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 0.1atm. 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,

The value of K_(p) for the reaction at 27^(@)C Br_(2)(l)+CI_(2)(g)hAr

1.	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 0.1atm. 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,
Answer» `(10)/(6) MOL es` `(5)/(6)mol es` `(15)/(6)mol es` `2mol es` SOLUTION :`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`.