Question

The value of \( \Delta G^\circ \) for the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. The value of \( K_C \) at 298 K is:

• A. \( 7.72 \times 10^{-4} \)
• B. \( 5.62 \times 10^{-3} \)
• C. \( 4.81 \times 10^{-3} \)
• D. \( 3.81 \times 10^{-3} \)

Hint:

For reactions at equilibrium, the Gibbs free energy and equilibrium constant are related through the equation \( \Delta G^\circ = -RT \ln K_C \). Remember that the temperature should be in Kelvin, and \( R \) is the gas constant.

Answer 1

The Correct Option is D

The relation between \( \Delta G^\circ \) and the equilibrium constant \( K_C \) is given by the following equation: \[ \Delta G^\circ = -RT \ln K_C \] Where: - \( \Delta G^\circ \) is the standard Gibbs free energy change, - \( R \) is the universal gas constant \( 8.314 \, \text{J/mol·K} \), - \( T \) is the temperature in Kelvin, - \( K_C \) is the equilibrium constant. Given that \( \Delta G^\circ = 13.8 \, \text{kJ/mol} = 13800 \, \text{J/mol} \) and \( T = 298 \, \text{K} \), we can rearrange the equation to solve for \( K_C \): \[ K_C = e^{-\frac{\Delta G^\circ}{RT}} \] Substituting the known values: \[ K_C = e^{-\frac{13800}{8.314 \times 298}} = e^{-5.56} = 3.81 \times 10^{-3} \] Thus, the correct answer is Option (D).