You need to rewrite the question and include all of the details.
If the initial concentration of AB is .250 M, and the reaction mixture initially contains no products, what are the concentrations of A and B after 80s, k=5.6*10^-2 ( second order)
AB-A + B
Isn't the second order equation something like this?
[1/(A)]-[1/(A)o] = kt
You don't have any units on k in the problem. Make sure the units for k and unit for t agree.
[1/(A)]-[1/(A)o] = kt
k=5.6*10^-2 M^-1 s^-1
My answer is still wrong. Can you explain in more details
If I remember correctly. Your intial AB is .250 your k is 5.6 *10^-2 and t is 80s plug this all into the problem to.get the final equilibrium of concentration AB so you'll have 1/AB=whatever answer is then divide 1 by your answer to get AB by itself then you have to subtract that answer from the initial .250 and that will be your a and your b...they will be the same.....think.of the equation like this .... [1/AB]-[1/ABo]=Kt
