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A1 273 K and I atm, I Lof N(2)O(4(g)) de...

A1 273 K and I atm, I Lof `N_(2)O_(4(g))` decomposes to `NO_(2(g))` as given, `N_(2)O_(4(g)) 2NO_(2(g))`, At equilibrium . original volume is 25% lessthan the exisiting volume percentage decomposition of `N_(2)O_(4(g))` is thus,

A

0.25

B

0.5

C

0.666

D

0.3333

Text Solution

Verified by Experts

The correct Answer is:
D
PV=nRT, `V=(n)/(P)RT` at constant P & T , `V alpha n`
Thus, moles can be expressed in terms of volume
`underset(1-x)underset(1)(N_(2)O_(4(g))) harr underset(2x)underset(0)(2NO_(2(g)))`
Total volume at equilibrium = (1+x)
Given `(75)/(100)(1+x)=1 implies x=(1)/(3)=0.333` = 33.3% dissociation
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