$x$ mol $N_{2} O_{4}$ is taken at $P_{1}$ atm in a closed vessel & heated. When $75 % N_{2} O_{4}$ dissociated at equilibrium, total pressure is found to be $P_{2}$ atm. The relation between $P_{2} & P_{2}$ is:

P_{1} : P_{2} = 7 : 4

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P_{1} : P_{2} = 7 : 2

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P_{1} : P_{2} = 4 : 7

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P_{1} : P_{2} = 3 : 4

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Solution

The correct option is D $P_{1} : P_{2} = 4 : 7$
$N_{2} O_{4} = 3 N O_{2}$
at $t = 0$ $x$ $0$
at $t = e q m$ $n (1 - α)$ $2 x α$
$α = 0.75$ as $75$ dissociation
$∴$ at $t = e q m$ $x (1 - 0.75)$ $2 (x) (0.75)$
$= 0.25 x$ $= 1.5 x$
$∴$ total moles $= 0.25 x + 1.5 x = 1.75 x$
pressure $α$ no. of moles
$∴ \frac{P_{1}}{P_{2}} = \frac{x}{1.75 x}$
$∴ \frac{P_{1}}{P_{2}} = \frac{1}{1.75} = \frac{2}{3.5}$
$= \frac{4}{7}$

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x mol N2O4 is taken at P1 atm in a closed vessel & heated. When 75%N2O4 dissociated at equilibrium, total pressure is found to be P2 atm. The relation between P2&P2 is:

The correct option is D P1:P2=4:7 N2O4=3NO2at t=0 x 0at t=eqm n(1−α) 2xαα=0.75 as 75 dissociation∴ at t=eqm x(1−0.75) 2(x)(0.75) =0.25x =1.5x∴ total moles =0.25x+1.5x=1.75xpressure αno. of moles∴P1P2=x1.75x∴P1P2=11.75=23.5=47

$x$ mol $N_{2} O_{4}$ is taken at $P_{1}$ atm in a closed vessel & heated. When $75 % N_{2} O_{4}$ dissociated at equilibrium, total pressure is found to be $P_{2}$ atm. The relation between $P_{2} & P_{2}$ is: