JEE Main 2024 · 8 April · morning · Single correct · +4 / −1

For the given hypothetical reactions, the equilibrium constants are as follows :
X⇌ Y;K₁=1.0; Y⇌ Z;K₂=2.0; Z⇌ W;K₃=4.0
The equilibrium constant for the reaction X⇌ W is

  1. (a) 12.0
  2. (b) 8.0
  3. (c) 6.0
  4. (d) 7.0
Answer and solution

Answer (b) 8.0

Solution

Chemistry ma’am · 3 steps

Add the three steps and X goes all the way to W, so multiply the three K values together.

  1. Add the reactions. Add the three equations. Y and Z appear on both sides, so they cancel and you are left with X ⇌ W.
  2. Why K multiplies. Each K is a ratio: K₁ = [Y]/[X], K₂ = [Z]/[Y], K₃ = [W]/[Z]. Multiply them and [Y] and [Z] cancel, leaving [W]/[X], which is exactly K for X ⇌ W.

    K = K₁ × K₂ × K₃ = 1.0 × 2.0 × 4.0 = 8.0

  3. So. The equilibrium constant for X ⇌ W is 8.0, option (b).

The idea to keep. Adding reactions means multiplying their K values. Reversing a reaction inverts K, and multiplying a reaction by n raises K to the power n.

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 2024.