Question:
For the reaction $2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})$, when $\Delta \mathrm{S}=-176.0 \mathrm{JK}^{-1}$ and $\Delta \mathrm{H}=-57.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the magnitude of $\Delta \mathrm{G}$ at $298 \mathrm{~K}$ for the reaction is $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest integer)
Solution:
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$
$\Delta \mathrm{G}=57.8-\frac{298(-176)}{1000}$
$\Delta \mathrm{G}=-5.352 \mathrm{~kJ} / \mathrm{mole}$
$\mid$ Nearest integer value $\mid=5$
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