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Q.

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:

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a

0.2 atm

b

0.35 atm

c

0.19 atm

d

0.25 atm

answer is D.

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Detailed Solution

The reaction takes place as,

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

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

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

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

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An equilibrium mixture of NO2(g) and N2O4(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 NO2(g) at new equilibrium at 300 K will be: