5.16.At a high enough temperature, the equilibrium con-
stant is 4.00 for the gas-phase isotope exchange reaction
H 2 D 2 2 HDCalculate the equilibrium partial pressures if 0.50 atm of H 2
and 0.10 atm of D 2 were initially present in a closed system.
What is the extent of reaction at equilibrium?
5.17.If 0.50 atm of krypton were part of the equilibrium in
exercise 5.16, 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?
5.18.Nitrogen dioxide, NO 2 , dimerizes easily to form dinitro-
gen tetroxide, N 2 O 4 :
2NO 2 (g) N 2 O 4 (g)(a)Using data in Appendix 2, calculate (^) rxnG° and Kfor this
equilibrium.
(b)Calculate for this equilibrium if 1.00 mol NO 2 were pre-
sent initially and allowed to come to equilibrium with the
dimer in a 20.0-L system.
5.19.Another nitrogen-oxygen reaction of some importance is
2NO 2 (g) H 2 O (g) →HNO 3 (g) HNO 2 (g)
which is thought to be the primary reaction involved in the
production of acid rain. Determine (^) rxnG° and Kfor this re-
action.
5.20.Suppose the reaction in Example 5.5 occurred in a
20.0-L vessel. Would the amounts at equilibrium be different?
How about at equilibrium?
5.4 Solutions and Condensed Phases
5.21.Write proper expressions for the equilibrium constant
for the following reactions.
(a)PbCl 2 (s) Pb^2 (aq) 2Cl(aq)
(b)HNO 2 (aq) H(aq) NO 2 (aq)
(c)CaCO 3 (s) H 2 C 2 O 4 (aq)
CaC 2 O 4 (s) H 2 O () CO 2 (g)
5.22.The (^) fG° of diamond, a crystalline form of elemental
carbon, is 2.90 kJ/mol at 25.0°C. Give the equilibrium con-
stant for the reaction
C (s, graphite) C (s, diamond)
On the basis of your answer, speculate on the natural occur-
rence of diamond.
5.23.The densities of graphite and diamond are 2.25 and
3.51 g/cm^3 , respectively. Using the expression
(^) rxnG rxnG°RTln
a
a
g
d
r
ia
a
and equation 5.14, estimate the pressure necessary for (^) rxnG
to equal zero. What is the stable high-pressure solid phase of
carbon?
5.24.Buckminsterfullerene, C 60 , is a spherical molecule com-
posed of hexagons and pentagons of carbon atoms reminis-
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cent of a geodesic dome. It is currently the focus of muchscientific study. For C 60 , (^) fG° is 23.98 kJ/mol at 25.0°. Write
the balanced formation reaction for 1 mole of buckminster-
fullerene and calculate the equilibrium constant for the for-
mation reaction.
5.25.The bisulfate (or hydrogen sulfate) anion, HSO 4 , is a
weak acid. The equilibrium constant for the aqueous acid re-
action
HSO 4 HSO 42
is 1.2 10 ^2.
(a)Calculate G° for this equilibrium.
(b)At low concentrations, activity coefficients are approxi-
mately 1 and the activity of a dissolved solute equals its mo-
lality. Determine the equilibrium molalities of a 0.010-molal
solution of sodium hydrogen sulfate.
5.5 Changes in Equilibrium Constants
5.26.For the reaction
2Na (g) Na 2 (g)
the following values of Khave 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 (^) rxnH° for the reaction.
5.27.For a reaction whose standard enthalpy change is
100.0 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 20.0 kJ?
5.28.Consider the following equilibrium:
2SO 2 (g) O 2 (g) 2SO 3 (g)
What is the effect on the equilibrium of each of the following
changes? (You may need to calculate some standard enthalpy
or Gibbs free energy changes to answer these.) (a)The pres-
sure is increased by decreasing the volume. (b)The tempera-
ture is decreased. (c)The pressure is increased by the addi-
tion of nitrogen gas, N 2.
5.29.Show that equations 5.18 and 5.19 are equivalent.
5.6 Amino Acid Equilibria
5.30.Of the amino acids listed in Table 5.1, which one should
have an isoelectric point closest to 7, the pH of neutral water?
5.31.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. pK 1 2.34, pK 2 9.60. Do you need to
make any other assumptions to simplify the calculation?
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Exercises for Chapter 5 139