18.4 g of N_2O_4 was placed in a 1 L vessel | vedclass

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(B) Initial pressure $(p)$ of $N_2O_4$ can be calculated using the ideal gas equation $pV = nRT$: $p = \frac{nRT}{V} = \frac{(18.4 \ g / 92 \ g \ mol^

Vedclass · Accepted Answer

$24.24$

Vedclass · Answer

(B) Initial pressure $(p)$ of $N_2O_4$ can be calculated using the ideal gas equation $pV = nRT$: $p = \frac{nRT}{V} = \frac{(18.4 \ g / 92 \ g \ mol^{-1}) 	imes 0.083 \ L \ bar \ K^{-1} \ mol^{-1} 	imes 400 \ K}{1 \ L} = 0.2 \ mol 	imes 0.083 	imes 400 = 6.64 \ bar$. For the equilibrium reaction $N_2O_{4(g)} ightleftharpoons 2NO_{2(g)}$: Initial: $p$ atm,$0$ At equilibrium: $(p - p_i)$ atm,$2p_i$ atm Total pressure $(p_T)$ at equilibrium: $p_T = (p - p_i) + 2p_i = p + p_i$ $10.64 = 6.64 + p_i$ $p_i = 10.64 - 6.64 = 4.00 \ bar$. Partial pressures at equilibrium: $p_{N_2O_4} = p - p_i = 6.64 - 4.00 = 2.64 \ bar$ $p_{NO_2} = 2p_i = 2 	imes 4.00 = 8.00 \ bar$ Equilibrium constant $K_p$: $K_p = \frac{(p_{NO_2})^2}{p_{N_2O_4}} = \frac{(8.00)^2}{2.64} = \frac{64}{2.64} \approx 24.24$.

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