The Arrhenius Equation (HL) (DP IB Chemistry): Revision Note
The Arrhenius Equation
The rate equation shows how each of the reactants in a reaction affects the rate of the reaction and it includes the rate constant, k
However, k only remains constant if the concentration of the reactants is the only factor which is changed
If the temperature is changed or a catalyst is used or changed, then the rate constant, k, changes
At higher temperatures, a greater proportion of molecules have energy greater than the activation energy
Since the rate constant and rate of reaction are directly proportional to the fraction of molecules with energy equal or greater than the activation energy, then at higher temperatures:
The rate of reaction increases
The rate constant increases
What is the Arrhenius equation?
The relationship between the rate constant, the temperature and also the activation energy is given by the Arrhenius equation:
Where:
k = Rate constant
A = Arrhenius factor (also known as the frequency factor or pre-exponential factor) which is a constant that takes into account the frequency of collisions with proper orientations
Ea = Activation energy (J mol-1)
R = Gas constant (8.31 J K-1 mol-1)
T = Temperature (Kelvin, K)
e = Mathematical constant (can be found on your calculator - it has the approximate value of 2.718)
Ea and A are constants that are characteristic of a specific reaction
A does vary slightly with temperature but it can still be considered a constant
R is a fundamental physical constant for all reactions
k and T are the only variables in the Arrhenius equation
The Arrhenius equation is used to describe reactions that involve gases, reactions occurring in solution or reactions that occur on the surface of a catalyst
Using the Arrhenius Equation
The Arrhenius equation is easier to use if you take natural logarithms of each side of the equation, which results in the following equation:
The Arrhenius equation can be used to show the effect that a change in temperature has on the rate constant, k, and thus on the overall rate of the reaction
An increase in temperature (higher value of T) gives a greater value of ln k and therefore a higher value of k
Since the rate of the reaction depends on the rate constant, k, an increase in k also means an increased rate of reaction
The equation can also be used to show the effect of increasing the activation energy on the value of the rate constant, k
An increase in the activation energy, Ea, means that the proportion of molecules which possess at least the activation energy is less
This means that the rate of the reaction, and therefore the value of k, will decrease
The values of k and T for a reaction can be determined experimentally
These values of k and T can then be used to calculate the activation energy for a reaction
This is the most common type of calculation you will be asked to do on this topic
Examiner Tips and Tricks
In the exam, you could be asked to calculate any part of the Arrhenius equation
Using the equation in its natural logarithm form makes this easier.
Worked Example
Calculate the activation energy of a reaction which takes place at 400 K, where the rate constant of the reaction is 6.25 x 10-4 s-1.
A = 4.6 x 1013 and R = 8.31 J K-1 mol-1.
Answer:
Rearrange the Arrhenius equation for Ea:
Insert values from the question:
Convert Ea to kJ:
Graphing the Arrhenius Equation
Finding the activation energy and Arrhenius factor
A graph of experimental data can be used to determine the activation energy and the Arrhenius factor
Arrhenius equation graph
A graph of ln k against 1/T can be plotted, and then used to calculate Ea
This gives a line which follows the form y = mx + c
Graph of ln k against 1/T

The graph of ln k against 1/T is a straight line with gradient -Ea/R
From the graph, the equation is in the form of y = mx + c is as follows:
ln k | = | + | ln A | ||
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|
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y | = | m | x | + | c |
Where:
y = ln k
x =
m =
(the gradient)
c = ln A (the y-intercept)
Ea can be calculated by:
Ea = - gradient x R
As the slope has a negative gradient, the minus signs cancel giving Ea as a positive value
The Arrhenius factor, A, can be calculated by:
A = ey-intercept
Examiner Tips and Tricks
If the x-axis does not start at the origin, you cannot use the y-intercept to find A
Instead, take a point from the graph and substitute the values of ln k, 1/T and the gradient into the logarithmic Arrhenius equation to find ln A, and use this to calculate A.
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