question_answer 1)

  \[PC{{l}_{5}},PC{{l}_{3}}\] 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 \[PC{{l}_{5}}(g)\underset{{}}{\leftrightarrows}PC{{l}_{3}}(g)+C{{l}_{2}}(g)\]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) \[0.55\times {{10}^{4}}mol{{L}^{-1}}\]

Answer:

  (b)\[{{K}_{c}}=\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}=\frac{1.2\times {{10}^{3}}\times 1.2\times {{10}^{-3}}}{0.8\times {{10}^{-3}}}\] \[=1.8\times {{10}^{-3}}mol\,{{L}^{-1}}\]