## Chapter 13

## 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.

## Step-by-Step

## 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 10^3Because 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 10^3You 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.