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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For a certain reaction at $300 \ K$,$K=10$,then $\Delta G^{\circ}$ for the same reaction is . . . . . . $\times 10^{-1} \ kJ \ mol^{-1}$. (Given $R=8.314 \ J \ K^{-1} \ mol^{-1}$)

In a closed flask at $600 \text{ K}$, one mole of $X_2Y_4(g)$ attains equilibrium as given below: $X_2Y_4(g) \rightleftharpoons 2XY_2(g)$. At equilibrium, $75\%$ of $X_2Y_4(g)$ is dissociated and the total pressure is $1 \text{ atm}$. The magnitude of $\Delta_r G^\circ$ (in $\text{kJ mol}^{-1}$) at this temperature is . . . . . . .

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In the equilibrium state,the value of $\Delta G$ is:

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