For the equilibrium N-2O-4-g- - 2NO-2-g- mark the correct option that shows the correct relationship of the degree of dissociation - - - with D-d- where D is vapour density of N-2O-4 and d is vapour density of equilibrium mixture-

The correct option is D m_N_2O_4/m_mix = D_N_2O_4/d_m = 1 + α nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; (N_2O_ ...

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Byju's Answer

Standard XII

Chemistry

Boyle's Law

For the equil...

Question

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.

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Solution

The correct option is D
$\frac{m_{N_{2} O_{4}}}{m_{m i x}} = \frac{D_{N_{2} O_{4}}}{d_{m}} = 1 + α$

$(N_{2} O_{4} ⇌ 2 N O_{2})$
$10$
$1 - α 2 α$

$∴ 1 + α = \frac{D}{d}$ ....(i)
From the equation, it can be determined that the correct graph is shown by option (D).

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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:

Q. Before equilibrium is set-up for the chemical reaction

N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)

, vapour density

d

of the gaseous mixture was measured. If

D

is the theoretical value of vapour density, variation of

α

with

D / d

is given by the graph. What is the value of

D / d

at point

A

Q. The numerical value of

d / D

when

α = 0

for

P C l_{5} ⇌ P C l_{3} + C l_{2}

; where

d

are vapour density of equilibrium mixture and

D

is vapour density of

P C l_{5}

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?

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For the equilibrium N2O4(g)⇌2NO2(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 N2O4 and d is vapour density of equilibrium mixture.

The correct option is D mN2O4mmix=DN2O4dm=1+α (N2O4⇌2NO2) 10 1−α2α ∴1+α=Dd....(i) From the equation, it can be determined that the correct graph is shown by option (D).

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.

The correct option is D
$\frac{m_{N_{2} O_{4}}}{m_{m i x}} = \frac{D_{N_{2} O_{4}}}{d_{m}} = 1 + α$

$(N_{2} O_{4} ⇌ 2 N O_{2})$
$10$
$1 - α 2 α$

$∴ 1 + α = \frac{D}{d}$ ....(i)
From the equation, it can be determined that the correct graph is shown by option (D).