# 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?

Feb 2, 2015

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

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

$\ln \left(\frac{\left[V\right]}{V} _ 0\right) = - k \cdot t$, where

${\left[V\right]}_{0}$ - the initial concentration of the virus;
$\left[V\right]$ - the concentration after the first minute;
$k$ - the rate constant;
$t$ - the time that passed - in your case, $t = \text{1 minute}$.

So, you start with an initial viral concentration of ${\left[V\right]}_{0}$. After the first minute of the experiment, $\text{2.0%}$ of the virus was found to be inactive. This means that the viral concentration was reduced by $\text{2.0%}$ after the first minute, so $\left[V\right]$ wil be

$\left[V\right] = {\left[V\right]}_{0} - 0.02 \cdot {\left[V\right]}_{0} = \left(1 - 0.02\right) \cdot {\left[V\right]}_{0} = 0.98 \cdot {\left[V\right]}_{0}$

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)