Question

At a particular temperature, K = 1.00×10² for the Reaction

${\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{I}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\stackrel{}{\mathbf{\rightleftharpoons }}\mathbf{2}\mathbf{HI}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

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

Short answer

The equilibrium concentrations of H₂, I₂ and HI are 0.25 M, 0.25 M and 2.50 M respectively.

Step-by-step solution

Equillibrium constant

The equilibrium constant expression for this reaction is as follows:

$\text{K=}\frac{{\left[\text{HI}\right]}^{\text{2}}}{\left[{\text{H}}_{\text{2}}\right]\left[{\text{I}}_{\text{2}}\right]}$ …(1)

The initial moles of H₂, I₂ and HI are 1.00 mol for each. The volume of the container is 1.00 L. Hence, the initial concentrations of H₂, I₂ and HI are 1.00 M for each.

Change in concentration

Assume x will be the change in concentration.

Setup the ICE table:

$\begin{array}{cccccc}& {\text{H}}_{\text{2}}\left(\text{g}\right)& \text{+}& {\text{I}}_{\text{2}}\left(\text{g}\right)& \stackrel{}{\rightleftharpoons }& \text{2HI}\left(\text{g}\right)\\ \text{Iinitial}\left(\text{M}\right)& \text{1.00}& & \text{1.00}& & \text{1.00}\\ \text{Change}\left(\text{M}\right)& \text{\hspace{0.17em}–\hspace{0.17em}x}& & \text{\hspace{0.17em}–\hspace{0.17em}x}& & \text{+2x}\\ \text{Equilibrium}\left(\text{M}\right)& \text{1.00\hspace{0.17em}–\hspace{0.17em}x}& & \text{1.00\hspace{0.17em}–\hspace{0.17em}x}& & \text{1.00+2x}\end{array}$

Now insert these equilibrium concentrations and the equilibrium constant value into equation (1) as shown below.

${\text{1.00×10}}^{\text{2}}\text{=}\frac{{\left(\text{1.00+2x}\right)}^{\text{2}}}{\left(\text{1.00\hspace{0.17em}–\hspace{0.17em}x}\right)\left(\text{1.00\hspace{0.17em}–\hspace{0.17em}x}\right)}$

$\text{100=}\frac{{\left(\text{1.00+2x}\right)}^{\text{2}}}{{\left(\text{1.00\hspace{0.17em}–\hspace{0.17em}x}\right)}^2}$

$\sqrt{\text{100}}\text{=}\frac{\left(\text{1.00+2x}\right)}{\left(\text{1.00\hspace{0.17em}–\hspace{0.17em}x}\right)}$

Solve for x.

$\text{x=0.75}$

Therefore, the change in concentration, x is 0.75 M.

Equillibrium concentration

The equilibrium concentrations of H₂, I₂ and HI can be calculated as follows:

$\begin{array}{c}\left[{\text{H}}_{\text{2}}\right]\text{=1.00\hspace{0.17em}–\hspace{0.17em}x}\\ \text{=1.00\hspace{0.17em}–\hspace{0.17em}0.75}\\ \text{=0.25\hspace{0.17em}M}\end{array}$

$\begin{array}{c}\left[{\text{I}}_{\text{2}}\right]\text{=1.00\hspace{0.17em}–\hspace{0.17em}x}\\ \text{=1.00\hspace{0.17em}–\hspace{0.17em}0.75}\\ \text{=0.25\hspace{0.17em}M}\end{array}$

$\begin{array}{c}\left[\text{HI}\right]\text{=1.00+2x}\\ \text{=1.00\hspace{0.17em}+2×\hspace{0.17em}0.75}\\ \text{=1.00\hspace{0.17em}+1.50}\\ \text{=2.50\hspace{0.17em}M}\end{array}$

Hence, the equilibrium concentrations of H₂, I₂ and HI are 0.25 M, 0.25 M and 2.50 M respectively.