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If a mixture 0.4 mole H2 and 0.2 mole Br...

If a mixture 0.4 mole `H_2` and 0.2 mole `Br_2` is heated at 700 K 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`)

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The correct Answer is:
80
`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`
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