Question

Consider the following inter-conversion, which occurs in glycolysis: Fructose-6-phosphate ↔ Glucose-6-phosphate Keq’ = 1.97 What is ΔG°’ for the reaction (Keq’ measured at 25°C; and R=8.314 J•mol-1•K-1)? (b) If the concentration of fructose-6-phosphate is adjusted to 1.5 M and that of glucose- 6-phosphate is adjusted to 0.50 M, what is ΔG? (c) Why are ΔG°’ and ΔG different?

Answer 1

Explanation:

(a)   The relation between $\Delta G^{o}$ and equilibrium constant is as follows.

                $\Delta G^{o} = -RT ln K$

As it is given that,

              T = $25^{o}C$ = (25 + 273) K = 298 K

              K = 1.97

              R = 8.314 J/mol K

Hence, putting the given values into the above formula as follows.

                     $\Delta G^{o} = -RT ln K$

                                  = $-8.314 J/mol K \times 298 K \times ln (1.97)$

                                  = -1679.88 J/mol

                                  = 1.679 kJ/mol    (as 1 kJ = 1000 J)

Therefore, the value of $\Delta G^{o}$ is 1.679 kJ/mol.

(b)  Expression for $\Delta G$ is as follows.

         $\Delta G = \Delta G^{o} + RT ln \frac{\text{Glucose-6-phosphate}}{\text{Fructose-6-phosphate}}$

                       = $-1679.88 J/mol + 8.314 J/mol K \times 298 K \times \n \frac{0.5}{1.5}$

                        = -4401.8 J/mol

                        = 44.01 kJ/mol

Hence, the value for $\Delta G$ is 44.01 kJ/mol.

(c)    The value of both $\Delta G$ and $\Delta G^{o}$ are different because $\Delta G$ is measured at any concentration whereas $\Delta G^{o}$ is measured at equilibrium concentration.