A viral preparation was inactivated in a chemical bath. The activation process was found to be first order in virus concentration. The beginning of the experiment 2% of the virus was found to be inactivated per minute.

Question

[i] Evaluate k for the inactivation process in units (1/s).

[ii] How much time would be required for the virus to become 50% inactivated?

[iii] How much time would be required for the virus to become 75% inactivated?

Answer 1

ln(No/N) = kt
No = 100
N = 98
k is to be found
time is 1 min or 60 seconds depending upon how the answer is to be displayed.

For #2, N is 50

For #3 N = 25.