Q32 / 120 · VITEEE 2010 · Chemistry

The standard Gibb's free energy change, $\Delta \mathrm{G}^{\circ}$ is related to equilibrium constant, $\mathrm{K}_{\mathrm{p}}$ as

  • A
    $K_{p}=-R T \ln \Delta G^{\circ}$
  • B
    $K_{p}=\left[\frac{e}{R T}\right]^{\Delta G^{\circ}}$
  • C
    $\mathrm{K}_{\mathrm{p}}=-\frac{\Delta \mathrm{G}}{\mathrm{RT}}$
  • Correct answer
    D
    $K_{p}=e^{-\Delta G^{0} / R T}$

Solution

$\Delta G^{\circ}$ and $K_{p}$ are related as $$ \begin{array}{l} \Delta G^{\circ}=-R T \ln K_{p} \\ \Rightarrow \ln K_{p}=\frac{\Delta G^{\circ}}{R T} \Rightarrow K_{p}=e^{-\Delta G^{\circ} / R T} \end{array} $$

Hence, the correct answer is option: 'D'.

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