Question

A sample of milk splits after \[60\]min. at \[300\]\[K\] and after \[40\] min. at \[400\]\[K\] when the population of lactobacillus acidophilus in it doubles. The activation energy (in \[KJ/mol\]) for this process is closest to _____________.
(Given, \[R = 8.3\]\[J\;mo{l^{ - 1}}\;{K^{ - 1}}\],\[\ln \left( {\dfrac{2}{3}} \right) = 0.4\],\[{e^{ - 3}} = 4.0\])

Answer 1

Hint: The Arrhenius equation used to calculate activation energies. The activation energy can be found algebraically by substituting two rate constants and the two corresponding reaction temperatures into the Arrhenius Equation. The rate of a reaction always increases with increasing temperature so the activation energy is always positive.

Complete answer:
The Arrhenius equation at two different absolute temperatures is
\[\ln \left( {\dfrac{{{k_2}}}{{{k_1}}}} \right) = \dfrac{{{E_a}}}{R}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]\]----(1)
there \[{k_1}\] and \[{k_2}\] are the rate constants for a single reaction at two different absolute temperatures (\[{T_1}\] and \[{T_2}\] ).
Given \[{k_1} = 60\], \[{T_1} = 300\], \[{k_2} = 40\] , and \[{T_2} = 400\]
Then the equation (1) becomes
\[\ln \left( {\dfrac{{40}}{{60}}} \right) = \dfrac{{{E_a}}}{{8.3}}\left[ {\dfrac{1}{{300}} - \dfrac{1}{{400}}} \right]\]
\[ \Rightarrow \]\[\ln \left( {\dfrac{2}{3}} \right) = \dfrac{{{E_a}}}{{8.3}}\left[ {\dfrac{{100}}{{300 \times 400}}} \right]\]--(2)
On converting the equation (2) by bring \[{E_a}\]by separately , we get
\[{E_a} = \ln \left( {\dfrac{2}{3}} \right) \times 1200 \times 8.3\]
\[ \Rightarrow \]\[{E_a} = 0.4 \times 1200 \times 8.3\]
\[ \Rightarrow \]\[{E_a} = 3984\]\[J/mol\]
The activation energy (in \[KJ/mol\]) for this process is closest to \[3984\]\[J/mol\].

Note:
Arrhenius equation shows the relationship between the rate constant \[k\] and the temperature \[T\] in kelvins and is typically written as
\[K = A{e^{ - \dfrac{{{E_a}}}{{RT}}}}\]
where \[R\] is the gas constant (\[8.314\] \[J/mol.K\]), \[A\] is a constant called the frequency factor, and \[{E_a}\] is the activation energy for the reaction.
When the Arrhenius equation is rearranged as a linear equation with the form \[y = mx + c\] where \[y = \ln (k)\], \[x = \dfrac{1}{T}\], and m is \[m = - \dfrac{{{E_a}}}{R}\]. The activation energy for the reaction can be determined by finding the slope of the line.