Answer:
(a) \[{{K}_{c}}\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}=8.3\times
{{10}^{-3}}mol\,{{L}^{-1}}\]
(b)\[PC{{l}_{3}}(g)+C{{l}_{2}}(g)\rightleftharpoons
PC{{l}_{5}}(g)\]
\[{{K}_{c}}=\frac{[PC{{l}_{5}}]}{[PC{{l}_{3}}][C{{l}_{2}}]}=\frac{1}{8.3\times
{{10}^{-3}}}\]
\[=120.48L\,mo{{l}^{-1}}\]
(c) (i) \[{{K}_{c}}\]
is unaffected by addition of \[PC{{l}_{5}}\].
(ii) The reaction is endothermic, hence the equilibrium constant\[{{K}_{c}}\]
will increase with increase in temperature.
You need to login to perform this action.
You will be redirected in
3 sec