Chemical Kinetics - Result Question 61
####61. The rate of first order reaction is $0.04 mol L^{-1} s^{-1}$ at $10 min$ and $0.03 mol L^{-1} s^{-1}$ at $20 min$ after initiation. Find the half-life of the reaction.
(2001, 5M)
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Answer:
Correct Answer: 61. $\left(3.26 \times 10^{-2} min^{-1}\right)$
Solution:
$$ \begin{aligned} & & R & =k[A] \ \Rightarrow & & R _1 & =k[A] _1 \text { and } \quad R _2=k[A] _2 \ & & \frac{R _1}{R _2} & =\frac{4}{3}=\frac{[A] _1}{[A] _2} \ & \text { Also } & k\left(t _2-t _1\right) & =\ln \frac{[A] _1}{[A] _2}=\ln \frac{4}{3} \ \Rightarrow & & \frac{\ln 2}{t _{1 / 2}} \times 10 & =\ln \frac{4}{3} \ \Rightarrow & & t _{1 / 2} & =\frac{10 \log 3}{\log 4-\log 3}=\frac{3}{0.6-0.48}=25 min \end{aligned} $$
