At 273K and one atm, ‘a’ litre of N _ 2 O _ 4 decomposes to N O _ 2 as : N _ 2 O _ 4 (g) ⇌ 2N O _ 2 (g) . To what extent has the decomposition proceeded when he original volume is 25% less then hat of existing volume?

Correct answer: 33.33 (∵ Moles ∝ volume a consant P and T ) _ array ll InitialMoles Molesatequilibrium array N _ 2 O _ 4 _ array ll a (a-x) array ⇌ 2NO _ 2 ( g ) _ array ll 0 2x array Given original volume = (75)/(100) of existing volume at equilibrium or initial moles =( 75 100 )× moles at equilibrium ∴ a = (75)/(100) ( a + x ) or x = (25)/(75)a = 0 · 33.33a ∴ Degreeofdissociation, α = (x)/(a) = 0.3333a a = 0.3333or33.33%

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

At 273K and one atm, ‘a’ litre of N _ 2 O _ 4 decomposes to N O _ 2...

273K

and one atm,

‘a’

litre of

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

decomposes to

N{O}_{2}

as :

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

To what extent has the decomposition proceeded when he original volume is 25% less then hat of existing volume?
[Report your answer up to two decimal places.]

Correct answer

33.33

Step-by-step solution

\left(∵ Moles ∝ volume\text{a consant P\text{ and } T}\right)

\mathop{}_{\begin{\text{ array }}{\text{ ll }}\text{I}\text{n}\text{i}\text{t}\text{i}\text{a}\text{l}\text{M}\text{o}\text{l}\text{e}\text{s} \\ \text{M}\text{o}\text{l}\text{e}\text{s}\text{a}\text{t}\text{e}\text{q}\text{u}\text{i}\text{l}\text{i}\text{b}\text{r}\text{i}\text{u}\text{m}\end{\text{ array }}}\mathop{{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}}_{

\begin{array}{ll}\text{a} \\ \text{(}\text{a}\text{-}\text{x}\text{)}\end{array}

} ⇌ \mathop{{\text{2NO}}_{\text{2}} ( g )}_{

\begin{array}{ll}\text{0} \\ 2x\end{array}

}

Given original volume

= \dfrac{75}{100}

of existing volume at equilibrium
or initial moles

\text{=}\left(\dfrac{\text{75}}{\text{100}}\right)\text{\times moles at equilibrium}

∴ a = \dfrac{75}{100} ( a + x )

x = \dfrac{25}{75}a = 0 \cdot 33.33a

∴ Degreeofdissociation, \alpha = \dfrac{x}{a}

= \dfrac{0.3333\text{a}}{\text{a}} = 0.3333\text{or}33.33\text{%}

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