Question

The isomerization of 3-phosphoglycerate (3PG) to 2-phosphoglycerate (2PG) has a ΔG°´ = +4.4 kJ/mol. If [2PG] = 4.0 mM and [3PG] = 22.2 mM, in which direction will the reaction move to reach equilibrium at 37°C?

Answer 1

The reaction will move to the right, that is, the formation of 2PG

Step-by-step explanation:

Data:

• gas constant, R = 8.3144 J/(mol K)

• temperature, T = 37 °C = 310 K

• standard free energy of reaction, ΔG° = 4.4 kJ/mol = 4400 J/mol

• 2-phosphoglycerate concentration [2PG] = 4.0 mM

• 3-phosphoglycerate concentration [3PG] = 22.2 mM

Reaction:

3PG ↔ 2 PG

This variables are related by the following equation:

ΔG° = -R*T*ln(Kp)

Replacing with data:

4400/(-8.3144*310) = ln(Kp)

Kp = e^(-1,71)

Kp = 0,1813

For the given reaction the equilibrium constant is:

Kp = [2PG]/[3PG]

Replacing with data:

Kp' = 4/22.2 = 0.18018

which is barely less than the equilibrium constant at the temperature of interest. Then, the concentration of 3PG must be reduced and the concentration of 2PG must be increased.