JEE Main 2020 · 4 September · evening · Single correct · +4 / −1

If the equilibrium constant for
A ⇌ B + C is K_(eq)⁽¹⁾ and that of
B + C ⇌ P is K_(eq)⁽²⁾, the equilibrium
constant for A ⇌ P is :

  1. (a) K_(eq)⁽¹⁾/K_(eq)⁽²⁾
  2. (b) K_(eq)⁽¹⁾ + K_(eq)⁽²⁾
  3. (c) K_(eq)⁽²⁾ - K_(eq)⁽¹⁾
  4. (d) K_(eq)⁽¹⁾ K_(eq)⁽²⁾
Answer and solution

Answer (d) K_(eq)⁽¹⁾ K_(eq)⁽²⁾

Solution

Chemistry ma’am · 3 steps

Add the two reactions and B + C cancels out, so you multiply the two K values.

  1. Write both K. By the law of mass action, K_(eq)⁽¹⁾ = [B][C]/[A] and K_(eq)⁽²⁾ = [P]/[B][C].
  2. Add the reactions. Adding A ⇌ B + C and B + C ⇌ P gives A ⇌ P, because B and C appear on both sides and cancel. The same cancellation happens when you multiply the two expressions.

    K_(eq)⁽¹⁾ × K_(eq)⁽²⁾ = [B][C]/[A] × [P]/[B][C] = [P]/[A] = K_(eq)

  3. So. The equilibrium constant for A ⇌ P is K_(eq)⁽¹⁾ K_(eq)⁽²⁾, option (d).

The idea to keep. When reactions are added, their K values are multiplied. Reversing a reaction gives 1/K, and multiplying a reaction by n gives Kⁿ.

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

Study Chemical Equilibrium

Asked about 90 times in JEE Main since 2020. This question is from 2020.