Question

A viral preparation was inactive in a chemical broth. The inactivation process was found to be 1st order in viral concentration. At the beginning of the experiment 2.0% of virus was found to be inactive/min. Evaluate the k for the inactivation process?

Answer 1

The rate constant for the inactivation process is equal to `"k" = 3.37 * 10^(-4) "s"^(-1)`.

You know that the inactivation process is first order in viral concentration. Mathematically, you can express this by

`ln(([V])/[V]_0) = - k* t`, where

`[V]_0` - the initial concentration of the virus;
`[V]` - the concentration after the first minute;
`k` - the rate constant;
`t` - the time that passed - in your case, `t = "1 minute"`.

So, you start with an initial viral concentration of `[V]_0`. After the first minute of the experiment, `"2.0%"` of the virus was found to be inactive. This means that the viral concentration was reduced by `"2.0%"` after the first minute, so `[V]` wil be

`[V] = [V]_0 - 0.02 * [V]_0 = (1 - 0.02) * [V]_0 = 0.98 * [V]_0`

Now just plug in this value into the equation shown above

`ln((0.98 * [V]_0)/([V]_0)) = -k * "60.0 s" => k = -1/("60 s") * ln(0.98)`

Therefore, the rate constant will be equal to

`k = -0.0202 * (-1/("60 s")) = 3.37 * 10^(-4)"s"^(-1)`