Question

$X_2(g) + Y_2(g) \rightleftharpoons 2Z(g)$. Equilibrium moles of $X_2, Y_2, Z$ are 3, 3, 9 mol (in 1 L). 10 mol of Z(g) is added. New equilibrium moles of Z(g) is ___. (Nearest integer)

Hint:

If $Q_c>K_c$, the reaction proceeds towards the reactants to re-establish equilibrium.

Answer 1

Correct Answer: 15

Calculate the equilibrium constant $K_c$ (volume $V=1 \text{ L}$):
$K_c = \frac{[Z]^2}{[X_2][Y_2]} = \frac{9^2}{3 \times 3} = 9$.
Disturbance: 10 mol of Z is added. Initial concentrations for the shift: $[X_2]=3, [Y_2]=3, [Z]=19$.
$Q_c = \frac{19^2}{3 \times 3} = \frac{361}{9} \approx 40.11$. Since $Q_c>K_c$, the reaction shifts to the left.
$X_2 + Y_2 \xleftarrow{k} 2Z$
Initial: $3 \quad 3 \quad 19$
Change: $+x \quad +x \quad -2x$
Equil: $3+x \quad 3+x \quad 19-2x$
$K_c = \frac{(19-2x)^2}{(3+x)^2} = 9$.
Taking square root: $\frac{19-2x}{3+x} = 3$.
$19 - 2x = 9 + 3x \implies 5x = 10 \implies x = 2$.
New equilibrium moles of Z $= 19 - 2x = 19 - 4 = 15$.