At `700 K`, hydrogen and bromine react to form hydrogen bromine. The value of equilibrium constant for this reaction is `5xx10^(8)`. Calculate the amount of the `H_(2), Br_(2)` and `HBr` at equilibrium if a mixture of `0.6 mol` of `H_(2)` and `0.2 mol` of `Br_(2)` is heated to `700K`.

Explanation

For the given reaction of `H_(2)` and `Br_(2)` to give `HBr`, the value of equilibrium constant `K_(c)=5xx10^(8)` and thus, exceptionally higher value of `K_(c)` suggests that reaction goes almost to completion i.e.,
`{:(,H_(2),+,Br_(2),hArr,2HBr),("Before reaction",0.6,,0.2,,0),("At equilibrium",0.4,,0,,0.4):}`
Thus, `H_(2)=0.4 "mole"`
`Br_(2)=0.0`
`HBr=0.4 "mole"`