Question

Nitric oxide is a colourless gas. It is used in the preparation of nitric acid from ammonia. One way of producing it is by treating ammonia with oxygen as shown below:
4NH3(s) + 5O2(g) Û 4NO(g) + 6H2O(g)

A. 0.35 mol of NH3(g) is treated with 0.35 mol of O2(g) in a rigid 5.0 L container. 0.25 mol of NO(g) is obtained at a specific temperature at equilibrium. Calculate the equilibrium constant and the composition of the equilibrium mixture at the same temperature.

Answer 1

Let the extent of reaction be ξ. Since 4 mol NO are produced per reaction, ξ = n(NO)/4 = 0.25/4 = 0.0625 mol.

Equilibrium moles:
- NH3: 0.35 − 4ξ = 0.35 − 0.25 = 0.10 mol
- O2: 0.35 − 5ξ = 0.35 − 0.3125 = 0.0375 mol
- NO: given 0.25 mol
- H2O: 6ξ = 6 × 0.0625 = 0.375 mol

Concentrations in the 5.0 L container (mol L−1):
- [NH3] = 0.10/5.0 = 0.020 M
- [O2] = 0.0375/5.0 = 0.00750 M
- [NO] = 0.25/5.0 = 0.0500 M
- [H2O] = 0.375/5.0 = 0.0750 M

Equilibrium constant (Kc) for 4NH3 + 5O2 ⇌ 4NO + 6H2O:
Kc = [NO]^4 [H2O]^6 / ([NH3]^4 [O2]^5)
≈ (0.0500^4)(0.0750^6) / (0.020^4 · 0.00750^5)
≈ 2.9 × 10^5

So Kc ≈ 2.9 × 10^5. The equilibrium composition (moles) is NH3 0.100, O2 0.0375, NO 0.250, H2O 0.375 (concentrations given above).