Calculate the value of $\Delta G$ for following reaction at $300 \mathrm{~K}$. $\begin{aligned} &…

Calculate the value of $\Delta G$ for following reaction at $300 \mathrm{~K}$. $\begin{aligned} & \mathrm{H}_2 \mathrm{O}_{(\mathrm{s})} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(l)} \\ & \left(\Delta \mathrm{H}=7 \mathrm{~kJ}, \Delta \mathrm{S}=24.8 \mathrm{~J} \mathrm{~K}^{-1}\right)\end{aligned}$
  1. $0.74 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  2. $-0.82 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  3. $0.21 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  4. $-0.44 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

$\begin{aligned} \Delta \mathrm{H} & =7 \mathrm{~kJ} \\ \Delta \mathrm{S} & =24.8 \mathrm{~J} \mathrm{~K}^{-1}=24.8 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \\ \mathrm{~T} & =300 \mathrm{~K} \\ \Delta \mathrm{G} & =\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \\ \therefore \quad \Delta \mathrm{G} & =7 \mathrm{~kJ}-\left(300 \mathrm{~K} \times 24.8 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1}\right) \\ & =7 \mathrm{~kJ}-7.44 \mathrm{~kJ}=-0.44 \mathrm{~kJ}\end{aligned}$

Asked in: MHT CET 2023 (09 May Shift 1)

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