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Standard XII

Chemistry

Arrhenius Acid

If vapour den...

Question

If vapour density of equilibrium $N_{2} O_{4} (g) ⇋ 2 N O_{2} (g),$ is found to be 31 then the degree of dissociation of $N_{2} O_{4}$ will be:

0.48

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0.25

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0.65

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0.35

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Solution

The correct option is C 0.48
$N_{2} O_{4} ⇌ {N O}_{2}$
$1$ $-$
$1 - x$ $2 x$
$\frac{n_{i n i t i a l}}{n_{f i n a l}} = \frac{M_{f i n a l}}{M_{i n i t i a l}} M_{f i n a l} = 2 \times V O = 32$
$\frac{1}{1 + x} = \frac{62}{92} \Rightarrow x = 0.48$

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Similar questions

For the equilibrium $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ . The vapour density at $60^{\circ} C$ was found to be 30.15.The degree of dissociation of $N_{2} O_{4}$ is

N_{2} O_{4}

dissociates as

N_{2} O_{4 (g)} ⇌ 2 N O_{2 (g)}

273 K

and

2 a t m

pressure. The equilibrium mixture has a density of

41

. What will be the degree of dissociation?

Q. For the equilibrium

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

, mark the correct option that shows the correct relationship of the degree of dissociation -

^{'} α^{'}

with

D / d

; where

D

is vapour density of

N_{2} O_{4}

and

d

is vapour density of equilibrium mixture.

Q. For the reaction

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

, if percentage dissociation of

N_{2} O_{4}

are

20 %, 45 %, 65 %

and

80 %

, then the sequence of observed vapour densities will be:

Vapour density of the equilibrium mixture of

N O_{2}

and

N_{2} O_{4}

is found to be

38.33

. For the equilibrium:

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

Calculate:

a. Abnormal molecular weight.

b. Degree of dissociation.

c. Percentage of

N O_{2}

in the mixture.

K_{P}

for the reaction if total pressure is 2 atm.

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If vapour density of equilibrium N2O4(g)⇋2NO2(g), is found to be 31 then the degree of dissociation of N2O4 will be:

The correct option is C 0.48N2O4⇌NO2 1 −1−x 2xninitialnfinal=MfinalMinitialMfinal=2×VO=3211+x=6292⇒x=0.48

If vapour density of equilibrium $N_{2} O_{4} (g) ⇋ 2 N O_{2} (g),$ is found to be 31 then the degree of dissociation of $N_{2} O_{4}$ will be:

The correct option is C 0.48
$N_{2} O_{4} ⇌ {N O}_{2}$
$1$ $-$
$1 - x$ $2 x$
$\frac{n_{i n i t i a l}}{n_{f i n a l}} = \frac{M_{f i n a l}}{M_{i n i t i a l}} M_{f i n a l} = 2 \times V O = 32$
$\frac{1}{1 + x} = \frac{62}{92} \Rightarrow x = 0.48$