At 700 k, hydrogen and bromine react to form bromine react to form hydrogen bromide. The value of equilibrium constant for this reaction is5 xx 10^(8). .Claculate the amount of H_(2), Br_(2) and HBr " at equilibrium it a mixture of" 0*6" mole of "H_(2) and 0*2 " mole of "Br_(2)is heated to700 K.

At 700 k, hydrogen and bromine react to form bromine react to form hyd

1.	At 700 k, hydrogen and bromine react to form bromine react to form hydrogen bromide. The value of equilibrium constant for this reaction is5 xx 10^(8). .Claculate the amount of H_(2), Br_(2) and HBr " at equilibrium it a mixture of" 06" mole of "H_(2) and 02 " mole of "Br_(2)is heated to700 K.
Answer» Solution :`{:(,PCl_(5)(g),hArr,PCl_(3) (g) ,+,Cl_(2)(g)),("Initial",300 " mol ",,0,,0),("At. eqm.",(30 - x)" mol ",,x " mol" ,,x " mol" ),("MOLAR CONC.",(30-x),,x,,x):}` `{:(,H_(2),+,Br_(2),hArr,2 HBr),("Intial amounts",06"mole",,02 "mole",,),("Amount at eqm.",(06-x),,(02 - x),,2" x moles"),("Molar cones at eqm.",((06-x)/V),,(02 -x)/V,,(2x)/V mol L^(-1)):}` ( V= Volumeof reaction mixture) `K= ((2x//V)^(2))/((06-x)/V xx(02 -x)/V)=(4x^(2))/((06-x)(02-x))= 5 xx 10^(8)` or `(4x^(2))/((012- 08 x +x^(2)))=5 xx 10^(8) or (x^(2) - 08 x + 012) 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) - - 08 x + 0* 12 = 0` ` x= ( 08 pm sqrt((08)^(2) - 4 xx 012))/2=(08 pm 0693)/2= 07465 and 00535` ` x= 0 7465 " is impossible ( because initially", H_(2) = 06 "mole")` Hence, `x= 00535` `:. "Amounts at equilibrium will be " ` `H_(2) = 06 - 00535 = 05465" mole"` `Br_(2) = 02 - 00535 = 01465 " mole " ` `HBr = 2 xx 00535 = 01070 " mole"`