Chapter 17 – Entropy, Free Energy, and Equilibrium
68. Using the thermodynamic data provided below, calculate the standard change in entropy
when one mole of sodium nitrate is dissolved in water.
Will the solubility of sodium nitrate increase or decrease if the temperature of the system is
increased?
Chapter 17 – Entropy, Free Energy, and Equilibrium
69.
Using the thermodynamic data provided below, calculate the standard change in entropy when one mole of sodium sulfate is
dissolved in water.
Will the solubility of sodium nitrate increase or decrease if the temperature of the system is increased?
Chapter 17 – Entropy, Free Energy, and Equilibrium
70.
Select True or False: For the reaction CuS(s) + H2(g) H2S(g) + Cu(s),
Gf (CuS) = –53.6 kJ/mol
Gf (H2S) = –33.6 kJ/mol
Hf (CuS) = –53.1 kJ/mol
Hf (H2S) = – 20.6 kJ/mol
This reaction proceeds spontaneously at 298 K and 1 atm pressure.
71. For the reaction CuS(s) + H2(g) H2S(g) + Cu(s),
Gf (CuS) = –53.6 kJ/mol
Gf (H2S) = –33.6 kJ/mol
Hf (CuS) = –53.1 kJ/mol
Hf (H2S) = – 20.6 kJ/mol
Calculate the value of the equilibrium constant (Kp) for this reaction at 298 K.
Chapter 17 – Entropy, Free Energy, and Equilibrium
72. For the reaction CuS(s) + H2(g) H2S(g) + Cu(s),
Gf (CuS) = –53.6 kJ/mol
Gf (H2S) = –33.6 kJ/mol
Hf (CuS) = –53.1 kJ/mol
Hf (H2S) = –20.6 kJ/mol
Calculate G at 798 K and 1 atm pressure (assume S and H do not change with
temperature).
73. For the reaction CuS(s) + H2(g) H2S(g) + Cu(s),
Gf (CuS) = –53.6 kJ/mol
Gf (H2S) = –33.6 kJ/mol
Hf (CuS) = –53.1 kJ/mol
Hf (H2S) = –20.6 kJ/mol
Calculate the value of the equilibrium constant (Kp) at 798 K and 1 atm pressure.
Chapter 17 – Entropy, Free Energy, and Equilibrium
74.
Select True or False: For the reaction SbCl5(g) SbCl3(g) + Cl2(g),
Gf (SbCl5) = –334.34 kJ/mol
Gf (SbCl3) = –301.25 kJ/mol
Hf (SbCl5) = –394.34 kJ/mol
Hf (SbCl3) = –313.80 kJ/mol
This reaction proceeds spontaneously at 298 K and 1 atm pressure.
75. For the reaction SbCl5(g) SbCl3(g) + Cl2(g),
Gf (SbCl5) = –334.34 kJ/mol
Gf (SbCl3) = –301.25 kJ/mol
Hf (SbCl5) = –394.34 kJ/mol
Hf (SbCl3) = –313.80 kJ/mol
Chapter 17 – Entropy, Free Energy, and Equilibrium
76. For the reaction SbCl5(g) SbCl3(g) + Cl2(g),
Gf (SbCl5) = –334.34 kJ/mol
Gf (SbCl3) = –301.25 kJ/mol
Hf (SbCl5) = –394.34 kJ/mol
Hf (SbCl3) = –313.80 kJ/mol
Calculate G at 800 K and 1 atm pressure (assume S and H do not change with
77. For the reaction SbCl5(g) SbCl3(g) + Cl2(g),
Gf (SbCl5) = –334.34 kJ/mol
Gf (SbCl3) = –301.25 kJ/mol
Hf (SbCl5) = –394.34 kJ/mol
Hf (SbCl3) = –313.80 kJ/mol
Calculate the value of the equilibrium constant (Kp) at 800 K and 1 atm pressure.
Chapter 17 – Entropy, Free Energy, and Equilibrium
78. Assuming S and H do not vary with temperature, at what temperature will the
reaction shown below become spontaneous?
C(s) + H2O(g) → H2(g) + CO(s) (S = 133.6 J/K·mol; H = 131.3 kJ/mol)
79. Rubidium has a heat of vaporization of 69.0 kJ/mol at its boiling point (686C). Calculate
S for this process, Rb(l) → Rb(g), at 1 atm and 686C.
Chapter 17 – Entropy, Free Energy, and Equilibrium
80. The free energy of formation of nitric oxide, NO, at 1000 K (roughly the temperature in
an automobile engine during ignition) is about 78 kJ/mol. Calculate the equilibrium constant
Kp for the reaction N2(g) + O2(g) 2NO(g) at this temperature.
81. Predict the signs (–, +, or 0) of H and S, in that order, for the reaction: O2(g) → 2O(g).
Chapter 17 – Entropy, Free Energy, and Equilibrium
82. Select True or False: Under the conditions of high temperatures only, the reaction O2(g)
Chapter 17 – Entropy, Free Energy, and Equilibrium
83. Predict the signs (–, +, or 0) of H and S, in that order, for the expansion of an ideal gas
into a vacuum.
84. Select True or False: The expansion of an ideal gas into a vacuum is expected to be
Chapter 17 – Entropy, Free Energy, and Equilibrium
85. Predict the signs (–, +, or 0) of H and S, in that order, for the process: H2O(l) →
H2O(s).
86. Select True or False: The process: H2O(l) → H2O(s) expected to be spontaneous at low
temperatures only.
Chapter 17 – Entropy, Free Energy, and Equilibrium
87. Predict the signs (–, +, or 0) of H and S, in that order, for the reaction: 6CO2(g) +
6H2O(g) → C6H12O6(g) + 6O2(g).
88. Select True or False: The reaction: 6CO2(g) + 6H2O(g) → C6H12O6(g) + 6O2(g) is never
Chapter 17 – Entropy, Free Energy, and Equilibrium
89. What is the free energy change for the reaction SiO2(s) + Pb(s) → PbO2(s) + Si(s)?
Gf (PbO2) = –217 kJ/mol
Gf (SiO2) = –856 kJ/mol
90. Select True or False: The reaction SiO2(s) + Pb(s) → PbO2(s) + Si(s) is spontaneous.
Gf (PbO2) = –217 kJ/mol
Gf (SiO2) = –856 kJ/mol
Chapter 17 – Entropy, Free Energy, and Equilibrium
91. For a certain reaction, G = 87 kJ/mol, H = 100 kJ/mol at STP. At what temperature,
in K, is the reaction at equilibrium, assuming that S and H are temperature-independent?
92. The heat of vaporization of water is 2.27 kJ/g. What is Svap per mole at the normal
boiling point?
Chapter 17 – Entropy, Free Energy, and Equilibrium
93. Calculate the free energy of formation of NaBr(s) given the following information:
NaBr(s) → Na(s) + 1/2Br2(l), G = 349 kJ/mol
94. The following reaction is nonspontaneous at 25C:
Cu2O(s) → 2Cu(s) + 1/2O2(g), G = 141 kJ/mol
If S = 75.8 J/K·mol, what is the lowest temperature at which the reaction will be
spontaneous?
Chapter 17 – Entropy, Free Energy, and Equilibrium
95. For the reaction 3H2(g) + N2(g) 2NH3(g), Kc = 9.0 at 350C. Calculate G at
350C.
96.
Select True or False: The reaction 3H2(g) + N2(g) 2NH3(g),(Kc = 9.0 at 350C) proceeds from right to left at 350C
under standard state conditions.
Chapter 17 – Entropy, Free Energy, and Equilibrium
97. For the reaction 3H2(g) + N2(g) 2NH3(g), Kc = 9.0 at 350C. What is the value of
G at this temperature when 1.0 mol NH3, 5.0 mol N2, and 5.0 mol H2 are mixed in a 2.5 L
reactor?
98.
Select True or False: The reaction 3H2(g) + N2(g) 2NH3(g), Kc = 9.0 at 350C proceeds from right to left when 1.0
mol NH3, 5.0 mol N2, and 5.0 mol H2 are mixed in a 2.5 L reactor.
Chapter 17 – Entropy, Free Energy, and Equilibrium
99. Consider the reaction CO(g) + 2H2(g) CH3OH(l) at 25C.
Gf (CO) = –137.3 kJ/mol
Gf (CH3OH) = –166.3 kJ/mol
Hf (CO) = –110.5 kJ/mol
Hf (CH3OH) = –238.7 kJ/mol
S(CO) = 197.9 J/K·mol
S(CH3OH) = 126.8 J/K·mol
Calculate G at 25C.