In the following reaction started only with `A_(B), 2A_(B)(g) hArr3A_(2)(g)+A_(4)(g)` mole fraction of `A_(2)` is found to `0.36` at a total pressure of `100 atm` at equilibrium. The mole fraction of `A_(B)(g)` at equlibrium is `:`

Options

A. `0.28`
B. `0.72`
C. `0.18`
D. None of these

Answer

A

Explanation

`{:(,2A_(8),hArr,2A_(3),+,3A_(2),+,A_(4)),(t=0,2,,0,,0,,0),(t=t_(eq),2-2alpha,,2alpha,,3alpha,,alpha):}`
`n_(T)=2+4alpha`
given mole fraction of `A_(2) "is"=0.36`.
`0.36=(3alpha)/(2+4alpha)`
`alpha=0.46`
Mole fraction of `A_(3)=(2-2alpha)/(2+4alpha)=(2-2xx0.46)/(2+4xx0.46)=0.28`