Calculate $\Delta \mathrm{S}_{\text {total }}$ for the following reaction at $300 \mathrm{~K}$.…

Calculate $\Delta \mathrm{S}_{\text {total }}$ for the following reaction at $300 \mathrm{~K}$. $\begin{aligned} & \mathrm{NH}_4 \mathrm{NO}_{3(\mathrm{~s})} \longrightarrow \mathrm{NH}_{(\mathrm{zq})}^{+}+\mathrm{NO}_{3(\mathrm{uq})}^{-} \\ & \left(\Delta \mathrm{H}=28.1 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{S}_{\mathrm{sys}}=108.7 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\end{aligned}$
  1. $15.1 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
  2. $93.6 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
  3. $84.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
  4. $202.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$

Solution

$\begin{aligned} \Delta \mathrm{S}_{\text {surr }} & =\frac{Q_{\text {rvv }}}{T}=\frac{-\Delta H}{T} \\ & =\frac{-28.1 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{300 \mathrm{~K}} \\ & =-93.67 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\ \Delta \mathrm{S}_{\text {total }} & =\Delta \mathrm{S}_{\text {sys }}+\Delta \mathrm{S}_{\text {surr }} \\ & =108.7+(-93.67) \\ & =15.03 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\end{aligned}$

Asked in: MHT CET 2023 (11 May Shift 2)

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