Chem 102 Dr. Naleway Discussion 13 KEY
Chem 102 – Discussion 13 – Dr. Naleway
Thermodynamics
ANSWER KEY
1) Draw a Diagram illustrating the influence of Temperature on Free Energy (∆G); Note
the FOUR possible Scenarios! (Assume that H and S do not change with
temperature.)
2) Under what conditions would a reaction that is spontaneous at HIGH temperatures
and becomes non-spontaneous at LOWER temperatures? (Assume that H and S do
not change with temperature.)
If ΔH is positive, the reaction is endothermic, which is not as favorable as a reaction
which is exothermic. However, if the ΔS is positive (reaction has increased entropy and
is more disordered) an endothermic reaction can be spontaneous at higher
temperatures. Red square on graph above.
3) What are the appropriate conditions NECESSARY for each of the following
thermodynamic factors?
Equilibrium
Spontaneous
Non-Spontaneous
∆Suniv
0
∆G
0
E(EMF)
0
∆Ssystem
∆Ssystem + ∆Ssurr = 0
+
-
+
+
∆Ssystem + ∆Ssurr > 0
-
∆Ssystem + ∆Ssurr < 0
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Chem 102 Dr. Naleway Discussion 13 KEY
4) Use the following standard-state free energy of formation data to calculate the
acid dissociation equilibrium constant (pKa) for Carbonic acid.
ΔG° (kJ/mol)
-372.3
0.0
-351.0
Compound
H2CO3 (aq)
H+ (aq)
HCO3- (aq)
H2CO3  H+ + HCO3-351 – (-372.3) = +21.3 kJ/mol
ΔG° = - RT ln Keq
+21.3 kJ/mol = -(0.008314 kJ/K)(298 K)(ln Keq)
-8.597 = ln Keq
Keq = INV ex -8.597
Keq (Ka) = 1.85 x 10-4
pKa = -log (1.85 x 10-4) = 3.73
5) Calculate ∆H and ∆S for the following reaction. Is this reaction Endothermic or
Exothermic? Then using these values; determine the ∆Go for this reaction at 25 °C
AND illustrate why NH4NO3 spontaneously dissolves is water at room temperature.
NH4NO3(s) + H2O(l) <==> NH4+ (aq) + NO3- (aq)
Compound
NH4NO3 (s)
NH4+ (aq)
NO3- (aq)
Hfo(kJ/mol)
-365.56
-132.51
-205.0
So(J/molK)
151.08
113.40
146.40
ΔS = [0.11340 + 0.14640] – [0.15108] = 0.10872 kJ/mol K
ΔH = [-132.51 + -205] – [-365.56] = +28.05 kJ/mol
ΔG = ΔH – TΔS = 28.05 kJ/mol – (298 K)(0.10872 kJ/mol K) = -4.35 kJ/mol
ΔG° = - RT ln Keq
-4.35 kJ/mol = -(0.008314 kJ/K)(298 K)ln Keq
Keq = INV ex 1.756 = 5.79
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Chem 102 Dr. Naleway Discussion 13 KEY
Since Keq > 1 this means the reaction goes spontaneously towards products.
6) Using a standard-state enthalpy of formation and absolute entropy data table for the
reaction
2NO2 <==> N2O4
Hfo(kJ/mol)
So(J/mol-K)
NO2(g)
33.18
240.0
N2O4 (g)
9.16
RXN
-57.2
304.2
- 0.1758 kJ/mol
K
ΔG° (kJ/mol)
At 298 K
= -4.8
At 373 K
= 8.37
What is the Equilibrium Constant at a temperature of 25 oC AND at 100 oC ?
9.16 – (2 x 33.18) = -57.2 kJ/mol (ΔH)
0.3042 - (2 x 0.240) = - 0.1758 kJ/mol K (ΔS)
ΔG° = ΔH° - T(ΔS°)
ΔG° (at 298 K)
- 57.2 – (298)(-0.1758) = - 4.8 kJ/mol
ΔG° = -RT ln Keq
-4.8 = (-0.008314)(298) ln Keq
ln Keq = 1.937
ex 1.937 = 6.94 (Keq)
ΔG° (at 373 K)
- 57.2 – (373)(-0.1758) = 8.3734
8.3734 = (-0.008314)(373) ln Keq
ln Keq = - 2.7
ex -2.7 = 0.067 (Keq)
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Chem 102 Dr. Naleway Discussion 13 KEY
7) For the Oxidation of Iron to Rust
4 Fe + 3 O2  2 Fe2O3
∆H rxn0
∆S rxn0
-1648.4 kJ/mol
-543.7 J/(mol K)
What is the standard Gibbs free energy(∆G rxn0 ) for this reaction ?
ΔG° = -1648.4 – (298)(-0.5437) = -1486.4 kJ/mol
At what Temperature do the forward and reverse rusting reactions occur in
equilibrium (assuming that H0 and S0 are temperature independent?)
At equilibrium a reaction’s ΔG° = 0, so plug zero in for the equation ΔG° = ΔH° - TΔS°
and plug in the ΔH° and ΔS° to solve for T.
0 = -1648.4 – (T)(-0.5437)
1648.4 = T (0.5437)
T = 1648.4 / 0.5437 = 3032 K
8) Based on the following thermodynamic data; calculate the boiling point of ethanol in
degrees Celsius.
Substance
∆Hf0 (kJ/mol)
∆S0 (J/K mol)
C2H5OH(l)
-277.7
160.6
C2H5OH(g)
-235.1
282.6
C2H5OH (l) -----> C2H5OH (g)
0.2826 – 0.1606 = 0.122 kJ/mol K (ΔS°)
-235.1 – (-277.7) = 42.6 kJ/mol
When the liquid and vapor are in equilibrium, ΔG° will be zero
0 = 42.6 – (T)(0.122)
-42.6 = (-T)(0.122)
T = -42.6 / -0.122 = 349 K
349 K – 273 = 76 °C
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