The value of `k_2` for the reaction at `27^@C`
`Br_2(l)+Cl_2(g)hArr2BrCl(g)`
is '1 atm'.At equilibrium in a closed container partial pressure of BrCl gas is 0.1 atm 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 `Cl_2` , initially to get above equilibrium situation :

Options

A. `10/6` moles
B. `5/6` moles
C. `15/6` moles
D. `2` moles

Answer

C

Explanation

`{:(Br_2(l)+,Cl_2(g)" "hArr,2BrCl(g)),(t=0,1,0),(,(1-x),2x):}`
`k_p=((P_(BrCl)))^2/P_(Cl_2)=1` so, `P_(Cl_2)=(P_(BrCl))^2=0.01` atm
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` moles
Moles of `Br_2(l)` required for maintaining vapour pressure of 0.1 atm `=2xx5/6 "moles" =10/6"moles"="moles of" BrCl(g)`
Moles required for taking part in reaction=moles of `Cl_2` used up `=5/6` moles .
Therefore required total mole `=10/6+5/6=15/6` mole