Question

The K${}_{\rho }$ for the reaction $N_2{O}_4\rightleftharpoons 2N{O}_2$ is 640 mm at 775 K. The percentage dissociation of $N_2{O}_4$ at equilibrium pressure of 160 mm is:

• A. 80%
• B. 30%
• C. 50%
• D. 70%

Answer 1

The correct option is D 70%

Mole before equilibrium $N_2{O}_4\rightleftharpoons 2N{O}_2$
Mole at equilibrium 1
(1 - x) 2x
$K_{\rho }=\frac{4x^2}{(1-x)}\times {\left[\frac{P}{\sum n}\right]}^{\Delta n}$
$640=\frac{4x^2}{(1-x)}\frac{160}{(1+x)}$
$4=\frac{4x^2}{(1-x)}$ or $1-x^2=x^2$ or $2x^2=1$
$\therefore x^2=1/2$ or $x=0.707=70.7$%