`PCl_(5)` is `10%` dissociated at `1 atm`. What is `%` dissociation at 4 atm . `PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)`

Options

A. `40%`
B. `2.5%`
C. `5%`
D. `10%`

Answer

3

Explanation

`{:(PCl_(5)(g),hArr,PCl_(3)(g),+,Cl_(2)(g)),(P_(1),,-,,-),(P_(1)(1-alpha),,P_(1)alpha,,P_(1)alpha),(0.9P_(1),,0.1P_(1),,0.1P_(1)):}`
`K_(p)=((0.1P_(1))^(2))/(0.9P_(1))=(P_(1))/(90)`
`P_(1)=(1)/(1.1)`
So, `K_(p)=(1)/(99)`
For new condition,
`{:(PCl_(5)(g),hArr,PCl_(2)(g),+,Cl_(2)(g)),(P_(2),,-,,-),(P_(2)(1-alpha),,P_(2)alpha,,P_(2)alpha):}`
`rArrP_(2)(1+alpha)=4`
`K_(p)=(P_(2)alpha^(2))/(1-alpha)=(1)/(99)`
`alpha^(2)=(1)/(397)" "rArr" "alpha=0.05`