Chapter 18: Kinetics

18.5 Collision Theory and the Effect of Temperature on Reaction Rate

Learning Outcomes

  • Use the postulates of collision theory to explain the effects of physical state, temperature, and concentration on reaction rates
  • Define the concepts of activation energy and transition state
  • Use the Arrhenius equation in calculations relating rate constants to temperature

We should not be surprised that atoms, molecules, or ions must collide before they can react with each other. Atoms must be close together to form chemical bonds. This simple premise is the basis for a very powerful theory that explains many observations regarding chemical kinetics, including factors affecting reaction rates.

Collision theory is based on the following postulates:

  1. The rate of a reaction is proportional to the rate of reactant collisions: [latex]\text{reaction rate}\propto \dfrac{\#\text{collisions}}{\text{time}}[/latex]
  2. The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product.
  3. The collision must occur with adequate energy to permit mutual penetration of the reacting species’ valence shells so that the electrons can rearrange and form new bonds (and new chemical species).

We can see the importance of the two physical factors noted in postulates 2 and 3, the orientation and energy of collisions, when we consider the reaction of carbon monoxide with oxygen:

[latex]\ce{2CO}(g)+\ce{O2}(g)\rightarrow \ce{2CO2}(g)[/latex]

Carbon monoxide is a pollutant produced by the combustion of hydrocarbon fuels. To reduce this pollutant, automobiles have catalytic converters that use a catalyst to carry out this reaction. It is also a side reaction of the combustion of gunpowder that results in muzzle flash for many firearms. If carbon monoxide and oxygen are present in sufficient quantity, the reaction is spontaneous at high temperature and pressure.

The first step in the gas-phase reaction between carbon monoxide and oxygen is a collision between the two molecules:


Although there are many different possible orientations the two molecules can have relative to each other, consider the two presented in Figure 18.5.1. In the first case, the oxygen side of the carbon monoxide molecule collides with the oxygen molecule. In the second case, the carbon side of the carbon monoxide molecule collides with the oxygen molecule. The second case is clearly more likely to result in the formation of carbon dioxide, which has a central carbon atom bonded to two oxygen atoms [latex]\ce{(O=C=O)}[/latex]. This is a rather simple example of how important the orientation of the collision is in terms of creating the desired product of the reaction.

A diagram is shown that illustrates two possible collisions between C O and O subscript 2. In the diagram, oxygen atoms are represented as red spheres and carbon atoms are represented as black spheres. The diagram is divided into upper and lower halves by a horizontal dashed line. At the top left, a C O molecule is shown striking an O subscript 2 molecule such that the O atom from the C O molecule is at the point of collision. Surrounding this collision are a mix of molecules of C O, and O subscript 2 of varying sizes. At the top middle region of the figure, two separated O atoms are shown as red spheres with the label, “Oxygen to oxygen,” beneath them. To the upper right, “No reaction” is written. Similarly in the lower left of the diagram, a C O molecule is shown striking an O subscript 2 molecule such that the C atom from the C O molecule is at the point of collision. Surrounding this collision are a mix of molecules of C O, and O subscript 2 of varying sizes. At the lower middle region of the figure, a black sphere and a red spheres are shown with the label, “Carbon to oxygen,” beneath them. To the lower right, “More C O subscript 2 formation” is written and three models of C O subscript 2 composed each of a single central black sphere and two red spheres in a linear arrangement are shown.
Figure 18.5.1. Illustrated are two collisions that might take place between carbon monoxide and oxygen molecules. The orientation of the colliding molecules partially determines whether a reaction between the two molecules will occur.

If the collision does take place with the correct orientation, there is still no guarantee that the reaction will proceed to form carbon dioxide. In addition to a proper orientation, the collision must also occur with sufficient energy to result in product formation. When reactant species collide with both proper orientation and adequate energy, they combine to form an unstable species called an activated complex or a transition state. These species are very short lived and usually undetectable by most analytical instruments. In some cases, sophisticated spectral measurements have been used to observe transition states.

Collision theory explains why most reaction rates increase as concentrations increase. With an increase in the concentration of any reacting substance, the chances for collisions between molecules are increased because there are more molecules per unit of volume. More collisions mean a faster reaction rate, assuming the energy of the collisions is adequate.

Activation Energy and the Arrhenius Equation

The minimum energy necessary to form a product during a collision between reactants is called the activation energy (Ea). 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.

A graph is shown with the label, “Extent of reaction,” bon the x-axis and the label, “Energy,” on the y-axis. Above the x-axis, a portion of a 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 red 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 red 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.”
Figure 18.5.3. Reaction diagram for the exothermic reaction A + B [latex]\longrightarrow[/latex] C + D

Figure 18.5.3 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: [latex]A+B\rightarrow C+D[/latex].

These reaction diagrams are widely used in chemical kinetics to illustrate various properties of the reaction of interest. Viewing the diagram from left to right, the system initially comprises reactants only, A + B. Reactant molecules with sufficient energy can collide to form a high-energy activated complex or transition state. The unstable transition state can then subsequently decay to yield stable products, C + D. The diagram depicts the reaction’s activation energy, Ea, as the energy difference between the reactants and the transition state. Using a specific energy, the enthalpy (see chapter on thermochemsitry), the enthalpy change of the reaction, [latex]\Delta[/latex]H, is estimated as the energy difference between the reactants and products. In this case, the reaction is exothermic ([latex]\Delta[/latex]H < 0) since it yields a decrease in system enthalpy.

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


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, Ea 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.

Postulates of collision theory are nicely accommodated by the Arrhenius equation. The frequency factor, A, reflects how well the reaction conditions favor properly oriented collisions between reactant molecules. An increased probability of effectively oriented collisions results in larger values for A and faster reaction rates.

The exponential term, [latex]{e}^{-E_{\text{a}}\text{/}RT}[/latex], describes the effect of activation energy on reaction rate. According to kinetic molecular theory (see the chapter on gases), the temperature of matter is a measure of the average kinetic energy of its constituent atoms or molecules. The distribution of energies among the molecules composing a sample of matter at any given temperature is described by the plot shown in Figure 18.5.4(a). Two shaded areas under the curve represent the numbers of molecules possessing adequate energy (RT) to overcome the activation barriers (Ea). A lower activation energy results in a greater fraction of adequately energized molecules and a faster reaction.

The exponential term also describes the effect of temperature on reaction rate. A higher temperature represents a correspondingly greater fraction of molecules possessing sufficient energy (RT) to overcome the activation barrier (Ea), as shown in Figure 18.5.4(b). This yields a greater value for the rate constant and a correspondingly faster reaction rate.

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.
Figure 18.5.4. Molecular energy distributions showing numbers of molecules with energies exceeding (a) two different activation energies at a given temperature, and (b) a given activation energy at two different temperatures.

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

[latex]\begin{array}{ccc}\hfill \text{ln }k& =& \left(\dfrac{{-}{E}_{\text{a}}}{R}\right)\left(\dfrac{1}{T}\right)+\text{ln }A\hfill \\ \hfill y& =& mx+b\hfill \end{array}[/latex]

A plot of ln(k) versus [latex]\dfrac{1}{T}[/latex] is linear with the slope [latex]\dfrac{{-}{E}_{\text{a}}}{R}[/latex], from which Ea may be determined. The intercept gives the value of ln(A).

Example 18.5.1: Determination of Ea

The variation of the rate constant with temperature for the decomposition of [latex]\ce{HI}[/latex](g) to [latex]\ce{H2}[/latex](g) and [latex]\ce{I2}[/latex](g) is given below. What is the activation energy for the reaction?


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
Show Solution

Values of [latex]\frac{1}{T}[/latex] and ln k are:

[latex]\frac{1}{\text{T}}\left({\text{K}}^{-1}\right)[/latex] 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

Figure 18.5.5 is a graph of lnk versus [latex]\frac{1}{T}.[/latex]

In practice, the equation of the line (slope and y-intercept) that best fits these plotted data points would be derived using a statistical process called regression. This is helpful for most experimental data because a perfect fit of each data point with the line is rarely encountered. For the data here, the fit is nearly perfect and the slope may be estimated using any two of the provided data pairs. Using the first and last data points permits estimation of the slope.

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.”
Figure 18.5.5. This graph shows the linear relationship between lnk and 1/T for the reaction 2HI → H2 + I2 according to the Arrhenius equation.

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

[latex]\begin{array}{cc}\hfill \text{Slope}& =\dfrac{\Delta\left(\mathrm{ln}k\right)}{\Delta\left(\frac{1}{T}\right)}\hfill \\ & =\dfrac{\left(-14.447\right)-\left(-2.593\right)}{\left(1.78\times {10}^{-3}{\text{K}}^{-1}\right)-\left(1.25\times {10}^{-3}{\text{K}}^{-1}\right)}\hfill \\ & =\dfrac{-11.854}{0.53\times {10}^{-3}{\text{K}}^{-1}}=2.2\times {10}^{4}\text{K}\hfill \\ & =-\dfrac{{E}_{\text{a}}}{R}\hfill \end{array}[/latex]


Ea = –slope × R = –(–2.2 × 104 K × 8.314 J mol–1 K–1)

Ea = 1.8 × 105 J mol–1

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:


can be rearranged as shown to give:




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


Using the experimental data presented below, 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) [latex]\frac{1}{T}\left({\text{K}}^{-1}\right)[/latex] lnk
555 3.52 × 10–7 1.80 × 10–3 –14.860
781 3.95 × 10–2 1.28 × 10–3 –3.231

After calculating [latex]\dfrac{1}{T}[/latex] and ln k, we can substitute into the equation:

[latex]{E}_{\text{a}}=-8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\left(\dfrac{-3.231-\left(-14.860\right)}{1.28\times {10}^{-3}{\text{K}}^{-1}-1.80\times {10}^{-3}{\text{K}}^{-1}}\right)[/latex]

and the result is Ea = [latex]1.8\times{10}^{5}\text{ J Mol}^{-1}[/latex] or [latex]180\text{ kJ/mol}[/latex].

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

Check Your Learning

Key Concepts and Summary

Chemical reactions require collisions between reactant species. These reactant collisions must be of proper orientation and sufficient energy in order to result in product formation. Collision theory provides a simple but effective explanation for the effect of many experimental parameters on reaction rates. The Arrhenius equation describes the relation between a reaction’s rate constant and its activation energy, temperature, and dependence on collision orientation.

Key Equations

  • [latex]k=A{e}^{{-}{E}_{\text{a}}\text{/}RT}[/latex]
  • [latex]\text{ln}k=\left(\dfrac{{-}{E}_{\text{a}}}{R}\right)\left(\dfrac{1}{T}\right)+\text{ln}A[/latex]
  • [latex]\text{ln}\dfrac{{k}_{1}}{{k}_{2}}=\dfrac{{E}_{\text{a}}}{R}\left(\dfrac{1}{{T}_{2}}-\dfrac{1}{{T}_{1}}\right)[/latex]

Try It

  1. The rate constant for the decomposition of acetaldehyde, [latex]\ce{CH3CHO}[/latex], to methane, [latex]\ce{CH4}[/latex], and carbon monoxide, [latex]\ce{CO}[/latex], in the gas phase is 1.1 × 10−2 L/mol/s at 703 K and 4.95 L/mol/s at 865 K. Determine the activation energy for this decomposition.
Show Selected Solutions



activated complex: (also, transition state) unstable combination of reactant species representing the highest energy state of a reaction system

activation energy (Ea): energy necessary in order for a reaction to take place

Arrhenius equation: mathematical relationship between the rate constant and the activation energy of a reaction

collision theory: model that emphasizes the energy and orientation of molecular collisions to explain and predict reaction kinetics

frequency factor (A): proportionality constant in the Arrhenius equation, related to the relative number of collisions having an orientation capable of leading to product formation

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