Consider the gas-phase reaction H2(g) + I2(g) ⇄ 2 HI(g)

Question

Suppose that a sealed flask containing H2 and I2 has been heated to 425 °C and the initial concentrations of H2 and I2 were each 0.0175 mol/L. With time, the concentrations of H2 and I2 decline, and the concentration of HI increases. At equilibrium [HI] = 0.0276 mol/L. Use this experimental information to calculate the equilibrium constant.

Answer 1

K = [HI]² / [H₂][I₂]

H₂(g) + I₂(g) <==> 2HI(g)

0.0175...0.0175........0.........Initial

-x...........-x.............+2x........Change

0.0175-x..0.0175-x..2x........Equilibrium

We are told that @ equilibrium HI = 0.0276 M so this means 2x = 0.0276 and x = 0.0138

At equilibrium the following concentrations are present:

[HI] = 0.0276 M

[H₂] = 0.0175 - 0.0138 = 0.0037

[I₂] = 0.0037

Solving for K we have

K = (0.0276)² / (0.0037)²

K = 55.6