Question
Derive the formula for determining activation energy from Arrhenius equation using rate constant at different temperatures. And also explain how activation energy can be determined using graph.
✓
Answer
Activation energy using graph:
$\rightarrow$ Taking natural logarithm on both the side of Arrhenius equation $k = Ae ^{-\frac{ E _{ a }}{ RT }}$
$\rightarrow \ln k =-\frac{ E _{ a }}{ RT }+\ln A \ \ \ \ldots \ldots$ Eq. $(1)$
$\rightarrow$ The plot of $\ln k$ vs $1 / T$ gives a straight line as shown in figure.
$\rightarrow$ In figure, slope $=-\frac{E_{ a }}{R}$ and intercept $=\ln A$.
So, we can calculate activation energy $( E _{ a } )$ and Arrhenius constant $A$ using these values.
Formula of activation energy:
$\rightarrow$ It has been found from Arrhenius equation that increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in rate constant.
$\rightarrow$ Thus, at temperature $T_1$, equation $(1)$ is
$\ln k _1=-\frac{ E _{ a }}{ RT _1}+\ln A ........Eq. (2)$
$\rightarrow$ at temperature $T _2,$ equation $(1)$
is $\ln k _2=-\frac{ E _{ a }}{ RT _2}+\ln A ........ Eq. (3)$
$($since $A$ is constant for given reaction$)$
$k _1$ and $k _2$ are rate constant at temperatures $T _1$ and $T _2$ respectively.
$\rightarrow$ Subtracting equation $(2)$ from $(3),$ we obtain
$\ln k _2-\ln k _1=\frac{- E _{ a }}{ RT _2}+\frac{ E _{ a }}{ RT _1}$
$\ln \frac{ k _2}{ k _1}=\frac{ E _{ a }}{ R }\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
$\log \frac{ k _2}{ k _1}=\frac{ E _{ a }}{2.303 R }\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
$\log \frac{ k _2}{ k _1}=\frac{ E _{ a }}{2.303 R }\left[\frac{ T _2- T _1}{T_1 T_2}\right]$
$\rightarrow$ From above formula activation energy can be calculate using measured values of rate constants at different temperatures.
$\rightarrow$ Taking natural logarithm on both the side of Arrhenius equation $k = Ae ^{-\frac{ E _{ a }}{ RT }}$
$\rightarrow \ln k =-\frac{ E _{ a }}{ RT }+\ln A \ \ \ \ldots \ldots$ Eq. $(1)$
$\rightarrow$ The plot of $\ln k$ vs $1 / T$ gives a straight line as shown in figure.
$\rightarrow$ In figure, slope $=-\frac{E_{ a }}{R}$ and intercept $=\ln A$.
So, we can calculate activation energy $( E _{ a } )$ and Arrhenius constant $A$ using these values.
Formula of activation energy:
$\rightarrow$ It has been found from Arrhenius equation that increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in rate constant.
$\rightarrow$ Thus, at temperature $T_1$, equation $(1)$ is
$\ln k _1=-\frac{ E _{ a }}{ RT _1}+\ln A ........Eq. (2)$
$\rightarrow$ at temperature $T _2,$ equation $(1)$
is $\ln k _2=-\frac{ E _{ a }}{ RT _2}+\ln A ........ Eq. (3)$
$($since $A$ is constant for given reaction$)$
$k _1$ and $k _2$ are rate constant at temperatures $T _1$ and $T _2$ respectively.
$\rightarrow$ Subtracting equation $(2)$ from $(3),$ we obtain
$\ln k _2-\ln k _1=\frac{- E _{ a }}{ RT _2}+\frac{ E _{ a }}{ RT _1}$
$\ln \frac{ k _2}{ k _1}=\frac{ E _{ a }}{ R }\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
$\log \frac{ k _2}{ k _1}=\frac{ E _{ a }}{2.303 R }\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
$\log \frac{ k _2}{ k _1}=\frac{ E _{ a }}{2.303 R }\left[\frac{ T _2- T _1}{T_1 T_2}\right]$
$\rightarrow$ From above formula activation energy can be calculate using measured values of rate constants at different temperatures.
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$iii.$ What precautions he should take while bottling so that his product does not lose fizz during storage and handling across long distances?
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Give reason :
$(i)$ The boiling point of ethanol is higher than that of hydroxymethane.
$(ii)$ Phenol is a stronger acid than alcohol.
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$(ii)$ Phenol is a stronger acid than alcohol.
$(iii)$ Ethanol easily dissolve in water.