## Q. 13.9

Predicting the Direction of Reaction

A mixture of 1.57 mol of $N_2$, 1.92 mol of $H_2$, and 8.13 mol of $NH_3$ is introduced into a 20.0 L reaction vessel at 500 K. At this temperature, the equilibrium constant $K_c$ for the reaction $\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)$ is 1.7 × 10² . Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

STRATEGY

To determine whether the reaction mixture is at equilibrium, we need to calculate the value of the reaction quotient $Q_c$ and then compare it with the equilibrium constant $K_c$. If the mixture is not at equilibrium, the relative values of $Q_c$ and $K_c$ tell us the direction of the net reaction. Because we are given amounts in moles, we must first convert moles to molar concentrations before substituting into the expression for $Q_c$.

## Verified Solution

The initial concentration of $N_2$ is (1.57 mol)/(20.0 L) = 0.0785 M. Similarly, [ $H_2$ ] = 0.0960 M and [ $NH_3$ ] = 0.406 M. Substituting these concentrations into the equilibrium-constant expression gives

$Q_{\mathrm{c}}=\frac{\left[\mathrm{NH}_3\right]_t^2}{\left[\mathrm{~N}_2\right]_t\left[\mathrm{H}_2\right]_t^3}=\frac{(0.406)^2}{(0.0785)(0.0960)^3}=2.37 \times 10^3$

Because $Q_c$ does not equal $K_c (1.7 \times 10^2)$, the reaction mixture is not at equilibrium. Because $Q_c$ is greater than $K_c$, net reaction will proceed from right to left, decreasing the concentration of $NH_3$ and increasing the concentrations of $N_2$ and $H_2$ until $Q_c = K_c = 1.7 \times 10^2$ .

BALLPARK CHECK

Approximate initial concentrations can be calculated by dividing rounded values of the number of moles of each substance by the volume; [ $N_2$ ]≈(1.6 mol)/(20 L) ≈ 0.08 M, [ $H_2$ ]≈(2 mol)/(20 L)≈0.1 M, and [ $NH_3$ ]≈(8 mol)/(20 L)≈0.4 M. Substituting these concentrations into the expression for $Q_c$ gives a ballpark estimate of $Q_c$:

$Q_{\mathrm{c}}=\frac{\left[\mathrm{NH}_3\right]_t^2}{\left[\mathrm{~N}_2\right]_t\left[\mathrm{H}_2\right]_t^3}=\frac{(0.4)^2}{(0.08)(0.1)^3}=2 \times 10^3$

You can calculate this value without a calculator because it equals (16 × $10^{-2}$ )/(8 × $10^{-5}$ ). The ballpark estimate of $Q_c$, like the more exact value (2.37 × 10³ ), exceeds $K_c$, so the reaction mixture is not at equilibrium.