Chemical Kinetics Question 325
Question: The half-life for the virus inactivation if in the beginning 1.5% of the virus is inactivated per minute is (Given: The reaction is of first order)
Options:
A) 76 min
B) 66 min
C) 56 min
D) 46 min
Show Answer
Answer:
Correct Answer: D
Solution:
[d] For the first order reaction for small finite change $ k_1=\frac{1}{[A]}\frac{\Delta [A]}{\Delta t}\Rightarrow \frac{\Delta [A]/[A]}{\Delta t}=1.5 $ % $ {{\min }^{-1}} $
$ =0.015mi{n^{-1}} $
$ {t_{1/2}}=\frac{0.693}{0.015{{\min }^{-1}}}=46.2min\approx 46mm $
