_Hydrogen_Cyanide_to_Aldehydes_and_Ketones |
This page gives you the facts and simple, uncluttered mechanisms for the nucleophilic addition reactions between carbonyl compounds (specifically aldehydes and ketones) and hydrogen cyanide, HCN. Hydrogen cyanide adds across the carbon-oxygen double bond in aldehydes and ketones to produce compounds known as hydroxynitriles. For example, with ethanal (an aldehyde) you get 2-hydroxypropanenitrile: With propanone (a ketone) you get 2-hydroxy-2-methylpropanenitrile: The reaction isn't normally done using hydrogen cyanide itself, because this is an extremely poisonous gas. Instead, the aldehyde or ketone is mixed with a solution of sodium or potassium cyanide in water to which a little sulphuric acid has been added. The pH of the solution is adjusted to about 4 - 5, because this gives the fastest reaction. The solution will contain hydrogen cyanide (from the reaction between the sodium or potassium cyanide and the sulphuric acid), but still contains some free cyanide ions. This is important for the mechanism. These are examples of nucleophilic addition. The carbon-oxygen double bond is highly polar, and the slightly positive carbon atom is attacked by the cyanide ion acting as a nucleophile. In the first stage, there is a nucleophilic attack by the cyanide ion on the slightly positive carbon atom. The negative ion formed then picks up a hydrogen ion from somewhere - for example, from a hydrogen cyanide molecule. The hydrogen ion could also come from the water or the H O ions present in the slightly acidic solution. You don't need to remember all of these. One equation is perfectly adequate. As before, the reaction starts with a nucleophilic attack by the cyanide ion on the slightly positive carbon atom. It is completed by the addition of a hydrogen ion from, for example, a hydrogen cyanide molecule. When 2-hydroxypropanenitrile is made in this last mechanism, it occurs as a racemic mixture - a 50/50 mixture of two optical isomers. It is possible that you might be exected to explain how this arises. Optical isomerism occurs in compounds which have four different groups attached to a single carbon atom. In this case, the product molecule contains a CH , a CN, an H and an OH all attached to the central carbon atom. The reason for the formation of equal amounts of two isomers lies in the way the ethanal gets attacked. Ethanal is a planar molecule, and attack by a cyanide ion will either be from above the plane of the molecule, or from below. There is an equal chance of either happening. Attack from one side will lead to one of the two isomers, and attack from the other side will lead to the other. All aldehydes will form a racemic mixture in this way. Unsymmetrical ketones will as well. (A ketone can be unsymmetrical in the sense that there is a different alkyl group either side of the carbonyl group.) What matters is that the product molecule must have four different groups attached to the carbon which was originally part of the carbon-oxygen double bond. Jim Clark ( ) | 3,029 | 1,222 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/11%3A_Reactions_and_Other_Chemical_Processes/11.08%3A_The_Thermodynamic_Equilibrium_Constant |
\( \newcommand{\tx}[1]{\text{#1}} % text in math mode\)
\( \newcommand{\subs}[1]{_{\text{#1}}} % subscript text\)
\( \newcommand{\sups}[1]{^{\text{#1}}} % superscript text\)
\( \newcommand{\st}{^\circ} % standard state symbol\)
\( \newcommand{\id}{^{\text{id}}} % ideal\)
\( \newcommand{\rf}{^{\text{ref}}} % reference state\)
\( \newcommand{\units}[1]{\mbox{$\thinspace$#1}}\)
\( \newcommand{\K}{\units{K}} % kelvins\)
\( \newcommand{\degC}{^\circ\text{C}} % degrees Celsius\)
\( \newcommand{\br}{\units{bar}} % bar (\bar is already defined)\)
\( \newcommand{\Pa}{\units{Pa}}\)
\( \newcommand{\mol}{\units{mol}} % mole\)
\( \newcommand{\V}{\units{V}} % volts\)
\( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\)
\( \newcommand{\per}{^{-1}} % minus one power\)
\( \newcommand{\m}{_{\text{m}}} % subscript m for molar quantity\)
\( \newcommand{\CVm}{C_{V,\text{m}}} % molar heat capacity at const.V\)
\( \newcommand{\Cpm}{C_{p,\text{m}}} % molar heat capacity at const.p\)
\( \newcommand{\kT}{\kappa_T} % isothermal compressibility\)
\( \newcommand{\A}{_{\text{A}}} % subscript A for solvent or state A\)
\( \newcommand{\B}{_{\text{B}}} % subscript B for solute or state B\)
\( \newcommand{\bd}{_{\text{b}}} % subscript b for boundary or boiling point\)
\( \newcommand{\C}{_{\text{C}}} % subscript C\)
\( \newcommand{\f}{_{\text{f}}} % subscript f for freezing point\)
\( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\)
\( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\)
\( \newcommand{\mi}{_{\text{m},i}} % subscript m,i (m=molar)\)
\( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\)
\( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\)
\( \newcommand{\xbB}{_{x,\text{B}}} % x basis, B\)
\( \newcommand{\xbC}{_{x,\text{C}}} % x basis, C\)
\( \newcommand{\cbB}{_{c,\text{B}}} % c basis, B\)
\( \newcommand{\mbB}{_{m,\text{B}}} % m basis, B\)
\( \newcommand{\kHi}{k_{\text{H},i}} % Henry's law constant, x basis, i\)
\( \newcommand{\kHB}{k_{\text{H,B}}} % Henry's law constant, x basis, B\)
\( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\)
\( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\)
\( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\)
\( \newcommand{\eq}{\subs{eq}} % equilibrium state\)
\( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\)
\( \newcommand{\sys}{\subs{sys}} % system property\)
\( \newcommand{\sur}{\sups{sur}} % surroundings\)
\( \renewcommand{\in}{\sups{int}} % internal\)
\( \newcommand{\lab}{\subs{lab}} % lab frame\)
\( \newcommand{\cm}{\subs{cm}} % center of mass\)
\( \newcommand{\rev}{\subs{rev}} % reversible\)
\( \newcommand{\irr}{\subs{irr}} % irreversible\)
\( \newcommand{\fric}{\subs{fric}} % friction\)
\( \newcommand{\diss}{\subs{diss}} % dissipation\)
\( \newcommand{\el}{\subs{el}} % electrical\)
\( \newcommand{\cell}{\subs{cell}} % cell\)
\( \newcommand{\As}{A\subs{s}} % surface area\)
\( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)
\( \newcommand{\allni}{\{n_i \}} % set of all n_i\)
\( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\)
\( \newcommand{\solmB}{\tx{(sol,$\,$$m\B$)}}\)
\( \newcommand{\dil}{\tx{(dil)}}\)
\( \newcommand{\sln}{\tx{(sln)}}\)
\( \newcommand{\mix}{\tx{(mix)}}\)
\( \newcommand{\rxn}{\tx{(rxn)}}\)
\( \newcommand{\expt}{\tx{(expt)}}\)
\( \newcommand{\solid}{\tx{(s)}}\)
\( \newcommand{\liquid}{\tx{(l)}}\)
\( \newcommand{\gas}{\tx{(g)}}\)
\( \newcommand{\pha}{\alpha} % phase alpha\)
\( \newcommand{\phb}{\beta} % phase beta\)
\( \newcommand{\phg}{\gamma} % phase gamma\)
\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
\( \newcommand{\bph}{^{\beta}} % beta phase superscript\)
\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
\( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\)
\( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\)
\( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)
\( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)
\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
\( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)
\( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)
\( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)
\( \newcommand{\Del}{\Delta}\)
\( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)
\( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)
\( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)
\( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)
\( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)
\( \newcommand{\dotprod}{\small\bullet}\)
\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
\( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)
\( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)
\( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\)
\( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\)
\( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)
\( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)
\( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)
\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) Equation 10.1.9 gives the general relation between the chemical potential \(\mu_i\) and the activity \(a_i\) of species \(i\) in a phase of electric potential \(\phi\): \begin{equation} \mu_i=\mu_i\st + RT\ln a_i + z_i F\phi \tag{11.8.1} \end{equation} The electric potential affects \(\mu_i\) only if the charge number \(z_i\) is nonzero, i.e., only if species \(i\) is an ion. Consider a reaction in which any reactants and products that are ions are in a single phase of electric potential \(\phi'\), or in several phases of equal electric potential \(\phi'\). Under these conditions, substitution of the expression above for \(\mu_i\) in \(\Delsub{r}G = \sum_i\!\nu_i\mu_i\) gives \begin{gather} \s{ \Delsub{r}G = \sum_i\nu_i\mu_i\st + RT \sum_i\nu_i\ln a_i + F\phi'\sum_i\nu_i z_i } \tag{11.8.2} \cond{(all ions at \(\phi{=}\phi'\))} \end{gather} The first term on the right side of Eq. 11.8.2 is the , or standard molar Gibbs energy of reaction: \begin{equation} \Delsub{r}G\st\defn \sum_i\nu_i \mu_i\st \tag{11.8.3} \end{equation} Since the standard chemical potential \(\mu\st_i\) of each species \(i\) is a function only of \(T\), the value of \(\Delsub{r}G\st\) for a given reaction as defined by the reaction equation depends only on \(T\) and on the choice of a standard state for each reactant and product. The last term on the right side of Eq. 11.8.2 is the sum \(\sum_i\!\nu_i z_i\). Because charge is conserved during the advancement of a reaction in a closed system, this sum is zero. With these substitutions, Eq. 11.8.2 becomes \begin{gather} \s{ \Delsub{r}G = \Delsub{r}G\st + RT\sum_i\nu_i\ln a_i } \tag{11.8.4} \cond{(all ions at same \(\phi\))} \end{gather} This relation enables us to say that for a reaction at a given temperature in which any charged reactants or products are all in the same phase, or in phases of equal electric potential, the value of \(\Delsub{r}G\) and \(\sum_i\!\nu_i\mu_i\) depends only on the activities of the reactants and products and is independent of what the electric potentials of any of the phases might happen to be. Unless a reaction involving ions is carried out in a galvanic cell, the ions are usually present in a single phase, and this will not be shown as a condition of validity in the rest of this chapter. The special case of a reaction in a galvanic cell will be discussed in Sec. 14.3. We may use properties of logarithms to write the sum on the right side of Eq. 11.8.4 as follows: \begin{equation} \sum_i\nu_i\ln a_i = \sum_i\ln \left( a_i^{\nu_i} \right) = \ln \prod_i a_i^{\nu_i} \tag{11.8.5} \end{equation} The symbol \(\prod\) stands for a continued product. If, for instance, there are three species, \(\prod_i a_i^{\nu_i}\) is the product \((a_1^{\nu_1})(a_2^{\nu_2})(a_3^{\nu_3})\). The product \(\prod_i a_i^{\nu_i}\) is called the or activity quotient, \(Q\subs{rxn}\): \begin{equation} Q\subs{rxn} \defn \prod_i a_i ^{\nu_i} \tag{11.8.6} \end{equation} \(Q\subs{rxn}\) consists of a factor for each reactant and product. Each factor is the activity raised to the power of the stoichiometric number \(\nu_i\). Since the value of \(\nu_i\) is positive for a product and negative for a reactant, \(Q\subs{rxn}\) is a quotient in which the activities of the products appear in the numerator and those of the reactants appear in the denominator, with each activity raised to a power equal to the corresponding stoichiometric coefficient in the reaction equation. Such a quotient, with quantities raised to these powers, is called a . The reaction quotient is a proper quotient of activities. For instance, for the ammonia synthesis reaction N\(_2\)(g) + 3 H\(_2\)(g)\(\arrow\)2 NH\(_3\)(g) the reaction quotient is given by \begin{equation} Q\subs{rxn} = \frac{a\subs{NH\(_3\)}^2}{a\subs{N\(_2\)}a\subs{H\(_2\)}^3} \tag{11.8.7} \end{equation} \(Q\subs{rxn}\) is a dimensionless quantity. It is a function of \(T\), \(p\), and the mixture composition, so its value changes as the reaction advances. The expression for the molar reaction Gibbs energy given by Eq. 11.8.4 can now be written \begin{equation} \Delsub{r}G = \Delsub{r}G\st + RT\ln Q\subs{rxn} \tag{11.8.8} \end{equation} The value of \(Q\subs{rxn}\) under equilibrium conditions is the , \(K\). The general definition of \(K\) is \begin{equation} K \defn \prod_i (a_i)\eq^{\nu_i} \tag{11.8.9} \end{equation} where the subscript eq indicates an equilibrium state. Note that \(K\), like \(Q\subs{rxn}\), is dimensionless. The IUPAC Green Book (E. Richard Cohen et al, , 3rd edition, RSC Publishing, Cambridge, 2007, p. 58) gives \(K^{\small \unicode{x29B5}}\) as an alternative symbol for the thermodynamic equilibrium constant, the appended superscript denoting “standard.” An IUPAC Commission on Thermodynamics (M. B. Ewing et al, , , 533–552, 1994) has furthermore recommended the name “standard equilibrium constant,” apparently because its value depends on the choice of standard states. Using this alternative symbol and name could cause confusion, since the quantity defined by Eq. 11.8.9 does not refer to reactants and products in their standard states but rather to reactants and products in an state. Substituting the equilibrium conditions \(\Delsub{r}G = 0\) and \(Q\subs{rxn} = K\) in Eq. 11.8.8 gives an important relation between the standard molar reaction Gibbs energy and the thermodynamic equilibrium constant: \begin{equation} \Delsub{r}G\st = -RT\ln K \tag{11.8.10} \end{equation} We can solve this equation for \(K\) to obtain the equivalent relation \begin{equation} K = \exp \left( -\frac{\Delsub{r}G\st}{RT} \right) \tag{11.8.11} \end{equation} We have seen that the value of \(\Delsub{r}G\st\) depends only on \(T\) and the choice of the standard states of the reactants and products. This being so, Eq. 11.8.11 shows that the value of \(K\) for a given reaction depends only on \(T\) and the choice of standard states. No other condition, neither pressure nor composition, can affect the value of \(K\). We also see from Eq. 11.8.11 that \(K\) is less than \(1\) if \(\Delsub{r}G\st\) is positive and greater than \(1\) if \(\Delsub{r}G\st\) is negative. At a fixed temperature, reaction equilibrium is attained only if and only if the value of \(Q\subs{rxn}\) becomes equal to the value of \(K\) at that temperature. The thermodynamic equilibrium constant \(K\) is the proper quotient of the activities of species in reaction equilibrium. At typical temperatures and pressures, an activity cannot be many orders of magnitude greater than \(1\). For instance, a partial pressure cannot be greater than the total pressure, so at a pressure of \(10\br\) the activity of a gaseous constituent cannot be greater than about \(10\). The molarity of a solute is rarely much greater than \(10\units{mol dm\(^{-3}\)}\), corresponding to an activity (on a concentration basis) of about \(10\). Activities can, however, be extremely small. These considerations lead us to the conclusion that in an equilibrium state of a reaction with a very value of \(K\), the activity of at least one of the must be very small. That is, if \(K\) is very large then the reaction goes practically to completion and at equilibrium a limiting reactant is essentially entirely exhausted. The opposite case, a reaction with a very value of \(K\), must have at equilibrium one or more with very small activities. These two cases are the two extremes of the trends shown in Fig. 11.16. Equation 11.8.10 correctly relates \(\Delsub{r}G\st\) and \(K\) only if they are both calculated with the same standard states. For instance, if we base the standard state of a particular solute species on molality in calculating \(\Delsub{r}G\st\), the activity of that species appearing in the expression for \(K\) (Eq. 11.8.9) must also be based on molality. If a reaction takes place in a gaseous mixture, the standard state of each reactant and product is the pure gas behaving ideally at the standard pressure \(p\st\) (Sec. 9.3.3). In this case, each activity is given by \(a_i\gas = \fug_i/p\st = \phi_i p_i/p\st\) where \(\phi_i\) is a fugacity coefficient (Table 9.5). When we substitute this expression into Eq. 11.8.9, we find we can express the thermodynamic equilibrium constant as the product of three factors: \begin{gather} \s{ K = \left[\prod_i (\phi_i)\eq^{\nu_i}\right] \left[\prod_i (p_i)\eq^{\nu_i}\right] \left[(p\st)^{- \sum_i\nu_i}\right] } \tag{11.8.12} \cond{(gas mixture)} \end{gather} On the right side of this equation, the first factor is the proper quotient of fugacity coefficients in the mixture at reaction equilibrium, the second factor is the proper quotient of partial pressures in this mixture, and the third factor is the power of \(p\st\) needed to make \(K\) dimensionless. The proper quotient of equilibrium partial pressures is an , \(K_p\): \begin{gather} \s{ K_p = \prod_i (p_i)\eq^{\nu_i} } \tag{11.8.13} \cond{(gas mixture)} \end{gather} Note that \(K_p\) is dimensionless only if \(\sum_i\!\nu_i\) is equal to zero. The value of \(K_p\) can vary at constant temperature, so \(K_p\) is not a thermodynamic equilibrium constant. For instance, consider what happens when we take an ideal gas mixture at reaction equilibrium and compress it isothermally. As the gas pressure increases, the fugacity coefficient of each constituent changes from its low pressure value of \(1\) and the gas mixture becomes nonideal. In order for the mixture to remain in reaction equilibrium, and the product of factors on the right side of Eq. 11.8.12 to remain constant, there must be a change in the value of \(K_p\). In other words, the reaction equilibrium as we increase \(p\) at constant \(T\), an effect that will be considered in more detail in Sec. 11.9. As an example of the difference between \(K\) and \(K_p\), consider again the ammonia synthesis \(\ce{N2}\tx{(g)} + \ce{3H2}\tx{(g)} \arrow \ce{2NH3}\tx{(g)}\) in which the sum \(\sum_i\!\nu_i\) equals \(-2\). For this reaction, the expression for the thermodynamic equilibrium constant is \begin{equation} K = \left( \frac{\phi\subs{NH\(_3\)}^2}{\phi\subs{N\(_2\)}\phi\subs{H\(_2\)}^3} \right)\eq K_p (p\st)^2 \tag{11.8.14} \end{equation} where \(K_p\) is given by \begin{equation} K_p = \left( \frac{p\subs{NH\(_3\)}^2}{p\subs{N\(_2\)}p\subs{H\(_2\)}^3} \right)\eq \tag{11.8.15} \end{equation} If any of the reactants or products are solutes in a solution, the value of \(K\) depends on the choice of the solute standard state. For a given reaction at a given temperature, we can derive relations between values of \(K\) that are based on different solute standard states. In the limit of infinite dilution, each solute activity coefficient is unity, and at the standard pressure each pressure factor is unity. Under these conditions of infinite dilution and standard pressure, the activities of solute B on a mole fraction, concentration, and molality basis are therefore \begin{equation} a\xbB=x\B\qquad a\cbB=c\B/c\st \qquad a\mbB=m\B/m\st \tag{11.8.16} \end{equation} In the limit of infinite dilution, the solute composition variables approach values given by the relations in Eq. 9.1.14: \(x\B = V\A^*c\B = M\A m\B\). Combining these with \(a\xbB=x\B\) from Eq. 11.8.16, we write \begin{equation} a\xbB = V\A^*c\B = M\A m\B \tag{11.8.17} \end{equation} Then, using the relations for \(a\cbB\) and \(a\mbB\) in Eq. 11.8.16, we find that the activities of solute B at infinite dilution and pressure \(p\st\) are related by \begin{equation} a\xbB = V\A^* c\st a\cbB = M\A m\st a\mbB \tag{11.8.18} \end{equation} The expression \(K=\prod_i(a_i)\eq^{\nu_i}\) has a factor \((a\B)\eq^{\nu\B}\) for each solute B that is a reactant or product. From Eq. 11.8.18, we see that for solutes at infinite dilution at pressure \(p\st\), the relations between the values of \(K\) based on different solute standard states are \begin{equation} K\tx{(\(x\) basis)} = \prod_\tx{B}(V\A^*c\st)^{\nu\B} K\tx{(\(c\) basis)} = \prod_\tx{B}(M\A m\st)^{\nu\B} K\tx{(\(m\) basis)} \tag{11.8.19} \end{equation} For a given reaction at a given temperature, and with a given choice of solute standard state, the value of \(K\) is not affected by pressure or dilution. The relations of Eq. 11.8.19 are therefore valid under all conditions. The relation \(K=\exp (-\Delsub{r}G\st/RT)\) (Eq. 11.8.11) gives us a way to evaluate the thermodynamic equilibrium constant \(K\) of a reaction at a given temperature from the value of the standard molar reaction Gibbs energy \(\Delsub{r}G\st\) at that temperature. If we know the value of \(\Delsub{r}G\st\), we can calculate the value of \(K\). One method is to calculate \(\Delsub{r}G\st\) from values of the \(\Delsub{f}G\st\) of each reactant and product. These values are the standard molar reaction Gibbs energies for the formation reactions of the substances. To relate \(\Delsub{f}G\st\) to measurable quantities, we make the substitution \(\mu_i = H_i - TS_i\) (Eq. 9.2.46) in \(\Delsub{r}G = \sum_i\!\nu_i\mu_i\) to give \(\Delsub{r}G = \sum_i\!\nu_i H_i - T \sum_i\!\nu_i S_i\), or \begin{equation} \Delsub{r}G = \Delsub{r}H - T\Delsub{r}S \tag{11.8.20} \end{equation} When we apply this equation to a reaction with each reactant and product in its standard state, it becomes \begin{equation} \Delsub{r}G\st = \Delsub{r}H\st - T\Delsub{r}S\st \tag{11.8.21} \end{equation} where the standard molar reaction entropy is given by \begin{equation} \Delsub{r}S\st = \sum_i\nu_i S_i\st \tag{11.8.22} \end{equation} If the reaction is the reaction of a substance, we have \begin{equation} \Delsub{f}G\st = \Delsub{f}H\st - T\sum_i\nu_i S_i\st \tag{11.8.23} \end{equation} where the sum over \(i\) is for the reactants and product of the formation reaction. We can evaluate the standard molar Gibbs energy of formation of a substance, then, from its standard molar enthalpy of formation and the standard molar entropies of the reactants and product. Extensive tables are available of values of \(\Delsub{f}G\st\) for substances and ions. An abbreviated version at the single temperature \(298.15\K\) is given in Appendix H. For a reaction of interest, the tabulated values enable us to evaluate \(\Delsub{r}G\st\), and then \(K\), from the expression (analogous to Hess’s law) \begin{equation} \Delsub{r}G\st = \sum_i\nu_i\Delsub{f}G\st(i) \tag{11.8.24} \end{equation} The sum over \(i\) is for the reactants and products of the reaction of interest. Recall that the standard molar enthalpies of formation needed in Eq. 11.8.23 can be evaluated by calorimetric methods (Sec. 11.3.2). The absolute molar entropy values \(S_i\st\) come from heat capacity data or statistical mechanical theory by methods discussed in Sec. 6.2. Thus, it is entirely feasible to use nothing but calorimetry to evaluate an equilibrium constant, a goal sought by thermodynamicists during the first half of the 20th century. (Another method, for a reaction that can be carried out reversibly in a galvanic cell, is described in Sec. 14.3.3.) For , the values of \(S\m\st\) and \(\Delsub{f}G\st\) found in Appendix H are based on the reference values \(S\m\st=0\) and \(\Delsub{f}G\st = 0\) for H\(^+\)(aq) at all temperatures, similar to the convention for \(\Delsub{f}H\st\) values discussed in Sec. 11.3.2. For a reaction with aqueous ions as reactants or products, these values correctly give \(\Delsub{r}S\st\) using Eq. 11.8.22, or \(\Delsub{r}G\st\) using Eq. 11.8.24. Note that the values of \(S\m\st\) in Appendix H for some ions, unlike the values for substances, are ; this simply means that the standard molar entropies of these ions are less than that of H\(^+\)(aq). The relation of Eq. 11.8.23 does not apply to an ion, because we cannot write a formation reaction for a single ion. Instead, the relation between \(\Delsub{f}G\st\), \(\Delsub{f}H\st\) and \(S\m\st\) is more complicated. Consider first a hypothetical reaction in which hydrogen ions and one or more elements form H\(_2\) and a cation M\(^{z_+}\) with charge number \(z_+\): \[ z_+\tx{H\(^+\)(aq)}+\tx{elements} \arrow (z_+/2)\tx{H\(_2\)(g)}+\tx{M\(^{z_+}\)(aq)} \] For this reaction, using the convention that \(\Delsub{f}H\st\), \(S\m\st\), and \(\Delsub{f}G\st\) are zero for the aqueous H\(^+\) ion and the fact that \(\Delsub{f}H\st\) and \(\Delsub{f}G\st\) are zero for the elements, we can write the following expressions for standard molar reaction quantities: \begin{equation} \Delsub{r}H\st = \Delsub{f}H\st(\tx{M\(^{z_+}\)}) \tag{11.8.25} \end{equation} \begin{equation} \Delsub{r}S\st = (z_+/2)S\m\st(\tx{H\(_2\)}) +S\m\st(\tx{M\(^{z_+}\)})- \! \sum_{\tx{elements}}\!\!\!S_i\st \tag{11.8.26} \end{equation} \begin{equation} \Delsub{r}G\st = \Delsub{f}G\st(\tx{M\(^{z_+}\)}) \tag{11.8.27} \end{equation} Then, from \(\Delsub{r}G\st=\Delsub{r}H\st-T\Delsub{r}S\st\), we find \begin{equation} \Delsub{f}G\st(\tx{M\(^{z_+}\)}) = \Delsub{f}H\st(\tx{M\(^{z_+}\)}) \quad -T\left[ S\m\st(\tx{M\(^{z_+}\)}) -\sum_{\tx{elements}}\!\!\!S_i\st + (z_+/2)S\m\st(\tx{H\(_2\)}) \right] \tag{11.8.28} \end{equation} For example, the standard molar Gibbs energy of the aqueous mercury(I) ion is found from \begin{equation} \textstyle \Delsub{f}G\st(\tx{Hg\(_2\)\(^{2+}\)}) = \Delsub{f}H\st(\tx{Hg\(_2\)\(^{2+}\)}) - TS\m\st(\tx{Hg\(_2\)\(^{2+}\)}) + 2TS\m\st(\tx{Hg}) - \frac{2}{2}TS\m\st(\tx{H\(_2\)}) \tag{11.8.29} \end{equation} For an anion X\(^{z_-}\) with negative charge number \(z_-\), using the hypothetical reaction \[ |z_-/2| \tx{H\(_2\)(g)}+\tx{elements} \arrow |z_-| \tx{H\(^+\)(aq)}+\tx{X\(^{z_-}\)(aq)} \] we find by the same method \begin{equation} \Delsub{f}G\st(\tx{X\(^{z_-}\)}) = \Delsub{f}H\st(\tx{X\(^{z_-}\)}) -T\left[ S\m\st(\tx{X\(^{z_-}\)})- \! \sum_{\tx{elements}}\!\!\!S_i\st -|z_-/2| S\m\st(\tx{H\(_2\)}) \right] \tag{11.8.30} \end{equation} For example, the calculation for the nitrate ion is \begin{equation} \textstyle \Delsub{f}G\st(\tx{NO\(_3\)\(^-\)}) = \Delsub{f}H\st(\tx{NO\(_3\)\(^-\)}) - TS\m\st(\tx{NO\(_3\)\(^-\)}) + \frac{1}{2}TS\m\st(\tx{N\(_2\)}) + \frac{3}{2}TS\m\st(\tx{O\(_2\)}) + \frac{1}{2}TS\m\st(\tx{H\(_2\)}) \tag{11.8.31} \end{equation} | 25,600 | 1,223 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/22%3A_Arenes_Electrophilic_Aromatic_Substitution/22.11%3A_Sources_and_Uses_of_Aromatic_Hydrocarbons |
Benzene and many of its derivatives are manufactured on a large scale for use in high-octane gasolines and in the production of polymers, insecticides, detergents, dyes, and many miscellaneous chemicals. Prior to World War II, coal was the only important source of aromatic hydrocarbons, but during the war and thereafter, the demand for benzene, methylbenzene, and the dimethylbenzenes rose so sharply that other sources had to be found. Today, most of the benzene and almost all of the methylbenzene and the dimethylbenzenes produced in the United States are derived from petroleum. Coal tar, which is a distillate obtained in the coking of coal ( ), is a source of an amazing number of aromatic compounds. Some of these are listed in Table 22-7, which includes nitrogen, oxygen, and sulfur compounds, as well as hydrocarbons. Although petroleum from some locations contains fairly substantial amounts of aromatic hydrocarbons, it is not a principal source for such compounds. Rather, aromatic compounds are synthesized from the \(\ce{C_6}\)-\(\ce{C_{10}}\) gasoline fraction from petroleum refining by a process referred to in the petroleum industry as or . This involves heating a \(\ce{C_6}\)-\(\ce{C_{10}}\) fraction with hydrogen in the presence of a catalyst to modify the molecular structure of its components. Some amazing transformations take place, and the \(\ce{C_6}\)-\(\ce{C_7}\) alkanes can be converted to cycloalkanes, which, in turn, are converted to arenes. Benzene, methylbenzene (toluene), and the dimethylbenzenes (xylenes) are produced primarily in this way: and (1977) | 1,615 | 1,224 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Radiative_Decay/Chemiluminescence |
Most chemiluminescence methods involve only a few chemical components to actually generate light. Luminol chemiluminescence (Nieman, 1989), which has been extensively investigated, and peroxyoxalate chemiluminescence (Given and Schowen, 1989; Orosz et al., 1996) are both used in bioanalytical methods and will be the subject of this primer on chemiluminescence. In each system, a "fuel" is chemically oxidized to produced an excited state product. In many it is this excited product that emits the light for the signal. In p , the initial excited state product does not emit light at all and instead it reacts with another compound. Chemiluminescence takes its place among other spectroscopic techniques because of its inherent sensitivity and selectivity. It requires: Maybe this list should be entitled "What chemiluminescent systems do not require." Although not as widely applicable as excitation spectroscopy, the detection limits for chemiluminescent methods can be 10 to 100 times lower than other luminescence techniques. Most chemiluminescence methods involve only a few chemical components to actually generate light. Luminol chemiluminescence (Nieman, 1989), which has been extensively investigated, and peroxyoxalate chemiluminescence (Given and Schowen, 1989; Orosz et al., 1996) are both used in bioanalytical methods and will be the subject of this primer on chemiluminescence. In each system, a "fuel" is chemically oxidized to produced an excited state product. In many luminol methods it is this excited product that emits the light for the signal. In peroxyoxalate chemiluminescence, the initial excited state product does not emit light at all and instead it reacts with another compound, often a compound also viable as a fluorescent dye, and it is this fluorophore which becomes excited and emits light. That said, the oxalate reactions, to have practical applicability in, for instance HPLC, require a mixed solvent system (buffer/organic solvent) to assure solubility of the reagents, optimized pH, and allow compatibility with the analytes. A general discussion of these two methods, their applicability as reported in some of the recent literature, and a discussion of the emission spectra of each--complete with movies that show short experiments with each--will be presented. One of the suggested reaction sequences in the reaction of peroxyoxalates, of which bis(2,4,6-trichlorophenyl)oxlate (TCPO) is the most prominent example, follows. It involves the fuel (TCPO) plus the oxidant (H O ) reacting to produce a proposed intermediate, in this example shown as a dioxetane; although, this reaction probably produces many intermediates, and others, such as hydroperoxyoxalate, have been proposed (Milofsky and Birks, 1991; Choksi et al., 1990). The intermediate, shown here as 1,2-dioxetanedione, excites a fluorophore. In the included movie demonstrating TCPO chemiluminescence, 9,10-diphenylanthracene acts as the fluorophore; its lambda max is 425 nm in the solvent used, tetrahydrofuran. Its reaction with the intermediate produces the excited state product which quickly emits light. The process of transferring the energy of the initial reaction, the chemical reaction of hydrogen peroxide with TCPO, to light emission from the excited state fluorophore (fluorophore*) can be sidetracked along the way by loses in each step of the process: the initial oxidation to produce the intermediate, the reaction of the intermediate with a fluorophore, and the reaction of the excited fluorophore to produce light (Orosz et al., 1996). The initial oxidation can yield the high energy intermediate or The high energy intermediate can react to excite the fluorophore or Finally the excited fluorophore can loose energy by emission of light or In normal chromatographic (HPLC) procedures, these alternate mechanistic routes can be effected by solvent and buffers (Orosz, 1989; Jennings and Capomacchia, 1988); pH (de Jong et al., 1986); catalyst (Orlovic et al., 1989; Alverez et al., 1986); and type of fuel (Orlovic et al., 1989; Orosz, 1989), oxidant (Orlovic et al., 1989), and fluorophore concentration and identify. Possibly most important for chromatographers, eluent and reagent flows (Givens and Schowen, 1989; Kwakman and Brinkman, 1992), detector volume and geometry (de Jong et al., 1990; Grayeski and Weber, 1984), and mixing parameters (Kobayashi and Imai, 1980; Sugiura et al., 1993) can all effect this method's light production. This, therefore, sets the stage for analytical methods whereby manipulating the appropriate parameter allows for the sensitive determination of hydrogen peroxide (Pontén et al., 1996; Stigbrand et al., 1994) or fluorophore content. Recently, for example, Hamachi et al. (1999) determined the concentration of propentofylline in hypocampus extracts from rats by derivitizing the analyte to create a fluorophore which would chemiluminesce with another peroxyoxalate, TDPO [bis(2-(3,6,9-trioadecanyloxycarbonyl)-4-nitrophenyl)oxalate, and hydrogen peroxide following HPLC. Propentofylline is a reported inhibitor of dopamine released during low oxygenation events in the cerebellum. The derivatization of propentofylline was carried out in trifluoracetic acid/acetonitrile solution using DBD-H (a benzoaxadiazole). The detection limit for the analyte, 31 fg/injection, was about 200 times better than comparable HPLC-UV methods. A solution of TCPO and 9,10-diphenylanthracene (DPA; Aldrich Chemicals Co., Milwaukee, WI USA) both in the 1 x 10 M concentration range dissolved in tetrahydrofuran (THF) were mixed with a dilute solution of H O in THF (~0.3%) at ~25 C. The resulting emission spectrum was recorded on a fluorescence spectrometer (Hitachi F-4500; 1 cm quartz cell) in chemiluminescence mode (with no excitation source). The slit and PMT voltage were adjusted to allow for the detection of a strong signal without overloading the detector. The components were mixed and the emission spectrum scanned immediately (1200 nm/min). As the Figure below shows, the emission was centered around 425 nm. This is, of course, similar to DPA's "normal" fluorescent emission. The movie included here involves that same solution, TCPO and 9,10-diphenylanthracene dissolved in THF. If you look closely you may be able to see the milky consistency of the slightly yellow, initial mixture--shown under fluorescent lights, before hydrogen peroxide was added. Without a mixed solvent system, the solubility of each of these components is relatively low and so the solution is basically saturated with each of these reagents (but still in the low millimolar concentration range). In the dark, a solution of ~0.3% H O in THF was added dropwise to approximately 8 mL of the fuel + fluorophore in THF (~25 C) in an open-topped vial. The reaction(s) immediately produces light from the excited fluorophore. The emission is relatively short lived but since H O is apparently limiting, a second and third dropwise addition of the oxidant yields additional bursts of light. If you will look carefully at the end of the movie you will see a clear--yet still yellow--solution in which all precipitates have dissolved. Also apparent to the experimenter, but undetectable in the movie, was the formation of a gas produced by the reaction; this appeared as a bubbling that could be seen while the reaction was still producing light yet which stopped as the reaction reached completion, about 30 seconds after the last (excess) H O addition. This kind of gas production has been used as evidence for the production of CO as a product from the 1,2-dioxetandione intermediate as detailed in the figure above. Further peroxide addition does not yield more bubbling so this is not simply H O decomposition. The process of filming this reaction is described below. Luminol is also widely used as a chemiluminescent reagent, but unlike the peroxyoxalate systems does not require an organic/mixed solvent system. The chemiluminescent emitter is a "direct descendent" of the oxidation of luminol (or an isomer like isoluminol) by an oxidant in basic aqueous solution. Probably the most useful oxidant is also hydrogen peroxide similar to peroxyoxalate chemiluminescence; however, other oxidants have been used such as perborate, permanganate (Lu and Lu, 1992), hypochlorite (Cunningham et al., 1998), and iodine (Seitz, 1981). If the fuel is luminol, the emitting species is 3-aminophthalate (see below); however, luminol-derivatized analytes allow for determination of compounds that would not normally chemiluminescence in this system and presumably have slightly different emitters (Edwards et al., 1995; Kawasaki et al., 1985; Lippman, 1980; Nakazone et al., 1992; Pontén et al., 1996). The presence of a catalyst is paramount to this chemiluminescent method as an analytical tool. Many metal cations catalyze the reaction of luminol, H O , and OH in aqueous solution to increase light emission or at least to increase the speed of the oxidation to produce the emitter and therefore the onset/intensity of light production. [Some metals, however, repress chemiluminescence at different concentrations (Yuan and Shiller, 1999; see below.] This therefore can be the foundation of significantly different analytical determinations. For instance, this system can be used: This last is particularly powerful feature of this system because many compounds complex metallic cations and thereby make themselves "known." Amino acids (Koerner and Nieman, 1987), fructose and tagatose (Valeri et al., 1997), glycerol (Robards and Worsfold, 1992), thiols (Sano and Nakamura, 1998), and serum albumin (Tie et. al., 1995) among many others have been determined using luminol chemiluminescence. Most recently, Yuan and Shiller (1999) report a subnanomolar detection limit for H O using luminol chemiluminescence. Their method, which was used to determine hydrogen peroxide content in sea water, was based on the cobalt(II) catalytic oxidation of luminol. While Co is the most sensitive luminol metal catalyst, it is also present in sea water at very low concentrations. The pH of the luminol solution used in this work was 10.15, and interferences from seven different metals were investigated. Interestingly some metals interfered positively and some negatively, and Fe(III) interfered positively at one concentration and negatively at another. Finally, very low concentrations of iron(II) showed a significant positive interference in determination of H O , but the authors used the relatively short half life of Fe(II) in marine water as a means of eliminating Fe(II) interference in the determination of hydrogen peroxide in their analysis by storing samples for over 1 hr before analysis. Approximately 15 ml of a solution containing luminol, copper catalyst, and pH controllers were placed in a glass vial at ~25 C (1 x 10 M luminol; 0.05 M sodium carbonate; 0.3 M sodium bicarbonate; 5 x 10 M ammonium carbonate; 1.5 x 10 M Cu(II) added as sulfate salt). An aqueous solution of approximately 0.25% H O was added dropwise. The emission spectrum was taken as before using a fluorescence spectrometer with the excitation source off. The light intensity-time decay data were taken immediately after mixing the reagents and for 60 seconds. The lambda max is at approximately 445 nm, slightly longer wavelength than the TCPO/DPA system described above. Online presentations of the light intensity-time decay aspects of the luminol reaction with hydrogen peroxide and differing concentrations of Cu(II) as catalyst are also available elsewhere (Iwata and Locker, 1998); however, with this reagent mixture the onset of emission was almost instantaneous and reached a maximum within a few seconds. As the figure shows the light intensity decayed to approximately 50% of maximum at about 8 seconds. Iwata and Locker found that both the initial intensity and rate of decay in this kind of system was dependent on Cu(II) content. In the TCPO system described above, Orosz et al. (1996) reported that decay rate, rise constant, maximal light intensity, and quantum efficiency depended on hydrogen peroxide concentration. These authors present a comprehensive review of efforts to model the optimization of reagent flow rates and concentrations on HPLC detector responses with the TCPO reaction(s). The luminol reaction described above was carried out by placing approximately 15 mL of a solution containing the fuel (luminol), Cu (1.5 x 10 M as the sulfate), and buffers detailed above in a open-topped glass vial (~25 C). The initial solution is visible at the movie's beginning as light blue in color under the laboratory's fluorescent light due to aqueous copper cations. In the dark, aqueous hydrogen peroxide (~0.3%) was added dropwise four times (small 1 or 2 mL squirts is probably a better description). The light emission is also, as before with TCPO, almost simultaneous upon mixing. The light produced appears white/blue and, as in the TCPO/DPA movie, since fuel is initially in excess, multiple injections of the limiting H O reagent are necessary to take the reaction nearer to completion. Finally, after the fourth addition, the mixture was allowed to decay undisturbed and the light intensity drops off rather quickly (see the time decay data above). Approximately 80 seconds after the initial mixing began, the overhead fluorescent light were turned on and the final frame shows that solution. The light blue solution then appear green with a finely dispersed, black precipitate. Table 1: Analytically Useful Chemiluminescent Emitters | 13,603 | 1,225 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/02%3A_Gas_Laws/2.14%3A_Gas_Mixtures_-_Dalton's_Law_of_Partial_Pressures |
Thus far, our discussion of the properties of a gas has implicitly assumed that the gas is pure. We turn our attention now to mixtures of gases—gas samples that contain molecules of more than one compound. Mixtures of gases are common, and it is important to understand their behavior in terms of the properties of the individual gases that make it up. The ideal-gas laws we have for mixtures are approximations. Fortunately, these approximations are often very good. When we think about it, this is not surprising. After all, the distinguishing feature of a gas is that its molecules do not interact with one another very much. Even if the gas is composed of molecules of different kinds, the unimportance of molecule—molecule interactions means that the properties of one kind of molecules should be nearly independent of the properties of the other kinds. Consider a sample of gas that contains a fixed number of moles of each of two or more compounds. This sample has a pressure, a volume, a temperature, and a specified composition. Evidently, the challenge here is to describe the pressure, volume, and temperature of the mixture in terms of measurable properties of the component compounds. There is no ambiguity about what we mean by the pressure, volume, and temperature of the mixture; we can measure these properties without difficulty. Given the nature of temperature, it is both reasonable and unambiguous to say that the temperature of the sample and the temperature of its components are the same. However, we cannot measure the pressure or volume of an individual component in the mixture. If we hope to describe the properties of the mixture in terms of properties of the components, we must first define some related quantities that we can measure. The concepts of a component partial pressure and a component partial volume meet this need. We define the of a component of a gas mixture as the pressure exerted by the same number of moles of the pure component when present in the volume occupied by the mixture, \(V_{mixture}\), at the temperature of the mixture. In a mixture of \(n_A\) moles of component \(A\), \(n_B\) moles of component \(B\), etc., it is customary to designate the partial pressure of component \(A\) as \(P_A\). It is important to appreciate that the partial pressure of a real gas can only be determined by experiment. We define the of a component of a gas mixture as the volume occupied by the same number of moles of the pure component when the pressure is the same as the pressure of the mixture, \(P_{mixture}\), at the temperature of the mixture. In a mixture of components \(A\), \(B\), etc., it is customary to designate the partial volume of component \(A\) as \(V_A\). The partial volume of a real gas can only be determined by experiment. . That is, for a mixture of components A, B, C, etc., the pressure of the mixture is \[P_{mixture}=P_A+P_B+P_C+\dots \label{Dalton}\] Under conditions in which the ideal gas law is a good approximation to the behavior of the individual components, Dalton’s law is usually a good approximation to the behavior of real gas mixtures. For mixtures of ideal gases, it is exact. To see this, we recognize that, for an ideal gas, the definition of partial pressure becomes \[P_A=\frac{n_ART}{V_{mixture}}\] The ideal-gas mixture contains \(n_{mixture}=n_A+n_B+n_C+\dots \text{moles}\), so that \[ \begin{align*} P_{mixture} &=\frac{n_{mixture}RT}{V_{mixture}} \\[4pt] &=\frac{\left(n_A+n_B+n_C+\dots \right)RT}{V_{mixture}} \\[4pt] &=\frac{n_ART}{V_{mixture}}+\frac{n_BRT}{V_{mixture}}+\frac{n_CRT}{V_{mixture}}+\dots \\[4pt] &=P_A+P_B+P_C+\dots \end{align*}\] Applied to the mixture, the ideal-gas equation yields Dalton’s law (Equation \ref{Dalton}). When \(x_A\) is the mole fraction of A in a mixture of ideal gases, \[P_A=x_AP_{mixture}.\] | 3,844 | 1,226 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.03%3A_Lewis_Diagrams/5.3.01%3A_Lewis_Diagrams_and_Biological_and_Chemical_Properties |
A first step toward gaining some insight into the biological properties of atoms or compounds is to draw their . For example, we'll explore lithium fluoride and sodium chloride below. Lithium fluoride and sodium chloride are similar in some ways: The both contain 1+ alkali metal ions and 1- halogen ions; they both are high melting, white, crystalline solids; and they both dissolve in water to make conducting solutions. But as we have come to suspect, the biological properties are quite different, and must be considered along with chemical properties. Lithium Fluoride is very toxic Because of the lithium ion's small size, it causes ion (electrolyte) imbalances, and, of course, is a strong sedative. Fluoride ion forms hydrofluoric acid in the stomach which is very corrosive, and disrupts metabolism . Sodium chloride (table salt) is much less toxic, with high doses leading to chronic high blood pressure. G.N. Lewis used simple diagrams (now called "Lewis Diagrams") to keep track of how many electrons were present in the outermost, or valence, shell of a given atom. The of the atom, i.e., the nucleus together with the inner electrons, is represented by the chemical symbol, and only the valence electrons are drawn as dots surrounding the chemical symbol. Thus the three atoms shown here from can be represented by the following Lewis diagrams: If the atom is a noble-gas atom, two alternative procedures are possible. Either we can consider the atom to have zero valence electrons or we can regard the outermost filled shell as the valence shell. The first three noble gases can thus be written as Draw Lewis diagrams for an atom of each of the following elements: Li N F Na We find from the periodic table inside the front cover that Li has an atomic number of 3. It thus contains three electrons, one more than the noble gas He. This means that the outermost, or valence, shell contains only one electron, and the Lewis diagram is Notice from the preceding example that the Lewis diagrams of the are identical except for their chemical symbols. This agrees nicely with the very similar chemical behavior of the alkali metals. Similarly, Lewis diagrams for all elements in other groups, such as the or , look the same. The Lewis diagrams may also be used to predict of the elements. Lewis suggested that the number of valences of an atom was equal to the number of electrons in its valence shell or to the number of electrons which would have to be added to the valence shell to achieve the electronic shell structure of the next noble gas. As an example of this idea, consider the elements Be and O. Their Lewis diagrams and those of the noble gases He and Ne are Comparing Be with He, we see that the former has two more electrons and therefore should have a valence of 2. The element O might be expected to have a valence of 6 or a valence of 2 since it has six valence electrons—two less than Ne. Using rules of valence developed in this way, Lewis was able to account for the regular increase and decrease in the subscripts of the compounds in the table below. For example, he reasoned that lithium would lose an electron to fluorine, forming toxic Li F : Or that chlorine would gain an electron from sodium to form Na Cl : Similarly, oxygen would gain one electron from each of two sodium ions to form nontoxic Na O , another of the successfully predicted compounds in the table. In addition he was able to account for more than 50 percent of the formulas in the table. (Those that agree with his ideas are shaded in color or gray in the table. You may wish to refer to that table now and verify that some of the indicated formulas follow Lewis’ rules.) Lewis’ success in this connection gave a clear indication that electrons were the most important factor in holding atoms together when molecules formed. \(\Page {1}\) Compounds of Hydrogen, Oxygen, and Chlorine
Despite these successes, there are also difficulties to be found in Lewis’ theories, in particular for elements beyond calcium in the periodic table. The element Br ( = 35), for example, has 17 more electrons than the noble-gas Ar ( = 18). This leads us to conclude that Br has 17 valence electrons, which makes it awkward to explain why Br resembles Cl and F so closely even though these two atoms have only seven valence electrons. From ChemPRIME: | 4,380 | 1,228 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/7%3A_Bonding_in_Organic_Molecules/7.3%3A_The_Alkenes_and_Alkynes |
As noted before, alkenes are hydrocarbons with carbon-to-carbon double bonds (R C=CR ) and alkynes are hydrocarbons with carbon-to-carbon triple bonds (R–C≡C–R). Collectively, they are called unsaturated hydrocarbons because they have fewer hydrogen atoms than does an alkane with the same number of carbon atoms, as is indicated in the following general formulas: Some representative alkenes—their names, structures, and physical properties—are given in Table \(\Page {1}\). We used only condensed structural formulas in Table \(\Page {1}\). Thus, CH =CH stands for The double bond is shared by the two carbons and does not involve the hydrogen atoms, although the condensed formula does not make this point obvious. Note that the molecular formula for ethene is C H , whereas that for ethane is C H . The first two alkenes in Table \(\Page {1}\), ethene and propene, are most often called by their common names—ethylene and propylene, respectively (Figure \(\Page {1}\)). Ethylene is a major commercial chemical. The chemical industry produces about 25 billion kilograms of ethylene annually, more than any other synthetic organic chemical. More than half of this ethylene goes into the manufacture of polyethylene, one of the most familiar plastics. Propylene is also an important industrial chemical. It is converted to plastics, isopropyl alcohol, and a variety of other products. Although there is only one alkene with the formula C H (ethene) and only one with the formula C H (propene), there are several alkenes with the formula C H . Here are some basic rules for naming alkenes from the International Union of Pure and Applied Chemistry (IUPAC): Name each compound. Name each compound. Just as there are cycloalkanes, there are . These compounds are named like alkenes, but with the prefix - attached to the beginning of the parent alkene name. Draw the structure for each compound. Now place the methyl group on the third carbon atom and add enough hydrogen atoms to give each carbon atom a total of four bonds. Draw the structure for each compound. | 2,086 | 1,230 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/03%3A_The_First_Law_of_Thermodynamics/3.07%3A_Bond_Energies_and_Enthalpies |
In the absence of standard formation enthalpies, reaction enthalpies can be estimated using average bond enthalpies. This method is not perfect, but it can be used to get ball-park estimates when more detailed data is not available. A \(D\) is defined by \[XY(g) \rightarrow X(g) + Y(g)\] \(\Delta H \equiv D(X-Y)\) In this process, one adds energy to the reaction to break bonds, and extracts energy for the bonds that are formed. \[\Delta H_{rxn} = \sum (\text{bonds broken}) - \sum (\text{bonds formed})\] As an example, consider the combustion of ethanol: In this reaction, five C-H bonds, one C-C bond, and one C-O bond, and one O=O bond must be broken. Also, four C=O bonds, and one O-H bond are formed. The reaction enthalpy is then given by \[ \begin{align} \Delta H_c = \, &5(413 \,kJ/mol) + 1(348\, kJ/mol) + 1(358 \,kJ/mol) \nonumber \\ & + 1(495\, kJ/mol) - 4(799 \,kJ/mol) – 2(463\, kJ/mol) \nonumber \\ =\,& -856\, kJ/mol \end{align}\] Because the bond energies are defined for gas-phase reactants and products, this method does not account for the enthalpy change of condensation to form liquids or solids, and so the result may be off systematically due to these differences. Also, since the bond enthalpies are averaged over a large number of molecules containing the particular type of bond, the results may deviate due to the variance in the actual bond enthalpy in the specific molecule under consideration. Typically, reaction enthalpies derived by this method are only reliable to within ± 5-10%. | 1,543 | 1,231 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/29%3A_Chemical_Kinetics_II-_Reaction_Mechanisms/29.05%3A_Rate_Laws_Do_Not_Imply_Unique_Mechanism |
Because a proposed mechanism can only be valid if it is consistent with the rate law found experimentally, the rate law plays a central role in the investigation of chemical reaction mechanisms. The discussion above introduces the problems and methods associated with collecting rate data and with finding an empirical rate law that fits experimental concentration- -time data. We turn now to finding the rate laws that are consistent with a particular proposed mechanism. For simplicity, we consider reactions in closed constant-volume systems. In principle, numerical integration can be used to predict the concentration at any time of each of the species in any proposed reaction mechanism. This prediction can be compared to experimental observations to see whether they are consistent with the proposed mechanism. To do the numerical integration, it is necessary to know the initial concentrations of all of the chemical species and to know, or assume, values of all of the rate constants. The initial concentrations are known from the procedure used to initiate the reaction. However, the rate constants must be determined by some iterative procedure in which initial estimates of the rate constants are used to predict concentration- -time data that can be compared to the experimental results to produce refined estimates. In practice, we tailor our choice of reaction conditions so that we can use various approximations to test whether a proposed mechanism can explain the data. We now consider the most generally useful of these approximations. In this discussion, we assume that the overall reaction goes to completion; that is, at equilibrium the concentration of the reactant whose concentration is limiting has become essentially zero. If the overall reaction involves more than one elementary step, then an intermediate compound is involved. A valid mechanism must include this intermediate, and more than one differential equation may be needed to characterize the time rate of change of all of the species involved in the reaction. We focus on conditions and approximations under which the rate of appearance of the final products in a multi-step reaction mechanism can be described by a single differential equation, the rate law. We examine the application of these approximations to a particular reaction mechanism. When we understand the application of these approximations to this mechanism, the ways in which they can be used in other situations are clear. Consider the following sequence of elementary steps \[\ce{ A + B <=>[k_1,k_2] C ->[k_3] D}\] whose kinetics are described by the following simultaneous differential equations: \[\begin{align*} \frac{d[A]}{dt}=\frac{d[B]}{dt}&={-k}_1[A,B]+k_2\left[C\right] \\[4pt] \frac{d\left[C\right]}{dt} &=k_1[A,B]-k_2\left[C\right]-k_3\left[C\right] \\[4pt] \frac{d\left[D\right]}{dt} &=k_3\left[C\right] \end{align*}\] The general analytical solution for this system of coupled differential equations can be obtained, but it is rather complex, because \(\left[C\right]\) increases early in the reaction, passes through a maximum, and then decreases at long times. In principle, experimental data could be fit to these equations. The numerical approach requires that we select values for \(k_1\), \(k_2\), \(k_3\), \({[A]}_0\), \({[B]}_0\), \({\left[C\right]}_0\), and \({\left[D\right]}_0\), and then numerically integrate to get \([A]\), \([B]\), \(\left[C\right]\), and \(\left[D\right]\) as functions of time. In principle, we could refine our estimates of \(k_1\), \(k_2\), and \(k_3\) by comparing the calculated values of one or more concentrations to the experimental ones. In practice, the approximate treatments we consider next are more expedient. When we begin a kinetic study, we normally have a working hypothesis about the reaction mechanism, and we design our experiments to simplify the differential equations that apply to it. For the present example, we will assume that we always arrange the experiment so that \({\left[C\right]}_0=0\) and \({\left[D\right]}_0=0\). In consequence, at all times: \[{[A]}_0=[A]+\left[C\right]+\left[D\right].\] Also, we restrict our considerations to experiments in which \({[B]}_0\gg {[A]}_0\). This exemplifies the use of . The practical effect is that the concentration of \(B\) remains effectively constant at its initial value throughout the entire reaction, which simplifies the differential equations significantly. In the present instance, setting \({[B]}_0\gg {[A]}_0\) means that the rate-law term \(k_1[A,B]\) can be replaced, to a good approximation, by \(k_{obs}[A]\), where \(k_{obs}=k_1{[B]}_0\). Once we have decided upon the reaction conditions we are going to use, whether the resulting concentration- -time data can be described by a single differential equation depends on the relative magnitudes of the rate constants in the several steps of the overall reaction. Particular combinations of relationships that lead to simplifications are often referred to by particular names; we talk about a combination that has a , or one that involves a , or one in which a is applicable. To see what we mean by these terms, let us consider some particular relationships that can exist among the rate constants in the mechanism above. Suppose that \(k_1[A,B]\gg k_2\left[C\right]\) and \(k_3\gg k_2\). We often describe this situation by saying, rather imprecisely, that the reaction to convert \(C\) to \(D\) is very fast and that the reaction to convert \(C\) back to \(A\) and \(B\) is very slow—compared to the reaction that forms \(C\) from \(A\) and \(B\). When \(C\) is produced in these circumstances, it is converted to \(D\) so rapidly that we never observe a significant concentration of \(C\) in the reaction mixture. The formation of a molecule of \(C\) is tantamount to the formation of a molecule of \(D\), and the reaction produces \(D\) at essentially the same rate that it consumes \(A\) or \(B\). We say that the first step, \(A+B\to C\), is the rate-determining step in the reaction. We have \[-\frac{d[A]}{dt}=\ -\frac{d[B]}{dt}\approx \frac{d\left[D\right]}{dt}\] The assumption that \(k_1[A,B]\gg k_2\left[C\right]\) means that we can neglect the smaller term in the equation for \({d[A]}/{dt}\), giving the approximation \[\frac{d[A]}{dt}=\ \frac{d[B]}{dt}=\ -\frac{d\left[D\right]}{dt}=\ -k_1[A,B]\] Letting \(\left[D\right]=x\) and recognizing that our assumptions make \(\left[C\right]\approx 0\), the mass-balance condition, \({[A]}_0=[A]+\left[C\right]+\left[D\right]\), becomes \([A]={[A]}_0-x\). Choosing \({[B]}_0\gg {[A]}_0\) means that \(k_1[B]\approx k_1{[B]}_0=k_{I,obs}\). The rate equation becomes first-order: \[\frac{dx}{dt}=k_{I,obs}\left(\ {[A]}_0-x\right)\] Since \(k_{I,obs}\) is not strictly constant, it is a pseudo-first-order rate constant. The disappearance of \(A\) is said to follow a pseudo-first-order rate equation. The concept of a rate-determining step step is an . In general, the consequence we have in mind when we invoke this approximation is that no intermediate species can accumulate to a significant concentration if it is produced by the rate-determining step or by a step that occurs after the rate-determining step. We do not intend to exclude the accumulation of a species that is at equilibrium with another product. Thus, in the mechanism \[\ce{ A ->[k] B <=> C}\] we suppose that the conversion of \(A\) to \(B\) is rate-determining and that the interconversion of \(B\) and \(C\) is so rapid that their concentrations always satisfy the equilibrium relationship \[K=\dfrac{[C]}{[B]}.\] For the purpose at hand, we do not consider \(B\) to be an intermediate; \(B\) is a product that happens to be at equilibrium with the co-product, \(C\). Suppose that \(k_1[A,B]\gg k_3\left[C\right]\). In this case \(A+B\to C\) is fast compared to the rate at which \(C\) is converted to \(D\), and we say that \(C\to D\) is the rate-determining step. We can now distinguish three sub-cases depending upon the way \(\left[C\right]\) behaves during the course of the reaction. Suppose that \(k_1[A,B]\gg k_3\left[C\right]\) and \(k_3\gg k_2\). Then \(A+B\to C\) is rapid and essentially quantitative. That is, within a short time of initiating the reaction, all of the stoichiometrically limiting reactant is converted to \(C\). Letting \(\left[D\right]=x\) and recognizing that our assumptions make \([A]\approx 0\), the mass-balance condition, \[{[A]}_0=[A]+\left[C\right]+\left[D\right]\] becomes \[\left[C\right]={[A]}_0-x.\] After a short time, the rate at which \(D\) is formed becomes \[\frac{d\left[D\right]}{dt}=k_3\left[C\right]\] or \[\frac{dx}{dt}=k_3\left(\ {[A]}_0-x\right)\] The disappearance of \(C\) and the formation of \(D\) follow a first-order rate law. If the forward and reverse reactions in the first elementary process are rapid, then this process may be effectively at equilibrium during the entire time that \(D\) is being formed. (This is the case that \(k_1[A,B]\gg k_3\left[C\right]\) and \(k_2\gg k_3\).) Then, throughout the course of the reaction, we have \[K_{eq}={\left[C\right]}/{[A,B]}\] Letting \(\left[D\right]=x\) and making the further assumption that \([A]\gg \left[C\right]\approx 0\) throughout the reaction, the mass-balance condition, \({[A]}_0=[A]+\left[C\right]+\left[D\right]\), becomes \([A]={[A]}_0-x\). Substituting into the equilibrium-constant expression, we find \[\left[C\right]=K_{eq}{[B]}_0\ \left({[A]}_0-x\right)\] Substituting into \({d\left[D\right]}/{dt}=k_3\left[C\right]\) we have \[\frac{dx}{dt}=k_3K_{eq}{[B]}_0\ \left({[A]}_0-x\right)=k_{IIa,obs}\left({[A]}_0-x\right)\] where \(k_{IIa,obs}=k_3K_{eq}{[B]}_0\). The disappearance of A and the formation of D follow a pseudo-first-order rate equation. The pseudo-first-order rate constant is a composite quantity that is directly proportional to \({[B]}_0\). If we suppose that the first step is effectively at equilibrium during the entire time that \(D\) is being produced (as in case IIb) but that \(\left[C\right]\) is not negligibly small compared to \([A]\), we again have \(K_{eq}={\left[C\right]}/{[A,B]}\). With \(\left[D\right]=x\), the mass-balance condition becomes \([A]={[A]}_0-\left[C\right]-x\). Eliminating \([A]\) between the mass-balance and equilibrium-constant equations gives \[\left[C\right]=\frac{K_{eq}{[B]}_0\left({[A]}_0-x\right)}{1+K_{eq}{[B]}_0}\] so that \({d\left[D\right]}/{dt}=k_3\left[C\right]\) becomes \[\frac{dx}{dt}=\left(\frac{{k_3K}_{eq}{[B]}_0}{1+K_{eq}{[B]}_0}\right)\left({[A]}_0-x\right)=k_{IIc,obs}\left({[A]}_0-x\right)\] The formation of \(D\) follows a pseudo-first-order rate equation. (The disappearance of \(A\) is also pseudo-first-order, but the pseudo-first-order rate constant is different.) As in Case IIb, the pseudo-first-order rate constant, \(k_{IIc,obs}\), is a composite quantity, but now its dependence on \({[B]}_0\) is more complex. The result for Case IIc reduces to that for Case IIb if \(K_{eq}{[B]}_0\ll 1\). In the cases above, we have assumed that one or more reactions are intrinsically much slower than others are. The differential equations for this mechanism can also become much simpler if all three reactions proceed at similar rates, but do so in such a way that the concentration of the intermediate is always very small, \(\left[C\right]\approx 0\). If the concentration of \(C\) is always very small, then we expect the graph of \(\left[C\right]\) versus time to have a slope, \({d\left[C\right]}/{dt}\), that is approximately zero. In this case, we have \[\frac{d\left[C\right]}{dt}=k_1[A,B]-k_2\left[C\right]-k_3\left[C\right]\approx 0\] so that \[\left[C\right]=\frac{k_1[A,B]}{k_2+k_3}\] With \(\left[D\right]=x\), \({d\left[D\right]}/{dt}=k_3\left[C\right]\) becomes \[\frac{dx}{dt}=\left(\frac{k_1k_3{[B]}_0}{k_2+k_3}\right)\left({[A]}_0-x\right)=k_{III,obs}\left({[A]}_0-x\right)\] As in the previous cases, the disappearance of \(A\) and the formation of \(D\) follow a pseudo-first-order rate equation. The pseudo-first-order rate constant is again a composite quantity, which depends on \({[B]}_0\) and the values of all of the rate constants. Case III illustrates the , in which we assume that the concentration of an intermediate species is much smaller than the concentrations of other species that affect the reaction rate. Under these circumstances, we can be confident that the time-derivative of the intermediate’s concentration is negligible compared to the reaction rate, so that it is a good approximation to set it equal to zero. The idea is simply that, if the concentration is always small, its time-derivative must also be small. If the graph of the intermediate’s concentration versus time is always much lower than that of other participating species, then its slope will be much less. Equating the time derivative of the steady-state intermediate’s concentration to zero produces an algebraic expression that involves the intermediate’s concentration. Solving this expression for the concentration of the steady-state intermediate makes it possible to greatly simplify the set of simultaneous differential equations that is predicted by the mechanism. When there are multiple intermediates to which the approximation is applicable, remarkable simplifications can result. This often happens when the mechanism involves free-radical intermediates. The name “steady-state approximation” is traditional. When we use it, we do so on the understanding that the “state” which is approximately “steady” is the concentration of the intermediate, not the state of the system. Since a net reaction is occurring, the state of the system is distinctly not constant. | 13,689 | 1,232 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.10%3A_Enthalpy_of_Fusion_and_Enthalpy_of_Vaporization/10.10.02%3A_Lecture_Demonstrations |
A small mass of water at 0 C is added to a measured mass of liquid nitrogen, and the amount that evaporates is compared to the mass that evaporates when a larger mass of water at 95 C is added to liquid nitrogen. This demonstration requires knowledge of both specific heat and heat capacity. Calculate Enthalpy of Fusion of Ice (assuming heat capacity prerequisite) Dip a computer-interfaced thermistor probe in 100g of water in a styrofoam cup calorimeter. Add 3-5 g of ice to a paper towel on a balance, and record the total mass. Start temperature acquisition 1 sample/second, 3 minutes total, and after a few readings, remove ~2 g of ice from the balance and add it to the calorimeter. Record the final mass on the balance and calculate the mass of ice. Display the T vs. time plot . Record the final temperature. q (cal) + q (water ) = q (water from ice) + q (ice) q = 581 J ΔH = 581J/1.90 g x (1kg/1000 J) x (18 g/mol) = 5.5 kJ/mol (6.07 kJ/mol true value) Ed Vitz (Kutztown University), (University of | 1,032 | 1,233 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Fundamentals_of_Thermodynamics/State_vs._Path_Functions |
A is a property whose value does not depend on the path taken to reach that specific value. In contrast, functions that depend on the path from two values are call . Both path and state functions are often encountered in thermodynamics. Whenever compounds or chemical reactions are discussed, one of the first things mentioned is the state of the specific molecule or compound. "State" refers to temperature, pressure, and the amount and type of substance present. Once the state has been established, state functions can be defined. State functions are values that depend on the state of the substance, and not on how that state was reached. For example, density is a state function, because a substance's density is not affected by how the substance is obtained. Consider a quantity of H O: it does not matter whether that H O is obtained from the tap, from a well, or from a bottle, because as long as all three are in the same state, they have the same density. When deciding whether a certain property is a state function or not, keep this rule in mind: is this property or value affected by the path or way taken to establish it? If the answer is no, then it is a state function, but if the answer is yes, then it is not a state function. Another way to think of state functions is as integrals. Integrals depend on only three things: the function, the lower limit and the upper limit. Similarly, state functions depend on three things: the property, the initial value, and the final value. In other words, integrals illustrate how state functions depend only on the final and initial value and not on the object's history or the path taken to get from the initial to the final value. Here is an example of the integral of enthalpy, \(H\), where \(t_0\) represents the initial state and \(t_1\) represents the final state. \[ \displaystyle \int_{t_o}^{t_1} \; H(t) dt = H(t_1)-H(t_o) \] This is equivalent to a familiar definition of enthalpy: \[ \Delta H = H_{final} - H_{initial}\] As represented by the solution to the integral, enthalpy is a state function because it only depends on the initial and final conditions, and not on the path taken to establish these conditions. Therefore, the integral of state functions can be taken using only two values: the final and initial values. On the other hand, multiple integrals and multiple limits of integration are required take the integral of a path function. If an integral of a certain property can be calculated using just the property and it's initial and final value, the property is a state function. State functions are defined by comparing them to path functions. As stated before, a state function is a property whose value does not depend on the path taken to reach that specific function or value. In essence, if something is not a path function, it is probably a state function. To better understand state functions, first define path functions and then compare path and state functions. Path functions are functions that depend on the path taken to reach that specific value. For example, suppose you have $1000 in your savings account. Suppose you want to deposit some money to this account. The amount you deposit is a path function because it is dependent upon the path taken to obtain that money. In other words, the amount of money you will deposit in your savings account is dependent upon the path or way taken to obtain that money. If you work as a CEO of a company for a week versus working at a gas station for a week, you would receive two different amounts of money at the end of the week. Thus, a path function is a property or value that is dependent on the path taken to establish that value. State functions do not depend on the path taken. Using the same example, suppose you have $1000 in your savings account. You withdraw $500 from your savings account. It does not matter whether you withdraw the $500 in one shot or whether you do so at a rate of $50. At the end when you receive your monthly statement, you will notice a net withdrawal of $500 and will see your resulting balance as $500. Thus, the bank balance is a state function because it does not depend on the path or way taken to withdraw or deposit money. In the end whether you do so in one lump or in multiple transactions, your bank balance will stay the same. The figure below illustrates state functions in the form of enthalpy: In this figure, two different steps are shown to form \(NaCl_{(s)}\). : The first path takes only a single step with an enthalpy of formation of -411 kJ/mol: \[Na^+_{(g)} + Cl^-_{(g)} \rightarrow NaCl_{(s)}\] : The second path takes five steps to form \(NaCl_{(s)}\) \[Na_{(s)} + 1/2 \;Cl_{(g)} \rightarrow Na_{(g)} + 1/2\; Cl_{(g)} \tag{1: sublimation}\] \[Na_{(g)} + 1/2 \;Cl_{(g)} \rightarrow Na_{(g)} + Cl_{(g)} \tag{2: atomization}\] \[Na_{(g)} + Cl_{(g)} \rightarrow Na^+_{(g)} + Cl_{(g)} \tag{3: ionization}\] \[Na^+_{(g)} + Cl_{(g)} \rightarrow Na^+_{(g)} + Cl^-_{(g)} \tag{4: electron affinity}\] \[Na^+_{(g)} + Cl^-_{(g)} \rightarrow NaCl_{(s)} \tag{5: lattice formation}\] When enthalpies of all these steps are added, the enthalpy of formation of \(NaCl_{(s)}\) is still -411 kJ/mol. This is a perfect example of a state function: no matter which path is taken to form \(NaCl_{(s)}\), it results the same enthalpy of formation of -411 kJ/mol. The last comparison made is a generalization that does not necessarily hold for all aspects and calculations involved in chemistry. The main point to remember when trying to identify a state function is to determine whether the path taken to reach the function affects the value. The analogy below illustrates how to tell whether a certain property is a state function. Every morning, millions of people must decide how to reach their offices. Some opt for taking the stairs, whereas others take the elevator. In this situation, ∆y, or change in vertical position is the same whether a person take the stairs or the elevator. The distance from the office lobby to the office stays the same, irrespective of the path taken to get to your office. As a result, ∆y is a state function because its value is independent of the path taken to establish its value. In the same situation, time, or ∆t, is not a state function. If someone takes the longer way of getting to the office (climbing the stairs), ∆t would be greater, whereas ∆t would be smaller if the elevator is taken. In this analogy, ∆t is not a state function because its value is dependent on the path. State functions are commonly encountered in thermodynamics; many of the equations involved with thermodynamics, such as \(\Delta U\) and \(\Delta H\), are state functions. Additionally, state functions are crucial in thermodynamics because they make calculations simple and allow one to calculate data that could otherwise only be obtained through experiments. More specifically, state functions facilitate the use of , which allows the manipulation (addition, subtraction, multiply etc.) of the enthalpies of half reactions when adding multiple half reactions to form a full reaction. Hess's Law is dependent upon the fact that enthalpy is a state function. If enthalpy was not a state function, Hess's Law would be much more complicated, because the enthalpies of half reactions could not be added. Instead, several additional calculations would be required. Furthermore, state functions and Hess's Law helps one calculate the enthalpy of complex reactions without having to actually replicate these reactions in a laboratory. All that is required is to write out and sum the enthalpy of the half reactions or of the hypothetical steps leading to the chemical reaction. State functions are also encountered in many other equations involved with thermodynamics such as internal energy (∆U), Gibb's free energy, enthalpy, and entropy. | 7,857 | 1,235 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/18%3A_Partition_Functions_and_Ideal_Gases/18.02%3A_Most_Atoms_are_in_the_Ground_Electronic_State |
Writing the electronic energies as \(E_1, E_2 ,E_3, ...\) with corresponding degeneracies \(g_1, g_2, g_3, \ldots\). The electronic partition function is then given by the following summation: \[ q_{el}(T) = g_1 e^{-E_1/kT} + g_2 e^{-E_2/kT} + g_3 e^{-E_3/kT} + \ldots \label{Q1} \] Usually, the differences in electronic energies are significantly greater than thermal energy \(kT\): \[ k T \ll E_1 < E_2 < E_3 \nonumber \] If we treat the lowest energy electronic state \(E_1\) as the reference value of zero of energy, the electronic partition function (Equation \ref{Q1}) can be approximated as: \[q_{elec}(T) = g_1 + g_2 e^{-E_2/kT} + g_3 e^{-E_3/kT} + \ldots \label{3.24} \] Typically electronic states are tens of thousands of wave numbers above the ground state. For example, the first excited electronic state of nitric oxide (NO) is ~40,000 cm . Using this value: \[ \frac{E_2}{kT} = \frac{40000\text{ cm}^{-1}}{kT} = \frac{10^4\text{ K}}{T} \nonumber \] That means, that even at 1,000 K, the value of the second term in {\reference{3.24}} is: \[ g_2 e^{-10} = g_2 4.5\times 10^{-5} \nonumber \] The result is that the higher electronic states are not accessible under ordinary temperatures. There are some cases, however, where the first excited state lies much closer to the ground state, but these are the exception rather than the rule. Find the electronic partition of \(\ce{H_2}\) at 300 K. The lowest electronic energy level of \(\ce{H_2}\) is near \(- 32\; eV\) and the next level is about \(5\; eV\) higher. Taking -32 eV as the zero (or reference value of energy), then \[q_{el} = e_0 + e^{-5 eV/ kT} + ... \nonumber \] At 300 K, T = 0.02\; eV and \[ \begin{align*} q_{el} &= 1 + e^{-200} +... \\[4pt] &\approx 1.0 \end{align*} \nonumber \] Where all terms other than the first are essentially 0. This implies that \(q_{el} = 1\). The physical meaning of the result from Example 18.2.1
is that only the ground electronic state is generally thermally accessible at room temperature. | 2,020 | 1,236 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/04%3A_Biological_and_Synthetic_Dioxygen_Carriers/4.16%3A_Selected_Chemistry_of_Dioxygen_Iron_Copper_and_Cobalt |
Dioxygen is a powerful oxidant, capable of oxidizing all but the noble metals and of converting many low-valent metal complexes to higher-valent states. As will be detailed in this section, the binding of dioxygen to metals is most usefully considered as an oxidative addition process. The nature of the interaction is determined by the metal, its oxidation state, and its ligands that modulate the redox properties of the metal center. In biological and nonbiological oxygen carriers, several factors allow reversible binding of O to occur, even though this process is metastable with respect to (irreversible) oxidation of the metal, or its ligands, or other species that may be present. Later in this section the bioinorganic chemistry of iron, copper, and cobalt is described. For a wider perspective on the coordination chemistry of these metals, see comprehensive texts on inorganic chemistry. Many techniques have been used to probe the metal-dioxygen moiety. A summary of these techniques, key concepts, and results is presented in Table 4.3. UV-visible spectroscopy usually characterizes the oxidation state of the metal and in favorable cases the number, geometry, and ligand field strength of ligands. The O—O and M—O stretching modes may be investigated with infrared spectroscopy, provided that the complex is not a centrosymmetric dimer, for then the O—O stretch for the \(\mu\)-dioxygen species is infrared-inactive. Resonance Raman techniques complement infrared spectroscopy. Not only are the selection rules different in Raman spectroscopy, but a suitable choice of the irradiating wavelength (to coincide approximately with an M-L electronic transition) can amplify those vibrational modes that are coupled, or in resonance, with the electronic transition. This technique is particularly suited as a probe of the metal-ligand environment of metalloproteins, since the many solely protein vibrational modes disappear into background noise. Geometric information on the orientation of the CO moiety with respect to the heme normal has been obtained by examining polarization behavior of infrared bands following photolysis of the Fe—CO bond by linearly polarized light. Spin and oxidation states of mononuclear iron-porphyrin systems may be assigned directly from magnetic susceptibility measurements and indirectly from Mössbauer spectroscopy. Variable temperature susceptibility measurements are particularly useful for detecting dinuclear systems that share at least one ligand in common if there is (or ferromagnetic) coupling of the electron spin of one metal center with that of a second. Definitive characterization of the stereochemistry is usually provided by x-ray diffraction data when single crystals are available. In general, the level of resolution and precision available from protein crystal structures leads to tantalizing uncertainties over the geometry of the M—O species and of the structural changes occurring on oxygenation that are the origin of cooperativity. Precise structural data are more readily obtained from small-molecule model systems. The relevance of these to biological systems is established through congruence of spectroscopic and functional properties. X-ray diffraction techniques also provide important information on the environment beyond the immediate surroundings of the metal center: this information is usually unobtainable from other techniques, although recent developments in two-dimensional NMR spectroscopy can provide this information for diamagnetic systems. Limited information may be obtained with the use of spin labels or, if the metal center is paramagnetic, with EPR techniques. Two other techniques that selectively probe the immediate environment of the metal center are EXAFS (Extended X-ray Absorption Fine Structure) and XANES (X-ray Absorption Near-Edge Structure). The former may yield information on the number and type of bonded atoms and their radial separation from the metal center. The latter technique may reveal the oxidation state and, in principle, may yield geometric information, although in its present state of development some interpretations are contentious. Both techniques have the advantage of not requiring crystalline material. The structural information is more reliable if definitive model systems are available for comparison. X-ray (and, less frequently, neutron) diffraction techniques on single crystals give absolute structural information* and thus provide the basis for interpretation of data obtained from these other techniques that yield relative structural information. * In favorable situations, sophisticated NMR techniques have been applied successfully to detennine the polypeptide folding (e.g., in metallothionein). | 4,763 | 1,238 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/02%3A_Spectroscopy-_how_we_know_what_we_know_about_the_structure_of_matter |
From your earlier studies, you will recall that at the atomic/molecular level, the energies of atoms and molecules are quantized: that is, there are discrete, separate energy states with nothing in between. This applies not only to the energies of electrons, but also to the energies of vibration of bonds and rotation around bonds. Nuclear energies are also quantized. It is possible to switch from one energy state to another by absorbing or emitting an amount of electromagnetic energy (a photon) that corresponds to the energy difference between the two states. Photons with the energy that does not correspond to differences between two energy states are not absorbed or emitted, which means that we can use the energies of the photons absorbed or emitted to tell us about the energy differences between quantum states in atoms and molecules. The differences between these quantized energy levels is highly dependent on the identity and environment of the atoms and molecules, and therefore we can the photons emitted to identify particular species. For example, in isolated atoms the energy differences between levels often correspond to electromagnetic energy in the ultraviolet or visible. This results in the atomic absorption or emission spectra that allow us to determine what elements are present in (for example) stars, and interstellar space.
| 1,371 | 1,239 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/17%3A_Boltzmann_Factor_and_Partition_Functions/17.05%3A_Pressure_in_Terms_of_Partition_Functions |
Pressure can also be derived from the canonical partition function. The average pressure is the sum of the probability times the pressure: \[ \langle P \rangle = \sum_i{P_i(N,V,T)P_i(N,V)} \nonumber \] The pressure of a macroscopic system is: \[ P(N,V) = -\left(\dfrac{\partial E_i}{\partial V}\right)_N \nonumber \] So we can write the average pressure as: \[\begin{split} \langle P \rangle &= \sum_i{P_i(N,V,\beta)\left(-\dfrac{\partial E_i}{\partial V}\right)_N} \\ &= \sum_i{\left(-\dfrac{\partial E_i}{\partial V}\right)_N\dfrac{e^{-E_i(N,V)/kT}}{Q(N,V,T)}} \end{split} \nonumber \] In a few steps we can show that the temperature can be expressed in terms of the partition function: \[ Q(N,V,T) = \sum_i{e^{-E_i(N,V)/kT}} \nonumber \] Writing in terms of \(\beta\) instead of temperature: \[ Q(N,V,\beta) = \sum_i{e^{-\beta E_i(N,V)}} \nonumber \] The derivative of the partition function with respect to volume is: \[ \left(\dfrac{\partial Q}{\partial V}\right)_{N,\beta} = -\beta \sum_i{\left(\dfrac{\partial E_i}{\partial V}\right)_{N}e^{-\beta E_i(N,V)}} \nonumber \] The average pressure can then be written as: \[ \langle P \rangle = \dfrac{kT}{Q(N,V,\beta}\left(\dfrac{\partial Q}{\partial V}\right)_{N,\beta} \nonumber \] Which shows that the pressure can be expressed solely terms of the partition function: \[ \langle P \rangle = kT\left(\dfrac{\partial \ln{Q}}{\partial V}\right)_{N,\beta} \nonumber \] We can use this result to derive the ideal gas law. For \(N\) particles of an ideal gas: \[ Q(N,V,\beta) = \dfrac{[q(V,\beta)]^N}{N!} \nonumber \] where: \[ q(V,\beta) = \left(\dfrac{2\pi m}{h^2\beta}\right)^{3/2}V \nonumber \] is the translational partition function. The utility of expressing the pressure as a logarithm is clear from the fact that we can write: \[\begin{split} \ln{Q} &= N\ln{q}-\ln{N!} \\ &= -\dfrac{3N}{2}\ln{\left(\dfrac{2\pi m}{h^2\beta}\right)}+N\ln{V}-\ln{N!} \end{split} \nonumber \] We have used the property of logarithms that \(\ln{(AB)} = \ln{(A)} + \ln{(B)}\) and \(\ln{(X^Y)} = Y\ln{(X)}\). Only one term in the ln \(Q\) depends on \(V\). Taking the derivative of \(N\ln{V}\) with respect to \(V\) gives: \[ \left(\dfrac{\partial \ln{Q}}{\partial V}\right)_{N,\beta} = \dfrac{N}{V} \nonumber \] Substituting this into the above equation for the pressure gives: \[ P = \dfrac{NkT}{V} \nonumber \] which is the ideal gas law. Recall that \(Nk = nR\) where \(N\) is the number of molecules and \(n\) is the number of moles. \(R\) is the universal gas constant (8.314 J/mol-K) which is nothing more than \(k\) multiplied by Avagadro’s number. \(N_Ak = R\) converts the constant from a "per molecule" to a "per mole" basis. Let us consider a simple thought experiment, which is illustrated in the figure above: A system of \(N\) particles is compressed by a piston pushing in the positive \(z\) direction. Since this is a classical thought experiment, we think in terms of forces. The piston exerts a constant force of magnitude \(F\) on the system. The direction of the force is purely in the positive \(z\) direction, so that we can write the force vector \(\bf{F}\) as \(\bf{F} = \begin{pmatrix} 0, 0, F \end{pmatrix}\). At equilibrium (the piston is not moving), the system exerts an equal and opposite force on the piston of the form \(\begin{pmatrix} 0, 0, -F \end{pmatrix}\). If the energy of the system is \(E\), then the force exerted by the system on the piston will be given by the negative change in \(E\) with respect to \(z\): \[-F = -\dfrac{dE}{dz} \label{Eq3.29} \] or: \[F = \dfrac{dE}{dz} \label{Eq3.30} \] The force exerted by the system on the piston is manifest as an observable pressure \(P\) equal to the force \(F\) divided by the area \(A\) of the piston, \(P=F/A\). Given this, the observed pressure is just: \[P = \dfrac{dE}{Adz} \label{Eq3.31} \] Since the volume decreases when the system is compressed, we see that \(Adz = -dV\). Hence, we can write the pressure as \(P = -dE/dV\). Of course, the relation \(P = -dE/dV\) is a thermodynamic one, but we need a function of \(x\) that we can average over the ensemble. The most natural choice is: \[p(x) = -\dfrac{d \mathcal{E} (x)}{dV} \label{Eq3.32} \] so that \(P = \langle p(x) \rangle\). Setting up the average, we obtain: \[\begin{align} P &= -\dfrac{C_N}{Q(N, V, T)} \int \dfrac{\partial \mathcal{E}}{\partial V} e^{-\beta \mathcal{E} (x)} \\ &= \dfrac{C_N}{Q(N, V, T)} \dfrac{1}{\beta} \int \dfrac{\partial}{\partial V} e^{-\beta \mathcal{E} (x)} \\ &= \dfrac{kT}{Q(N, V, T} \dfrac{\partial}{\partial V} C_N \int e^{-\beta \mathcal{E} (x)} \\ &= kT \left( \dfrac{\partial \: \text{ln} \: Q(N, V, T)}{\partial V} \right) \end{align} \label{Eq3.33} \] Recall that the mechanical energy for an ideal gas is: \[\mathcal{E} (x) = \sum_{i=1}^N \dfrac{\textbf{p}_i^2}{2m} \label{Eq3.36} \] where all particles are identical and have mass \(m\). Thus, the expression for the canonical partition function \(Q(N, V, T)\) is: \[Q(N, V, T) = \dfrac{1}{N!h^{3N}} \int dx \: e^{-\beta \sum_{i=1}^N \textbf{p}_i^2/2m} \nonumber \] Note that this can be expressed as: \[Q(N, V, T) = \dfrac{1}{N!h^{3N}}V^N \left[ \int dp \: e^{-\beta p^2/2m} \right]^{3N} \nonumber \] Evaluating the Gaussian integral gives us the final result immediately: \[Q(N, V, T) = \dfrac{1}{N!} \left[ \dfrac{V}{h^3} \left( \dfrac{2 \pi m}{\beta} \right)^{3/2} \right]^N \nonumber \] The expressions for the energy: \[E = -\dfrac{\partial}{\partial \beta} \: \text{ln} \: Q(N, V, T) \nonumber \] which gives: \[E = \dfrac{3}{2}NkT = \dfrac{3}{2}nRT \label{Eq3.37} \] and pressure: \[P = kT \left( \dfrac{\partial \: \text{ln} \: Q(N, V, T)}{\partial V} \right) \nonumber \] giving: \[P = \dfrac{NkT}{V} = \dfrac{nRT}{V} \label{Eq3.38} \] which is the ideal gas law. | 5,774 | 1,240 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Chemiluminescence/4%3A_Instrumentation/4.03%3A_Sequential_Injection_Analysis_(SIA)%3A_lab_on_a_valve |
SIA, like FIA, is based on reproducible sample handling and controlled dispersion of sample and reagents into a carrier stream. Unlike FIA, it makes use of a computer-controlled multiposition valve and pump, usually peristaltic and operated synchronously with the valve.
Morphine is solvent-extracted from opium poppies on an industrial scale. Barnett et al. have used SIA to determine the drug in aqueous and non-aqueous process streams, with chemiluminescence detection involving oxidation with acidified potassium permanganate in the presence of sodium hexametaphosphate. Figure D3.1 shows a suitable SIA manifold for carrying out this determination. The process streams contain several related alkaloids and a range of other organic compounds as well as both dissolved and suspended solids. It is a good indication of the effectiveness of SIA-chemiluminescence that in these conditions the results correlated well with high performance liquid chromatography, a standard methodology that suffers the defect a much lower sample throughput. In SIA, sample and reagents are aspirated into the holding coil by operating the pump in reverse so that carrier is returned to the reservoir. Restoration of forward pumping is synchronised with the opening of the valve port leading to the detector. The flow reversal leads to a mixing of the stack of sample and reagent zones to form a product zone which is transported to the detector. The pump tubing comes into contact only with the carrier, the samples and reagents being aspirated (instead of pumped) into the holding coil. This is a very useful characteristic of SIA when using samples /reagents that would attack PVC pump tubing, such as those containing non-aqueous solvents. | 1,748 | 1,241 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/04%3A_Reaction_Mechanisms/4.12%3A_Steady-State_Approximation |
When a reaction mechanism has several steps of comparable rates, the rate-determining step is often not obvious. However, there is an intermediate in some of the steps. An is a species that is neither one of the reactants, nor one of the products. The is a method used to derive a rate law. The method is based on the assumption that one intermediate in the reaction mechanism is consumed as quickly as it is generated. Its concentration remains the same in a duration of the reaction. An is a species that is neither one of the reactants, nor one of the products. It transiently exists during the course of the reaction. When a reaction involves one or more intermediates, the concentration of one of the intermediates remains constant at some stage of the reaction. Thus, the system has reached a . The concentration of one of the intermediates, \([Int]\), varies with as shown in Figure \(\Page {1}\). At the start and end of the reaction, [Int] does vary with time. \[ \dfrac{d[Int]}{dt}= 0 \nonumber \] When a reaction mechanism has several steps with comparable rates, the rate-determining step is not obvious. However, there is an intermediate in some of the steps. The steady-state approximation implies that you select an intermediate in the reaction mechanism, and calculate its concentration by assuming that it is consumed as quickly as it is generated. In the following, an example is given to show how the steady-state approximation method works. Use the steady-state approximation to derive the rate law for this reaction \[\ce{2 N2O5 \rightarrow 4 NO2 + O2}\nonumber \] assuming it follows the following three-step mechanism: \[\begin{align} \ce{N_2O_5} &\underset{\Large{k_{\textrm b}}}{\overset{\Large{k_{\textrm f}}}\rightleftharpoons} \ce{NO_2 + NO_3} \tag{step 1} \\[4pt] \ce{NO3 + NO2} &\ce{->[\large{k_2}] NO + NO2 + O2} \tag{step 2} \\[4pt] \ce{NO3 + NO} & \ce{->[\Large{k_3}] 2 NO2} \tag{step 3} \end{align} \] In these steps, \(\ce{NO}\) and \(\ce{NO3}\) are intermediates. You have \(\ce{production\: rate\: of\: NO} = k_2 \ce{[NO3] [NO2]}\)
\(\ce{consumption\: rate\: of\: NO} = k_3 \ce{[NO3] [NO]}\) A steady-state approach makes use of the assumption that the rate of production of an intermediate is equal to the rate of its consumption. Thus, we have \(k_2 \ce{[NO3] [NO2]} = k_3 \ce{[NO3] [NO]}\) and solving for \(\ce{[NO]}\) gives the result, \(\ce{[NO]} = \dfrac{k_2 \ce{[NO3] [NO2]}}{k_3 \ce{[NO3]}} \tag{1}\) For the other intermediate \(\ce{NO3}\), \(\ce{production\: rate\: of\: NO3} = k_{\ce f} \ce{[N2O5]}\)
\(\ce{consumption\: rate\: of\: NO3} = k_2\ce{[NO3] [NO2]} + k_3\ce{[NO3] [NO]} + k_{\ce b}\ce{[NO3] [NO2]}\) Applying the steady-state assumption gives: \(k_{\ce f} \ce{[N2O5]} = k_2\ce{[NO3] [NO2]} + k_3\ce{[NO3] [NO]} + k_{\ce b}\ce{[NO3] [NO2]}\) Thus, \(\ce{[NO3]} = \dfrac{k_{\ce f} \ce{[N2O5]}}{k_2\ce{[NO2]} + k_3\ce{[NO]} + k_{\ce b}\ce{[NO2]}}\tag{2}\) Let's review the three equations (steps) in the mechanism:
Step i. is at equilibrium and thus can not give a rate expression.
Step ii. leads to the production of some products, and the active species \(\ce{NO}\) causes further reaction in step iii. This consideration led to a rate expression from step ii. as: \(\ce{\dfrac{d[O2]}{dt}} = k_2 \ce{[NO3] [NO2]} \tag{3}\) Substituting (1) in (2) and then in (3) gives \(\ce{\dfrac{d[O2]}{dt}}= \dfrac{k_{\ce f} k_2 \ce{[N2O5]}}{k_{\ce b} + 2 k_2} = \ce{k [N2O5]}\) where \(\ce{k} = \dfrac{k_{\ce f} k_2}{k_{\ce b} + 2 k_2}\). This is the differential rate law, and it agrees with the experimental results. Carry out the above manipulation yourself on a piece of paper. Simply reading the above will not lead to solid learning yet. This page gives another example to illustrate the technique of deriving rate laws using the steady-state approximation. The reaction considered here is between \(\ce{H2}\) and \(\ce{I2}\) gases. For the reaction: \(\ce{H_{2\large{(g)}} + I_{2\large{(g)}} \rightarrow 2 HI_{\large{(g)}}}\) what mechanisms might be appropriate? Derive a rate law from the proposed mechanism. Well, this question does not have a simple answer, and there is no way to prove one over another for its validity. Beginning chemistry students will not be asked to propose a mechanism, but you will be asked to derive the rate law from the proposed mechanism. First of all, you should be able to express the rate of reaction in terms of the concentration changes, \(rate = - \ce{\dfrac{d[H2]}{dt}} = - \ce{\dfrac{ d[I2]}{dt}} = \ce{\dfrac{1}{2}\dfrac{d[HI]}{dt}}\) Look at the overall reaction equation again to see its relationship and the rate expressions. In order to propose a mechanism, we apply the following reasoning. Since the bonding between \(\ce{I-I}\) is weak, we expect \(\ce{I2}\) to dissociate into atoms or radicals. These radicals are active, and they react with \(\ce{H2}\) to produce the products. Thus we propose the three-step mechanism: Since only step iii. gives the real products, we expect you to recognize that step iii. hints the rate law to be: \(rate = k_3 \ce{[H2] [I]^2}\) However, this is not a proper rate law, because \(\ce{I}\) is an intermediate, not a reactant. So, you have to express \(\ce{[I]}\) or \(\ce{[I]^2}\) in terms of the concentration of reactants. To do this, we use the steady-state approximation and write out the following relationships: \(\textrm{rate of producing I} = 2 k_1 \ce{[I2]}\)
\(\textrm{rate of consuming I} = 2 k_2 \ce{[I]^2} + 2 k_3 \ce{[H2] [I]^2}\)
\(\textrm{producing rate of I} = \textrm{consuming rate of I}\) Thus, \(\ce{[I]^2} = \dfrac{k_1 \ce{[I2]}}{k_2 + k_3 \ce{[H2]}}\) Substituting this for \(\ce{[I]^2}\) into the expression, you have \(\begin{align}
rate &= k_3 \ce{[H2]} \dfrac{k_1 \ce{[I2]}}{k_2 + k_3 \ce{[H2]}}\\
&= \dfrac{k_1 k_3 \ce{[H2] [I2]}}{k_2 + k_3 \ce{[H2]}}
\end{align}\) \(\ce{rate} = \ce{k [H2] [I2]}\), where \(\ce{k} = \dfrac{k_1 k_3}{k_2}\). Since the rate law is first order with respect to both reactants, one may argue that the rate law also supports a one-step mechanism, \(\ce{H_{2\large{(g)}} + I_{2\large{(g)}} \rightarrow 2 HI}\) This elementary step is the same as the overall reaction. Suppose we use a large quantity of \(\ce{H2}\) compared to \(\ce{I2}\), then the change in \(\ce{[H2]}\) is insignificant. For example, if \(\mathrm{[H_2] = 10}\), and \(\mathrm{[I_2] = 0.1}\) initially, \(\ce{[H2]}\) remains essentially 10 (9.9 with only one significant figure). In other words, \(\ce{[H2]}\) hardly changed when the reaction ended. Thus, \(k_3 \ce{[H2]} >> k_3\) and the rate law becomes: \(rate = k_1 \ce{[I2]}\). Thus, the reaction is a , due to the large quantity of one reactant. The results suggest iii. a fast step (due to large quantity of \(\ce{H2}\)), and i. the rate determining step. | 6,818 | 1,242 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Lewis_Theory_of_Bonding/Violations_of_the_Octet_Rule |
Three cases can be constructed that do not follow the , and as such, they are known as the exceptions to the Octet Rule. Following the Octet Rule for leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. However, it is hard to imagine that one rule could be followed by all molecules. There is always an exception, and in this case, three exceptions. The Octet Rule is violated in these three scenarios: The first exception to the Octet Rule is when there are an odd number of valence electrons. An example of this would be the nitrogen (II) oxide molecule (\(NO\)). Nitrogen atom has 5 valence electrons while the oxygen atom has 6 electrons. The total would be 11 valence electrons to be used. The Octet Rule for this molecule is fulfilled in the above example, however that is with 10 valence electrons. The last one does not know where to go. The lone electron is called an unpaired electron. But where should the unpaired electron go? The unpaired electron is usually placed in the Lewis Dot Structure so that each element in the structure will have the possible. The formal charge is . The formula to find a formal charge is: No formal charge at all is the most ideal situation. An example of a stable molecule with an odd number of valence electrons would be nitrogen monoxide. Nitrogen monoxide has 11 valence electrons (Figure 1). If you need more information about formal charges, see Lewis Structures. If we were to consider the nitrogen monoxide cation (\(NO^+\) with ten valence electrons, then the following Lewis structure would be constructed: Nitrogen normally has five valence electrons. In Figure 1, it has two lone pair electrons and it participates in two bonds (a double bond) with oxygen. This results in nitrogen having a formal charge of +1. Oxygen normally has six valence electrons. In Figure 1, oxygen has four lone pair electrons and it participates in two bonds with nitrogen. Oxygen therefore has a formal charge of 0. The overall molecule here has a formal charge of +1 (+1 for nitrogen, 0 for oxygen. +1 + 0 = +1). However, if we add the eleventh electron to nitrogen (because we want the molecule to have the total formal charge), it will bring both the nitrogen and the molecule's overall charges to zero, the most ideal formal charge situation. That is exactly what is done to get the correct Lewis structure for nitrogen monoxide: There are actually very few stable molecules with odd numbers of electrons that exist, since that unpaired electron is willing to react with other unpaired electrons. Most odd electron species are highly reactive, which we call . Because of their instability, free radicals bond to atoms in which they can take an electron from in order to become stable, making them very chemically reactive. Radicals are found as both reactants and products, but generally react to form more stable molecules as soon as they can. To emphasize the existence of the unpaired electron, radicals are denoted with a dot in front of their chemical symbol as with \({\cdot}OH\), the hydroxyl radical. An example of a radical you may by familiar with already is the gaseous chlorine atom, denoted \({\cdot}Cl\). Interestingly, molecules with an odd number of Valence electrons will be paramagnetic. The second exception to the Octet Rule is when there are too few valence electrons that results in an incomplete Octet. There are even more occasions where the octet rule does not give the most correct depiction of a molecule or ion. This is also the case with incomplete octets. Species with incomplete octets are pretty rare and generally are only found in some beryllium, aluminum, and boron compounds including the boron hydrides. Let's take a look at one such hydride, \(BH_3\) (Borane). If one was to make a Lewis structure for \(BH_3\) following the basic strategies for drawing Lewis structures, one would probably come up with this structure (Figure 3): The problem with this structure is that boron has an incomplete octet; it only has around it. Hydrogen atoms can naturally only have only 2 electrons in their outermost shell (their version of an octet), and as such there are no spare electrons to form a double bond with boron. One might surmise that the failure of this structure to form complete octets must mean that this bond should be ionic instead of covalent. However, boron has an electronegativity that is very similar to hydrogen, meaning there is likely very little ionic character in the hydrogen to boron bonds, and as such this Lewis structure, though it does not fulfill the octet rule, is likely the best structure possible for depicting BH with Lewis theory. One of the things that may account for BH 's incomplete octet is that it is commonly a transitory species, formed temporarily in reactions that involve multiple steps. Let's take a look at another incomplete octet situation dealing with boron, BF (Boron trifluorine). Like with BH , the initial drawing of a Lewis structure of BF will form a structure where boron has only six electrons around it (Figure 4). If you look Figure 4, you can see that the fluorine atoms possess extra lone pairs that they can use to make additional bonds with boron, and you might think that all you have to do is make one lone pair into a bond and the structure will be correct. If we add one double bond between boron and one of the fluorines we get the following Lewis Structure (Figure 5): Each fluorine has eight electrons, and the boron atom has eight as well! Each atom has a perfect octet, right? Not so fast. We must examine the formal charges of this structure. The fluorine that shares a double bond with boron has six electrons around it (four from its two lone pairs of electrons and one each from its two bonds with boron). This is one less electron than the number of valence electrons it would have naturally (Group seven elements have seven valence electrons), so it has a formal charge of +1. The two flourines that share single bonds with boron have seven electrons around them (six from their three lone pairs and one from their single bonds with boron). This is the same amount as the number of valence electrons they would have on their own, so they both have a formal charge of zero. Finally, boron has four electrons around it (one from each of its four bonds shared with fluorine). This is one more electron than the number of valence electrons that boron would have on its own, and as such boron has a formal charge of -1. This structure is supported by the fact that the experimentally determined bond length of the boron to fluorine bonds in BF is less than what would be typical for a single bond (see However, this structure contradicts one of the major rules of formal charges: Negative formal charges are supposed to be found on the more electronegative atom(s) in a bond, but in the structure depicted in Figure 5, a formal charge is found on fluorine, which not only is the most electronegative element in the structure, but the most electronegative element in the entire periodic table (\(\chi=4.0\)). Boron on the other hand, with the much lower electronegativity of 2.0, has the negative formal charge in this structure. This formal charge-electronegativity disagreement makes this double-bonded structure impossible. However the large electronegativity difference here, as opposed to in BH , signifies significant polar bonds between boron and fluorine, which means there is a high ionic character to this molecule. This suggests the possibility of a semi-ionic structure such as seen in Figure 6: None of these three structures is the "correct" structure in this instance. The most "correct" structure is most likely a of all three structures: the one with the incomplete octet (Figure 4), the one with the double bond (Figure 5), and the one with the ionic bond (Figure 6). The most contributing structure is probably the incomplete octet structure (due to Figure 5 being basically impossible and Figure 6 not matching up with the behavior and properties of BF ). As you can see even when other possibilities exist, incomplete octets may best portray a molecular structure. As a side note, it is important to note that BF frequently bonds with a F ion in order to form BF rather than staying as BF . This structure completes boron's octet and it is more common in nature. This exemplifies the fact that incomplete octets are rare, and other configurations are typically more favorable, including bonding with additional ions as in the case of BF . Draw the Lewis structure for boron trifluoride (BF ). 1. Add electrons (3*7) + 3 = 2. Draw connectivities: 3. Add octets to outer atoms: 4. Add extra electrons (24-24=0) to central atom: 5. Does central electron have octet? 6. The central Boron now has an octet (there would be three resonance Lewis structures) More common than incomplete octets are expanded octets where the central atom in a Lewis structure has more than eight electrons in its valence shell. In expanded octets, the central atom can have ten electrons, or even twelve. , which those terminal atoms bond to. For example, \(PCl_5\) is a legitimate compound (whereas \(NCl_5\)) is not: Expanded valence shells are observed for elements in period 3 (i.e. n=3) and beyond The 'octet' rule is based upon available n and n orbitals for valence electrons (2 electrons in the orbitals, and 6 in the orbitals). Beginning with the n=3 principle quantum number, the d orbitals become available ( =2). The orbital diagram for the valence shell of phosphorous is: Hence, the third period elements occasionally exceed the octet rule by using their empty d orbitals to accommodate additional electrons. Size is also an important consideration: There is currently much scientific exploration and inquiry into the reason why expanded valence shells are found. The top area of interest is figuring out where the extra pair(s) of electrons are found. Many chemists think that there is not a very large energy difference between the 3p and 3d orbitals, and as such it is plausible for extra electrons to easily fill the 3d orbital when an expanded octet is more favorable than having a complete octet. This matter is still under hot debate, however and there is even debate as to what makes an expanded octet more favorable than a configuration that follows the octet rule. One of the situations where expanded octet structures are treated as more favorable than Lewis structures that follow the octet rule is when the formal charges in the expanded octet structure are smaller than in a structure that adheres to the octet rule, or when there are less formal charges in the expanded octet than in the structure a structure that adheres to the octet rule. The sulfate ion, SO . is an ion that prefers an expanded octet structure. A strict adherence to the octet rule forms the following Lewis structure: If we look at the formal charges on this molecule, we can see that all of the oxygen atoms have seven electrons around them (six from the three lone pairs and one from the bond with sulfur). This is one more electron than the number of valence electrons then they would have normally, and as such each of the oxygen atoms in this structure has a formal charge of -1. has four electrons around it in this structure (one from each of its four bonds) which is two electrons fewer than the number of valence electrons it would have normally, and as such it carries a formal charge of +2. If instead we made a structure for the sulfate ion with an expanded octet, it would look like this: Looking at the formal charges for this structure, the sulfur ion has six electrons around it (one from each of its bonds). This is the same amount as the number of valence electrons it would have naturally. This leaves sulfur with a formal charge of zero. The two oxygens that have double bonds to sulfur have six electrons each around them (four from the two lone pairs and one each from the two bonds with sulfur). This is the same amount of electrons as the number of valence electrons that oxygen atoms have on their own, and as such both of these oxygen atoms have a formal charge of zero. The two oxygens with the single bonds to sulfur have seven electrons around them in this structure (six from the three lone pairs and one from the bond to sulfur). That is one electron more than the number of valence electrons that oxygen would have on its own, and as such those two oxygens carry a formal charge of -1. Remember that with formal charges, the goal is to keep the formal charges (or the difference between the formal charges of each atom) as small as possible. The number of and values of the formal charges on this structure (-1 and 0 (difference of 1) in Figure 12, as opposed to +2 and -1 (difference of 3) in Figure 12) is significantly lower than on the structure that follows the octet rule, and as such an expanded octet is plausible, and even preferred to a normal octet, in this case. Draw the Lewis structure for \(ICl_4^-\) ion. 1. Count up the valence electrons: 7+(4*7)+1 = electrons 2. Draw the connectivities: 3. Add octet of electrons to outer atoms: 4. Add extra electrons (36-32= ) to central atom: 5. The ICl ion thus has 12 valence electrons around the central Iodine (in the 5 orbitals) Expanded Lewis structures are also plausible depictions of molecules when experimentally determined bond lengths suggest partial double bond characters even when single bonds would already fully fill the octet of the central atom. Despite the cases for expanded octets, as mentioned for incomplete octets, it is important to keep in mind that, in general, the octet rule applies. or 1. 2. 3. 10 (Sodium and higher) 4. 5. ( ) | 13,865 | 1,243 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Mass_Spectrometry/Introductory_Mass_Spectrometry/High_Resolution_vs_Low_Resolution |
Two common categories of mass spectrometry are high resolution mass spectrometry (HRMS) and low resolution mass spectrometry (LRMS). Not all mass spectrometers simply measure molecular weights as whole numbers. High resolution mass spectrometers can measure mass so accurately that they can detect the minute differences in mass between two compounds that, on a regular low-resolution instrument, would appear to be identical. The reason is because atomic masses are not exact multiples of the mass of a proton, as we might usually think. As a result, on a high resolution mass spectrometer, 2-octanone, C H O, has a molecular weight of 128.12018 instead of 128. Naphthalene, C H , has a molecular weight of 128.06264. Thus a high resolution mass spectrometer can supply an exact molecular formula for a compound because of the unique combination of masses that result. HRMS relies on the fact that the mass of an individual atom does not correspond to an integral number of atomic mass units. Calculate the high-resolution molecular weights for the following formulae. , | 1,084 | 1,246 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/12%3A_Cycloalkanes_Cycloalkenes_and_Cycloalkynes/12.02%3A_Spectroscopic_Properties_of_Cyclohexanes |
The spectroscopic properties of cycloalkanes are considerably similar to those of alkanes. We mentioned previously the main features of their infrared spectra (Section 9-7D), and that their lack of ultraviolet absorption at wavelengths down to 200 nm makes them useful solvents for the determination of ultraviolet spectra of other substances (Section 9-9B). Some spectoscopic properties of cycloalkanes has been notable in the proton NMR spectra. The proton NMR spectra of alkanes and cycloalkanes are characteristic but difficult to interpret because the chemical shifts between the various kinds of protons are usually small. Although proton spectra of simple cycloalkanes, \(\ce{(CH_2)}_n\), show one sharp line at room temperature, when alkyl substituents are present, differences in chemical shifts between the ring hydrogens occur and, with spin-spin splitting, provide more closely spaced lines than normally can be resolved. The complexity so introduced can be seen by comparing the proton spectra of cyclooctane and methylcyclohexane shown in Figure 12-1. For methyl-substituted cycloalkanes the methyl resonances generally stand out as high-field signals centered on \(0.9 \: \text{ppm}\), and the area of these signals relative to the other \(\ce{C-H}\) signals may be useful in indicating how many methyl groups there are (see Section 9-10K, especially Figure 9-46). However, in cyclopropanes the ring protons have abnormally small chemical shifts (\(\delta = 0.22\) for cyclopropane), which often overlap with the shifts of methyl groups \(\left( \delta \cong 0.9 \: \text{ppm} \right)\). Although spectroscopic properties of cycloalkanes in proton spectra are not very useful for identification purposes, \(\ce{^{13}C}\) NMR spectra are very useful. Chain-branching and ring-substitution normally cause quite large chemical-shift changes, and it is not uncommon to observe \(\ce{^{13}C}\) shifts in cycloalkanes spanning \(35 \: \text{ppm}\). Some special features of application of \(\ce{^{13}C}\) NMR spectra to conformational analysis of cycloalkanes are described in Section 12-3D. and (1977) | 2,132 | 1,247 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/02%3A_Gas_Laws/2.07%3A_The_Ideal_Gas_Constant_and_Boltzmann's_Constant |
Having developed the ideal gas equation and analyzed experimental results for a variety of gases, we will have found the value of . It is useful to have expressed using a number of different energy units. Frequently useful values are \[ \begin{aligned} R & = 8.314 \text{ Pa m}^{3} \text{ K}^{-1} \text{ mol}^{-1} \\ ~ & = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \\ ~ & = 0.08314 \text{ L bar K}^{-1} \text{ mol}^{-1} \\ ~ & = 1.987 \text{ cal K}^{-1} \text{ mol}^{-1} \\ ~ & = 0.08205 \text{ L atm K}^{-1} \text{ mol}^{-1} \end{aligned}\] We also need the gas constant expressed per molecule rather than per mole. Since there is Avogadro’s number of molecules per mole, we can divide any of the values above by \(\overline{N}\) to get \(R\) on a per-molecule basis. Traditionally, however, this constant is given a different name; it is , usually given the symbol \(k\). \[k={R}/{\overline{N}}=1.381\times {10}^{-23}\ \mathrm{J}\ {\mathrm{K}}^{-1}\ {\mathrm{molecule}}^{-1}\] This means that we can also write the ideal gas equation as \(PV=nRT=n\overline{N}kT\). Because the number of molecules in the sample, \(N\), is \(N=n\overline{N}\), we have \[PV=NkT.\] | 1,178 | 1,248 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Fluorescence_and_Phosphorescence |
Fluorescence and phosphorescence are types of molecular luminescence methods. A molecule of analyte absorbs a photon and excites a species. The emission spectrum can provide qualitative and quantitative analysis. The term fluorescence and phosphorescence are usually referred as photoluminescence because both are alike in excitation brought by absorption of a photon. Fluorescence differs from phosphorescence in that the electronic energy transition that is responsible for fluorescence does not change in electron spin, which results in short-live electrons (<10 s) in the excited state of fluorescence. In phosphorescence, there is a change in electron spin, which results in a longer lifetime of the excited state (second to minutes). Fluorescence and phosphorescence occurs at longer wavelength than the excitation radiation. Fluorescence can occur in gaseous, liquid, and solid chemical systems. The simple kind of fluorescence is by dilute atomic vapors. A fluorescence example would be if a 3s electron of a vaporized sodium atom is excited to the 3p state by absorption of a radiation at wavelength 589.6 and 589.0 nm. After 10 s, the electron returns to ground state and on its return it emits radiation of the two wavelengths in all directions. This type of fluorescence in which the absorbed radiation is remitted without a change in frequency is known as resonance fluorescence. Resonance fluorescence can also occur in molecular species. Molecular fluorescence band centers at wavelengths longer than resonance lines. The shift toward longer wavelength is referred to as the Stokes Shift. Understanding the difference between fluorescence and phosphorescence requires the knowledge of electron spin and the differences between singlet and triplet states. states that two electrons in an atom cannot have the same four quantum numbers (\(n\), \(l\), \(m_l\), \(m_s\)) and only two electrons can occupy each orbital where they must have opposite spin states. These opposite spin states are called spin pairing. Because of this spin pairing, most molecules do not exhibit a magnetic field and are diamagnetic. In diamagnetic molecules, electrons are not attracted or repelled by the static electric field. Free radicals are paramagnetic because they contain unpaired electrons have magnetic moments that are attracted to the magnetic field. Singlet state is defined when all the electron spins are paired in the molecular electronic state and the electronic energy levels do not split when the molecule is exposed into a magnetic field. A doublet state occurs when there is an unpaired electron that gives two possible orientations when exposed in a magnetic field and imparts different energy to the system. A singlet or a triplet can form when one electron is excited to a higher energy level. In an excited singlet state, the electron is promoted in the same spin orientation as it was in the ground state (paired). In a triplet excited stated, the electron that is promoted has the same spin orientation (parallel) to the other unpaired electron. The difference between the spins of ground singlet, excited singlet, and excited triplet is shown in Figure \(\Page {1}\). Singlet, doublet and triplet is derived using the equation for multiplicity, 2S+1, where S is the total spin angular momentum (sum of all the electron spins). Individual spins are denoted as spin up (s = +1/2) or spin down (s = -1/2). If we were to calculated the S for the excited singlet state, the equation would be 2(+1/2 + -1/2)+1 = 2(0)+1 = 1, therefore making the center orbital in the figure a singlet state. If the spin multiplicity for the excited triplet state was calculated, we obtain 2(+1/2 + +1/2)+1 = 2(1)+1 =3, which gives a triplet state as expected. The difference between a molecule in the ground and excited state is that the electrons is diamagnetic in the ground state and paramagnetic in the triplet state.This difference in spin state makes the transition from singlet to triplet (or triplet to singlet) more improbable than the singlet-to-singlet transitions. This singlet to triplet (or reverse) transition involves a change in state. For this reason, the lifetime of the triplet state is longer the singlet state by approximately 10 seconds fold difference.The radiation that induced the transition from ground to excited triplet state has a low probability of occurring, thus their absorption bands are less intense than singlet-singlet state absorption. The excited triplet state can be populated from the excited singlet state of certain molecules which results in phosphorescence. These spin multiplicities in ground and excited states can be used to explain transition in photoluminescence molecules by the Jablonski diagram. The Jablonski diagram that drawn below is a partial energy diagram that represents the energy of photoluminescent molecule in its different energy states. The lowest and darkest horizontal line represents the ground-state electronic energy of the molecule which is the singlet state labeled as \(S_o\). At room temperature, majority of the molecules in a solution are in this state. The upper lines represent the energy state of the three excited electronic states: S and S represent the electronic singlet state (left) and T represents the first electronic triplet state (right). The upper darkest line represents the ground vibrational state of the three excited electronic state.The energy of the triplet state is lower than the energy of the corresponding singlet state. There are numerous vibrational levels that can be associated with each electronic state as denoted by the thinner lines. Absorption transitions (blues lines in Figure \(\Page {2}\)) can occur from the ground singlet electronic state (S ) to various vibrational levels in the singlet excited vibrational states. It is unlikely that a transition from the ground singlet electronic state to the triplet electronic state because the electron spin is parallel to the spin in its ground state (Figure \(\Page {1}\)). This transition leads to a change in multiplicity and thus has a low probability of occurring which is a Molecules also go through vibration relaxation to lose any excess vibrational energy that remains when excited to the electronic states (\(S_1\) and \(S_2\)) as demonstrated in wavy lines in Figure \(\Page {2}\). The knowledge of forbidden transition is used to explain and compare the peaks of absorption and emission. The table below compares the absorption and emission rates of fluorescence and phosphorescence.The rate of photon absorption is very rapid. Fluorescence emission occurs at a slower rate.Since the triplet to singlet (or reverse) is a forbidden transition, meaning it is less likely to occur than the singlet-to-singlet transition, the rate of triplet to singlet is typically slower. Therefore, phosphorescence emission requires more time than fluorescence. A molecule that is excited can return to the ground state by several combinations of mechanical steps that will be described below and shown in Figure \(\Page {2}\).The deactivation process of fluorescence and phosphorescence involve an emission of a photon radiation as shown by the straight arrow in Figure \(\Page {2}\). The wiggly arrows in Figure \(\Page {2}\) are deactivation processes without the use of radiation. The favored deactivation process is the route that is most rapid and spends less time in the excited state.If the rate constant for fluorescence is more favorable in the radiationless path, the fluorescence will be less intense or absent. After discussing all the possible deactivation processes, variable that affect the emissions to occur. Molecular structure and its chemical environment influence whether a substance will fluoresce and the intensities of these emissions. The or is used to measure the probability that a molecule will fluoresce or phosphoresce. For fluorescence and phosphorescence is the ratio of the number of molecules that luminescent to the total number of excited molecules. For highly fluoresce molecules, the quantum efficiency approaches to one.Molecules that do not fluoresce have quantum efficiencies that approach to zero. Fluorescence quantum yield (\( \phi \)) for a compound is determined by the relative rate constants (k) of various deactivation processes by which the lowest excited singlet state is deactivated to the ground state. The deactivation processes including fluorescence (kf), intersystem crossing (\(k_i\)), internal conversion (kic), predissociation (kpd), dissociation (kd), and external conversion (kec) allows one to qualitatively interpret the structural and environmental factors that influence the intensity of the fluorescence. They are related by the quantum yield equation given below: \[ \dfrac{k_f}{k_f+k_i+k_{ec}+k_{ic}+k_{pd}+k_d} \] Using this equation as an example to explain fluorescence, a high fluorescence rate (k ) value and low values of the all the other relative rate constant terms (k +k +k +k +k +k ) will give a large \( \phi \), which suggest that fluorescence is enhanced. The magnitudes of kf , kd, and kpd depend on the chemical structure, while the rest of the constants ki, kec, and kic are strongly influenced by the environment. Fluorescence rarely results from absorption of ultraviolet radiation of wavelength shorter than 250 nm because radiation at this wavelength has sufficient energy to deactivate the electron in the excited state by predissociation or dissociation. The bond of some organic molecules would rupture at 140 kcal/mol, which corresponds to 200-nm of radiation. For this reason, \(\sigma \rightarrow \sigma^{*}\) transition in fluorescence are rarely observed. Instead, emissions from the less energetic transition will occur which are either \(\pi^{*} \rightarrow \pi \) or \(\pi^{*} \rightarrow n \) transition. Molecules that are excited electronically will return to the lowest excited state by rapid vibrational relaxation and internal conversion, which produces no radiation emission. Fluorescence arises from a transition from the lowest vibrational level of the first excited electronic state to one of the vibrational levels in the electronic ground state. In most fluorescent compounds, radiation is produced by a \(\pi^{*} \rightarrow \pi \) or \(\pi^{*} \rightarrow n \) transition depending on which requires the least energy for the transition to occur. Fluorescence is most commonly found in compounds in which the lowest energy transition is \(\pi \rightarrow \pi^{*} \) (excited singlet state) than \(n \rightarrow \pi^{*} \) which suggest that the quantum efficiency is greater for \(\pi \rightarrow \pi^{*} \) transitions. The reason for this is that the molar absorptivity, which measures the probability that a transition will occur, of the \(\pi \rightarrow \pi^{*} \) transition is 100 to 1000 fold greater than \(n \rightarrow \pi^{*} \) process. The lifetime of \(\pi \rightarrow \pi^{*} \) (10 to 10 s) is shorter than the lifetime of \(n \rightarrow \pi^{*} \) (10 to 10 ). Phosphorescent quantum efficiency is the opposite of fluorescence in that it occurs in the \(n \rightarrow \pi^{*} \) excited state which tends to be short lived and less suceptable to deactivation than the \(\pi \rightarrow \pi^{*} \) triplet state. Intersystem crossing is also more probable for \(\pi \rightarrow \pi^{*} \) excited state than for the \(n \rightarrow \pi^{*} \)state because the energy difference between the singlet and triplet state is large and spin-orbit coupling is less likely to occur. The most intense fluorescence is found in compounds containing aromatic group with low-energy \(\pi \rightarrow \pi^{*} \) transitions. A few aliphatic, alicyclic carbonyl, and highly conjugated double-bond structures also exhibit fluorescence as well. Most unsubstituted aromatic hydrocarbons fluoresce in solution too. The quantum efficiency increases as the number of rings and the degree of condensation increases. Simple heterocycles such as the structures listed below do not exhibit fluorescence. Pyridine Pyrrole Furan Thiophene With nitrogen heterocyclics, the lowest energy transitions is involved in \(n \rightarrow \pi^{*} \) system that rapidly converts to the triplet state and prevents fluorescence. Although simple heterocyclics do not fluoresce, fused-ring structures do. For instance, a fusion of a benzene ring to a hetercyclic structure results in an increase in molar absorptivity of the absorption band. The lifetime of the excited state in fused structure and fluorescence is observed. Examples of fluorescent compounds is shown below. quinoline Benzene ring substitution causes a shift in the absorption maxima of the wavelength and changes in fluorescence emission. The table below is used to demonstrate and visually show that as benzene is substituted with increasing methyl addition, the relative intensity of fluorescence increases. Compound Structure Wavelength of Fluorescence (nm) Relative intensity of Fluorescence Benzene 270-310 10 Toluene 270-320 17 Propyl Benzene 270-320 17 The relative intensity of fluorescence increases as oxygenated species increases in substitution. The values for such increase is demonstrated in the table below. Compound Structure Wavelength of Fluorescence (nm) Relative intensity of Fluorescence Phenol 285-365 18 Phenolate ion 310-400 10 Anisole 285-345 20 Influence of a halogen substitution decreases fluorescence as the molar mass of the halogen increases. This is an example of the “heavy atom effect” which suggest that the probability of intersystem crossing increases as the size of the molecule increases. As demonstrated in the table below, as the molar mass of the substituted compound increases, the relative intensity of the fluorescence decreases. Compound Structure Wavelength of Fluorescence (nm) Relative intensity of Fluorescence Fluorobenzene 270-320 10 Chlorobenzene 275-345 7 Bromobenzene 290-380 5 In heavy atom substitution such as nitro derivatives or heavy halogen substitution such as iodobenzene, the compounds are subject to predissociation. These compounds have bonds that easily rupture that can then absorb excitation energy and go through internal conversion. Therefore, the relative intensity of fluorescence and fluorescent wavelength is not observed and this is demonstrated in the table below. Compound Structure Wavelength of Fluorescence (nm) Relative intensity of Fluorescence Iodobenzene None 0 Anilinium ion None 0 Nitrobenzene None 0 Carboxylic acid or carbonyl group on aromatic ring generally inhibits fluorescence since the energy of the \(n \rightarrow \pi^*\) transition is less than \(\pi \rightarrow \pi^*\) transition. Therefore, the fluorescence yield from \(n \rightarrow \pi^*\) transition is low. Compound Structure Wavelength of Fluorescence (nm) Relative intensity of Fluorescence Benzoic Acid 310-390 3 Fluorescence is particularly favored in molecules with rigid structures. The table below compares the quantum efficiencies of fluorine and biphenyl which are both similar in structure that there is a bond between the two benzene group. The difference is that fluorene is more rigid from the addition methylene bridging group. By looking at the table below, rigid fluorene has a higher quantum efficiency than unrigid biphenyl which indicates that fluorescence is favored in rigid molecules. Compound Structure Quantum Efficiency Fluorene 1.0 Biphenyl 0.2 This concept of rigidity was used to explain the increase in fluorescence of organic chelating agent when the compound is complexed with a metal ion. The fluorescence intensity of 8-hydroxyquinoline is much less than its zinc complex. vs 8-hydroxyquinoline 8-hydroxyquinoline with Zinc complexed The explanation for lower quantum efficiency or lack of rigidity in caused by the enhanced internal conversion rate (k ) which increases the probability that there will be radiationless deactivation. Nonrigid molecules can also undergo low-frequency vibration which accounts for small energy loss. Quantum efficiency of Fluorescence decreases with increasing temperature. As the temperature increases, the frequency of the collision increases which increases the probability of deactivation by external conversion. Solvents with lower viscosity have higher possibility of deactivation by external conversion. Fluorescence of a molecule decreases when its solvent contains heavy atoms such as carbon tetrabromide and ethyl iodide, or when heavy atoms are substituted into the fluorescing compound. Orbital spin interaction result from an increase in the rate of triplet formation, which decreases the possibility of fluorescence. Heavy atoms are usually incorporated into solvent to enhance phosphorescence. The fluorescence of aromatic compound with basic or acid substituent rings are usually pH dependent. The wavelength and emission intensity is different for protonated and unprotonated forms of the compound as illustrated in the table below: Compound Structure Wavelength of Fluorescence (nm) Relative intensity of Fluorescence aniline 310-405 20 Anilinium ion None 0 The emission changes of this compound arises from different number of resonance structures associated with the acidic and basic forms of the molecule.The additional resonance forms provides a more stable first excited state, thus leading to fluorescence in the ultraviolet region.The resonance structures of basic aniline and acidic anilinium ion is shown below: basic Aniline Fluorescence of certain compounds have been used a detection of end points in acid-base titrations.An example of this type of fluorescence seen in compound as a function of pH is the phenolic form of 1-naphthol-4-sulfonic acid.This compound is not detectable with the eye because it occurs in the ultraviolet region, but with an addition of a base, it becomes converted to a phenolate ion, the emission band shifts to the visible wavelength where it can be visually seen. Acid dissociation constant for excited molecules differs for the same species in the ground state.These changes in acid or base dissociation constant differ in four or five orders of magnitude. Dissolved oxygen reduces the intensity of fluorescence in solution, which results from a photochemically induced oxidation of fluorescing species.Quenching takes place from the paramagnetic properties of molecular oxygen that promotes intersystem crossing and conversion of excited molecules to triplet state.Paramagnetic properties tend to quench fluorescence. The power of fluorescence emission \(F\) is proportional to the radiant power is proportional to the radiant power of the excitation beam that is absorbed by the system. The equation below best describes this relationship. (1) Since \(\phi_f K”\) is constant in the system, it is represented at K’. The table below defines the variables in this equation. Variable Definition F Power of fluorescence emission P0 Power of incident beam on solution P Power after transversing length b in medium K” Constant dependent on geometry and other factors f Quantum efficiency Fluorescence emission (\(F\)) can be related to concentration (\(c\)) using Beer’s Law stating: \[ F = \epsilon b c \label{2}\] where \( \epsilon \) is the molar absorptivity of the molecule that is fluorescing. Rewriting Equation 2 gives: \[ P =P_o 10^{-\epsilon b c} \label{3}\] Plugging this Equation \(\ref{3}\) into Equation \(\ref{1}\) and factoring out \(P_ gives us this equation: \[F=K^{\prime} P_{0}\left(1-10^{-\varepsilon b c}\right)\] The MacLaurin series could be used to solved the exponential term. \[F=K^{\prime} P_{0}\left[2.303 \varepsilon b c-\frac{(2.303 \varepsilon b c)^{2}}{2 !}+\frac{(2.303 \varepsilon b c)^{3}}{3 !}+\frac{(2.303 \varepsilon b c)^{4}}{4 !}+\ldots \frac{(2.303 \varepsilon b c)^{n}}{n !}\right]_{1}\] Given that \((2.303 \epsilon b c = \text{Absorbance} <0.05\), all the subsequent terms after the first can be dropped since the maximum error is 0.13%. Using only the first term, Equation \(\ref{5}\) can be rewritten as: \[F=K^{\prime} P_{0} 2.303 \varepsilon b c\] Equation \(\ref{6}\) can be expanded to the equation below and simplified to compare the fluorescence emission F with concentration. If the equation below were to be plotted with F versus c, a linear relation would be observed. \[F=\phi_{f} K^{\prime \prime} P_{0} 2.303 \varepsilon b c\] If \(c\) becomes so great that the absorbance > 0.05, the higher terms start to become taken into account and the linearity is lost. F then lies below the extrapolation of the straight-line plot. This excessive absorption is the Another cause of this negative downfall of linearity is the when the wavelength of emission overlaps the absorption band. This occurs when the emission transverse the solution and gets reabsorbed by other molecules by analyte or other species in the solution, which leads to a decrease in fluorescence. Dynamic Quenching is a nonradiative energy transfer between the excited and the quenching agent species (Q).The requirements for a successful dynamic quenching are that the two collision species the concentration must be high so that there is a higher possibility of collision between the two species.Temperature and quenching agent viscosity play a role on the rate of dynamic quenching.Dynamic quenching reduces fluorescence quantum yield and the fluorescence lifetime. Dissolved oxygen in a solution increases the intensity of the fluorescence by photochemically inducing oxidation of the fluorescing species.Quenching results from the paramagnetic properties of molecular oxygen that promotes intersystem crossing and converts the excited molecules to triplet state.Paramagnetic species and dissolved oxygen tend to quench fluorescence and quench the triplet state. Static quenching occurs when the quencher and ground state fluorophore forms a dark complex.Fluorescence is usually observed from unbound fluorophore.Static quenching can be differentiated from dynamic quenching in that the lifetime is not affected in static quenching.In long range (Förster) quenching, energy transfer occurs without collision between molecules, but dipole-dipole coupling occurs between excited fluorophore and quencher. One of the ways to visually distinguish the difference between each photoluminescence is to compare the relative intensities of emission/excitation at each wavelength. An example of the three types of photoluminescence (absorption, fluorescence and phosphorescence) is shown for phenanthrene in the spectrum below.In the spectrum, the luminescent intensity is measure in a wavelength is fixed while the excitation wavelength is varied. The spectrum in red represents the excitation spectrum, which is identical to the absorption spectrum because in order for fluorescence emission to occur, radiation needs to be absorbed to create an excited state.The spectrum in blue represent fluorescence and green spectrum represents the phosphorescence. Fluorescence and Phosphorescence occur at wavelengths that are longer than their absorption wavelengths.Phosphorescence bands are found at a longer wavelength than fluorescence band because the excited triplet state is lower in energy than the singlet state.The difference in wavelength could also be used to measure the energy difference between the singlet and triplet state of the molecule. As the wavelength increases, the energy of the molecule decrease and vice versa. | 23,488 | 1,249 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/11%3A_Reactions_and_Other_Chemical_Processes/11.07%3A_Gibbs_Energy_and_Reaction_Equilibrium |
\( \newcommand{\tx}[1]{\text{#1}} % text in math mode\)
\( \newcommand{\subs}[1]{_{\text{#1}}} % subscript text\)
\( \newcommand{\sups}[1]{^{\text{#1}}} % superscript text\)
\( \newcommand{\st}{^\circ} % standard state symbol\)
\( \newcommand{\id}{^{\text{id}}} % ideal\)
\( \newcommand{\rf}{^{\text{ref}}} % reference state\)
\( \newcommand{\units}[1]{\mbox{$\thinspace$#1}}\)
\( \newcommand{\K}{\units{K}} % kelvins\)
\( \newcommand{\degC}{^\circ\text{C}} % degrees Celsius\)
\( \newcommand{\br}{\units{bar}} % bar (\bar is already defined)\)
\( \newcommand{\Pa}{\units{Pa}}\)
\( \newcommand{\mol}{\units{mol}} % mole\)
\( \newcommand{\V}{\units{V}} % volts\)
\( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\)
\( \newcommand{\per}{^{-1}} % minus one power\)
\( \newcommand{\m}{_{\text{m}}} % subscript m for molar quantity\)
\( \newcommand{\CVm}{C_{V,\text{m}}} % molar heat capacity at const.V\)
\( \newcommand{\Cpm}{C_{p,\text{m}}} % molar heat capacity at const.p\)
\( \newcommand{\kT}{\kappa_T} % isothermal compressibility\)
\( \newcommand{\A}{_{\text{A}}} % subscript A for solvent or state A\)
\( \newcommand{\B}{_{\text{B}}} % subscript B for solute or state B\)
\( \newcommand{\bd}{_{\text{b}}} % subscript b for boundary or boiling point\)
\( \newcommand{\C}{_{\text{C}}} % subscript C\)
\( \newcommand{\f}{_{\text{f}}} % subscript f for freezing point\)
\( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\)
\( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\)
\( \newcommand{\mi}{_{\text{m},i}} % subscript m,i (m=molar)\)
\( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\)
\( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\)
\( \newcommand{\xbB}{_{x,\text{B}}} % x basis, B\)
\( \newcommand{\xbC}{_{x,\text{C}}} % x basis, C\)
\( \newcommand{\cbB}{_{c,\text{B}}} % c basis, B\)
\( \newcommand{\mbB}{_{m,\text{B}}} % m basis, B\)
\( \newcommand{\kHi}{k_{\text{H},i}} % Henry's law constant, x basis, i\)
\( \newcommand{\kHB}{k_{\text{H,B}}} % Henry's law constant, x basis, B\)
\( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\)
\( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\)
\( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\)
\( \newcommand{\eq}{\subs{eq}} % equilibrium state\)
\( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\)
\( \newcommand{\sys}{\subs{sys}} % system property\)
\( \newcommand{\sur}{\sups{sur}} % surroundings\)
\( \renewcommand{\in}{\sups{int}} % internal\)
\( \newcommand{\lab}{\subs{lab}} % lab frame\)
\( \newcommand{\cm}{\subs{cm}} % center of mass\)
\( \newcommand{\rev}{\subs{rev}} % reversible\)
\( \newcommand{\irr}{\subs{irr}} % irreversible\)
\( \newcommand{\fric}{\subs{fric}} % friction\)
\( \newcommand{\diss}{\subs{diss}} % dissipation\)
\( \newcommand{\el}{\subs{el}} % electrical\)
\( \newcommand{\cell}{\subs{cell}} % cell\)
\( \newcommand{\As}{A\subs{s}} % surface area\)
\( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)
\( \newcommand{\allni}{\{n_i \}} % set of all n_i\)
\( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\)
\( \newcommand{\solmB}{\tx{(sol,$\,$$m\B$)}}\)
\( \newcommand{\dil}{\tx{(dil)}}\)
\( \newcommand{\sln}{\tx{(sln)}}\)
\( \newcommand{\mix}{\tx{(mix)}}\)
\( \newcommand{\rxn}{\tx{(rxn)}}\)
\( \newcommand{\expt}{\tx{(expt)}}\)
\( \newcommand{\solid}{\tx{(s)}}\)
\( \newcommand{\liquid}{\tx{(l)}}\)
\( \newcommand{\gas}{\tx{(g)}}\)
\( \newcommand{\pha}{\alpha} % phase alpha\)
\( \newcommand{\phb}{\beta} % phase beta\)
\( \newcommand{\phg}{\gamma} % phase gamma\)
\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
\( \newcommand{\bph}{^{\beta}} % beta phase superscript\)
\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
\( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\)
\( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\)
\( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)
\( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)
\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
\( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)
\( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)
\( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)
\( \newcommand{\Del}{\Delta}\)
\( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)
\( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)
\( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)
\( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)
\( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)
\( \newcommand{\dotprod}{\small\bullet}\)
\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
\( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)
\( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)
\( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\)
\( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\)
\( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)
\( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)
\( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)
\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) This section begins by examining the way in which the Gibbs energy changes as a chemical process advances in a closed system at constant \(T\) and \(p\) with expansion work only. A universal criterion for reaction equilibrium is derived involving the molar reaction Gibbs energy. Applying the general definition of a molar differential reaction quantity (Eq. 11.2.15) to the Gibbs energy of a closed system with \(T\), \(p\), and \(\xi\) as the independent variables, we obtain the definition of the or molar Gibbs energy of reaction, \(\Delsub{r}G\): \[\begin{equation} \Delsub{r}G \defn \sum_i\nu_i \mu_i \label{11.7.1} \end{equation}\] Equation 11.2.16 shows that this quantity is also given by the partial derivative \[\begin{gather} \s{ \Delsub{r}G = \Pd{G}{\xi}{T,p} } \label{11.7.2} \cond{(closed system)} \end{gather}\] The total differential of \(G\) is then \[\begin{gather} \s{ \dif G = -S\dif T + V\difp + \Delsub{r}G\dif\xi } \label{11.7.3} \cond{(closed system)} \end{gather}\] In Sec. 5.8, we found that the spontaneous direction of a process taking place in a closed system at constant \(T\) and \(p\), with expansion work only, is the direction of decreasing \(G\). In the case of a chemical process occurring at constant \(T\) and \(p\), \(\Delsub{r}G\) is the rate at which \(G\) changes with \(\xi\). Thus if \(\Delsub{r}G\) is positive, \(\xi\) spontaneously decreases; if \(\Delsub{r}G\) is negative, \(\xi\) spontaneously increases. During a spontaneous process \(\dif\xi\) and \(\Delsub{r}G\) have opposite signs. Sometimes reaction spontaneity at constant \(T\) and \(p\) is ascribed to the “driving force” of a quantity called the , defined as the negative of \(\Delsub{r}G\). \(\xi\) increases spontaneously if the affinity is positive and decreases spontaneously if the affinity is negative; the system is at equilibrium when the affinity is zero. Note how the equality of Equation \ref{11.7.3} agrees with the inequality \(\dif G<-S\dif T+V\difp\), a criterion of spontaneity in a closed system with expansion work only (Eq. 5.8.6). When \(\dif\xi\) and \(\Delsub{r}G\) have opposite signs, \(\Delsub{r}G\dif\xi\) is negative and \(\dif G=(-S\dif T+V\difp+\Delsub{r}G\dif\xi)\) is less than \((-S\dif T+V\difp)\). If the system is closed and contains at least one phase that is a mixture, a state of reaction equilibrium can be approached spontaneously at constant \(T\) and \(p\) in either direction of the reaction; that is, by both positive and negative changes of \(\xi\). In this equilibrium state, therefore, \(G\) has its minimum value for the given \(T\) and \(p\). Since \(G\) is a smooth function of \(\xi\), its rate of change with respect to \(\xi\) is zero in the equilibrium state. The condition for , then, is that \(\Delsub{r}G\) must be zero: \begin{gather} \s{ \Delsub{r}G=\sum_i\nu_i\mu_i=0 } \tag{11.7.4} \cond{(reaction equilibrium)} \end{gather} It is important to realize that this condition is independent of whether or not reaction equilibrium is approached at constant temperature and pressure. It is a universal criterion of reaction equilibrium. The value of \(\Delsub{r}G\) is equal to \(\sum_i\!\nu_i\mu_i\) and depends on the state of the system. If the state is such that \(\Delsub{r}G\) is positive, the direction of spontaneous change is one that, under the existing constraints, allows \(\Delsub{r}G\) to decrease. If \(\Delsub{r}G\) is negative, the spontaneous change increases the value of \(\Delsub{r}G\). When the system reaches reaction equilibrium, whatever the path of the spontaneous process, the value of \(\Delsub{r}G\) becomes zero. We can obtain the condition of reaction equilibrium given by Eq. 11.7.4 in a more general and rigorous way by an extension of the derivation of Sec. 9.2.7, which was for equilibrium conditions in a multiphase, multicomponent system. Consider a system with a reference phase, \(\pha'\), and optionally other phases labeled by \(\pha\ne\pha'\). Each phase contains one or more species labeled by subscript \(i\), and some or all of the species are the reactants and products of a reaction. The total differential of the internal energy is given by Eq. 9.2.37: \begin{equation} \begin{split} \dif U & = T\aphp\dif S\aphp - p\aphp\dif V\aphp + \sum_i\mu_i\aphp\dif n_i\aphp \cr & \quad + \sum_{\pha\ne\pha'}\left(T\aph\dif S\aph - p\aph\dif V\aph + \sum_i\mu_i\aph\dif n_i\aph\right) \end{split} \tag{11.7.5} \end{equation} The conditions of isolation are \begin{equation} \dif U = 0 \qquad \tx{(constant internal energy)} \tag{11.7.6} \end{equation} \begin{equation} \dif V\aphp + \sum_{\pha\ne\pha'}\dif V\aph = 0 \qquad \tx{(no expansion work)} \tag{11.7.7} \end{equation} \begin{equation} \begin{split} &\tx{For each species \(i\):} \cr &\dif n_i\aphp + \sum_{\pha\ne\pha'}\dif n_i\aph = \nu_i\dif\xi \qquad \tx{(closed system)} \end{split} \tag{11.7.8} \end{equation} In Eq. 11.7.8, \(\dif n^{\pha''}_{i'}\) should be set equal to zero for a species \(i'\) that is excluded from phase \(\pha''\), and \(\nu_{i''}\) should be set equal to zero for a species \(i''\) that is not a reactant or product of the reaction. We use these conditions of isolation to substitute for \(\dif U\), \(\dif V\aphp\), and \(\dif n_i\aphp\) in Eq. 11.7.5, and make the further substitution \(\dif S\aphp = \dif S - \sum_{\pha\ne\pha'}\dif S\aph\). Solving for \(\dif S\), we obtain \begin{equation} \begin{split} \dif S & = \sum_{\pha\ne\pha'}\frac{(T\aphp-T\aph)}{T\aphp}\dif S\aph - \sum_{\pha\ne\pha'}\frac{(p\aphp-p\aph)}{T\aphp}\dif V\aph \cr & \quad + \sum_i\sum_{\pha\ne\pha'}\frac{(\mu_i\aphp-\mu_i\aph)}{T\aphp}\dif n_i\aph - \frac{ \sum_i\nu_i\mu_i\aphp}{T\aphp}\dif\xi \end{split} \tag{11.7.9} \end{equation} The equilibrium condition is that the coefficient multiplying each differential on the right side of Eq. 11.7.9 must be zero. We conclude that at equilibrium the temperature of each phase is equal to that of phase \(\pha'\); the pressure of each phase is equal to that of phase \(\pha'\); the chemical potential of each species, in each phase containing that species, is equal to the chemical potential of the species in phase \(\pha'\); and the quantity \( \sum_i\!\nu_i\mu_i\aphp\) (which is equal to \(\Delsub{r}G\)) is zero. In short, . Consider a chemical process in which each reactant and product is in a separate pure phase. For example, the decomposition of calcium carbonate, \[\ce{CaCO3(s) -> CaO(s) + CO2(g)}\] involves three pure phases if no other gas is allowed to mix with the \(\ce{CO2}\). The situation is different when the number of molecules changes during the reaction. Consider the reaction A\(\arrow\)2 B in an ideal gas mixture. As this reaction proceeds to the right at constant \(T\), the volume increases if the pressure is held constant and the pressure increases if the volume is held constant. Figure 11.17 shows how \(G\) depends on both \(p\) and \(V\) for this reaction. Movement along the horizontal dashed line in the figure corresponds to reaction at constant \(T\) and \(p\). The minimum of \(G\) along this line is at the volume indicated by the open circle. At this volume, \(G\) has an even lower minimum at the pressure indicated by the filled circle, where the vertical dashed line is tangent to one of the contours of constant \(G\). The condition needed for reaction equilibrium, however, is that \(\Delsub{r}G\) must be zero. This condition is satisfied along the vertical dashed line only at the position of the open circle. This example demonstrates that for a reaction occurring at constant temperature and in which the pressure changes, the point of reaction equilibrium is not the point of minimum \(G\). Instead, the point of reaction equilibrium in this case is at the minimum of the Helmholtz energy \(A\) (Sec. 11.7.5). | 15,126 | 1,250 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.6%3A_Acid-Base_Titration_Curves |
In an acid–base titration, a is used to deliver measured volumes of an acid or a base solution of known concentration (the titrant) to a flask that contains a solution of a base or an acid, respectively, of unknown concentration (the unknown). If the concentration of the titrant is known, then the concentration of the unknown can be determined. The following discussion focuses on the pH changes that occur during an acid–base titration. Plotting the pH of the solution in the flask against the amount of acid or base added produces a titration curve. The shape of the curve provides important information about what is occurring in solution during the titration. Figure \(\Page {1a}\) shows a plot of the pH as 0.20 M \(\ce{HCl}\) is gradually added to 50.00 mL of pure water. The pH of the sample in the flask is initially 7.00 (as expected for pure water), but it drops very rapidly as \(\ce{HCl}\) is added. Eventually the pH becomes constant at 0.70—a point well beyond its value of 1.00 with the addition of 50.0 mL of \(\ce{HCl}\) (0.70 is the pH of 0.20 M HCl). In contrast, when 0.20 M \(\ce{NaOH}\) is added to 50.00 mL of distilled water, the pH (initially 7.00) climbs very rapidly at first but then more gradually, eventually approaching a limit of 13.30 (the pH of 0.20 M NaOH), again well beyond its value of 13.00 with the addition of 50.0 mL of \(\ce{NaOH}\) as shown in Figure \(\Page {1b}\). As you can see from these plots, the titration curve for adding a base is the mirror image of the curve for adding an acid. Suppose that we now add 0.20 M \(\ce{NaOH}\) to 50.0 mL of a 0.10 M solution of \(\ce{HCl}\). Because \(\ce{HCl}\) is a strong acid that is completely ionized in water, the initial \([H^+]\) is 0.10 M, and the initial pH is 1.00. Adding \(\ce{NaOH}\) decreases the concentration of H+ because of the neutralization reaction (Figure \(\Page {2a}\)): \[\ce{OH^{−} + H^{+} <=> H_2O}. \nonumber \] Thus the pH of the solution increases gradually. Near the equivalence point, however, the point at which the number of moles of base (or acid) added equals the number of moles of acid (or base) originally present in the solution, the pH increases much more rapidly because most of the \(\ce{H^{+}}\) ions originally present have been consumed. For the titration of a monoprotic strong acid (\(\ce{HCl}\)) with a monobasic strong base (\(\ce{NaOH}\)), we can calculate the volume of base needed to reach the equivalence point from the following relationship: \[moles\;of \;base=(volume)_b(molarity)_bV_bM_b= moles \;of \;acid=(volume)_a(molarity)_a=V_aM_a \label{Eq1} \] If 0.20 M \(\ce{NaOH}\) is added to 50.0 mL of a 0.10 M solution of \(\ce{HCl}\), we solve for \(V_b\): \[V_b(0.20 Me)=0.025 L=25 mL \nonumber \] At the equivalence point (when 25.0 mL of \(\ce{NaOH}\) solution has been added), the neutralization is complete: only a salt remains in solution (NaCl), and the pH of the solution is 7.00. Adding more \(\ce{NaOH}\) produces a rapid increase in pH, but eventually the pH levels off at a value of about 13.30, the pH of 0.20 M \(NaOH\). As shown in Figure \(\Page {2b}\), the titration of 50.0 mL of a 0.10 M solution of \(\ce{NaOH}\) with 0.20 M \(\ce{HCl}\) produces a titration curve that is nearly the mirror image of the titration curve in Figure \(\Page {2a}\). The pH is initially 13.00, and it slowly decreases as \(\ce{HCl}\) is added. As the equivalence point is approached, the pH drops rapidly before leveling off at a value of about 0.70, the pH of 0.20 M \(\ce{HCl}\). The titration of either a strong acid with a strong base or a strong base with a strong acid produces an S-shaped curve. The curve is somewhat asymmetrical because the steady increase in the volume of the solution during the titration causes the solution to become more dilute. Due to the leveling effect, the shape of the curve for a titration involving a strong acid and a strong base depends on only the concentrations of the acid and base, not their identities. The shape of the titration curve involving a strong acid and a strong base depends only on their concentrations, not their identities. Calculate the pH of the solution after 24.90 mL of 0.200 M \(\ce{NaOH}\) has been added to 50.00 mL of 0.100 M \(\ce{HCl}\). : volumes and concentrations of strong base and acid : pH : Because 0.100 mol/L is equivalent to 0.100 mmol/mL, the number of millimoles of \(\ce{H^{+}}\) in 50.00 mL of 0.100 M \(\ce{HCl}\) can be calculated as follows: \[ 50.00 \cancel{mL} \left ( \dfrac{0.100 \;mmol \;HCl}{\cancel{mL}} \right )= 5.00 \;mmol \;HCl=5.00 \;mmol \;H^{+} \nonumber \] The number of millimoles of \(\ce{NaOH}\) added is as follows: \[ 24.90 \cancel{mL} \left ( \dfrac{0.200 \;mmol \;NaOH}{\cancel{mL}} \right )= 4.98 \;mmol \;NaOH=4.98 \;mmol \;OH^{-} \nonumber \] Thus \(\ce{H^{+}}\) is in excess. To completely neutralize the acid requires the addition of 5.00 mmol of \(\ce{OH^{-}}\) to the \(\ce{HCl}\) solution. Because only 4.98 mmol of \(OH^-\) has been added, the amount of excess \(\ce{H^{+}}\) is 5.00 mmol − 4.98 mmol = 0.02 mmol of \(H^+\). The final volume of the solution is 50.00 mL + 24.90 mL = 74.90 mL, so the final concentration of \(\ce{H^{+}}\) is as follows: \[ \left [ H^{+} \right ]= \dfrac{0.02 \;mmol \;H^{+}}{74.90 \; mL}=3 \times 10^{-4} \; M \nonumber \] Hence, \[pH \approx −\log[\ce{H^{+}}] = −\log(3 \times 10^{-4}) = 3.5 \nonumber \] This is significantly less than the pH of 7.00 for a neutral solution. Calculate the pH of a solution prepared by adding \(40.00\; mL\) of \(0.237\; M\) \(HCl\) to \(75.00\; mL\) of a \(0.133 M\) solution of \(NaOH\). 11.6 pH after the addition of 10 ml of Strong Base to a Strong Acid: pH at the Equivalence Point in a Strong Acid/Strong Base Titration: In contrast to strong acids and bases, the shape of the titration curve for a weak acid or a weak base depends dramatically on the identity of the acid or the base and the corresponding \(K_a\) or \(K_b\). As we shall see, the pH also changes much more gradually around the equivalence point in the titration of a weak acid or a weak base. As you learned previously, \([\ce{H^{+}}]\) of a solution of a weak acid (HA) is not equal to the concentration of the acid but depends on both its \(pK_a\) and its concentration. Because only a fraction of a weak acid dissociates, \([\(\ce{H^{+}}]\) is less than \([\ce{HA}]\). Thus the pH of a solution of a weak acid is greater than the pH of a solution of a strong acid of the same concentration. Figure \(\Page {3a}\) shows the titration curve for 50.0 mL of a 0.100 M solution of acetic acid with 0.200 M \(\ce{NaOH}\) superimposed on the curve for the titration of 0.100 M \(\ce{HCl}\) shown in part (a) in Figure \(\Page {2}\). Below the equivalence point, the two curves are very different. Before any base is added, the pH of the acetic acid solution is greater than the pH of the \(\ce{HCl}\) solution, and the pH changes more rapidly during the first part of the titration. Note also that the pH of the acetic acid solution at the equivalence point is greater than 7.00. That is, at the equivalence point, the solution is basic. In addition, the change in pH around the equivalence point is only about half as large as for the \(\ce{HCl}\) titration; the magnitude of the pH change at the equivalence point depends on the \(pK_a\) of the acid being titrated. Above the equivalence point, however, the two curves are identical. Once the acid has been neutralized, the pH of the solution is controlled only by the amount of excess \(\ce{NaOH}\) present, regardless of whether the acid is weak or strong. The shape of the titration curve of a weak acid or weak base depends heavily on their identities and the \(K_a\) or \(K_b\). The titration curve in Figure \(\Page {3a}\) was created by calculating the starting pH of the acetic acid solution before any \(\ce{NaOH}\) is added and then calculating the pH of the solution after adding increasing volumes of \(NaOH\). The procedure is illustrated in the following subsection and Example \(\Page {2}\) for three points on the titration curve, using the \(pK_a\) of acetic acid (4.76 at 25°C; \(K_a = 1.7 \times 10^{-5}\). As explained discussed, if we know \(K_a\) or \(K_b\) and the initial concentration of a weak acid or a weak base, we can calculate the pH of a solution of a weak acid or a weak base by setting up a table (i.e, initial concentrations, changes in concentrations, and final concentrations). In this situation, the initial concentration of acetic acid is 0.100 M. If we define \(x\) as \([\ce{H^{+}}]\) due to the dissociation of the acid, then the table of concentrations for the ionization of 0.100 M acetic acid is as follows: \[\ce{CH3CO2H(aq) <=> H^{+}(aq) + CH3CO2^{−}} \nonumber \] In this and all subsequent examples, we will ignore \([H^+]\) and \([OH^-]\) due to the autoionization of water when calculating the final concentration. However, you should use Equation 16.45 and Equation 16.46 to check that this assumption is justified. Inserting the expressions for the final concentrations into the equilibrium equation (and using approximations), \[ \begin{align*} K_a &=\dfrac{[H^+,CH_3CO_2^-]}{[CH_3CO_2H]} \\[4pt] &=\dfrac{(x)(x)}{0.100 - x} \\[4pt] &\approx \dfrac{x^2}{0.100} \\[4pt] &\approx 1.74 \times 10^{-5} \end{align*} \nonumber \] Solving this equation gives \(x = [H^+] = 1.32 \times 10^{-3}\; M\). Thus the pH of a 0.100 M solution of acetic acid is as follows: \[pH = −\log(1.32 \times 10^{-3}) = 2.879 \nonumber \] pH at the Start of a Weak Acid/Strong Base Titration: Now consider what happens when we add 5.00 mL of 0.200 M \(\ce{NaOH}\) to 50.00 mL of 0.100 M \(CH_3CO_2H\) (part (a) in Figure \(\Page {3}\)). Because the neutralization reaction proceeds to completion, all of the \(OH^-\) ions added will react with the acetic acid to generate acetate ion and water: \[ CH_3CO_2H_{(aq)} + OH^-_{(aq)} \rightarrow CH_3CO^-_{2\;(aq)} + H_2O_{(l)} \label{Eq2} \] All problems of this type must be solved in two steps: a stoichiometric calculation followed by an equilibrium calculation. In the first step, we use the stoichiometry of the neutralization reaction to calculate the amounts of acid and conjugate base present in solution after the neutralization reaction has occurred. In the second step, we use the equilibrium equation to determine \([\ce{H^{+}}]\) of the resulting solution. To determine the amount of acid and conjugate base in solution after the neutralization reaction, we calculate the amount of \(\ce{CH_3CO_2H}\) in the original solution and the amount of \(\ce{OH^{-}}\) in the \(\ce{NaOH}\) solution that was added. The acetic acid solution contained \[ 50.00 \; \cancel{mL} (0.100 \;mmol (\ce{CH_3CO_2H})/\cancel{mL} )=5.00\; mmol (\ce{CH_3CO_2H}) \nonumber \] The \(\ce{NaOH}\) solution contained 5.00 mL=1.00 mmol \(NaOH\) Comparing the amounts shows that \(CH_3CO_2H\) is in excess. Because \(OH^-\) reacts with \(CH_3CO_2H\) in a 1:1 stoichiometry, the amount of excess \(CH_3CO_2H\) is as follows: 5.00 mmol \(CH_3CO_2H\) − 1.00 mmol \(OH^-\) = 4.00 mmol \(CH_3CO_2H\) Each 1 mmol of \(OH^-\) reacts to produce 1 mmol of acetate ion, so the final amount of \(CH_3CO_2^−\) is 1.00 mmol. The stoichiometry of the reaction is summarized in the following ICE table, which shows the numbers of moles of the various species, not their concentrations. \[\ce{CH3CO2H(aq) + OH^{−} (aq) <=> CH3CO2^{-}(aq) + H2O(l)} \nonumber \] This ICE table gives the initial amount of acetate and the final amount of \(OH^-\) ions as 0. Because an aqueous solution of acetic acid always contains at least a small amount of acetate ion in equilibrium with acetic acid, however, the initial acetate concentration is not actually 0. The value can be ignored in this calculation because the amount of \(CH_3CO_2^−\) in equilibrium is insignificant compared to the amount of \(OH^-\) added. Moreover, due to the autoionization of water, no aqueous solution can contain 0 mmol of \(OH^-\), but the amount of \(OH^-\) due to the autoionization of water is insignificant compared to the amount of \(OH^-\) added. We use the initial amounts of the reactants to determine the stoichiometry of the reaction and defer a consideration of the equilibrium until the second half of the problem. To calculate \([\ce{H^{+}}]\) at equilibrium following the addition of \(NaOH\), we must first calculate [\(\ce{CH_3CO_2H}\)] and \([\ce{CH3CO2^{−}}]\) using the number of millimoles of each and the total volume of the solution at this point in the titration: \[ final \;volume=50.00 \;mL+5.00 \;mL=55.00 \;mL \nonumber \] \[ \left [ CH_{3}CO_{2}H \right ] = \dfrac{4.00 \; mmol \; CH_{3}CO_{2}H }{55.00 \; mL} =7.27 \times 10^{-2} \;M \nonumber \] \[ \left [ CH_{3}CO_{2}^{-} \right ] = \dfrac{1.00 \; mmol \; CH_{3}CO_{2}^{-} }{55.00 \; mL} =1.82 \times 10^{-2} \;M \nonumber \] Knowing the concentrations of acetic acid and acetate ion at equilibrium and \(K_a\) for acetic acid (\(1.74 \times 10^{-5}\)), we can calculate \([H^+]\) at equilibrium: \[ K_{a}=\dfrac{\left [ CH_{3}CO_{2}^{-} \right ]\left [ H^{+} \right ]}{\left [ CH_{3}CO_{2}H \right ]} \nonumber \] \[ \left [ H^{+} \right ]=\dfrac{K_{a}\left [ CH_{3}CO_{2}H \right ]}{\left [ CH_{3}CO_{2}^{-} \right ]} = \dfrac{\left ( 1.72 \times 10^{-5} \right )\left ( 7.27 \times 10^{-2} \;M\right )}{\left ( 1.82 \times 10^{-2} \right )}= 6.95 \times 10^{-5} \;M \nonumber \] Calculating \(−\log[\ce{H^{+}}]\) gives \[pH = −\log(6.95 \times 10^{−5}) = 4.158. \nonumber \] Comparing the titration curves for \(\ce{HCl}\) and acetic acid in Figure \(\Page {3a}\), we see that adding the same amount (5.00 mL) of 0.200 M \(\ce{NaOH}\) to 50 mL of a 0.100 M solution of both acids causes a much smaller pH change for \(\ce{HCl}\) (from 1.00 to 1.14) than for acetic acid (2.88 to 4.16). This is consistent with the qualitative description of the shapes of the titration curves at the beginning of this section. In Example \(\Page {2}\), we calculate another point for constructing the titration curve of acetic acid. pH Before the Equivalence Point of a Weak Acid/Strong Base Titration: What is the pH of the solution after 25.00 mL of 0.200 M \(\ce{NaOH}\) is added to 50.00 mL of 0.100 M acetic acid? Given: volume and molarity of base and acid : pH Ignoring the spectator ion (\(Na^+\)), the equation for this reaction is as follows: \[CH_3CO_2H_{ (aq)} + OH^-(aq) \rightarrow CH_3CO_2^-(aq) + H_2O(l) \nonumber \] The initial numbers of millimoles of \(OH^-\) and \(CH_3CO_2H\) are as follows: 25.00 mL(0.200 mmol OH−mL=5.00 mmol \(OH-\) \[50.00\; mL (0.100 CH_3CO_2 HL=5.00 mmol \; CH_3CO_2H \nonumber \] The number of millimoles of \(OH^-\) equals the number of millimoles of \(CH_3CO_2H\), so neither species is present in excess. Because the number of millimoles of \(OH^-\) added corresponds to the number of millimoles of acetic acid in solution, this is the equivalence point. The results of the neutralization reaction can be summarized in tabular form. \[CH_3CO_2H_{(aq)}+OH^-_{(aq)} \rightleftharpoons CH_3CO_2^{-}(aq)+H_2O(l) \nonumber \] C Because the product of the neutralization reaction is a weak base, we must consider the reaction of the weak base with water to calculate [H+] at equilibrium and thus the final pH of the solution. The initial concentration of acetate is obtained from the neutralization reaction: \[ [\ce{CH_3CO_2}]=\dfrac{5.00 \;mmol \; CH_3CO_2^{-}}{(50.00+25.00) \; mL}=6.67\times 10^{-2} \; M \nonumber \] The equilibrium reaction of acetate with water is as follows: \[\ce{CH_3CO^{-}2(aq) + H2O(l) <=> CH3CO2H(aq) + OH^{-} (aq)} \nonumber \] The equilibrium constant for this reaction is \[K_b = \dfrac{K_w}{K_a} \label{16.18} \] where \(K_a\) is the acid ionization constant of acetic acid. We therefore define x as \([\ce{OH^{−}}]\) produced by the reaction of acetate with water. Here is the completed table of concentrations: \[H_2O_{(l)}+CH_3CO^−_{2(aq)} \rightleftharpoons CH_3CO_2H_{(aq)} +OH^−_{(aq)} \nonumber \] We can obtain \(K_b\) by substituting the known values into Equation \ref{16.18}: \[ K_{b}= \dfrac{K_w}{K_a} =\dfrac{1.01 \times 10^{-14}}{1.74 \times 10^{-5}} = 5.80 \times 10^{-10} \label{16.23} \] Substituting the expressions for the final values from the ICE table into Equation \ref{16.23} and solving for \(x\): \[ \begin{align*} \dfrac{x^{2}}{0.0667} &= 5.80 \times 10^{-10} \\[4pt] x &= \sqrt{(5.80 \times 10^{-10})(0.0667)} \\[4pt] &= 6.22 \times 10^{-6}\end{align*} \nonumber \] Thus \([OH^{−}] = 6.22 \times 10^{−6}\, M\) and the pH of the final solution is 8.794 (Figure \(\Page {3a}\)). As expected for the titration of a weak acid, the pH at the equivalence point is greater than 7.00 because the product of the titration is a base, the acetate ion, which then reacts with water to produce \(\ce{OH^{-}}\). Calculate the pH of a solution prepared by adding 45.0 mL of a 0.213 M \(\ce{HCl}\) solution to 125.0 mL of a 0.150 M solution of ammonia. The \(pK_b\) of ammonia is 4.75 at 25°C. 9.23 As shown in part (b) in Figure \(\Page {3}\), the titration curve for NH3, a weak base, is the reverse of the titration curve for acetic acid. In particular, the pH at the equivalence point in the titration of a weak base is less than 7.00 because the titration produces an acid. The identity of the weak acid or weak base being titrated strongly affects the shape of the titration curve. Figure \(\Page {4}\) illustrates the shape of titration curves as a function of the \(pK_a\) or the \(pK_b\). As the acid or the base being titrated becomes weaker (its \(pK_a\) or \(pK_b\) becomes larger), the pH change around the equivalence point decreases significantly. With very dilute solutions, the curve becomes so shallow that it can no longer be used to determine the equivalence point. One point in the titration of a weak acid or a weak base is particularly important: the midpoint of a titration is defined as the point at which exactly enough acid (or base) has been added to neutralize one-half of the acid (or the base) originally present and occurs halfway to the equivalence point. The midpoint is indicated in Figures \(\Page {4a}\) and \(\Page {4b}\) for the two shallowest curves. By definition, at the midpoint of the titration of an acid, [HA] = [A−]. Recall that the ionization constant for a weak acid is as follows: \[K_a=\dfrac{[H_3O^+,A^−]}{[HA]} \nonumber \] If \([HA] = [A^−]\), this reduces to \(K_a = [H_3O^+]\). Taking the negative logarithm of both sides, \[−\log K_a = −\log[H_3O+] \nonumber \] From the definitions of \(pK_a\) and pH, we see that this is identical to \[pK_a = pH \label{16.52} \] Thus the pH at the midpoint of the titration of a weak acid is equal to the \(pK_a\) of the weak acid, as indicated in part (a) in Figure \(\Page {4}\) for the weakest acid where we see that the midpoint for \(pK_a\) = 10 occurs at pH = 10. Titration methods can therefore be used to determine both the concentration and the \(pK_a\) (or the \(pK_b\)) of a weak acid (or a weak base). The pH at the midpoint of the titration of a weak acid is equal to the \(pK_a\) of the weak acid. When a strong base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. The most acidic group is titrated first, followed by the next most acidic, and so forth. If the \(pK_a\) values are separated by at least three \(pK_a\) units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton. A titration of the triprotic acid \(H_3PO_4\) with \(\ce{NaOH}\) is illustrated in Figure \(\Page {5}\) and shows two well-defined steps: the first midpoint corresponds to \(pK_a\)1, and the second midpoint corresponds to \(pK_a\) . Because HPO is such a weak acid, \(pK_a\) has such a high value that the third step cannot be resolved using 0.100 M \(\ce{NaOH}\) as the titrant. The titration curve for the reaction of a polyprotic base with a strong acid is the mirror image of the curve shown in Figure \(\Page {5}\). The initial pH is high, but as acid is added, the pH decreases in steps if the successive \(pK_b\) values are well separated. Table E1 lists the ionization constants and \(pK_a\) values for some common polyprotic acids and bases. Calculate the pH of a solution prepared by adding 55.0 mL of a 0.120 M \(\ce{NaOH}\) solution to 100.0 mL of a 0.0510 M solution of oxalic acid (\(\ce{HO_2CCO_2H}\)), a diprotic acid (abbreviated as \(\ce{H2ox}\)). Oxalic acid, the simplest dicarboxylic acid, is found in rhubarb and many other plants. Rhubarb leaves are toxic because they contain the calcium salt of the fully deprotonated form of oxalic acid, the oxalate ion (\(\ce{O2CCO2^{2−}}\), abbreviated \(\ce{ox^{2-}}\)).Oxalate salts are toxic for two reasons. First, oxalate salts of divalent cations such as \(\ce{Ca^{2+}}\) are insoluble at neutral pH but soluble at low pH. As a result, calcium oxalate dissolves in the dilute acid of the stomach, allowing oxalate to be absorbed and transported into cells, where it can react with calcium to form tiny calcium oxalate crystals that damage tissues. Second, oxalate forms stable complexes with metal ions, which can alter the distribution of metal ions in biological fluids. : volume and concentration of acid and base : pH : : gives the \(pK_a\) values of oxalic acid as 1.25 and 3.81. Again we proceed by determining the millimoles of acid and base initially present: \[ 100.00 \cancel{mL} \left ( \dfrac{0.510 \;mmol \;H_{2}ox}{\cancel{mL}} \right )= 5.10 \;mmol \;H_{2}ox \nonumber \] \[ 55.00 \cancel{mL} \left ( \dfrac{0.120 \;mmol \;NaOH}{\cancel{mL}} \right )= 6.60 \;mmol \;NaOH \nonumber \] The strongest acid (\(H_2ox\)) reacts with the base first. This leaves (6.60 − 5.10) = 1.50 mmol of \(OH^-\) to react with Hox−, forming ox and H O. The reactions can be written as follows: \[ \underset{5.10\;mmol}{H_{2}ox}+\underset{6.60\;mmol}{OH^{-}} \rightarrow \underset{5.10\;mmol}{Hox^{-}}+ \underset{5.10\;mmol}{H_{2}O} \nonumber \] \[ \underset{5.10\;mmol}{Hox^{-}}+\underset{1.50\;mmol}{OH^{-}} \rightarrow \underset{1.50\;mmol}{ox^{2-}}+ \underset{1.50\;mmol}{H_{2}O} \nonumber \] In tabular form, B The equilibrium between the weak acid (\(\ce{Hox^{-}}\)) and its conjugate base (\(\ce{ox^{2-}}\)) in the final solution is determined by the magnitude of the second ionization constant, \(K_{a2} = 10^{−3.81} = 1.6 \times 10^{−4}\). To calculate the pH of the solution, we need to know \(\ce{[H^{+}]}\), which is determined using exactly the same method as in the acetic acid titration in Example \(\Page {2}\): \[\text{final volume of solution} = 100.0\, mL + 55.0\, mL = 155.0 \,mL \nonumber \] Thus the concentrations of \(\ce{Hox^{-}}\) and \(\ce{ox^{2-}}\) are as follows: \[ \left [ Hox^{-} \right ] = \dfrac{3.60 \; mmol \; Hox^{-}}{155.0 \; mL} = 2.32 \times 10^{-2} \;M \nonumber \] \[ \left [ ox^{2-} \right ] = \dfrac{1.50 \; mmol \; ox^{2-}}{155.0 \; mL} = 9.68 \times 10^{-3} \;M \nonumber \] We can now calculate [H+] at equilibrium using the following equation: \[ K_{a2} =\dfrac{\left [ ox^{2-} \right ]\left [ H^{+} \right ] }{\left [ Hox^{-} \right ]} \nonumber \] Rearranging this equation and substituting the values for the concentrations of \(\ce{Hox^{−}}\) and \(\ce{ox^{2−}}\), \[ \left [ H^{+} \right ] =\dfrac{K_{a2}\left [ Hox^{-} \right ]}{\left [ ox^{2-} \right ]} = \dfrac{\left ( 1.6\times 10^{-4} \right ) \left ( 2.32\times 10^{-2} \right )}{\left ( 9.68\times 10^{-3} \right )}=3.7\times 10^{-4} \; M \nonumber \] So \[ pH = -\log\left [ H^{+} \right ]= -\log\left ( 3.7 \times 10^{-4} \right )= 3.43 \nonumber \] This answer makes chemical sense because the pH is between the first and second \(pK_a\) values of oxalic acid, as it must be. We added enough hydroxide ion to completely titrate the first, more acidic proton (which should give us a pH greater than \(pK_{a1}\)), but we added only enough to titrate less than half of the second, less acidic proton, with \(pK_{a2}\). If we had added exactly enough hydroxide to completely titrate the first proton plus half of the second, we would be at the midpoint of the second step in the titration, and the pH would be 3.81, equal to \(pK_{a2}\). Piperazine is a diprotic base used to control intestinal parasites (“worms”) in pets and humans. A dog is given 500 mg (5.80 mmol) of piperazine (\(pK_{b1}\) = 4.27, \(pK_{b2}\) = 8.67). If the dog’s stomach initially contains 100 mL of 0.10 M \(\ce{HCl}\) (pH = 1.00), calculate the pH of the stomach contents after ingestion of the piperazine. pH=4.9 In practice, most acid–base titrations are not monitored by recording the pH as a function of the amount of the strong acid or base solution used as the titrant. Instead, an acid–base indicator is often used that, if carefully selected, undergoes a dramatic color change at the pH corresponding to the equivalence point of the titration. Indicators are weak acids or bases that exhibit intense colors that vary with pH. The conjugate acid and conjugate base of a good indicator have very different colors so that they can be distinguished easily. Some indicators are colorless in the conjugate acid form but intensely colored when deprotonated (phenolphthalein, for example), which makes them particularly useful. We can describe the chemistry of indicators by the following general equation: \[ \ce{ HIn (aq) <=> H^{+}(aq) + In^{-}(aq)} \nonumber \] where the protonated form is designated by \(\ce{HIn}\) and the conjugate base by \(\ce{In^{−}}\). The ionization constant for the deprotonation of indicator \(\ce{HIn}\) is as follows: \[ K_{In} =\dfrac{ [\ce{H^{+}} , \ce{In^{-}}]}{[\ce{HIn}]} \label{Eq3} \] The \(pK_{in}\) (its \(pK_a\)) determines the pH at which the indicator changes color. Many different substances can be used as indicators, depending on the particular reaction to be monitored. For example, red cabbage juice contains a mixture of colored substances that change from deep red at low pH to light blue at intermediate pH to yellow at high pH. Similarly, flowers can be blue, red, pink, light purple, or dark purple depending on the soil pH (Figure \(\Page {6}\)). Acidic soils will produce blue flowers, whereas alkaline soils will produce pinkish flowers. Irrespective of the origins, a good indicator must have the following properties: Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. Figure \(\Page {7}\) shows the approximate pH range over which some common indicators change color and their change in color. In addition, some indicators (such as thymol blue) are polyprotic acids or bases, which change color twice at widely separated pH values. It is important to be aware that an indicator does not change color abruptly at a particular pH value; instead, it actually undergoes a pH titration just like any other acid or base. As the concentration of HIn decreases and the concentration of In− increases, the color of the solution slowly changes from the characteristic color of HIn to that of In−. As we will see later, the [In−]/[HIn] ratio changes from 0.1 at a pH one unit below pKin to 10 at a pH one unit above pKin. Thus most indicators change color over a pH range of about two pH units. We have stated that a good indicator should have a pKin value that is close to the expected pH at the equivalence point. For a strong acid–strong base titration, the choice of the indicator is not especially critical due to the very large change in pH that occurs around the equivalence point. In contrast, using the wrong indicator for a titration of a weak acid or a weak base can result in relatively large errors, as illustrated in Figure \(\Page {8}\). This figure shows plots of pH versus volume of base added for the titration of 50.0 mL of a 0.100 M solution of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M \(NaOH\). The pH ranges over which two common indicators (methyl red, \(pK_{in} = 5.0\), and phenolphthalein, \(pK_{in} = 9.5\)) change color are also shown. The horizontal bars indicate the pH ranges over which both indicators change color cross the \(\ce{HCl}\) titration curve, where it is almost vertical. Hence both indicators change color when essentially the same volume of \(\ce{NaOH}\) has been added (about 50 mL), which corresponds to the equivalence point. In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Although the pH range over which phenolphthalein changes color is slightly greater than the pH at the equivalence point of the strong acid titration, the error will be negligible due to the slope of this portion of the titration curve. Just as with the \(\ce{HCl}\) titration, the phenolphthalein indicator will turn pink when about 50 mL of \(\ce{NaOH}\) has been added to the acetic acid solution. In contrast, methyl red begins to change from red to yellow around pH 5, which is near the midpoint of the acetic acid titration, not the equivalence point. Adding only about 25–30 mL of \(\ce{NaOH}\) will therefore cause the methyl red indicator to change color, resulting in a huge error. The graph shows the results obtained using two indicators (methyl red and phenolphthalein) for the titration of 0.100 M solutions of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M \(NaOH\). Due to the steepness of the titration curve of a strong acid around the equivalence point, either indicator will rapidly change color at the equivalence point for the titration of the strong acid. In contrast, the pKin for methyl red (5.0) is very close to the \(pK_a\) of acetic acid (4.76); the midpoint of the color change for methyl red occurs near the midpoint of the titration, rather than at the equivalence point. In general, for titrations of strong acids with strong bases (and vice versa), any indicator with a pKin between about 4.0 and 10.0 will do. For the titration of a weak acid, however, the pH at the equivalence point is greater than 7.0, so an indicator such as phenolphthalein or thymol blue, with pKin > 7.0, should be used. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with pKin < 7.0, should be used. The existence of many different indicators with different colors and pKin values also provides a convenient way to estimate the pH of a solution without using an expensive electronic pH meter and a fragile pH electrode. Paper or plastic strips impregnated with combinations of indicators are used as “pH paper,” which allows you to estimate the pH of a solution by simply dipping a piece of pH paper into it and comparing the resulting color with the standards printed on the container (Figure \(\Page {9}\)). pH Indicators: Plots of acid–base titrations generate titration curves that can be used to calculate the pH, the pOH, the \(pK_a\), and the \(pK_b\) of the system. The shape of a titration curve, a plot of pH versus the amount of acid or base added, provides important information about what is occurring in solution during a titration. The shapes of titration curves for weak acids and bases depend dramatically on the identity of the compound. The equivalence point of an acid–base titration is the point at which exactly enough acid or base has been added to react completely with the other component. The equivalence point in the titration of a strong acid or a strong base occurs at pH 7.0. In titrations of weak acids or weak bases, however, the pH at the equivalence point is greater or less than 7.0, respectively. The pH tends to change more slowly before the equivalence point is reached in titrations of weak acids and weak bases than in titrations of strong acids and strong bases. The pH at the midpoint, the point halfway on the titration curve to the equivalence point, is equal to the \(pK_a\) of the weak acid or the \(pK_b\) of the weak base. Thus titration methods can be used to determine both the concentration and the \(pK_a\) (or the \(pK_b\)) of a weak acid (or a weak base). Acid–base indicators are compounds that change color at a particular pH. They are typically weak acids or bases whose changes in color correspond to deprotonation or protonation of the indicator itself. | 32,404 | 1,252 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Examples/Examples_of_Catalysis/2._Examples_of_Catalysis_in_the_Inorganic_Chemical_Industry |
This page takes a brief look at the catalysts used in the Contact Process to manufacture sulphuric acid, in the Haber Process to manufacture ammonia, and in the conversion of ammonia into nitric acid. At the heart of the Contact Process is a reaction which converts sulphur dioxide into sulphur trioxide. Sulphur dioxide gas is passed together with air (as a source of oxygen) over a solid vanadium(V) oxide catalyst. This is therefore an example of heterogeneous catalysis. The fact that this is a reversible reaction makes no difference to the operation of the catalyst. It speeds up both the forward reaction and the back reaction by the same amount. The Haber Process combines hydrogen and nitrogen to make ammonia using an iron catalyst. This is another reversible reaction, and another example of heterogeneous catalysis. This is yet another example of heterogeneous catalysis. This process involves oxidation of the ammonia from the Haber Process by oxygen in the air in the presence of a platinum-rhodium catalyst. Large sheets of metal gauze are used in order to reduce expense and to maximise the surface area of the catalyst. Although in principle the sheets would last for ever because the metals are acting as a catalyst, in practice they do deteriorate over time and have to be replaced. The sheets of gauze are held at a temperature of about 900°C. The reaction is very exothermic, and once it starts the temperature is maintained by the heat evolved. The ammonia is oxidised to nitrogen monoxide gas. This is cooled. At ordinary temperatures and in the presence of excess air, it is oxidised further to nitrogen dioxide. The nitrogen dioxide (still in the presence of excess air) is absorbed in water where it reacts to give a concentrated solution of nitric acid. | 1,792 | 1,254 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/18%3A_Chemical_Kinetics/18.7%3A_Kinetics_of_Catalysis |
Catalysts are substances that increase the reaction rate of a chemical reaction without being consumed in the process. A catalyst, therefore, does not appear in the overall stoichiometry of the reaction it catalyzes, but it must appear in at least one of the elementary reactions in the mechanism for the catalyzed reaction. The catalyzed pathway has a lower , but the net change in energy that results from the reaction (the difference between the energy of the reactants and the energy of the products) is not affected by the presence of a catalyst ( e \(\Page {1}\)). Nevertheless, because of its lower , the reaction rate of a catalyzed reaction is faster than the reaction rate of the uncatalyzed reaction at the same temperature. Because a catalyst decreases the height of the energy barrier, its presence increases the reaction rates of both the forward and the reverse reactions by the same amount. In this section, we will examine the three major classes of catalysts: heterogeneous catalysts, homogeneous catalysts, and enzymes. A catalyst affects , not Δ . In , the catalyst is in a different phase from the reactants. At least one of the reactants interacts with the solid surface in a physical process called adsorption in such a way that a chemical bond in the reactant becomes weak and then breaks. Poisons are substances that bind irreversibly to catalysts, preventing reactants from adsorbing and thus reducing or destroying the catalyst’s efficiency. An example of heterogeneous catalysis is the interaction of hydrogen gas with the surface of a metal, such as Ni, Pd, or Pt. As shown in part (a) in \(\Page {2}\) the hydrogen–hydrogen bonds break and produce individual adsorbed hydrogen atoms on the surface of the metal. Because the adsorbed atoms can move around on the surface, two hydrogen atoms can collide and form a molecule of hydrogen gas that can then leave the surface in the reverse process, called desorption. Adsorbed H atoms on a metal surface are substantially more reactive than a hydrogen molecule. Because the relatively strong H–H bond (dissociation energy = 432 kJ/mol) has already been broken, the energy barrier for most reactions of H is substantially lower on the catalyst surface. \(\Page {2}\) shows a process called , in which hydrogen atoms are added to the double bond of an alkene, such as ethylene, to give a product that contains C–C single bonds, in this case ethane. Hydrogenation is used in the food industry to convert vegetable oils, which consist of long chains of alkenes, to more commercially valuable solid derivatives that contain alkyl chains. Hydrogenation of some of the double bonds in polyunsaturated vegetable oils, for example, produces margarine, a product with a melting point, texture, and other physical properties similar to those of butter. Several important examples of industrial heterogeneous catalytic reactions are in \(\Page {1}\). Although the mechanisms of these reactions are considerably more complex than the simple hydrogenation reaction described here, they all involve adsorption of the reactants onto a solid catalytic surface, chemical reaction of the adsorbed species (sometimes via a number of intermediate species), and finally desorption of the products from the surface. In , the catalyst is in the same phase as the reactant(s). The number of collisions between reactants and catalyst is at a maximum because the catalyst is uniformly dispersed throughout the reaction mixture. Many homogeneous catalysts in industry are transition metal compounds ( \(\Page {2}\)), but recovering these expensive catalysts from solution has been a major challenge. As an added barrier to their widespread commercial use, many homogeneous catalysts can be used only at relatively low temperatures, and even then they tend to decompose slowly in solution. Despite these problems, a number of commercially viable processes have been developed in recent years. High-density polyethylene and polypropylene are produced by homogeneous catalysis. Enzymes, catalysts that occur naturally in living organisms, are almost all protein molecules with typical molecular masses of 20,000–100,000 amu. Some are homogeneous catalysts that react in aqueous solution within a cellular compartment of an organism. Others are heterogeneous catalysts embedded within the membranes that separate cells and cellular compartments from their surroundings. The reactant in an enzyme-catalyzed reaction is called a . Because enzymes can increase reaction rates by enormous factors (up to 10 times the uncatalyzed rate) and tend to be very specific, typically producing only a single product in quantitative yield, they are the focus of active research. At the same time, enzymes are usually expensive to obtain, they often cease functioning at temperatures greater than 37 °C, have limited stability in solution, and have such high specificity that they are confined to turning one particular set of reactants into one particular product. This means that separate processes using different enzymes must be developed for chemically similar reactions, which is time-consuming and expensive. Thus far, enzymes have found only limited industrial applications, although they are used as ingredients in laundry detergents, contact lens cleaners, and meat tenderizers. The enzymes in these applications tend to be proteases, which are able to cleave the amide bonds that hold amino acids together in proteins. Meat tenderizers, for example, contain a protease called papain, which is isolated from papaya juice. It cleaves some of the long, fibrous protein molecules that make inexpensive cuts of beef tough, producing a piece of meat that is more tender. Some insects, like the bombadier beetle, carry an enzyme capable of catalyzing the decomposition of hydrogen peroxide to water ( \(\Page {3}\)). cause a decrease in the reaction rate of an enzyme-catalyzed reaction by binding to a specific portion of an enzyme and thus slowing or preventing a reaction from occurring. Irreversible inhibitors are therefore the equivalent of poisons in heterogeneous catalysis. One of the oldest and most widely used commercial enzyme inhibitors is aspirin, which selectively inhibits one of the enzymes involved in the synthesis of molecules that trigger inflammation. The design and synthesis of related molecules that are more effective, more selective, and less toxic than aspirin are important objectives of biomedical research. Catalysts participate in a chemical reaction and increase its rate. They do not appear in the reaction’s net equation and are not consumed during the reaction. Catalysts allow a reaction to proceed via a pathway that has a lower activation energy than the uncatalyzed reaction. In heterogeneous catalysis, catalysts provide a surface to which reactants bind in a process of adsorption. In homogeneous catalysis, catalysts are in the same phase as the reactants. Enzymes are biological catalysts that produce large increases in reaction rates and tend to be specific for certain reactants and products. The reactant in an enzyme-catalyzed reaction is called a substrate. Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction. | 7,265 | 1,255 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Thermodynamic_Cycles/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 | 1,258 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Basics_Thermodynamics_(General_Chemistry)/Gibbs_Energy |
Experience shows that, over time, the quotient tends to approach a constant for a given reaction in a closed system. Such a state is called , -> The constant depends on temperature and the nature of the reactants and products. Thus, is called the . This is known as the mass action law. In other words, there is a tendency for the reaction to reach a equilibrium such that Using activities covers a wider range of concentrations than using concentrations to define the reaction quotient or equilibrium constant . Concentrations were used in freshman chemistry for simplicity. Energy is the driving force for reactions. The tendency for a reaction to reach a equilibrium is driven by the Gibbs free energy, symbol and the relationship has been defined by the relationship D = - ln At the , activities of all reactants and products are unity (all equal to 1). In this system, = 1. If > 1, the the forward reaction is spontaneous, -> . The Gibb's free energy is a negative quantity for such a system. Since this is always true, we can generalize the condition. , 0 - ln as the system changes from standard condition to an equilibrium state. In contrast to the decrease in Gibb's energy, the entropy increases as an isolated system undergoes an spontaneous reaction. The generalized statement can be represented by a generalized Gibb's free energy change, , for a system not at standard condition, but whose reaction quotient is . Obviously, the formulation is D = D + ln . As the system strives to reach an equilibrium state, (no longer any net change), -> we have the following results, D = - ln
D + ln = 0
DG = 0. The previous discussion leads to the following conclusion. So far, the Gibb's free energy is defined as the driving force for a system to reach a chemical equilibrium. The energy comes from the enthalpy and entropy of reaction in the system, and has been define in terms of enthalpy and entropy changes, and , at temperature as: D = D - D . Since Gibb's energy, enthalpy, and entropy are state functions, they have been treated as the functions in thermodynamics, and as a result, the delta is omitted. The relation is simply, The generalized equation is very useful, and it can be differentiacted with respect to other thermodynamical variables. However, we will not discuss it any further at this point. For redox reactions, Gibb's energy is the , which, when properly setup in an electric cell, is the charge transferred ( in Coulomb) times the potential (in V). Each mole of electron has a charge of 1 faraday (1 = 96458 C), and moles of electron have a charge of . Since the voltage is usually a positive value, we have, D = -
Note units for the two quantities. The standard entropies in J/(mol K) are also the absolute entropies. Thus, = e = 2.0
Examples 1, 2, and 3 illustrate how can be derived from a table of thermodynamic data. In some tables, the value of are also given. Thus, = e
= 13.82*10 ,
a very large number indeed indicating a reaction to almost exhaust at least one of the reactants.
This example illustrates how you may use a thermodynamic data table. Again, we write the standard entropies below the formula 3/2 H (g) + ½ N (g) -> NH (g)
3/2*130.680 ½*191.609 192.77 = (products) - (reactants)
= 192.77 - ( / *130.680 + ½*191.609)
= -99.125 kJ/mol
The standard Gibb's free energy of formation is,
= -
= -45.94 - 298*(-0.099125) kJ/mol = -16.40 kJ/mol
Results from the previous and this examples are used in the next example. | 3,594 | 1,259 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/02%3A_Properties_of_Gases/2.09%3A_Graham's_Laws_of_Diffusion_and_Effusion |
Diffusion is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, effusion, is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space. The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship , after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy (recall that a result of the Kinetic Theory of Gases is that the temperature, in degrees Kelvin, is directly proportional to the average kinetic energy of the molecules.) . We can write the expression for the average kinetic energy of two gases with different molar masses: \[KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\label{2}\] Multiplying both sides by 2 and rearranging give \[\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\label{3}\] Taking the square root of both sides gives \[\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\label{4}\] Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation \(\ref{4}\) shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy. Heavy molecules effuse through a porous material more slowly than light molecules, as illustrated schematically in Figure \(\Page {1}\) for ethylene oxide \(C_2H_4O \) and helium \(He\). Helium ( = 4.00 g/mol) effuses much more rapidly than ethylene oxide ( = 44.0 g/mol). Because helium is less dense than air, helium-filled balloons “float” at the end of a tethering string. Unfortunately, rubber balloons filled with helium soon lose their buoyancy along with much of their volume. In contrast, rubber balloons filled with air tend to retain their shape and volume for a much longer time. Because helium has a molar mass of 4.00 g/mol, whereas air has an average molar mass of about 29 g/mol, pure helium effuses through the microscopic pores in the rubber balloon \(\sqrt{\frac{29}{4.00}}=2.7\) times faster than air. For this reason, high-quality helium-filled balloons are usually made of Mylar, a dense, strong, opaque material with a high molecular mass that forms films that have many fewer pores than rubber. Hence, mylar balloons can retain their helium for days. During World War II, scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of \(\ce{^{235}U}\). Naturally occurring uranium is only 0.720% \(\ce{^{235}U}\), whereas most of the rest (99.275%) is \(\ce{^{238}U}\), which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound \(UF_6\) (boiling point = 56°C). isotopic content of naturally occurring uranium and atomic masses of U and U ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure UF The molar mass of UF is The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law using Equation \(\ref{1}\):
\[\rm\dfrac{\text{rate }^{235}UF_6}{\text{rate }^{238}UF_6}=\sqrt{\dfrac{352.04\;g/mol}{349.03\;g/mol}}=1.0043\] Thus passing UF containing a mixture of the two isotopes through a single porous barrier gives an enrichment of 1.0043, so after one step the isotopic content is (0.720%)(1.0043) = 0.723% UF . In this case, 0.990 = (0.00720)(1.0043) , which can be rearranged to give
\[1.0043^n=\dfrac{0.990}{0.00720}=137.50\] Thus at least a thousand effusion steps are necessary to obtain highly enriched U. Figure \(\Page {2}\) shows a small part of a system that is used to prepare enriched uranium on a large scale. Helium consists of two isotopes: He (natural abundance = 0.000134%) and He (natural abundance = 99.999866%). Their atomic masses are 3.01603 and 4.00260, respectively. Helium-3 has unique physical properties and is used in the study of ultralow temperatures. It is separated from the more abundant He by a process of gaseous effusion. a. ratio of effusion rates = 1.15200; one step gives 0.000154% He; b. 96 steps Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 10 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line. Gaseous particles are in constant random motion. Gaseous particles tend to undergo diffusion because they have kinetic energy. Diffusion is faster at higher temperatures because the gas molecules have greater kinetic energy. Effusion refers to the movement of gas particles through a small hole. Graham's Law states that the effusion rate of a gas is inversely proportional to the square root of the mass of its particles. | 6,375 | 1,260 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/14%3A_Organohalogen_and_Organometallic_Compounds/14.11%3A_Preparation_of_Organometallic_Compounds |
The reaction of a metal with an organic halide is a convenient method for preparation of organometallic compounds of reasonably active metals such as lithium, magnesium, and zinc. Ethers, particularly diethyl ether and oxacyclopentane (tetrahydrofuran), provide inert, slightly polar media in which organometallic compounds usually are soluble. Care is necessary to exclude moisture, oxygen, and carbon dioxide, which would react with the organometallic compound. This can be accomplished by using an inert atmosphere of nitrogen or helium. The reactivity order of the halides is \(\ce{I} \: > \: \ce{Br} \: > \: \ce{Cl} \: \gg \ce{F}\). Whereas magnesium and lithium react well with chlorides, bromides, and iodides, zinc is satisfactory only with bromides and iodides. Mercury only reacts when amalgamated with sodium. Sodium and potassium present special problems because of the high reactivity of alkylsodium and alkylpotassium compounds toward ether and organic halides. Alkane solvents usually are necessary. Alkenyl, alkynyl, and aryl halides, like alkyl halides, can be converted to the corresponding magnesium and lithium compounds. However, the reaction conditions, such as choice of solvent, can be critical. Bromoethene, for instance, can be converted to ethenylmagnesium bromide in good yield if the solvent is oxacyclopentane [tetrahydrofuran, \(\ce{(CH_2)_4O}\)]: The more reactive allylic and benzylic halides present a problem - not so much in forming the organometallic derivative as in keeping it from reacting further with the starting halide. An often unwanted side reaction in the preparation of organometallic compounds is a displacement reaction, probably of the \(S_\text{N}2\) type: This problem can be lessened greatly by using a large excess of magnesium and dilute solutions of the allylic halide to minimize the coupling reaction. The same difficulty also occurs in the preparation of alkylsodium compounds. The starting halide \(\ce{RX}\) couples with \(\ce{RNa}\) (to give \(\ce{R-R}\) and \(\ce{NaX}\)) or is converted to an alkene. These reactions appear to involve radical intermediates undergoing combination and disproportionation ( ): In the absence of metallic sodium, ethylsodium probably still reacts with ethyl bromide by a radical reaction rather than \(S_\text{N}2\) or \(E2\). This happens because \(\ce{CH_3CH_2^-}\) tends to lose an electron easily and can act like metallic sodium to donate an electron to \(\ce{CH_3CH_2Br}\) to form an ethyl radical and itself become an ethyl radical: Reactions between the resulting radicals then produce butane, ethane, and ethene. The point at which one can expect \(S_\text{N}2\) and \(E2\) reactions to go faster than radical formation as the structures of the halides and the nature of the metal are changed is not yet clearly defined. However, it is becoming increasingly evident that there are substitution reactions of "unactivated" aryl halides that proceed without rearrangement by way of radical intermediates. The key step in these reactions is donation of an electron to one of the unfilled \(\pi\) orbitals of the ring and subsequent ejection of a halide ion: Such a mechanism probably is involved in the formation of organometallic compounds from aryl halides and metals. Brief descriptions follow of less general but very useful methods of forming organometallic compounds (also see Table 14-7). In each of these preparations the solvent must be inert to all of the organometallic compounds involved. \[\ce{RBr} + \ce{R'Li} \rightleftharpoons \ce{RLi} + \ce{R'Br}\] The equilibrium in these reactions favors formation of the organometallic compound with the metal attached to the more electronegative \(\ce{R}\) group. The method is mainly used in the preparation of organolithium compounds derived from unreactive halides such as aryl, ethenyl, or ethynyl halides. These halides do not always react readily with lithium metal, but may react well with butyllithium: \[\ce{R_2Hg} + 2 \ce{Na} \rightleftharpoons 2 \ce{RNa} + \ce{Hg}\] Here the equilibrium is such that the \(\ce{R}\) group favors attachment to the more electropositive metal. \[\ce{RMgCl} + \ce{HgCl_2} \rightleftharpoons \ce{RHgCl} + \ce{MgCl_2}\] \[\ce{RLi} + \ce{CuI} \rightleftharpoons \ce{RCu} + \ce{LiI}\] The equilibrium favors the products with \(\ce{R}\) connected to the less electropositive metal so the reaction tends to form a less reactive organometallic compound from a more reactive one. Some organometallic compounds are prepared best by the reaction of a strong base or an alkyl metal derivative with an acidic hydrocarbon, such as an alkyne: An especially important example is that of 1,3-cyclopentadiene, which is acidic because its conjugate base (cyclopentadienide anion) is greatly stabilized by electron delocalization. The anion is formed easily from the hydrocarbon and methyllithium: Diorganometallic compounds cannot be prepared from dihalides if the halogens are separated by three \(\ce{C-C}\) bonds or less because elimination or other reactions usually predominate. With active metals and 1,1-, 1,2-, or 1,3-dihalides, the following reactions normally occur: When the halogens are at least four carbons apart a diorganometallic compound can be formed: Carbenes, \(\ce{R_2C} :\) ( ) are produced by \(\alpha\) eliminations from polyhalogen compounds with organometallic reagents. The first step is halogen-metal exchange and this is followed by elimination of metal halide: Elimination reactions of this type can be useful in synthesis for the formation of carbon-carbon bonds. For example, if dibromocarbene is generated in the presence of an alkene, it will react by cycloaddition to give a cyclopropane derivative: A related example is the generation of benzyne from 1-bromo-2-fluorobenzene with magnesium in oxacyclopentane (tetrahydrofuran). If the temperature is kept around \(0^\text{o}\), 2-fluorophenylmagnesium bromide is formed. At higher temperatures, magnesium halide is eliminated and benzyne results: If a diene is present, the benzyne will react with it by a [4 + 2] cycloaddition as in the following example: and (1977) | 6,151 | 1,261 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Electrophilic_Substitution_Reactions/H._Some_Substitution_Reactions_of_Methylbenzene |
This page discusses the problems which arise if you try to write the mechanism for an electrophilic substitution reaction into a benzene ring which already has something else attached to it. There are two problems you might come across: If you substitute a nitro group, -NO , into the benzene ring in methylbenzene, you could possibly get any of the following products: In the case of methylbenzene, whatever you attach to the ring, you always get a mixture consisting mainly of the 2- and 4- isomers. The methyl group is said to be 2,4-directing, in the sense that it seems to "push" incoming groups into those positions. Some other groups which might already be on the ring (for example, the -NO group in nitrobenzene) "push" incoming groups into the 3- position. Reacting methylbenzene with a mixture of concentrated nitric and sulphuric acids gives both 2-nitromethylbenzene and 4-nitromethylbenzene. The mechanism is exactly the same as the nitration of benzene. You just have to be careful about the way that you draw the structure of the intermediate ion. This just shows the first step of the electrophilic substitution reaction. Notice that the partial delocalisation in the intermediate ion covers all the carbon atoms in the ring except for the one that the -NO group gets attached to. That is the only point of interest in this example - everything else is just the same as with the nitration of benzene. The hydrogen atom is then removed by an HSO ion - exactly as in the benzene case. Once again, the only point of interest is in the way the partial delocalisation in the intermediate ion is drawn - again, it covers all the carbon atoms in the ring apart from the one with the -NO group attached. This is a good example of a case where what is already attached to the ring can also get involved in the reaction. It is possible to get two quite different substitution reactions between methylbenzene and chlorine depending on the conditions used. The chlorine can substitute into the ring or into the methyl group. Here we are only interested in substitution into the ring. This happens in the presence of aluminium chloride or iron, and in the absence of UV light. Substituting into the ring gives a mixture of 2-chloromethylbenzene and 4-chloromethylbenzene. The mechanisms are exactly the same as the substitution of chlorine into benzene - although you would have to be careful about the way you draw the intermediate ion. For example, the complete mechanism for substitution into the 4- position is: Methyl groups direct new groups into the 2- and 4- positions, but a nitro group, -NO , already on the ring directs incoming groups into the 3- position. For example, if the temperature is raised above 50°C, the nitation of benzene doesn't just produce nitrobenzene - it also produces some 1,3-dinitrobenzene. A second nitro group is substituted into the ring in the 3- position. The mechanism is exactly the same as the nitration of benzene or of methylbenzene - you just have to be careful in drawing the intermediate ion. Draw the partial delocalisation to include all the carbons except for the one the new -NO group gets attached to. In the second stage, the hydrogen atom is then removed by an HSO ion - exactly as in the benzene case. This isn't shown because there's nothing new. Jim Clark ( ) | 3,337 | 1,262 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/05%3A_Dioxygen_Reactions/5.05%3A_Dioxygen_Toxicity |
Before we consider the enzymatically controlled reactions of dioxygen in living systems, it is instructive to consider the uncontrolled and deleterious reactions that must also occur in aerobic organisms. Life originally appeared on Earth at a time when the atmosphere contained only low concentrations of dioxygen, and was reducing rather than oxidizing, as it is today. With the appearance of photosynthetic organisms approximately 2.5 billion years ago, however, the conversion to an aerobic, oxidizing atmosphere exposed the existing anaerobic organisms to a gradually increasing level of oxidative stress. Modern-day anaerobic bacteria, the descendants of the original primitive anaerobic organisms, evolved in ways that enabled them to avoid contact with normal atmospheric concentrations of dioxygen. Modern-day aerobic organisms, by contrast, evolved by developing aerobic metabolism to harness the oxidizing power of dioxygen and thus to obtain usable metabolic energy. This remarkably successful adaptation enabled life to survive and flourish as the atmosphere became aerobic, and also allowed larger, multicellular organisms to evolve. An important aspect of dioxygen chemistry that enabled the development of aerobic metabolism is the relatively slow rate of dioxygen reactions in the absence of catalysts. Thus, enzymes could be used to direct and control the oxidation of substrates either for energy generation or for biosynthesis. Nevertheless, the balance achieved between constructive and destructive oxidation is a delicate one, maintained in aerobic organisms by several means, e.g.: compartmentalization of oxidative reactions in mitochondria, peroxisomes, and chloroplasts; scavenging or detoxification of toxic byproducts of dioxygen reactions; repair of some types of oxidatively damaged species; and degradation and replacement of other species. The classification "anaerobic" actually includes organisms with varying degrees of tolerance for dioxygen: strict anaerobes, for which even small concentrations of O are toxic; moderate anaerobes, which can tolerate low levels of dioxygen; and microaerophiles, which require low concentrations of O for growth, but cannot tolerate normal atmospheric concentrations, i.e., 21 percent O , 1 atm pressure. Anaerobic organisms thrive in places protected from the atmosphere, for example, in rotting organic material, decaying teeth, the colon, and gangrenous wounds. Dioxygen appears to be toxic to anaerobic organisms largely because it depletes the reducing equivalents in the cell that are needed for normal biosynthetic reactions. Aerobic organisms can, of course, live in environments in which they are exposed to normal atmospheric concentrations of O . Nevertheless, there is much evidence that O is toxic to these organisms as well. For example, plants grown in varying concentrations of O have been observed to grow faster in lower than normal concentrations of O . grown under 5 atm of O ceased to grow unless the growth medium was supplemented with branched-chain amino acids or precursors. High concentrations of O damaged the enzyme dihydroxy acid dehydratase, an important component in the biosynthetic pathway for those amino acids. In mammals, elevated levels of O are clearly toxic, leading first to coughing and soreness of the throat, and then to convulsions when the level of 5 atm of 100 percent O is reached. Eventually, elevated concentrations of O lead to pulmonary edema and irreversible lung damage, with obvious damage to other tissues as well. The effects of high concentrations of O on humans is of some medical interest, since dioxygen is used therapeutically for patients experiencing difficulty breathing, or for those suffering from infection by anaerobic organisms. The major biochemical targets of O toxicity appear to be lipids, DNA, and proteins. The chemical reactions accounting for the damage to each type of target are probably different, not only because of the different reactivities of these three classes of molecules, but also because of the different environment for each one inside the cell. Lipids, for example, are essential components of membranes and are extremely hydrophobic. The oxidative damage that is observed is due to free-radical autoxidation (see Reactions 5.16 to 5.21), and the products observed are lipid hydroperoxides (see Reaction 5.23). The introduction of the hydroperoxide group into the interior of the lipid bilayer apparently causes that structure to be disrupted, as the configuration of the lipid rearranges in order to bring that polar group out of the hydrophobic membrane interior and up to the membrane-water interface. DNA, by contrast, is in the interior of the cell, and its exposed portions are surrounded by an aqueous medium. It is particularly vulnerable to oxidative attack at the base or at the sugar, and multiple products are formed when samples are exposed to oxidants . Since oxidation of DNA may lead to mutations, this type of damage is potentially very serious. Proteins also suffer oxidative damage, with amino-acid side chains, particularly the sulfur-containing residues cysteine and methionine, appearing to be the most vulnerable sites. The biological defense systems protecting against oxidative damage and its consequences are summarized below. Some examples of small-molecule antioxidants are \(\alpha\)-tocopherol (vitamin E; 5.24), which is found dissolved in cell membranes and protects them against lipid peroxidation, and ascorbate (vitamin C; 5.25) and glutathione (5.26), which are found in the cytosol of many cells. Several others are known as well. \(\tag{5.24}\) \(\tag{5.25}\) \(\tag{5.26}\) The enzymatic antioxidants are (a) catalase and the various peroxidases, whose presence lowers the concentration of hydrogen peroxide, thereby preventing it from entering into potentially damaging reactions with various cell components (see Section VI and Reactions 5.82 and 5.83), and (b) the superoxide dismutases, whose presence provides protection against dioxygen toxicity that is believed to be mediated by the superoxide anion, O (see Section VII and Reaction 5.95). Some of the enzymatic and nonenzymatic antioxidants in the cell are illustrated in Figure 5.1. Redox-active metal ions are present in the cell in their free, uncomplexed state only in extremely low concentrations. They are instead sequestered by metal-ion storage and transport proteins, such as ferritin and transferrin for iron (see Chapter 1) and ceruloplasmin for copper. This arrangement prevents such metal ions from catalyzing deleterious oxidative reactions, but makes them available for incorporation into metalloenzymes as they are needed. In vitro experiments have shown quite clearly that redox-active metal ions such as Fe or Cu are extremely good catalysts for oxidation of sulfhydryl groups by O (Reaction 5.27). \[4RSH + O_{2} \xrightarrow{M^{n+}} 2RSSR + 2H_{2}O \tag{5.27}\] In addition, in the reducing environment of the cell, redox-active metal ions catalyze a very efficient one-electron reduction of hydrogen peroxide to produce hydroxyl radical, one of the most potent and reactive oxidants known (Reactions 5.28 to 5.30). \[M^{n+} + Red^{-} \rightarrow M^{(n-1)+} + Red \tag{5.28}\] \[M^{(n-1)+} + H_{2}O_{2} \rightarrow M^{n+} + OH^{-} + HO \cdotp \tag{5.29}\] \[Red^{-} + H_{2}O_{2} \rightarrow Red + OH^{-} + HO \cdotp \tag{5.30}\] \[(Red^{-} = reducing\; agent)\] Binding those metal ions in a metalloprotein usually prevents them from entering into these types of reactions. For example, transferrin, the iron-transport enzyme in serum, is normally only 30 percent saturated with iron. Under conditions of increasing iron overload, the empty iron-binding sites on transferrin are observed to fill, and symptoms of iron poisoning are not observed until after transferrin has been totally saturated with iron. Ceruloplasmin and metallothionein may playa similar role in preventing copper toxicity. It is very likely that both iron and copper toxicity are largely due to catalysis of oxidation reactions by those metal ions. Repair of oxidative damage must go on constantly, even under normal conditions of aerobic metabolism. For lipids, repair of peroxidized fatty-acid chains is catalyzed by phospholipase A , which recognizes the structural changes at the lipid-water interface caused by the fatty-acid hydroperoxide, and catalyzes removal of the fatty acid at that site. The repair is then completed by enzymatic reacylation. Although some oxidatively damaged proteins are repaired, more commonly such proteins are recognized, degraded at accelerated rates, and then replaced. For DNA, several multi-enzyme systems exist whose function is to repair oxidatively damaged DNA. For example, one such system catalyzes recognition and removal of damaged bases, removal of the damaged part of the strand, synthesis of new DNA to fill in the gaps, and religation to restore the DNA to its original, undamaged state. Mutant organisms that lack these repair enzymes are found to be hypersensitive to O , H O , or other oxidants. One particularly interesting aspect of oxidant stress is that most aerobic organisms can survive in the presence of normally lethal levels of oxidants if they have first been exposed to lower, nontoxic levels of oxidants. This phenomenon has been observed in animals, plants, yeast, and bacteria, and suggests that low levels of oxidants cause antioxidant systems to be induced . In certain bacteria, the mechanism of this induction is at least partially understood. A DNA-binding regulatory protein named OxyR that exists in two redox states has been identified in these systems. Increased oxidant stress presumably increases concentration of the oxidized form, which then acts to turn on the transcription of the genes for some of the antioxidant enzymes. A related phenomenon may occur when bacteria and yeast switch from anaerobic to aerobic metabolism. When dioxygen is absent, these microorganisms live by fermentation, and do not waste energy by synthesizing the enzymes and other proteins needed for aerobic metabolism. However, when they are exposed to dioxygen, the synthesis of the respiratory apparatus is turned on. The details of this induction are not known completely, but some steps at least depend on the presence of heme, the prosthetic group of hemoglobin and other heme proteins, whose synthesis requires the presence of dioxygen. What has been left out of the preceding discussion is the identification of the species responsible for oxidative damage, i.e., the agents that directly attack the various vulnerable targets in the cell. They were left out because the details of the chemistry responsible for dioxygen toxicity are largely unknown. In 1954, Rebeca Gerschman formulated the "free-radical theory of oxygen toxicity" after noting that tissues subjected to ionizing radiation resemble those exposed to elevated levels of dioxygen. Fourteen years later, Irwin Fridovich proposed that the free radical responsible for dioxygen toxicity was superoxide, O , based on his identification of the first of the superoxide dismutase enzymes. Today it is still not known if superoxide is the principal agent of dioxygen toxicity, and, if so, what the chemistry responsible for that toxicity is. There is no question that superoxide is formed during the normal course of aerobic metabolism, although it is difficult to obtain estimates of the amount under varying conditions, because, even in the absence of a catalyst, superoxide disproportionates quite rapidly to dioxygen and hydrogen peroxide (Reaction 5.4) and therefore never accumulates to any great extent in the cell under normal conditions of pH. One major problem in this area is that a satisfactory chemical explanation for the purported toxicity of superoxide has never been found, despite much indirect evidence from experiments that the presence of superoxide can lead to undesirable oxidation of various cell components and that such oxidation can be inhibited by superoxide dismutase. The mechanism most commonly proposed is production of hydroxyl radicals via Reactions (5.28) to (5.30) with Red = O , which is referred to as the "Metal-Catalyzed Haber-Weiss Reaction". The role of superoxide in this mechanism is to reduce oxidized metal ions, such as Cu or Fe , present in the cell in trace amounts, to a lower oxidation state. Hydroxyl radical is an extremely powerful and indiscriminate oxidant. It can abstract hydrogen atoms from organic substrates, and oxidize most reducing agents very rapidly. It is also a very effective initiator of free-radical autoxidation reactions (see Section II.C above). Therefore, reactions that produce hydroxyl radical in a living cell will probably be very deleterious. The problem with this explanation for superoxide toxicity is that the only role played by superoxide here is that of a reducing agent of trace metal ions. The interior of a cell is a highly reducing environment, however, and other reducing agents naturally present in the cell such as, for example, ascorbate anion can also act as Red in Reaction (5.28), and the resulting oxidation reactions due to hydroxyl radical are therefore no longer inhibitable by SOD. Other possible explanations for superoxide toxicity exist, of course, but none has ever been demonstrated experimentally. Superoxide might bind to a specific enzyme and inhibit it, much as cytochrome oxidase is inhibited by cyanide or hemoglobin by carbon monoxide. Certain enzymes may be extraordinarily sensitive to direct oxidation by superoxide, as has been suggested for the enzyme aconitase, an iron-sulfur enzyme that contains an exposed iron atom. Another possibility is that the protonated and therefore neutral form of superoxide, HO , dissolves in membranes and acts as an initiator of lipid peroxidation. It has also been suggested that superoxide may react with nitric oxide, NO, in the cell producing peroxynitrite, a very potent oxidant. One particularly appealing mechanism for superoxide toxicity that has gained favor in recent years is the "Site-Specific Haber-Weiss Mechanism." The idea here is that traces of redox-active metal ions such as copper and iron are bound to macromolecules under normal conditions in the cell. Most reducing agents in the cell are too bulky to come into close proximity to these sequestered metal ions. Superoxide, however, in addition to being an excellent reducing agent, is very small, and could penetrate to these metal ions and reduce them. The reduced metal ions could then react with hydrogen peroxide, generating hydroxyl radical, which would immediately attack at a site near the location of the bound metal ion. This mechanism is very similar to that of the metal complexes that cause DNA cleavage; by reacting with hydrogen peroxide while bound to DNA, they generate powerful oxidants that react with DNA with high efficiency because of their proximity to it (see Chapter 8). Although we are unsure what specific chemical reactions superoxide might undergo inside of the cell, there nevertheless does exist strong evidence that the superoxide dismutases play an important role in protection against dioxygen-induced damage. Mutant strains of bacteria and yeast that lack superoxide dismutases are killed by elevated concentrations of dioxygen that have no effect on the wild-type cells. This extreme sensitivity to dioxygen is alleviated when the gene coding for a superoxide dismutase is reinserted into the cell, even if the new SOD is of another type and from a different organism. In summary, we know a great deal about the sites that are vulnerable to oxidative damage in biological systems, about the agents that protect against such damage, and about the mechanisms that repair such damage. Metal ions are involved in all this chemistry, both as catalysts of deleterious oxidative reactions and as cofactors in the enzymes that protect against and repair such damage. What we still do not know at this time, however, is how dioxygen initiates the sequence of chemical reactions that produce the agents that attack the vulnerable biological targets | 16,226 | 1,263 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Exercises%3A_Physical_and_Theoretical_Chemistry/Data-Driven_Exercises/Adsorption_Isotherms_-_Methylene_Blue_on_Activated_Carbon |
: Data is presented for the adsorption of methylene blue onto activated carbon from aqueous solution. The results are analyzed within the context of two common adsorption models; the Langmuir and Freundlich adsorption isotherms. : An introductory knowledge of dynamic equilibria, rate laws, and using adsorption spectroscopy data to determine solute concentrations. : This exercise should be carried out within a software environment that is capable of data manipulation and which can generate a best-fit line for an x-y data set. You will also be graphing the data along with the fitted function. Chemical reactions are often classified as being either homogeneous or heterogeneous. In the later case, the reactants and catalyst (if any) exist in different phases, and consequently the reaction occurs at the phase boundary. Many important industrial reactions fall into this category, including the classic Haber ammonia synthesis, whereby nitrogen and hydrogen gas are catalytically combined on a metal surface at elevated pressures. Modeling the adsorption of various solutes on a solid substrate remains an active area of research which is advancing our understanding of catalytic processes and many biological reactions. One of the initial models for the adsorption of a species onto a simple surface was put forth by Irving Langmuir in 1916. Langmuir assumed that a surface consists of a given number of equivalent sites where a species can physically or chemically stick. Physical adsorption through van der Waals interactions is called physisorption, whereas chemical adsorption through the formation of a covalent bond is called . It is important to realize that the processes of adsorption and the opposite process ( ) is dynamic; a rate law can be written for each process, and when the rates become equal an equilibrium state will exist characterized by a constant fractional coverage of the original sites. Letting \(x_o\) represent the total concentration of available sites on a given amount of fresh solid substrate, we can define a fractional coverage (\(θ\)) as \[\theta = \dfrac{x}{x_o} \label{1}\] where x is the concentration of occupied sites. The rate of adsorption va will be proportional to the concentration of gas or liquid (c) above the surface and the fraction of the surface that is not covered \((1-θ)\), yielding a rate equation \[v_a = k_Ac(1-\theta) \label{2}\] where \(k_a\) is the rate constant for adsorption. The rate of desorption is simply proportional to the fraction of the surface that is already occupied, so the rate equation is \[v_d = k_d \theta \label{3}\] and kd is the rate constant for desorption. Setting Equations \ref{2} and \ref{3} equal yields an equilibrium statement that can be written as \[\dfrac{\theta }{1-\theta} = \dfrac{k_a}{k_d} c \label{4}\] The ratio of rate constants in Equation \ref{4} is equal to an equilibrium constant \[K = \dfrac{k_a}{k_d}.\] Upon substitution and further rearrangement, the fractional coverage is given by \[ \theta = \dfrac{Kc}{1+Kc} \label{5}\] Equation \ref{5} is plotted in Figure \(\Page {1}\) for an arbitrary value of \(K\), illustrating how the surface sites becomes saturated as the concentration rises. The magnitude of K quantifies the relative affinity that a given solute has for surface adsorption. Like all equilibrium constants, K is temperature dependent. A plot like the one shown in figure 1 is often called a Langmuir absorption isotherm. For many studies, it is more convenient to substitute \(x/x_o\) back in place of \(θ\) and rearrange Equation \ref{5} into the following form for the concentration of occupied sites; \[ x = \dfrac{x_oKc}{1+Kc} \label{6}\] In this case, \(K\) and \(x_o\) are both constants for a system. They are normally evaluated by writing Equation \ref{6} in the reciprocal form \[ \dfrac{1}{x} = \left( \dfrac{1}{x_mK} \right) \dfrac{1}{c} + \dfrac{1}{x_m} \label{7}\] \(x\) is measured experimentally at different values of \(c\) and a plot of \(1/x\) versus \(1/c\) yields linear data where the slope and intercept can be used to evaluate the constants \(K\) and \(x_o\). How well does the Langmuir model work? In cases where there is negligible intermolecular interaction between adsorbed solute particles and where only one monolayer of solute can potentially adsorb, the Langmuir model works quite well. As such, the model can be thought of as a limiting law (much like the ideal gas equation). Several more sophisticated models have since been developed that account for deviations from Langmuir adsorption behavior. One such model is called the Freundlich isotherm (proposed by Herbert Max Finlay Freundlich, 1880-1941) and is described by the equation \[ x =k c^{1/n} \label{8}\] where \(k\) and \(n\) are empirical constants (and x and c have the same meaning as before). While originally proposed by Freundlich as an empirical law, Equation \ref{8} can be arrived at theoretically by taking into account repulsive interactions between adsorbed solute particles and also accounting for surface heterogeneities. To determine the constants \(k\) and \(n\), Equation \ref{8} is rearranged into the form \[ \log_{10} x = \dfrac{1}{n} \log_{10} c + log_{10} k \label{9}\] A plot of \(\log_{10} x\) versus \(\log_{10} c\) should be linear and the slope and intercept yield the empirical constants.
In the exercise below, you will analyze data corresponding to the adsorption of dye called methylene blue onto activated carbon from solution using both the Langmuir and Freundlich adsorption isotherm models. The following data are based upon measurements described in J.H. Potgieter, Journal of Chemical Education, 68, 349 (1991). For each trial, a given amount of activated carbon (indicated below in milligrams) was measured into a small flask. 100 mL of a 25 mg/L methylene blue stock solution was then added to each flask. The flasks were agitated on a mechanical shaker for 72 hours at 25 °C, at which point the solutions were filtered and the amount of methylene blue remaining in solution was measured by spectrophotometry at 630 nm. The absorbance readings are reported in the table below relative to the absorbance of the methylene blue stock solution (which had an absorbance Ao). | 6,252 | 1,264 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_An_Introduction_to_the_Electronic_Structure_of_Atoms_and_Molecules_(Bader)/03%3A_The_Hydrogen_Atom/3.01%3A_Introduction |
The study of the hydrogen atom is more complicated than our previous example of an electron confined to move on a line. Not only does the motion of the electron occur in three dimensions but there is also a force acting on the electron. This force, the electrostatic force of attraction, is responsible for holding the atom together. The magnitude of this force is given by the product of the nuclear and electronic charges divided by the square of the distance between them. In the previous example of an electron confined to move on a line, the total energy was entirely kinetic in origin since there were no forces acting on the electron. In the hydrogen atom however, the energy of the electron, because of the force exerted on it by the nucleus, will consist of a potential energy (one which depends on the position of the electron relative to the nucleus) as well as a kinetic energy. The potential energy \(V(r)\) arising from the force of attraction between the nucleus and the electron is: \[V(r) = \dfrac{-e^2}{r}\] Let us imagine for the moment that the proton and the electron behave classically. Then, if the nucleus is held fixed at the origin and the electron allowed to move relative to it, the potential energy would vary in the manner indicated in . The potential energy is independent of the direction in space and depends only on the distance between the electron and the nucleus. Thus refers to any line directed from the nucleus to the electron. The -axis in the figure may be taken literally as a line through the nucleus. Whether the electron moves to the right or to the left the potential energy varies in the same manner. The potential energy is zero when the two particles are very far apart ( = ), and equals minus infinity when equals zero. We shall take the energy for = as our zero of energy. Every energy will be measured relative to this value. When a stable atom is formed, the electron is attracted to the nucleus, is less than infinity, and the energy will be negative. A negative value for the energy implies that energy must be supplied to the system if the electron is to overcome the attractive force of the nucleus and escape from the atom. The electron has again "fallen into a potential well." However, the shape of the well is no longer a simple square one as previously considered for an electron confined to move on a line, but has the shape shown in . This shape is a consequence of there being a force acting on the electron and hence a potential energy contribution which depends on the distance between the two particles. This is the nature of the problem. Now let us see what quantum mechanics predicts for the motion of the electron in such a situation. | 2,741 | 1,265 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/11%3A_Solutions/11.6%3A_Phase_Equilibrium_in_Solutions_-_Volatile_Solutes |
External pressure has very little effect on the solubility of liquids and solids. In contrast, the solubility of gases increases as the partial pressure of the gas above a solution increases. This point is illustrated in Figure \(\Page {1}\), which shows the effect of increased pressure on the dynamic equilibrium that is established between the dissolved gas molecules in solution and the molecules in the gas phase above the solution. Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures. The relationship between pressure and the solubility of a gas is described quantitatively by Henry’s law, which is named for its discoverer, the English physician and chemist, William Henry (1775–1836): \[C = k_HP \label{11.61}\] where \(C\) is the concentration of dissolved gas at equilibrium, \(P\) is the partial pressure of the gas, and \(k_H\) is the Henry’s law constant, which must be determined experimentally for each combination of gas, solvent, and temperature. Although the gas concentration may be expressed in any convenient units, we will use molarity exclusively. The units of the Henry’s law constant are therefore mol/(L·atm) = M/atm. Values of the Henry’s law constants for solutions of several gases in water at 20°C are listed in Table \(\Page {1}\). As the data in Table \(\Page {1}\) demonstrate, the concentration of a dissolved gas in water at a given pressure depends strongly on its physical properties. For a series of related substances, London dispersion forces increase as molecular mass increases. Thus among the elements of , the Henry’s law constants increase smoothly from \(\ce{He}\) to \(\ce{Ne}\) to \(\ce{Ar}\). The table also shows that \(\ce{O_2}\) is almost twice as soluble as \(\ce{N_2}\). Although are too weak to explain such a large difference, \(\ce{O_2}\) is and hence more polarizable than \(\ce{N_2}\), which explains its high solubility. Gases that react chemically with water, such as \(\ce{HCl}\) and the other hydrogen halides, \(\ce{H_2S}\), and \(\ce{NH_3}\), do not obey Henry’s law; all of these gases are much more soluble than predicted by Henry’s law. For example, \(\ce{HCl}\) reacts with water to give \(\ce{H^{+}(aq)}\) and \(\ce{Cl^{-}(aq)}\), not dissolved \(\ce{HCl}\) molecules, \[\ce{HCl(g) + H2O(l) \rightarrow H3O^{+}(aq) + Cl^{-}(aq)}
\nonumber\] The dissociation of \(\ce{HCl}\) into ions results in a much higher effective "solubility" than expected for a neutral molecule. Henry’s law has important applications. For example, bubbles of \(\ce{CO_2}\) form as soon as a carbonated beverage is opened because the drink was bottled under \(\ce{CO_2}\) at a pressure greater than 1 atm. When the bottle is opened, the pressure of \(\ce{CO_2}\) above the solution drops rapidly, and some of the dissolved gas escapes from the solution as bubbles. Henry’s law also explains why scuba divers have to be careful to ascend to the surface slowly after a dive if they are breathing compressed air. At the higher pressures under water, more \(\ce{N_2}\) from the air dissolves in the diver’s internal fluids. If the diver ascends too quickly, the rapid pressure change causes small bubbles of \(\ce{N_2}\) to form throughout the body, a condition known as “the bends.” These bubbles can block the flow of blood through the small blood vessels, causing great pain and even proving fatal in some cases. Due to the low Henry’s law constant for \(\ce{O_2}\) in water, the levels of dissolved oxygen in water are too low to support the energy needs of multicellular organisms, including humans. To increase the \(\ce{O_2}\) concentration in internal fluids, organisms synthesize highly soluble carrier molecules that bind \(\ce{O_2}\) reversibly. For example, human red blood cells contain a protein called hemoglobin that specifically binds \(\ce{O_2}\) and facilitates its transport from the lungs to the tissues, where it is used to oxidize food molecules to provide energy. The concentration of hemoglobin in normal blood is about 2.2 mM, and each hemoglobin molecule can bind four \(\ce{O_2}\) molecules. Although the concentration of dissolved \(\ce{O_2}\) in blood serum at 37°C (normal body temperature) is only 0.010 mM, the total dissolved \(\ce{O_2}\) concentration is 8.8 mM, almost a thousand times greater than would be possible without hemoglobin. Synthetic oxygen carriers based on fluorinated alkanes have been developed for use as an emergency replacement for whole blood. Unlike donated blood, these “blood substitutes” do not require refrigeration and have a long shelf life. Their very high Henry’s law constants for \(\ce{O_2}\) result in dissolved oxygen concentrations comparable to those in normal blood. The Henry’s law constant for \(\ce{O_2}\) in water at 25°C is \(1.27 \times 10^{-3} M/atm\), and the mole fraction of \(\ce{O_2}\) in the atmosphere is 0.21. Calculate the solubility of \(\ce{O_2}\) in water at 25°C at an atmospheric pressure of 1.00 atm. : Henry’s law constant, mole fraction of \(\ce{O_2}\), and pressure : : A According to Dalton’s law, the partial pressure of \(O_2\) is proportional to the mole fraction of \(O_2\): \[P_A = \chi_A P_t = (0.21)(1.00\; atm) = 0.21\; atm
\nonumber\] B From Henry’s law, the concentration of dissolved oxygen under these conditions is \[CO_2=k_H P_{O_2}=(1.27 \times 10^{-3}\; M/\cancel{atm}) (0.21\; \cancel{atm}) =2.7 \times 10^{-4}\; M
\nonumber\] To understand why soft drinks “fizz” and then go “flat” after being opened, calculate the concentration of dissolved \(CO_2\) in a soft drink: 0.17 M \(1 \times 10^{-5} M\) The solubility of most substances depends strongly on the temperature and, in the case of gases, on the pressure. The solubility of most solid or liquid solutes increases with increasing temperature. The components of a mixture can often be separated using fractional crystallization, which separates compounds according to their solubilities. The solubility of a gas decreases with increasing temperature. Henry’s law describes the relationship between the pressure and the solubility of a gas. | 6,231 | 1,266 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Enthalpy/Heat_of_Vaporization |
Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called vaporization or evaporation, generates a vapor pressure above the liquid. The Heat of Vaporization (also called the Enthalpy of Vaporization) is the heat required to induce this . Since vaporization requires heat to be added to the system and hence is an , therefore \( \Delta H_{vap} > 0\) as defined: \[ \Delta H_{vap} = H_{vapor} - H_{liquid}\] where Heat is absorbed when a liquid boils because molecules which are held together by and are jostled free of each other as the gas is formed. Such a separation requires energy (in the form of heat). In general the energy needed differs from one liquid to another depending on the magnitude of the intermolecular forces. We can thus expect liquids with strong intermolecular forces to have larger enthalpies of vaporization. The list of enthalpies of vaporization given in the bears this out. If a liquid uses 50 Joules of heat to vaporize one mole of liquid, then what would be the enthalpy of vaporization? The heat in the process is equal to the change of enthalpy, which involves vaporization in this case \[q_{tot} = \Delta_{vap}\] so \[q_{tot} = 50 \; J= \Delta_{vap}\] So the enthalpy of vaporization for one mole of substance is 50 J. The kinetic energy of the molecules in the gas and the silquid are the same since the vaporization process occues at constant temperature. However, the add thermal energy is used to break the potential energies of the intermolecular forces in the liquid, to generate molecules in the gas that are free of potential energy (for an ideal gass). Thus, while \(H_{vapor} > H_{liquid}\), the kinetic energies of the molecules are equal. Condensation is the opposite of vaporization, and therefore \( \Delta H_{condensation}\) is also the opposite of \( \Delta H_{vap}\). Because \( \Delta H_{vap}\) is an endothermic process, where heat is lost in a reaction and must be added into the system from the surroundings, \( \Delta H_{condensation}\) is an exothermic process, where heat is absorbed in a reaction and must be given off from the system into the surroundings. \[\begin{align} ΔH_{condensation} &= H_{liquid} - H_{vapor} \\[4pt] &= -ΔH_{vap} \end{align}\] Because \(ΔH_{condensation}\), also written as \(ΔH_{cond}\), is an exothermic process, its value is always negative. Moreover, \(ΔH_{cond}\) is equal in magnitude to \(ΔH_{vap}\), so the only difference between the two values for one given compound or element is the positive or negative sign. 2.055 liters of steam at 100°C was collected and stored in a cooler container. What was the amount of heat involved in this reaction? The \(ΔH_{vap}\) of water = 44.0 kJ/mol. First, convert 100°C to Kelvin. °C + 273.15 = K
100.0 + 273.15 = 373.15 K Find the amount involved (in moles). \[\begin{align*} n_{water} &= \dfrac{PV}{RT} \\[4pt] &= \dfrac{(1.0\; atm)(2.055\; L)}{(0.08206\; L\; atm\; mol^{-1} K^{-1})(373.15\; K)} \\[4pt] &= 0.0671\; mol \end{align*}\] Find \(ΔH_{cond}\) \[ΔH_{cond} = -ΔH_{vap} \nonumber\] so \[ΔH_{cond} = -44.0\; kJ/ mol \nonumber\] Using the \(ΔH_{cond}\) of water and the amount in moles, calculate the amount of heat involved in the reaction. To find kJ, multiply the \(ΔH_{cond}\) by the amount in moles involved. \[\begin{align*} (ΔH_{cond})(n_{water}) &= (-44.0\; kJ/mol)(0.0671\; mol) \\[4pt] &= -2.95\; kJ \end{align*} \] (or ) the transition of molecules from a liquid to a gaseous state; the molecules on a surface are usually the first to undergo a phase change. | 3,733 | 1,267 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Le_Chateliers_Principle/The_Effect_of_Changing_Conditions |
This page looks at the relationship between equilibrium constants and Le Chatelier's Principle. Students often get confused about how it is possible for the position of equilibrium to change as you change the conditions of a reaction, although the equilibrium constant may remain the same. Equilibrium constants are not changed if you change the concentrations of things present in the equilibrium. The only thing that changes an equilibrium constant is a change of temperature. The position of equilibrium is changed if you change the concentration of something present in the mixture. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. Suppose you have an equilibrium established between four substances A, B, C and D. \[ A + 2B \rightleftharpoons C + D\] According to Le Chatelier's Principle, if you decrease the concentration of C, for example, the position of equilibrium will move to the right to increase the concentration again. The equilibrium constant, \(K_c\) for this reaction looks like this: \[ K_c = \dfrac{[C,D]}{[A,B]^2}\] If you have moved the position of the equilibrium to the right (and so increased the amount of \(C\) and \(D\)), why hasn't the equilibrium constant increased? This is actually the wrong question to ask! We need to look at it the other way round. Let's assume that the equilibrium constant must not change if you decrease the concentration of \(C\) - because equilibrium constants are constant at constant temperature. Why does the position of equilibrium move as it does? If you decrease the concentration of \(C\), the top of the \(K_c\) expression gets smaller. That would change the value of \(K_c\). In order for that not to happen, the concentrations of \(C\) and \(D\) will have to increase again, and those of \(A\) and \(B\) decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before. The position of equilibrium moves - not because Le Chatelier says it must - but because of the need to keep a constant value for the equilibrium constant. If you decrease the concentration of \(C\): This only applies to systems involving at least one gas. Equilibrium constants are not changed if you change the pressure of the system. The only thing that changes an equilibrium constant is a change of temperature. The position of equilibrium may be changed if you change the pressure. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. That means that if you increase the pressure, the position of equilibrium will move in such a way as to decrease the pressure again - if that is possible. It can do this by favoring the reaction which produces the fewer molecules. If there are the same number of molecules on each side of the equation, then a change of pressure makes no difference to the position of equilibrium. Let's look at the same equilibrium we've used before. This one would be affected by pressure because there are three molecules on the left, but only two on the right. An increase in pressure would move the position of equilibrium to the right. \[A_{(g)} + 2B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)} \] Because this is an all-gas equilibrium, it is much easier to use \(K_p\): \[K_p = \dfrac{P_c\;P_D}{P_A\;P_B^2} \label{EqC1}\] Once again, it is easy to suppose that, because the position of equilibrium will move to the right if you increase the pressure, \(K_p\) will increase as well. Not so! To understand why, you need to modify the \(K_p\) expr ession. Remember the relationship between partial pressure, mole fraction and total pressure? \[ P_A = (\text{mole fraction of A} )( \text{total pressure})\] \[P_A = \chi_A P_{tot}\] Replacing all the partial pressure terms in \(\ref{EqC1}\) by mole fractions\( \chi_A \) and total pressure (\(P_{tot}\)) g ives you this: \[ K_p \dfrac{(\chi_C P_{tot} )( \chi_D P_{tot} )}{(\chi_A P_{tot} )(\chi_B P_{tot})^2}\]
Most of the "P"s cancel out, with one left at the bottom of the expression. \[ K_p \dfrac{\chi_C \chi_D}{\chi_A\chi_B^2 P_{tot}}\] Now, remember that \(K_p\) has got to stay constant because the temperature is unchanged. How can that happen if you increase P? To compensate, you would have to increase the terms on the top, \(\chi_C\) and \(\chi_D\), and decrease the terms on the bottom, \(\chi_A\) and \(\chi_B\). Increasing the terms on the top means that you have increased the mole fractions of the molecules on the right-hand side. Decreasing the terms on the bottom means that you have decreased the mole fractions of the molecules on the left. That is another way of saying that the position of equilibrium has moved to the right - exactly what Le Chatelier's Principle predicts. The position of equilibrium moves so that the value of \(K_p\) is kept constant. There are the same numbers of molecules on each side of the equation. In this case, the position of equilibrium is not affected by a change of pressure. Why not? \[A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)} \] Let's go through the same process as in Case 1: \[ K_p = \dfrac{ P_C P_D}{P_A P_B}\] Substituting mole fractions and total pressure: \[ K_p = \dfrac{ (\chi_C P_{tot}) (\chi_D P_{tot}) }{ (\chi_A P_{tot}) (\chi_B P_{tot})}\] Cancelling out as far as possible: \[ K_p = \dfrac{ \chi_C \chi_D }{ \chi_A \chi_B }\] There is not a single \(P_{tot}\) left in the expression so changing the pressure makes no difference to the \(K_p\) expression. The position of equilibrium doesn't need to move to keep \(K_p\) constant. Equilibrium constants are changed if you change the temperature of the system. \(K_c\) or \(K_p\) are constant at constant temperature, but they vary as the temperature changes. Look at the equilibrium involving hydrogen, iodine and hydrogen iodide: \[ H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)} \label{EqHI}\] with \(\Delta H = -10.4\; kJ/mol\). The \(K_p\) expression is: \[ P_p =\dfrac{P_{HI}^2}{P_{H_2}P_{(I_2)}}\] Two values for \(K_p\) a re: You can see that as the temperature increases, the value of \(K_p\) falls. This is typical of what happens with any equilibrium where the forward reaction is exothermic. Increasing the temperature decreases the value of the equilibrium constant. Where the forward reaction is endothermic, increasing the temperature increases the value of the equilibrium constant. The position of equilibrium also changes if you change the temperature. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made. If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again. It will do that by favoring the reaction which absorbs heat. In the equilibrium we've just looked at (\(\ref{EqHI}\), that will be the back reaction because the forward reaction is exothermic. So, according to Le Chatelier's Principle the position of equilibrium will move to the left with increasing temperature. Less hydrogen iodide will be formed, and the equilibrium mixture will contain more unreacted hydrogen and iodine. That is entirely consistent with a fall in the value of the equilibrium constant. Equilibrium constants are not changed if you add (or change) a catalyst. The only thing that changes an equilibrium constant is a change of temperature. The position of equilibrium is not changed if you add (or change) a catalyst. A catalyst speeds up both the forward and back reactions by exactly the same amount. Dynamic equilibrium is established when the rates of the forward and back reactions become equal. If a catalyst speeds up both reactions to the same extent, then they will remain equal without any need for a shift in position of equilibrium. Jim Clark ( ) | 7,923 | 1,268 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Enthalpy/Assorted_Definitions |
This page explains what an enthalpy change is, and then gives a definition and brief comment for three of the various kinds of enthalpy change that you will come across. Enthalpy change is the name given to the amount of heat evolved or absorbed in a reaction carried out at constant pressure. It is given the symbol ΔH, read as "delta H". Standard enthalpy changes refer to reactions done under standard conditions, and with everything present in their standard states. Standard states are sometimes referred to as "reference states". Standard conditions are: For a standard enthalpy change everything has to be present in its standard state. That is the physical and chemical state that you would expect to find it in under standard conditions. That means that the standard state for water, for example, is liquid water, H O(l) - not steam or water vapour or ice. Oxygen's standard state is the gas, O (g) - not liquid oxygen or oxygen atoms. For elements which have allotropes (two different forms of the element in the same physical state), the standard state is the most energetically stable of the allotropes. For example, carbon exists in the solid state as both diamond and graphite. Graphite is energetically slightly more stable than diamond, and so graphite is taken as the standard state of carbon. Similarly, under standard conditions, oxygen can exist as O (simply called oxygen) or as O (called ozone - but it is just an allotrope of oxygen). The O form is far more energetically stable than O , so the standard state for oxygen is the common O (g). The symbol for a standard enthalpy change is ΔH°, read as "delta H standard" or, perhaps more commonly, as "delta H nought". Remember that an enthalpy change is the heat evolved or absorbed when a reaction takes place at constant pressure. The standard enthalpy change of a reaction is the enthalpy change which occurs when equation quantities of materials react under standard conditions, and with everything in its standard state. That needs exploring a bit. Here is a simple reaction between hydrogen and oxygen to make water: Whenever a standard enthalpy change is quoted, standard conditions are assumed. If the reaction has to be done under different conditions, a different enthalpy change would be recorded. That has to be calculated back to what it would be under standard conditions. Fortunately, you don't have to know how to do that at this level. The standard enthalpy change of formation of a compound is the enthalpy change which occurs when one mole of the compound is formed from its elements under standard conditions, and with everything in its standard state. The equation showing the standard enthalpy change of formation for water is: When you are writing one of these equations for enthalpy change of formation, you must end up with 1 mole of the compound. If that needs you to write fractions on the left-hand side of the equation, that is OK. (In fact, it is not just OK, it is essential, because otherwise you will end up with more than 1 mole of compound, or else the equation won't balance!) The equation shows that 286 kJ of heat energy is given out when 1 mole of liquid water is formed from its elements under standard conditions. Standard enthalpy changes of formation can be written for any compound, even if you can't make it directly from the elements. For example, the standard enthalpy change of formation for liquid benzene is +49 kJ mol . The equation is: If carbon won't react with hydrogen to make benzene, what is the point of this, and how does anybody know what the enthalpy change is? What the figure of +49 shows is the relative positions of benzene and its elements on an energy diagram: How do we know this if the reaction doesn't happen? It is actually very simple to calculate it from other values which we can measure - for example, from enthalpy changes of combustion (coming up next). We will come back to this again when we look at calculations on another page. Knowing the enthalpy changes of formation of compounds enables you to calculate the enthalpy changes in a whole host of reactions and, again, we will explore that in a bit more detail on another page. And one final comment about enthalpy changes of formation: The standard enthalpy change of formation of an element in its standard state is zero. That's an important fact. The reason is obvious . . . For example, if you "make" one mole of hydrogen gas starting from one mole of hydrogen gas you aren't changing it in any way, so you wouldn't expect any enthalpy change. That is equally true of any other element. The enthalpy change of formation of any element has to be zero because of the way enthalpy change of formation is defined. The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one mole of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state. The enthalpy change of combustion will always have a negative value, of course, because burning always releases heat. Two examples: Notice: Similarly, if you are burning something like ethanol, which is a liquid under standard conditions, you must show it as a liquid in any equation you use. Jim Clark ( ) | 5,263 | 1,269 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.08%3A_Second-Order_Reactions |
Many important biological reactions, such as the formation of double-stranded DNA from two complementary strands, can be described using second order kinetics. In a second-order reaction, the sum of the exponents in the rate law is equal to two. The two most common forms of second-order reactions will be discussed in detail in this section. To describe how the rate of a second-order reaction changes with concentration of reactants or products, the differential (derivative) rate equation is used as well as the integrated rate equation. The differential rate law can show us how the rate of the reaction changes in time, while the integrated rate equation shows how the concentration of species changes over time. The latter form, when graphed, yields a linear function and is, therefore, more convenient to look at. Nonetheless, both of these equations can be derived from the above expression for the reaction rate. Plotting these equations can also help us determine whether or not a certain reaction is second-order. Two of the same reactant (\(\ce{A}\)) combine in a single elementary step. \[\begin{align} \ce{A} + \ce{A} &\ce{->} \ce{P} \label{case1a} \\[4pt] \ce{2A} &\ce{->} \ce{P} \label{case1b} \end{align} \] The reaction rate for this step can be written as \[\text{Rate} = - \dfrac{1}{2} \dfrac{d[A]}{dt} = + \dfrac{d[P]}{dt} \nonumber \] and the rate of loss of reactant \(\ce{A}\) \[\dfrac{dA}{dt}= -k[A,A] = -k[A]^2 \label{2ndlaw} \] where \(k\) is a second order rate constant with units of M min or M s . Therefore, doubling the concentration of reactant \(\ce{A}\) will quadruple the rate of the reaction. In this particular case, another reactant (\(B\)) could be present with \(A\); however, its concentration does not affect the rate of the reaction, i.e., the reaction order with respect to B is zero, and we can express the rate law as \(v = k[A]^2[B]^0\). Integration of Equation \ref{2ndlaw} yields \[ \dfrac{1}{[A]} = \dfrac{1}{[A]_0}+kt \nonumber \] which is easily rearranged into a form of the equation for a straight line and yields plots similar to the one shown below. The half-life is given by \[ t_{1/2}=\dfrac{1}{k[A_o]} \nonumber \] Notice that the half-life of a second-order reaction depends on the initial concentration, in contrast to first-order reactions. For this reason, the concept of half-life for a second-order reaction is far less useful. Reaction rates are discussed in more detail Reaction orders are defined Here are explanations of and order reactions. For reactions that follow Equation \ref{case1a} or \ref{case1b}, the rate at which \(\ce{A}\) decreases can be expressed using the . \[-\dfrac{d[A]}{dt} = k[A]^2 \nonumber \] The equation can then be rearranged: \[\dfrac{d[A]}{[A]^2} = -k\,dt \nonumber \] Since we are interested in the change in concentration of A over a period of time, we integrate between \(t = 0\) and \(t\), the time of interest. \[ \int_{[A]_o}^{[A]_t} \dfrac{d[A]}{[A]^2} = -k \int_0^t dt \nonumber \] To solve this, we use the following rule of integration (power rule): \[\int \dfrac{dx}{x^2} = -\dfrac{1}{x} + constant \nonumber \] We then obtain the . \[\dfrac{1}{[A]_t} - \dfrac{1}{[A]_o} = kt \nonumber \] Upon rearrangement of the integrated rate equation, we obtain an equation of the line: \[\dfrac{1}{[A]_t} = kt + \dfrac{1}{[A]_o} \nonumber \] The crucial part of this process is not understanding precisely how to derive the integrated rate law equation, rather it is important to understand how the equation directly relates to the graph which provides a linear relationship. In this case, and for all second order reactions, the linear plot of versus time will yield the graph below. This graph is useful in a variety of ways. If we only know the concentrations at specific times for a reaction, we can attempt to create a graph similar to the one above. If the graph yields a straight line, then the reaction in question must be second order. In addition, with this graph we can find the slope of the line and this slope is \(k\), the reaction constant. The slope can be found be finding the "rise" and then dividing it by the "run" of the line. For an example of how to find the slope, please see the example section below. The plot of versus time would result in a straight line if the reaction were zeroth order. It does, however, yield less information for a second order graph. This is because both the graphs of a first or second order reaction would look like exponential decays. The only obvious difference, as seen in the graph below, is that the concentration of reactants approaches zero more slowly in a second-order, compared to that in a first order reaction. Two different reactants (\(\ce{A}\) and \(\ce{B}\)) combine in a single elementary step: \[A + B \longrightarrow P \label{case2} \] The reaction rate for this step can be written as \[\text{Rate} = - \dfrac{d[A]}{dt}= - \dfrac{d[B]}{dt}= + \dfrac{d[P]}{dt} \nonumber \] and the rate of loss of reactant \(\ce{A}\) \[ \dfrac{d[A]}{dt}= - k[A,B] \nonumber \] where the reaction order with respect to each reactant is 1. This means that when the concentration of reactant A is doubled, the rate of the reaction will double, and quadrupling the concentration of reactant in a separate experiment will quadruple the rate. If we double the concentration of \(\ce{A}\) and quadruple the concentration of \(\ce{B}\) at the same time, then the reaction rate is increased by a factor of 8. This relationship holds true for any varying concentrations of \(\ce{A}\) or \(\ce{B}\). As before, the rate at which \(A\) decreases can be expressed using the differential rate equation: \[ \dfrac{d[A]}{dt} = -k[A,B] \nonumber \] Two situations can be identified. Situation 2a is the situation that the initial concentration of the two reactants are not equal. Let \(x\) be the concentration of each species reacted at time \(t\). Let \( [A]_0 =a\) and \([B]_0 =b\), then \([A]= a-x\) ;\( [B]= b-x\). The expression of rate law becomes: \[-\dfrac{dx}{dt} = -k([A]_o - x)([B]_o - x)\nonumber \] which can be rearranged to: \[\dfrac{dx}{([A]_o - x)([B]_o - x)} = kdt\nonumber \] We integrate between \(t = 0\) (when \(x = 0\)) and \(t\), the time of interest. \[ \int_0^x \dfrac{dx}{([A]_o - x)([B]_o - x)} = k \int_0^t dt \nonumber \] To solve this integral, we use the method of partial fractions. \[ \int_0^x \dfrac{1}{(a - x)(b -x)}dx = \dfrac{1}{b - a}\left(\ln\dfrac{1}{a - x} - \ln\dfrac{1}{b - x}\right)\nonumber \] Evaluating the integral gives us: \[ \int_0^x \dfrac{dx}{([A]_o - x)([B]_o - x)} = \dfrac{1}{[B]_o - [A]_o}\left(\ln\dfrac{[A]_o}{[A]_o - x} - \ln\dfrac{[B]_o}{[B]_o - x}\right) \nonumber \] Applying the , the equation simplifies to: \[\int _0^x \dfrac{dx}{([A]_o - x)([B]_o - x)} = \dfrac{1}{[B]_o - [A]_o} \ln \dfrac{[B,A]_o}{[A,B]_o} \nonumber \] We then obtain the (under the condition that [A] and [B] are not equal). \[ \dfrac{1}{[B]_o - [A]_o}\ln \dfrac{[B,A]_o}{[A,B]_o} = kt \nonumber \] Upon rearrangement of the integrated rate equation, we obtain: \[ \ln\dfrac{[B,A]_o}{[A,B]_o} = k([B]_o - [A]_o)t \nonumber \] Hence, from the last equation, we can see that a linear plot of \(\ln\dfrac{[A]_o[B]}{[A,B]_o}\) versus time is characteristic of second-order reactions. This graph can be used in the same manner as the graph in the section above or written in the other way: \[\ln\dfrac{[A]}{[B]} = k([A]_o - [B]_o)t+\ln\dfrac{[A]_o}{[B]_o}\nonumber \] in form \( y = ax + b\) with a slope of \(a= k([B]_0-[A]_0)\) and a y-intercept of \( b = \ln \dfrac{[A]_0}{[B]_0}\) Because \(A + B \rightarrow P\) Since \(A\) and \(B\) react with a 1 to 1 stoichiometry, \([A]= [A]_0 -x\) and \([B] = [B]_0 -x\) at any time \(t\), \([A] = [B]\) and the rate law will be, \[\text{rate} = k[A,B] = k[A,A] = k[A]^2.\nonumber \] Thus, it is assumed as the !!! at 200° C \[\ce{ 5E(g) -> 4K(g) + G(g)} \nonumber \] This reaction follows a second order rate law with regards to \(\ce{E}\). For this reaction suppose that the rate constant at 200° C is equivalent to \(4.0 \times 10^{-2} M^{-1}s^{-1}\) and the initial concentration is \(0.050\; M\). W . Start by defining the reaction rate in terms of the loss of reactants \[ \text{Rate (initial)} = - \dfrac{1}{5} \dfrac{d[E]}{dt}\nonumber \] and then use the rate law to define the rate of loss of \(\ce{E}\) \[ \dfrac{d[E]}{dt} = -k [A]_i^2 \nonumber \] We already know \(k\) and \([A]_i\) but we need to figure out \(x\). To do this look at the units of \(k\) and one sees it is M s which means the overall reaction is a second order reaction with \(x=2\). \[\begin{align*} \text{Initial rate} &= (4.0 \times 10^{-2} M^{-1}s^{-1})(0.050\,M)^2 \\[4pt] &= 1 \times 10^{-4} \, Ms^{-1}\end{align*} \nonumber \] Another characteristic used to determine the order of a reaction from experimental data is the (\(t_{1/2}\)). By definition, the half life of any reaction is the amount of time it takes to consume half of the starting material. For a second-order reaction, the half-life is inversely related to the initial concentration of the reactant (A). For a second-order reaction each half-life is twice as long as the life span of the one before. Consider the reaction \(2A \rightarrow P\): We can find an expression for the half-life of a second order reaction by using the previously derived integrated rate equation. \[\dfrac{1}{[A]_t} - \dfrac{1}{[A]_o} = kt \nonumber \] Since, \[[A]_{t_{1/2}} = \dfrac{1}{2}[A]_o \nonumber \] when \(t = t_{1/2} \). Our integrated rate equation becomes: \[\dfrac{1}{\dfrac{1}{2}[A]_o} - \dfrac{1}{[A]_o} = kt_{1/2} \nonumber \] After a series of algebraic steps, \[\begin{align*} \dfrac{2}{[A]_o} - \dfrac{1}{[A]_o} &= kt_{1/2} \\[4pt] \dfrac{1}{[A]_o} &= kt_{1/2} \end{align*} \] We obtain the equation for the half-life of a second order reaction: \[t_{1/2} = \dfrac{1}{k[A]_o} \label{2nd halflife} \] This inverse relationship suggests that as the initial concentration of reactant is increased, there is a higher probability of the two reactant molecules interacting to form product. Consequently, the reactant will be consumed in a shorter amount of time, i.e. the reaction will have a shorter half-life. This equation also implies that since the half-life is longer when the concentrations are low, species decaying according to second-order kinetics may exist for a longer amount of time if their initial concentrations are small. Note that for the second scenario in which \(A + B \rightarrow P\), the half-life of the reaction cannot be determined. As stated earlier, \([A]_o\) cannot be equal to \([B]_o\). Hence, the time it takes to consume one-half of A is not the same as the time it takes to consume one-half of B. Because of this, we cannot define a general equation for the half-life of this type of second-order reaction. \([A]_0=4.50 \times 10^{-5}\,M\) \(k=0.89 M^{-1}s^{-1}\). What is the half-life of this reaction? This is a direct application of Equation \ref{2nd halflife}. \[\begin{align*} \dfrac{1}{k[A]_0} &= \dfrac{1}{(4.50 \times 10^{-5} M)(0.89 M^{-1}{s^{-1})}} \\[4pt] &= 2.50 \times 10^4 \,s \end{align*} \] The graph below is the graph that tests if a reaction is second order. The reaction is second order if the graph has a straight line, as is in the example below. 1. Given the following information, determine the order of the reaction and the value of k, the reaction constant. *Hint: Begin by graphing
2. Using the following information, determine the half life of this reaction, assuming there is a single reactant. 3. Given the information from the previous problem, what is the concentration after 5 minutes? The slope can be found by taking the "rise" over the "run". This means taking two points, (10,1) and (20,2). The "rise" is the vertical distance between the points (2-1=1) and the "run" is the horizontal distance (20-10=10). Therefore the slope is 1/10=0.1. The value of k, therefore, is 0.1 M s .
| 11,959 | 1,270 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/04%3A_Biological_and_Synthetic_Dioxygen_Carriers/4.03%3A_Detailed_Structures_of_Hemoglobins_and_Model_Systems |
With thermodynamic background and general structural features relevant to ligand affinity enumerated, attention may now be turned to the detailed structural aspects of the active site and its surroundings. As was shown crudely in Figure 4.3, the ligand affinity of an iron porphyrin may be perturbed either by modulating the structure of the deoxy material or by modulating the structure and surroundings of the liganded material or both. The model systems provide the reference points against which the protein structures may be compared. The structure of the picket-fence porphyrin compound, Fe(PF)(2-MeIm), is shown in Figure 4.28. Minus the pickets, it is essentially a magnified view of the active site of deoxymyoglobin, shown in Figure 4.29. Some metrical details of these structures, of a very similar unsubstituted tetraphenylporphyrin, and of several other deoxyhemoglobins are listed in Table 4.7. In general they are all similar, but important differences exist. a) See Figure 4.25 for definition of symbols. b) From a difference refinement of CoHb vs. Hb, where the difference in metal-to-porphyrin-plane separation was 0.24(2) Å and the difference in M-N was 0.13(4) Å. Doming is similar to Hb. Resol. (Å) Fe-N (Å) Fe • • • Porp (Å) Doming (Å) Fe—N (Å) \(\phi\) (deg) Tilt (deg) HbA (\(\alpha\) • • • H O) In all structures, except deoxyerythrocruorin, the iron atom is displaced about 0.4 to 0.5 Å from the plane of the porphyrin toward the axial base. For deoxyerythrocruorin the displacement is less than half this, perhaps because the water molecule is weakly coordinated to the iron center. An imidazole group from a histidine residue—the distal histidine E7 in position 7 on helix labeled E—hovers over the binding site for most vertebrate hemoglobins, except for genetically engineered mutants of human hemoglobin (\(\beta\)E7His → Gly), pathological mutant hemoglobins, such as hemoglobin Zürich (\(\beta\)E7His → Arg), and some others, such as elephant hemoglobin. Long believed to be noncoordinating, this distal histidine may, in fact, coordinate weakly to the Fe center at low temperature. In the \(\alpha\) chains of human deoxyhemoglobin, hemoglobin A, a water molecule is found in the binding cavity. For many years the binding cavity has been referred to as the hydrophobic pocket—literally, water-hating. Although many hydrophobic groups, such as valine, leucine, isoleucine, and phenylalanine are positioned over the porphyrin, the immediate environment around the binding site is, in fact, polar, with the distal histidine and associated water molecules, as well as the heme group itself. As will be shown in the next section, the label "hydrophobic pocket" becomes more misleading when the interaction of coordinated ligands with distal groups is examined. The orientation of the axial base, angle \(\phi_{1}\), is similar for Fe(PF)(2-MeIm) and for several vertebrate deoxyhemoglobins. On the other hand, Fe(TPP)(2- Melm) and deoxyerythrocruorin have a similar eclipsed axial-base orientation. At least for five-coordinate species, where the iron center is substantially out of the porphyrin plane, orientation of the axial base does not invariably induce structural perturbations, e.g., doming, in the porphyrin skeleton. The conformation of the protein chain is such that the proximal histidine in deoxyhemoglobin coordinates in a slightly tilted manner, comparable to the tilt that the sterically active 2-methyl substituent induces in the synthetic systems. Clearly, coordination of the histidine to the heme in a symmetric manner, as would be expected in the absence of the protein constraints, does produce the conformation of lowest free energy for the molecule. | 3,729 | 1,271 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/10%3A_Alkenes_and_Alkynes_I_-_Ionic_and_Radical_Addition_Reactions/10.10%3A_Alkylation_of_Alkenes |
Addition of a saturated hydrocarbon \(\left( \ce{R-H} \right)\) to an alkene to yield a saturated hydrocarbon of higher molecular weight is known as : Such reactions are used by the petroleum industry to produce medium-molecular-weight hydrocarbons from smaller molecules. A particularly important example is afforded by the addition of 2-methylpropane to 2-methylpropene in the presence of sulfuric acid or anhydrous hydrogen fluoride to yield 2,2,4-trimethylpentane: The overall reaction appears to be different from any so far discussed, because it involves addition of a nonpolar reagent \(\left( \ce{RH} \right)\) to an alkene bond. The key to the mechanism of hydrocarbon alkylation was provided by the discovery by P. D. Bartlett, in 1940, that a carbocation can react rapidly with a hydrocarbon having a tertiary hydrogen to yield a new carbocation and a new hydrocarbon. Some of these "hydrogen-transfer" reactions are extraordinarily fast and may be complete in seconds or less. The hydrogen is transferred with bonding electrons \(\left( \ce{H}^\ominus \right)\). For example, With the knowledge that the hydrogen transfer is fast, the alkylation of 2-methylpropene with 2-methylpropane can be formulated as involving first polymerization of two 2-methylpropene molecules under the influence of the sulfuric acid catalyst to give the same octyl cation as was postulated for the dimerization of 2-methylpropene: The octyl cation then can undergo a hydrogen-transfer reaction with 2-methylpropane to furnish 2,2,4-trimethylpentane and a -butyl cation: Attack by the -butyl cation on another molecule of 2-methylpropene produces an eight-carbon tertiary cation, which them proceeds to another molecule of "alkylate": This is an important example of a reaction. and (1977) | 1,805 | 1,272 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Vibrational_Spectroscopy/Vibrational_Modes/Mode_Analysis |
Γ = Γ + Γ + Γ + Γ and Γ = Γ + Γ For easier understanding, there are steps immediately followed by examples using the D3h. First, add all the x, y and z rows on the . If x, y or z are in () on the far right then only count them once, otherwise count the row a second time (Keep the column separated). This is called Γ . Next, move the molecule with the designated column symbols and if an atom does not move then it is counted. Finally, multiply Γ and UMA and it will equal Γtotal. 6 3 2 4 1 4 -need to find the irreducible representation of Gamma total. Take Γtotal multiply by the number in front of the symbols (the order) and multiply by each number inside of the character table. Add up each row, and divide each row by the total order. For the D h the order is 12. Γ = 2A1' + A2' + 4E' + 3A2" + 2E" The irreducible form of Γtrans, one needs to look at the second to last column. look at the row that the x, y and z and take the irreducible representation. For instance the D3h would be Γ = E'+A2" One can find Γ the same way as Γ . Instead of looking at x,y,z, one would look at the Rx, Ry, Rz. For the D3h. The Γ = A2'+E" To find Γ , just take Γ -Γ -Γ = Γ . D3h example Only R , R , R , x, y, and z can be ir active. which means only A2', E', A2", and E" can be IR active bands for the D h. Next add up the number in front of the irreducible representation and that is how many IR active bonds. For instance for the same problem there are 3E'+2A2". There are 5 bands, three of them (meaning the E) are two fold degenerate. Only x +y , z , xy, xz, yz, x2-y2 can be Raman active. which means only A1', E', and E" can be raman active for the D3h. Next add up the number in front of the irreducible representation, and that is how many Raman active bonds there are. For instance for the same problem there are 3E' + E". There are four bands, 4 of which are two fold degenerate. Looking at the molecules point group, do each of the symmetry representation and count the number of unmoved bonds. Γ = Γ =Γ . next multiply unmoved bonds by the symmetry operations and then the numbers inside the character tables. Then add the rows up and divide by the order of the point group. 5 2 1 3 0 3 Γ = Γ = Γ = 2A1' + 1E'+ 1A2" Γ Γ = | 2,256 | 1,273 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Period/Period_3_Elements/Chlorides_of_Period_3_Elements |
This page discusses the structures of the chlorides of the Period 3 elements (sodium to sulfur), their physical properties and their reactions with water. Chlorine and argon are omitted—chlorine because it is meaningless to talk about "chlorine chloride", and argon because it is inert and does not form a chloride. The chlorides of interest are given in the table below: Sulfur forms three chlorides, but S Cl is most common. Aluminum chloride also exists under some conditions as a dimer, Al Cl . Sodium chloride is an ionic compound consisting of a giant array of sodium and chloride ions. A small representative portion of a sodium chloride lattice looks like this: This is normally drawn in an exploded form as: The strong attractions between the positive and negative ions require a large amount of heat energy to break, so sodium chloride has high melting and boiling points. The compound does not conduct electricity in the solid state because it has no mobile electrons, and the ions are constrained by the crystal lattice. However, when it melts it undergoes electrolysis. Sodium chloride dissolves in water to give a neutral solution. Like sodium chloride, magnesium chloride also forms an ionic solid, but with a more complicated crystal structure of the ions to accommodate twice as many chloride ions as magnesium ions. As with sodium chloride, large amounts of heat energy are needed to overcome the attractions between the ions (because of the high of the compound), so the melting and boiling points are also high. Solid magnesium chloride is a non-conductor of electricity because the ions are constrained. However, upon melting, the compound undergoes electrolysis. Magnesium chloride dissolves in water to give a slightly acidic solution (with a pH of approximately 6). When magnesium ions are solvated from the solid lattice, there is enough attraction between the 2+ ions and the water molecules to form coordinate (dative covalent) bonds between the magnesium ions and lone pairs on surrounding water molecules. Hexaaquamagnesium are formed, [Mg(H O) ] , as follows: \[ MgCl_{2 (s)} + 6H_2O \rightarrow [Mg(H_2O)_6]^{2+}_{(aq)} + 2Cl^-_{(aq)}\] Many complex ions are acidic, the degree of acidity depending on the attraction between the electrons in the water molecules and the metal at the center of the ion. The hydrogen atoms carry less electron density in this state, and are thus more easily removed by a base. For magnesium, \[ [Mg(H_2O)_6]^{2+} + H_2O_{(l)} \rightleftharpoons [Mg(H_2O)_5(OH)^{2+}]^+ +H_3O^+_{(aq)}\] The make the solution acidic. Few are formed (the equilibrium lies well to the left) because the solution is only weakly acidic. The previous equation can be simplified as follows: \[ [Mg(H_2O)^6]^{2+}_{(aq)} \rightleftharpoons [Mg(H_2O)_5(OH)]^+_{(aq)} + H^+_{(aq)}\] It is essential to include the state symbols if the equation is written this way. Electronegativity increases across the period; aluminum and chlorine do not differ enough in electronegativity to form a simple ionic bond. The structure of aluminum chloride changes with temperature. At room temperature, the aluminum is 6-coordinated (i.e. each aluminum is surrounded by 6 chlorine atoms). The structure is an , but it has a lot of covalent character. At atmospheric pressure, aluminum chloride sublimes at about 180°C. If the pressure is increased to just over 2 atmospheres, it melts instead at a temperature of 192°C. Both of these temperatures are far below the expected range for an ionic compound. They suggest comparatively weak attractions between molecules instead of strong attractions between ions. This is because the coordination of the aluminum changes at these temperatures. It becomes 4-coordinated—each aluminum is surrounded by 4 chlorine atoms rather than 6. The original lattice converts Al Cl In the conversion, all ionic character is lost, causing the aluminum chloride to vaporize or melt (depending on the pressure). These dimers and simple AlCl molecules exist in equilibrium. As the temperature increases further, the position of equilibrium shifts more and more to the right of the following system: \[ Al_2Cl_6 \rightleftharpoons 2AlCl_3\] Solid aluminum chloride does not conduct electricity at room temperature because the ions are not free to move. Molten aluminum chloride (only possible at increased pressures) is also nonconductive, because it has lost its ionic character. Aluminum chloride reacts dramatically with water. A drop of water placed onto solid aluminum chloride produces steamy clouds of hydrogen chloride gas. Solid aluminum chloride in an excess of water still splutters, but instead an acidic solution is formed. A solution of aluminum chloride of ordinary concentrations (around 1 mol dm , for example) has a pH around 2-3. More concentrated solutions have a lower pH. The aluminum chloride with the water rather than simply dissolving in it. In the first instance, hexaaquaaluminum complex ions and chloride ions are formed: \[ AlCl_3 (s) + 6H_2O (l) \rightarrow [Al(H_2O)_6]^{3+} (aq) + 3Cl^- (aq)\] This is very similar to the magnesium chloride equation given above—the only difference is the charge on the ion. The greater charge attracts electrons in the water molecules quite strongly toward the aluminum, making the hydrogen atoms more positive and therefore easier to remove from the ion. Hence, this ion is much more acidic than in the corresponding magnesium case. The acid-base equilibria for this reaction lie further to the right than those for magnesium, and so the solution formed is more acidic—more hydronium ions are formed, as shown: \[ [Al(H_2O)_6]^{3+} + H_2O \rightleftharpoons [Al(H_2O)_5(OH)]^{2+} + H_3O^+ \] or, more simply: \[ [Al(H_2O)_6]^{3+} (aq) \rightleftharpoons [Al(H_2O)_5(OH)]^{2+} (aq) + H^+ \] If there is little water present, hydrogen chloride gas is produced. Because of the heat produced in the reaction and the concentration of the solution formed, hydrogen ions and chloride ions in the mixture combine together as hydrogen chloride (\(HCl\)) molecules and are given off as a gas. In a large excess of water, the temperature is never high enough for this to happen; the ions remain solvated. Silicon tetrachloride is a simple no-messing-about covalent chloride. There isn't enough electronegativity difference between the silicon and the chlorine for the two to form ionic bonds. Silicon tetrachloride is a colorless liquid at room temperature which fumes in moist air. The only attractions between the molecules are van der Waals dispersion forces. It doesn't conduct electricity because of the lack of ions or mobile electrons. It fumes in moist air because it reacts with water in the air to produce hydrogen chloride. If you add water to silicon tetrachloride, there is a violent reaction to produce silicon dioxide and fumes of hydrogen chloride. In a large excess of water, the hydrogen chloride will, of course, dissolve to give a strongly acidic solution containing hydrochloric acid. \[ SiCl_4 + 2H_2O \rightarrow SiO_2 + 4HCl\] There are two phosphorus chlorides: phosphorus(III) chloride, PCl , and phosphorus(V) chloride, PCl . This simple covalent chloride exists as a fuming liquid at room temperature because there are only van der Waals dispersion forces and dipole-dipole attractions between the molecules. The liquid does not conduct electricity because of the lack of ions or mobile electrons. Phosphorus(III) chloride reacts violently with water to generate phosphorous acid, H PO , and hydrogen chloride fumes (or a solution containing hydrochloric acid in excess of water): \[ PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl\] Phosphorus(V) chloride is structurally more complicated than phosphorus(III) chloride. At room temperature, it forms a white solid which sublimes at 163°C. Increasing the temperature beyond its sublimation point dissociates (divides reversibly) more the phosphorus(V) chloride into phosphorus(III) chloride and chlorine: \[ PCl_5 \rightleftharpoons PCl_3 + Cl_2\] Phosphorus(V) chloride is an ionic solid. The formation of the ions involves two molecules of PCl . A chloride ion transfers from one of the original molecules to the other, leaving a positive ion, [PCl ] , and a negative ion, [PCl ] . At 163°C, the phosphorus(V) chloride converts to a molecular form containing PCl molecules. Because only van der Waals dispersion forces exist between these molecules, the species vaporizes. Solid phosphorus(V) chloride does not conduct electricity. Phosphorus(V) chloride reacts violently with water, producing hydrogen chloride fumes. As with the other covalent chlorides, if there is enough water present, these dissolve to give a hydrochloric acid solution. The reaction happens in two stages. The first takes place in cold water; phosphorus oxychloride, POCl , is produced along with HCl: \[ PCl_5 + 4H_2O \rightarrow POCl_3 + 2HCl\] As the solution is brought to a boil, the phosphorus(V) chloride reacts further to give phosphoric(V) acid and more HCl. Phosphoric(V) acid is also known as phosphoric acid or as orthophosphoric acid: \[ POCl_3 + 3H_2O \rightarrow H_3PO_4 + 3HCl\] Combining these equations gives the overall reaction in boiling water: \[ PCl_5 + 4H_2O \rightarrow H_3PO_4 + 5HCl\] Disulfur dichloride is one of three sulfur chlorides and is the species formed when chlorine reacts with hot sulfur. Disulfur dichloride is an orange, unpleasant-smelling covalent liquid. Its rather unusual structure is given below: The molecule's conformation indicates its possible intermolecular interactions: Disulfur dichloride reacts slowly with water to produce a complex mixture of hydrochloric acid, sulfur, hydrogen sulfide and various sulfur-containing acids and anions. Jim Clark ( ) | 9,784 | 1,274 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/06%3A_Dynamics_and_Kinetics/21%3A_Binding_and_Association/21.01%3A_Thermodynamics_and_Biomolecular_Reactions |
To begin, we recognize that binding and association processes are bimolecular reactions. Let’s describe the basics of this process. The simplest kinetic scheme for bimolecular association is \[ A+B \rightleftharpoons C\] and could be any two molecules that interact chemically or physically to result in a final bound state; for instance, an enzyme and its substrate, a ligand and receptor, or two specifically interacting proteins. From a mechanistic point of view, it is helpful to add an intermediate step: \[ A+B \rightleftharpoons AB \rightleftharpoons C \nonumber \] Here refers to transient encounter complex, which may be a metastable kinetic intermediate or a transition state. Then the initial step in this scheme reflects the rates of two molecules diffusing into proximity of their mutual target sites (including proper alignments). The second step is recognition and binding. It reflects the detailed chemical process needed to form specific contacts, execute conformational rearrangements, or perform activated chemical reactions. We separate these steps here to build a conceptual perspective, but in practice these processes may be intimately intertwined. Let’s start by reviewing the basic thermodynamics of bimolecular reactions, such as reaction scheme (21.1.1). The thermodynamics is described in terms of the chemical potential for the molecular species in the system (i = A,B,C) \[\mu_i = \left( \dfrac{\partial G}{\partial N_i} \right)_{p,T,\{ N_j,j\neq \}} \nonumber\] where is the number of molecules of species . The dependence of the chemical potential on the concentration can be expressed as \[\mu_i = \mu_i^0 +RT\ln \dfrac{c_i}{c^0} \] c is the concentration of reactant in mol L , and the standard state concentration is c = 1 mol L . So the molar reaction free energy for scheme (1) is \[\begin{aligned} \Delta \overline{G} &=\sum_i v_i\mu_i \\ &=\mu_C-\mu_A\mu_B , \\ &=\Delta \overline{G}^0+RT\ln K \end{aligned}\] v is the stoichiometric coefficient for component . is the reaction quotient \[K= \dfrac{(c_C/c^0)}{(c_A/c^0)(c_B/c^0)} \] At equilibrium, \(\Delta \overline{G} = 0\), so \[\Delta \overline{G}^0 = -RT\ln K_a \] where the association constant is the value of the reaction quotient under equilibrium conditions. Dropping , with the understanding that we must express concentration in M units: \[ K_a=\dfrac{c_C}{c_Ac_B} \] Since it is defined as a standard state quantity, is a fundamental constant independent of concentration and pressure or volume, and is only dependent on temperature. The inverse of is the equilibrium constant for the dissociation reaction \(C \rightleftharpoons A+B\). Experimentally one controls the total mass \(m_{TOT}=m_C+m_A+m_B\), or concentration \[c_{TOT}=c_C+c_A+c_B\] The composition of system can be described by the fraction of concentration due to species as \[\begin{aligned} \theta_i &=\dfrac{c_i}{c_{TOT}}\\ \theta_A +\theta_B + \theta_C &=1 \end{aligned} \] We can readily relate to , but it is practical to set some specific constraint on the composition here. If we constrain the composition to be 1:1, which is enforced either by initially mixing equal mole fractions of and , or by preparing the system initially with pure , then \[\begin{aligned} K_{a} &=\frac{4 \theta_{C}}{\left(1-\theta_{C}\right)^{2} c_{T O T}} \qquad \qquad (\theta_A=\theta_B) \\ &=\frac{\left(1-2 \theta_{A}\right)}{\theta_{A}^{2} c_{T O T}} \end{aligned}\] This expression might be used for mixing equimolar solutions of binding partners, such as complementary DNA oligonucleotides. Using eq. (21.1.6) (with ) and (21.1.7) here, we can obtain the composition as a function of total concentration fraction as a function of the total concentration \[\begin{array}{l}
\theta_{C}=\left(1+\frac{2}{K_{a} c_{T O T}}\right)-\sqrt{\left(1+\frac{2}{K_{a} c_{T O T}}\right)^{2}-1} \\
\theta_{A}=\frac{1}{2}\left(1-\theta_{C}\right)
\end{array} \] In the case where , applicable to homodimerization or hybridization of self-complementary oligonucleotides, we rewrite scheme (21.1.1) as the association of monomers to form a dimer \[2M \rightleftharpoons D \nonumber\] and find: \[\begin{aligned} K_a &=\theta_D/2(1-\theta_D)^2c_{TOT} \\ K_a &=(1-\theta_M)/2\theta_M^2c_{TOT} \end{aligned} \] \[ \theta_D =1+\dfrac{1}{4c_{TOT}K_a} \left( 1-\sqrt{1+8c_{TOT}K_a} \right) \] \[\theta_M = 1-\theta_D \] These expressions for the fraction of monomer and dimer, and the corresponding concentrations of monomer and dimer are shown below. An increase in the total concentration results in a shift of the equilibrium toward the dimer state. Note that = (9 ) = /9 at θ = θ = 0.5, For ligand receptor binding, ligand concentration will typically be much greater than that of the receptor, and we are commonly interested in fraction of receptors that have a ligand bound, θ . Re-writing our association reaction as \[L+R\rightleftharpoons LR \qquad\qquad K_a= \dfrac{c_{LR}}{c_Lc_R} \] we write the fraction bound as \[\begin{aligned} \theta_{bound} &= \dfrac{c_{LR}}{c_R+c_{LR}} \\ &= \dfrac{c_LK_a}{1+c_LK_a} \end{aligned} \] This is equivalent to a . The temperature dependence of is governed by eq. (21.1.4) and the fundamental relation \[ \Delta G^0(T)=\Delta H^0(T)-T\Delta S^0(T) \] Under the assumption that ΔH and ΔS are temperature independent, we find \[K_a(T) = exp \left[ -\dfrac{\Delta H_a^0}{RT}+ \dfrac{\Delta S_a^0}{R} \right] \] This allows us to describe the temperature-dependent composition of a system using the expressions above for θ . While eq. (12) allows you to predict a melting curve for a given set of thermodynamic parameters, it is more difficult to use it to extract those parameters from experiments because it only relates the value of at one temperature to another. Temperature is often used to thermally dissociate or melt dsDNA or proteins, and the analysis of these experiments requires that we define a reference temperature. In the case of DNA melting, the most common and readily accessible reference temperature is the melting temperature Tm defined as the point where the mole fractions of ssDNA (monomer) and dsDNA (dimer) are equal, θ = θ = 0.5. This definition is practically motivated, since DNA melting curves typically have high and low temperature limits that correspond to pure dimer or pure monomer. Then Tm is commonly associated with the inflection point of the melting curve or the peak of the first derivative of the melting curve. From eq. (21.1.9), we see that the equilibrium constants for the association and dissociation reaction are given by the total concentration of DNA: K (T ) = K (T )−1 = c and ΔG (T ) = ‒RT ln . Furthermore, eq. (21.1.12) implies T = ΔH /ΔS . The examples below show the dependence of melting curves on thermodynamic parameters, T , and concentration. These examples set a constant value of T (ΔH /ΔS ). The concentration dependence is plotted for ΔH = 15 kcal mol and ΔS = 50 cal mol K . For conformational changes in macromolecules, it is expected that the enthalpy and entropy will be temperature dependent. Drawing from the definition of the heat capacity, \[ C_p = \left( \dfrac{\partial H}{\partial T} \right)_{N,P} = T\left( \dfrac{\partial S}{\partial T} \right)_{N,P} \nonumber \] we can describe the temperature dependence of ΔH and ΔS by integrating from a reference temperature T to T. If ΔC is independent of temperature over a small enough temperature range, then we obtain a linear temperature dependence to the enthalpy and entropy of the form \[\Delta H^0 (T) = \Delta H^0 (T_0) + \Delta C_p[T-T_0] \] \[\Delta S^0 (T) = \Delta S^0(T_0) +\Delta C_p \left( \dfrac{T}{T_0} \right) \] These expressions allow us to relate values of ΔH , ΔS , and ΔG at temperature T to its value at the reference temperature T . From these expressions, we obtain a more accurate description of the temperature dependence of the equilibrium constant is \[ K_d(T) = exp \left[ -\dfrac{\Delta H_m^0}{RT} +\dfrac{\Delta S_m^0}{R}-\dfrac{C_p}{R} \left[ 1-\dfrac{T_m}{T}-\ln \left( \dfrac{T}{T_m} \right) \right] \right] \] where \(\Delta H_m^0 = \Delta H^0(T_m) \) and \(\Delta S_m^0 = \Delta S^0(T_m) \) are the enthalpy and entropy for the dissociation reaction evaluated at T . | 8,285 | 1,276 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/06%3A_Bonding_in_Organic_Molecules/6.05%3A_Atomic-Orbital_Models |
Saturated compounds such as the alkanes and their derivatives, which have normal tetrahedral angles for the bonds to carbon, can be formulated readily in terms of atomic orbitals with \(sp^3 \: \sigma\) bonds to carbon. An example is shown in Figure 6-11, which also shows how an atomic-orbital model can be drawn in abbreviated style. The lines in this drawing correspond to bonds and are labeled as \(sp^3\) with \(sp^3\) (the overlapping orbitals of the \(C-C\) bond) or as \(sp^3\) with \(s\) (the overlapping orbitals of the \(C-H\) bonds). Many important molecules such as ammonia, water, and hydrogen fluoride have atoms with unshared pairs of electrons: If we formulate each of these molecules in such a way to minimize repulsions between like charges, a basically tetrahedral arrangement will be expected because this will place the nuclei (and electron pairs) as widely separated as possible. The water molecule could be formulated this way, as in \(6\), with the oxygen at the center of the tetrahedron: The simple picture predicts that the \(H-O-H\) bond angle should be tetrahedral, \(109.5^\text{o}\). But actually it is \(104.5^\text{o}\). There are two schools of thought as to why the angle is \(104.5^\text{o}\). One idea is that the repulsion model is too simple and has to be modified to take into account that the repulsion is more severe between pairs of unshared electrons than between electrons in bonding orbitals on the same atom. This is because when a bond is formed between nuclei, the attraction of the nuclei for the electrons shrinks the orbitals available to the bonding electrons, thereby reducing their electrostatic repulsion with other pairs. The degree of repulsion between electron pairs diminishes in the sequence: unshared pairs . unshared pairs \(>\) unshared pairs . bonding pairs \(>\) bonding pairs . bonding pairs. From this, we expect that in water the \(H-O-H\) angle will be than tetrahedral, because the larger repulsion between the two unshared pairs will tend to push the bonding pairs closer together. A similar, but smaller, effect is expected for ammonia because now the repulsion is only between the one unshared pair and bonding pairs. The ammonia \(H-N-H\) angle is \(107.3^\text{o}\), which is only slightly smaller than the tetrahedral value of \(109.5^\text{o}\). The alternative point of view of why the bond angle of water is \(104.5^\text{o}\) starts with the premise that, in the simplest approximation, the angle should be \(90^\text{o}\)! To see how this comes about let us compare \(H:Be:H\) with \(H:\underset{\cdot \cdot}{\ddot{O}}:H\). You will recall that to form two bonds to \(Be\), we had to promote an electron and change the electronic configuration to the valence configuration, \(\left( 2s \right)^1 \left( 2p \right)^1\). The situation with \(H_2O\) is different in that the oxygen ground state and valence state are the same, \(\left( 2s \right)^2 \left( 2p_x \right)^1 \left( 2p_y \right)^1 \left( 2p_z \right)^1\). This means we could form two bonds to oxygen using the \(2p_y\) and \(2p_z\) orbitals at an angle of \(90^\text{o}\) (Figure 6-12). Now, to explain why the \(H-O-H\) bond angles are \(104.5^\text{o}\) instead of \(90^\text{o}\), we can say that the repulsion between the hydrogen nuclei is expected to widen the bond angle. An argument in favor of this formulation is provided by the bond angle in \(H_2S\), which is \(92.2^\text{o}\). This is much closer to the \(90^\text{o}\) expected for \(p\)-bond orbitals and the hydrogens in \(H_2S\) would not be expected to repel each other as much as in \(H_2O\) because sulfur is a larger atom than oxygen. Both ways of formulating the orbitals used in the bonding of water molecules are in current use. Arguments can be advanced in favor of both. Highly sophisticated quantum-mechanical calculations, which we will say more about later, suggest that oxygen in water molecules uses orbitals that are \(18\% \: s\) and \(82\% \: p\) in its bonds (\(sp^{4.5}\)), and furthermore, that the unshared pairs are in orbitals [ one pair as \(\left( 2s \right)^2\) and the other as \(\left( 2p \right)^2\)]. Each of the unshared electron-pair orbitals of oxygen in water is calculated to be about \(40\% \: s\) and \(60\% \: p\) (\(sp^{1.5}\)). The results are hardly clearcut, but the bonding orbitals are considerably closer to \(sp^3\) (\(25\% \: s\) and \(75\% \: p\)) than they are to \(100\% \: p\). We recommend that the bonding orbitals of nitrogen and oxygen be considered to be \(sp^3\) and the unshared pairs designated simply as \(\left( n \right)^2\). An abbreviated atomic orbital model of methanol, \(CH_3OH\), made on this basis is shown in Figure 6-13. Recall from Chapter 2 that bond angles in compounds with carbon-carbon double bonds such as ethene are closer to \(120^\text{o}\) than to the normal tetrahedral value of \(109.5^\text{o}\). There are several way sin which a carbon-carbon double bond can be formulated in terms of atomic-orbital models. One very popular approach is to consider that ethene has two \(sp^2\)-hybridized carbons that form one carbon-carbon \(\sigma\) bond and four carbon-hydrogen \(\sigma\) bonds by overlap of the six \(sp^2\) orbitals, as shown in Figure 6-14. The remaining carbon-carbon bond is formulated as arising from overlap of the two \(p\) orbitals, one on each carbon, that are not utilized in making the \(sp^2\) hybrids. Sidewise overlap of \(p\) orbitals is called to distinguish it from the endwise of the type we have discussed previously (Figure 6-15). The resulting differs from the in that electron density is concentrated in the regions above and below the bond axis rather than along the bond axis. Formulations of ethene in this way suggests that it should be a planar molecule with \(H-C-H\) angles of \(120^\text{o}\). Ethene is indeed planar, but its \(H-C-H\) angles are found to be \(117^\text{o}\), rather than the \(120^\text{o}\) predicted for \(sp^2\) bonds. An explanation of this discrepancy using further electron-repulsion arguments will be discussed later in the chapter. The simple elegance of the \(\sigma\)-\(\pi\) model of ethene should not be taken as proving that there actually are two different kinds of bonds between the carbons. The \(\sigma\)-\(\pi\) representation of double bonds is not really unique. Given \(sp^2\) hybridization of the carbons so there are \(sp^2\)-\(\sigma\) bonds to the hydrogens, it is possible to take the \(sp^2\) and \(p\) orbitals used for the \(\sigma\) and \(\pi\) bonds, rehybridize them, and so derive a new set of overlapping orbitals for the double bond. These orbitals are called \(\tau\) ( ) and can be represented by two banana-shaped orbitals between the carbons (Figure 6-16). The result is two completely equivalent \(C-C\) bonds. The \(\tau\) model has the advantage of offering a striking parallel to ball-and-stick models, whereas the \(\sigma\)-\(\pi\) model is of particular value as a basis for quantitative calculations, as will be discussed in Chapter 21. Using the \(\sigma\)-\(\pi\) model of double bonds, we conclude that the twisted configuration shown in Figure 6-17 should not be very stable. Here the \(p\) orbitals are not in position to overlap effectively in the \(\pi\) manner. The favored configuration is expected to have the axes of the \(p\)-\(\pi\) orbitals parallel. Because considerable energy would have to be expended to break the \(p\)-\(\pi\) double bond and to permit rotation about the remaining \(sp^2\)-\(\sigma\) bond, restricted rotation and stable cis-trans isomers are expected. Similar conclusions can be reached on the basis of the \(\tau\) model of the double bond. Ethyne, \(C_2H_2\), is an organic compound that usually is formulated with \(sp\) hybrid bonds. The carbon-hydrogen framework is built up through \(\sigma\) overlap of two \(sp\)-hybrid orbitals, one from each carbon atom, to form a \(C-C\) bond, and \(\sigma\) overlap of the remaining \(sp\) orbitals with the \(s\) orbital of two hydrogens to form \(C-H\) bonds. The remaining carbon-carbon bonds result through sidewise \(\pi\) overlap of the pure \(p\) orbitals, as shown in Figure 6-18. This model fits well with the properties of the ethyne molecule being linear (bond angles of \(180^\text{o}\). Also, the \(C-H\) bonds in ethyne are different from those in ethene or ethane, as judged by their \(C-H\) stretching and bending frequencies in the infrared (Chapter 9), their bond energies, (Table 4-6), and their acidities ( ). These differences in properties are in keeping with the different states of hybridization of the carbon orbitals that we have postulated for ethane, ethene, and ethyne. A summary of the directional character of the \(s\)-\(p\) hybrid atomic orbitals discussed so far is given in Table 6-2. By referring to this table, it usually is possible to deduce the nature of the bonding orbitals for most organic compounds from the molecular geometry, if this is known, Thus a tetrahedral molecule \(AX_4\) with four attached ligands uses \(sp^3\) hybrid orbitals localized on atom \(A\); a planar triangular molecule \(AX_3\) with three attached ligands at angles of \(120^\text{o}\) is \(sp^2\) hybridized at atom \(A\); a linear molecule \(AX_2\) with two ligands is \(sp\) hybridized at \(A\). Applying the converse of these rules, one should be able to predict molecular geometry by making reasonable assumptions as to the state of hybridization for each atom in the molecule. Obviously in doing this we have to take account of unshared electron pairs. Prediction is easy if unshared pairs are absent. Thus, four attached ligands, as in \(CH_4\), \(CCl_4\), or \(BF_4^\ominus\), imply \(sp^3\) hybridization at the central atom and therefore a tetrahedral arrangement of ligands. Three ligands, as bonded to carbon in \(CH_3^\oplus\) or to boron in \(BF_3\), imply \(sp^2\) hybridization for the central atom and a planar triangular arrangement if ligands. Two ligands, as in \(CO_2\), imply \(sp\) hybridization and linear geometry. In many of our later discussions of organic reactions, we will be concerned with cationic, radical, and anionic carbon species that are substitution products of \(CH_3^\oplus\), \(CH_3 \cdot\), and \(CH_3:^\ominus\). Because of the importance of these entities, you should know how to formulate them and related substances, such as \(^\ominus: \ddot{N}H_2\), with atomic orbitals. Perhaps the most straightforward way is to start from \(CH_4\) and see what changes in the \(C-H\) bonds we would expect as the result of the hypothetical processes: \(CH_4 \rightarrow CH_3:^\ominus + H^\oplus\), \(CH_4 \rightarrow CH_3^\oplus + H:^\ominus\), and \(CH_4 \rightarrow CH_3 \cdot + H \cdot\). Methane is tetrahedral with \(sp^3\) carbon bonding orbitals. Removal of \(H^\oplus\) gives \(CH_3:^\ominus\), which corresponds in electronic structure to \(H_3N:\) and, for the same reasons, should have a pyramidal shape with nearly tetrahedral \(H-C-H\) angles. Removal of \(H:^\ominus\) from \(CH_4\) to give \(CH_3^\oplus\) with bonding electrons, suggests a change to \(sp^2\) bonding orbitals for the carbon and planar geometry with \(H-C-H\) angles of \(120^\text{o}\). The radical, \(CH_3 \cdot\) presents a special problem. We can think of it as being formed by the loss of \(H \cdot\) from \(CH_4\), by adding an electron to \(CH_3^\oplus\), or by removing an electron from pyramidal \(CH_3:^\ominus\). We can formulate \(CH_3 \cdot\) with \(Sp^2\) orbitals for the \(C-H\) bonds and the extra electron in a \(p\) orbital, or with \(sp^3\) orbitals for the \(C-H\) bonds and the extra electron in an \(sp^3\) orbital: The actual structure of \(CH_3 \cdot\) has the hydrogens and carbons in a plane (left). Therefore it appears that the repulsions between the bonding electron pairs is greater than the repulsions between the extra electron and the bonding pairs. The actual structure corresponds to the one in which the bonding pairs are as far apart as possible. Molecules of the type \(AX_4\), which have four identical ligands on the central atom and no unshared electrons on \(A\) (e.g., \(CH_4\) and \(CCl_4\)), are expected to be, and are, tetrahedral. By the same reasoning, three electron pairs around one atom should seek a planar arrangement with \(120^\text{o}\) angles to minimize electron repulsion; accordingly, species of the type \(AX_3\), which have no unshared pairs on \(A\) (e.g., \(BF_3\) and \(CH_3^\oplus\)), have this geometry. With only two electron pairs, the preferred arrangement is linear. The bond angles of compounds with multiple bonds can be explained similarly. For example, in ethene the four electrons of the double bond occupy the region in space between the two carbon nuclei. The situation at either carbon is rather like the \(AX_3\) case, except that one of the ligands now has a double complement of bonding electrons: Therefore the carbon orbitals are expected to be directed in one plane to give bond angles that deviate somewhat from \(120^\text{o}\) because of the high density of electrons in the multiple bond. Thus the \(H-C-H\) angle shrinks to \(117^\text{o}\), whereas the \(H-C=C\) angles open up to \(122^\text{o}\), because repulsion between electrons in the \(H-C=C\) bonds is greater than between electrons in the \(H-C-H\) bonds. Electron-attracting power (or ) of the ligands also is important in determining bond angles. Thus for compounds of the type \(CH_3X\), in which \(X\) is a more electron-attracting group than carbon, the \(C-X\) bond is polarized in the sense \(H_3 \overset{\delta \oplus}{C} - - - \overset{\delta \ominus}{X}\), and the carbon then should have some of the character of \(CH_3^\oplus\). Thus the \(H-C-H\) angles are expected to be greater than \(109.5^\text{o}\), as in fact they are. In chloromethane, for example, the \(H-C-H\) angle is \(111^\text{o}\). Also, we can explain on the basis of electron repulsions why the bond angle in phosphine, \(:PH_3\) (\(93^\text{o}\)), is less than that in ammonia, \(:NH_3\) (\(107.3^\text{o}\)), and the bond angle in \(H: \underset{\cdot \cdot}{\ddot{S}} :H\) (\(92.2^\text{o}\)) is less than that in \(H: \underset{\cdot \cdot}{\ddot{O}}\) (\(104.5^\text{o}\)). The important point is that phosphorus and sulfur are larger atoms than nitrogen and oxygen. This means than the \(H-S-H\) and \(H-P-H\) bond angles can be about \(90^\text{o}\) without bringing the hydrogens and the bonding pairs as close together as they are in \(H_2O\) and \(NH_3\) where the bond angles are near to the tetrahedral value. and (1977) | 14,642 | 1,277 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Valence_Bond_Theory/Overview_of_Valence_Bond_Theory |
Valence Bond (VB) Theory looks at the interaction between atoms to explain chemical bonds. It is one of the two common theories that helps describe the bonding between atoms. The other theory is Molecular Orbital Theory. Take note that these are theories and should be treated as such; they are not always perfect. Valence Bond Theory has its roots in Gilbert Newton Lewis’s paper . Possibly unaware that Lewis’s model existed, Walter Heitler and Fritz London came up with the idea that resonance and wavefunctions contributed to chemical bonds, in which they used dihydrogen as an example. Their theory was equivalent to Lewis’s theory, with the difference of quantum mechanics being developed. Nonetheless, Heitler and London's theory proved to be successful, providing Linus Pauling and John C. Slater with an opportunity to assemble a general chemical theory containing all of these ideas. Valence Bond Theory was the result, which included the ideas of resonance, covalent-ionic superposition, atomic orbital overlap, and hybridization to describe chemical bonds. Bonds are formed between atoms because the atomic orbitals overlap and that the electrons in those orbitals are localized within that overlap. They have a higher probability of being found within that bond; we will return to this statement when we talk about wavefunctions. Dihydrogen (H ) is a simple diatomic gas that has been used to illustrate this idea; however, let's look at Cl as a simple example. Cl has seven valence electrons. From its Lewis structure, one can see that Cl has a radical. That sole radical indicates that Cl can bond once. As a general rule, the number of unpaired electrons denotes how many bonds that atom can make. Since there is only one unpaired electron in each Cl, those electrons interact to bond. In this case, the 3p orbitals overlap. Lone pairs can be seen as the orbitals not interacting with each other. The idea of atomic orbitals overlapping works well for simple molecules, such as diatomic gases, but more complex molecules cannot be explained simply by the overlap of atomic orbitals, especially if they defy the octet rule when drawing out Lewis structures and if they bond beyond their predicted amount. One small concept to add in here is that an orbital from one atoms can be overlap with the other orbital of the second atom. This will eventually give one of the two results. First, the two orbitals have correct symmetry to interact or "mix" (4). Second, the two orbitals do not have correct symmetry to interact or "mix" (4). In the case of two orbitals do not interact, there will be no bonding interaction, which means one of the atomic orbitals will not contribute into the bonding. Another word, it will not affected because of the presence of other atomic orbitals. For further explaination, when the two wavefunctions are knowns as constructive interaction, which means the orbitals do not interact with one another. And if the two orbitals are interacting, they are knowns as destructive interaction where their wavefunctions have opposite site (4). In the case of two orbitals do interact with each others, it will become molecular orbitals, and it is knowns as constructive interaction, resulting bonding orbitals. On the other hand, if destructive interaction, it will create an antibonding interaction. The two molecular orbitals are the result from adding and subtracting the two wavefunctions (4). Consider phosphorus pentaflouride (PF ). P, the central atom, has five valence electrons, with three lone electrons. Thus, P should only be able to form three bonds. However, in order for PF to exist with P as the central atoms, P must be able to bond five times. This event is described by . In order to create degenerate hybrid orbitals that allow atoms to bond well beyond their normal amount. Orbitals involved in hybridization are the s, p, and d atomic orbitals. There are certain principles that must be followed: Hybrid orbitals consist of sp, sp , sp , sp d, sp d, and sp d (3). Returning back to PF , let's see how hybrid orbitals can describe its bonds. Remember that P can only bond three times, but we need it to bond five times. In order to bond five times, five orbitals must be singularly filled in. Bonds occur when orbitals with only one electron are spin paired with the electron from another atom. The five orbitals can be acquired by sp d orbital hybridization. Starting off with a total of five orbitals will result in five hybrid orbitals. Note that since P is in period 3, it has d orbitals in the same energy level. Thus, 3d orbitals can be used to hybridize, even though electrons do not occupy it. The process is outlined in the original figure below. Remember to follow the , , and the when assigning electrons to their orbitals. Note that valence electrons are only shown. Other orbitals and their corresponding electrons are not hybridized and are not involved in bonding. Although they are not shown, they still exist. Now, P has five hybrid orbitals. Remember that P has five valence electrons. Electrons too are not lost or gained, so those electrons transfer over to the hybrid orbitals. Fill them in accordingly, following Hund's rule and the Pauli exclusion principle. Note that the hybrid orbitals are now degenerate, so the aufbau principle doesn't apply for this case. From sp d hybridization, P has five unpaired electrons and can now bond with five F. Generally, if two bonds are needed, then use an sp; for three, use sp ; the number of bonds needed equals the number of orbitals that need to be hybridized. Please take into account that lone pairs apply as well. If there is a lone pair when you complete the Lewis structure, then the number of required hybrid orbitals equals the number of lone pairs plus the number of bonds; this is also called the coordinate number. PF , although not shown, does not have lone pairs. The type of hybrid orbitals also corresponds to the molecular shape. In the case of PF , its shape as defined by is trigonal bypyramidal. If hybrid orbitals are involved, the shape can also be determined by the type of hybridization. As another example, sp is linear, and sp is trigonal planar. sp d is square planar. Think over carefully why the type of hybridization also determines the shape. An example problem will aid you in your thinking in the problems section. The wavefunction describes the state of an electron. From the name of the function, one can derive that the electron can behave like a wave. This wave-like behavior of the electrons defines the shapes of the orbitals. Thus, it makes sense that wavefunctions are related to the Valence Bond Theory. If orbitals overlap to create bonds, and orbital shapes and the state of an electron is described by the wavefunction, then it makes sense that the overlap of orbitals (the bonds) can be described by wavefunctions as well. Thus, covalent and ionic bonds can be described by wavefunctions. Covalent and ionic representations of a bond represent the same bond, but they differ in how the electro ns are placed. This is called resonance. Different intermolecular interactions give rise to different wavefunctions. Recall from Hund's Rule and the Pauli exclusion principle that electrons must be spin paired when the right conditions are met. Due to this, there are two separate ways to represent a covalent bond in terms of electron spin, which is related to the wavefunction. The figure shows the two different cases for two electrons in a bond. Thus, it is possible to write two different wavefunctions that describe each case, which can be superimposed to describe the overall covalent bond (2). The superposition of the covalent bond and ionic bond wavefunctions will result in an overall wavefunction that describes the state of the molecule. Due to this module being an overview of Valence Bond Theory, the full details will not be covered. The wavefunction squared will result in a probability density. The wavefunction alone has no physical significance; however, when the wavefunction is squared, the square wavefunction can determine where the electron is most likely located. Recall that when atomic orbitals overlap, the electrons are localized and more likely to be found within that overlap. In terms of electrostatic interactions, this bond will result in some form of equilibrium between all of the electrostatic forces (between the electrons and the electrons with each nucleus) (1). If the overlap is too far in, there is a net repulsion force; the electrons will also be forced apart. If the atomic orbital overlap is too small, the net attraction force is very small; the electron will have a smaller chance of remaining in the overlap since they will be more attracted to their own nucleus. If the overlap is just right, then the electrons are attracted to both nuclei and more likely to stay in the overlapped area. Thus, wavefunctions and electrostatic forces can both explain why electrons are localized within their bonds. As one can see, Valence Bond Theory can help describe how bonds are formed. However, there are some notable failures when it comes to Valence Bond Theory. One such failure is dioxygen. Valence Bond Theory fails to predict dioxygen's paramagnitism; it predicts that oxygen is diamagnetic. A species is paramagnetic if electrons are not spin paired and diamagnetic if the electrons are spin paired. Since Valence Bond theory begins with the basis that atomic orbitals overlap to create bonds and through that reasoning, one can see that electrons are spin paired when bonds overlap, dioxygen is indeed predicted to be diamagnetic if Valence Bond Theory is used. In reality, that is not the case. Also, sp d and sp both have a coordinate number of four. Thus, Valence Bond Theory cannot predict whether the molecule is a square planar or the other shape (3). One must correctly draw the Lewis structure and use VSEPR to determine the shape. HF: Hydrogen has one electron in the 1s orbital. Flourine has two electrons in the 2s orbital, and five in the 2p orbitals. These can be derived from the electron configurations. We are only interested in the valence electrons, so the lower shell can be ignored. Remember to fill in the orbitals accordingly. The orbitals availiable to bond are Hydrogen's 1s orbital and one of Flourine's 2p orbitals. Thus, the bond must take place between these two orbitals. This is a sigma bond. A bond of this type, or one of similar orientation, such as two s orbitals, are the bonds associated with hybrid orbitals when hybridization takes place. CO : C has four valence electrons; two are in 2s and two are in 2p. O has six valence electrons; two in 2s and four in 2p. Notice that there is a double bond. Now, hybridize to explain the bonds. When determining the coordinate number, double bonds and triple bonds count as one bond. Thus, the coordinate number of C is 2 and the corresponding hybridization is sp. The diagram below shows the sp hybridization process for C. Notice that two 2p orbital remains unhybridized. We fill in the hybrid orbitals with two of C's electrons so that C has space available to bond with the two O's. The hybrid orbitals describes two sigma bonds between C and the O's; one 2p orbital from each O is used in this process. Two electrons remain for C, which are allocated to the unhybridized orbitals. Now, C can form the other two bonds for a total of four which is what is expected due to the presence of two double bonds. The unhybridized orbitals overlap O's 2p orbitals and forms what is called a pi bond. There is one pi bond and one sigma bond between C and each O. The lone pairs of the O's are the filled 2s orbital and the final and filled 2p orbital. Thus, a double bond is explained by one sigma bond associated with the hybrid orbitals and a pi bond associated with unhybridized p orbitals. C H : C has four valence electrons (two in 2s and two in 2p) and H has one (one in 1s). There is a total of ten valence electrons. From the Lewis structure, one can see that there is a triple bond between the two C's. For each C, one can explain the bonds through sp hybridization (a triple bond and one single bond). This process is similar to CO . However, in this case, C's available unhybridized 2p orbitals bond together with the unhybridized p orbitals of the other C. Now, there are two pi bonds and one sigma bond between the C's; there is one sigma bond between each C and H bond. Thus, a triple bond is explained by one sigma bond associated with the hybrid orbitals and two pi bonds associated with unhybridized p orbitals. 2. I is the central atom. Usually, the least electronegative atom is the central atom. Drawing the Lewis structure reveals that I can only form one bond. However, with the use of hybridization, I can form more bonds. Showing that is the goal. First, find out how many valence electrons IBr has. This is simply derived by adding all of the valence electrons from each atom; all have 7 valence electrons, so (6)(7)=42. Then, use Lewis structure and allocate all of the electrons into the correct places. You will find that the Lewis structure looks like the picture below. Notice that there is a lone pair. Remember that the number of hybrid orbitals needed is the summation of lone pairs and bonds. This is a total of six; thus a sp d hybridization is needed. This process is shown below. Take note that the lone pair is the only spin paired electrons in the hybrid orbitals. Since the coordinate number is six, then the family shape of this molecule is an octahedral shape. Even the lone pairs are accounted for when hybridization takes place. It is because we used the coordinate number (determined by using Lewis dot structure first) to determiine the type of hybridization that hybridization does indeed give the correct family shape. There are of course exceptions, such as sp d and sp . We cannot predict the shape based on hybridization alone for these. 3. The shapes are tetrahedral, trigonal bypyramidal, and octahedral respectively. Not all molecules will have the shape associated with the hybridization type because lone pairs must be taken into account as well. For example, look at problem number two. The coordinate number is six. That tells you that the is an octahedral. However, the shape of IBr will only be that shape if the lone pair was another atom. The parent shape will not always match the actual shape of the molecule. It gives you a place to start and is only true if lone pairs do not exist. "Removing" one bond away from the octahedral shape will give the result of the square-base pyramidal shape. 4. One can solve this problem simply by counting the valence electrons. H has only one electron; it is not paired with any other electron and thus must be paramagnetic. H has two valence electrons; thus, those electrons are spin paired and H is diamagnetic. NO has eleven valence electrons; it is paramagnetic. In general, if an atom/molecule has an odd number of electrons, then that atom/molecule is paramagnetic. It is diamagnetic if it has an even number of electrons; however, like the case of O , this does not always work. 5. Valence Bond Theory looks at the interaction between orbitals to describe bonds. It can also be used to derive the shape of the molecule in question, as well as determining whether or not an atom/molecule is diamagnetic or paramagnetic; however Valence Bond Theory is not always reliable. It fails in some cases. One must always remember that this is a theory. | 15,567 | 1,278 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Statistical_Mechanics/Boltzmann_Average/Ideal_Gas_Partition_Function |
The canonical ensemble partition function, , for a system of identical particles each of mass is given by \[Q_{NVT}=\frac{1}{N!}\frac{1}{h^{3N}}\int\int d{\mathbf p}^N d{\mathbf r}^N \exp \left[ - \frac{H({\mathbf p}^N,{\mathbf r}^N)}{k_B T}\right]\] where is Planck's constant, is the temperature and \(k_B\) is the Boltzmann constant. When the particles are distinguishable then the factor disappears. \(H(p^N, r^N)\) is the Hamiltonian corresponding to the total energy of the system. is a function of the positions and momenta of the particles in the system. The Hamiltonian can be written as the sum of the kinetic and the potential energies of the system as follows \[H({\mathbf p}^N, {\mathbf r}^N)= \sum_{i=1}^N \frac{|{\mathbf p}_i |^2}{2m} + {\mathcal V}({\mathbf r}^N)\] Thus we have \[Q_{NVT}=\frac{1}{N!}\frac{1}{h^{3N}}\int d{\mathbf p}^N \exp \left[ - \frac{|{\mathbf p}_i |^2}{2mk_B T}\right] \int d{\mathbf r}^N \exp \left[ - \frac{{\mathcal V}({\mathbf r}^N)} {k_B T}\right]\] This separation is only possible if \({\mathcal V}({\mathbf r}^N)\) is independent of velocity (as is generally the case). The momentum integral can be solved analytically: \[\int d{\mathbf p}^N \exp \left[ - \frac{|{\mathbf p} |^2}{2mk_B T}\right]=(2 \pi m k_B T)^{3N/2}\] Thus we have \[Q_{NVT}=\frac{1}{N!} \frac{1}{h^{3N}} \left( 2 \pi m k_B T\right)^{3N/2} \int d{\mathbf r}^N \exp \left[ - \frac{{\mathcal V}({\mathbf r}^N)} {k_B T}\right]\] The integral over positions is known as the configuration integral, \(Z_{NVT}\) (from the German meaning "sum over states") \[Z_{NVT}= \int d{\mathbf r}^N \exp \left[ - \frac{{\mathcal V}({\mathbf r}^N)} {k_B T}\right]\] In an ideal gas there are no interactions between particles so \({\mathcal V}({\mathbf r}^N)=0\). Thus \(\exp(-{\mathcal V}({\mathbf r}^N)/k_B T)=1\) for every gas particle. The integral of 1 over the coordinates of each atom is equal to the volume so for particles the is given by \(V^N\) where is the volume. Thus we have \[Q_{NVT}=\frac{V^N}{N!}\left( \frac{2 \pi m k_B T}{h^2}\right)^{3N/2}\] If we define the de Broglie thermal wavelength as \(\Lambda\) where \[\Lambda = \sqrt{h^2 / 2 \pi m k_B T}\] one arrives at (Eq. 4-12 in ) \[Q_{NVT}=\frac{1}{N!} \left( \frac{V}{\Lambda^{3}}\right)^N = \frac{q^N}{N!}\] where \[q= \frac{V}{\Lambda^{3}}\] is the single particle translational partition function. Thus one can now write the partition function for a real system can be built up from the contribution of the ideal system (the momenta) and a contribution due to particle interactions, \[Q_{NVT}=Q_{NVT}^{\rm ideal} ~Q_{NVT}^{\rm excess}\] | 2,655 | 1,282 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/04%3A_Transport/17%3A_Directed_and_Active_Transport/17.4%3A_Polymerization_Ratchet_and_Translocation_Ratchet |
\[J(x) = -D \left( \dfrac{\partial P}{\partial x} + \dfrac{f}{k_BT}P \right) \] \[ \dfrac{\partial P}{\partial t} = -\dfrac{\partial J}{\partial x} \nonumber \] gener ly and irreversibly bind once the diffusing chain reaches an extension \(\Delta x\). (No back-stepping is allowed.) And we use the conservation statement: \[ \int_0^{\Delta x} dx P(x) = 1 \nonumber \] which says that a protein must be bound within the interval 0 to Δx. The steady-state probability distribution with these two boundary conditions is \[ P(x) = \dfrac{ \underset{ \sim }{f} \left[ \exp \left( \underset{ \sim }{f} (1-x/\Delta x) \right) -1 \right] }{\Delta x\left( 1+ \underset{ \sim }{f} - e^{ \underset{ \sim }{f} } \right) } \] \[ \underset{ \sim }{f} = \dfrac{f \Delta x}{k_BT} \nonumber \] \( \underset{\sim}{f}\) is a dimensionless constant that expresses the load force in units of k T opposing ratchet displacement by Δx. Substituting eq. (17.4.2) into eq. (17.4.1) allows us to solve for J. \[J(x) =\dfrac{-D\underset{ \sim }{f}^2 }{\Delta x^2\left( 1+ \underset{ \sim }{f}-e^{\underset{ \sim }{f}} \right) } \left( 1-2\exp \left[ \underset{ \sim }{f} \left( \dfrac{x}{\Delta x} -1 \right) \right] \right) \nonumber \] Now, the average velocity can be determined from \(\langle \nu \rangle = J\Delta x \). Evaluating the flux at x = Δx: \[ \langle \nu \rangle =\dfrac{2D}{\Delta x} \left[ \dfrac{\underset{ \sim }{f}^2/2}{e^{\underset{ \sim }{f}}-\underset{ \sim }{f}-1} \right] \nonumber \] Now look at low force limit \(f \rightarrow 0 \). Expand \(e^{\underset{ \sim }{f}}=1+\underset{ \sim }{f}+\underset{ \sim }{f}^2/2 \): \[ \langle \nu \rangle \rightarrow \dfrac{2D}{\Delta x} = v_{max} \nonumber \] Note that this is the maximum velocity for ideal ratchet, and it follows the expected behavior for pure diffusive motion. Now consider probability of the protein binding is governed by equilibrium between free and bound forms: \[ F \overset{k_a}{\underset{k_d} \rightleftharpoons} B \qquad \qquad K= \dfrac{k_a}{k_d} = \dfrac{\tau_B}{\tau_F} \nonumber \] Here k refers to the effecting quasi-first-order rate constant for binding at a chaperone concentration [chap]: \(k_a = k_a' [ chap ] \). \[ \begin{aligned} &\langle \nu \rangle = \dfrac{2D}{\Delta x} \left[ \dfrac{\underset{ \sim }{f}^2/2}{\dfrac{e^{\underset{ \sim }{f}}-1}{1-K(e^{\underset{ \sim }{f}}-1)}-\underset{ \sim }{f}} \right] \\ &\langle \nu \rangle_{max} = \dfrac{2D}{\Delta x} \left( \dfrac{1}{1+2K} \right) \end{aligned} \] \[ f_0 = \dfrac{k_BT}{\Delta x} \ln \left( 1+ \dfrac{1}{K} \right) \nonumber \] ________________________________ | 2,669 | 1,283 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/24%3A_Nuclear_Chemistry/24.04%3A_The_Interaction_of_Nuclear_Radiation_with_Matter |
Because nuclear reactions do not typically affect the valence electrons of the atom (although electron capture draws an electron from an orbital of the lowest energy level), they do not directly cause chemical changes. Nonetheless, the particles and the photons emitted during nuclear decay are very energetic, and they can indirectly produce chemical changes in the matter surrounding the nucleus that has decayed. For instance, an α particle is an ionized helium nucleus (He ) that can act as a powerful oxidant. In this section, we describe how radiation interacts with matter and the some of the chemical and biological effects of radiation. The effects of radiation on matter are determined primarily by the energy of the radiation, which depends on the nuclear decay reaction that produced it. Nonionizing radiation is relatively low in energy; when it collides with an atom in a molecule or an ion, most or all of its energy can be absorbed without causing a structural or a chemical change. Instead, the kinetic energy of the radiation is transferred to the atom or molecule with which it collides, causing it to rotate, vibrate, or move more rapidly. Because this energy can be transferred to adjacent molecules or ions in the form of heat, many radioactive substances are warm to the touch. Highly radioactive elements such as polonium, for example, have been used as heat sources in the US space program. As long as the intensity of the nonionizing radiation is not great enough to cause overheating, it is relatively harmless, and its effects can be neutralized by cooling. In contrast, ionizing radiation is higher in energy, and some of its energy can be transferred to one or more atoms with which it collides as it passes through matter. If enough energy is transferred, electrons can be excited to very high energy levels, resulting in the formation of positively charged ions:
\( atom + ionizing \; radiation \rightarrow ion^{+} + e^{-} \tag{24.3.1} \) Molecules that have been ionized in this way are often highly reactive, and they can decompose or undergo other chemical changes that create a cascade of reactive molecules that can damage biological tissues and other materials ( ). Because the energy of ionizing radiation is very high, we often report its energy in units such as megaelectronvolts (MeV) per particle: 1 MeV/particle = 96 billion J/mol. The effects of ionizing radiation depend on four factors: The relative abilities of the various forms of ionizing radiation to penetrate biological tissues are illustrated in . Because of its high charge and mass, α radiation interacts strongly with matter. Consequently, it does not penetrate deeply into an object, and it can be stopped by a piece of paper, clothing, or skin. In contrast, γ rays, with no charge and essentially no mass, do not interact strongly with matter and penetrate deeply into most objects, including the human body. Several inches of lead or more than 12 inches of special concrete are needed to completely stop γ rays. Because β particles are intermediate in mass and charge between α particles and γ rays, their interaction with matter is also intermediate. Beta particles readily penetrate paper or skin, but they can be stopped by a piece of wood or a relatively thin sheet of metal. Because of their great penetrating ability, γ rays are by far the most dangerous type of radiation when they come from a source the body. Alpha particles, however, are the most damaging if their source is the body because internal tissues absorb all of their energy. Thus danger from radiation depends strongly on the type of radiation emitted and the extent of exposure, which allows scientists to safely handle many radioactive materials if they take precautions to avoid, for example, inhaling fine particulate dust that contains alpha emitters. Some properties of ionizing radiation are summarized in Born in the Lower Rhine Province of Germany, Röntgen was the only child of a cloth manufacturer and merchant. His family moved to the Netherlands where he showed no particular aptitude in school, but where he was fond of roaming the countryside. Röntgen was expelled from technical school in Utrecht after being unjustly accused of drawing a caricature of one of the teachers. He began studying mechanical engineering in Zurich, which he could enter without having the credentials of a regular student, and received a PhD at the University of Zurich in 1869. In 1876 he became professor of physics. There are many different ways to measure radiation exposure, or the . The roentgen (R) , which measures the amount of energy absorbed by dry air, can be used to describe quantitative exposure. Damage to biological tissues, however, is proportional to the amount of energy absorbed by , not air. The most common unit used to measure the effects of radiation on biological tissue is the rad (radiation absorbed dose) ; the SI equivalent is the (Gy). The rad is defined as the amount of radiation that causes 0.01 J of energy to be absorbed by 1 kg of matter, and the gray is defined as the amount of radiation that causes 1 J of energy to be absorbed per kilogram:
\( 1 \; rad=0.010 \; J/kg \;\;\;\; 1 \; Gy=1 \; J/kg \tag{24.3.2} \) Thus a 70 kg human who receives a dose of 1.0 rad over his or her entire body absorbs 0.010 J/70 kg = 1.4 × 10 J, or 0.14 mJ. To put this in perspective, 0.14 mJ is the amount of energy transferred to your skin by a 3.8 × 10 g droplet of boiling water. Because the energy of the droplet of water is transferred to a relatively large area of tissue, it is harmless. A radioactive particle, however, transfers its energy to a single molecule, which makes it the atomic equivalent of a bullet fired from a high-powered rifle. Because α particles have a much higher mass and charge than β particles or γ rays, the difference in mass between α and β particles is analogous to being hit by a bowling ball instead of a table tennis ball traveling at the same speed. Thus the amount of tissue damage caused by 1 rad of α particles is much greater than the damage caused by 1 rad of β particles or γ rays. Thus a unit called the rem (roentgen equivalent in man) was devised to describe the amount of tissue damage caused by a given amount of radiation. The number of rems of radiation is equal to the number of rads multiplied by the (relative biological effectiveness) factor, which is 1 for β particles, γ rays, and x-rays and about 20 for α particles. Because actual radiation doses tend to be very small, most measurements are reported in (1 mrem = 10 rem). We are continuously exposed to measurable background radiation from a variety of natural sources, which, on average, is equal to about 150–600 mrem/yr ( ). One component of background radiation is , high-energy particles and γ rays emitted by the sun and other stars, which bombard Earth continuously. Because cosmic rays are partially absorbed by the atmosphere before they reach Earth’s surface, the exposure of people living at sea level (about 30 mrem/yr) is significantly less than the exposure of people living at higher altitudes (about 50 mrem/yr in Denver, Colorado). Every 4 hours spent in an airplane at greater than 30,000 ft adds about 1 mrem to a person’s annual radiation exposure. Data source: Office of Civilian Radioactive Waste Management A second component of background radiation is , produced by the interaction of cosmic rays with gases in the upper atmosphere. When high-energy cosmic rays collide with oxygen and nitrogen atoms, neutrons and protons are released. These, in turn, react with other atoms to produce radioactive isotopes, such as C:
\( _{7}^{14}N + _{0}^{1}n\rightarrow _{6}^{14}C + _{1}^{1}p \tag{24.3.4} \) The carbon atoms react with oxygen atoms to form CO , which is eventually washed to Earth’s surface in rain and taken up by plants. About 1 atom in 1 × 10 of the carbon atoms in our bodies is radioactive C, which decays by beta emission. About 5000 C nuclei disintegrate in your body during the 15 s or so that it takes you to read this paragraph. Tritium ( H) is also produced in the upper atmosphere and falls to Earth in precipitation. The total radiation dose attributable to C is estimated to be 1 mrem/yr, while that due to H is about 1000 times less. The third major component of background radiation is , which is due to the remnants of radioactive elements that were present on primordial Earth and their decay products. For example, many rocks and minerals in the soil contain small amounts of radioactive isotopes, such as Th and U, as well as radioactive daughter isotopes, such as Ra. The amount of background radiation from these sources is about the same as that from cosmic rays (approximately 30 mrem/yr). These isotopes are also found in small amounts in building materials derived from rocks and minerals, which significantly increases the radiation exposure for people who live in brick or concrete-block houses (60–160 mrem/yr) instead of houses made of wood (10–20 mrem/yr). Our tissues also absorb radiation (about 40 mrem/yr) from naturally occurring radioactive elements that are present in our bodies. For example, the average adult contains about 140 g of potassium as the K ion. Naturally occurring potassium contains 0.0117% K, which decays by emitting both a β particle and a γ ray. In the last 20 seconds, about the time it took you to read this paragraph, approximately 40,000 K nuclei disintegrated in your body. By far the most important source of background radiation is , the heaviest of the noble gases (group 18). Radon-222 is produced during the decay of U, and other isotopes of radon are produced by the decay of other heavy elements. Even though radon is chemically inert, all its isotopes are radioactive. For example, Rn undergoes two successive alpha-decay events to give Pb:
\( _{86}^{222}Rn \rightarrow _{84}^{218}Po + _{2}^{4}\alpha \rightarrow _{82}^{214}Pb + _{2}^{4}\alpha \tag{24.3.5} \) Because radon is a dense gas, it tends to accumulate in enclosed spaces such as basements, especially in locations where the soil contains greater-than-average amounts of naturally occurring uranium minerals. Under most conditions, radioactive decay of radon poses no problems because of the very short range of the emitted α particle. If an atom of radon happens to be in your lungs when it decays, however, the chemically reactive daughter isotope polonium-218 can become irreversibly bound to molecules in the lung tissue. Subsequent decay of Po releases an α particle directly into one of the cells lining the lung, and the resulting damage can eventually cause lung cancer. The Po isotope is also readily absorbed by particles in cigarette smoke, which adhere to the surface of the lungs and can hold the radioactive isotope in place. Recent estimates suggest that radon exposure is a contributing factor in about 15% of the deaths due to lung cancer. Because of the potential health problem radon poses, many states require houses to be tested for radon before they can be sold. By current estimates, radon accounts for more than half of the radiation exposure of a typical adult in the United States. In addition to naturally occurring background radiation, humans are exposed to small amounts of radiation from a variety of artificial sources. The most important of these are the x-rays used for diagnostic purposes in medicine and dentistry, which are photons with much lower energy than γ rays. A single chest x-ray provides a radiation dose of about 10 mrem, and a dental x-ray about 2–3 mrem. Other minor sources include television screens and computer monitors with cathode-ray tubes, which also produce x-rays. Luminescent paints for watch dials originally used radium, a highly toxic alpha emitter if ingested by those painting the dials. Radium was replaced by tritium ( H) and promethium ( Pr), which emit low-energy β particles that are absorbed by the watch crystal or the glass covering the instrument. Radiation exposure from television screens, monitors, and luminescent dials totals about 2 mrem/yr. Residual fallout from previous atmospheric nuclear-weapons testing is estimated to account for about twice this amount, and the nuclear power industry accounts for less than 1 mrem/yr (about the same as a single 4 h jet flight). Calculate the annual radiation dose in rads a typical 70 kg chemistry student receives from the naturally occurring K in his or her body, which contains about 140 g of potassium (as the K ion). The natural abundance of K is 0.0117%. Each 1.00 mol of K undergoes 1.05 × 10 decays/s, and each decay event is accompanied by the emission of a 1.32 MeV β particle. mass of student, mass of isotope, natural abundance, rate of decay, and energy of particle annual radiation dose in rads Calculate the number of moles of K present using its mass, molar mass, and natural abundance. Determine the number of decays per year for this amount of K. Multiply the number of decays per year by the energy associated with each decay event. To obtain the annual radiation dose, use the mass of the student to convert this value to rads. The number of moles of K present in the body is the total number of potassium atoms times the natural abundance of potassium atoms present as K divided by the atomic mass of K: \( moles \; ^{40}K= 140 \cancel{gK}\times \dfrac{0.0117 \; mol ^{40}K}{100 \; \cancel{mol\;K}} \times \dfrac{1\;\cancel{mol\;K}}{40.0\;\cancel{gK}} = 4.10 \times 10^{-4} \;mol\;^{40}K \) We are given the number of atoms of K that decay per second in 1.00 mol of K, so the number of decays per year is as follows: \( \dfrac{decays}{year}= 4.10 \times 10^{-4} \;mol\;^{40}K \times \dfrac{1.05 \times 10^{7} \; decays/\cancel{s}}{1 \; \cancel{mol\;^{40}K}} \times \dfrac{60\;\cancel{s}}{1\;\cancel{min}}\times \dfrac{60\;\cancel{min}}{1\;\cancel{h}} \times \dfrac{24\;\cancel{h}}{1\; \cancel{day}}\times \dfrac{365\;\cancel{day}}{1\;yr} ) The total energy the body receives per year from the decay of K is equal to the total number of decays per year multiplied by the energy associated with each decay event: \( total \; energy \; per \; year= \dfrac{1.36 \times 10^{11} \; \cancel{decays}}{yr} \times \dfrac{1.32 \; \cancel{MeV}}{\cancel{decay}}\times \dfrac{10^{6} \; \cancel{eV}}{\cancel{MeV}} \times \dfrac{1.602 \times 10^{-19} \; J}{\cancel{eV}} \) \( =2.87 \times 10^{-2}\;J/yr \) We use the definition of the rad (1 rad = 10 J/kg of tissue) to convert this figure to a radiation dose in rads. If we assume the dose is equally distributed throughout the body, then the radiation dose per year is as follows: \( radiation \; dose \; per \; year= \dfrac{2.87 \times 10^{-2} \; \cancel{J}/yr}{70 \; \cancel{kg}} \times \dfrac{1 \; rad}{1 \times 10^{-2} \; \cancel{J}/\cancel{kg}} \) \( =4.10 \times 10^{-2}\;rad/yr = 41\; mrad/yr \) This corresponds to almost half of the normal background radiation most people experience. Exercise Because strontium is chemically similar to calcium, small amounts of the Sr ion are taken up by the body and deposited in calcium-rich tissues such as bone, using the same mechanism that is responsible for the absorption of Ca . Consequently, the radioactive strontium ( Sr) found in fission waste and released by atmospheric nuclear-weapons testing is a major health concern. A normal 70 kg human body has about 280 mg of strontium, and each mole of Sr undergoes 4.55 × 10 decays/s by the emission of a 0.546 MeV β particle. What would be the annual radiation dose in rads for a 70 kg person if 0.10% of the strontium ingested were Sr? 5.7 × 10 rad/yr (which is 10 times the fatal dose) One of the more controversial public policy issues debated today is whether the radiation exposure from artificial sources, when combined with exposure from natural sources, poses a significant risk to human health. The effects of single radiation doses of different magnitudes on humans are listed in . Because of the many factors involved in radiation exposure (length of exposure, intensity of the source, and energy and type of particle), it is difficult to quantify the specific dangers of one radioisotope versus another. Nonetheless, some general conclusions regarding the effects of radiation exposure are generally accepted as valid. Radiation doses of 600 rem and higher are invariably fatal, while a dose of 500 rem kills half the exposed subjects within 30 days. Smaller doses (≤ 50 rem) appear to cause only limited health effects, even though they correspond to tens of years of natural radiation. This does not, however, mean that such doses have no ill effects; they may cause long-term health problems, such as cancer or genetic changes that affect offspring. The possible detrimental effects of the much smaller doses attributable to artificial sources (< 100 mrem/yr) are more difficult to assess. The tissues most affected by large, whole-body exposures are bone marrow, intestinal tissue, hair follicles, and reproductive organs, all of which contain rapidly dividing cells. The susceptibility of rapidly dividing cells to radiation exposure explains why cancers are often treated by radiation. Because cancer cells divide faster than normal cells, they are destroyed preferentially by radiation. Long-term radiation-exposure studies on fruit flies show a linear relationship between the number of genetic defects and both the magnitude of the dose and the exposure time. In contrast, similar studies on mice show a much lower number of defects when a given dose of radiation is spread out over a long period of time rather than received all at once. Both patterns are plotted in , but which of the two is applicable to humans? According to one hypothesis, mice have very low risk from low doses because their bodies have ways of dealing with the damage caused by natural radiation. At much higher doses, however, their natural repair mechanisms are overwhelmed, leading to irreversible damage. Because mice are biochemically much more similar to humans than are fruit flies, many scientists believe that this model also applies to humans. In contrast, the linear model assumes that exposure to radiation is intrinsically damaging and suggests that stringent regulation of low-level radiation exposure is necessary. Which view is more accurate? The answer—while yet unknown—has extremely important consequences for regulating radiation exposure. The effects of radiation on matter depend on the energy of the radiation. is relatively low in energy, and the energy is transferred to matter in the form of heat. is relatively high in energy, and when it collides with an atom, it can completely remove an electron to form a positively charged ion that can damage biological tissues. Alpha particles do not penetrate very far into matter, whereas γ rays penetrate more deeply. Common units of radiation exposure, or dose, are the , the amount of energy absorbed by dry air, and the , the amount of radiation that produces 0.01 J of energy in 1 kg of matter. The measures the actual amount of tissue damage caused by a given amount of radiation. Natural sources of radiation include , consisting of high-energy particles and γ rays emitted by the sun and other stars; , which is produced by the interaction of cosmic rays with gases in the upper atmosphere; and , from radioactive elements present on primordial Earth and their decay products. The risks of ionizing radiation depend on the intensity of the radiation, the mode of exposure, and the duration of the exposure. : 1 rad = 0.01 J/kg Why are many radioactive substances warm to the touch? Why do many radioactive substances glow? Describe the differences between nonionizing and ionizing radiation in terms of the intensity of energy emitted and the effect each has on an atom or molecule after collision. Which nuclear decay reactions are more likely to produce ionizing radiation? nonionizing radiation? Would you expect nonionizing or ionizing radiation to be more effective at treating cancer? Why? Historically, concrete shelters have been used to protect people from nuclear blasts. Comment on the effectiveness of such shelters. Gamma rays are a very high-energy radiation, yet α particles inflict more damage on biological tissue. Why? List the three primary sources of naturally occurring radiation. Explain the factors that influence the dose that one receives throughout the year. Which is the largest contributor to overall exposure? Which is the most hazardous? Because radon is a noble gas, it is inert and generally unreactive. Despite this, exposure to even low concentrations of radon in air is quite dangerous. Describe the physical consequences of exposure to radon gas. Why are people who smoke more susceptible to these effects? Most medical imaging uses isotopes that have extremely short half-lives. These isotopes usually undergo only one kind of nuclear decay reaction. Which kind of decay reaction is usually used? Why? Why would a short half-life be preferred in these cases? Which would you prefer: one exposure of 100 rem, or 10 exposures of 10 rem each? Explain your rationale. Ionizing radiation is higher in energy and causes greater tissue damage, so it is more likely to destroy cancerous cells. Ten exposures of 10 rem are less likely to cause major damage. A 2.14 kg sample of rock contains 0.0985 g of uranium. How much energy is emitted over 25 yr if 99.27% of the uranium is U, which has a half-life of 4.46 × 10 yr, if each decay event is accompanied by the release of 4.039 MeV? If a 180 lb individual absorbs all of the emitted radiation, how much radiation has been absorbed in rads? There is a story about a “radioactive boy scout” who attempted to convert thorium-232, which he isolated from about 1000 gas lantern mantles, to uranium-233 by bombarding the thorium with neutrons. The neutrons were generated via bombarding an aluminum target with α particles from the decay of americium-241, which was isolated from 100 smoke detectors. Write balanced nuclear reactions for these processes. The “radioactive boy scout” spent approximately 2 h/day with his experiment for 2 yr. Assuming that the alpha emission of americium has an energy of 5.24 MeV/particle and that the americium-241 was undergoing 3.5 × 10 decays/s, what was the exposure of the 60.0 kg scout in rads? The intrepid scientist apparently showed no ill effects from this exposure. Why? | 22,522 | 1,284 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/20%3A_Periodic_Trends_and_the_s-Block_Elements/20.03%3A_The_Chemistry_of_Hydrogen |
We now turn from an overview of periodic trends to a discussion of the s-block elements, first by focusing on hydrogen, whose chemistry is sufficiently distinct and important to be discussed in a category of its own. Most versions of the periodic table place hydrogen in the upper left corner immediately above lithium, implying that hydrogen, with a 1s electron configuration, is a member of group 1. In fact, the chemistry of hydrogen does not greatly resemble that of the metals of Group 1. Indeed, some versions of the periodic table place hydrogen above fluorine in because the addition of a single electron to a hydrogen atom completes its valence shell. Although hydrogen has an ns electron configuration, its chemistry does not resemble that of the . Hydrogen, the most abundant element in the universe, is the ultimate source of all other elements by the process of nuclear fusion. Table \(\Page {1}\) "The Isotopes of Hydrogen" compares the three isotopes of hydrogen, all of which contain one proton and one electron per atom. The most common isotope is ( H or H), followed by ( H or D), which has an additional neutron. The rarest isotope of hydrogen is ( H or T), which is produced in the upper atmosphere by a nuclear reaction when cosmic rays strike nitrogen and other atoms; it is then washed into the oceans by rainfall. Tritium is radioactive, decaying to He with a half-life of only 12.32 years. Consequently, the atmosphere and oceans contain only a very low, steady-state level of tritium. The term hydrogen and the symbol H normally refer to the naturally occurring mixture of the three isotopes. The different masses of the three isotopes of hydrogen cause them to have different physical properties. Thus H , D , and T differ in their melting points, boiling points, densities, and heats of fusion and vaporization. In 1931, Harold Urey and coworkers discovered deuterium by slowly evaporating several liters of liquid hydrogen until a volume of about 1 mL remained. When that remaining liquid was vaporized and its emission spectrum examined, they observed new absorption lines in addition to those previously identified as originating from hydrogen. The natural abundance of tritium, in contrast, is so low that it could not be detected by similar experiments; it was first prepared in 1934 by a nuclear reaction. Urey won the Nobel Prize in Chemistry in 1934 for his discovery of deuterium ( H). Urey was born and educated in rural Indiana. After earning a BS in zoology from the University of Montana in 1917, Urey changed career directions. He earned his PhD in chemistry at Berkeley with G. N. Lewis and subsequently worked with Niels Bohr in Copenhagen. During World War II, Urey was the director of war research for the Atom Bomb Project at Columbia University. In later years, his research focused on the evolution of life. In 1953, he and his graduate student, Stanley Miller, showed that organic compounds, including amino acids, could be formed by passing an electric discharge through a mixture of compounds thought to be present in the atmosphere of primitive Earth. Because the normal boiling point of D O is 101.4°C (compared to 100.0°C for H O), evaporation or fractional distillation can be used to increase the concentration of deuterium in a sample of water by the selective removal of the more volatile H O. Thus bodies of water that have no outlet, such as the Great Salt Lake and the Dead Sea, which maintain their level solely by evaporation, have significantly higher concentrations of deuterated water than does lake or seawater with at least one outlet. A more efficient way to obtain water highly enriched in deuterium is by prolonged electrolysis of an aqueous solution. Because a deuteron (D ) has twice the mass of a proton (H ), it diffuses more slowly toward the electrode surface. Consequently, the gas evolved at the cathode is enriched in H, the species that diffuses more rapidly, favoring the formation of H over D or HD. Meanwhile, the solution becomes enriched in deuterium. Deuterium-rich water is called heavy water because the density of D O (1.1044 g/cm at 25°C) is greater than that of H O (0.99978 g/cm ). Heavy water was an important constituent of early nuclear reactors. Because deuterons diffuse so much more slowly, D O will not support life and is actually toxic if administered to mammals in large amounts. The rate-limiting step in many important reactions catalyzed by enzymes involves proton transfer. The transfer of D is so slow compared with that of H because bonds to D break more slowly than those to H, so the delicate balance of reactions in the cell is disrupted. Nonetheless, deuterium and tritium are important research tools for biochemists. By incorporating these isotopes into specific positions in selected molecules, where they act as labels, or tracers, biochemists can follow the path of a molecule through an organism or a cell. Tracers can also be used to provide information about the mechanism of enzymatic reactions. The 1s electron configuration of hydrogen indicates a single valence electron. Because the 1s orbital has a maximum capacity of two electrons, hydrogen can form compounds with other elements in three ways (Figure \(\Page {1}\)): Hydrogen can also act as a bridge between two atoms. One familiar example is the , an electrostatic interaction between a hydrogen bonded to an electronegative atom and an atom that has one or more lone pairs of electrons (Figure \(\Page {2}\)). An example of this kind of interaction is the hydrogen bonding network found in water (Figure \(\Page {2}\)). Hydrogen can also form a , in which a hydride bridges two electropositive atoms. Compounds that contain hydrogen bonded to boron and similar elements often have this type of bonding. The B–H–B units found in boron hydrides cannot be described in terms of localized electron-pair bonds. Because the H atom in the middle of such a unit can accommodate a maximum of only two electrons in its 1s orbital, the B–H–B unit can be described as containing a hydride that interacts simultaneously with empty sp orbitals on two boron atoms (Figure \(\Page {3}\)). In these bonds, only two bonding electrons are used to hold three atoms together, making them electron-deficient bonds. You encountered a similar phenomenon in the discussion of π bonding in ozone and the nitrite ion. Recall that in both these cases, we used the presence of two electrons in a π molecular orbital extending over three atoms to explain the fact that the two O–O bond distances in ozone and the two N–O bond distances in nitrite are the same, which otherwise can be explained only by the use of resonance structures. Hydrogen can lose its electron to form H , accept an electron to form H , share its electron, hydrogen bond, or form a three-center bond. The first known preparation of elemental hydrogen was in 1671, when Robert Boyle dissolved iron in dilute acid and obtained a colorless, odorless, gaseous product. Hydrogen was finally identified as an element in 1766, when Henry Cavendish showed that water was the sole product of the reaction of the gas with oxygen. The explosive properties of mixtures of hydrogen with air were not discovered until early in the 18th century; they partially caused the spectacular explosion of the hydrogen-filled dirigible Hindenburg in 1937 (Figure \(\Page {4}\)). Due to its extremely low molecular mass, hydrogen gas is difficult to condense to a liquid (boiling point = 20.3 K), and solid hydrogen has one of the lowest melting points known (13.8 K). The most common way to produce small amounts of highly pure hydrogen gas in the laboratory was discovered by Boyle: reacting an active metal (M), such as iron, magnesium, or zinc, with dilute acid: \[M_{(s)} + 2H^+_{(aq)} \rightarrow H_{2(g)} + M^{2+}_{(aq)} \label{21.1}\] Hydrogen gas can also be generated by reacting metals such as aluminum or zinc with a strong base: \[\mathrm{Al(s)}+\mathrm{OH^-(aq)}+\mathrm{3H_2O(l)}\rightarrow\frac{3}{2}\mathrm{H_2(g)}+\mathrm{[Al(OH)_4]^-(aq)} \label{21.2}\] Solid commercial drain cleaners such as Drano use this reaction to generate gas bubbles that help break up clogs in a drainpipe. Hydrogen gas is also produced by reacting ionic hydrides with water. Because ionic hydrides are expensive, however, this reaction is generally used for only specialized purposes, such as producing HD gas by reacting a hydride with D O: \[MH_{(s)} + D_2O(l) \rightarrow HD_{(g)} + M^+(aq) + OD^−_{(aq)} \label{21.3}\] On an industrial scale, H is produced from methane by means of catalytic steam reforming, a method used to convert hydrocarbons to a mixture of CO and H known as synthesis gas, or syngas. The process is carried out at elevated temperatures (800°C) in the presence of a nickel catalyst: \[\mathrm{CH_4(g)}+\mathrm{H_2O(g)}\xrightarrow{\mathrm{Ni}}\mathrm{CO(g)}+\mathrm{3H_2(g)} \label{21.4}\] Most of the elements in the periodic table form binary compounds with hydrogen, which are collectively referred to as hydrides. Binary hydrides in turn can be classified in one of three ways, each with its own characteristic properties. Covalent hydrides contain hydrogen bonded to another atom via a covalent bond or a polar covalent bond. Covalent hydrides are usually molecular substances that are relatively volatile and have low melting points. Ionic hydrides contain the hydride ion as the anion with cations derived from electropositive metals. Like most ionic compounds, they are typically nonvolatile solids that contain three-dimensional lattices of cations and anions. Unlike most ionic compounds, however, they often decompose to H (g) and the parent metal after heating. Metallic hydrides are formed by hydrogen and less electropositive metals such as the transition metals. The properties of metallic hydrides are usually similar to those of the parent metal. Consequently, metallic hydrides are best viewed as metals that contain many hydrogen atoms present as interstitial impurities. Covalent hydrides are relatively volatile and have low melting points; ionic hydrides are generally nonvolatile solids in a lattice framework. Hydrogen can lose an electron to form a proton, gain an electron to form a hydride ion, or form a covalent bond or polar covalent electron-pair bond. The three isotopes of hydrogen—protium ( H or H), deuterium ( H or D), and tritium ( H or T)—have different physical properties. Deuterium and tritium can be used as tracers, substances that enable biochemists to follow the path of a molecule through an organism or a cell. Hydrogen can form compounds that contain a proton (H ), a hydride ion (H ), an electron-pair bond to H, a hydrogen bond, or a three-center bond (or electron-deficient bond), in which two electrons are shared between three atoms. Hydrogen gas can be generated by reacting an active metal with dilute acid, reacting Al or Zn with a strong base, or industrially by catalytic steam reforming, which produces synthesis gas, or syngas. | 11,046 | 1,285 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/07%3A_Acids_bases_and_ions_in_aqueous_solution/7.08%3A_Amphoteric_Oxides_and_Hydroxides/7.8A%3A_Amphoteric_Behavior |
An early classification of substances arose from the differences observed in their solubility in acidic and basic solutions. This led to the classification of oxides and hydroxides as being either or . Acidic oxides or hydroxides either reacted with water to produce an acidic solution or were soluble in aqueous base. Basic oxides and hydroxides either reacted with water to produce a basic solution or readily dissolved in aqueous acids. The diagram below shows there is strong correlation between the acidic or basic character of oxides (E O ) and the position of the element, E, in the periodic table.
Oxides of metallic elements are generally basic oxides, and oxides of nonmetallic elements acidic oxides. Take for example, the reactions with water of calcium oxide, a metallic oxide, and carbon dioxide, a nonmetallic oxide:
There is a gradual transition from basic oxides to acidic oxides from the lower left to the upper right in the periodic table.
This reactivity can be used to separate different cations, such as zinc(II), which dissolves in base, from manganese(II), which does not dissolve in base.
Aluminium hydroxide is another amphoteric species: | 1,197 | 1,286 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/IV._Nucleophilic_Substitution_Reactions/F._Substitution_Reactions_Involving_Ammonia |
This page gives you the facts and simple, uncluttered mechanisms for the nucleophilic substitution reactions between halogenoalkanes and ammonia to produce primary amines. The halogenoalkane is heated with a concentrated solution of ammonia in ethanol. The reaction is carried out in a sealed tube. You couldn't heat this mixture under reflux, because the ammonia would simply escape up the condenser as a gas. We'll talk about the reaction using 1-bromoethane as a typical primary halogenoalkane. The reaction happens in two stages. In the first stage, a salt is formed - in this case, ethylammonium bromide. This is just like ammonium bromide, except that one of the hydrogens in the ammonium ion is replaced by an ethyl group. \[ CH_3CH_2Br + NH_3 \rightarrow CH_3CH_2NH_3^+Br^-\] There is then the possibility of a reversible reaction between this salt and excess ammonia in the mixture. \[ CH_3CH_2NH_3^+Br^- + NH_3 \rightleftharpoons CH_3CH_2NH_2 + NH_4Br^- \] The ammonia removes a hydrogen ion from the ethylammonium ion to leave a primary amine - ethylamine. The more ammonia there is in the mixture, the more the forward reaction is favored as predicted by Le Chatelier's principle. The mechanism involves two steps. The first is a simple nucleophilic substitution reaction: Because the mechanism involves collision between two species in this slow step of the reaction, it is known as an S 2 reaction. In the second step of the reaction an ammonia molecule may remove one of the hydrogens on the -NH . An ammonium ion is formed, together with a primary amine - in this case, ethylamine. This reaction is, however, reversible. Your product will therefore contain a mixture of ethylammonium ions, ammonia, ethylamine and ammonium ions. Your major product will only be ethylamine if the ammonia is present in very large excess. Unfortunately the reaction doesn't stop here. Ethylamine is a good nucleophile, and goes on to attack unused bromoethane. This gets so complicated that it is dealt with on a separate page. You will find a link at the bottom of this page. The facts of the reactions are exactly the same as with primary halogenoalkanes. The halogenoalkane is heated in a sealed tube with a solution of ammonia in ethanol. For example: Followed by: This mechanism involves an initial ionisation of the halogenoalkane: followed by a very rapid attack by the ammonia on the carbocation (carbonium ion) formed: This is again an example of nucleophilic substitution. This time the slow step of the reaction only involves one species - the halogenoalkane. It is known as an S 1 reaction. There is a second stage exactly as with primary halogenoalkanes. An ammonia molecule removes a hydrogen ion from the -NH group in a reversible reaction. An ammonium ion is formed, together with an amine. It is very unlikely that any of the current UK-based syllabuses for 16 - 18 year olds will ask you about this. In the extremely unlikely event that you will ever need it, secondary halogenoalkanes use both an S 2 mechanism and an S 1. Make sure you understand what happens with primary and tertiary halogenoalkanes, and then adapt it for secondary ones should ever need to. Jim Clark ( ) | 3,204 | 1,287 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Radiative_Decay/Phosphorescence |
Unlike florescence, phosphorescence does not re-emit the light immediately. Instead, phosphorescence releases light very slowly in the dark due to its energy transition state. When light such as ultraviolet light is shined upon a glow in dark object, the object emits light, creating phosphorescence. There is a similarity between the phosphorescent and the fluorescent materials. They both contain substances with identical atoms. It is very important to understand the impurity state energy band, which is located between the conduction and valence energy bands. In a phosphorescence event, the absorbed energy usually goes through a high energy state which happens to be triplet state. The energy gets trapped in the triplet state because its physical situation forbids the transition to return to lower energy state, also as known from impurity to valence band. In order to change the energy of valence band, electrons must regain the energy they had lost during the impurity band transitional process. If the quantum yield of the phosphorescence is high enough, a great amount of light will be released and thus making the object glow in the dark. Most compounds have the ground state of singlet S . When it absorbs light, the electrons in the molecule may move to excited state of S , S , S and so on. There are also triplet states T and T . The energy of the T state is typically below the S state, while T is between S and S state. The wavelength of the radiation can determine which state the electron will move to. It is possible for the electron to return from excited state back to the ground state. An example is phosphorescence, where the emitting of radiation demotes the electrons from the excited state of T to ground state S . The molecule of phosphorescence has long life time, it loses energy easily, so it is hard to observe phosphorescence. Materials that can produce phosphorescence often contain zinc sulfide, sodium fluorescein, rhodamine, or strontium. The majority of phosphorescence is often used in drugs in pharmaceutical field. Some common drugs that have phosphorescence property include Aspirin, benzoic acid, morphine, and dopamine. Phosphorescence is also used to analyze water, air and chemical pollutions. | 2,268 | 1,290 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Vibrational_Spectroscopy/Vibrational_Modes/Normal_Modes |
Normal modes are used to describe the different vibrational motions in molecules. Each mode can be characterized by a different type of motion and each mode has a certain symmetry associated with it. Group theory is a useful tool in order to determine what symmetries the normal modes contain and predict if these modes are IR and/or Raman active. Consequently, IR and Raman spectroscopy is often used for vibrational spectra. In general, a normal mode is an independent motion of atoms in a molecule that occurs without causing movement to any of the other modes. Normal modes, as implied by their name, are orthogonal to each other. In order to discuss the quantum-mechanical equations that govern molecular vibrations it is convenient to convert Cartesian coordinates into so called normal coordinates. Vibrations in polyatomic molecules are represented by these normal coordinates. There exists an important fact about normal coordinates. Each of these coordinates belongs to an irreducible representation of the point the molecule under investigation. Vibrational wavefunctions associated with vibrational energy levels share this property as well. The normal coordinates and the vibration wavefunction can be categorized further according to the point group they belong to. From the character table predictions can be made for which symmetries can exist. The irreducible representation offers insight into the IR and/or Raman activity of the molecule in question. 3N where N represents the number of nuclei present in the molecule is the total number of coordinates needed to describe the location of a molecule in 3D-space. 3N is most often referred to as the total number of degrees of freedom of the molecule being investigated. The total number of degrees of freedom, can be divided into: Ethane, \(C_2H_6\) has eight atoms (\N=8\) and is a nonlinear molecule so of the \(3N=24\) degrees of freedom, three are translational and three are rotational. The remaining 18 degrees of freedom are internal (vibrational). This is consistent with: \[3N -6 =3(8)-6=18\] Carbon Dioxide, \(CO_2\) has three atoms (\(N=3\) and is a linear molecule so of the \(3N=9\) degrees of freedom, three are translational and two are rotational. The remaining 4 degrees of freedom are vibrational. This is consistent with: \[3N - 5 = 3(3)-5 = 4\] If there is no external field present, the energy of a molecule does not depend on its orientation in space (its translational degrees of freedom) nor its center of mass (its rotational degrees of freedom). The potential energy of the molecule is therefore made up of its vibrational degrees of freedom only of \(3N-6\) (or \(3N-5\) for linear molecules). The difference in potential energy is given by: \[\begin{align} \Delta V &= V(q_1,q_2,q_3,...,q_n) - V(0,0,0,...,0) \tag{1} \\[4pt] &= \dfrac{1}{2} \sum_{i=1}^{N_{vib}} \sum_{j=1}^{N_{vib}} \left(\dfrac{\partial^2 V}{\partial q_i\partial q_j} \right) q_iq_j \tag{2} \\[4pt] &= \dfrac{1}{2}\sum_{i=1}^{N_{vib}} \sum_{j=1}^{N_{vib}} f_{ij} q_iq_j \tag{3} \end{align}\] where For simplicity, the terms are neglected in this equation (consequently there are no higher order terms present). A theorem of classical mechanics states that the cross terms can be eliminated from the above equation (the details of the theorem are very complex and will not be discussed in detail). By using matrix algebra a new set of coordinates {Q } can be found such that \[\Delta{V} = \dfrac{1}{2} \sum_{j=1}^{N_{vib}}{F_jQ_j^2} \tag{4}\] Note that there are no cross terms in this new expression. These new coordinates are called or . With these new normal coordinates in hand, the Hamiltonian operator for vibrations can be written as follows: \[\hat{H}_{vib} = -\sum_{j=1}^{N_{vib}} \dfrac{\hbar^2}{2\mu_i} \dfrac{d^2}{dQ_j^2} + \dfrac{1}{2} \sum_{j=1}^{N_{vib}}F_jQ_j^2 \tag{5}\] The total wavefunction is a product of the individual wavefunctions and the energy is the sum of independent energies. This leads to: \[ \hat{H}_{vib} = \sum_{j=1}^{N_{vib}} \hat{H}_{vib,j} = \sum_{j=1}^{N_{vib}} \left( \dfrac{-\hbar^2}{2 \mu_j}\dfrac{d^2}{dQ_i^2} + \dfrac{1}{2}\sum_{j=1}^{N_{vib}} F_jQ_j^2 \right) \tag{6}\] and the wavefunction is then \[ \psi_{vib} = Q_1,Q_2, Q_3 ..., Q_{vib} = \psi_{vib,1}(Q_1) \psi_{vib,2}(Q_2) \psi_{vib,3}(Q_3) , ..., \psi_{vib,N_{vib}}(Q_{N_{vib}}) \tag{7}\] and the total vibrational energy of the molecule is \[E_{vib} = \sum_{j=1}^{N_{vin}} h\nu_j \left (v_j + \dfrac{1}{2}\right) \tag{8}\] where \(v_j= 0,1,2,3...\) The consequence of the result stated in the above equations is that each vibrational mode can be treated as a harmonic oscillator approximation. There are \(N_{vib}\) harmonic oscillators corresponding to the total number of vibrational modes present in the molecule. In the ground vibrational state the energy of the molecule is equal to (1/2) . The ground state energy is referred to as zero point energy. A vibration transition in a molecule is induced when it absorbs a quantum of energy according to The first excited state is separated from the ground state by E = (3/2)h since v = 1, the next energy level separation is (5/2)h , etc... The harmonic oscillator is a good approximation, but it does not take into account that the molecule, once it has absorbed enough energy to break the vibrating bond, does dissociate. A better approximation is the which takes into account anharmonicity. The Morse potential also accounts for bond dissociation as well as energy levels getting closer together at higher energies. The normal coordinate q is used to follow the path of a normal mode of vibration. As shown in Figure 2 the displacement of the C atom, denoted by , and the displacement of the O atom, denoted by , occur at the same frequency. The displacement of atoms is measured from the equilibrium distance in ground vibrational state, r . In direct correlation with symmetry, subscripts s (symmetric), as (asymmetric) and d (degenerate) are used to further describe the different modes. A normal mode corresponding to an asymmetric stretch can be best described by a harmonic oscillator: As one bond lengthens, the other bond shortens. A normal mode that corresponds can be best described by a Morse potential well: As the bond length increases the potential energy increases and levels off as the bond length gets further away from the equilibrium. It is important to realize that every normal mode has a certain type of symmetry associated with it. Identifying the point group of the molecule is therefore an important step. With this in mind it is not surprising that every normal mode forms a basis set for an irreducible representation of the point group the molecule belongs to. For a molecule such as water, having a structure of XY , three normal coordinates can be determined. The two stretching modes are equivalent in symmetry and energy. The figure below shows the three normal modes for the water molecule:
y convention, with nonlinear molecules, the symmetric stretch is denoted v whereas the asymmetric stretch is denoted v . Bending motions are v . With linear molecules, the bending motion is v whereas asymmetric stretch is v . The water molecule has C symmetry and its symmetry elements are E, C , σ(xz) and σ(yz). In order to determine the symmetries of the three vibrations and how they each transform, symmetry operations will be performed. As an example, performing C operations using the two normal mode v and v gives the following transformation: Once all the symmetry operations have been performed in a systematic manner for each modes the symmetry can be assigned to the normal mode using the for C : Water has three normal modes that can be grouped together as the reducible representation \[Γ_{vib}= 2a_1 + b_2.\] Determination of normal modes becomes quite complex as the number of atoms in the molecule increases. Nowadays, computer programs that simulate molecular vibrations can be used to perform these calculations. The example of [PtCl ] shows the increasing complexity. The molecule has five atoms and therefore 15 degrees of freedom, 9 of these are vibrational degrees of freedom. The nine normal modes are exemplified below along with the irreducible representation the normal mode belongs to (D point group). A , b and e are stretching vibrations whereas b , a , b and e are bending vibrations. A transition from \(\ce{v -> v'}\) is IR active if the transition moment integral contains the totally symmetric irreducible representation of the point group the molecule belongs to. The transition moment integral is derived from the one-dimensional harmonic oscillator. Using the definition of dipole moment the integral is: \[M\left(v \rightarrow v^{\prime}\right)=\int_{-\infty}^{\infty} \psi^{*}\left(v^{\prime}\right) \mu \psi(v) d x\] If , the dipole moment, would be a constant and therefore independent of the vibration, it could be taken outside the integral. Since v and v' are mutually orthogonal to each other, the integral will equal zero and the transition will not be allowed. In order for the integral to be nonzero, must change during a vibration. This selection rule explains why homonuclear diatomic molecules do not produce an IR spectrum. There is no change in dipole moment resulting in a transition moment integral of zero and a transition that is forbidden. For a transition to be Raman active the same rules apply. The transition moment integral must contain the totally symmetric irreducible representation of the point group. The integral contains the polarizability tensor (usually represented by a square matrix): \[M\left(v \rightarrow v^{\prime}\right)=\int_{-\infty}^{\infty} \psi^{*}\left(v^{\prime}\right) \alpha \psi(v) d x\] \(α\) must be nonzero in order for the transition to be allowed and show Raman scattering. For a molecule to be IR active the dipole moment has to change during the vibration. For a molecule to be Raman active the polarizability of the molecule has to change during the vibration. The reducible representation can also be found by determining the reducible representation of the 3N degrees of freedom of H O, . By applying Group Theory it is straightforward to find as well as UMA (number of unmoved atoms). Again, using water as an example with C symmetry where 3N = 9, can be determined: Note that contains nine degrees of freedom consistent with 3N = 9. contains , as well as . can be obtained by finding the irreducible representations corresponding to x,y and z in the right side of the character table, by finding the ones corresponding to R , R and R . can be obtained by - - . (H O) = (3a + a + 2b + 3b ) - (a + b + b ) - (a + b + b ) = In order to determine which modes are IR active, a simple check of the irreducible representation that corresponds to x,y and z and a cross check with the reducible representation is necessary. If they contain the same irreducible representation, the mode is IR active.
For H O, z transforms as a , x as b and y as b . The modes a1 and b2 are IR active since Γvib contains 2a + b . In order to determine which modes are Raman active, the irreducible representation that corresponds to z , x -y , xy, xz and yz is used and again cross checked with Γvib. For H O, z and x -y transform as a , xy as a , xz as b and yz as b .The modes a and b are also Raman active since Γ contains both these modes. The IR spectrum of H2O does indeed have three bands as predicted by Group Theory. The two symmetric stretches v1 and v2 occur at 3756 and 3657 cm-1 whereas the bending v3 motion occurs at 1595 cm-1. In order to determine which normal modes are stretching vibrations and which one are bending vibrations, a stretching analysis can be performed. Then the stretching vibrations can be deducted from the total vibrations in order to obtain the bending vibrations. A double-headed arrow is drawn between the atom as depicted below: Then a determination of how the arrows transform under each symmetry operation in C2v symmetry will yield the following results: = - = 2a + b -a - b = a H O has two stretching vibrations as well as one bending vibration. This concept can be expanded to complex molecules such as PtCl4 . Four double headed arrows can be drawn between the atoms of the molecule and determine how these transform in D symmetry. Once the irreducible representation for has been worked out, can be determined by = - . The transition from v=0 (ground state) -> v=1 (first excited state) is called the fundamental transition. This transition has the greatest intensity. The transition from v=0 --> v=2 is is referred to as the first overtone, from v=0 --> v=3 is called the second overtone, etc. Ovetones occur when a mode is excited above the v = 1 level. The harmonic oscillator approximation supports the prediction that the transition to a second overtone will be twice as energetic as a fundamental transition. Most molecules are in their zero point energy at room temperature. Therefore, most transitions do originate from the v=0 state. Some molecules do have a significant population of the v=1 state at room temperature and transitions from this thermally excited state are called bands. can occur if more than one vibration is excited by the absorption of a photon. The overall energy of a combination band is the result of the sum of individual transitions. | 13,450 | 1,291 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/11%3A_Reactions_and_Other_Chemical_Processes/11.04%3A__Enthalpies_of_Solution_and_Dilution |
\( \newcommand{\tx}[1]{\text{#1}} % text in math mode\)
\( \newcommand{\subs}[1]{_{\text{#1}}} % subscript text\)
\( \newcommand{\sups}[1]{^{\text{#1}}} % superscript text\)
\( \newcommand{\st}{^\circ} % standard state symbol\)
\( \newcommand{\id}{^{\text{id}}} % ideal\)
\( \newcommand{\rf}{^{\text{ref}}} % reference state\)
\( \newcommand{\units}[1]{\mbox{$\thinspace$#1}}\)
\( \newcommand{\K}{\units{K}} % kelvins\)
\( \newcommand{\degC}{^\circ\text{C}} % degrees Celsius\)
\( \newcommand{\br}{\units{bar}} % bar (\bar is already defined)\)
\( \newcommand{\Pa}{\units{Pa}}\)
\( \newcommand{\mol}{\units{mol}} % mole\)
\( \newcommand{\V}{\units{V}} % volts\)
\( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\)
\( \newcommand{\per}{^{-1}} % minus one power\)
\( \newcommand{\m}{_{\text{m}}} % subscript m for molar quantity\)
\( \newcommand{\CVm}{C_{V,\text{m}}} % molar heat capacity at const.V\)
\( \newcommand{\Cpm}{C_{p,\text{m}}} % molar heat capacity at const.p\)
\( \newcommand{\kT}{\kappa_T} % isothermal compressibility\)
\( \newcommand{\A}{_{\text{A}}} % subscript A for solvent or state A\)
\( \newcommand{\B}{_{\text{B}}} % subscript B for solute or state B\)
\( \newcommand{\bd}{_{\text{b}}} % subscript b for boundary or boiling point\)
\( \newcommand{\C}{_{\text{C}}} % subscript C\)
\( \newcommand{\f}{_{\text{f}}} % subscript f for freezing point\)
\( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\)
\( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\)
\( \newcommand{\mi}{_{\text{m},i}} % subscript m,i (m=molar)\)
\( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\)
\( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\)
\( \newcommand{\xbB}{_{x,\text{B}}} % x basis, B\)
\( \newcommand{\xbC}{_{x,\text{C}}} % x basis, C\)
\( \newcommand{\cbB}{_{c,\text{B}}} % c basis, B\)
\( \newcommand{\mbB}{_{m,\text{B}}} % m basis, B\)
\( \newcommand{\kHi}{k_{\text{H},i}} % Henry's law constant, x basis, i\)
\( \newcommand{\kHB}{k_{\text{H,B}}} % Henry's law constant, x basis, B\)
\( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\)
\( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\)
\( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\)
\( \newcommand{\eq}{\subs{eq}} % equilibrium state\)
\( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\)
\( \newcommand{\sys}{\subs{sys}} % system property\)
\( \newcommand{\sur}{\sups{sur}} % surroundings\)
\( \renewcommand{\in}{\sups{int}} % internal\)
\( \newcommand{\lab}{\subs{lab}} % lab frame\)
\( \newcommand{\cm}{\subs{cm}} % center of mass\)
\( \newcommand{\rev}{\subs{rev}} % reversible\)
\( \newcommand{\irr}{\subs{irr}} % irreversible\)
\( \newcommand{\fric}{\subs{fric}} % friction\)
\( \newcommand{\diss}{\subs{diss}} % dissipation\)
\( \newcommand{\el}{\subs{el}} % electrical\)
\( \newcommand{\cell}{\subs{cell}} % cell\)
\( \newcommand{\As}{A\subs{s}} % surface area\)
\( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)
\( \newcommand{\allni}{\{n_i \}} % set of all n_i\)
\( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\)
\( \newcommand{\solmB}{\tx{(sol,$\,$$m\B$)}}\)
\( \newcommand{\dil}{\tx{(dil)}}\)
\( \newcommand{\sln}{\tx{(sln)}}\)
\( \newcommand{\mix}{\tx{(mix)}}\)
\( \newcommand{\rxn}{\tx{(rxn)}}\)
\( \newcommand{\expt}{\tx{(expt)}}\)
\( \newcommand{\solid}{\tx{(s)}}\)
\( \newcommand{\liquid}{\tx{(l)}}\)
\( \newcommand{\gas}{\tx{(g)}}\)
\( \newcommand{\pha}{\alpha} % phase alpha\)
\( \newcommand{\phb}{\beta} % phase beta\)
\( \newcommand{\phg}{\gamma} % phase gamma\)
\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
\( \newcommand{\bph}{^{\beta}} % beta phase superscript\)
\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
\( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\)
\( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\)
\( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)
\( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)
\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
\( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)
\( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)
\( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)
\( \newcommand{\Del}{\Delta}\)
\( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)
\( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)
\( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)
\( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)
\( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)
\( \newcommand{\dotprod}{\small\bullet}\)
\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
\( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)
\( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)
\( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\)
\( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\)
\( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)
\( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)
\( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)
\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) The processes of solution (dissolution) and dilution are related. The IUPAC Green Book (E. Richard Cohen et al, , 3rd edition, RSC Publishing, Cambridge, 2007, Sec. 2.11.1) recommends the abbreviations sol and dil for these processes. For an electrolyte solute, a plot of \(\Del H\m\solmB\) versus \(m\B\) has a limiting slope of \(+\infty\) at \(m\B{=}0\), whereas the limiting slope of \(\Del H\m\solmB\) versus \(\sqrt{m\B}\) is finite and can be predicted from the Debye–Hückel limiting law. Accordingly, a satisfactory procedure is to plot \(\Del H\m\solmB\) versus \(\sqrt{m\B}\), perform a linear extrapolation of the experimental points to \(\sqrt{m\B}{=}0\), and then shift the origin to the extrapolated intercept. The result is a plot of \(\varPhi_L\) versus \(\sqrt{m\B}\). An example for aqueous NaCl solutions is shown in Fig. 11.10(a). We can also evaluate \(\varPhi_L\) from experimental enthalpies of dilution. From Eqs. 11.4.10 and 11.4.22, we obtain the relation \begin{equation} \varPhi_L(m\B'')-\varPhi_L(m\B') = \Del H\m(\tx{dil, \(m\B'{\ra}m\B''\)}) \tag{11.4.25} \end{equation} We can measure the enthalpy changes for diluting a solution of initial molality \(m\B'\) to various molalities \(m\B''\), plot the values of \(\Del H\m(\tx{dil, \(m\B'{\ra}m\B''\)})\) versus \(\sqrt{m\B}\), extrapolate the curve to \(\sqrt{m\B}{=}0\), and shift the origin to the extrapolated intercept, resulting in a plot of \(\varPhi_L\) versus \(\sqrt{m\B}\). In order to be able to use Eq. 11.4.23, we need to relate the derivative \(\dif\varPhi_L/\dif m\B\) to the slope of the curve of \(\varPhi_L\) versus \(\sqrt{m\B}\). We write \begin{equation} \dif \sqrt{m\B} = \frac{1}{2\sqrt{m\B}}\dif m\B \qquad \dif m\B = 2\sqrt{m\B} \dif\sqrt{m\B} \tag{11.4.26} \end{equation} Substituting this expression for \(\dif m\B\) into Eq. 11.4.23, we obtain the following operational equation for evaluating \(L\B\) from the plot of \(\varPhi_L\) versus \(\sqrt{m\B}\): \begin{gather} \s{ L\B = \varPhi_L + \frac{\sqrt{m\B}}{2} \frac{\dif\varPhi_L}{\dif\sqrt{m\B}} } \tag{11.4.27} \cond{(constant \(T\) and \(p\))} \end{gather} The value of \(\varPhi_L\) goes to zero at infinite dilution. When the solute is an electrolyte, the dependence of \(\varPhi_L\) on \(m\B\) in solutions dilute enough for the Debye–Hückel limiting law to apply is given by \begin{gather} \s{ \varPhi_L = C_{\varPhi_L}\sqrt{m\B} } \tag{11.4.28} \cond{(very dilute solution)} \end{gather} For aqueous solutions of a 1:1 electrolyte at \(25\units{\(\degC\)}\), the coefficient \(C_{\varPhi_L}\) has the value \begin{equation} C_{\varPhi_L} = 1.988\timesten{3}\units{J kg\(^{1/2}\) mol\(^{-3/2}\)} \tag{11.4.29} \end{equation} (The fact that \(C_{\varPhi_L}\) is positive means, according to Eq. 11.4.25, that dilution of a very dilute electrolyte solution is an exothermic process.) \(C_{\varPhi_L}\) is equal to the limiting slope of \(\varPhi_L\) versus \(\sqrt{m\B}\), of \(\Del H\m\solmB\) versus \(\sqrt{m\B}\), and of \(\Del H\m(\tx{dil, \(m\B'{\ra}m\B''\)})\) versus \(\sqrt{m'\B}\). The value given by Eq. 11.4.29 can be used for extrapolation of measurements at \(25\units{\(\degC\)}\) and low molality to infinite dilution. Equation 11.4.28 can be derived as follows. For simplicity, we assume the pressure is the standard pressure \(p\st\). At this pressure \(H\B^\infty\) is the same as \(H\B\st\), and Eq. 11.4.17 becomes \(L\B=H\B-H\B\st\). From Eqs. 12.1.3 and 12.1.6 in the next chapter, we can write the relations \begin{equation} H\B=-T^2\bPd{(\mu\B/T)}{T}{p,\allni} \qquad H\B\st=-T^2\frac{\dif(\mu\mbB\st/T)}{\dif T} \tag{11.4.30} \end{equation} Subtracting the second of these relations from the first, we obtain \begin{equation} H\B-H\B\st = -T^2\bPd{(\mu\B-\mu\mbB\st)/T}{T}{p,\allni} \tag{11.4.31} \end{equation} The solute activity on a molality basis, \(a\mbB\), is defined by \(\mu\B-\mu\mbB\st=RT\ln a\mbB\). The activity of an electrolyte solute at the standard pressure, from Eq. 10.3.10, is given by \(a\mbB = (\nu_{+}^{\nu_{+}} \nu_{-}^{\nu_{-}}) \g_{\pm}^{\nu} (m\B/m\st)^{\nu}\). Accordingly, the relative partial molar enthalpy of the solute is related to the mean ionic activity coefficient by \begin{equation} L\B=-RT^2\nu\Pd{\ln\g_{\pm}}{T}{\!\!p,\allni} \tag{11.4.32} \end{equation} We assume the solution is sufficiently dilute for the mean ionic activity coefficient to be adequately described by the Debye–Hückel limiting law, Eq. 10.4.8: \(\ln\g_{\pm} = -A\subs{DH}\left|z_+z_-\right|\sqrt{I_m}\), where \(A\subs{DH}\) is a temperature-dependent quantity defined in Sec. 10.4. Then Eq. 11.4.32 becomes \begin{gather} \s{ L\B=RT^2\nu\left|z_+z_-\right|\sqrt{I_m}\Pd{A\subs{DH}}{T}{\!\!p,\allni} } \tag{11.4.33} \cond{(very dilute solution)} \end{gather} Substitution of the expression given by Eq. 10.4.9 for \(I_m\) in a solution of a single completely-dissociated electrolyte converts Eq. 11.4.33 to \begin{gather} \s{ L\B = \left[ \frac{RT^2}{\sqrt{2}}\Pd{\rho\A^*A\subs{DH}}{T}{p,\allni} \left(\nu\left|z_+z_-\right|\right)^{3/2} \right]\sqrt{m\B} = C_{L\B}\sqrt{m\B} } \tag{11.4.34} \cond{(very dilute solution)} \end{gather} The coefficient \(C_{L\B}\) (the quantity in brackets) depends on \(T\), the kind of solvent, and the ion charges and number of ions per solute formula unit, but not on the solute molality. Let \(C_{\varPhi_L}\) represent the limiting slope of \(\varPhi_L\) versus \(\sqrt{m\B}\). In a very dilute solution we have \(\varPhi_L = C_{\varPhi_L}\sqrt{m\B}\), and Eq. 11.4.27 becomes \begin{equation} L\B = \varPhi_L + \frac{\sqrt{m\B}}{2} \frac{\dif\varPhi_L}{\dif\sqrt{m\B}} = C_{\varPhi_L}\sqrt{m\B} + \frac{\sqrt{m\B}}{2} C_{\varPhi_L} \tag{11.4.35} \end{equation} By equating this expression for \(L\B\) with the one given by Eq. 11.4.34 and solving for \(C_{\varPhi_L}\), we obtain \(C_{\varPhi_L}=(2/3)C_{L\B}\) and \(\varPhi_L = (2/3)C_{L\B}\sqrt{m\B}\). | 13,284 | 1,292 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.08%3A_Thermochemical_Equations/3.8.03%3A_Foods-_Energy_from_Fats_and_Sugars |
Energy from Fats & Sugars Earlier we discussed the nature of fats and mentioned that fats typically provide 9 Cal/g of food energy, while sugars provide about 4 Cal/g. So in order to store the energy in 10 lb of fat, your body would need to store 22.5 lb of carbohydrates or sugars; but it's more extreme than that. Because sugars carry about their own weight of associated water in the body, 67.5 lb (31 kg) of hydrated glycogen has the energy equivalent of 10 lb (5 kg) of fat! The food energy in various food types is given approximately in the following table , and you can find the fat content (as well as all other nutritional information) about nearly all foods in the United States Department of Agriculture's Bulletin #8 which has a searchable USDA Nutrient database. In the database, fats are list under "lipids"/"Fatty Acids" and then under "saturated" and "18:0", indicating the number of carbon atoms (18) in the fatty acid, and the number of double bonds (0) (see Example 1). These caloric values are measured in a "bomb calorimeter" like the one in the Figure. A 1-2 g sample of food is sealed in a heavy walled steel cylinder (about 4" in diameter and 7" high), shown in the center of the Figure, which is then filled with pure oxygen at 30-40 atmospheres pressure, and immersed in a few liters of water. The sample is ignited electrically, and the heat released is determined by measuring the temperature increase of water that surrounds the "bomb".
Let's investigate the basis for these caloric values in terms of the chemical reactions that provide the energy. Energy changes which accompany chemical reactions are almost always expressed by . The combustion of stearic acid, which is the main component of saturated fats, is written: C H O ( ) + 24 O ( ) → 18 CO ( ) + 18 H O( ) (25°C, 1 atm pressure) Δ = –11 407 kJ (1) Here the Δ (delta subscript m) tells us whether heat energy is released or absorbed when the reaction occurs and also enables us to find the actual quantity of energy involved. By convention, if Δ is , heat is by the reaction; i.e., it is . More commonly, Δ is as in Eq. (1), indicating that heat energy is rather than absorbed by the reaction, and that the reaction is . This convention as to whether Δ is positive or negative looks at the heat change in terms of the matter actually involved in the reaction rather than its surroundings. In the reaction in Eq. (1), the C, H, and O atoms have collectively lost energy and it is this loss which is indicated by a negative value of Δ . It is important to notice that Δ is the enthalpy for the The quantity of heat released or absorbed by a reaction is proportional to the amount of each substance consumed or produced by the reaction. Thus Eq. (1) tells us that 890.4 kJ of heat energy is given off C H O which is consumed. Alternatively, it tells us that 11407 kJ is released H O produced, or for every 18 mol of carbon dioxide produced, or 24 mol of oxygen consumed. Seen in this way, Δ is a conversion factor enabling us to calculate the heat absorbed when a given amount of substance is consumed or produced. If is the quantity of heat absorbed and is the amount of substance involved, then \(\Delta H_{\text{m}}=\frac{q}{n}\) (2) In the molecular model, each bend in the structure is occupied by a carbon atom, and each carbon atom has 4 bonds; missing bonds are to hydrogen atoms, which are not shown. a. How much heat energy is obtained when 1 g of C H O , is burned in oxygen according to the equation above? The molar mass of steric acid is 284.48 g/mol. b. What is the caloric value of 1 g of stearic acid, given that Δ = –11407 kJ for equation (1)? a. The mass of C H O is easily converted to the amount of C H O from which the heat energy is easily calculated by means of Eq. (2). The value of Δ is –11407 kJ per mole of C H O . The road map is \(m_{\text{C}_{\text{18}}\text{H}_{\text{36}}\text{O}_{\text{2}}}\text{ }\xrightarrow{M}\text{ }n_{\text{C}_{\text{18}}\text{H}_{\text{36}}\text{O}_2}\text{ }\xrightarrow{\Delta H_{m}}\text{ }q\) so that \(q=\text{1.0 g C}_{\text{18}}\text{H}_{\text{36}}\text{O}_{2}\text{ }\times \text{ }\frac{\text{1 mol C}_{\text{18}}\text{H}_{\text{36}}\text{O}_2}{\text{284}\text{.48 g C}_{\text{18}}\text{H}_{\text{36}}\text{O}_{2}}\text{ }\times \text{ }\frac{-\text{11407 kJ}}{\text{1 mol C}_{\text{182}}\text{H}_{\text{36}}\text{O}_{2}}=\text{-40.09 kJ}\) b. \(\text{-40.09 kJ}~\times~\frac{\text{1 kcal}}{\text{4.184 kJ}}~\times~\frac{\text{1 Cal}}{\text{1 kcal}}~=~-\text{9.84 Cal}\) Note: By convention a negative value of corresponds to a release of heat energy by the matter involved in the reaction. This is close to the estimated 9 Cal/g for fats. We saw earlier that most fats are triglycerides, that is, they would have 3 fatty acid substituents (like stearic acid) attached to a glycerol "backbone" in a fat like glyceryl tristearate ("stearin") (C H O , M = 891.48). Stearin has a heat of combustion of -35 663 kJ/mol, so 1 g produces (35 663 kJ/mol) / (891.48 g/mol) x (1 Cal / 4.184 kJ) = 9.57 Cal. When energy is required by our body, triglycerides are converted free fatty acids, and transported by serum albumin in the blood to cells where energy is required. Serum albumin is necessary because the solubility of fatty acids is low in water-based blood. In comparison, sucrose (C H O ) has a molar mass of 342.3 g/mol and a heat of combustion of -5645 kJ/mol, so it produces 16.49 kJ/g or 3.94 Cal/g, very close to the estimated value, by the combustion: C H O ( ) + 12 O ( ) → 12 CO ( ) + 11 H O( ) (25°C, 1 atm pressure) Δ = –5 645 kJ mol The quantity Δ is the . In this context the symbol Δ (delta) signifies change in” while is the symbol for the quantity being changed, namely the enthalpy. We will deal with the enthalpy in some detail in Chap. 15. For the moment we can think of it as a property of matter which increases when matter absorbs energy and decreases when matter releases energy. It is important to realize that the value of Δ given in thermochemical equations like (1) depends on the physical state of both the reactants and the products. Thus, if water were obtained as a liquid instead of a gas in the reaction in Eq. (1), the value of Δ would be different from -890.4 kJ. It is also necessary to specify both the temperature and pressure since the value of Δ depends very slightly on these variables. If these are not specified [as in Eq. (3)] they usually refer to 25°C and to normal atmospheric pressure. Two more characteristics of thermochemical equations arise from the law of conservation of energy. The first is that For example, H O( ) → H O( ) Δ = 44 kJ (3 ) tells us that when a mole of liquid water vaporizes, 44 kJ of heat is absorbed. This corresponds to the fact that heat is absorbed from your skin when perspiration evaporates, and you cool off. Condensation of 1 mol of water vapor, on the other hand, gives off exactly the same quantity of heat. H O( ) → H O( ) Δ = –44 kJ (3 ) To see why this must be true, suppose that Δ [Eq. (4a)] = 44 kJ while Δ [Eq. (4b)] = –50.0 kJ. If we took 1 mol of liquid water and allowed it to evaporate, 44 kJ would be absorbed. We could then condense the water vapor, and 50.0 kJ would be given off. We could again have 1 mol of liquid water at 25°C but we would also have 6 kJ of heat which had been created from nowhere! This would violate the law of conservation of energy. The only way the problem can he avoided is for Δ of the reverse reaction to be equal in magnitude but opposite in sign from Δ of the forward reaction. That is, Δ forward = –Δ reverse | 7,663 | 1,293 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/14%3A_Organohalogen_and_Organometallic_Compounds/14.10%3A_Properties_of_Organometallic_Compounds |
How carbon-metal bonds are formed depends on the metal that is used. Conditions that are suitable for one metal may be wholly unsuited for another. Some organometallic compounds react very sluggishly even toward acids, whereas others react avidly with water, oxygen, carbon dioxide, and almost all solvents but the alkanes themselves. Reactivity increases with increasing polarity of the carbon-metal bond, which is determined by the electropositivity of the metal. Strongly electropositive metals, such as sodium and potassium, form largely ionic bonds to carbon, as we have mentioned in the case of alkynide salts, \(\ce{RC \equiv C}^\ominus \ce{Na}^\oplus\) ( ). Estimates of the ionic character of various carbon-metal bonds are given in Table 14-3, and it will be seen that . Many organosodium and organopotassium compounds burn spontaneously when exposed to air and react violently with water and carbon dioxide. As might be expected from their saltlike character, they are nonvolatile and do not dissolve readily in nonpolar solvents. In contrast, the more covalent, less ionic, organometallic compounds, such as \(\ce{(CH_3)_2Hg}\), are far less reactive; they are stable in air, quite volatile, and dissolve in nonpolar solvents. All of these compounds must be handled with great care because some are dangerously reactive and others are very toxic. They seldom are isolated from the solutions in which they are prepared, but are used immediately in other reactions. and (1977) | 1,506 | 1,294 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Fundamentals_of_Spectroscopy/Electromagnetic_Radiation |
As you read the print off this computer screen now, you are reading pages of fluctuating energy and magnetic fields. Light, electricity, and magnetism are all different forms of electromagnetic radiation. Electromagnetic radiation is a form of energy that is produced by oscillating electric and magnetic disturbance, or by the movement of electrically charged particles traveling through a vacuum or matter. The electric and magnetic fields come at right angles to each other and combined wave moves perpendicular to both magnetic and electric oscillating fields thus the disturbance. electric and magnetic waves travel perpendicular to each other and have certain characteristics, including amplitude, wavelength, and frequency. General Properties of all electromagnetic radiation: Amplitude is the distance from the maximum vertical displacement of the wave to the middle of the wave. This measures the magnitude of oscillation of a particular wave. In short, the amplitude is basically the height of the wave. Larger amplitude means higher energy and lower amplitude means lower energy. Amplitude is important because it tells you the intensity or brightness of a wave in comparison with other waves. Wavelength (\(\lambda\)) is the distance of one full cycle of the oscillation. Longer wavelength waves such as radio waves carry low energy; this is why we can listen to the radio without any harmful consequences. Shorter wavelength waves such as x-rays carry higher energy that can be hazardous to our health. Consequently lead aprons are worn to protect our bodies from harmful radiation when we undergo x-rays. This wavelength frequently relationship is characterized by: \[ c = \lambda\nu \] where Shorter wavelength means greater frequency, and greater frequency means higher energy. Wavelengths are important in that they tell one what type of wave one is dealing with. Remember, and Frequency is defined as the number of cycles per second, and is expressed as sec or Hertz (Hz). Frequency is directly proportional to energy and can be express as: \[ E = h\nu \] where Period (T) is the amount of time a wave takes to travel one wavelength; it is measured in seconds (s). The velocity of wave in general is expressed as: \[ velocity = \lambda\nu \] For Electromagnetic wave, the velocity in vacuum is \(2.99 \times 10^8\;m/s\) or \(186,282\) miles/second. As a wave’s wavelength increases, the frequency decreases, and as wave’s wavelength decreases, the frequency increases. When electromagnetic energy is released as the energy level increases, the wavelength decreases and frequency decreases. Thus, electromagnetic radiation is then grouped into categories based on its wavelength or frequency into the electromagnetic spectrum. The different types of electromagnetic radiation shown in the electromagnetic spectrum consists of radio waves, microwaves, infrared waves, visible light, ultraviolet radiation, X-rays, and gamma rays. The part of the electromagnetic spectrum that we are able to see is the visible light spectrum. Fig. 6: Electromagnetic Spectrum with Radiation Types are approximately 10 m in wavelength. As the name implies, radio waves are transmitted by radio broadcasts, TV broadcasts, and even cell phones. Radio waves have the lowest energy levels. Radio waves are used in remote sensing, where hydrogen gas in space releases radio energy with a low frequency and is collected as radio waves. They are also used in radar systems, where they release radio energy and collect the bounced energy back. Especially useful in weather, radar systems are used to can illustrate maps of the surface of the Earth and predict weather patterns since radio energy easily breaks through the atmosphere. ; can be used to broadcast information through space, as well as warm food. They are also used in remote sensing in which microwaves are released and bounced back to collect information on their reflections. Microwaves can be measured in centimeters. They are good for transmitting information because the energy can go through substances such as clouds and light rain. Short microwaves are sometimes used in Doppler radars to predict weather forecasts. can be released as heat or thermal energy. It can also be bounced back, which is called near infrared because of its similarities with visible light energy. Infrared Radiation is most commonly used in remote sensing as infrared sensors collect thermal energy, providing us with weather conditions. This picture represents a snap shot in mid-infrared light. is the only part of the electromagnetic spectrum that humans can see with an unaided eye. This part of the spectrum includes a range of different colors that all represent a particular wavelength. Rainbows are formed in this way; light passes through matter in which it is absorbed or reflected based on its wavelength. Thus, some colors are reflected more than other, leading to the creation of a rainbow. Fig. 7: The color regions of the Visible Spectrum Dispersion of Light Through A Prism Ultraviolet, Radiation, X-Rays, and Gamma Rays are all related to events occurring in space. UV radiation is most commonly known because of its severe effects on the skin from the sun, leading to cancer. X-rays are used to produce medical images of the body. Gamma Rays can used in chemotherapy in order to rid of tumors in a body since it has such a high energy level. The shortest waves, Gamma rays, are approximately 10 m in wavelength. Out this huge spectrum, the human eyes can only detect waves from 390 nm to 780 nm. The mathematical description of a wave is: \[ y = A\sin(kx - \omega{t}) \] where A is the amplitude, k is the wave number, x is the displacement on the x-axis. \[ k = \dfrac{2\pi}{\lambda} \] where \(\lambda\) is the wavelength. Angular frequency described as: \[ \omega = 2\pi \nu = \dfrac{2\pi}{T} \] where \(\nu\) is frequency and period (T) is the amount of time for the wave to travel one wavelength. An important property of waves is the ability to combine with other waves. There are two type of interference: constructive and destructive. Constructive interference occurs when two or more waves are in phase and and their displacements add to produce a higher amplitude. On the contrary, destructive interference occurs when two or more waves are out of phase and their displacements negate each other to produce lower amplitude. Interference can be demonstrated effectively through the double slit experiment. This experiment consists of a light source pointing toward a plate with one slit and a second plate with two slits. As the light travels through the slits, we notice bands of alternating intensity on the wall behind the second plate. The banding in the middle is the most intense because the two waves are perfectly in phase at that point and thus constructively interfere. The dark bands are caused by out of phase waves which result in destructive interference. This is why you observe nodes on figure 4. In a similar way, if electrons are used instead of light, electrons will be represented both as waves and particles. Electromagnetic radiation can either acts as a wave or a particle, a photon. As a wave, it is represented by velocity, wavelength, and frequency. Light is an EM wave since the speed of EM waves is the same as the speed of light. As a particle, EM is represented as a photon, which transports energy. When a photon is absorbed, the electron can be moved up or down an energy level. When it moves up, it absorbs energy, when it moves down, energy is released. Thus, since each atom has its own distinct set of energy levels, each element emits and absorbs different frequencies. Photons with higher energies produce shorter wavelengths and photons with lower energies produce longer wavelengths. Electromagnetic Radiation is also categorized into two groups based, ionizing and non-ionizing, on the severity of the radiation. Ionizing radiation holds a great amount of energy to remove electrons and cause the matter to become ionized. Thus, higher frequency waves such as the X-rays and gamma-rays have ionizing radiation. However, lower frequency waves such as radio waves, do not have ionizing radiation and are grouped as non-ionizing. Electromagnetic radiation released is related to the temperature of the body. Stephan-Boltzmann Law says that if this body is a black body, one which perfectly absorbs and emits radiation, the radiation released is equal to the temperature raised to the fourth power. Therefore, as temperature increases, the amount of radiation released increases greatly. Objects that release radiation very well also absorb radiation at certain wavelengths very well. This is explained by the . Wavelengths are also related to temperature. As the temperature increases, the wavelength of maximum emission decreases. | 8,864 | 1,296 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/04%3A_Lewis_Formulas_Structural_Isomerism_and_Resonance_Structures/4.02%3A_Common_Bonding_Patterns_for_1st_and_2nd_Row_Elements |
Once we write enough Lewis formulas containing the elements of interest in organic chemistry, which are mostly the second row elements, we find that certain bonding patterns occur over and over. Learning these patterns is useful when trying to write Lewis formulas because they provide a convenient starting point. For example, in several of the structures given in the previous section, we find that the carbon bonded to three hydrogens is a unit that occurs quite frequently. It is called the methyl group, represented by CH . It is so common that it is valid to write it as such in Lewis formulas, even though it is in fact an abbreviated form, because everybody knows what it stands for. Other common bonding patterns are shown below. Usually forms only bond. Forms four bonds when neutral, but it can also have only three bonds by bearing a positive or a negative charge. When it bears a negative charge it should also carry a pair of unshared electrons. Neutral carbon A has a central carbon with an incomplete octet and a formal +1 charge. A has a central carbon with an unshared electron pair and a formal -1 charge. Forms three bonds and carries a lone pair of electrons when neutral. It can also form four bonds by bearing a positive charge, in which case it carries no unshared electrons. Finally, it can also form two bonds as it carries two unshared electron pairs and a negative charge. Neutral nitrogen Positively charged nitrogen Negatively charged nitrogen Forms two bonds and carries two lone pairs when neutral. It can form three bonds with a positive charge, or one bond with a negative charge. In each case it must carry the appropriate number of unshared electron pairs to complete the octet. Form one bond and carry three electron pairs when neutral. Can carry a negative charge with no bonds. They are rarely seen with positive charges. They behave like their second-row counterparts, except that they can expand their valence shells if needed. Electron pairs on oxygen are not shown for clarity. One thing worth noting is that, in the second row, only boron and carbon can form relatively stable species in which they bond with an incomplete octet. Examples have already been discussed. Boron has no choice but to be electron deficient. Carbon can bond with a complete octet or with an incomplete octet. Obviously bonding with a complete octet provides higher stability. Boron has no choice but to have an incomplete octet An electron deficient carbon in a carbocation It is however very rare to observe species where nitrogen or oxygen bond with incomplete octets. Their high electronegativity renders such situation high energy, and therefore very unstable. For all intents and purposes, avoid writing formulas where oxygen or nitrogen are shown with incomplete octets, even if they carry a positive charge. To write this structure without the lone pair of electrons on oxygen is unacceptable. This structure is unacceptable and indeed it looks quite awkward This species might exist in the high energy environment of a mass spectrometer, but it is not frequently observed in common organic reactions. | 3,149 | 1,298 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/7%3A_Bonding_in_Organic_Molecules/7.4%3A_Aromatic_Hydrocarbons |
Next we consider a class of hydrocarbons with molecular formulas like those of unsaturated hydrocarbons, but which, unlike the alkenes, do not readily undergo addition reactions. These compounds comprise a distinct class, called aromatic hydrocarbons, with unique structures and properties. We start with the simplest of these compounds. Benzene (C H ) is of great commercial importance, but it also has noteworthy health effects. The formula C H seems to indicate that benzene has a high degree of unsaturation. (Hexane, the saturated hydrocarbon with six carbon atoms has the formula C H —eight more hydrogen atoms than benzene.) However, despite the seeming low level of saturation, benzene is rather unreactive. It does not, for example, react readily with bromine, which, is a test for unsaturation. Benzene is a liquid that smells like gasoline, boils at 80°C, and freezes at 5.5°C. It is the aromatic hydrocarbon produced in the largest volume. It was formerly used to decaffeinate coffee and was a significant component of many consumer products, such as paint strippers, rubber cements, and home dry-cleaning spot removers. It was removed from many product formulations in the 1950s, but others continued to use benzene in products until the 1970s when it was associated with leukemia deaths. Benzene is still important in industry as a precursor in the production of plastics (such as Styrofoam and nylon), drugs, detergents, synthetic rubber, pesticides, and dyes. It is used as a solvent for such things as cleaning and maintaining printing equipment and for adhesives such as those used to attach soles to shoes. Benzene is a natural constituent of petroleum products, but because it is a known carcinogen, its use as an additive in gasoline is now limited. To explain the surprising properties of benzene, chemists suppose the molecule has a cyclic, hexagonal, planar structure of six carbon atoms with one hydrogen atom bonded to each. We can write a structure with alternate single and double bonds, either as a full structural formula or as a line-angle formula: However, these structures do not explain the unique properties of benzene. Furthermore, experimental evidence indicates that all the carbon-to-carbon bonds in benzene are equivalent, and the molecule is unusually stable. Chemists often represent benzene as a hexagon with an inscribed circle: The inner circle indicates that the valence electrons are shared equally by all six carbon atoms (that is, the electrons are , or spread out, over all the carbon atoms). It is understood that each corner of the hexagon is occupied by one carbon atom, and each carbon atom has one hydrogen atom attached to it. Any other atom or groups of atoms substituted for a hydrogen atom must be shown bonded to a particular corner of the hexagon. We use this modern symbolism, but many scientists still use the earlier structure with alternate double and single bonds. Most of the benzene used commercially comes from petroleum. It is employed as a starting material for the production of detergents, drugs, dyes, insecticides, and plastics. Once widely used as an organic solvent, benzene is now known to have both short- and long-term toxic effects. The inhalation of large concentrations can cause nausea and even death due to respiratory or heart failure, while repeated exposure leads to a progressive disease in which the ability of the bone marrow to make new blood cells is eventually destroyed. This results in a condition called , in which there is a decrease in the numbers of both the red and white blood cells. | 3,606 | 1,299 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/10%3A_Multi-electron_Atoms/Electron_Configuration |
The of an atomic species (neutral or ionic) allows us to understand the shape and energy of its electrons. Many general rules are taken into consideration when assigning the "location" of the electron to its prospective energy state, however these assignments are arbitrary and it is always uncertain as to which electron is being described. Knowing the electron configuration of a species gives us a better understanding of its , and other chemical properties. The is the standard notation used to describe the electronic structure of an atom. Under the orbital approximation, we let each electron occupy an orbital, which can be solved by a single wavefunction. In doing so, we obtain three quantum numbers (n, ,m ), which are the same as the ones obtained from solving the Schrödinger's equation for Bohr's hydrogen atom. Hence, many of the rules that we use to describe the electron's address in the hydrogen atom can also be used in systems involving multiple electrons. When assigning electrons to orbitals, we must follow a set of three rules: the Aufbau Principle, the Pauli-Exclusion Principle, and Hund's Rule. The wavefunction is the solution to the Schrödinger equation. By solving the Schrödinger equation for the hydrogen atom, we obtain three quantum numbers, namely the principal quantum number (n), the orbital angular momentum quantum number ( ), and the magnetic quantum number (m ). There is a fourth quantum number, called the spin magnetic quantum number (m ), which is not obtained from solving the Schrödinger equation. Together, these four quantum numbers can be used to describe the location of an electron in Bohr's hydrogen atom. These numbers can be thought of as an electron's "address" in the atom. To help describe the appropriate notation for electron configuration, it is best to do so through example. For this example, we will use the iodine atom. There are two ways in which electron configuration can be written: I: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p or I: [Kr]5s 4d 5p In both of these types of notations, the order of the energy levels must be written by increased energy, showing the number of electrons in each subshell as an exponent. In the short notation, you place brackets around the noble gas element followed by the valence shell electron configuration. The periodic table shows that kyrpton (Kr) is the previous noble gas listed before iodine. The noble gas configuration encompases the energy states lower than the valence shell electrons. Therefore, in this case [Kr]=1s 2s 2p 3s 3p 4s 3d 4p . The principal quantum number indicates the shell or energy level in which the electron is found. The value of can be set between 1 to , where is the value of the outermost shell containing an electron. This quantum number can only be positive, non-zero, and integer values. That is, =1,2,3,4,.. For example, an Iodine atom has its outmost electrons in the 5p orbital. Therefore, the principle quantum number for Iodine is 5. The orbital angular momentum quantum number, , indicates the subshell of the electron. You can also tell the shape of the atomic orbital with this quantum number. An subshell corresponds to =0, a subshell = 1, a subshell = 2, a subshell = 3, and so forth. This quantum number can only be positive and integer values, although it can take on a zero value. In general, for every value of n, there are n values of . Furthermore, the value of ranges from 0 to n-1. For example, if n=3, =0,1,2. So in regards to the example used above, the values of Iodine for n = 5 are = 0, 1, 2, 3, 4. The magnetic quantum number, m , represents the orbitals of a given subshell. For a given , m can range from to . A p subshell ( =1), for instance, can have three orbitals corresponding to m = -1, 0, +1. In other words, it defines the p , p and p orbitals of the p subshell. (However, the m numbers don't necessarily correspond to a given orbital. The fact that there are three orbitals simply is indicative of the three orbitals of a p subshell.) In general, for a given , there are 2 +1 possible values for m ; and in a principal shell, there are orbitals found in that energy level. Continuing on from out example from above, the m values of Iodine are m = -4, -3, -2, -1, 0 1, 2, 3, 4. These arbitrarily correspond to the 5s, 5p , 5p , 5p 4d , 4d , 4d , 4d , and 4d orbitals. The spin magnetic quantum number can only have a value of either +1/2 or -1/2. The value of 1/2 is the spin quantum number, s, which describes the electron's spin. Due to the spinning of the electron, it generates a magnetic field. In general, an electron with a m =+1/2 is called an alpha electron, and one with a m =-1/2 is called a beta electron. No two paired electrons can have the same spin value. Out of these four quantum numbers, however, Bohr postulated that only the principal quantum number, n, determines the energy of the electron. Therefore, the 3s orbital ( =0) has the same energy as the 3p ( =1) and 3d ( =2) orbitals, regardless of a difference in values. This postulate, however, holds true only for Bohr's hydrogen atom or other hydrogen-like atoms. When dealing with multi-electron systems, we must consider the electron-electron interactions. Hence, the previously described postulate breaks down in that the energy of the electron is now determined by both the principal quantum number, n, and the orbital angular momentum quantum number, . Although the Schrödinger equation for many-electron atoms is extremely difficult to solve mathematically, we can still describe their electronic structures via electron configurations. There are a set of general rules that are used to figure out the electron configuration of an atomic species: Aufbau Principle, Hund's Rule and the Pauli-Exclusion Principle. Before continuing, it's important to understand that each orbital can be occupied by electrons of opposite spin (which will be further discussed later). The following table shows the number of electrons that can occupy each orbital in a given subshell. 14 Using our example, iodine, again, we see on the periodic table that its atomic number is 53 (meaning it contains 53 electrons in its neutral state). Its complete electron configuration is 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p . If you count up all of these electrons, you will see that it adds up to 53 electrons. Notice that each subshell can only contain the max amount of electrons as indicated in the table above. The word 'Aufbau' is German for 'building up'. , also called the building-up principle, states that electron's occupy orbitals in order of increasing energy. The order of occupation is as follows: Another way to view this order of increasing energy is by using :
This order of occupation roughly represents the increasing energy level of the orbitals. Hence, electrons occupy the orbitals in such a way that the energy is kept at a minimum. That is, the 7s, 5f, 6d, 7p subshells will not be filled with electrons unless the lower energy orbitals, 1s to 6p, are already fully occupied. Also, it is important to note that although the energy of the 3d orbital has been mathematically shown to be lower than that of the 4s orbital, electrons occupy the 4s orbital first before the 3d orbital. This observation can be ascribed to the fact that 3d electrons are more likely to be found closer to the nucleus; hence, they repel each other more strongly. Nonetheless, remembering the order of orbital energies, and hence assigning electrons to orbitals, can become rather easy when related to the . To understand this principle, let's consider the bromine atom. Bromine (Z=35), which has 35 electrons, can be found in Period 4, Group VII of the periodic table. Since bromine has 7 valence electrons, the 4s orbital will be completely filled with 2 electrons, and the remaining five electrons will occupy the 4p orbital. Hence the full or expanded electronic configuration for bromine in accord with the Aufbau Principle is 1s 2s 2p 3s 3p 4s 3d 4p . If we add the exponents, we get a total of 35 electrons, confirming that our notation is correct. Hund's Rule states that when electrons occupy degenerate orbitals (i.e. same n and quantum numbers), they must first occupy the empty orbitals before double occupying them. Furthermore, the most stable configuration results when the spins are parallel (i.e. all alpha electrons or all beta electrons). Nitrogen, for example, has 3 electrons occupying the 2p orbital. According to Hund's Rule, they must first occupy each of the three degenerate p orbitals, namely the 2p orbital, 2p orbital, and the 2p orbital, and with parallel spins (Figure 2). The configuration below is incorrect because the third electron occupies does not occupy the empty 2p orbital. Instead, it occupies the half-filled 2p orbital. This, therefore, is a violation of Hund's Rule (Figure 2).
Wolfgang Pauli postulated that each electron can be described with a unique set of four quantum numbers. Therefore, if two electrons occupy the same orbital, such as the 3s orbital, their spins must be paired. Although they have the same principal quantum number (n=3), the same orbital angular momentum quantum number (l=0), and the same magnetic quantum number (m =0), they have different spin magnetic quantum numbers (m =+1/2 and m =-1/2). The way we designate electronic configurations for cations and anions is essentially similar to that for neutral atoms in their ground state. That is, we follow the three important rules: Aufbau Principle, Pauli-exclusion Principle, and Hund's Rule. The electronic configuration of cations is assigned by removing electrons first in the outermost p orbital, followed by the s orbital and finally the d orbitals (if any more electrons need to be removed). For instance, the ground state electronic configuration of calcium (Z=20) is 1s 2s 2p 3s 3p 4s . The calcium ion (Ca ), however, has two electrons less. Hence, the electron configuration for Ca is 1s 2s 2p 3s 3p . Since we need to take away two electrons, we first remove electrons from the outermost shell (n=4). In this case, all the 4p subshells are empty; hence, we start by removing from the s orbital, which is the 4s orbital. The electron configuration for Ca is the same as that for Argon, which has 18 electrons. Hence, we can say that both are isoelectronic. The electronic configuration of anions is assigned by adding electrons according to Aufbau Principle. We add electrons to fill the outermost orbital that is occupied, and then add more electrons to the next higher orbital. The neutral atom chlorine (Z=17), for instance has 17 electrons. Therefore, its ground state electronic configuration can be written as 1s 2s 2p 3s 3p . The chloride ion (Cl ), on the other hand, has an additional electron for a total of 18 electrons. Following Aufbau Principle, the electron occupies the partially filled 3p subshell first, making the 3p orbital completely filled. The electronic configuration for Cl can, therefore, be designated as 1s 2s 2p 3s 3p . Again, the electron configuration for the chloride ion is the same as that for Ca and Argon. Hence, they are all isoelectronic to each other. 1. Which of the princples explained above tells us that electrons that are paired cannot have the same spin value? 2. Find the values of n, , m , and m for the following: a. Mg b. Ga c. Co 3. What is a possible combination for the quantum numbers of the 5d orbital? Give an example of an element which has the 5d orbital as it's most outer orbital. 4. Which of the following cannot exist (there may be more than one answer): a. n = 4; = 4; m = -2; m = +1/2 b. n = 3; = 2; m = 1; m = 1 c. n = 4; = 3; m = 0; m = +1/2 d. n = 1; = 0; m = 0; m = +1/2 e. n = 0; = 0; m = 0; m = +1/2 5. Write electron configurations for the following: a. P b. S c. Zn 1. Pauli-exclusion Principle 2. a. n = 3; = 0, 1, 2; m = -2, -1, 0, 1, 2; m can be either +1/2 or -1/2 b. n = 4; = 0, 1, 2, 3; m = -3, -2, -1, 0, 1, 2, 3; m can be either +1/2 or -1/2 c. n = 3; = 0, 1, 2; m = -2, -1, 0, 1, 2, 3; m can be either +1/2 or -1/2 3. n = 5; = 3; m = 0; m = +1/2. Osmium (Os) is an example. 4. a. The value of cannot be 4, because ranges from (0 - n-1) b. m can only be +1/2 or -1/2 c. Okay d. Okay e. The value of n cannot be zero. 5. a. 1s 2s 2p 3s 3p b. 1s 2s 2p 3s 3p c. 1s 2s 2p 3s 3p 4s 3d | 12,441 | 1,301 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/01%3A_Matter-_Its_Properties_And_Measurement/1.5%3A_Density_and_Percent_Composition_-_Their_Use_in_Problem_Solving |
Density and percent composition are important properties in chemistry. Each have basic components as well as broad applications. Components of density are: mass and volume, both of which can be more confusing than at first glance. An application of the concept of density is determining the volume of an irregular shape using a known mass and density. Determining Percent Composition requires knowing the mass of entire object or molecule and the mass of its components. Which one weighs more, a kilogram of feathers or a kilogram of bricks? Though many people will say that a kilogram of bricks is heavier, they actually weigh the same! However, many people are caught up by the concept of density \(\rho\), which causes them to answer the question incorrectly. A kilogram of feathers clearly takes up more space, but this is because it is less "dense." But what is density, and how can we determine it? In the laboratory, density can be used to identify an element, while percent composition is used to determine the amount, by mass, of each element present in a chemical compound. In daily life, density explains everything from why boats float to why air bubbles will try to escape from soda. It even affects your health because bone density is very important. Similarly, percent composition is commonly used to make animal feed and compounds such as the baking soda found in your kitchen. is a physical property found by dividing the mass of an object by its volume. Regardless of the sample size, density is always constant. For example, the density of a pure sample of tungsten is always 19.25 grams per cubic centimeter. This means that whether you have one gram or one kilogram of the sample, the density will never vary. The equation, as we already know, is as follows: \[ \text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \label{1.5.4}\] or just \[\rho = \dfrac{m}{v} \label{1.5.5}\] \[\text{Weight} = \text{mass} \times \text{gravity} \label{1.5.6}\] \(Weight = {m}{g} \label{1.5.7}\) Another important difference between mass and weight is how they are measured. Weight is measured with a scale, while mass must be measured with a balance. Just as people confuse mass and weight, they also confuse scales and balances. A balance counteracts the effects of gravity while a scale incorporates it. There are two types of balances found in the laboratory: electronic and manual. With a manual balance, you find the unknown mass of an object by adjusting or comparing known masses until equilibrium is reached. With an electronic balance, which is what you will work with in the UC Davis laboratory, the mass is found by electronic counterbalancing with little effort from the user. An electronic balance can be far more accurate than any other balance and is easier to use, but they are expensive. Also, keep in mind that all instruments used in the laboratory have . Volume describes the quantity of three dimensional space than an object occupies. The SI unit for volume is meters cubed (m ), but milliliters (mL), centimeters cubed (cm ), and liters (L) are more common in the laboratory. There are many equations to find volume. Here are just a few of the easy ones: Volume = (length) or Volume = (length)(width)(height) or Volume = (base area)(height) We know all of density's components, so let's take a closer look at density itself. \(\rho = \frac{7\; g}{5\;cm^3}= 1.4\; g/cm^3\). STP corresponds to a temperature of 273 K (0° Celsius) and 1 atmosphere of pressure. As can be seen from the table, the most dense element is Osmium (Os) with a density of 22.6 g/cm . The least dense element is Hydrogen (H) with a density of 0.09 g/cm . Density generally decreases with increasing temperature and likewise increases with decreasing temperatures. This is because volume differs according to temperature. Volume increases with increasing temperature. Below is a table showing the density of pure water with differing temperatures. As can be seen from Table \(\Page {2}\), the density of water decreases with increasing temperature. Liquid water also shows an exception to this rule from 0 degrees Celsius to 4 degrees Celsius, where it increases in density instead of decreasing as expected. Looking at the table, you can also see that ice is less dense than water. This is unusual as solids are generally denser than their liquid counterparts. Ice is less dense than water due to hydrogen bonding. In the water molecule, the hydrogen bonds are strong and compact. As the water freezes into the hexagonal crystals of ice, these hydrogen bonds are forced farther apart and the volume increases. With this volume increase comes a decrease in density. This explains why ice floats to the top of a cup of water: the ice is less dense. Even though the rule of density and temperature has its exceptions, it is still useful. For example, it explains how hot air balloons work. Density increases with increasing pressure because volume decreases as pressure increases. And since density=mass/volume , the lower the volume, the higher the density. This is why all density values in the Periodic Table are recorded at STP, as mentioned in the section "Density and the Periodic Table." The decrease in volume as related to pressure is explained in Boyle's Law: \(P_1V_1 = P_2V_2\) where P = pressure and V = volume. This idea is explained in the figure below. The Greek scientist Archimedes made a significant discovery in 212 B.C. The story goes that Archimedes was asked to find out for the King if his goldsmith was cheating him by replacing his gold for the crown with silver, a cheaper metal. Archimedes did not know how to find the volume of an irregularly shaped object such as the crown, even though he knew he could distinguish between elements by their density. While meditating on this puzzle in a bath, Archimedes recognized that when he entered the bath, the water rose. He then realized that he could use a similar process to determine the density of the crown! He then supposedly ran through the streets naked shouting "Eureka," which means "I found it!" in Latin. Archimedes then tested the king's crown by taking a genuine gold crown of equal mass and comparing the densities of the two. The king's crown displaced more water than the gold crown of the same mass, meaning that the king's crown had a greater volume and thus had a smaller density than the real gold crown. The king's "gold" crown, therefore, was not made of pure gold. Of course, this tale is disputed today because Archimedes was not precise in all his measurements, which would make it hard to determine accurately the differences between the two crowns. Archimedes' Principle states that if an object has a greater density than the liquid that it is placed into, it will sink and displace a volume of liquid equal to its own. If it has a smaller density, it will float and displace a mass of liquid equal to its own. If the density is equal, it will not sink or float (Figure \(\Page {2}\)). The principle explains why balloons filled with helium float. Balloons, as we learned in the section concerning density and temperature, float because they are less dense than the surrounding air. Helium is less dense than the atmospheric air, so it rises. Archimedes' Principle can also be used to explain why boats float. Boats, including all the air space, within their hulls, are far less dense than water. Boats made of steel can float because they displace their mass in water without submerging all the way. The table below gives the densities of a few liquids to put things into perspective. Percent composition is very simple. tells you by mass what percent of each element is present in a compound. A is the combination of two or more elements. If you are studying a chemical compound, you may want to find the percent composition of a certain element within that chemical compound. The equation for percent composition is \[\text{Percent Composition} = \dfrac{\text{Total mass of element present}}{\text{Molecular mass}} \times 100\% \label{1.5.3}\] If you want to know the percent composition of the elements in an compound, follow these steps: Steps to Solve: Tips for solving: These steps are outlined in the figure below. For another example, if you wanted to know the percent composition of hydrochloric acid (HCl), first find the molar mass of Hydrogen. . Now find the molecular mass of HCl: . Follow steps 3 and 4: Now just subtract to find the percent by mass of Chlorine in the compound: Therefore, HCl is Hydrogen and Chlorine by mass. Percent composition plays an important role in everyday life. It is more than just the amount of chlorine in your swimming pool because it concerns everything from the money in your pocket to your health and how you live. The next two sections describe percent composition as it relates to you. The nutrition label found on the container of every bit of processed food sold by the local grocery store employs the idea of percent composition. On all nutrition labels, a known serving size is broken down in five categories: Total Fat, Cholesterol, Sodium, Total Carbohydrate, and Protein. These categories are broken down into further subcategories, including Saturated Fat and Dietary Fiber. The mass for each category, except Protein, is then converted to percent of Daily Value. Only two subcategories, Saturated Fat and Dietary Fiber are converted to percent of Daily Value. The Daily Value is based on a the mass of each category recommended per day per person for a 2000 calorie diet. The mass of protein is not converted to percent because their is no recommended daily value for protein. Following is a picture outlining these ideas. For example, if you wanted to know the percent by mass of the daily value for sodium you are eating when you eat one serving of the food with this nutrition label, then go to the category marked sodium. Look across the same row and read the percent written. If you eat one serving of this food, then you will have consumed about 9% of your daily recommended value for sodium. To find the percent mass of fat in the whole food, you could divide 3.5 grams by 15 grams, and see that this snack is 23.33% fat. The penny should be called "the lucky copper coin." The penny has not been made of solid copper since part of 1857. After 1857, the US government started adding other cheaper metals to the mix. The penny, being only one cent, is literally not worth its weight in copper. People could melt copper pennies and sell the copper for more than the pennies were worth. After 1857, nickel was mixed with the more expensive copper. After 1864, the penny was made of bronze. Bronze is 95% copper and 5% zinc and tin. For one year, 1943, the penny had no copper in it due to the expense of the World War II. It was just zinc coated steel. After 1943 until 1982, the penny went through periods where it was brass or bronze. The percent composition of a penny may actually affect health, particularly the health of small children and pets. Since the newer pennies are made mainly of zinc instead of copper, they are a danger to a child's health if ingested. Zinc is very susceptible to acid. If the thin copper coating is scratched and the hydrochloric acid present in the stomach comes into contact with the zinc core it could cause ulcers, anemia, kidney and liver damage, or even death in severe cases. Three important factors in penny ingestion are time, pH of the stomach, and amount of pennies ingested. Of course, the more pennies swallowed, the more danger of an overdose of zinc. The more acidic the environment, the more zinc will be released in less time. This zinc is then absorbed and sent to the liver where it begins to cause damage. In this kind of situation, time is of the essence. The faster the penny is removed, the less zinc is absorbed. If the penny or pennies are not removed, organ failure and death can occur. Below is a picture of a scratched penny before and after it had been submerged in lemon juice. Lemon juice has a similar pH of 1.5-2.5 when compared to the normal human stomach after food has been consumed. Time elapsed: 36 hours. As you can see, the copper is vastly unharmed by the lemon juice. That's why pennies made before 1982 with mainly copper (except the 1943 penny) are relatively safe to swallow. Chances are they would pass through the digestive system naturally before any damage could be done. Yet, it is clear that the zinc was partially dissolved even though it was in the lemon juice for only a limited amount of time. Therefore, the percent composition of post 1982 pennies is hazardous to your health and the health of your pets if ingested. Following are examples of different types of percent composition and density problems. These problems are meant to be easy in the beginning and then gradually become more challenging. Unless otherwise stated, answers should be in or the equivalent . These problems will follow the same pattern of difficulty as those of density. Here are the solutions to the listed practice problems. | 13,100 | 1,302 |
Subsets and Splits