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

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

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

Let the change in concentration be x

Therefore, equilibrium concentration of all the species will be

\(\begin{array}{*{20}{c}}{[HOCl] = 2x}\\{\,\,\,\,\,\,\left[ {{H_2}O} \right] = 1 - x}\\{\,\,\,\,\,\left[ {C{l_2}O} \right] = 1 - x}\end{array}\)

Equilibrium constant can be written as:

\(\begin{array}{*{20}{c}}{\,\,\,\,\,\,\,\,\,{K_c} = \frac{{{{[HOCl]}^2}}}{{\left[ {{H_2}O} \right] \times \left[ {C{l_2}O} \right]}}}\\{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.0900 = \frac{{{{(2x)}^2}}}{{(1.00 - x) \times (1.00 - x)}}}\\{0.0900 = \frac{{4{x^2}}}{{1.00 - 2x + {x^2}}}}\end{array}\)

\(\begin{array}{*{20}{c}}{4{x^2} = 0.09 - 0.18x + 0.09{x^2}}\\{\,\,\,\,\,0 = 3.91{x^2} + 0.18x - 0.09}\\{{\rm{\;Using equation solver, we get\;}}}\\{x = 0.13{\rm{M}}}\end{array}\)

Therefore, equilibrium concentrations are:

\(\begin{array}{*{20}{c}}{[{\rm{HOCl}}] = 2{\rm{x}} = 0.26{\rm{M}}}\\{\,\,\,\,\,\,\left[ {{{\rm{H}}_2}{\rm{O}}} \right] = 1 - {\rm{x}} = 0.87{\rm{M}}}\\{\,\,\,\,\left[ {{\rm{C}}{{\rm{l}}_2}{\rm{O}}} \right] = 1 - {\rm{x}} = 0.87{\rm{M}}}\end{array}\)