A mixture of 0.438 M H2, 0.444 M I2, and 0.895 M HI is enclosed in a vessel and heated to 430 °C. H2 (g) + I2 (g) 2 HI (g) Kc = 54.3 at 430∘C Calculate the equilibrium concentrations of each gas at 430∘C.

[H₂] = 0.178 M

[I₂] = 0.184 M

[HI] = 1.415 M

Explanation:

For the equilibrium:

H₂ (g) + I₂ (g) ⇄ 2 HI (g)

the equilibrium constant is given by the equation:

Kc = [ HI]² / [H₂][I₂]

Lets use first the reaction quotient which has the same expression as the equilibrium constant to predict the direction the reaction will take, i. e towards reactants or product side.

Q = (0.895) ² / (0.438) (0.444) = 4.12

Q is less than Kc so the reaction will favor the product side.

We can set up the following table to account for all the species at equilibrium:

H₂ I₂ HI

initial 0.438 0.444 0.895

change - x - x + 2x

equilibrium 0.438 - x 0.444 - x 0.895 + 2x

Now we are in position to express these concentrations in terms of the equilibrium conctant, Kc

54.3 = (0.895 + 2x) ² / (0.438 - x) (0.444 - x)

performing the calculatiopns will result in a quadratic equation:

0.801 + 3.580x + 4x² = (0.194 - 0.882x + x²) x 54.3

Upon rearrangement and some algebra, we have

0.801 + 3.580 x + 4x² = 10.534 - 47.893x + 54.3 x²

0 = 9.733 - 51.473 x + 54.3 x²

This equation has two roots X₁ = 0.687 and X₂ = 0.26

The first is physically impossible since it will imply that more 0.687 will make the quantity at equilibrium for both H₂ and I₂ negative.

Therefore the concentrations at equilibrium of each gas are:

[H₂] = (0.438 - 0.260) = 0.178 M

[I₂] = (0.444 - 0.260) M = 0.184 M

[HI] = [0.895 + 2x (0.260) ] M = 1.415 M

Note if we plug these values into the equilibrium expression we get 61 which is due to the rounding errors propagating in the quadratic equation.