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

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(C) The reaction is: $AB_{3(g)} \rightleftharpoons AB_{2(g)} + \frac{1}{2} B_{2(g)}$Let the initial pressure of $AB_3$ be $P_i = 800 \ torr$.Let the c

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$25$

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(C) The reaction is: $AB_{3(g)} ightleftharpoons AB_{2(g)} + \frac{1}{2} B_{2(g)}$Let the initial pressure of $AB_3$ be $P_i = 800 \ torr$.Let the change in pressure due to dissociation be $x$.At equilibrium:$P_{AB_3} = 800 - x$$P_{AB_2} = x$$P_{B_2} = \frac{x}{2}$The total pressure at equilibrium is given as $900 \ torr$:$(800 - x) + x + \frac{x}{2} = 900$$800 + \frac{x}{2} = 900$$\frac{x}{2} = 100$$x = 200 \ torr$The degree of dissociation $(\alpha)$ is the fraction of initial pressure dissociated:$\alpha = \frac{x}{P_i} = \frac{200}{800} = 0.25$Percentage of dissociation = $0.25 	imes 100 = 25 \%$.

$AB_{3(g)}$ 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 developed at equilibrium is $900 \ torr$,what percentage of $AB_{3(g)}$ is dissociated?

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$AB_{3(g)}$ 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 developed at equilibrium is $900 \ torr$,what percentage of $AB_{3(g)}$ is dissociated?