Collision Theory

LEARNING OBJECTIVES

By the end of this module, you will be able to:

  • 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 \frac{\#\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]2\text{CO(}g\text{)}+{\text{O}}_{2}\text{(}g\text{)}\rightarrow 2{\text{CO}}_{2}\text{(}g\text{)}[/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:

[latex]\text{CO(}g\text{)}+{\text{O}}_{2}\text{(}g\text{)}\rightarrow{\text{CO}}_{2}\text{(}g\text{)}+\text{O(}g\text{)}[/latex]

Although there are many different possible orientations the two molecules can have relative to each other, consider the two presented in Figure 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 (O=C=O). 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 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. Every reaction requires a certain amount of activation energy for it to proceed in the forward direction, yielding an appropriate activated complex along the way. As Figure 2 demonstrates, even a collision with the correct orientation can fail to form the reaction product. In the study of reaction mechanisms, each of these three arrangements of atoms is called a proposed activated complex or transition state.

This figure shows three rows of structures. In the first row, an O atom on the left is connected to a C atom on its right with a double bond indicated by a pair of short parallel line segments. To the right of the C atom are three dots in a horizontal row followed by an O atom double bonded to another O atom on its right. In the second row, an O atom is followed by three dots in a horizontal row, which are followed by a C atom and a second grouping of three dots. To the right is an O atom double bonded to another O atom. In the third row, an O atom on the left is connected to a C atom on its right with a double bond indicated by a pair of short parallel line segments. To the right of the C atom are three dots in a horizontal row followed by an O atom followed by another grouping of three dots to another O atom on its right.

Figure 2. Possible transition states (activated complexes) for carbon monoxide reacting with oxygen to form carbon dioxide. Solid lines represent covalent bonds, while dotted lines represent unstable orbital overlaps that may, or may not, become covalent bonds as product is formed. In the first two examples in this figure, the O=O double bond is not impacted; therefore, carbon dioxide cannot form. The third proposed transition state will result in the formation of carbon dioxide if the third “extra” oxygen atom separates from the rest of the molecule.

In most circumstances, it is impossible to isolate or identify a transition state or activated complex. In the reaction between carbon monoxide and oxygen to form carbon dioxide, activated complexes have only been observed spectroscopically in systems that utilize a heterogeneous catalyst. The gas-phase reaction occurs too rapidly to isolate any such chemical compound.

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.

Figure 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]

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 Ea, the activation energy. Thus, the sum of the kinetic energies of A and B must be equal to or greater than Ea 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 Figure 3, ΔH represents the difference in enthalpy between the reactants (A and B) and the products (C and D). The sum of Ea and ΔH represents the activation energy for the reverse reaction:

[latex]C+D\rightarrow A+B[/latex]

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.”

Figure 3. This graph shows the potential energy relationships for the reaction A + B → 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 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 Ea. 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 is represented the transition state.

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

[latex]k=A{e}^{\text{-}{E}_{\text{a}}\text{/}RT}[/latex]

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.

Both postulates of the collision theory of reaction rates 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, [latex]{e}^{\text{-}{E}_{\text{a}}\text{/}RT},[/latex] 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 Ea results in a smaller value for [latex]{e}^{\text{-}{E}_{\text{a}}\text{/}RT},[/latex] reflecting the smaller fraction of molecules with enough energy to react. Alternatively, the reaction with the smaller Ea has a larger fraction of molecules with enough energy to react. This will be reflected as a larger value of [latex]{e}^{\text{-}{E}_{\text{a}}\text{/}RT},[/latex] 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 (Figure 4), as indicated by an increase in the value of [latex]{e}^{\text{-}{E}_{\text{a}}\text{/}RT}.[/latex] 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.

Figure 4. (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 Ea, as shown by the yellow shaded area.

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

Thus, a plot of lnk versus [latex]\frac{1}{T}[/latex] gives a straight line with the slope [latex]\frac{\text{-}{E}_{\text{a}}}{R},[/latex] from which Ea may be determined. The intercept gives the value of ln A.

Example 1

Determination of Ea

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

[latex]\text{2HI}\left(g\right)\rightarrow{\text{H}}_{2}\left(g\right)+{\text{I}}_{2}\left(g\right)[/latex]
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 [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 5 is a graph of lnk versus [latex]\frac{1}{T}.[/latex] To determine the slope of the line, we need two values of ln k, which are determined from the line at two values of [latex]\frac{1}{T}[/latex] (one near each end of the line is preferable). For example, the value of lnk determined from the line when [latex]\frac{1}{T}=1.25\times {10}^{-3}[/latex] is –2.593; the value when [latex]\frac{1}{T}=1.78\times {10}^{-3}[/latex] 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.”

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

Thus:

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:

[latex]\text{ln}k=\left(\frac{\text{-}{E}_{\text{a}}}{R}\right)\left(\frac{1}{T}\right)+\text{ln}A[/latex]

can be rearranged as shown to give:

[latex]\frac{\Delta\left(\text{ln}k\right)}{\Delta\left(\frac{1}{T}\right)}=-\frac{{E}_{\text{a}}}{R}[/latex]

or

[latex]\text{ln}\frac{{k}_{1}}{{k}_{2}}=\frac{{E}_{\text{a}}}{R}\left(\frac{1}{{T}_{2}}-\frac{1}{{T}_{1}}\right)[/latex]

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

[latex]{E}_{\text{a}}=-R\left(\frac{\text{ln}{k}_{2}-\text{ln}{k}_{1}}{\left(\frac{1}{{T}_{2}}\right)-\left(\frac{1}{{T}_{1}}\right)}\right)[/latex]

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]\frac{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(\frac{-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 = 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.

Check Your Learning

The rate constant for the rate of decomposition of N2O5 to NO and O2 in the gas phase is 1.66 L/mol/s at 650 K and 7.39 L/mol/s at 700 K:

[latex]{\text{2N}}_{2}{\text{O}}_{5}\left(g\right)\rightarrow 4\text{NO}\left(g\right)+{\text{3O}}_{2}\left(g\right)[/latex]

Assuming the kinetics of this reaction are consistent with the Arrhenius equation, calculate the activation energy for this decomposition.

Answer: 113,000 J/mol

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

Chemistry End of Chapter Exercises

  1. Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?
  2. When every collision between reactants leads to a reaction, what determines the rate at which the reaction occurs?
  3. What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction?
  4. Account for the relationship between the rate of a reaction and its activation energy.
  5. Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures.
  6. How does an increase in temperature affect rate of reaction? Explain this effect in terms of the collision theory of the reaction rate.
  7. The rate of a certain reaction doubles for every 10 °C rise in temperature.
    1. (a) How much faster does the reaction proceed at 45 °C than at 25 °C?
    2. (b) How much faster does the reaction proceed at 95 °C than at 25 °C?
  8. In an experiment, a sample of NaClO3 was 90% decomposed in 48 min. Approximately how long would this decomposition have taken if the sample had been heated 20 °C higher?
  9. The rate constant at 325 °C for the decomposition reaction [latex]{\text{C}}_{4}{\text{H}}_{8}\rightarrow{2\text{C}}_{2}{\text{H}}_{4}[/latex] is 6.1 × 10-8 s-1, and the activation energy is 261 kJ per mole of C4H8. Determine the frequency factor for the reaction.
  10. The rate constant for the decomposition of acetaldehyde, CH3CHO, to methane, CH4, and carbon monoxide, CO, 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.
  11. An elevated level of the enzyme alkaline phosphatase (ALP) in the serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from 30 °C to 37 °C. What is the activation energy for the ALP–catalyzed conversion of PNPP to PNP and phosphate?
  12. In terms of collision theory, to which of the following is the rate of a chemical reaction proportional?
    1. the change in free energy per second
    2. the change in temperature per second
    3. the number of collisions per second
    4. the number of product molecules
  13. Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H2, and iodine, I2. The value of the rate constant, k, for the reaction was measured at several different temperatures and the data are shown here:
    Temperature (K) k (M-1 s-1)
    555 6.23 × 10-7
    575 2.42 × 10-6
    645 1.44 × 10-4
    700 2.01 × 10-3

    What is the value of the activation energy (in kJ/mol) for this reaction?

  14. The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data:
    T (K) k (s-1)
    293 0.054
    298 0.100
  15. The hydrolysis of the sugar sucrose to the sugars glucose and fructose,[latex]{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}+{\text{H}}_{2}\text{O}\rightarrow{\text{C}}_{6}{\text{H}}_{12}{\text{O}}_{6}+{\text{C}}_{6}{\text{H}}_{12}{\text{O}}_{6}[/latex]follows a first-order rate equation for the disappearance of sucrose: [latex]\text{Rate}=k\left[{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\right][/latex] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)
    1. In neutral solution, k = 2.1 × 10-11 s-1 at 27 °C and 8.5 × 10-11 s-1 at 37 °C. Determine the activation energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
    2. When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is 1.65 × 10-7M. How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
    3. Why does assuming that the reaction is irreversible simplify the calculation in part (b)?
  16. Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first [latex]A+BC\rightarrow AB+C[/latex] reaction (A is yellow, B is purple, and C is navy blue). Using the “straight shot” default option, try launching the A atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why?
  17. Use the PhET Reactions & Rates interactive simulation to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first [latex]A+BC\rightarrow AB+C[/latex] reaction (A is yellow, B is purple, and C is navy blue). Using the “angled shot” option, try launching the A atom with varying angles, but with more Total energy than the transition state. What happens when the A atom hits the BC molecule from different directions? Why?

Selected Answers

1. The reactants either may be moving too slowly to have enough kinetic energy to exceed the activation energy for the reaction, or the orientation of the molecules when they collide may prevent the reaction from occurring.

3. The activation energy is the minimum amount of energy necessary to form the activated complex in a reaction. It is usually expressed as the energy necessary to form one mole of activated complex.

5. After finding k at several different temperatures, a plot of lnk versus [latex]\frac{\text{-}{E}_{\text{a}}}{R}[/latex] gives a straight line with the slope [latex]\frac{1}{T},[/latex] from which Ea may be determined.

7. (a) The rate doubles for each 10 °C rise in temperature; 45 °C is a 20 °C increases over 25 °C. Thus, the rate doubles two times, or 22 (rate at 25 °C) = 4-times faster.

(b) 95 °C is a 70 °C increases over 25 °C. Thus the rate doubles seven times, or 27 (rate at 25 °C) = 128-times faster.

9. The rate constant k is related to the activation energy Ea by a relationship known as the Arrhenius equation. Its form is:

[latex]k=A\times {10}^{-\left({E}_{\text{a}}\text{/}2.303RT\right)}=A\times {e}^{-\left({E}_{\text{a}}\text{/}RT\right)}[/latex]

where A is the frequency factor. Using the data provided, and converting kilojoules to joules:

[latex]\begin{array}{cc}\hfill 6.1\times {10}^{-8}{\text{s}}^{-1}& =A\times {10}^{-\left[+261.000\text{J}\text{/}2.303\text{(}8.314\text{J}{\text{K}}^{-1}\text{)}\text{(}325+\text{273)K}\right]}\\ & =A\times {10}^{-22.8}\hfill \end{array}[/latex]

[latex]A=\frac{6.1\times {10}^{-8}{\text{s}}^{-1}}{1.58\times {10}^{-23}}=3.9\times {10}^{15}{\text{s}}^{-1}[/latex]

11. Note that [latex]{e}^{\text{-}x}={10}^{\text{-}x\text{/}2.303}.[/latex] Changes in rate brought about by temperature changes are governed by the Arrhenius equation: [latex]k=A\times {10}^{\text{-}{E}_{\text{a}}\text{/}2.303RT}.[/latex] In this particular reaction, k increases by 1.47 as T changes from 30 °C (303 K). The Arrhenius equation may be solved for A under both sets of conditions and then A can be eliminated between the two equations. Eliminating k from both sides, taking logs, and rearranging gives:

[latex]\begin{array}{l}\frac{\text{-}{E}_{\text{a}}}{2.303\times 8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\left(310\text{K}\right)}=\mathrm{log}1.47-\frac{{E}_{\text{a}}}{2.303\times 8.314\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}\left(303\text{K}\right)}\\ \frac{\text{-}{E}_{\text{a}}}{5935.6\text{J}{\text{mol}}^{-1}}=0.1673-\frac{{E}_{\text{a}}}{5801.6\text{J}{\text{mol}}^{-1}}\\ \frac{{E}_{\text{a}}}{5801.6}-\frac{{E}_{\text{a}}}{5935.6}=0.1673\text{J}{\text{mol}}^{-1}\end{array}[/latex]

Ea(1.72366 × 10-4 – 1.68474 × 10-4) = 0.1673 J/mol

3.892 × 10-6Ea = 0.1673 J/mol

Ea = 42986 J/mol = 43.0 kJ/mol

13.  Ea may be determined from a plot of ln k against [latex]\frac{1}{T}[/latex] that gives a straight line whose slope is [latex]\frac{\text{-}{E}_{\text{a}}}{R}:[/latex]

T (K) [latex]\frac{1}{T}\times {10}^{3}[/latex] k (M-1 s-1) lnk
555 1.802 6.23 × 10-7 -14.289
575 1.739 2.42 × 10-6 -12.932
645 1.550 1.44 × 10-4 -8.846
700 1.429 2.42 × 10-3 -6.210

A plot of this data shows a straight line. Two points marked by an X are picked for convenience of reading and are used to determine the slope of the line:


A graph is shown with the label, “1 divided by T times 10 superscript 3 ( K superscript negative 1 ),” on the x-axis and, “ln k,” on the y-axis. The x-axis has markings beginning at 1.4, increasing by 0.10 up to and including 1.80. The y-axis shows markings of negative 14 through negative 6 at intervals of 1. A decreasing linear trend line is drawn through four points represented as black dots at the coordinates: (1.429, negative 6.210), (1.550, negative 8.846), (1.739, negative 12.932), and (1.802, negative 14.289). A vertical dashed line is drawn from the point (negative 6.75, 1.45) which is just right of the data point nearest the y-axis. Similarly, a horizontal dashed line is draw from the point (negative 14.75, 1.825) which is just below 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.”

[latex]\begin{array}{lll}\hfill \text{slope}& =& \frac{-14.750-\left(-6.750\right)}{1.825\times {10}^{-3}-1.450\times {10}^{-3}}=\frac{-8.000}{3.75\times {10}^{-4}}=-2.13\times {10}^{4}\hfill \\ \hfill \frac{\text{-}{E}_{\text{a}}}{R}& =& -2.13\times {10}^{4}\hfill \end{array}[/latex]

Ea = -2.13 × 104 × 8.314 J/mol = 177 kJ/mol

15.  (a) The text demonstrates that the value of Ea may be determined from a plot of logk against [latex]\frac{1}{T}[/latex] that gives a straight line whose slope is [latex]\frac{\text{-}{E}_{\text{a}}}{2.303R}.[/latex] This relationship is based on the equation [latex]\text{ln}k=\text{ln}A-\frac{{E}_{\text{a}}}{RT}[/latex] or [latex]\text{log}k=\text{log}A-\frac{{E}_{\text{a}}}{2.303RT}[/latex] where [latex]\text{In}k=2.303\text{log}k.[/latex] Only two data points are given, and these must determine a straight line when logk is plotted against 1/T. The values needed are:

k1 = 2.1 × 10-11

log k1 = -10.6778

k2 = 8.5 × 10-11

log k2 = -10.0706

T1 = 27 °C = 300 K

[latex]\frac{1}{{T}_{1}}[/latex] = 3.3333 × 10-3

T2 = 37 °C = 310 K

[latex]\frac{1}{{T}_{2}}[/latex] = 3.2258 × 10-3

The slope of the line determined by these points is given by:

[latex]\begin{array}{cc}\hfill \text{Slope}& =\frac{\Delta\left(\text{log}k\right)}{\Delta\frac{1}{T}}=\frac{\left(-10.0706\right)-\left(-10.6778\right)}{\left(3.2258\times {10}^{-3}\right)-\left(3.3333\times {10}^{-3}\right)}\hfill \\ & =\frac{0.6072}{-0.1075\times {10}^{-3}}=-5648\hfill \end{array}[/latex]

Ea = 2.303(8.314 J/mol)(-5648) = 108,100 J = 108 kJ

Whenever differences of very small numbers are taken, such as the reciprocals of T provided, an inherent problem occurs. To have accurate differences, a larger number of significant figures than justified by the data must be used. Thus five figures were used to obtain the value Ea = 108 kJ. This difficulty may be alleviated by the following approach.

For only two data points, the Arrhenius equation [latex]k=A\times {10}^{\text{-}{E}_{\text{a}}\text{/}2.303RT}[/latex] may be used in an equally accurate, analytical solution for Ea. This application is possible because the value of A will be the same throughout the course of the reaction. Once the value of Ea is determined, the value of A may be determined from either Equation (1) or (2). Then k at 47 °C may be determined using the value of Ea and A so determined. The procedure is as follows:

[latex]\begin{array}{cccc}\hfill k& =& A\times {10}^{\text{-}{E}_{\text{a}}\text{/}2.303RT}\hfill & \\ \hfill 2.1\times {10}^{-11}{\text{s}}^{-1}& =& A\times {10}^{\text{-}{E}_{\text{a}}\text{/}2.303\left(8.314\text{J}{\text{K}}^{-1}\right)\text{(300 K)}}\hfill & \text{(equation 1)}\hfill \\ \hfill 8.5\times {10}^{-11}{\text{s}}^{-1}& =& A\times {10}^{\text{-}{E}_{\text{a}}\text{/}2.303\left(8.314\text{J}{\text{K}}^{-1}\right)\text{(300 K)}}\hfill & \text{(equation 2)}\hfill \end{array}[/latex]

Equating the values of A as solved from equations (1) and (2):

[latex]2.1\times {10}^{-11}{\text{s}}^{-1}\times {10}^{+{E}_{\text{a}}\text{/}2.303\left(8.314\text{J}{\text{K}}^{-1}\right)\text{(300 K)}}=8.5\times {10}^{-11}{\text{s}}^{-1}\times {10}^{+{E}_{\text{a}}\text{/}2.303\left(8.314\text{J}{\text{K}}^{-1}\right)\text{(300 K)}}[/latex] or [latex]2.1\times {10}^{-11}{\text{s}}^{-1}\times {10}^{+{E}_{\text{a}}\text{/}5744}=8.5\times {10}^{-11}{\text{s}}^{-1}\times {10}^{+{E}_{\text{a}}\text{/}5936}.[/latex]

Taking common logs of both sides gives:

[latex]\begin{array}{l}\\ \\ \left(\mathrm{log}2.1\times {10}^{-11}\right)+\frac{{E}_{\text{a}}}{5744}=\left(\mathrm{log}8.5\times {10}^{-11}\right)+\frac{{E}_{\text{a}}}{5936}-10.68+\frac{{E}_{\text{a}}}{5744}=-10.07+\frac{{E}_{\text{a}}}{5936}\\ {E}_{\text{a}}\left(\frac{1}{5744}-\frac{1}{5936}\right)=-10.07+10.68\\ {E}_{\text{a}}\left(1.741\times {10}^{-4}-1.685\times {10}^{-4}\right)=0.61\\ {E}_{\text{a}}=\frac{0.61}{0.056\times {10}^{-4}}=109\text{kJ}\end{array}[/latex]

The value of A may be found from either equation (1) or (2). Using equation (1):

[latex]2.1\times {10}^{-11}{\text{s}}^{-1}=A\times {10}^{\text{-109,000/2.303(8.314)(300)}}=A\times {10}^{-18.98}[/latex]

A = 2.1 × 10-11 s-1 × 10+18.91 = 2.1 × 10-11(9.55 × 1018 s-1) = 2.0 × 108 s-1

The value of k at 47°C may be determined from the Arrhenius equation now that the values of Ea and A have been calculated:

[latex]\begin{array}{cc}\hfill k& =A\times {10}^{\text{-}{E}_{\text{a}}\text{/}2.303RT}\hfill \\ & =2.0\times {10}^{8}{\text{s}}^{-1}\times {10}^{\text{-109,000 J/2.303(8.314 J}{\text{K}}^{-1}\text{)}\text{(320 K)}}\hfill \end{array}[/latex]

= 2.0 × 108 s-1 × 10-17.79 = 2.0 × 108(1.62 × 10-18) = 3.2 × 10-10 s-1

Using the earlier value of Ea = 108 kJ, the calculated value of A is 1.3 × 108 s-1, and k = 3.1 × 10-10 s-1. Either answer is acceptable.

(b) Since this is a first-order reaction we can use the integrated form of the rate equation to calculate the time that it takes for a reactant to fall from an initial concentration [A]0 to some final concentration [A]:

[latex]\text{ln}\frac{{\left[\text{A}\right]}_{0}}{\left[\text{A}\right]}-kt[/latex]

At 27 °C. k = 2.1 × 10-11 s-1.

In this case, the initial concentration is 0.150 M and the final concentration is 1.65 × 10-7M. We can now solve for the time t:

[latex]\begin{array}{l}\\ \\ \text{ln}\frac{\left[0.150\text{M}\right]}{\left[1.65\times {10}^{-7}\text{M}\right]}=\left(2.1\times {10}^{-11}{\text{s}}^{-1}\right)\left(t\right)\\ t=\frac{13.720}{2.1\times {10}^{-11}{\text{s}}^{-1}}=6.5\times {10}^{11}\text{s}\end{array}[/latex]

or 1.81 × 108 h or 7.6 × 106 day. (c) Assuming that the reaction is irreversible simplifies the calculation because we do not have to account for any reactant that, having been converted to product, returns to the original state.

17. The A atom has enough energy to react with BC; however, the different angles at which it bounces off of BC without reacting indicate that the orientation of the molecule is an important part of the reaction kinetics and determines whether a reaction will occur.

Glossary

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