37.8 g N_2O_5 was taken in a 1 L reaction v | vedclass

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(A) Initial moles of $N_2O_5 = \frac{37.8 \ g}{108 \ g/mol} = 0.35 \ mol$.Initial pressure $P_0 = \frac{nRT}{V} = \frac{0.35 	imes 0.082 	imes 500}{

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$962$

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(A) Initial moles of $N_2O_5 = \frac{37.8 \ g}{108 \ g/mol} = 0.35 \ mol$.Initial pressure $P_0 = \frac{nRT}{V} = \frac{0.35 	imes 0.082 	imes 500}{1} = 14.35 \ bar$.Reaction: $2N_2O_{5(g)} ightleftharpoons 2N_2O_{4(g)} + O_{2(g)}$At $t=0$: $P_0 = 14.35 \ bar$,$0$,$0$At equilibrium: $(14.35 - 2P)$,$2P$,$P$Total pressure at equilibrium: $(14.35 - 2P) + 2P + P = 14.35 + P = 18.65 \ bar$.Therefore,$P = 18.65 - 14.35 = 4.3 \ bar$.Equilibrium partial pressures:$P_{N_2O_5} = 14.35 - 2(4.3) = 14.35 - 8.6 = 5.75 \ bar$.$P_{N_2O_4} = 2(4.3) = 8.6 \ bar$.$P_{O_2} = 4.3 \ bar$.$K_p = \frac{(P_{N_2O_4})^2 	imes (P_{O_2})}{(P_{N_2O_5})^2} = \frac{(8.6)^2 	imes 4.3}{(5.75)^2} = \frac{73.96 	imes 4.3}{33.0625} = \frac{318.028}{33.0625} \approx 9.619$.$K_p = 9.619 = 961.9 	imes 10^{-2} \approx 962 	imes 10^{-2}$.

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