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

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

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

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?

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.

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

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.

To calculate the half-life (t1/2) and the rate constant (k) for a second-order reaction, we can use the equation:

t1/2 = 1 / (k * [A]0)
k = (2 / t1/2) * ([A]0 / ([A]0 - [A]))

First, let's fill in the blanks and calculate the missing values:
A (concentration) t (time)
20 0 sec
10 ?
5 40 sec
3 ?
? 200 sec

To find the missing concentration at 0 sec, we already have it as 20.

Next, to find the missing concentration at t = 40 sec, we can calculate it using the equation for a second-order reaction:

k = (2 / t) * ([A]0 / ([A]0 - [A]))

Substituting the known values:

k = (2 / 40) * (20 / (20 - 5))
k = 0.05 M^-1s^-1

Now, we can calculate the missing concentration at t = 40 sec:

k = (2 / t) * ([A]0 / ([A]0 - [A]))
0.05 = (2 / 40) * (20 / (20 - 5))
[A] = 7.14 M

Similarly, we can find the missing concentration at t = 200 sec:

k = (2 / t) * ([A]0 / ([A]0 - [A]))
0.05 = (2 / 200) * (20 / (20 - [A]))
[A] = 0.5 M

Finally, we can calculate the half-life:

t1/2 = 1 / (k * [A]0)
t1/2 = 1 / (0.05 * 20)
t1/2 = 1 / 1
t1/2 = 1 sec

Therefore:
t1/2 = 1 sec
k = 0.05 M^-1s^-1