Question

Rate = k[ab]2 and k = 0.20 l/mol·s. if the initial concentration of ab is 1.50 m, what is [ab] after 12.9 s?

Answer 1

AB---->  A  +  B
rate=  k(ab)^2
by  use  of   integrated  second  order  rate  law
1(/Ab)t  = 1/AB)0  +  Kt
t= time
k=constant
(AB)0   =initial  concentration
(AB)t=  final  concetration

1/(ab)t=  1/1.5  +  0.20  x12.9 =  3.35
1/(ab)t  =  3.25  multiply   both  side   by ( ab)t  
1=  3.25(ab)t  divide  both  side  by  3.25
(ab)t  is   therefore=0.308