Question

Consider the following metabolic reaction:

Answer 1

Answer:

ΔG = -1.53 kJ/mol

Explanation:

The given reaction is:

3-Phosphoglycerate → 2-Phosphoglycerate

The standard Gibbs free energy, ΔG°=+4.40 kJ

[2-Phosphoglycerate] = 0.290 mM

[3-Phosphoglycerate] = 2.90 mM

Temperature T = 37 C = 310 K

The standard Gibbs free energy, ΔG° is related to the free energy change ΔG at a given temperature by the following equation:

$\Delta G =\Delta G^{0}+RTlnQ$

In this reaction:

$\Delta G =\Delta G^{0}+RTln\frac{[2-Phosphoglycerate]}{[3-Phosphoglycerate]}$

$\Delta G = 4.40kJ/mol +0.008314 kJ/mol-K*310Kln\frac{[0.290]}{[2.90]}=-1.53 kJ/mol$