Consider the reaction,P(aq) rightleftharpoons | vedclass

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(B) Initial equilibrium state: $P(aq) \rightleftharpoons Q(aq)$ Initial: $[P] = 2 \ M, [Q] = 0 \ M$ At equilibrium: $[P] = 2 - x, [Q] = x$ $K = \frac{

Vedclass · Accepted Answer

$0.96$

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(B) Initial equilibrium state: $P(aq) \rightleftharpoons Q(aq)$ Initial: $[P] = 2 \ M, [Q] = 0 \ M$ At equilibrium: $[P] = 2 - x, [Q] = x$ $K = \frac{[Q]}{[P]} = \frac{x}{2 - x} = 1.5$ $x = 1.5(2 - x)$ $\Rightarrow x = 3 - 1.5x$ $\Rightarrow 2.5x = 3$ $\Rightarrow x = 1.2 \ M$ So,at initial equilibrium: $[P] = 0.8 \ M, [Q] = 1.2 \ M$. After removing half of $P$: New $[P] = 0.8 / 2 = 0.4 \ M$,$[Q] = 1.2 \ M$. Reaction quotient $Q_c = \frac{[Q]}{[P]} = \frac{1.2}{0.4} = 3$. Since $Q_c > K$ $(3 > 1.5)$,the reaction shifts backward. Let $y$ be the amount of $Q$ converted to $P$: New equilibrium: $[P] = 0.4 + y, [Q] = 1.2 - y$. $K = \frac{1.2 - y}{0.4 + y} = 1.5$ $1.2 - y = 1.5(0.4 + y) \Rightarrow 1.2 - y = 0.6 + 1.5y$ $0.6 = 2.5y \Rightarrow y = 0.24 \ M$ Final concentration of $Q = 1.2 - 0.24 = 0.96 \ M$.

$T_{1} = 25^{\circ}C$	$K_{1} = 100$
$T_{2} = 100^{\circ}C$	$K_{2} = 100$

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