Ch 16 Problem Set 5:
Reaction Spontaneity (∆H, ∆S & ∆G)
Enthalpy change
Entropy change
Free energy change
Gibbs-Helmholtz equation
1.
!H° = # !Hf° products " # !Hf° reactants
!S° = # S° products " # S° reactants
!G ° = # !Gf° products " # !Gf° reactants
!G ° = !H° " T!S°
Is the entropy (degree of disorder) increasing or decreasing in these reactions?
a.
H2(g) + Br2(l) → 2HBr(g)
DECREASING
b. CuSO4·5H2O(s) → CuSO4(s) + 5H2O(g)
INCREASING
c.
INCREASING
2XeO3(s) →2Xe(g) + 3O2(g)
2. Classify each of these systems as always spontaneous (A), never spontaneous (N), or depends on
the relative magnitude of the heat and entropy changes (D).
a.
Heat is released; entropy decreases
ΔH=(-)
ΔS=(-)
D
b. Heat is absorbed; entropy decreases
ΔH=(+)
ΔS=(-)
N
c.
ΔH=(-)
ΔS=(+)
D
ΔH=(-)
ΔS=(+)
A
Heat is absorbed; entropy increases
d. Heat is released; entropy increases
3. Calculate the standard entropy change associated with each reaction:
a.
2H2O 2(l) → 2H2O(l) + O2(g)
!S° = # S° products " # S° reactants
[2(69.4 J/mol•K) + 205.0 J/mol•K] – [2(92.0 J/mol•K)] = 160.9 J/mol•K
b. I2(g) →I2(s)
!S° = # S° products " # S° reactants
(117 J/mol•K) – (260.6 J/mol•K) = -144 J/mol•K
c.
2CO(g) + O2(g) → 2CO2(g)
!S° = # S° products " # S° reactants
[2(213.6 J/mol•K)] – [2(197.9 J/mol•K) + (205.0 J/mol•K)] = 173.6 J/mol•K
4. A reaction is endothermic (positive H) and has a positive entropy. Would this reaction more
likely be spontaneous at high or low temperatures? Justify your answer.
!G ° = !H° " T!S° = [(positive number) – T(positive number)]
More likely to be negative (spontaneous) at high temperatures
5. A reaction has a S of -122 J/K·mol and a H of -78 kJ/mol at 285°C.
a.
Calculate G for the above reaction.
!G ° = !H° " T!S°
ΔG = (-78 kJ/mol) – (558 K)(-0.122 kJ/K·mol)
ΔG = -9.92 kJ/mol ≈ -10. kJ/mol
b. Is this reaction spontaneous?
YES (because ΔG = neg.)
6. Calculate the standard free energy change for the reaction between iron (III) oxide and carbon
(graphite).
2Fe2O 3(s) + 3C(s) → 4Fe(s) + 3CO2(g)
ΔG = [3(-394.4 kJ/mol)] – [2(-741.2 kJ/ mol)]
ΔG = 299.2 kJ/mol
7. A student warned his friend not to swim in a river close to an electric plant. He claimed that the
ozone produced by the plant turned the river water to hydrogen peroxide, which would bleach
hair. The reaction is
O3(g) + H2O(l) → H2O2(aq) + O2(g)
Assuming that the river water is at 25°C and all species are at standard concentrations, show by
calculation whether his claim is plausible. Take !Gf° O3 (g) at 25°C to be +163.2kJ/mol and
!Gf° H2O2 (aq) = -134 kJ/mol.
!G ° = # !Gf° products " # !Gf° reactants
ΔG = [(-134.0 kJ/mol) + (0.0 kJ/mol)] – [(163.2 kJ/ mol) + (-237.2 kJ/mol)]
ΔG = -60.0 kJ/mol
YES; since the reaction is spontaneous (Δ G = -) at this temperature, the friend’s
claim IS PLAUSIBLE.