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At 57°C, gaseous dinitrogen tetraoxide is 50% dissociated. Calculate the standard free energy change per mole of $N_{2} O_{4} (g)$ at this temperature and at 1 atm. $R = 8.3 J K^{- 1} {mol}^{- 1}, ln 10 = 2.3, \log 2 = 0.3, \log 3 = 0.48$

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$- 856 {J mol}^{-1}$

$- {456J mol}^{-1}$

$- 756 {J mol}^{-1}$

$- 656 {J mol}^{-1}$

answer is A.

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Detailed Solution

$N_{2} O_{4} (g) ⇌ {2NO}_{2} (g)$

$\begin{matrix} \begin{array}{l} t = 0 1 \\ t = eq . 1 - 0.5 \\ P_{N_{2} O_{4}} = \frac{0.5}{1.5} atm \end{array} & \begin{array}{l} 0 \\ 2 \times 0.5 \\ P_{{NO}_{2}} = \frac{1}{1.5} atm \end{array} \end{matrix}$

$K_{p} = \frac{{(\frac{1}{1.5})}^{2}}{(\frac{0.5}{1.5})} = \frac{4}{3}$

$Δ G ° = - 2.3 \times 8.3 \times 330 \times \log (\frac{4}{3})$

$= - 2.3 \times 8.3 \times 330 \times (0.6 - 0.48)$

$= - 756 {J mol}^{-1}$

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