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Equilibrium constant Kc: Class 11 practice

For aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ/([A]ᵃ[B]ᵇ) using equilibrium concentrations. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Powers
  • Concentration

The idea behind equilibrium constant kc

For aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ/([A]ᵃ[B]ᵇ) using equilibrium concentrations. Pure solids and liquids are left out. Reversing the reaction gives 1/Kc.

Step-by-step method

  1. Write the Kc expression from the balanced equation.
  2. Substitute equilibrium concentrations (raise to coefficients).
  3. Simplify; for the reverse reaction take the reciprocal.

Worked example 1 (Basic)

For A + B ⇌ C at equilibrium [A] = 0.9, [B] = 0.4, [C] = 0.1 M. Find Kc.

  1. Kc = [C]/([A][B]) = 0.1/(0.9 × 0.4)
  2. = 0.278
  3. Answer: 0.278

Worked example 2 (Standard)

For N₂ + 3H₂ ⇌ 2NH₃, [N₂] = 0.3, [H₂] = 0.7, [NH₃] = 0.8 M. Find Kc.

  1. Kc = [NH₃]²/([N₂][H₂]³)
  2. = 0.6400000000000001/(0.3 × 0.343) = 6.22
  3. Answer: 6.22

Worked example 3 (Challenge)

Kc for A + B ⇌ 2C is 49. Find Kc for 2C ⇌ A + B and for C ⇌ ½A + ½B.

  1. Reverse: 1/49
  2. Halve the equation: (1/49)^½ = 0.143
  3. Answer: 1/49 and 0.143

Common mistakes and misconceptions

  • Using initial concentrations
  • Forgetting powers from coefficients

Practice questions

  1. (Basic) For A + B ⇌ C at equilibrium [A] = 0.8, [B] = 0.1, [C] = 0.2 M. Find Kc.
  2. (Basic) For A + B ⇌ C at equilibrium [A] = 0.5, [B] = 0.7, [C] = 0.1 M. Find Kc.
  3. (Standard) For N₂ + 3H₂ ⇌ 2NH₃, [N₂] = 0.2, [H₂] = 0.4, [NH₃] = 0.1 M. Find Kc.
  4. (Standard) For N₂ + 3H₂ ⇌ 2NH₃, [N₂] = 0.9, [H₂] = 0.7, [NH₃] = 0.2 M. Find Kc.
  5. (Challenge) Kc for A + B ⇌ 2C is 37. Find Kc for 2C ⇌ A + B and for C ⇌ ½A + ½B.
  6. (Challenge) Kc for A + B ⇌ 2C is 17. Find Kc for 2C ⇌ A + B and for C ⇌ ½A + ½B.

Answer key with full working

  • Q1: Kc = [C]/([A][B]) = 0.2/(0.8 × 0.1) → = 2.5 → Answer: 2.5
  • Q2: Kc = [C]/([A][B]) = 0.1/(0.5 × 0.7) → = 0.286 → Answer: 0.286
  • Q3: Kc = [NH₃]²/([N₂][H₂]³) → = 0.010000000000000002/(0.2 × 0.064) = 0.781 → Answer: 0.781
  • Q4: Kc = [NH₃]²/([N₂][H₂]³) → = 0.04000000000000001/(0.9 × 0.343) = 0.13 → Answer: 0.13
  • Q5: Reverse: 1/37 → Halve the equation: (1/37)^½ = 0.164 → Answer: 1/37 and 0.164
  • Q6: Reverse: 1/17 → Halve the equation: (1/17)^½ = 0.243 → Answer: 1/17 and 0.243

Syllabus note

This skill is listed under "Equilibrium" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is equilibrium constant kc in?

"Equilibrium", Class 11 Chemistry (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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