A) 0.532
B) 0.266
C) 0.133
D) 0.174
E) 0.25
Correct Answer: B
Solution :
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) \[[PC{{l}_{3}}]=\frac{1.2}{2}=0.6\,M{{L}^{-1}}\] \[[PC{{l}_{3}}]=[C{{l}_{2}}]=\frac{0.8}{2}=0.4\,M{{L}^{-1}}\] \[\therefore \] \[{{K}_{c}}=\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}=\frac{0.4\times 0.4}{0.6}\] \[=0.267\text{ }mol/d{{m}^{3}}\]You need to login to perform this action.
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