• # question_answer In the reaction $PC{{l}_{5(g)}}$$\rightleftharpoons$ $PC{{l}_{3(g)}}$$+C{{l}_{2(g)}}.$ The equilibrium concentrations of $PC{{l}_{5}}$ and $PC{{l}_{3}}$ are 0.4 and 0.2 mole/litre respectively. If the value of ${{K}_{c}}$is 0.5 what is the concentration of $C{{l}_{2}}$ in moles/litre A) 2.0 B) 1.5 C) 1.0 D) 0.5

[c] ${{K}_{c}}=\frac{[PC{{l}_{3}}]\,\,[C{{l}_{2}}]}{[PC{{l}_{5}}]}=\frac{0.2\times x}{0.4}=0.5$, $x=1$