If the standard enthalpy change $\left(\Delta_{\mathrm{r}} \mathrm{H}^\thetaight)$ for a certain reaction at…
If the standard enthalpy change $\left(\Delta_{\mathrm{r}} \mathrm{H}^\thetaight)$ for a certain reaction at $298 \mathrm{~K}$ and constant pressure is $-1860 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$, the standard entropy change $\left(\Delta_{\text {sys }} \mathrm{s}^\thetaight)$ of the same reaction is $-550 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$, which one of the following statements is correct?
$\left(\Delta_{\text {sys }} S^\thetaight)+\Delta_{\text {surr }} S^\theta=-7692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, the reaction is spontaneous
$\left(\Delta_{\text {sys }} \mathrm{S}^\thetaight)+\Delta_{\text {surr }} \mathrm{S}^\theta=-5692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, the reaction is non-spontaneous
$\left(\Delta_{\text {sys }} \mathrm{S}^\thetaight)+\Delta_{\text {surr }} \mathrm{S}^\theta=+5692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, the reaction is spontaneous
$\left(\Delta_{\text {sys }} S^\thetaight)+\Delta_{\text {surr }} S^\theta=+7692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, the reaction is non-spontaneous
Solution
$$
\begin{aligned}
& \text { } \Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{s}^{\circ} \\
& =\left(-1860 \times 10^3ight)-298(-550) \\
& =-1696.1 \mathrm{~kJ} \mathrm{Mol}^{-1}
\end{aligned}
$$
Thus, the reaction is Spontaneous as $\Delta \mathrm{G}^{\circ}$ is negative. Now, Since the system loses the heat, the surrounding must gain it and therefore,
$$
\begin{aligned}
& \Delta \mathrm{H}^{\circ}{ }_{\text {Surrounding }}=+1860 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1} . \\
& \Rightarrow \Delta \mathrm{S}_{\text {surrounding }}^{\circ}=\frac{\Delta \mathrm{H}_{\text {surrounding }}^{\circ}}{\mathrm{T}}
\end{aligned}
$$
$\begin{aligned} & =\frac{+1860 \times 10^3}{298} \\ & =6241.6 \mathrm{~J} \\ & \Rightarrow \Delta S_{\text {total }}^{\circ}=\Delta S_{\text {system }}^{\circ}+\Delta S_{\text {surrounding }}^{\circ} \\ & =(-550)+(6241.6) \\ & =+5692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} .\end{aligned}$
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