A mixture of 0.482 M H2, 0.314 M I2, and 0.879 M HI is enclosed in a vessel and heated to 430 °C.

Question

A mixture of 0.482 M H2, 0.314 M I2, and 0.879 M HI is enclosed in a vessel and heated to 430 °C.

H2(g)+I2(g)↽−−⇀2HI(g)𝐾c=54.3 at 430 ∘C

Calculate the equilibrium concentrations of each gas at 430 ∘C430⁢ ∘C .

[H2]=

[I2]=

[HI]=

Answer 1

Set up an ICE table:

H₂(g) + I₂(g) ===> 2HI(g) ... Kc = 54.3

0.482.....0.314..........0.879.......Initial

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

0.482-x..0.314-x........2x...........Equilibrium

Kc = [HI]² / [H₂] [I₂]

54.3 = (2x)² / (0.482-x)(0.314-x)

Solve for x using the quadratic formula

[H2] = 0.482 - x

[I2] = 0.314 - x

[HI] = 2x