Question

The equilibrium constant $ K_{c} $ value for equilibrium at $ 227^{\circ}C $ is $ 1 \times 10^{10} $ . If the standard enthalpy change for this is $ -97.95 \, kJ\, mol^{-1} $ , the standard entropy change in $ J\, K^{-1} \, mol^{-1} $ for this equilibrium at the same temperature is $ (R=8.3 \, J\, K^{-1}\, mol^{-1} $ and $ 2.303 \simeq$ 2.3)

• A. $ -5 $
• B. $ +5 $
• C. $ +10 $
• D. $ -10 $

Answer 1

Answer: (A)

$\Delta G^{\circ}=-2.303RT \,log\,K$
$=-2.3\times8.3\times500\,log\left(1\times10^{10}\right)=-95450\,J $
Now, $\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S$
$\Rightarrow -95450=-97950-500\,\Delta S$
$\Rightarrow \Delta S=-5\,J \, K^{-1}\, mol^{-1}$