# 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?

Feb 22, 2015

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

$\left[H I\right] = \text{1.87 M}$
$\left[{H}_{2}\right] = \text{0.107 M}$
$\left[{I}_{2}\right] = \text{0.00660 M}$

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 $\text{1.00-L}$ vessel, the starting concentrations of ${H}_{2}$ and ${I}_{2}$ will be

${C}_{{H}_{2}} = {n}_{{H}_{2}} / V = \text{0.200 moles"/"1.00 L" = "0.200 M}$, and

${C}_{{I}_{2}} = {n}_{{I}_{2}} / V = \text{0.100 moles"/"1.00 L" = "0.100 M}$

The initial concentration of the hydrogen iodide will be zero.

.......${H}_{2 \left(g\right)} + {I}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s 2 H {I}_{\left(g\right)}$
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

${K}_{c} = \frac{{\left[H I\right]}^{2}}{\left[{H}_{2}\right] \cdot \left[{I}_{2}\right]} = \frac{{\left(2 x\right)}^{2}}{\left(0.200 - x\right) \cdot \left(0.100 - x\right)}$

${K}_{c} = \frac{4 {x}^{2}}{\left(0.200 - x\right) \cdot \left(0.100 - x\right)} = 49.5$

$45.5 {x}^{2} - 14.85 x + 0.99 = 0$
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 $\text{0.0934}$. The equilibrium concentrations will be
$\left[H I\right] = 2 \cdot 0.0934 = \text{1.87 M}$
$\left[{H}_{2}\right] = \text{0.200 - 0.0934 = 0.107 M}$
$\left[{I}_{2}\right] = \text{0.100 - 0.0934 = 0.00660 M}$