Question

Two moles of $PC{{l}_{5}}$ is heated in a closed vessel of $2L$ capacity. When the equilibrium is attained 40% of it has been found to be dissociated. What is the ${{K}_{c}}$ in $mol/d{{m}^{3}}$ ?

• A. 0.532
• B. 0.266
• C. 0.133
• D. 0.174

Answer 1

Answer: (B)

0.266

Answer 2

Number of moles of $PC{{l}_{5}}$
dissociated at equilibrium $=2\times 40/100=0.8$
$\underset{2\,mol}{\mathop{PC{{l}_{5}}}}\,\underset{0}{\mathop{PC{{l}_{3}}}}\,+\underset{0}{\mathop{C{{l}_{2}}}}\,\underset{(initially)}{\mathop{{}}}\,$
(2-0.8) mol 0.8 mol 0.8 mol (at equilibrium) $[PC{{l}_{5}}]=\frac{1.2}{2}=0.6\,M{{L}^{-1}}$ $[PC{{l}_{3}}]=[C{{l}_{2}}]=\frac{0.8}{2}=0.4\,M{{L}^{-1}}$
$\therefore$ ${{K}_{c}}=\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}=\frac{0.4\times 0.4}{0.6}$
$=0.267\text{ }mol/d{{m}^{3}}$