Under cellular conditions (37 C), what is the actual free energy for the reactio

Question

Under cellular conditions (37 C), what is the actual free energy for the reaction shown below when the ratio of glucose-6-phosphate to fructose-6-phosphate is 0.8:0.2?
glucose-6-phosphate -> fructose-6-phosphate
Standard Gibbs Free Energy change = 1.7 kJ/mol. Under cellular conditions (37 C), what is the actual free energy for the reaction shown below when the ratio of glucose-6-phosphate to fructose-6-phosphate is 0.8:0.2?
glucose-6-phosphate -> fructose-6-phosphate
Standard Gibbs Free Energy change = 1.7 kJ/mol.
glucose-6-phosphate -> fructose-6-phosphate
Standard Gibbs Free Energy change = 1.7 kJ/mol.

Explanation / Answer

Q = Products/Reactants

Q = 0.2/0.8 = 0.25

then

dG = dG° + RT*ln(Q)

dG° = 1.7 kJ/mol = 1700 J/mol

dG = 1700 + 8.314*(37+273) * ln(0.25)

dG = -1872.95 J/mol

dG = -1.87 kJ/mol, this is favouring forward direction

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