Consider the following gaseous equilibrium in | vedclass

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(A) The initial equilibrium is $P_{2(g)} + Q_{2(g)} \rightleftharpoons 2PQ_{(g)}$.At equilibrium, $n(P_2) = 2, n(Q_2) = 2, n(PQ) = 2$.$K_c = \frac{[PQ

Vedclass · Accepted Answer

$2.67, 2.67, 2.67$

Vedclass · Answer

(A) The initial equilibrium is $P_{2(g)} + Q_{2(g)} ightleftharpoons 2PQ_{(g)}$.At equilibrium, $n(P_2) = 2, n(Q_2) = 2, n(PQ) = 2$.$K_c = \frac{[PQ]^2}{[P_2][Q_2]} = \frac{(2/V)^2}{(2/V)(2/V)} = 1$.After adding $1 	ext{ mole}$ each of $P_2$ and $Q_2$, the new initial moles are $n(P_2) = 3, n(Q_2) = 3, n(PQ) = 2$.Let $x$ be the number of moles of $P_2$ reacted to reach the new equilibrium.$P_{2(g)} + Q_{2(g)} ightleftharpoons 2PQ_{(g)}$Initial: $3, 3, 2$Equilibrium: $(3-x), (3-x), (2+2x)$$K_c = 1 = \frac{(2+2x)^2}{(3-x)^2}$.Taking the square root on both sides: $1 = \frac{2+2x}{3-x}$.$3-x = 2+2x \implies 3x = 1 \implies x = 1/3$.At new equilibrium:$n(P_2) = 3 - 1/3 = 8/3 \approx 2.67$.$n(Q_2) = 3 - 1/3 = 8/3 \approx 2.67$.$n(PQ) = 2 + 2(1/3) = 8/3 \approx 2.67$.

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