Question

For the reaction:

2HI(g) ↔ H2(g) + I2(g) Keq = 0.016

Initially a container contains 0.60 M HI, 0.038 M H2, and 0.15 M I2 at equilibrium. What is the new equilibrium concentration of H2, if the H2 concentration is increased by 0.276 M?

Answer 1

Initial concentration HI = 0.60 M

initial concentration H₂ = 0.038 M + 0.276 M = 0.314 M

initial concentration I₂ = 0.15 M

K_eq = 0.016

ICE table	2 HI (g)		H2 (g)	I2 (g)
Initial conc.	0.60 M		0.314 M	0.15 M
Change	-2x		+x	+x
Equilibrium conc.	0.60 M - 2x		0.314 M + x	0.15 M + x

K_eq = [H₂]_eq[I₂]_eq / [HI]_eq²

0.016 = (0.314 M + x) * (0.15 M + x) / (0.60 M - 2x)²

Solving for x, x = -0.101 M

equilibrium concentration of H₂ = 0.314 M + x

equilibrium concentration of H₂ = 0.314 M + (-0.101 M)

equilibrium concentration of H₂ = 0.213 M