P.Chem: Rate of chemical reactions; Half-life

A second- order reaction of the type A + B -->P was carried out in a solution that was initially 0.075 mol dm^-3 in A and 0.050 mol dm^-3 in B. After 1.0 h the concentration of A had fallen to 0.020 mol dm^-3. a) Calculate the Rate constant. b) Solve for the half- life of each of the reactants.
Hint: Answers are a) 16.22 dm^3/mol*h
b) 5.1 × 10^3 s, 2.1 × 10^3 s

My Attempt:
[A]intial= 0.075 mol dm^-3
[A]= 0.20
x=[A]initial-[A]=0.075-0.20= 0.055

intial=0.050 mol dm^-3
=0.050-x= 0.050-0.055= -0.005

this is where I'm confused because I get a negative for . When I plug that into the integrated rate law, I get a nonreal number. The integrated rate law is
kt(initial-[A]initial)=ln([A]intial/[A]intial)

Any idea?

Borek
Mentor
IMHO error in the question.

Forget about kinetics, just try to solve 0.075M and 0.050M question as a limiting reagent problem - no way you are going to get 0.020M A as a final result.

epenguin
Homework Helper
Gold Member
Agree with Borek. Sure you have copied the question correctly?

The integrated rate law is recognisably something like what you quote but not that.

Also second order reactions do not have such a thing as a half-life.