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)

If a mixture 0.4 "mole" H_(2) "and" 0.2 mole Br_(2) is heated at 700K

1.	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» <BR> SOLUTION :`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`