For the reaction 2P_{(g)} + Q_{(g)} rightleft | vedclass

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(B) The reaction is $2P_{(g)} + Q_{(g)} \rightleftharpoons R_{(g)} + S_{(g)}$.Initial moles: $P=2, Q=1, R=7, S=3$.At equilibrium: $P=(2-2x), Q=(1-x),

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$4 	imes 10^{-4} \ M$

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(B) The reaction is $2P_{(g)} + Q_{(g)} ightleftharpoons R_{(g)} + S_{(g)}$.Initial moles: $P=2, Q=1, R=7, S=3$.At equilibrium: $P=(2-2x), Q=(1-x), R=(7+x), S=(3+x)$.$K_C = \frac{[R][S]}{[P]^2 [Q]} = \frac{(7+x)(3+x)}{(2-2x)^2 (1-x)} = 10^{12}$.Since $K_C$ is very large,the reaction proceeds almost to completion. $Q$ is the limiting reagent,so $x \approx 1$.Let $x' = (1-x)$,then $[Q] = x'$ and $[P] = 2(1-x) = 2x'$.Substituting into the expression: $\frac{(7+1)(3+1)}{(2x')^2 (x')} = 10^{12}$.$\frac{32}{4x'^3} = 10^{12} \Rightarrow x'^3 = 8 	imes 10^{-12}$.$x' = 2 	imes 10^{-4}$.$[P] = 2x' = 2 	imes 2 	imes 10^{-4} = 4 	imes 10^{-4} \ M$.

For the reaction $2P_{(g)} + Q_{(g)} \rightleftharpoons R_{(g)} + S_{(g)}$; $K_C = 10^{12}$. If the initial moles of $P, Q, R, S$ are $2, 1, 7$ and $3$ moles respectively in a $1 \ L$ vessel,find the equilibrium concentration of $P$.

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