R2.2.12 Arrhenius equation (Higher Level Only)

The Arrhenius Equation: Understanding Temperature's Role in Reaction Rates

The Arrhenius Equation and Its Components

The Arrhenius equation is expressed as:

$k = A e^{- \frac{E_{a}}{R T}}$

Where:

$k$ : The rate constant, which determines the speed of the reaction.
$A$ : The Arrhenius factor(or frequency factor), representing the frequency of collisions with the correct orientation for a reaction to occur.
$E_{a}$ : The activation energy, the energy barrier that must be overcome for a reaction to proceed (measured in joules per mole, ${J mol}^{- 1}$ ).
$R$ : The gas constant, $8.31 {J mol}^{- 1} K^{- 1}$ .
$T$ : The absolute temperature in kelvin (K).
$e$ : The base of natural logarithms ( $e \approx 2.718$ ).

Note

This equation tells us two key things:

As temperature increases ( $T$ ), the term $- \frac{E_{a}}{R T}$ becomes less negative.
- This means $e^{- \frac{E_{a}}{R T}}$ increases, leading to a larger $k$ , and thus a faster reaction.
A higher activation energy ( $E_{a}$ ) reduces $k$ .
- Reactions with large activation energies are slower because fewer particles have enough energy to overcome the barrier.

Tip

The exponential term, $e^{- \frac{E_{a}}{R T}}$ , represents the fraction of particles with enough energy to overcome the activation energy barrier at a given temperature.

Exponential graph of the Arrhenius equation.

The Linear Form of the Arrhenius Equation

The Arrhenius equation can also be expressed in a linear form for easier analysis:

$\ln k = \ln A - \frac{E_{a}}{R T}$

This equation is in the form of a straight-line equation: $y = m x + c$ , where:

$y = \ln k$
$x = \frac{1}{T}$
The slope ( $m$ ) is $- \frac{E_{a}}{R}$ .
The y-intercept ( $c$ ) is $\ln A$ .

By plotting $\ln k$ against $\frac{1}{T}$ , you can determine the activation energy ( $E_{a}$ ) and the frequency factor ( $A$ ) from the graph:

The slope of the line gives $- \frac{E_{a}}{R}$ . Rearranging, $E_{a} = - slope \times R$ .
The y-intercept gives $\ln A$ , so $A = e^{y-intercept}$ .

Example

Determining Activation Energy

Imagine a reaction with the following data:

$\begin{array}{cc} Temperature (T) in K & Rate Constant (k) \\ 300 & 2.5 s^{- 1} \\ 350 & 6.8 s^{- 1} \end{array}$

Convert the data to $\ln k$ and $\frac{1}{T}$ :

For $T = 300$ : $\ln k = \ln (2.5) = 0.916$ , $\frac{1}{T} = \frac{1}{300} = 0.00333$ .
For $T = 350$ : $\ln k = \ln (6.8) = 1.918$ , $\frac{1}{T} = \frac{1}{350} = 0.00286$ .

Plot $\ln k$ against $\frac{1}{T}$ to obtain a straight line. The slope of the line is $- \frac{E_{a}}{R}$ .
Calculate $E_{a}$ :
- Slope = $\frac{\ln k_{2} - \ln k_{1}}{\frac{1}{T_{2}} - \frac{1}{T_{1}}} = \frac{1.918 - 0.916}{0.00286 - 0.00333} = - 2100 K$ .
- $E_{a} = - slope \times R$ $= - (- 2100) \times 8.31 = 17451 {J mol}^{- 1} = 17.45 {kJ mol}^{- 1}$ .

Why Does Temperature Affect Reaction Rates?

To understand why temperature has such a significant impact on reaction rates, we turn to the Maxwell-Boltzmann distribution.

This distribution describes the spread of kinetic energies among particles in a sample.

Key Points:

At any temperature, most particles have a kinetic energy close to the average, but some have much higher energies.
As temperature increases:
- The average kinetic energy of particles increases.
- The distribution curve flattens and shifts to the right, increasing the number of particles with energy greater than $E_{a}$ .

This explains the exponential relationship between temperature and the rate constant in the Arrhenius equation.

Analogy

Think of a crowd trying to jump over a hurdle. At low temperatures, only a few people have enough energy to clear it. As the temperature rises, more people can jump over, increasing the success rate.

Determining Activation Energy Experimentally

The activation energy of a reaction can be determined by measuring the rate constant ( $k$ ) at different temperatures and using the linear form of the Arrhenius equation.

Steps:

Conduct the reaction at several temperatures and determine $k$ at each temperature.
Calculate $\ln k$ and $\frac{1}{T}$ for each data point.
Plot $\ln k$ against $\frac{1}{T}$ .
Determine the slope ( $- \frac{E_{a}}{R}$ ) and calculate $E_{a}$ .

Tip

When performing experiments, ensure that all other variables (e.g., concentration, pressure) remain constant to isolate the effect of temperature.

Common Mistakes and Misconceptions

Common Mistake

Misinterpreting the Arrhenius factor ( $A$ ):

Some students assume $A$ is always large.
However, $A$ depends on the orientation and frequency of collisions.
Complex molecules often have smaller $A$ values due to the need for specific orientations.

Common Mistake

Forgetting units:

Ensure consistent units for $E_{a}$ ( ${J mol}^{- 1}$ ), $R$ ( ${J mol}^{- 1} K^{- 1}$ ), and $T$ ( $K$ ) when using the Arrhenius equation.

Reflection and Practice

Self review

What happens to the rate constant ( $k$ ) if the temperature is doubled?
How would you determine the activation energy of a reaction using experimental data?
Why does a higher activation energy result in a slower reaction?

Theory of Knowledge

How does the temperature dependence of reaction rates influence our understanding of climate change and its impact on chemical processes in the atmosphere?

You've read 2/2 free chapters this week.

Upgrade to PLUS or PRO to unlock all notes, for every subject.