Question

For the reaction : $N_2{O}_{3\left(g\right)}\rightleftharpoons {NO}_{\left(g\right)}+{NO}_{2\left(g\right)};totalpressure=P,degreeofdissociation=$ 50%. Then $K_{p}wouldbe-$

(A) $3P$

(B) $2P$

(C) $\frac{P}{3}$

(D) $\frac{P}{2}$

Answer 1

Solution:- (V) $\frac{P}{3}$

$\underset{\text{At equilibrium}}{\underset{\text{Initially}}{}}\underset{\underset{1-\alpha }{1}}{{N_2{O}_3}_{\left(g\right)}}\rightleftharpoons \underset{\underset{\alpha }{0}}{{NO}_{\left(g\right)}}+\underset{\underset{\alpha }{0}}{{N{O}_2}_{\left(g\right)}}$

Given:-

Total pressure $=P$

Degree of dissociation $=50\%\Rightarrow \alpha =0.5$

Final moles $=1-\alpha +\alpha +\alpha =1+\alpha =1+0.5=1.5$

Therefore, at equilibrium-

$P_{N_2{O}_3}=P\left(\frac{0.5}{1.5}\right)=\frac{P}{3}$

$P_{NO}=P\left(\frac{0.5}{1.5}\right)=\frac{P}{3}$

$P_{N{O}_2}=P\left(\frac{0.5}{1.5}\right)=\frac{P}{3}$

$K_p=\frac{P_{N{O}_2}\times P_{NO}}{P_{N_2{O}_3}}$

$\Rightarrow K_p=\frac{\left(\frac{P}{3}\right)\times \left(\frac{P}{3}\right)}{\left(\frac{P}{3}\right)}$

$\Rightarrow K_p=\frac{P}{3}$