Wize University Chemistry Textbook > Kinetics

Arrhenius Equation

Arrhenius Equation

Rates of reaction are highly temperature dependent! Imagine the speed of ice melting at 0°C vs 40°C
Svante Arrhenius discovered a simple equation that relates the rate of a chemical reaction to the temperature which it is performed at. The Arrhenius equation is shown below.

$\ln k=\ln A-\frac{Ea}{RT}$

$k=Ae^{-E_a/RT}$

k is still the rate constant
A is called the pre-exponential factor, it is unique for all reactions and is constant across all temperatures. It is related to the probability of a successful collision.
T is temperature and R is the Gas constant
Ea is the activation energy, which is shown below graphically

Activation energy, Ea, is the amount of energy it takes to get to the products
The Arrhenius equation on its own is not very useful because of the pre-exponential factor A is difficult to calculate. However by dividing one rate by another and rearranging we end up with the following equation. Notice that A has been cancelled out and is no longer necessary.

$\ln \bigg(\cfrac{k_1}{k_2}\bigg)=-\cfrac{E_a}{R}\bigg(\cfrac{1}{T_1}-\cfrac{1}{T_2}\bigg)$

We can use this equation to compare the rate of a reaction at two different temperatures.

a. A chemical reaction increases its rate by a factor of 15 when it is warmed from room temperature to 100oC. What is the activation energy of this process? R = 8.314 J/Kmol

$\ln \bigg(\cfrac{k_1}{k_2}\bigg)=\cfrac{-E_a}{R}\bigg(\cfrac{1}{T_1}-\cfrac{1}{T_2}\bigg)$

$E_a=\cfrac{-\ln\bigg(\cfrac{k_1}{k_2}\bigg)R}{\bigg(\cfrac{1}{T_1}-\cfrac{1}{T_2}\bigg)}=\cfrac{\ln \bigg(\cfrac{1}{15}\bigg)(8.314 JK^{-1}mol^{-1})}{\bigg(\cfrac{1}{298K}-\cfrac{1}{373K}\bigg)}=33.4kJmol^{-1}$

If the activation energy of a reaction is measured to be 59.6 kJ/mol, by what factor will the rate of this reaction be increased when the temperature is raised from 305 K to 405 K?

We start with the adapted Arrhenius equation,

\ln \bigg(\cfrac{k_1}{k_2}\bigg)=-\cfrac{E_a}{R}\bigg(\cfrac{1}{T_1}-\cfrac{1}{T_2}\bigg)

the question is asking for the ratio of k2 to k1

\frac{k_1}{k_2}=e^{-5.8}=3.02\times 10^{-3}

\frac{k_2}{k_1}=\frac{1}{3.02\times10^{-3}}=331

The reaction will be 331 times faster at 405 K than it will be at 305 K. That's quite a bit!

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A Schiff base reaction between 3-propanone and aniline proceeds with rate constants of 0.0065 M-1 s -1 and 0.0084 M-1 s -1 at 30oC and 80oC respectively. Determine the activation energy and preexponential factor for this reaction.

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Extra Practice

Kinetics: Arrhenius Equation

What is the activation energy of a reaction if the rate constant at 40oC is double the value of the rate constant at 25oC?

In the gas phase, carbon can react with oxygen in two different ways as shown below.

$C+O_2 \xrightarrow{k_1} CO_2 \hspace{20pt} E_a=24.6kJ/mol$

$C+0.5O_2 \xrightarrow{k_2} CO \hspace{20pt} E_a=19.6kJ/mol$

A first order gas phase reaction N2O4 → 2 NO2 was carried out isothermally and the activation barrier was found to be 23.5 kJ/mol. Initially p(N2O4) = 2.6 atm and p(NO2) = 0. After 1 hour p(NO2) = 2.0 atm and p(N2O4) =1.6 atm. Assume that both NO2 and N2O4 are ideal gases.

Determine the first order rate constant k, and the half life t1/2.
The above reaction was performed at 220oC. Calculate the pre-exponential factor.

In the gas phase, carbon can react with oxygen as shown below.

\rm C\quad+\quad O_2\quad\overset{k_1}{\overrightarrow{\hspace{1.5cm}}}\quad CO_2\qquad E_a=24.6\ kJ/mol

a) If the reaction is run at room temperature the rate constant is found to be $0.04\ atm\ s^{-1}$ . Determine the pre-exponential factor $A$ .

Report your answer in kJ/mol to two decimal places.

Kinetics: Arrhenius Equation

What is the activation energy of a reaction if the rate constant at 40oC is double the value of the rate constant at 25oC?

Arrhenius Equation

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