A) 0.013
B) 0.050
C) 0.033
D) 0.067
E) 0.045
Correct Answer: C
Solution :
\[PC{{l}_{5}}PC{{l}_{3}}+C{{l}_{2}}\] \[\begin{matrix} 1.6 & 0 & \,\,\,\,\,\,\,\,\,0 & Initially \\ (1.6-x) & x & \,\,\,\,\,\,\,\,\,x & At\,equilibrium \\ \left( \frac{1.6-x}{4} \right)\,\,\, & \frac{x}{4} & \,\,\,\,\,\,\,\,\,\frac{x}{4} & {} \\ \end{matrix}\] Given, \[1.6-x=1.2\] \[x=0.4\] \[[PC{{l}_{5}}]=\frac{1.2}{4}\] \[[PC{{l}_{3}}]=\frac{0.4}{4}\] \[[C{{l}_{2}}]=\frac{0.4}{4}\] \[{{K}_{c}}=\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}\] \[=\frac{\left( \frac{0.4}{4} \right)\left( \frac{0.4}{4} \right)}{\left( \frac{1.2}{4} \right)}\] \[=\frac{0.1\times 0.1}{0.3}\] \[=0.033\]
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