In your lab you are studying the kinetics of the degradation of a pain killer in
ID: 921811 • Letter: I
Question
In your lab you are studying the kinetics of the degradation of a pain killer in the human liver. You are monitoring the concentration of the pain killer over a period of time. The initial concentration of the pain killer in your experiment was 1.57 M. After 29.08 hours the concentration was found to be 0.785 M. In another 58.17 hours the concentration was found to be 0.3925 M (t = 87.25 hours overall).
If another experiment were to be set up where the initial concentration of the pain killer was 0.463 M, how long would it take for the pain killer to reach 0.0209 M?
Explanation / Answer
All you have to do is compute a plot of all these concentrations, for a first and a second order reaction. Once you get the data, you will fin the time for the concentration given.
For a first order reaction: LnA = LnAo - kt
For a second order reaction: 1/A = 1/Ao + kt
So, let's plot LnA vs t, to get the value of k, and if the reaction is of first order (r2 near 1), we can calculate the time for the given concentration. If it's not of first order, we'll compute 1/A vs t:
For a first order reaction, the data obtained, by plotting lnA vs t:
r2 = 0.99999999999
the expression of y= a + bx, the slope "b" would be k in the above expression. As the value of r2 was 0.9999999 we can assume this is a first order reaction, so:
a = lnAo = 0.451036
b = k = -0.0238
so LnA = 0.451036 - 0.0238t
Now, for the concentration given, solve for t:
ln (0.463) = 0.451036 - 0.0238t
-0.770028 - 0.451036 = -0.0238t
t = 51.2368 h
Hope this helps