The correct relationship between the standard free energy change in a reaction and the corresponding equilibrium constant $K_c$ is:

  • A
    $\Delta G = RT \ln K_c$
  • B
    $-\Delta G = RT \ln K_c$
  • C
    $\Delta G^o = RT \ln K_c$
  • D
    $-\Delta G^o = RT \ln K_c$

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Similar Questions

The $\Delta G^o$ for the reaction $X + Y \rightleftharpoons Z$ is $-4.606 \ kcal$. The value of the equilibrium constant of the reaction at $227 \ ^oC$ is $(R = 2.0 \ cal \ mol^{-1} K^{-1})$.

For an equilibrium reaction,if $\Delta G^{\circ} = 0$,the equilibrium constant $K$ is equal to:

Calculate the standard Gibbs free energy change $\Delta G^o$ at $298 \ K$ for the conversion of oxygen to ozone,given by the reaction: $\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)}$. The equilibrium constant $K_p$ for this conversion is $3 \times 10^{-29}$.

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At $60^{\circ} C$,dinitrogen tetroxide is $50 \%$ dissociated. Find its standard free energy change at this temperature and $1 \ atm$. [ Given: $\log 1.33 = 0.1239 ]$

At $298 \ K$,$\Delta_r G^{\ominus}$ for the following reaction is $165.469 \ kJ \ mol^{-1}$. What is the equilibrium constant for this reaction? $(R = 8.3 \ J \ mol^{-1} \ K^{-1})$
$\frac{3}{2} O_{2(g)} \longrightarrow O_{3(g)}$

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