The following concentrations were observed at 500 K for the formation of NH₃ from N₂ and H₂. At equilibrium ; [N₂]=2× 10⁻²M,[H₂]=3× 10⁻²M and [NH₃]=1.5× 10⁻²M. Equilibrium constant for the reaction is ________.

Solution

Chemistry ma’am · 4 steps

Write the balanced equation first, because the coefficients become the powers in K_c.

1. Balanced equation. Ammonia forms as N₂ + 3H₂ ⇌ 2NH₃. By the law of mass action, each concentration is raised to the power of its coefficient.

   K_c = [NH₃]²/[N₂][H₂]³

2. Work out each part. Numerator: (1.5 × 10⁻²)² = 2.25 × 10⁻⁴. Denominator: (2 × 10⁻²) × (3 × 10⁻²)³ = (2 × 10⁻²) × (27 × 10⁻⁶) = 5.4 × 10⁻⁷.
3. Divide. K_c = (2.25 × 10⁻⁴)/(5.4 × 10⁻⁷) = 416.7 L² mol⁻².
4. So. The equilibrium constant is about 417.

The common slip: 0.375. Leaving out the cube on [H₂] gives (2.25 × 10⁻⁴)/(6 × 10⁻⁴) = 0.375, not 417.

The idea to keep. In K_c, the stoichiometric coefficients of the balanced equation become the powers on the concentrations. Balance first, then raise each term to its power.

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