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(D) $1$. Initial moles of $N_2O_5 = 1 	ext{ mol}$.
$2$. At equilibrium, $N_2O_5$ dissociated is $50\%$, so remaining $N_2O_5 = 1 - 0.5 = 0.5 	ext{

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$2500$

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(D) $1$. Initial moles of $N_2O_5 = 1 	ext{ mol}$.
$2$. At equilibrium, $N_2O_5$ dissociated is $50\%$, so remaining $N_2O_5 = 1 - 0.5 = 0.5 	ext{ mol}$.
$3$. According to stoichiometry, $N_2O_4$ formed = $0.5 	ext{ mol}$ and $O_2$ formed = $0.25 	ext{ mol}$.
$4$. Total moles at equilibrium = $0.5 + 0.5 + 0.25 = 1.25 	ext{ mol}$.
$5$. Partial pressures: $P_{N_2O_5} = (0.5 / 1.25) 	imes 2 = 0.8 	ext{ atm}$, $P_{N_2O_4} = (0.5 / 1.25) 	imes 2 = 0.8 	ext{ atm}$, $P_{O_2} = (0.25 / 1.25) 	imes 2 = 0.4 	ext{ atm}$.
$6$. Equilibrium constant $K_p = (P_{N_2O_4}^2 	imes P_{O_2}) / P_{N_2O_5}^2 = (0.8^2 	imes 0.4) / 0.8^2 = 0.4$.
$7$. Standard free energy change $\Delta G^\circ = -RT \ln K_p$.
$8$. Given $T = 50 + 273 = 323 	ext{ K}$ and $R = 8.314 	ext{ J K}^{-1} 	ext{ mol}^{-1}$.
$9$. $\Delta G^\circ = -(8.314 	imes 323) 	imes \ln(0.4) = -2685.422 	imes (-0.9163) \approx 2460.6 	ext{ J mol}^{-1}$.
$10$. The magnitude $x \approx 2460 	ext{ J mol}^{-1}$, which is closest to $2500 	ext{ J mol}^{-1}$.

For the following reaction at $50^\circ$ $C$ and at $2 \text{ atm}$ pressure, $2N_2O_5(g) \rightleftharpoons 2N_2O_4(g) + O_2(g)$. $N_2O_5$ is $50\%$ dissociated. The magnitude of standard free energy change at this temperature is $x$. $x = . . . . . . \text{ J mol}^{-1}$.

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