In a closed flask at 600 text{ K},one mole of | vedclass

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(A) Initial moles: $X_2Y_4 = 1$,$XY_2 = 0$.At equilibrium,$75\%$ of $X_2Y_4$ dissociates,so $X_2Y_4 = 1 - 0.75 = 0.25$ moles.$XY_2$ produced = $2 	im

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(A) Initial moles: $X_2Y_4 = 1$,$XY_2 = 0$.At equilibrium,$75\%$ of $X_2Y_4$ dissociates,so $X_2Y_4 = 1 - 0.75 = 0.25$ moles.$XY_2$ produced = $2 	imes 0.75 = 1.5$ moles.Total moles at equilibrium = $0.25 + 1.5 = 1.75$ moles.Partial pressures at $P_{total} = 1 	ext{ atm}$:$P_{X_2Y_4} = (0.25 / 1.75) 	imes 1 = 1/7 	ext{ atm}$.$P_{XY_2} = (1.5 / 1.75) 	imes 1 = 6/7 	ext{ atm}$.Equilibrium constant $K_p = (P_{XY_2})^2 / P_{X_2Y_4} = (6/7)^2 / (1/7) = (36/49) 	imes 7 = 36/7 \approx 5.14$.Standard Gibbs free energy change $\Delta_r G^\circ = -RT \ln K_p$.Using $R = 8.314 	imes 10^{-3} 	ext{ kJ K}^{-1} 	ext{ mol}^{-1}$ and $T = 600 	ext{ K}$:$\Delta_r G^\circ = -(8.314 	imes 10^{-3}) 	imes 600 	imes \ln(5.14) = -4.9884 	imes 1.637 \approx -8.16 	ext{ kJ mol}^{-1}$.The magnitude is approximately $8.16 	ext{ kJ mol}^{-1}$.

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