Given are the following informations at 298 K N2O4(g)⇌2NO2(g)ΔfH∘/kJmol−1 9.16 33.18ΔfG∘/kJmol−1 97.54 51.31The reaction is carried out in a closed vessel starting from N2O4.The change in entropy of the reaction is

Given are the following informations at 298 K $\begin{array}{llr} N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g) \\ Δ_{f} H^{\circ} / kJ {mol}^{- 1} 9.16 33.18 \\ Δ_{f} G^{\circ} / kJ {mol}^{- 1} 97.54 51.31 \end{array}$
The reaction is carried out in a closed vessel starting from $N_{2} O_{4}$ .
The change in entropy of the reaction is

A
$174.9 {JK}^{- 1} {mol}^{- 1}$
B
$- 174.9 {JK}^{- 1} {mol}^{- 1}$
C
$262.3 {JK}^{- 1} {mol}^{- 1}$
D
$- 262.3 {JK}^{- 1} {mol}^{- 1}$

Solution:

$Δ_{r} S^{\circ} = (Δ_{r} H^{\circ} - Δ_{r} G^{\circ}) / T = (57.20 - 5.08) kJ {mol}^{- 1} / 298 K = 0.1749 {kJK}^{- 1} {mol}^{- 1} = 174.9 {JK}^{- 1} {mol}^{- 1}$

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