What are the concentrations of \(PC{l_5}\), \(PC{l_3}\), and \(C{l_2}\) in an equilibrium mixture produced by the decomposition of a sample of pure \(PC{l_5}\) with \([PC{l_5}] = 2.00M?\) \(PC{l_5}(g) \rightleftharpoons PC{l_3}(g) + {\mathbf{C}}{{\mathbf{l}}_{\mathbf{2}}}(g)\,\,\,\,\,\,\,\;{\mathbf{Kc}} = {\mathbf{0}}.{\mathbf{0}}211\)

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

\(\begin{array}{*{20}{c}}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{PC}}{{\rm{l}}_5}} \right] = 2.00 - {\rm{x}}}\\{\left[ {{\rm{PC}}{{\rm{l}}_3}} \right] = {\rm{x}}}\\{\,\,\,\left[ {{\rm{C}}{{\rm{l}}_2}} \right] = {\rm{x}}}\end{array}\)

\(\begin{array}{*{20}{c}}{{K_c} = \frac{{\left[ {PC{l_3}} \right] \times \left[ {C{l_2}} \right]}}{{\left[ {PC{l_5}} \right]}}}\\{0.0211 = \frac{{x \times x}}{{2.00 - x}}}\end{array}\)

\(\begin{array}{*{20}{c}}{{x^2} = 0.0422 - 0.0211x}\\{0 = {x^2} + 0.0211x - 0.0422}\end{array}\)

Using equation solver, we get,

\(x = 0.195{\rm{M}}\)

Therefore, equilibrium concentrations are:

\(\begin{array}{*{20}{c}}{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\rm{PC}}{{\rm{l}}_5}} \right] = 2.00 - {\rm{x}} = 1.805{\rm{M}}}\\{\left[ {{\rm{PC}}{{\rm{l}}_3}} \right] = {\rm{x}} = 0.195{\rm{M}}}\\{\,\,\,\left[ {{\rm{C}}{{\rm{l}}_2}} \right] = {\rm{x}} = 0.195{\rm{M}}}\end{array}\)