Equilibrium Pressures

Question

3 H₂ (g) + N₂ (g) ↔ 2NH₃ (g), K_p = 1.64 x 10^-4

In a reaction vessel containing only NH3 at 400°C, the initial pressure of NH3 is 0.850

In a reaction vessel containing only NH3 at 400°C, the initial pressure of NH3 is 0.850 atm. Calculate the pressures of each of the three gasses at equilibrium.

Answer 1

3H₂ (g) + N₂ (g) <==> 2NH₃ (g) ... K_p = 1.64 x 10^-4

For these types of problems, it is probably best to use an ICE table. In the present example, we note that the Kp is << 1 telling us that the reaction favors the reactants. So, when only NH₃ is present, the reaction will proceed to the left until equilibrium is reached. We can re-write the reaction as

2NH₃(g) <==> 3H₂(g) + N₂(g) .. Kp = 1/1.64x10^-4 = 6098

.....0.085...............0.............0.........Initial

.....-2x.................+3x.........+x.........Change

0.850-2x..............3x............x.........Equilibrium

Kp = 6098 = (H₂)³(N₂) / (NH₃)²

6098 = (3x)³(x) / (0.085-2x)

6098 = 27x⁴ / 4x² - 3.4x + 0.723

From here, you would need to solve for x (quartic) and then substitute the value for x, back into the equilibrium line of the ICE table. Thus, at equilibrium, the pressures of the three gases would be...

(NH₃) = 0.850 - 2x

(H₂) = 3x

(N₂) = x