Question Bank [2022] — Chemistry STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryQuestion Bank [2022]3 Marks
Question
Derive an integrated rate law expression for first order reaction: A → B + C
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Answer
Consider first order reaction, $A \rightarrow B+C$
The differential rate law is given by $ \text { rate }=-\frac{ d [ A ]}{ dt }= k [ A ] \cdots(1) $ where, $[A]$ is the concentration of reactant at time $t$.
Rearranging Eq. (1) $ \frac{ d [ A ]}{[ A ]}=- kdt \cdots(2) $
Let $[A]_0$ be the initial concentration of the reactant $A$ at time $t=0$.
Suppose $[A]_t$ is the concentration of $A$ at time $=t$
The equation (2) is integrated between limits $[A]=[A]_0$ at $t=0$ and $[A]=[A]_t$ at $t=t$ $ \int_{[ A ]_0}^{[ A ]_{ t }} \frac{ d [ A ]}{[ A ]}=- k \int_0^{ t } dt $
On integration, $ [\ln [ A ]]_{[ A ]_0}^{[ A ]_{ t }}=- k ( t )_0^{ t } $
Substitution of limits gives $ \ln [A]_t-\ln [A]_0=-k t $ $\operatorname{or} \ln \frac{[ A ]_{ t }}{[ A ]_0}=- kt$ or $k =\frac{1}{ t } \ln \frac{[ A ]_0}{[ A ]_{ t }}$
Converting In to $\log _{10}$, we write $ k =\frac{2.303}{ t } \log _{10} \frac{[ A ]_0}{[ A ]_{ t }} \cdots(4) $ Eq. (4) gives the integrated rate law for the first order reactions.
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