Question

A viral preparation was inactivated in a chemical bath. The inactivation process was found to be of first-order in virus concentration, and at the beginning of the experiment $2.0$% of the virus was found to be inactivated per minute. Evaluate '$k$' for the inactivation process.

Answer 1

$a=100$, $a-x=98$, $t=60$ sec
$k=\frac{2.303}{t}{\log }_{10}\left(\frac{a}{a-x}\right)$
$=\frac{2.303}{60}{\log }_{10}\left(\frac{100}{98}\right)=3.3\times {10}^{-4}{sec}^{-1}$