If the standard enthalpy change $\left(\Delta_{\mathrm{r}} \mathrm{H}^\theta\right)$ for a certain reaction…

If the standard enthalpy change $\left(\Delta_{\mathrm{r}} \mathrm{H}^\theta\right)$ for a certain reaction at $298 \mathrm{~K}$ and constant pressure is $-1860 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$, the standard entropy change $\left(\Delta_{\text {sys }} \mathrm{s}^\theta\right)$ of the same reaction is $-550 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$, which one of the following statements is correct?
  1. $\left(\Delta_{\text {sys }} S^\theta\right)+\Delta_{\text {surr }} S^\theta=-7692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, the reaction is spontaneous
  2. $\left(\Delta_{\text {sys }} \mathrm{S}^\theta\right)+\Delta_{\text {surr }} \mathrm{S}^\theta=-5692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, the reaction is non-spontaneous
  3. $\left(\Delta_{\text {sys }} \mathrm{S}^\theta\right)+\Delta_{\text {surr }} \mathrm{S}^\theta=+5692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, the reaction is spontaneous
  4. $\left(\Delta_{\text {sys }} S^\theta\right)+\Delta_{\text {surr }} S^\theta=+7692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, 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 10^3\right)-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} $ $\begin{aligned} & =\frac{+1860 \times 10^3}{298} \\ & =6241.6 \mathrm{~J} \\ & \Rightarrow \Delta S_{\text {total }}^{\circ}=\Delta S_{\text {system }}^{\circ}+\Delta S_{\text {surrounding }}^{\circ} \\ & =(-550)+(6241.6) \\ & =+5692 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} .\end{aligned}$

Asked in: AP EAMCET 2023 (15 May Shift 1)

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