Question

Iodine and bromine react to give iodine monobromide, IBr

I2+Br2 is in equilibrium to yield 2IBr

What is the equilibrium composition of a mixture at 115°C that initially contained 1.30 x 10 to the negative third 1.30 mol each of iodine and bromine in a 5.00 L vessel? The equilibrium constant Kc for this reaction at 115°C is 55.8.

I2 = M ?

Br2= M ?

IBr= M ?

Answer 1

            I2   +    Br2 ------------->2IBr

I         0.0013   1.3                     0

C         -x            -x                    2x

E        0.0013-x   1.3-x               2x

[I2] = 0.0013-x/5

[Br2]   = 1.3-x/5

[IBr]   = 2x/5

       Kc   = [IBr]^2/[Br2][I2]

        55.8   = (2x)^2/(0.0013-x)(1.3-x)

       55.8*(0.0013-x)(1.3-x)   = 4x^2

            x   = 0.00129

      [I2] = 0.0013-x/5   = 0.0013-0.00129/5   = 0.000002M

      [Br2]   = 1.3-x/5      = 1.3-0.00129/5   = 0.259M

       [IBr]   = 2x/5          = 2*0.00129/5        = 0.000516M