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. $(25 $ $min)$
Solution:
$ \begin{aligned} & & R & =k[A] \\ \Rightarrow & & R _1 & =k[A] _1 \text { and } \quad R _2=k[A] _2 \\ & & \dfrac{R _1}{R _2} & =\dfrac{4}{3}=\dfrac{[A] _1}{[A] _2} \\ & \text { Also } & k\left(t _2-t _1\right) & =\ln \dfrac{[A] _1}{[A] _2}=\ln \dfrac{4}{3} \\ \Rightarrow & & \dfrac{\ln 2}{t _{1 / 2}} \times 10 & =\ln \dfrac{4}{3} \\ \Rightarrow & & t _{1 / 2} & =\dfrac{10 \log 3}{\log 4-\log 3}=\dfrac{3}{0.6-0.48}=25 min \end{aligned} $
BETA
