At `273K` and `1` atm `a L` of `N_(2)O_(4)` decomposes to `NO_(2)` according to equation `N_(2)O_(4)(g) hArr 2NO_(2)(g)`. To what extent ha the decomposition proceeded when the original volume is `25%` less than that of exisiting 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%`