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 $



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