Chemical Kinetics Question 7
Question 7 - 2024 (30 Jan Shift 1)
The rate of first order reaction is $0.04 mol L^{-1} s^{-1}$ at 10 minutes and $0.03 mol L^{-1} s^{-1}$ at 20 minutes after initiation. Half life of the reaction is ___________ minutes. (Given $\log 2=0.3010, \log 3=0.4771$ )
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Answer (24)
Solution
$ 0.04=k[A] _0 e ^{-k \times 10 \times 60} ……{1}$
$0.03=k[A] _0 e ^{-k \times 20 \times 60} ……{2}$
$(1) /(2)$
Solutions
$ \frac{4}{3}=e^{600 k(2-1)} $
$ \frac{4}{3}=e^{600 k} $
$\ln \frac{4}{3}=600 k$
$\ln \frac{4}{3}=600 \times \frac{\ln 2}{t _{1 / 2}}$
$ t _{1 / 2}=600 \frac{\ln 2}{\ln \frac{4}{3}} sec $
$t _{1 / 2}=600 \times \frac{\log 2}{\log 4-\log 3} sec .=10 \times \frac{0.3010}{0.6020-0.477} \min$
$t _{1 / 2}=24.08 min$
Ans. 24
