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Question 13.9: Predicting the Direction of Reaction A mixture of 1.57 mol o......

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
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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.

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