Question

What are all concentrations after a mixture that contains $\left[{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\right]=\mathbf{1}.\mathbf{00}\mathbf{M}\mathbf{a}\mathbf{n}\mathbf{d}\left[\mathbf{C}{\mathbf{l}}_{\mathbf{2}}\mathbf{O}\right]=\mathbf{1}.\mathbf{00}\mathbf{M}$ comes to equilibrium at ${\mathbf{25}}^{\circ }\mathbf{C}$?

${\mathbf{H}}_{\mathbf{2}}\mathbf{O}(g)+\mathbf{C}{\mathbf{l}}_{\mathbf{2}}\mathbf{O}(g)\rightleftharpoons \mathbf{2}\mathbf{HOCl}(g);\mathbf{Kc}=\mathbf{0}.\mathbf{0900}$

Short answer

Equilibrium concentration of all species are:

$\begin{array}{c}[HOCl]=2x=0.26M\\ \left[H_{2}O\right]=1-x=0.87M\\ \left[C{l}_{2}O\right]=1-x=0.87M\end{array}$

Step-by-step solution

Define concentration.

It is a constituent divided by total volume of a mixture.

Find equilibrium concentration

Let the change in concentration be x

Therefore, equilibrium concentration of all the species will be

$\begin{array}{c}[HOCl]=2x\\ \left[H_{2}O\right]=1-x\\ \left[C{l}_{2}O\right]=1-x\end{array}$

Equilibrium constant can be written as:

$\begin{array}{c}{K}_c=\frac{{[HOCl]}^2}{\left[H_{2}O\right]\times \left[C{l}_{2}O\right]}\\ 0.0900=\frac{{(2x)}^2}{(1.00-x)\times (1.00-x)}\\ 0.0900=\frac{4x^2}{1.00-2x+x^2}\end{array}$

$\begin{array}{c}4x^2=0.09-0.18x+0.09x^2\\ 0=3.91x^2+0.18x-0.09\\ \mathrm{U}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{r},\mathrm{w}\mathrm{e}\mathrm{g}\mathrm{e}\mathrm{t}\\ x=0.13\mathrm{M}\end{array}$

Therefore, equilibrium concentrations are:

$\begin{array}{c}[\mathrm{H}\mathrm{O}\mathrm{C}\mathrm{l}]=2\mathrm{x}=0.26\mathrm{M}\\ \left[{\mathrm{H}}_2\mathrm{O}\right]=1-\mathrm{x}=0.87\mathrm{M}\\ \left[\mathrm{C}{\mathrm{l}}_2\mathrm{O}\right]=1-\mathrm{x}=0.87\mathrm{M}\end{array}$