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The activation energy, represented by E a , is the minimum energy required for a chemical reaction to take place , that is, it is the energy barrier that must be overcome so that the reactants can become products.
Activation energy is related to the kinetics of a reaction, that is, the rate at which products are formed or reactants are consumed. This relationship is due to the fact that the reactions occur when the molecules of the reactants collide with each other in the proper orientation and with a minimum kinetic energy.
When the activation energy is high, this means that the molecules must collide at high speed, or rather, with high kinetic energy, for the collision to be effective and for the reaction to take place. In this situation, if the temperature is not very high, most of the collisions do not result in the formation of products, so the reaction as a whole proceeds slowly.
On the other hand, when the activation energy is small, many of the collisions that occur generate products, so the reaction proceeds rapidly.
How is activation energy determined?
The activation energy of a reaction is related to the reaction rate through the rate constant. This relationship is given by the Arrhenius equation that relates the rate constant ( k ) to the absolute temperature ( T ), the activation energy (E a ) and a proportionality constant called the Arrhenius pre-exponential factor or collision factor (A ):
This equation can be used in two different ways to determine the activation energy:
Algebraic method to determine the activation energy
The simplest way to determine the activation energy is to determine the rate constant experimentally at two different temperatures, and then solve a system of two equations with two unknowns. The two equations consist of the Arrhenius equation applied at the two temperatures:
This system of equations is easily solved by dividing one of the equations by the other to eliminate the constant A, and then solving the resulting equation to obtain the activation energy.
Graphic method to determine the activation energy
Despite being very simple, the algebraic method is very sensitive to experimental errors in determining the rate constant. The effect of these random errors can be compensated if the rate constant is measured at a larger number of temperatures.
In these cases, instead of the algebraic method to determine the activation energy, a graph of all the data is carried out, which is adjusted to the best straight line using statistical methods. The result of this process is the activation energy that best fits all of the experimental data, rather than just two of them.
This method is also based on the Arrhenius equation, but written in a slightly different way. If we apply logarithm to both sides of the Arrhenius equation and then apply the properties of logarithms, we can rewrite it as:
This equation has the mathematical form of a straight line where ln( k ) is the y -coordinate , 1/T represents x, ln(A) is the y-intercept, and –E a /T is the slope. To determine the activation energy, first determine the constant at different temperatures, then plot ln( k) versus 1/T and obtain the activation energy from the slope of the line.
Below are two examples of problems determining the activation energy by both methods.
Example 1. Determination of the activation energy by the algebraic method
statement
In two different experiments the rate constant of a second order reaction was determined, one at 27°C and the other at 97°C. The rate constant at the first temperature was 4.59.10 -3 L.mol -1 s -1 while at the second it was 8.46.10 -2 L.mol -1 .s -1 . Determine the activation energy of this reaction in kcal.mol -1 .
Solution
The first thing we need to do is extract the data from the statement. In this case, we have two temperatures and two rate constants. Temperatures must be transformed to Kelvin, since the Arrhenius equation, like most equations in chemistry, uses absolute temperature.
T 1 = 27 °C + 273.15 = 300.15 K
k 1 = 4.59.10 -3 L.mol -1 s -1
T 2 = 97 °C + 273.15 = 370.15 K
k 2 = 8.46.10 -2 L.mol -1 s -1
Step 1: Write the system of equations
These four data are related to each other through the Arrhenius equation, giving rise to two equations with two unknowns:
Step 2: Divide both equations
Now we divide equation 2 by equation 1 , to obtain:
Step 3: Solve for E a
The third step is to solve this equation to obtain the activation energy. To do this, we first apply natural logarithm to both sides of the equation, to obtain:
Then we rearrange the factors to get the activation energy. The result is:
Step 4: We substitute the data and calculate the activation energy
Therefore, the reaction has an activation energy of 9,190 kcal.mol -1 .
Example 2. Determination of the activation energy by the graphical method
statement
The rate constant for a first order reaction at ten different temperatures between 25 °C and 250 °C was determined. The results are presented in the following table:
| Temperature (°C) | 25 | fifty | 75 | 100 | 125 | 150 | 175 | 200 | 225 | 250 |
| k (s -1 ) | 1,67.10 -9 | 5.95.10 -8 | 4,169.10 -7 | 1,061.10 -5 | 1,915.10 -4 | 7,271.10 -4 | 5,704.10 -3 | 6,863.10 -3 | 0.1599 | 0.3583 |
Determine the activation energy of the reaction in kJ/mol.
Solution
This problem must be solved using the graphical method since there are multiple determinations of the rate constant at different temperatures.
Step 1: Convert temperatures to Kelvin
It is not necessary in this case to extract the data because it is already organized in a table. However, it is necessary to transform all temperatures to Kelvin. The result is presented later.
Steps 2 and 3: Calculate the inverses of the temperature and the natural logarithms of the rate constants
In the graphical method, a plot of ln(k) versus 1/T is constructed, so these values must be determined for each temperature. The temperatures in Kelvin, as well as their inverses and the natural logarithms of the constants are presented in the following table.
| T(K) | 1/T (K-1) | ln(k) |
| 298.15 | 0.003354 | -20.21 |
| 323.15 | 0.003095 | -16.64 |
| 348.15 | 0.002872 | -14.69 |
| 373.15 | 0.002680 | -11.45 |
| 398.15 | 0.002512 | -8,561 |
| 423.15 | 0.002363 | -7,226 |
| 448.15 | 0.002231 | -5,167 |
| 473.15 | 0.002113 | -4,982 |
| 498.15 | 0.002007 | -1,833 |
| 523.15 | 0.001911 | -1026 |
Step 4: Construct a graph of ln(k) versus 1/T and obtain the equation of the line
Once we have the values of the inverses of temperature and logarithms of the constants, we proceed to build a scatter plot with these data. This can be done by hand using graph paper or by using a spreadsheet or calculator that has the linear regression function.
Once all the points are located on the graph, we proceed to draw the best line, the one that passes as close as possible to all the points. This is easier to do in the spreadsheet as it only involves adding a trend line.
It is also necessary to obtain the equation of this line in the form of a slope cut, since from there the activation energy will be obtained. The best straight line is the one determined by the method of least squares. Spreadsheets do this automatically, but it can also be done easily on a scientific calculator, even if it doesn’t have graphing functionality. All you have to do is enter all the points in linear regression mode and then find the cutoff and slope of the line among the linear regression results.
The following figure shows the graph of the previous data made in the Google Sheets spreadsheet. The least-squares-fitted equation of the line is shown at the top of the graph area.
Step 5: Calculate the activation energy from the slope
The slope of the line is related to the activation energy by means of the following equation:
From where it is obtained that:
Substituting the value of the slope presented in the graph (which has units of K) we obtain the activation energy:
Finally, the reaction has an activation energy of 110.63 kJ.mol -1 .
References
Atkins, P., & dePaula, J. (2014). Atkins’ Physical Chemistry (rev. ed.). Oxford, United Kingdom: Oxford University Press.
Chang, R. (2008). Physical Chemistry (3rd ed.). New York City, New York: McGraw Hill.
Arrhenius equation: reaction rate and temperature | Chemtube. (nd). Retrieved from https://www.quimitube.com/videos/cinetica-quimica-teoria-8-ecuacion-de-arrhenius/
Jorge-Mario, P. (2019, June). Methodology for calculating the pre-exponential factor using the isoconversional principle for the numerical simulation of the air injection process. Retrieved from http://www.scielo.org.co/scielo.php?script=sci_arttext&pid=S0122-53832019000100037#f9
The Arrhenius Law – Pre-exponential Factors. (2020, September 22). Retrieved from https://chem.libretexts.org/@go/page/1448
