If, for a second-order reaction, the elapsed time after the first half-life is 338.871 seconds, the elapsed time after four half-lives would be (in minutes)...

t1/2= 1/k(A).
k= 1/t1/2*(A)=1/338.871 s(1)
(I let (A)=1).
k = ??

After 4 half lives, beginning of the 5th, (A) = 1/24 = 1/16 of the original.

then t1/2=[1/k(1/16)] =
16*1/k=16*338.871 = ??
Change to minutes.
Check my thinking.

So t1/2 =

To find the elapsed time after four half-lives, we first need to determine the value of k by using the formula k = 1/t1/2 * (A). In this case, the given half-life is t1/2 = 338.871 seconds.

So, we calculate k as follows:
k = 1/338.871 s(1) = 0.0029494 s⁻¹ (approx.)

Now that we have the value of k, we can find the elapsed time after four half-lives. At the beginning of the fifth half-life, the concentration (A) would be 1/2^4 = 1/16 of the original concentration.

So, the elapsed time, t1/2, after four half-lives is given by the formula:

t1/2 = [1/k(1/16)]

Plugging in the value of k:
t1/2 = 16 * 1/k

Calculating this value:
t1/2 = 16 * 0.0029494 s⁻¹ ≈ 0.04719 s⁻¹

To convert this from seconds to minutes, we divide by 60, since there are 60 seconds in a minute:

t1/2 ≈ 0.04719 s⁻¹ / 60 ≈ 0.0007865 min⁻¹

Hence, the elapsed time after four half-lives would be approximately 0.0007865 minutes.