Q. The value of equilibrium constant of the reaction HI(g) ⇌ ½H₂(g) + ½I₂(g) is 8.0. The equilibrium constant of the reaction H₂(g) + I₂(g) ⇌ 2HI(g) will be
A) 1/8
B) 1/16
C) 1/64
D) 16
Explanation
[Solution from source] Doubling the given reaction gives 2HI ⇌ H₂ + I₂ with K = 8² = 64. Reversing it gives H₂ + I₂ ⇌ 2HI with K = 1/64. 1/8 only reverses the reaction and forgets that doubling the equation squares K. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.