If a mixture `0.4 "mole" H_(2) "and" 0.2` mole `Br_(2)` is heated at `700K` at equilibrium, the value of equilibrium constant is `0.25xx10^(10)` then find out the ratio of concentrations of `(Br)(2)) "and" (HBr)` (Report your answer as `(Br_(2))/(HBr)xx10^(11)`

Answer

80

Explanation

`H_(2)+Br_(2)hArr2HBr`
`{:(t=0,0.4,0.2,-),(t=t_(eq),0.2,y,0.4):}`
`="negligible"=y`
`because (1)/(4)xx10^(10)=(0.4xx0.4)/(0.2xxy)`
`y=3.2xx10^(-10)`
`(Br_(2))/(HBr)xx10^(11)=(3.2)/(0.2)xx10^(-10)xx10^(11)=80`