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 :

The value of k_2 for the reaction at 27^@C Br_2(l)+Cl_2(g)hArr2BrCl(g

1.	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 :
Answer» `10/6` MOLES `5/6` moles `15/6` moles `2` moles Solution :`{:(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