Question
Derive integrated rate equation for first order reaction and also explain how than rate constant can be determine with help of graph.
✓
Answer
$\rightarrow$ "First order reaction means that the rate reaction is proportional to the first power the concentration of the reactant$ R$."
$\rightarrow$ For example, consider the following reaction
$R \rightarrow P$
$\rightarrow$ Rate of reaction for this reaction can be expressed as
$ \begin{array}{l} \text { Rate }=-\frac{ d [ R ]}{ dt }= k [ R ] \\ \text { Or } \frac{ d [ R ]}{[ R ]}=- kdt \end{array} $
$\rightarrow$ Integrating this equation, we get
$ \ln [R]=-k t+I .. ..Eq. (1) $
where, $I$ is the constant of integration and its value can be determined easily.
$\rightarrow$ When $t =0, R =[ R ]_0$, where $[ R ]_0$ is the initia concentration of the reactant.
$\rightarrow$ Therefore, equation $(1$) can be written as
$\ln [R]_0=- k \times 0+ I$
$\ln [ R ]_0= I$
$\rightarrow$ Substituting the value of $I$ in equation $(1)$
$\ln [ R ]=- kt +\ln [ R ]_0 \quad \ldots \quad \ldots \text { Eq. (2) }$
$\rightarrow$ Rearranging this equation
$\ln \frac{[ R ]}{[ R ]_0}=- kt$
or $k =\frac{1}{ t } \ln \frac{[ R ]_0}{[ R ]}.... $ Eq. $(3)$
$\rightarrow$ At time $t_1$ from equation $(1)$
$\ln [R]_1=-k t_1+\ln [R]_0 \ldots \ldots$ Eq. $(4)$
At time $t _2$
$\ln [R]_2=- kt _2+\ln [ R ]_0 \ldots \ldots$ Eq. $(5)$
where, $[ R ]_1$ and $[ R ]_2$ are the concentration the reactants at time $t_1$ and $t_2$ respectively.
$\rightarrow \rightarrow \rightarrow$ Subtracting Eq. $(5)$ from $(4)$
$\begin{array}{l} \ln [ R ]_1-\ln [ R ]_2=- kt _1-\left(- kt _2\right) \\ \ln \frac{[ R ]_1}{[ R ]_2}= k \left( t _2- t _1\right) \\ k =\frac{1}{\left( t _2- t _1\right)} \ln \frac{[ R ]_1}{[ R ]_2} \quad \ldots \ldots \text { Eq. (6) }
\end{array}$
$\rightarrow$ So, equation $(2)$ can also be written as
$ \ln \frac{[ R ]}{[ R ]_0}=- kt $
Taking antilog of both the side
$[ R ]=[ R ]_0 e ^{- kt } .. ...$ Eq. $(7)$
$\rightarrow$ Comparing equation $(2)$ with $y = mx + c$, if we plot $\ln [R]$ against $t,$
we get a straight line with slope $=- k$ and intercept equal to $\ln [ R ]_0$
$\rightarrow$ The first order rate equation $(3)$ can also be written in the form
$k =\frac{2.303}{ t } \log \frac{[ R ]_0}{[ R ]} \quad \ldots \ldots \text { Eq. (8) }$
or $\log \frac{[ R ]_0}{[ R ]}=\frac{ kt }{2.303}$
$\rightarrow$ If we plot a graph between $\log \frac{[R]_0}{[R]}$ vs $t$, the slope $=\frac{ k }{2.303}$
$\rightarrow$ For example, consider the following reaction
$R \rightarrow P$
$\rightarrow$ Rate of reaction for this reaction can be expressed as
$ \begin{array}{l} \text { Rate }=-\frac{ d [ R ]}{ dt }= k [ R ] \\ \text { Or } \frac{ d [ R ]}{[ R ]}=- kdt \end{array} $
$\rightarrow$ Integrating this equation, we get
$ \ln [R]=-k t+I .. ..Eq. (1) $
where, $I$ is the constant of integration and its value can be determined easily.
$\rightarrow$ When $t =0, R =[ R ]_0$, where $[ R ]_0$ is the initia concentration of the reactant.
$\rightarrow$ Therefore, equation $(1$) can be written as
$\ln [R]_0=- k \times 0+ I$
$\ln [ R ]_0= I$
$\rightarrow$ Substituting the value of $I$ in equation $(1)$
$\ln [ R ]=- kt +\ln [ R ]_0 \quad \ldots \quad \ldots \text { Eq. (2) }$
$\rightarrow$ Rearranging this equation
$\ln \frac{[ R ]}{[ R ]_0}=- kt$
or $k =\frac{1}{ t } \ln \frac{[ R ]_0}{[ R ]}.... $ Eq. $(3)$
$\rightarrow$ At time $t_1$ from equation $(1)$
$\ln [R]_1=-k t_1+\ln [R]_0 \ldots \ldots$ Eq. $(4)$
At time $t _2$
$\ln [R]_2=- kt _2+\ln [ R ]_0 \ldots \ldots$ Eq. $(5)$
where, $[ R ]_1$ and $[ R ]_2$ are the concentration the reactants at time $t_1$ and $t_2$ respectively.
$\rightarrow \rightarrow \rightarrow$ Subtracting Eq. $(5)$ from $(4)$
$\begin{array}{l} \ln [ R ]_1-\ln [ R ]_2=- kt _1-\left(- kt _2\right) \\ \ln \frac{[ R ]_1}{[ R ]_2}= k \left( t _2- t _1\right) \\ k =\frac{1}{\left( t _2- t _1\right)} \ln \frac{[ R ]_1}{[ R ]_2} \quad \ldots \ldots \text { Eq. (6) }
\end{array}$
$\rightarrow$ So, equation $(2)$ can also be written as
$ \ln \frac{[ R ]}{[ R ]_0}=- kt $
Taking antilog of both the side
$[ R ]=[ R ]_0 e ^{- kt } .. ...$ Eq. $(7)$
$\rightarrow$ Comparing equation $(2)$ with $y = mx + c$, if we plot $\ln [R]$ against $t,$
we get a straight line with slope $=- k$ and intercept equal to $\ln [ R ]_0$
$\rightarrow$ The first order rate equation $(3)$ can also be written in the form
$k =\frac{2.303}{ t } \log \frac{[ R ]_0}{[ R ]} \quad \ldots \ldots \text { Eq. (8) }$
or $\log \frac{[ R ]_0}{[ R ]}=\frac{ kt }{2.303}$
$\rightarrow$ If we plot a graph between $\log \frac{[R]_0}{[R]}$ vs $t$, the slope $=\frac{ k }{2.303}$
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