An equilibrium mixture of ${NO}_{2} (g)$ and $N_{2} O_{4} (g)$ is present in a closed container at 300 K With pressures 0.4 atm and 0.2 atm respectively. On doubling the volume of container > the presaure of ${NO}_{2} (g)$ at new equilibrium at 300 K will be:

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

0.2 atm

0.35 atm

0.19 atm

0.25 atm

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

The reaction takes place as,

$N_{2} O_{4} ⇌ 2 {NO}_{2} K_{P_{1}} = \frac{(0.4)^{2}}{0.2} = 0.8$

$\begin{array}{lcc} Old eq. & 0.2 & 0.4 \\ New eq. & 0.1 - x & 0.2 + 2 x \end{array}$

$\frac{(0.2 + 2 x)^{2}}{(0.1 - x)} = 0.8$

$P_{{NO}_{2}} = 0.25 atm$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE