Gaseous N _ 2 O _ 4 dissociates into gaseous N O _ 2 according to the reaction N _ 2 O _ 4 (g) ⇌ 2N O _ 2 (g) At 300K and 1atm pressure, the degree of dissociation of N _ 2 O _ 4 is 0 . 2 . If one mole of N _ 2 O _ 4

Correct answer: A. 3 . 11g / L cases N _ 2 O _ 4 & ⇌ & 2N O _ 2 (1 - α) & & 2α cases Total moles at equilibrium = 1 - α + 2α = 1 + α = 1.2 M _ avg for equilibrium mixture = (0.8 × 92 + 0.4 × 46)/(1.2) = 76 . 67g / mol d _ average = P M _ avg RT = (1 × 76.67)/(0.082 × 300) = (76.67)/(24.6) = 3.11g L ^ - 1

JEE Main 201510 AprilOnline

Exam: JEE Main
Year: 2015
Date: 10 April
Mode: Online
Subject: Chemistry
Chapter: Chemical Equilibrium

JEE Main 2015 · 10 April · Online · Chemistry · Chemical Equilibrium

Gaseous N _ 2 O _ 4 dissociates into gaseous N O _ 2 according to...

Gaseous

{N}_{2}{O}_{4}

dissociates into gaseous

N{O}_{2}

according to the reaction

{N}_{2}{O}_{4}\left(g\right) ⇌ 2N{O}_{2}\left(g\right)

300K

and

1atm

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:

Options

A $3 . 11g / L$
B $1 . 56g / L$
C $4 . 56g / L$
D $6 . 22g / L$

Correct answer

A. $3 . 11g / L$

Step-by-step solution

\begin{cases}{N}_{2}{O}_{4} & ⇌ & 2N{O}_{2} \\ \left(1 - \alpha\right) & & 2\alpha\end{cases}

Total moles at equilibrium

= 1 - \alpha + 2\alpha = 1 + \alpha = 1.2

{M}_{avg}

for equilibrium mixture

= \dfrac{0.8 \times 92 + 0.4 \times 46}{1.2}

= 76 . 67g / mol

{d}_{average} = \dfrac{P{M}_{avg}}{RT} = \dfrac{1 \times 76.67}{0.082 \times 300} = \dfrac{76.67}{24.6}

= 3.11g{L}^{- 1}

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