At constant temperature and 1 atm pressure, PCl5 dissociate 2% then at what pressure PCl5 dissociate 4% :-
`(Use PCl_(5)(g) hArr PCl_(3) (g) + Cl_(2)(g)) `

Options

A. `1/8 ` atm
B. `1/(16) atm`
C. `(1)/(2) atm`
D. `1/4 atm`

Answer

D

Explanation

`{:(PCl5 , hArr, PCl 3, + ,Cl2),( 1,,0,,0),(1-alpha , ,alpha ,, alpha ):}`
`KP =( alpha^(2))/((1-alpha ))xx[(p)/((1+alpha))]^(1)`
`KP= (alpha ^(2) P)/(1-alpha ^(2)) ~~- alpha ^(2) P `
` alpha _(1) ^(2) P_(1) = alpha _(2) ^(2) P_(2) `
`( 0.02 ) xx 1 = ( 0.04 ) ^(2) .Pa`
`implies P2 = 0.25 atm `
`=(1)/(4) atm`