Question

How do you rearrange the Arrhenius equation for activation energy?

Answer 1

Hint: The Arrhenius equation is a combination of two concepts: the activation energy and the Boltzmann law. The Arrhenius equation relates the rate constant of the chemical equation with the absolute temperature and the pre exponential factor.

Complete step by step answer:
In 1899, the Swedish chemist Svante Arrhenius combined the two most important concepts of activation energy and the Boltzmann distribution law to give a new equation known as Arrhenius equation.
The Arrhenius equation is the expression which shows the relation between rate constant, absolute temperature, and A factor.
The expression for Arrhenius equation is shown below.
$k = A{e^{ - Ea/RT}}$
Where,
k is the rate constant of the chemical reaction.
A is the pre-exponential factor.
e is the base of natural logarithm
${E_a}$ is the activation energy
R is the universal gas constant.
T denotes the absolute temperature.
The Arrhenius equation can also be written as shown below.
$k = zp{e^{ - Ea/RT}}$
Where,
Z is the collision factor.
P is the steric factor.
A = zp
The Arrhenius equation can be rewritten in an non-exponential form which is more convenient to use and applied for the interpretation of the graph.
Logarithm is taken on both sides and by separating the exponential and pre-exponential terms the equation obtained is shown below.
$ \Rightarrow \ln K = \ln (A{e^{ - Ea/RT}})$
$ \Rightarrow \ln K = \operatorname{lnA} + ln({e^{ - Ea/RT}})$
$ \Rightarrow \ln K = \ln A + \dfrac{{ - {E_a}}}{{RT}}$
$ \Rightarrow \ln K = \left( {\dfrac{{ - {E_a}}}{{RT}}} \right)\left( {\dfrac{1}{T}} \right) + \ln A$
This equation is the equation of the straight line whose slope is $ - {E_a}/R$. This helps to determine the activation energy from the value of K which is noted at different temperatures, by plotting the lnK as a function of 1/T.

Note: For the first order reaction the unit for the rate constant of chemical equation is second inverse ${s^{ - 1}}$ and the unit for the second order reaction is Molarity inverse second inverse${M^{ - 1}}{s^{ - 1}}$. The molarity is the mol per liter.