Question

The equilibrium constant for the equilibrium $PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$ at a particular temperature is $2\times {10}^{-2}mol{L}^{-1}$. The number of moles of $PC{l}_5$ 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. $2.0$
• C. $1.6$
• D. $0.2$
• E. $0.1$

Answer 1

The correct option is A $2.2$

$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
Moles initial $x00$
Moles at equilibrium $x-0.20.20.2$

So, equilibrium constant,

$K=\frac{\left[PC{l}_3\right]\left[C{l}_2\right]}{\left[PC{l}_5\right]}$ [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

Hence, the correct option is $\text{A}$