1.

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)

Answer»


Solution :`{:(,H_2+," "Br_2" "hArr,2HBr),(t=0,0.4," "0.2,-),(t=t_(EQ),0.2,UNDERSET("=negligible")(y),underset(=y)(0.4)):}`
`because 1/4xx10^10=(0.4xx0.4)/(0.2xxy)IMPLIES y=3.2xx10^(-10)implies (Br_2)/(HBr)xx10^11=3.2/0.2xx10^(-10)xx10^(11)=80`


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