JEE Main 2019 · Chemistry · Chemical Equilibrium · medium

N2O4(g) rightleftharpoons 2NO2(g) is 20% dissociated at a total pressure of 1 atm at a fixed temperature. If the total pressure is lowered to 0.25…

Single correct option

\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO_2(g)}

20\%

dissociated at a total pressure of

1\ \mathrm{atm}

at a fixed temperature. If the total pressure is lowered to

0.25\ \mathrm{atm}

at the same temperature, the degree of dissociation becomes nearly

Show answer and solution

Answer: (B) $0.38$

With initial 1 mol and degree

\alpha

, total moles

=1+\alpha

, so

K_p = \dfrac{(2\alpha P/(1+\alpha))^2}{(1-\alpha)P/(1+\alpha)} = \dfrac{4\alpha^2 P}{1-\alpha^2}

. At

P = 1

\alpha = 0.2

K_p = 4(0.04)/0.96 = 1/6

. At

P=0.25

\dfrac{\alpha^2}{1-\alpha^2} = \dfrac{1/6}{1} = \dfrac16

, so

\alpha^2 = 1/7

and

\alpha = 0.378 \approx 0.38

. Lower pressure favours the side with more gas moles, so

\alpha

must rise above

0.20

Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake