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Snowcat [4.5K]
4 years ago
10

Consider the following metabolic reaction:

Chemistry
1 answer:
Klio2033 [76]4 years ago
5 0

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

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