Exam

NTA Abhyas

Year

2020

Subject

Chemistry

Chapter

Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

For the reaction N _ 2 O _ 4 ⇌ 2N O _ 2(g) , the degree of...

For the reaction ${N}_{2}{O}_{4} ⇌ 2N{O}_{2\left(g\right)}$$, the degree of dissociation of${N}_{2}{O}_{4}$is$0.2$at$1atm . Then the ${K}_{p}$ of $2N{O}_{2} ⇌ {N}_{2}{O}_{4}$ is

Step-by-step solution

${\text{K}}_{\text{p}} = \dfrac{{\left[{\text{P}}_{{\text{NO}}_{\text{2}}}\right]}^{2}}{{\text{P}}_{{\text{N}}_{2} {\text{O}}_{\text{4}}}}$

${\text{K}}_{\text{p}} = \dfrac{\text{4}{\alpha}^{2}}{1 - {\alpha}^{2}} \text{p}$

$⇒ {\text{K}}_{\text{p}} = \dfrac{4 \times {\left(0.2\right)}^{2}}{1 - {\left(0.2\right)}^{2}} = \dfrac{0.16}{0.96}$

$\text{for}2{\text{NO}}_{2} \overset{}{⇌} {\text{N}}_{\text{2}} {\text{O}}_{4} ; {\text{K}}_{\text{p}} = \dfrac{0.96}{0.16} = 6$