DG 5 2 278 J Would this be considered a spontaneous process? Because the pressure is not kept
5.4 Solutions and Condensed Phases 5.31. Write proper expressions for the equilibrium constant
(a) PbCl2 (s) m Pb21 (aq) 1 2Cl2 (aq) (b) HNO2 (aq) m H1 (aq) 1 NO22 (aq)
(c) CaCO3 (s) 1 H2C2O4 (aq) m CaC2O4 (s) 1 H2O (,) 1 CO2 (g) 5.32. For the reaction
AgCl (s) m Ag1 (aq) 1 Cl2 (aq)
K 5 1.8 3 10210. (a) Using DfG° [AgCl (s)] in Appendix 2, deter- mine DfG [Ag1 (aq) 1 Cl2 (aq)]. (b) If DfG [Cl2 (aq)] 5 2131.3 kJ, what is DfG [Ag1 (aq)]?
5.33. Use the data in Appendix 2 to calculate K for CaCO3 (s, arag) m Ca21 (aq) 1 CO322 (aq)
5.34. The DfG° of diamond, a crystalline form of elemental carbon, is 12.90 kJ/mol at 25.0°C. Give the equilibrium constant for the reaction
C (s, graphite)mC (s, diamond)
On the basis of your answer, speculate on the natural occur- rence of diamond.
5.35. The densities of graphite and diamond are 2.25 and 3.51 g/cm3, respectively. Using the expression
DrxnG 5 DrxnG° 1 RT ln aadia
gra
and equation 5.14, estimate the pressure necessary for DrxnG to equal zero. What is the stable high-pressure solid phase of carbon?
5.36. Buckminsterfullerene, C60, is a spherical molecule com- posed of hexagons and pentagons of carbon atoms remi- niscent of a geodesic dome. It is currently the focus of much scientific study. For C60, DfG° is 23.98 kJ/mol at 25.0°C. Write the balanced formation reaction for 1 mole of buckminster- fullerene and calculate the equilibrium constant for the for- mation reaction.
5.37. Buckminsterfullerene (see previous exercise) has a density of 1.65 g/cm3. What is the activity of C60 at 100°C and 1500 bar?
5.38. At what pressure does H2O have an activity of 2.00?
Assume a temperature of 25.0°C and a molar volume of 18.07 cm3.
5.39. The bisulfate (or hydrogen sulfate) anion, HSO42, is a weak acid. The equilibrium constant for the aqueous acid reaction
HSO42 (aq) m H1 (aq) 1 SO422 (aq) is 1.2 3 1022.
(a) Calculate DG° for this equilibrium. Assume a tempera- ture of 25.0ºC.
(b) At low concentrations, activity coefficients are approxi- mately 1 and the activity of a dissolved solute equals its molality. Determine the equilibrium molalities of a 0.010-molal solution of sodium hydrogen sulfate.
5.40. Activities of ionic solutes are defined a bit differently from other solutes (see Chapter 8), but the concept is the 5.22. True or false: If K 5 1 for a gas-phase reaction, then all
equilibrium partial pressures of reactants and products must equal 1.00 atm. Explain your choice.
5.23. The balanced chemical reaction for the formation of ammonia from its elements is
N2 1 3H2 (g) m 2NH3 (g)
(a) What is DrxnG° for this reaction? (b) What is DrxnG for this reaction if all species have a partial pressure of 0.500 bar at 25°C? Assume that the fugacities are equal to the partial pressures.
5.24. The answers in exercise 5.23 should show that chang- ing the partial pressure changes the instantaneous DrxnG even though the ratio of partial pressures stays the same (that is, 1:1:1 for standard pressure conditions is equal to 0.5:0.5:0.5 for the given conditions). This suggests the interesting possibility that at some equal partial pressure p of all components, the reaction reverses; that is, the instantaneous DrxnG becomes negative. Determine p for this equilibrium. (You will have to use the properties of logarithms as mentioned in the chapter to find the answer.) Is your answer of value to those who work with gases at high pressures, or at low pressures? What is your reasoning?
5.25. At a high enough temperature, the equilibrium constant is 4.00 for the gas-phase isotope exchange reaction
H2 + D2 m 2 HD
Calculate the equilibrium partial pressures if 0.50 atm of H2
and 0.10 atm of D2 were initially present in a 20.0-L system at 488 K. What is the extent of reaction at equilibrium?
5.26. If 0.50 atm of krypton were part of the equilibrium in exercise 5.25, would the value of the equilibrium constant be the same or different if the volume were kept the same? Is this case different from Examples 5.6 and 5.11?
5.27. Nitrogen dioxide, NO2, dimerizes easily to form dinitrogen tetroxide, N2O4:
2NO2 (g) m N2O4 (g)
(a) Using data in Appendix 2, calculate DrxnG° and K for this equilibrium.
(b) Calculate j for this equilibrium if 1.00 mol NO2 were present initially and allowed to come to equilibrium with the dimer in a 20.0-L system.
5.28. Another nitrogen-oxygen reaction of some impor- tance is
2NO2 (g) 1 H2O (g) m HNO3 (g) 1 HNO2 (g)
which is thought to be the primary reaction involved in the production of acid rain. Determine DrxnG° and K for this reaction.
5.29. Suppose the reaction in Example 5.5 occurred in a 20.0-L vessel. Would the amounts at equilibrium be different?
How about j at equilibrium?
5.30. Is it appropriate to define a Kmix from DmixG (see exer- cise 4.62) as
DmixG 5 2RT ln Kmix
Why or why not?
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EXERCISES FOR CHAPTER 5 153
5.46. For a reaction whose standard enthalpy change is 2100.0 kJ, what temperature is needed to double the equi- librium constant from its value at 298 K? What temperature is needed to increase the equilibrium constant by a factor of 10?
What if the standard enthalpy change were 220.0 kJ?
5.47. For the reaction
2NO2 (g) m N2O4 (g)
K(0°C) 5 58 and K(100°C) 5 0.065. Estimate DrxnH° for this reaction.
5.48. The isotope exchange reaction
H2 (g) 1 D2 (g) m 2HD (g) DrxnH° 5 0.64 kJ has an equilibrium constant of 4.00 at 1000 K. Estimate the temperature at which K 5 1.00.
5.49. Consider the following equilibrium:
2SO2 (g) 1 O2 (g) m 2SO3 (g)
What is the effect on the equilibrium of each of the following changes? (You may need to calculate some standard enthalpy or Gibbs energy changes to answer these.) (a) The pressure is increased by decreasing the volume. (b) The temperature is decreased. (c) The pressure is increased by the addition of nitrogen gas, N2.
5.50. For the equilibrium
Br2 (g) m 2Br (g)
equilibrium partial pressures at 755 K are pBr2 5 0.668 bar and pBr 5 0.226 bar. If the volume of the system is suddenly doubled so that pBr2 5 0.334 bar and pBr 5 0.113 bar, what are the new partial pressures when equilibrium is reestablished?
Is this consistent with Le Chatelier’s principle?
5.51. The decomposition of NaHCO3, used in kitchens to extinguish fires, is
2NaHCO3 (s) m Na2CO3 (s) 1 CO2 (g) 1 H2O (g) (a) Using the data in Appendix 2, calculate DrxnG° and
DrxnH°.
(b) Calculate K at 298 K.
(c) What are pCO2 and pH2O at equilibrium at 298 K?
(d) In a grease fire, the temperature can be around 1150°C.
What are pCO2 and pH2O at equilibrium at 1150°C?
5.52. For the equilibrium
Br2 (g) m 2Br (g)
at 1000 K, K = 4.55. In a 1.00-L flask, 0.012 mol of Br2 is intro- duced. (a) Calculate the equilibrium partial pressures of both species. (b) After equilibrium has been established, the volume of the flask doubles to 2.00 L. Calculate the new equi- librium partial pressures after the new equilibrium has been established.
5.53. Show that equations 5.18 and 5.19 are equivalent.
5.54. For the reaction
3O2 (g) m 2O3 (g)
the equilibrium constant at 1600 K is 0.235. (a) If initially, 1.00 atm of O2 were introduced into a chamber at 1600 K, what are the partial pressures present at equilibrium? (Hint: you will same. This table lists some concentrations and activities of
NaCl solutions:
Molality Activity
0.001 0.000996
0.002 0.00191
0.005 0.00465
0.010 0.00904
0.020 0.0175
0.050 0.0412
0.100 0.0780
0.200 0.146
0.500 0.340
1.00 0.660
Assuming that NaCl obeys equation 5.16, calculate the activ- ity coefficient g for NaCl at each concentration and explain any trend you see.
5.41. Write the equilibrium constant expression for each reaction in terms of activities, simplifying where appropriate.
(a) C (s) 1 O2 (g) m CO2 (g) (b) P4 (s) 1 5O2 (g) m P4O10 (s)
(c) 2HNO2 (g) 1 3Cl2 (g) m 2NCl3 (g) 1 H2 (g) 1 2O2 (g) 5.5 Changes in Equilibrium Constants
5.42. For the reaction
2Na (g) m Na2 (g)
the following values of K have been determined (C. T. Ewing et al., J. Chem. Phys. 1967, 71, 473):
T (K) K
900 1.32
1000 0.47
1100 0.21
1200 0.10
From these data, estimate DrxnH° for the reaction.
5.43. For a given chemical equilibrium, these data are deter- mined experimentally:
T K
350 K 3.76 3 1022 450 K 1.86 3 1021 Determine DH° for this reaction.
5.44. Biological standard states include specifying a reference temperature of 37.0°C rather than the common reference temperature of 25.0°C. For a chemical reaction in which DH 5 224.3 kJ (see equation 2.62), by what percent does K change?
5.45. (a) At 25.0°C, Kw for the autoionization of water is 1.01 3 10214, while at 100.0°C it is 5.60 3 10213. What is DH°
for the autoionization of water? (b) For D2O, the values are 1.10 3 10215 and 7.67 3 10214, respectively. What is the DH°
for the autoionization of D2O?
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154 Chapter 5 | Introduction to Chemical Equilibrium
amino acids. A model for this process is the dimerization of cysteine (Cys):
2 cysteine h Cys-S-S-Cys 1 H2
For this reaction, DH 5 173.4 kJ and DS 5 369.9 J/K. Calculate DG and K at 25.0°C. What do these values imply about disulfide bridge formation in proteins?
SymBoliC matH ExErCiSES
5.61. Consider the balanced chemical reaction CH4 (g) 1 2Br2 (g) hCH2Br2 (g) 1 2HBr (g)
A system starts with 10.0 mol CH4 and 3.75 mol Br2, and 0.00 mol of the two products. Plot j versus amount of each product and reactant. Comment on the differences in the plot.
5.62. For the gas-phase reaction
2H2 1 O2 h2H2O
DrxnG° is 2457.18 kJ. What does a graph of DG versus ln Q (ln Q varying from 250 to 150) look like at 25°C? Change the tem- perature and find out if the graph looks substantially different at different temperatures.
5.63. Simple equilibrium problems can get mathematically complicated when the coefficients are different small whole numbers. For the balanced reaction
2SO3 (g) hS2 (g) 1 3O2 (g)
the equilibrium constant has a value of 4.33 3 1022 at some elevated temperature. Calculate the equilibrium concentra- tions of all species if the initial amount of SO3 were (a) 0.150 atm, (b) 0.100 atm, (c) 0.001 atm.
5.64. Plot DG(substances), DG(mixing), and the sum of the two DG values (see Figure 5.4 for an example) for the reaction ASB where DfG(A) 5 265 kJ/mol, DfG(B) 5 269 kJ/mol, nA 1 nB 5 1 mol, and the temperature is 298 K. Estimate j at equilibrium. You may have to expand the y axis to find j.
have to solve a cubic equation. There are cubic equation solv- ers available on the Internet.) (b) In a second experiment, 1.00 atm of O2 is introduced into a chamber at 1600 K, along with 0.50 atm of Kr. Now what are the partial pressures pres- ent at equilibrium?
5.55. Use the van’t Hoff equation to argue which side of a chemical reaction is favored as the temperature increases for (a) an exothermic process and (b) an endothermic process. Do the directions conform with Le Chatelier’s principle?
5.6 Amino Acid Equilibria
5.56. Of the amino acids listed in Table 5.1, which one should have an isoelectric point closest to 7, the pH of neu- tral water?
5.57. Determine the concentrations of the three ionic forms of glycine present if 1.0 mol of glycine is used to make 1.00 L of aqueous solution. pK1 5 2.34, pK2 5 9.60.
Do you need to make any other assumptions to simplify the calculation?
5.58. The formation of zwitterionic glycine, CH2(NH31)(COO2), from its fully acid form CH2(NH31)(COOH) and its fully basic form CH2(NH2)(COO2) is
CH2(NH31)(COOH) 1 CH2(NH2)(COO2) m
2CH2(NH31)(COO2) Use the data in Table 5.1 to determine the equilibrium con- stant for this reaction. What does your answer imply about the relative amount of the zwitterion in solution?
5.59. Monosodium glutamate, or MSG, is the sodium salt of the amino acid glutamic acid. Its formula (with ions shown) is Na1C5H8NO42. Calculate the concentration of the zwitter- ionic glutamic acid in a 0.010-m solution of MSG. Assume the molalities are equal to the activities. Hint: See equation 5.21 and Table 5.1.
5.60. An important process in proteins is the formation of disulfide bridges between nearby S atoms on different
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6
The previous chapter introduced some of the concepts of equilibrium. This chapter and the next will expand on those concepts as we apply them to certain types of chemical systems. Here, we focus on the simplest of systems, those that consist of a single chemical component. It may seem strange that we would spend much effort on such simple systems, but there is a reason. The ideas we develop using simple systems apply to more complicated systems. The more thoroughly the basic concepts are developed, the more easily they can be applied to real systems.
Equilibria of the type we are about to consider are largely macroscopic phenomena. Because of this, there will be less microscopic-level considerations in this chapter than we saw in previous ones.