At 298 K
N₂ (g)+3H₂ (g)⇌ 2NH₃ (g), K₁=4× 10⁵
N₂ (g)+O₂ (g)⇌ 2NO (g), K₂=1.6× 10¹²
H₂ (g)+1/2O₂ (g)⇌ H₂O (g), K₃=1.0× 10⁻¹³
Based on above equilibria, then equilibrium constant of the reaction, 2NH₃(g)+5/2O₂ (g)⇌ 2NO (g)+3H₂O (g) is ____________ × 10⁻³³ (Nearest integer).

Solution

Chemistry ma’am · 4 steps

Build the target reaction out of the three given ones, and the K values follow the same moves.

1. Rules for K. Reversing a reaction turns K into 1/K. Multiplying a reaction by n raises K to the power n. Adding reactions multiplies their K values.
2. Assemble the target. NH₃ is a reactant in the target, so reverse reaction 1: 2NH₃ ⇌ N₂ + 3H₂. Add reaction 2 to make 2NO. Add 3 times reaction 3 to make 3H₂O. The N₂ and 3H₂ cancel, and the oxygen adds to 1 + 3/2 = 5/2 O₂, which matches the target.

   K = (K₂ × K₃³)/K₁ = ((1.6 × 10¹²) × (1.0 × 10⁻¹³)³)/(4 × 10⁵)

3. Put in numbers. K₃³ = 1.0 × 10⁻³⁹, so the numerator is 1.6 × 10⁻²⁷. Dividing by 4 × 10⁵ gives 0.4 × 10⁻³² = 4 × 10⁻³³.
4. So. The equilibrium constant is 4 × 10⁻³³, so the answer is 4.

The idea to keep. Match the target reaction step by step. A reversed reaction gives 1/K, a reaction multiplied by n gives Kⁿ, and added reactions multiply their K values.

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