1. Which of these species would you expect to have the lowest standard entropy (S°)?
a. Br2(l) or Br2 (g)
b. Cl2(g) or 2Cl(g)
c. F2(g) at 1 atm or F2 (g) at 2 atm
d. H2O(l) at 25C or H2O(l) at 45C
e. O2 at 298 K or O3 at 298K
2. Predict the sign of ΔS for the following processes:
a. Sublimation of Naphthalene solid  gas ΔS>0
b. sugar dissolving in water: sugar(s)  sugar(aq) ΔS >0
c. Ag+(aq) + Cl-(aq)  AgCl(s) ΔS <0
d. H+(aq) + OH-(aq)  H2O(l) ΔS < 0
e. 4 NH3 (g) + 3O2 (g)  2 N2 (g) + 6 H2O (l) ΔS <0
why? On the reactant side we have 7 moles of gases. On the product side, we only have 2
moles of gases. Therefore entropy goes down.
3. Calculate ΔS°, ΔH° and ΔG° at 298 K for the following reaction using the
thermodynamic data provided below.
2 CsCl(aq) + H2SO4(aq)  Cs2SO4(aq) + 2HCl(aq)
S°(J/K·mol)
Hf (kJ/mol)
CsCl(aq)
188
-415
H2SO4(aq)
17.2
-908
HCl(aq)
56.5
-167
Cs2SO4(aq)
283
-1400
ΔS° = (396 – 393.2) J/K= 2.8 J/K
ΔH° = (-1734 – -1738)kJ = 4 kJ= 4000 J
ΔG° = ΔH° - TΔS° =4000J –(298K x 2.8J/K) = 3166 J = 3.17 kJ
4. . Hydrogen peroxide (H2O2) decomposes according to the equation
H2O2(l) → H2O(l) + 1/2O2(g).
Calculate K for this reaction at 25°C given that ΔH° = –98.2 kJ/mol, ΔS° = 70.1 J/K·mol.
ΔG° = ΔH°- T ΔS° = –98.2 x 1000J/mol – 298K x 70.1 J/K·mol = -1.19 x 105J/mol
ΔG°=-RTln(K); -1.19 x 105J/mol = -8.314 J/K-mol x 298 K x ln(K)
ln(K) = 1.19 x 105J/mol/(8.314 J/K-mol x 298 K) = 48.1;
K=e48.1 = 7.75 x 1020
5. The equilibrium constant at 427°C for the reaction N2(g) + 3H2(g)  2NH3(g) is
Kp = 9.4 × 10–5.
a. Calculate the value of ΔG° for the reaction under these conditions.
b. Calculate the free energy change for the reaction when the partial pressures are 3.00
atm for N2, 2.00 atm for H2 and 6.00 atm for NH3.
c. Is the reaction spontaneous under these conditions?
ΔG°=-RTln(K) = -8.314 J/K-mol x 700 K x ln(9.4 × 10–5) = 53962 J/mol =
54.0 kJ/mol
b.
ΔG =ΔG°+RTln(Q); Q = 6.002/{3.00 x 2.003} = 1.50; ΔG =53962 J/mol +
8.314 J/K-mol x 700K x ln(1.50) = 56322 J/mol = 56.3 kJ/mol
c.
Not spontaneous because ΔG >0
a.
6.a. ΔG° at 298 K for the following reaction using the thermodynamic data provided
below.
ZnCl2(aq) + H2SO4(aq)  ZnSO4(aq) + 2HCl(aq)
Gf (kJ/mol)
ZnCl2 (aq)
-410
H2SO4(aq)
-742
HCl(aq)
-131
ZnSO4(aq)
-892
a.
ΔG°={-892 + (2 x -131)}kJ-{-410+-742}kJ = -2 kJ =-2000 J/mol
b.
Calculate the equilibrium constant K for the reaction at this temperature.
ln(K) = - ΔG°/RT = 2000 J/(8.314 J/K-mol x298K) = 0.807
K = e0.807 = 2.24
7. Specify if the following statements are true or false?
a. Hf = 0 for I2(g) False; The stable of I2 is I2(s) not I2(g)
b.
Gf = 0 for F2(g); True; The stable of F2 is F2(g)
c.
S = 0 for diamond at any Temperature; False; The S = 0 for a perfect crystal
such as diamond only at 0 K (III law of Thermodynamics)
d.
All exothermic reactions are spontaneous; False; H alone does not
determine spontaneity. G=H-TS should be <0 for spontaneity not H < 0.
e. At equilibrium, ΔG° = 0; False; At equilibrium G = 0; ΔG° = 0 implies that K=1