Question

In an experiment, 0.200 mol H2 and 0.100 mol I2 were placed in a 1.00 L vessel where the following equilibrium was established: H2 + I2 <-----> 2HI For this reaction, Kc = 49.5, What were the equilibrium concentration for H2, I2 and HI?

Answer 1

The equilibrium concentrations for all the species involved in the reaction are as follows:

`[HI] = "1.87 M"`
`[H_2] = "0.107 M"`
`[I_2] = "0.00660 M"`

So, start with the balanced chemical equation. You'll need to use the ICE chart method (more here: http://en.wikipedia.org/wiki/RICE_chart) to determine the equalibrium concentrations for all the species involved.

SInce you're dealing with a `"1.00-L"` vessel, the starting concentrations of `H_2` and `I_2` will be

`C_(H_2) = n_(H_2)/V = "0.200 moles"/"1.00 L" = "0.200 M"`, and

`C_(I_2) = n_(I_2)/V = "0.100 moles"/"1.00 L" = "0.100 M"`

The initial concentration of the hydrogen iodide will be zero.

.......`H_(2(g)) + I_(2(g)) rightleftharpoons 2HI_((g))`
I..0.200.........0.100...........0
C..(-x)................(-x)...............(+2x)
E..(0.200-x).....(0.100-x)......(2x)

The expression for the reaction's equilibrium constant is

`K_c = ([HI]^(2))/([H_2] * [I_2]) = ((2x)^(2))/((0.200-x) * (0.100-x))`

`K_c = (4x^(2))/((0.200-x) * (0.100-x)) = 49.5`

Rearrange to quadratic equation form

`45.5x^(2) - 14.85x + 0.99 = 0`

Solving for `x` will get you two values, `x_1 = 0.233` and `x_2 = 0.0934`. Look at the concentrations of the `H_2` and `I_2`. The value you chose for `x` must not produce a negative equilibrium concentration for neither species.

As a result, the suitable value for `x` will be `"0.0934"`. The equilibrium concentrations will be

`[HI] = 2 * 0.0934 = "1.87 M"`
`[H_2] = "0.200 - 0.0934 = 0.107 M"`
`[I_2] = "0.100 - 0.0934 = 0.00660 M"`