Chemical Equilibrium Part 2

Concept:For a gaseous reaction where total moles of reactants equals total moles of products, the equilibrium constant $K_c$ expressed in terms of moles is independent of container volume.Chemical Equation / Formula:$\mathrm{P}_2(g) + \mathrm{Q}_2(g) \rightleftharpoons 2\mathrm{PQ}(g)$$K_c = \frac{[\mathrm{PQ}]^2}{[\mathrm{P}_2][\mathrm{Q}_2]} = \frac{n_{\mathrm{PQ}}^2}{n_{\mathrm{P}_2}\, n_{\mathrm{Q}_2}}$Explanation:Initially at equilibrium: $n_{\mathrm{P}_2}=2$, $n_{\mathrm{Q}_2}=2$, $n_{\mathrm{PQ}}=2$.Calculate $K_c$: $K_c = \frac{2^2}{2 \times 2} = 1$.After adding $1$ mole each of $\mathrm{P}_2$ and $\mathrm{Q}_2$, the amounts just after addition are: $n_{\mathrm{P}_2}=3$, $n_{\mathrm{Q}_2}=3$, $n_{\mathrm{PQ}}=2$.Let the forward reaction go to new equilibrium by an extent $x$.New equilibrium moles: $n_{\mathrm{P}_2}=3-x$, $n_{\mathrm{Q}_2}=3-x$, $n_{\mathrm{PQ}}=2+2x$.Apply $K_c=1$: $\frac{(2+2x)^2}{(3-x)^2}=1$.Take square root (positive): $\frac{2+2x}{3-x}=1$.Solve: $2+2x=3-x$ ⇒ $3x=1$ ⇒ $x=\frac13$.Thus new moles: $n_{\mathrm{P}_2}=3-\frac13=\frac83=2.67$, $n_{\mathrm{Q}_2}=2.67$, $n_{\mathrm{PQ}}=2+2\times\frac13=\frac83=2.67$.Answer:The number of moles at new equilibrium are $2.67$, $2.67$, $2.67$, which corresponds to Option B.