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

We start with the equilibrium reaction $X_2(g) + Y_2(g) \rightleftharpoons 2Z(g)$. Initial equilibrium moles are $[X_2] = 3\,\text{mol}$, $[Y_2] = 3\,\text{mol}$, and $[Z] = 9\,\text{mol}$. When 10 mol of $Z$ is added, the initial moles of $Z$ become $19\,\text{mol}$.

To find the new equilibrium, we set up the equation for the change in moles due to the reaction shifting to re-establish equilibrium: Suppose $\Delta$ mol of $Z$ reacts back to form $X_2$ and $Y_2$. Thus, $Z$ will decrease by $2\Delta$ mol, whereas $X_2$ and $Y_2$ will each increase by $\Delta$ mol.

The new equilibrium moles are $[X_2] = 3+\Delta$, $[Y_2] = 3+\Delta$, and $[Z] = 19-2\Delta$.

The equilibrium constant expression is $K_c = \frac{[Z]^2}{[X_2][Y_2]}$ and remains unchanged. Initially, $K_c = \frac{9^2}{3 \cdot 3} = 9$. Setting the new equilibrium $K_c$ calculation to this value:

$$\frac{(19-2\Delta)^2}{(3+\Delta)(3+\Delta)} = 9$$.

Solving, $(19-2\Delta)^2 = 9(3+\Delta)^2$. Expanding both sides:

$361 - 76\Delta + 4\Delta^2 = 81 + 54\Delta + 9\Delta^2$.

Reorganizing terms, $4\Delta^2 - 9\Delta^2 - 76\Delta - 54\Delta + 361 - 81 = 0$, simplifies to $-5\Delta^2 - 130\Delta + 280 = 0$.

Dividing by -5 gives $\Delta^2 + 26\Delta - 56 = 0$.

Solving this quadratic equation, $\Delta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 26$, $c = -56$.

Calculating the discriminant: $26^2 + 4\cdot 56 = 676 + 224 = 900$.

Thus, $\Delta = \frac{-26 \pm 30}{2}$, giving $\Delta = 2$ or $\Delta = -28$ (discarded as it doesn't make sense physically).

The equilibrium moles of $Z$ are $[Z] = 19 - 2\Delta = 19 - 4 = 15\,\text{mol}$.

This value aligns with the provided range [15, 15]. Therefore, the new equilibrium moles of $Z$ are $\mathbf{15\,\text{mol}}$.