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Standard XII

Chemistry

Reaction Quotient

The equilibri...

Question

The equilibrium constant $K_{p}$ for the reaction $N_{2} O_{4} (g) ⇌ {2 N O}_{2} (g)$ is $4.5$ . What would be the average molar mass (in g/mol) of an equilibrium mixture of $N_{2} O_{4}$ and ${N O}_{2}$ formed by the dissociation of pure $N_{2} O_{4}$ at a total pressure of $2$ atm?

69

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57.5

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80.5

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85.5

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Solution

The correct option is B $57.5$
$N_{2} O_{4} (g) ⇌ {2 N O}_{2} (g)$
At eqm. $1 - a$ $2 a$

So, the total number of moles $= 1 + a$ ;

$K_{p} = \frac{P_{N O_{2}}^{2}}{P_{N_{2} O_{4}}} = \frac{4 a^{2} P_{t}}{(1 + a) (1 - a)} = \frac{4 a^{2} \times 2}{(1 - a^{2})} = 4.5;$

$a = 0.60$ .

So, the average molar mass of the mixture $= \frac{(1 - 0.6)}{1.6} \times 92 + \frac{2 \times 0.6}{1.6} \times 46 = 57.5$ .

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Similar questions

Q. The equilibrium constant

K_{P}

for the reaction

N_{2} O_{4} (g) ⇌ 2 N O_{2}

(g) is

4.5

.
What would be the average molar mass (in g⁄mol) of an equilibrium mixture of

N_{2} O_{4}

and

N O_{2}

formed by the dissociation of pure

N_{2} O_{4}

at a total pressure of 2 atm?

Q. The equilibrium constant

K_{p}

for the reaction

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

4.5

What would be the average molar mass (in g/mol) of an equilibrium mixture of

N_{2} O_{4}

and

N O_{2}

formed by the dissociation of pure

N_{2} O_{4}

at a total pressure of

2

atm?

Q. At the equilibrium of the reaction,

N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)

, the observed molar mass of

N_{2} O_{4}

77.70 g

. The percentage dissociation of

N_{2} O_{4}

is:

Q. Gaseous

N_{2} O_{4}

dissociates into gaseous

{N O}_{2}

according to the reaction

N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)

. At

300 K

and

1 a t m

pressure, the degree of dissociation of

N_{2} O_{4}

0.2

. If one mole of

N_{2} O_{4}

gas is contained in a vessel, then the density of the equilibrium mixture is:

9.2 g r a m s

N_{2} O_{4} (g)

is taken in a closed one-litre vessel and heated till the following equilibrium is reached.

N_{2} O_{4} (g) ⇌ 2 {N O}_{2} (g)

At equilibrium,

50

N_{2} O_{4} (g)

is dissociated. What is the equilibrium constant? (in mol litre

^{- 1}

)

[Molecular weight of

N_{2} O_{4} = 92

]

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The equilibrium constant Kp for the reaction N2O4(g)⇌2NO2(g) is 4.5. What would be the average molar mass (in g/mol) of an equilibrium mixture of N2O4 and NO2 formed by the dissociation of pure N2O4 at a total pressure of 2 atm?

The correct option is B 57.5 N2O4(g)⇌2NO2(g) At eqm. 1−a 2aSo, the total number of moles =1+a; Kp=P2NO2PN2O4=4a2Pt(1+a)(1−a)=4a2×2(1−a2)=4.5; a=0.60.So, the average molar mass of the mixture =(1−0.6)1.6×92+2×0.61.6×46=57.5.

The equilibrium constant $K_{p}$ for the reaction $N_{2} O_{4} (g) ⇌ {2 N O}_{2} (g)$ is $4.5$ . What would be the average molar mass (in g/mol) of an equilibrium mixture of $N_{2} O_{4}$ and ${N O}_{2}$ formed by the dissociation of pure $N_{2} O_{4}$ at a total pressure of $2$ atm?