For the equilibrium N 2 O 4 g - 2 N O 2 g - The vapour density at 60- C was found to be 30-15 - The degree of dissociation of N 2 O 4 isA- 0-525B- 0-75 C- 0-3015D- 0-92

The correct option is A 0.525 For an hypothetical reaction A_n ⇌ nA Degree of dissociation α = D d/d(n 1) where d = observed vapour density D = calcu ...

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Standard X

Chemistry

Industrial Preparation of Ammonia

For the equil...

Question

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

0.525

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0.75

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0.3015

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0.92

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Solution

The correct option is A
0.525

For an hypothetical reaction

$A_{n} ⇌ n A$

Degree of dissociation $α = \frac{D - d}{d (n - 1)}$

where d = observed vapour density

D = calculated vapour density

For the reaction

$N_{2} O_{4} (g) ⇌ 2 N O_{2}; n = 2$

$D_{N_{2} O_{4}} = \frac{m o l a r m a s s o f N_{2} O_{4}}{2} = \frac{92}{2} = 46$

$α = \frac{46 - 3015}{30.15 (2 - 1)} = 0.525$

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Similar questions

Q. 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:

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. The degree of dissociation of

N_{2} O_{4}

obeying the following equilibrium is approximately related to the pressure at equilibrium by:

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

Q. Vapour density of

N_{2} O_{4}

which dissciates according to teh equilibrium

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

25.67

100^{o} C

and a pressure of

1 a t m

. Calculate the degree dissociatrion for the reaction.

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For the equilibrium N2O4(g)⇌2NO2(g). The vapour density at 60∘C was found to be 30.15.The degree of dissociation of N2O4 is

The correct option is A 0.525 For an hypothetical reaction An⇌nA Degree of dissociation α=D−dd(n−1) where d = observed vapour density D = calculated vapour density For the reaction N2O4(g)⇌2NO2 ; n=2 DN2O4=molar mass of N2O42=922=46 α=46−301530.15(2−1)=0.525

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