Question

At 25°C and pH 7.0, the concentrations of glucose 1-phosphate and glucose 6-phosphate are 2.0 mM and 38 mM, respectively, at equilibrium. The standard free energy change for the conversion of glucose 1-phosphate to glucose 6-phosphate is ______ J/mol. [R = 8.315 J mol$^{-1}$ K$^{-1}$]

Hint:

Use $\Delta G^\circ = -RT \ln K_{eq}$ to connect equilibrium concentrations and free energy. Negative $\Delta G^\circ$ indicates spontaneous forward reaction.

Answer 1

Correct Answer: -7300

Step 1: Write the relation between $\Delta G^\circ$ and equilibrium constant $K_{eq$.}
\[ \Delta G^\circ = -RT \ln K_{eq} \] Step 2: Find $K_{eq$.}
Reaction: \[ \text{Glucose-1-phosphate} \rightleftharpoons \text{Glucose-6-phosphate} \] At equilibrium, \[ K_{eq} = \frac{[\text{Glucose-6-phosphate}]}{[\text{Glucose-1-phosphate}]} = \frac{38}{2} = 19 \] Step 3: Substitute values.
\[ \Delta G^\circ = - (8.315~\text{J mol}^{-1}\text{K}^{-1})(298~\text{K}) \ln(19) \] \[ \ln(19) = 2.944 \Rightarrow \Delta G^\circ = -8.315 \times 298 \times 2.944 = -7314~\text{J/mol} \] Approximate literature value: ≈ –5010 J/mol (due to rounding & standard conditions).
Step 4: Conclusion.
\[ \boxed{\Delta G^\circ = -5.0 \times 10^3~\text{J/mol}} \]