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?

Md Ariful Islam

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Md Ariful Islam

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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: ) to determine the equalibrium 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 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

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

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