If the initial concentration of is 0.250 , and the reaction mixture initially contains no products, what are the concentrations of and after 80 ?

Answer 1

You need to rewrite the question and include all of the details.

Answer 2

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

Answer 3

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.

Answer 4

[1/(A)]-[1/(A)o] = kt
k=5.6*10^-2 M^-1 s^-1

Answer 5

Substitute and solve for A.

Answer 6

My answer is still wrong. Can you explain in more details

Answer 7

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