Calculate the value of $\Delta \mathrm{G}$ for the following reaction. $\mathrm{N}_2…

Calculate the value of $\Delta \mathrm{G}$ for the following reaction. $\mathrm{N}_2 \mathrm{O}_{4(\mathrm{~g})} \longrightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}$ if $\Delta \mathrm{H}=57.44 \mathrm{~kJ}$ and $\Delta \mathrm{S}=176 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.
  1. 120.20 kJ
  2. -110.24 kJ
  3. $\quad-46.4 \mathrm{~kJ}$
  4. $\quad 4.64 \mathrm{~kJ}$

Solution

$\begin{aligned} \Delta \mathrm{H} & =57.44 \mathrm{~kJ} \\ \Delta \mathrm{~S} & =176 \mathrm{~J} \mathrm{~K} \\ \mathrm{~T} & =300 \mathrm{~K} \\ \Delta \mathrm{G} & =\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \\ \therefore \quad \Delta \mathrm{G} & =57.44 \mathrm{~kJ}-\left(300 \mathrm{~K} \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1}\right. \\ & =57.44 \mathrm{~kJ}-52.8 \mathrm{~kJ}=4.64 \mathrm{~kJ}\end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 1)

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