Question

For given reaction $X_2(g) + Y_2(g) \rightleftharpoons 2Z (g)$. Moles of $X_2, Y_2 & Z$ are at equilibrium are $3 \text{ mole}, 3 \text{ mole} & 9 \text{ mole}$ respectively. If $10 \text{ moles}$ of $Z$ are added at constant $T$ then find moles of $Z$ at re-established equilibrium.

• A. 15
• B. 18
• C. 19
• D. 13

Hint:

Le Chatelier's Principle dictates that adding product ($Z$) shifts equilibrium backward. When the number of moles of gaseous reactants equals gaseous products ($\Delta n_g = 0$), $K_c$ calculated using moles is equivalent to $K_p$ calculated using partial pressures.

Answer 1

The Correct Option is A

Initial equilibrium constant $K_c$: $K_c = \frac{[Z]^2}{[X_2][Y_2]}$. (Volume $V$ cancels out).
$K_c = \frac{9^2}{3 \times 3} = 9$.
When $10 \text{ moles}$ of $Z$ are added, initial concentrations are: $X_2=3, Y_2=3, Z=19$.
Since $Q_c = \frac{19^2}{3 \times 3}>9$, the reaction shifts left, consuming $Z$. Let $2x$ moles of $Z$ react.
\begin{tabular}{lccc} & $X_2$ & $Y_2$ & $2Z$
Initial: & 3 & 3 & 19
Change: & $+x$ & $+x$ & $-2x$
Equilibrium: & $3+x$ & $3+x$ & $19-2x$
\end{tabular}
$K_c = 9 = \frac{(19-2x)^2}{(3+x)(3+x)} = \left(\frac{19-2x}{3+x}\right)^2$.
Taking square root: $3 = \frac{19-2x}{3+x}$.
$3(3+x) = 19 - 2x$.
$9 + 3x = 19 - 2x$.
$5x = 10 \implies x = 2$.
Moles of $Z$ at new equilibrium $= 19 - 2x = 19 - 2(2) = 15 \text{ moles}$.