XY_2 dissociates as XY_{2(g)} rightleftharpoo | vedclass

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(B) The dissociation reaction is: $XY_{2(g)} \rightleftharpoons XY_{(g)} + Y_{(g)}$Initial pressure: $P_0 = 600 \ mm \ Hg$,$P_{XY} = 0$,$P_Y = 0$At eq

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(B) The dissociation reaction is: $XY_{2(g)} ightleftharpoons XY_{(g)} + Y_{(g)}$Initial pressure: $P_0 = 600 \ mm \ Hg$,$P_{XY} = 0$,$P_Y = 0$At equilibrium: $P_{XY_2} = 600 - x$,$P_{XY} = x$,$P_Y = x$Total pressure at equilibrium: $P_{total} = (600 - x) + x + x = 600 + x$Given $P_{total} = 800 \ mm \ Hg$,so $600 + x = 800$,which gives $x = 200 \ mm \ Hg$Equilibrium partial pressures are: $P_{XY_2} = 600 - 200 = 400 \ mm \ Hg$,$P_{XY} = 200 \ mm \ Hg$,$P_Y = 200 \ mm \ Hg$$K_p = \frac{P_{XY} 	imes P_Y}{P_{XY_2}} = \frac{200 	imes 200}{400} = 100 \ mm \ Hg$

$XY_2$ dissociates as $XY_{2(g)} \rightleftharpoons XY_{(g)} + Y_{(g)}$. When the initial pressure of $XY_2$ is $600 \ mm \ Hg$,the total equilibrium pressure is $800 \ mm \ Hg$. Calculate $K_p$ for the reaction,assuming the volume of the system remains unchanged.

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