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Raoult's Law

For the react...

Question

For the reaction $N_{2} O_{4} ⇌ 2 N O_{2}$ , equilibrium mixture contains $N O_{2}$ at $P = 1.1 a t m$ & $N_{2} O_{4}$ at $P = 0.28 a t m$ at 350 K. The volume of the container is doubled. Calculate the equilibrium pressures of the two gases when the system reaches new equilibrium :

A

P_{N O_{2}} = 0.32 a t m, P_{N_{2} O_{4}} = 0.095 a t m

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B

P_{N O_{2}} = 0.64 a t m, P_{N_{2} O_{4}} = 0.049 a t m

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C

P_{N O_{2}} = 0.64 a t m, P_{N_{2} O_{4}} = 0.05 a t m

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D

P_{N O_{2}} = 0.32 a t m, P_{N_{2} O_{4}} = 0.049 a t m

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Solution

The correct option is C $P_{N O_{2}} = 0.64 a t m, P_{N_{2} O_{4}} = 0.05 a t m$
For the first equilibrium, the total pressure is $1.1 + 0.28 = 1.38 a t m$
The value of the equilibrium constant will be $K_{p} = \frac{P_{N O_{2}}^{2}}{P_{N_{2} O_{4}}} = \frac{{1.1}^{2}}{0.28} = 4.32$
For the second equilibrium, since the volume of the container is doubled, the total pressure will be one half i.e $\frac{1.38}{2} = 0.69 a t m$
The partial pressures of $N_{2} O_{4}$ and $N O_{2}$ at equilibrium will be $0.69 - x$ atm and $x$ atm respectively.
The value of the equilibrium constant will be $K_{p} = \frac{P_{N O_{2}}^{2}}{P_{N_{2} O_{4}}} = \frac{x^{2}}{0.69 - x} = 4.32$
$P_{N O_{2}} = x = 0.64 a t m$ and $P_{N_{2} O_{4}} = 0.69 - x = 0.05 a t m$

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