Question

For the reaction,
$A(g) + B(g) \rightarrow C(g) + D(g), \Delta H^{\circ}$ and $\Delta S^{\circ}$ are, respectively, $-29.8 \, kJ \, mol^{-1}$ and $-0.100 \, kJ \, K^{-1} \, mol^{-1}$ at $298 \,K$. The equilibrium constant for the reaction at $298 \,K$ is :

• A. $1.0 \times 10^{-10}$
• B. $1.0 \times 10^{10}$
• C. 10
• D. 1

Answer 1

Answer: (D)

$\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}$
$\Delta G^{\circ}=-29.8\,KJ /mol +0.1\times298\,KJ\,mol^{-1}$
$\Delta G^{\circ}=-2.303\,RT\,log\,K$
If log $K=0 \,\therefore K=1$