Question 15.10: For the Haber process, N2(g) + 3 H2(g) ⇌ 2 NH3(g), Kp = 1.41...
For the Haber process, N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g), K_{p} = 1.41 × 10^{-9}, at 500 °C. In an equilibrium mixture of the three gases at 500 °C, the partial pressure of H_{2} is 94.03 bar and that of N_{2} is 43.77 bar. What is the partial pressure of NH_{3} in this equilibrium mixture?
Analyze We are given an equilibrium constant, K_{p}, and the equilibrium partial pressures of two of the three substances in the equation (N_{2} and H_{2}), and we are asked to calculate the equilibrium partial pressure for the third substance (NH_{3}).
Plan We can set K_{p} equal to the equilibrium expression and substitute in the partial pressures that we know. Then we can solve for the only unknown in the equation.
Solve
We tabulate the equilibrium pressures:
Because we do not know the equilibrium pressure of NH_{3},we represent it with x. At equilibrium, the pressures must satisfy the equilibrium expression:
K_{p} =\frac{(P_{NH_{3}})²}{P_{N_{2}}(P_{H_{2}})³ }=\frac{x²}{(43.77)(94.03)³}= 1.41 × 10^{-9}We now rearrange the equation to solve for x:
x² = (1.41 × 10^{-9})(43.77)(94.03)³ = 5.14 × 10^{-2}
x = \sqrt{ 5.14 × 10^{-2}} = 0.227 bar = P_{NH_{3}}
Check We can always check our answer by using it to recalculate the value of the equilibrium constant:
K_{p} =\frac{(0.227)²}{(43.77)(94.03)³} = 1.41 × 10^{-9}