Exercise on ΔG |
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Exercise on ΔG°′ and Keq
Constants and equations: ΔG = G°′ + R T ln ([products]/[reactants]) |
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R = 8.315 J/mol-K |
T = 298 K |
ΔG°′ = – R T ln Keq |
The citric acid cycle, which we will study at the end of the semester, has reactions with the following values for ΔG°′, going clockwise around the cycle.
Reaction |
A |
B |
C |
D |
E |
F |
G |
H |
ΔG°′ (kJ/mol) |
– 32.2 |
+ 13.3 |
– 20.9 |
– 33.5 |
– 2.9 |
0 |
– 3.8 |
+ 29.7 |
1. What is ΔG°′ when reactions B and C are coupled?
2. Which three reactions are definitely reversible?
3. What is Keqfor reaction E?
4. What is Keqfor reaction D?
5. Since this is a cycle, reaction H is followed by reaction A (that is, the product of H is a reactant in A). What is ΔG?' for the coupled H-A reaction?
6. For reaction H alone (not coupled), what must the ([products]/[reactants]) ratio be when ΔG = 0?
7. Which reaction has the largest Keq? Which has the smallest?
8. Choose a reaction with ΔG°′ > 0 and explain how this reaction can occur in the cell.
9. Try to answer this without doing the calculations. Which reaction is most favorable?
A has Keq= 23; B has Keq= 0.0023; C has Keq= 0.23;
D has ΔG°′ = – 23 kJ/mol; E has ΔG°′ = + 2.3 J/mol; F has ΔG°′ = + 2.3 kJ/mol.
Answers to the Exercise
1. – 17.6 kJ/mol
2. E, F, and G (smallest values for ΔG°′)
3. – 2.9 kJ/mol = – (8.315 J/mol-K) (298 K) ln Keq
– 2900 J/mol = – (2477.87 J/mol) ln Keq
ln Keq= 1.17
Keq= 3.2
4. 7.4 x 105
5. – 2.5 kJ/mol
6. 0 = 29.7 kJ/mol + R T ln ratio
– 29,700 J/mol = 2477.87 J/mol (ln ratio)
– 11.986 = ln ratio
ratio = 6.2 x 10–6
7. Largest is D (most favorable); smallest is H (least favorable)
8. H can occur when it is coupled with A, and it can also occur when [products]<<<[reactants].
9. D is most favorable, and B is least favorable.
Results of calculations:
Reaction |
A |
B |
C |
D |
E |
F |
Keq |
23 |
0.0023 |
0.23 |
1.07 x 104 |
0.999 |
0.395 |
ΔG°′ (kJ/mol) |
– 7.8 |
+ 15.1 |
+ 3.6 |
– 23 |
+ 0.0023 |
+ 2.3 |