Question

AG' for the isomerization reaction glucose-1-phosphate (GIP) $ glucose-6-phosphate (G6P) is -7.1 kJ/mol. Calculate the equilibrium ratio of [G1P] to (G6P) at 25°C. Read this carefully to make sure you solve for the correct ratio! SHOW WORK! [G1P][G6P] =

Answer 1

Answer:

The ratio [G1P]/[G6P] = 5.7 . 10⁻².

Explanation:

Let us consider the reaction G1P ⇄ G6P, with ΔG° = -7.1 kJ/mol. According to Hess's Law, we can write the inverse reaction, and Gibbs free energy would have an opposite sign.

G6P ⇄ G1P        ΔG° = 7.1 kJ/mol

This is the reaction for which we want to find the equilibrium constant (the equilibrium ratio of [G1P] to [G6P]):

$Kc=\frac{[G1P]}{[G6P]}$

The equilibrium constant and Gibbs free energy are related by the following expression:

$Kc=e^{-\Delta G\si{\textdegree}/R.T } } =e^{-7.1kJ/mol/8.314.10^{-3}kJ/mol.K.298K} } }=5.7.10^{-2}$

where,

R is the ideal gas constant (8.314 . 10⁻3 kJ/mol.K)

T is the absolute temperature (in kelvins)