Question
Consider the following gaseous equilibrium in a closed container of volume "V" at T(K).
$$P_{2}(\text{ }g) + Q_{2}(\text{ }g) \rightleftharpoons 2PQ(g) $$2 moles each of $P_{2}(\text{ }g),Q_{2}(\text{ }g)$ and $PQ(g)$ are present at equilibrium. Now one mole each of ' $P_{2}$ ' and ' $Q_{2}$ ' are added to the equilibrium keeping the temperature at $T(K)$. The number of moles of $P_{2}$, $Q_{2}$ and PQ at the new equilibrium, respectively, are -
Options
Solution
$\ P_{2}(\text{ }g) + Q_{2}(\text{ }g) \rightleftharpoons 2PQ(g)$
$t = t_{\text{eq~}}\ 2$ mole $\ 2$ mole $\ 2$ mole
$$K_{eq} = \frac{2^{2}}{2.2} = 1 $$Now 1 mole of each $P_{2}$ and $Q_{2}$ is added
So reaction will move in forward direction
$$\begin{matrix} & \ P_{2}(g) + Q_{2}(g) \rightleftharpoons 2PQ(g) \\ & t = t_{\text{eq.~}}'\ 3 - x\ 3 - x \\ & K_{C} = 1 = \frac{(2 + 2x)^{2}}{(3 - x)(3 - x)} \\ & \frac{2 + 2x}{3 - x} = 1 \\ & 2 + 2x = 3 - x \\ & x = \frac{1}{3} \end{matrix}$$
At new equilibrium :
Moles of $P_{2} = \frac{8}{3} = 2.67$
Moles of $Q_{2} = \frac{8}{3} = 2.67$
Moles of $PQ = \frac{8}{3} = 2.67$
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