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?

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"#