2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
In an equilibrium mixture, the partial pressures are
P SO₃ = 43 kPa; P O₂ = 530 Pa and P SO₂ = 45 kPa. The equilibrium constant K P = ___________ × 10⁻². (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Put all three pressures in kPa first, then square the ones with coefficient 2.

1. Same units first. Two pressures are in kPa but p_(O₂) is in Pa. Convert it: 530 Pa = 0.53 kPa. The key 172 matches K_P in kPa⁻¹, so all three pressures stay in kPa.
2. Write the expression. By the law of mass action, each partial pressure is raised to its coefficient in the balanced equation. Products go on top and reactants go below.

   K_P = p_(SO₃)²/(p_(SO₂)² p_(O₂)) = (43)²/((45)² × 0.53)

3. Put in numbers. The numerator is 43² = 1849. The denominator is 45² × 0.53 = 2025 × 0.53 = 1073.25. So K_P = 1849/1073.25 = 1.72 kPa⁻¹.
4. So. K_P = 1.72 = 172 × 10⁻², so the answer is 172.

The common slip: 180. Forgetting to square the SO₃ and SO₂ pressures gives 43/(45 × 0.53) = 1.80, that is 180, not 172.

The idea to keep. Before you use the law of mass action, bring every pressure to one unit. Then raise each pressure to its stoichiometric coefficient. A mixed Pa and kPa set will throw the answer off by powers of 1000.

Answer from the NTA final answer key. Solution by PrepWiser.