Answer to Question #145982 in General Chemistry for Flexus

Question

Consider the reaction: 2HI(g) ⇋ H2(g) + I2(g). The equilibrium concentrations for these reaction at
445℃ are [HI] = 2.0 M, [H2] = 0.50 M, [I2] = 0.10 M. Calculate the equilibrium constant (Kc).

Answer 1

The reaction given,

2HI(g) ⇋ H₂(g) + I₂(g)

Equilibrium constant,

Kc = [H₂][I₂]/[HI]²

Given,

At temperature = 445℃

[HI] = 2.0 M

[H₂] = 0.50 M

[I₂] = 0.10 M

So,

Kc = [H₂][I₂]/[HI]²

Or, Kc = 0.50×0.10/(2.0)2

Or, Kc = 0.0125

Hence, Equilibrium constant = 0.0125