JEE Main 2024 Chemistry question
The following concentrations were observed at 500K for the formation of NH₃ from N₂ and H₂. At equilibrium ; [N₂] = 2 × 10⁻² M, [H₂] = 3 × 10⁻² M and [NH₃] = 1.5 × 10⁻² M. Equilibrium constant for the reaction is .
Answer: 417
Solution
N₂ + 3H₂ ⇌ 2NH₃, = [NH₃]²/([N₂][H₂]³) = (1.5 × 10⁻²)²/((2 × 10⁻²)(3 × 10⁻²)³) = 416.67 ≈ 417.