
How do you calculate the pre-exponential factor from the Arrhenius equation?
Answer
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Hint: To answer this question we should be aware of the Arrhenius equation. If we understand how to calculate the pre-exponential factor from the Arrhenius equation we can understand the definition of pre-exponential factor better.
Complete answer:
Arrhenius proposed an empirical equation for calculating the energy of activation of a reaction having rate constant K at temperature T.
\[K=A{{e}^{-\dfrac{{{E}_{a}}}{RT}}}\]…………equation 1
Where, ${{E}_{a}}$ = Arrhenius activation energy
A = Pre exponential factor
R = Gas constant
The exponential factor is dimensionless. The rate constant K and the pre exponential factor have the same units.
${{E}_{a}}$ and A are known as Arrhenius parameters.
Taking log on both sides
\[\ln (K)=\ln (A{{e}^{-\dfrac{{{E}_{a}}}{RT}}})\]
Solving this,
\[\ln K=\ln A-\dfrac{{{E}_{a}}}{RT}\]……………equation 2
We can represent equation 2 like this too,
\[\ln K=-\dfrac{{{E}_{a}}}{R}\dfrac{1}{T}+\ln A\]………..equation 3
Equation 3 can correlate with Y = mx + c, which is an equation to represent a straight line.
We can calculate pre-exponential factors by plotting graph lnK against corresponding reciprocal temperature.
When we plot lnK against$\dfrac{1}{T}$,
A straight line is obtained with a slope equal to $\dfrac{{{E}_{a}}}{R}$ and an intercept which is equal to lnA.
If we take the exponential of lnA which is equal to A (pre exponential factor).
Thus, with reference to equation 3 a graph is plotted against lnK against the reciprocal of corresponding temperature for which lnA is obtained as incept.
Note:
The pre-exponential factor is also known as frequency factor, this is because the units of K and A are the same. The units of K for a first order reaction are ${{s}^{-1}}$, which is the unit of frequency. So, it is also known as frequency factor. 'A' represents the frequency of collision between reactants at constant concentration which varies constantly with its corresponding temperature.
Complete answer:
Arrhenius proposed an empirical equation for calculating the energy of activation of a reaction having rate constant K at temperature T.
\[K=A{{e}^{-\dfrac{{{E}_{a}}}{RT}}}\]…………equation 1
Where, ${{E}_{a}}$ = Arrhenius activation energy
A = Pre exponential factor
R = Gas constant
The exponential factor is dimensionless. The rate constant K and the pre exponential factor have the same units.
${{E}_{a}}$ and A are known as Arrhenius parameters.
Taking log on both sides
\[\ln (K)=\ln (A{{e}^{-\dfrac{{{E}_{a}}}{RT}}})\]
Solving this,
\[\ln K=\ln A-\dfrac{{{E}_{a}}}{RT}\]……………equation 2
We can represent equation 2 like this too,
\[\ln K=-\dfrac{{{E}_{a}}}{R}\dfrac{1}{T}+\ln A\]………..equation 3
Equation 3 can correlate with Y = mx + c, which is an equation to represent a straight line.
We can calculate pre-exponential factors by plotting graph lnK against corresponding reciprocal temperature.
When we plot lnK against$\dfrac{1}{T}$,
A straight line is obtained with a slope equal to $\dfrac{{{E}_{a}}}{R}$ and an intercept which is equal to lnA.
If we take the exponential of lnA which is equal to A (pre exponential factor).
Thus, with reference to equation 3 a graph is plotted against lnK against the reciprocal of corresponding temperature for which lnA is obtained as incept.
Note:
The pre-exponential factor is also known as frequency factor, this is because the units of K and A are the same. The units of K for a first order reaction are ${{s}^{-1}}$, which is the unit of frequency. So, it is also known as frequency factor. 'A' represents the frequency of collision between reactants at constant concentration which varies constantly with its corresponding temperature.
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