Calculate the concentrations of all species when equilibrium is reached.?

Question

At a particular temperature, K = 1.00e2 for the following reaction.

H2(g) + I2(g) reverse reaction arrow 2 HI(g)

In an experiment, 1.35 mol H2, 1.35 mol I2, and 1.35 mol HI are introduced into a 1.00-L container. Calculate the concentrations of all species when equilibrium is reached.

Answer 1

I think Steve O is wrong. He forgot to add in the initial concentration of HI. Back to the question. Since they gave you the equilibrium constant you can determine the equilibrium concentrations with the help of a chemical equation and equilibrium expression. To do this problem you set up a reaction table.

................... H2(g)...... +....... I2(g) .........--> .......... 2HI(g)

Initial ........ 1.35M .............. 1.35M .................... 1.35 + 2x

Reacting ..... -x ...................... -x ............................. +2x

Equilibrium 1.35-x .............1.35-x ....................... 1.35+2x

[H2] = 1.35-x              [I2] = 1.35-x                  [HI] = 1.35+2x

Notice how the reactants have a negative x and the products have a positive x. This is because the equilibrium constant is large so it favors products. We can assume that reactants are decreasing and products are increasing.

K = ([HI]^2)/([H2][I2])   K = 100

Plug in the values and it turns into a quadratic equation.

100 = ((1.35+2x)^2)/(1.35-x)^2

100(1.8225 - 2.7x + x^2) = 1.8225 + 5.4x + 4x^2

182.25 - 270x + 100x^2 = 1.8225 + 5.4x + 4x^2

180.4275 - 275.4x + 96x^2 = 0

x = 1.0125

[H2] = [I2] = 1.35 - x = 1.35 - 1.0125 = .3375M

[HI] = 1.35 + 2x = 1.35 + 2(1.0125) = 3.375M