JEE Main Chemistry question
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:
Answer: (A)
Solution
K(target) = [1/K(given)]² = (1/8)² = 1/64.
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:
K(target) = [1/K(given)]² = (1/8)² = 1/64.