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`.

Answer

A, B, D

Explanation

`H_(2)+Br_(2) hArr 2HBr`
`{:("Initial",0.6,0.2,0),("At equilibrium",(0.6-x)/V,(0.2-x)/V,(2x)/V "mole"),("mol. conc.",,,):}`
`K=(((2x)/v)^(2))/(((0.6-x)/v)((0.2-x)/v))`
`(4x^(2))/((0.6-x)(0.2-x))=5xx10^(8)`
`(x^(2)-0.8x-0.12)xx5xx10^(8)=4x^(2)`
Neglecting `4x^(2)` in comparison to `5xx10^(8)x^(2)`, we get
`x^(2)-0.8-0.12=0`
`x=(0.8 +- sqrt((0.8)^(2)-4xx0.12))/2`

`=(0.8 +- 0.693)/(2)`
`=0.7465` and `0.0535`
`x=0.7465` is impossible
hence, `x=0.0535`
`[H_(2)]_(eq)=0.6-0.0535=0.5465` mol
`[Br_(2)]_(eq)=0.2-0.0535=0.1465` mol
`[HBr]_(eq)=2xx0.0535=0.1070` mol