The value of `K_(p)` for the reaction at `27^(@)C`
`Br_(2)(l)+CI_(2)(g)hArr2BrCI(g)`
is `1atm`. At equilibrium in a closed container partial pressure of `BrCI` gas `0.1atm` and at this temperature the vapour pressure of `Br_(2)(l)` is also `0.1`atm. Then what will be minimum moles of `Br_(2)(l)` to be added to `1` mole of `CI_(2)`, initially, to get above equilibrium stiuation,

Options

A. `(10)/(6) mol es`
B. `(5)/(6)mol es`
C. `(15)/(6)mol es`
D. `2mol es`

Answer

C

Explanation

`Br_(2)(l)+CI_(2)(g)hArr2BrCI(g)`
`t=0 1 0`
`(1-x) 2x`
`K_(p)=((P_(BrCl)^(2)))/(P_(Cl_(2)))=1 "so", P_(Cl_(2))=(P_(BrCl))^(2))=0.01atm`
then at equilibrium, `(n_(BrCl))/(n_(Cl_(2))=(0.1)/(0.01)=10=(2x)/(1-x)`
So, `10-10x=2x "or" x=(10)/(12)=(5)/(6)mol es`
Mole of `Br_(2)(l)` required for maintaining vapour pressure of `01atm`
`=2xx(5)/(6)mol es=(10)/(6)mol es="moles of" BrCl(g)`.
Moles required for taking part in reaction`="moles of" Cl_(2)` used up`=(5)/(6)mol es`.
Hence total moles required`=(5)/(6)+(10)/(6)=(15)/(6)mol es`.