COMEDK · Chemistry · 7. Chemical Equilibrium
The value of \(\Delta G^{\circ}\) for the phosphorylation of glucose in glycolysis is \(13.8 \mathrm{~kJ} / \mathrm{mol}\). The value of \(K_{C}\) at \(298 \mathrm{~K}\) is
- A \(7.72 \times 10^{-4}\)
- B \(5.62 \times 10^{-4}\)
- C \(4.81 \times 10^{-3}\)
- D \(3.81 \times 10^{-3}\)
Answer & Solution
Correct Answer
(D) \(3.81 \times 10^{-3}\)
Step-by-step Solution
Detailed explanation
Given, \(\Delta G^{\circ}=13.8 \mathrm{~kJ} / \mathrm{mol}=13.8 \times 10^{3} \mathrm{~J} / \mathrm{mol}\) \[ \begin{aligned} \because & \Delta G^{\circ} &=-R T \ln K_{C} \\ & & 13800 &=-8.314 \times 298 \ln K_{C} \\ \therefore & K_{C} &=3.81 \times 10^{-3} \end{aligned} \]
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