Question

The Kc of the reaction H2 (g)+Br2 (g)=2HBr (g) is 2.18×10+6. If the initial concentration of HBr in 12.0L vessel is 3.20 moles, calculate the concentration of H2, Br2 and HBr at equilibrium. Use ICE table. Show calculations and all work & may have to use quadratic equation.

Answer 1

    H₂ (g)    +   Br₂ (g) =    2HBr (g)      Kc   = 2.18×10⁺⁶

for the reverse equilibrium

       2HBr (g) =   H₂ (g)    +   Br₂ (g)         kc _new = 1/Kc   = 1/2.18×10⁺⁶

       3.2/12             0                  0

       0.266 -x          x                   x

    kc _new= {[H2] x [Br2]}/[HBr]² = (x X x )/(0.266-x)² = 1/2.18×10⁺⁶

solving the quadratic

x/(0.266-x) = 1/1.47×10⁺³

x ~ 0.266/1.47x10³ = 1.8 x 10^-4

[H2 ] = [Br2 ] = 1.8 x 10^-4

[HBr ] = 0.266 -x ~ 0.266