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Reaction Quotient

A vessel cont...

Question

A vessel contains $N_{2} O_{4}$ and $N O_{2}$ in 2:3 molar ratio at 10 atm under equlibrium. Now $K_{p}$ for $N_{2} O_{4} ⇌ 2 N O_{2}$ is:

A

9 a t m^{- 1}

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B

9 a t m

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C

4.5 a t m^{2}

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D

10 a t m

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Solution

The correct option is B $9 a t m$
The given reaction is :-
$N_{2} O_{4} ⇌ 2 {N O}_{2}$
Given that eqm pressure is $10$ atm and $N_{2} O_{4}$ and ${N O}_{2}$ are in $2 : 3$ molar ratio.
So, Pressure of $N_{2} O_{4} = \frac{2}{5} \times 10 = 4 a t m$
Pressure of ${N O}_{2} = \frac{3}{5} \times 10 = 6 a t m$
$K_{P} = \frac{{(p {N O}_{2})}^{2}}{p N_{2} O_{4}} = \frac{6 \times 6}{4} = 9 a t m$

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Similar questions

Q. A vessel contains

N_{2} O_{4}

N O_{2}

in 2:3 molar ratio at 10 atm under equilibrium. now

K_{p}

N_{2} O_{4} ⇌ 2 N O_{2}

Q. A vessel contains

N_{2} O_{4}

{N O}_{2}

2 : 3

10

atm under equilibrium, now

K_{p}

N_{2} O_{4} ⇌ 2 {N O}_{2}

Q. Find moles of

N_{2} O_{4}

N O_{2}

at equilibrium are

1

2

respectively total pressure at equilibrium is

9 a t m

K_{P}

for the reaction

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

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N_{2}, N O_{2}

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. On heating to a temperature at which all the

N_{2} O_{4}

dissociated as :

N_{2} O_{4} \to 2 N O_{2}

The mean molar mass of mixture becomes

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. The molar ratio of

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2 N O_{2} ⇌ N_{2} O_{4} K_{p} = 6.8 a t m^{- 1}

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