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.