Question

For the reaction : $P\rightleftharpoons Q+R$. Initially, 2 moles of P was taken. Up to equilibrium $0.5$ moles of P was dissociated. What would be the degree of dissociation?

• A. $0.5$
• B. $1$
• C. $0.25$
• D. $4.2$

Answer 1

The correct option is C $0.25$

$\underset{\text{At equillibrium}:}{\underset{\text{Initially}:}{}}\underset{2(1-\alpha )M}{\underset{2}{2P}}\rightleftharpoons \underset{2\alpha }{\underset{0}{2Q}}+\underset{2\alpha }{\underset{0}{2R}}$

Given that at equillibrium, 0.5 mole of $P$ was dissociated.

$\therefore$ No. of moles of $P$ left at equillibrium $=2-0.5=1.5=2(1-\alpha )$

$\Rightarrow 2-2\alpha =1.5$

$\Rightarrow 2\alpha =2-1.5$

$\Rightarrow \alpha =\frac{0.5}{2}=0.25$

Hence the degree of dissociation is $0.25$.