AB_{3(g)} dissociates as AB_{3(g)} rightlefth | vedclass

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(C) The dissociation reaction is: $AB_{3(g)} \rightleftharpoons AB_{2(g)} + \frac{1}{2}B_{2(g)}$Let the degree of dissociation be $x$ (in terms of pre

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$25$

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(C) The dissociation reaction is: $AB_{3(g)} ightleftharpoons AB_{2(g)} + \frac{1}{2}B_{2(g)}$Let the degree of dissociation be $x$ (in terms of pressure).Initial pressure: $P_{AB_3} = 800 \ torr$,$P_{AB_2} = 0$,$P_{B_2} = 0$.At equilibrium: $P_{AB_3} = 800 - x$,$P_{AB_2} = x$,$P_{B_2} = \frac{x}{2}$.Total pressure at equilibrium: $(800 - x) + x + \frac{x}{2} = 900 \ torr$.$800 + \frac{x}{2} = 900 \implies \frac{x}{2} = 100 \implies x = 200 \ torr$.Fraction of $AB_{3(g)}$ dissociated $= \frac{x}{P_{initial}} 	imes 100 = \frac{200}{800} 	imes 100 = 25 \%$.

$AB_{3(g)}$ dissociates as $AB_{3(g)} \rightleftharpoons AB_{2(g)} + \frac{1}{2}B_{2(g)}$. When the initial pressure of $AB_3$ is $800 \ torr$ and the total pressure at equilibrium is $900 \ torr$,what fraction of $AB_{3(g)}$ is dissociated (in $\%$)?

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