Question

The equilibrium constant for the equilibrium $PCl _{5}(g) \rightleftharpoons PCl _{3}(g)+ Cl _{2}(g)$ at a particular temperature is $2 \times 10^{-2}\, mol L ^{-1}$. The number of moles of $PCl _{5}$ that must be taken in a one litre flask at the same temperature to obtain a concentration of $0.20$ mole of chlorine at equilibrium is

• A. 2.2
• B. 2
• C. 1.8
• D. 0.2

Answer 1

Answer: (A)

2.2

Answer 2

So, equilibrium constant, $K=\frac{\left[ PCl _{3}\right]\left[ Cl _{2}\right]}{\left[ PCl _{5}\right]}$ [given, $K=2 \times 10^{-2} mol / L$] $\therefore 2 \times 10^{-2}= \frac{0.2 \times 0.2}{x-0.2}$ $\Rightarrow x=2.2$ moles