Consider the reaction: 2 HI (g) ⇌ H₂ (g) + I₂ (g) Kc = 7.21 If at equilibrium the reaction vessel contains 1.20 M of H₂ and 1.70 M I₂, the what is the equilibrium concentration of HI?

Answer 1

2HI(g) <===> H₂(g) + I₂(g) Kc = 7.21

Kc = [H₂][I₂] / [HI]²

7.21 = [1.20][1.70] / [HI]²

[HI]² = 2.04 / 7.21 = 0.283

[HI] = 0.532