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
- (a) 12.0
- (b) 8.0
- (c) 6.0
- (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.
- Add the reactions. Add the three equations. Y and Z appear on both sides, so they cancel and you are left with X ⇌ W.
- 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
- 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.