
How do you solve the Arrhenius equation for activation energy?
Answer
527.4k+ views
Hint: To solve this question, first we will discuss the formula of Arrhenius equation for activation energy. And, then we will discuss the Arrhenius equation theoretically. We will also discuss the constants and variables that we use in the formula.
Complete step by step solution:
Generally, it can be done by graphing. The Arrhenius equation is:
$ k = A{e^{\dfrac{{ - Ea}}{{RT}}}} $
where:
$ k $ is the rate constant, in units that depend on the rate law. For instance, if
$ r(t) = k{[A]^2} $ , then has units of $ \dfrac{M}{S}.\dfrac{1}{{{M^2}}} = \dfrac{1}{{M.s}} $ .
$ A $ is the "pre-exponential factor", which is merely an experimentally-determined constant correlating with the frequency of properly oriented collisions.
$ {E_a} $ is the activation energy in units of, says, $ J $ .
$ R $ is the universal gas constant.
$ T $ is temperature in .
Or, if we meant literally solve for it, we would get:
$ \begin{gathered}
\ln (\dfrac{k}{A}) = - \dfrac{{{E_a}}}{{RT}} \\
\Rightarrow {E_a} = - RT\,\ln (\dfrac{k}{A}) \\
\end{gathered} $
So knowing the temperature, rate constant, and $ A $ , you can solve for $ {E_a} $ .
However, since $ A $ is experimentally determined, we shouldn't anticipate knowing ahead of time (unless the reaction has been done before), so the first method is more foolproof. Furthermore, using $ k $ and $ T $ for one trial is not very good science. It's better to do multiple trials and be more sure.
Note:
This chemical kinetics video tutorial focuses on solving activation energy problems using the Arrhenius Equation. It explains how to calculate the activation energy, the rate constant K, and the temperature given the rate constant and activation energy.
Complete step by step solution:
Generally, it can be done by graphing. The Arrhenius equation is:
$ k = A{e^{\dfrac{{ - Ea}}{{RT}}}} $
where:
$ k $ is the rate constant, in units that depend on the rate law. For instance, if
$ r(t) = k{[A]^2} $ , then has units of $ \dfrac{M}{S}.\dfrac{1}{{{M^2}}} = \dfrac{1}{{M.s}} $ .
$ A $ is the "pre-exponential factor", which is merely an experimentally-determined constant correlating with the frequency of properly oriented collisions.
$ {E_a} $ is the activation energy in units of, says, $ J $ .
$ R $ is the universal gas constant.
$ T $ is temperature in .
Or, if we meant literally solve for it, we would get:
$ \begin{gathered}
\ln (\dfrac{k}{A}) = - \dfrac{{{E_a}}}{{RT}} \\
\Rightarrow {E_a} = - RT\,\ln (\dfrac{k}{A}) \\
\end{gathered} $
So knowing the temperature, rate constant, and $ A $ , you can solve for $ {E_a} $ .
However, since $ A $ is experimentally determined, we shouldn't anticipate knowing ahead of time (unless the reaction has been done before), so the first method is more foolproof. Furthermore, using $ k $ and $ T $ for one trial is not very good science. It's better to do multiple trials and be more sure.
Note:
This chemical kinetics video tutorial focuses on solving activation energy problems using the Arrhenius Equation. It explains how to calculate the activation energy, the rate constant K, and the temperature given the rate constant and activation energy.
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