A homogeneous ideal gaseous reaction AB_{2(g) | vedclass

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(B) The reaction is $AB_{2(g)} \rightleftharpoons A_{(g)} + 2B_{(g)}$.Initial moles: $AB_{2} = 1$,$A = 0$,$B = 0$.Moles at equilibrium: $AB_{2} = 1-\a

Vedclass · Accepted Answer

$74$

Vedclass · Answer

(B) The reaction is $AB_{2(g)} ightleftharpoons A_{(g)} + 2B_{(g)}$.Initial moles: $AB_{2} = 1$,$A = 0$,$B = 0$.Moles at equilibrium: $AB_{2} = 1-\alpha$,$A = \alpha$,$B = 2\alpha$.Total moles at equilibrium $n_{T} = (1-\alpha) + \alpha + 2\alpha = 1 + 2\alpha$.Using the ideal gas law $PV = nRT$ at equilibrium:$1.9 	imes 25 = n_{T} 	imes 0.0821 	imes 300$.$n_{T} = \frac{1.9 	imes 25}{0.0821 	imes 300} \approx 1.93$.$1 + 2\alpha = 1.93 \implies 2\alpha = 0.93 \implies \alpha = 0.465$.Equilibrium moles: $n_{AB_{2}} = 0.535$,$n_{A} = 0.465$,$n_{B} = 0.93$.Partial pressures: $P_{i} = \frac{n_{i}}{n_{T}} 	imes P_{total}$.$P_{AB_{2}} = \frac{0.535}{1.93} 	imes 1.9 = 0.526 \ atm$.$P_{A} = \frac{0.465}{1.93} 	imes 1.9 = 0.458 \ atm$.$P_{B} = \frac{0.93}{1.93} 	imes 1.9 = 0.916 \ atm$.$K_{P} = \frac{P_{A} 	imes (P_{B})^{2}}{P_{AB_{2}}} = \frac{0.458 	imes (0.916)^{2}}{0.526} \approx 0.73 \ atm^{2}$.$K_{P} = 73 	imes 10^{-2}$. Thus,$x = 73$ (closest integer option is $74$ due to rounding variations).

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