Question
$$PC{l_5},PC{l_3}$$ and $$C{l_2}$$ are at equilibrium at $$500\,K$$ in a closed container and their concentrations are $$0.8 \times {10^{ - 3}}\,mol\,{L^{ - 1}},$$ $$1.2 \times {10^{ - 3}}\,mol\,{L^{ - 1}}$$ and $$1.2 \times {10^{ - 3}}mol\,{L^{ - 1}}$$ respectively. The value of $${K_c}$$ for the reaction : $$PC{l_{5\left( g \right)}} \rightleftharpoons PC{l_{3\left( g \right)}} + C{l_{2\left( g \right)}}$$ will be
A.
$$1.8 \times {10^3}\,mol\,{L^{ - 1}}$$
B.
$$1.8 \times {10^{ - 3}}\,mol\,{L^{ - 1}}$$
C.
$$1.8 \times {10^{ - 3}}\,L\,mo{l^{ - 1}}$$
D.
$$0.55 \times {10^4}\,L\,mo{l^{ - 1}}$$
Answer :
$$1.8 \times {10^{ - 3}}\,mol\,{L^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{For the reaction:}} \cr
& {\text{PC}}{{\text{l}}_{5\left( g \right)}} \rightleftharpoons PC{l_{3\left( g \right)}} + C{l_{2\left( g \right)}} \cr
& {K_c} = \frac{{\left[ {PC{l_3}} \right]\left[ {C{l_2}} \right]}}{{\left[ {PC{l_5}} \right]}} \cr
& \,\,\,\,\,\,\,\,\, = \frac{{{{\left( {1.2 \times {{10}^{ - 3}}} \right)}^2}}}{{0.8 \times {{10}^{ - 3}}}} \cr
& \,\,\,\,\,\,\,\,\, = 1.8 \times {10^{ - 3}}\,mol\,{L^{ - 1}} \cr} $$