$\Delta G$ for a reaction at $300\mathrm{ }\mathrm{K}$ is $-16\mathrm{kcal}$ and $\Delta H$ is $-10\mathrm{kcal}$. The entropy of the reaction is

Given data:

We have $\Delta G=-16\mathrm{kcal}$ and $\Delta H$ is $-10\mathrm{kcal}$.

Formula/concept used: $\Delta G=\Delta H-T\Delta S$

Detailed explanation:

We know that,

$\Delta G=\Delta H-T\Delta S$

On putting the given values of $\Delta G,\Delta H$ and $\mathrm{T}$, we get,

$$\begin{gathered} -16=-10-\left(300\right)\Delta S \\ \Delta S=0.02\mathrm{kcal}/\mathrm{deg} \\ \Delta S=20\mathrm{cal}/\mathrm{deg} \end{gathered}$$

Thus the entropy of the given reaction is $20\mathrm{cal}/\mathrm{deg}$.

Hence, option (A) is correct.