JEE Main 2022 · 30 June · morning · Single correct · +4 / −1

The equilibrium constant for the reversible reaction
2A(g) ⇌ 2B(g) + C(g) is K₁
3/2A(g) ⇌ 3/2B(g) + 3/4C(g) is K₂.
K₁ and K₂ are related as :

  1. (a) K₁=√K₂
  2. (b) K₂=√K₁
  3. (c) K₂=K₁^(3/4)
  4. (d) K₁=K₂^(3/4)
Answer and solution

Answer (c) K₂=K₁^(3/4)

Solution

Chemistry ma’am · 3 steps

Find the factor that turns the first equation into the second; K gets raised to that same power.

  1. Compare coefficients. Divide each coefficient of the second equation by the matching one in the first: (3/2)/2 = 3/4 for A, (3/2)/2 = 3/4 for B and (3/4)/1 = 3/4 for C. So the second reaction is the first one multiplied by 3/4.
  2. Write both constants. By the law of mass action, K₁ = [B]²[C]/[A]². Each power in K₂ is 3/4 of the matching power in K₁.

    K₂ = [B]^(3/2)[C]^(3/4)/[A]^(3/2) = ([B]²[C]/[A]²)^(3/4) = K₁^(3/4)

  3. So. K₂ = K₁^(3/4), which is option (c).

The idea to keep. Multiplying a balanced equation by n raises its equilibrium constant to the power n. Reversing the equation gives 1/K, and adding two equations multiplies their constants.

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 2022.