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?

1 Answer
Stefan V.
Feb 2, 2015

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)#

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