Exercise on ΔG
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Exercise on ΔG°′ and Keq

Constants and equations: ΔG = G°′ + R T ln ([products]/[reactants])

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

 

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