Question

A viral preparation was inactivated in a chemical bath. The inactivation process was found to be first order in virus concentrations. At the beginning of the experiment, 2.0% of the virus was found to be inactivated per minute. The k for inactivation process is:

• A. $3\times {10}^{-4}{s}^{-1}$
• B. $4\times {10}^{-4}{s}^{-1}$
• C. $5\times {10}^{-4}{s}^{-1}$
• D. $6\times {10}^{-4}{s}^{-1}$

Answer 1

The correct option is A $3\times {10}^{-4}{s}^{-1}$

2.0% of the virus was found to be inactivated per minute. The expression for the specific rate constant is $k=\frac{2.303}{t}\log \frac{a}{(a-x)}=\frac{2.303}{1\times 60}\log \frac{100}{(100-2)}=3\times {10}^{-4}{s}^{-1}$.