Question

Consider the following metabolic reaction:

3-Phosphoglycerate → 2-Phosphoglycerate ΔG°’ = +4.40 kJ/mol

What is the ΔG for this reaction when the concentration of 2-phosphoglycerate is 0.290 mM and the concentration of 3-phosphoglycerate is 2.90 mM at 37°C?

Answer 1

ΔG = -1.53 kJ/mol

Step-by-step 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([2-Phosphoglycerate])/([3-Phosphoglycerate])`

`\Delta G = 4.40kJ/mol +0.008314 kJ/mol-K*310Kln([0.290])/([2.90])=-1.53 kJ/mol`