Activation Energy and the Arrhenius Equation

The minimum energy necessary to form a product during a collision between reactants is called the activation energy (E_a). The kinetic energy of reactant molecules plays an important role in a reaction because the energy necessary to form a product is provided by a collision of a reactant molecule with another reactant molecule. (In single-reactant reactions, activation energy may be provided by a collision of the reactant molecule with the wall of the reaction vessel or with molecules of an inert contaminant.)

If the activation energy is much larger than the average kinetic energy of the molecules, the reaction will occur slowly: Only a few fast-moving molecules will have enough energy to react. If the activation energy is much smaller than the average kinetic energy of the molecules, the fraction of molecules possessing the necessary kinetic energy will be large; most collisions between molecules will result in reaction, and the reaction will occur rapidly.

The figure below shows the energy relationships for the general reaction of a molecule of A with a molecule of B to form molecules of C and D:

\(A+B\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}C+D\)

The figure shows that the energy of the transition state is higher than that of the reactants A and B by an amount equal to E_a, the activation energy. Thus, the sum of the kinetic energies of A and B must be equal to or greater than E_a to reach the transition state. After the transition state has been reached, and as C and D begin to form, the system loses energy until its total energy is lower than that of the initial mixture. This lost energy is transferred to other molecules, giving them enough energy to reach the transition state.

The forward reaction (that between molecules A and B) therefore tends to take place readily once the reaction has started. In the figure below, ΔH represents the difference in enthalpy between the reactants (A and B) and the products (C and D). The sum of E_a and ΔH represents the activation energy for the reverse reaction:

\(C+D\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}A+B\)

A graph is shown with the label, “Extent of reaction,” bon the x-axis and the label, “Potential energy,” on the y-axis. Above the x-axis, a portion of a dashed curve is labeled “A plus B.” From the right end of this region, the concave down curve continues upward to reach a maximum near the height of the y-axis. The peak of this curve is labeled, “Transition state.” A double sided arrow extends from a dashed horizontal line that originates at the y-axis at a common endpoint with the curve to the peak of the curve. This arrow is labeled “E subscript a.” A second horizontal dashed line segment is drawn from the right end of the black curve left to the vertical axis at a level significantly lower than the initial “A plus B” labeled end of the curve. The end of the curve that is shared with this segment is labeled, “C plus D.” The curve, which was initially dashed, continues as a solid curve from the maximum to its endpoint at the right side of the diagram. A second double sided arrow is shown. This arrow extends between the two dashed horizontal lines and is labeled, “capital delta H.”

This graph shows the potential energy relationships for the reaction \(A+B\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}C+D.\) The dashed portion of the curve represents the energy of the system with a molecule of A and a molecule of B present, and the solid portion represents the energy of the system with a molecule of C and a molecule of D present. The activation energy for the forward reaction is represented by E_a. The activation energy for the reverse reaction is greater than that for the forward reaction by an amount equal to ΔH. The curve’s peak represents the transition state.

We can use the Arrhenius equation to relate the activation energy and the rate constant, k, of a given reaction:

\(k=A{e}^{\text{−}{E}_{\text{a}}\text{/}RT}\)

In this equation, R is the ideal gas constant, which has a value 8.314 J/mol/K, T is temperature on the Kelvin scale, E_a is the activation energy in joules per mole, e is the constant 2.7183, and A is a constant called the frequency factor, which is related to the frequency of collisions and the orientation of the reacting molecules.

The postulates of collision theory are accommodated in the Arrhenius equation. The frequency factor A is related to the rate at which collisions having the correct orientation occur. The exponential term, \({e}^{\text{−}{E}_{\text{a}}\text{/}RT},\) is related to the fraction of collisions providing adequate energy to overcome the activation barrier of the reaction.

At one extreme, the system does not contain enough energy for collisions to overcome the activation barrier. In such cases, no reaction occurs. At the other extreme, the system has so much energy that every collision with the correct orientation can overcome the activation barrier, causing the reaction to proceed. In such cases, the reaction is nearly instantaneous.

The Arrhenius equation describes quantitatively much of what we have already discussed about reaction rates. For two reactions at the same temperature, the reaction with the higher activation energy has the lower rate constant and the slower rate. The larger value of E_a results in a smaller value for \({e}^{\text{−}{E}_{\text{a}}\text{/}RT},\) reflecting the smaller fraction of molecules with enough energy to react.

Alternatively, the reaction with the smaller E_a has a larger fraction of molecules with enough energy to react. This will be reflected as a larger value of \({e}^{\text{−}{E}_{\text{a}}\text{/}RT},\) a larger rate constant, and a faster rate for the reaction. An increase in temperature has the same effect as a decrease in activation energy. A larger fraction of molecules has the necessary energy to react (see the figure below), as indicated by an increase in the value of \({e}^{\text{−}{E}_{\text{a}}\text{/}RT}.\) The rate constant is also directly proportional to the frequency factor, A. Hence a change in conditions or reactants that increases the number of collisions with a favorable orientation for reaction results in an increase in A and, consequently, an increase in k.

Two graphs are shown each with an x-axis label of “Kinetic energy” and a y-axis label of “Fraction of molecules.” Each contains a positively skewed curve indicated in red that begins at the origin and approaches the x-axis at the right side of the graph. In a, a small area under the far right end of the curve is shaded orange. An arrow points down from above the curve to the left end of this region where the shading begins. This arrow is labeled, “Higher activation energy, E subscript a.” In b, the same red curve appears, and a second curve is drawn in black. It is also positively skewed, but reaches a lower maximum value and takes on a broadened appearance as compared to the curve in red. In this graph, the red curve is labeled, “T subscript 1” and the black curve is labeled, “T subscript 2.” In the open space at the upper right on the graph is the label, “T subscript 1 less than T subscript 2.” As with the first graph, the region under the curves at the far right is shaded orange and a downward arrow labeled “E subscript a” points to the left end of this shaded region.

(a) As the activation energy of a reaction decreases, the number of molecules with at least this much energy increases, as shown by the shaded areas. (b) At a higher temperature, T₂, more molecules have kinetic energies greater than E_a, as shown by the yellow shaded area.

A convenient approach for determining E_a for a reaction involves the measurement of k at different temperatures and using an alternate version of the Arrhenius equation that takes the form of a linear equation

\(\begin{array}{ccc}\hfill \text{ln}\phantom{\rule{0.2em}{0ex}}k& =& \left(\cfrac{\text{−}{E}_{\text{a}}}{R}\right)\left(\cfrac{1}{T}\right)+\text{ln}\phantom{\rule{0.2em}{0ex}}A\hfill \\ \hfill y& =& mx+b\hfill \end{array}\)

Thus, a plot of ln k versus \(\cfrac{1}{T}\) gives a straight line with the slope \(\cfrac{\text{−}{E}_{\text{a}}}{R},\) from which E_a may be determined. The intercept gives the value of ln A.

Example

Determination of E_a

The variation of the rate constant with temperature for the decomposition of HI(g) to H₂(g) and I₂(g) is given here. What is the activation energy for the reaction?

\(\text{2HI}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)\)

T (K)	k (L/mol/s)
555	3.52 \(×\) 10⁻⁷
575	1.22 \(×\) 10⁻⁶
645	8.59 \(×\) 10⁻⁵
700	1.16 \(×\) 10⁻³
781	3.95 \(×\) 10⁻²

Solution

Values of \(\cfrac{1}{T}\) and ln k are:

\(\cfrac{1}{\text{T}}\phantom{\rule{0.4em}{0ex}}\left({\text{K}}^{-1}\right)\)	ln k
1.80 \(×\) 10⁻³	−14.860
1.74 \(×\) 10⁻³	−13.617
1.55 \(×\) 10⁻³	−9.362
1.43 \(×\) 10⁻³	−6.759
1.28 \(×\) 10⁻³	−3.231

The figure below is a graph of ln k versus \(\cfrac{1}{T}.\) To determine the slope of the line, we need two values of ln k, which are determined from the line at two values of \(\cfrac{1}{T}\) (one near each end of the line is preferable). For example, the value of ln k determined from the line when \(\cfrac{1}{T}\phantom{\rule{0.1em}{0ex}}=1.25\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\) is −2.593; the value when \(\cfrac{1}{T}\phantom{\rule{0.1em}{0ex}}=1.78\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\) is −14.447.

A graph is shown with the label “1 divided by T ( K superscript negative 1 )” on the x-axis and “l n k” on the y-axis. The horizontal axis has markings at 1.4 times 10 superscript 3, 1.6 times 10 superscript 3, and 1.8 times 10 superscript 3. The y-axis shows markings at intervals of 2 from negative 14 through negative 2. A decreasing linear trend line is drawn through five points at the coordinates: (1.28 times 10 superscript negative 3, negative 3.231), (1.43 times 10 superscript negative 3, negative 6.759), (1.55 times 10 superscript negative 3, negative 9.362), (1.74 times 10 superscript negative 3, negative 13.617), and (1.80 times 10 superscript negative 3, negative 14.860). A vertical dashed line is drawn from a point just left of the data point nearest the y-axis. Similarly, a horizontal dashed line is draw from a point just above the data point closest to the x-axis. These dashed lines intersect to form a right triangle with a vertical leg label of “capital delta l n k” and a horizontal leg label of “capital delta 1 divided by T.”

This graph shows the linear relationship between ln k and \(\cfrac{1}{T}\) for the reaction \(\text{2HI}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}+{\text{I}}_{2}\) according to the Arrhenius equation.

The slope of this line is given by the following expression:

\(\begin{array}{cc}\hfill \text{Slope}& =\phantom{\rule{0.1em}{0ex}}\cfrac{\text{Δ}\left(\text{ln}\phantom{\rule{0.2em}{0ex}}k\right)}{\text{Δ}\left(\cfrac{1}{T}\right)}\hfill \\ & =\phantom{\rule{0.1em}{0ex}}\cfrac{\left(-14.447\right)\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\left(-2.593\right)}{\left(1.78\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{-1}\right)\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\left(1.25\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{-1}\right)}\hfill \\ & =\phantom{\rule{0.1em}{0ex}}\cfrac{-11.854}{0.53\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{-1}}\phantom{\rule{0.1em}{0ex}}=2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{K}\hfill \\ & =-\cfrac{{E}_{\text{a}}}{R}\hfill \end{array}\)

Thus:

\({E}_{\text{a}}=\text{−slope}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}R=-\left(-2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{K}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\text{8.314 J mol}}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{-1}\right)\)

\({E}_{\text{a}}=1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}{\phantom{\rule{0.2em}{0ex}}}^{\text{J mol}}\)

In many situations, it is possible to obtain a reasonable estimate of the activation energy without going through the entire process of constructing the Arrhenius plot. The Arrhenius equation:

\(\text{ln}\phantom{\rule{0.2em}{0ex}}k=\left(\cfrac{\text{−}{E}_{\text{a}}}{R}\right)\phantom{\rule{0.2em}{0ex}}\left(\cfrac{1}{T}\right)+\text{ln}\phantom{\rule{0.2em}{0ex}}A\)

can be rearranged as shown to give:

\(\cfrac{\text{Δ}\left(\text{ln}\phantom{\rule{0.2em}{0ex}}k\right)}{\text{Δ}\left(\cfrac{1}{T}\right)}\phantom{\rule{0.1em}{0ex}}=-\cfrac{{E}_{\text{a}}}{R}\)

\(\text{ln}\phantom{\rule{0.2em}{0ex}}\cfrac{{k}_{1}}{{k}_{2}}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\cfrac{{E}_{\text{a}}}{R}\phantom{\rule{0.2em}{0ex}}\left(\cfrac{1}{{T}_{2}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{T}_{1}}\right)\)

This equation can be rearranged to give a one-step calculation to obtain an estimate for the activation energy:

\({E}_{\text{a}}=\text{−}R\phantom{\rule{0.2em}{0ex}}\left(\cfrac{\text{ln}\phantom{\rule{0.2em}{0ex}}{k}_{2}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}{k}_{1}}{\left(\cfrac{1}{{T}_{2}}\right)\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\left(\cfrac{1}{{T}_{1}}\right)}\right)\)

Using the experimental data presented here, we can simply select two data entries. For this example, we select the first entry and the last entry:

T (K)	k (L/mol/s)	\(\cfrac{1}{T}\phantom{\rule{0.4em}{0ex}}\left({\text{K}}^{-1}\right)\)	ln k
555	3.52 \(×\) 10⁻⁷	1.80 \(×\) 10⁻³	−14.860
781	3.95 \(×\) 10⁻²	1.28 \(×\) 10⁻³	−3.231

After calculating \(\cfrac{1}{T}\) and ln k, we can substitute into the equation:

\({E}_{\text{a}}=-8.314\phantom{\rule{0.2em}{0ex}}\text{J}\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{-1}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{-1}\left(\cfrac{-3.231-\left(-14.860\right)}{1.28\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{-1}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}1.80\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{-1}}\right)\)

and the result is E_a = 185,900 J/mol.

This method is very effective, especially when a limited number of temperature-dependent rate constants are available for the reaction of interest.

Activation Energy and the Arrhenius Equation