MCQ
The standard Gibbs free energy change $\Delta {G^o}$ is related to equilibrium constant ${K_p}$ as
  • A
    ${K_p} = - RT\,\ln \,\Delta {G^o}$
  • B
    ${K_p} = {\left( {\frac{e}{{RT}}} \right)^{\Delta {G^o}}}$
  • C
    ${K_p} = - \frac{{\Delta {G^o}}}{{RT}}$
  • ${K_p} = {e^{ - \,\frac{{\Delta {G^o}}}{{RT}}}}$

Answer

Correct option: D.
${K_p} = {e^{ - \,\frac{{\Delta {G^o}}}{{RT}}}}$
(d) ${K_p} = {e^{ - \Delta {G^o}/RT}}$.

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