If the initial concentration of ab is 0.290 m , and the reaction mixture initially contains no products, what are the concentrations of a and b after 80 s ?

According to the integrated second order rate law is:

1/[AB] = 1/[AB]o + Kt

when K is constant (given) = 5.5 x 10^-2

and t is the time = 80 s

and [AB]o is the initial concentration = 0.29 m

so by substitution:

1/[AB] = 1/0.29 + (5.5 x 10^-2) * 80

∴[AB] = 0.127 m

from the reaction equation, we can see the molar ratio between AB& B & A is 1:1:1

the [AB] lost = 0.29 m - 0.127m 

                     = 0.163 m

so both A & B will gain the same number of moles 

∴∴[A]= [B] = 0.163 m