At 273 K and 1atm , 10 litre of `N_(2)O_(4)` decompose to `NO_(4)` decompoes to ` NO_(2)` according to equation
`N_(2)O_(4)(g)hArr2NO_(@)(G)`
What is degree of dissociation `(alpha)` when the original volume is 25% less then that os existing volume?

Answer

A, C

Explanation

`{:(,N_(2)O_(4)(g),hArr,2NO_(2)(g)),("moles before",a,,0),("equilibrium",,,),("moles at equilibrium",(a-x),,2x):}`
Given that original volume `=75/100xx` existing volume at equilibrium
Since, moles `prop` volume (at constant P and T)
`:.` Initial moles `=75/100xx` moles at equilibrium
`a=75/100(a+x) :. X=25/75 a=0.33a`
Now, `%` decomposition `(alpha)=x/a=(0.33a)/a=0.33` or `33%`