For a second order reaction A <--> B, calculate the t1/2 (half life) and the k(constant) and also fill in the blanks.

Question

A (concentration) t(time)
20 0 sec
10 ?
5 40 sec
3 ?
? 200 sec
t1/2=? k=?

Answer 1

20 at t = 0
5 at t = 40

A = C e^kt
5 = 20 e^40 k
.25 = e^40 k
ln .25 = 40 k
-1.39 = 40 k
k = - .03466
so
A = 20 e^-.03466 t
when A = 10, we have not only the blank filled but the half life
.5 = e^-.03466 t
ln .5 = -.6931 = -.03466t
t = 20
I think you can do it now

Answer 2

Thanks man that makes so much clearer. i posted another question similar to this but its a first order reaction. would that change how you work it or is it done the same?

Answer 3

I believe Damon has worked this as a first order reaction but the problem says it is a second order equation.
I think it should be as follows:
(1/A) - (1/Ao) = kt
(1/5) - (1/20) = 40k
0.2-0.05 = 40k
and k = 0.15/40 = 0.00375 and not 0.03466.

I assume you can do the remainder of the problem.

Answer 4

ok thank you, whats the formula to find the time of the missing blanks in the question?

Answer 5

It's what I have above.
(1/A) - (1/Ao) = kt.
You know k now, the variables are A, Ao (not really a variable I guess) and t. Solve for the missing one.