The equilibrium constant for the reaction `Br_2(l)+Cl_(2)(g)hArr2Br Cl(g)` at `27^@C` is `k_P=1` atm in a closed container of volume 164 L initially 1

To maintain the mentioned equilibrium there must be some (negligible) mass of `Br_2(l)` at equilibrium so there will be an equilibrium
`Br_2(l)hArrBr_(2)(g) " " K_(p)=0.25` atm
so some `Br_2(l)` will be required for this conversion (into vapour) moles of `Br_2(l)` required for above equilibrium
`n=(PV)/(RT)=((0.25 "atm")(164 L))/((0.082 L "atm" "mole"^(-1)K^(-1))(300 K)^(-))=5/3` moles
Now for equilibrium
`{:(,Br_2(l)+,Cl_2(g)hArr,2BrCl(g),),(t=0,x,10,0,),(at eq.,0,10-x,2x,"Total gaseous moles"=(10+x)):}`
{Let x be moles of `Br_2(l)` required just to maintain above equilibrium}
we have `P_(Cl_2)+P_(BrCl)=2 "atm" " " So, P_(Cl_2)=(10-x)/(10+x)(2"atm")`
`P_(BrCl)=((2x)/(10+x))(2 atm) " " So, K_(P)=P_(BrCl)^2/P_(Cl_2)=(4x^2(10+x))/((10+x)^2(10-x))(2atm)`
`implies (8x^2)/(100-x^2)=1implies 8x^2=100-x^2implies 9x^2=100. " " So, x=10/3`
Hence total moles of `Br_2(l)` required to maintain both of above equilibria `=5/3+10/3=15/3=5` moles