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InterviewSolution
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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,A. `(10)/(6) mol es`B. `(5)/(6)mol es`C. `(15)/(6)mol es`D. `2mol es` |
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Answer» Correct Answer - C `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`. |
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