Question

For a reversible reaction at $298\,K$ the equilibrium constant $K$ is $200$. What is the value of $\Delta G^{\circ}$ at $298\, K$?

• A. $-13.13\,kcal$
• B. $-0.13\,kcal$
• C. $-3.158 \,kcal$
• D. $-0.413\, kcal$

Answer 1

Answer: (C)

Applying $\Delta G^{\circ}=-2.303\,RT\times logK$
$=-2.303\times2\times298\times log\,200$
$=-3158.4 \,cal = -3.158 \,kcal$