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`.

Explanation

`{:(,PCl_(5)(g),hArr,PCl_(3) (g) ,+,Cl_(2)(g)),("Initial",3*00 " mol ",,0,,0),("At. eqm.",(3*0 - x)" mol ",,x " mol" ,,x " mol" ),("Molar conc.",(3*0-x),,x,,x):}`
`{:(,H_(2),+,Br_(2),hArr,2 HBr),("Intial amounts",0*6"mole",,0*2 "mole",,),("Amount at eqm.",(0*6-x),,(0*2 - x),,2 " x moles"),("Molar cones at eqm.",((0*6-x)/V),,(0*2 -x)/V,,(2x)/V mol L^(-1)):}` ( V= Volume of reaction mixture)
`K= ((2x//V)^(2))/((0*6-x)/V xx(0*2 -x)/V)=(4x^(2))/((0*6-x)(0*2-x))= 5 xx 10^(8)`
or `(4x^(2))/((0*12- 0*8 x +x^(2)))=5 xx 10^(8) or (x^(2) - 0*8 x + 0*12) xx 5 xx 10^(8) = 4 x^(2)`
Neglecting `4x^(2) " in comparison to " 5 xx 10^(8) x^(2), (i.e., "taking 4 "x^(2)=0) ` we get
`x^(2) - - 0*8 x + 0* 12 = 0`
` x= ( 0*8 pm sqrt((0*8)^(2) - 4 xx 0*12))/2=(0*8 pm 0*693)/2= 0*7465 and 0*0535`
` x= 0* 7465 " is impossible ( because initially", H_(2) = 0*6 "mole")`
Hence, `x= 0*0535`
`:. "Amounts at equilibrium will be " `
`H_(2) = 0*6 - 0*0535 = 0*5465" mole"`
`Br_(2) = 0*2 - 0*0535 = 0*1465 " mole " `
`HBr = 2 xx 0*0535 = 0*1070 " mole"`