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Chemistry Textbook

Arrhenius Equation and Catalysis

Activation Energy and the Arrhenius Equation

Recall the postulate of collision theory which states that reactant molecules must collide with adequate energy in order to change to products (assuming they collide with proper orientation). The minimum energy necessary to form a product during a collision between reactants is called the activation energy (Ea). How this energy compares to the kinetic energy provided by colliding reactant molecules is a primary factor affecting the rate of a chemical reaction. If the activation energy is much larger than the average kinetic energy of the molecules, the reaction will occur slowly since 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, a large fraction of molecules will be adequately energetic and the reaction will proceed rapidly.

Figure 10.12 shows how the energy of a chemical system changes as it undergoes a reaction converting reactants to products according to the equation

A+BC+DA+BC+D

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 unstable species called 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. The enthalpy change of the reaction, ΔH, is estimated as the energy difference between the reactants and products. In this case, the reaction is exothermic (ΔH < 0) since the products have lower enthalpy than reactants.

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 10.12 Reaction diagram for the exothermic reaction A+BC+D.A+BC+D.

The Arrhenius equation relates the activation energy and the rate constant, k, for many chemical reactions:

k=AeEa/RTk=AeEa/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, 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, e−Ea/RT, describes the effect of activation energy on reaction rate. According to kinetic molecular theory (see 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 10.13(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 10.13(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 10.13 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 for determining Ea for a reaction involves the measurement of k at two or more different temperatures and using an alternate version of the Arrhenius equation that takes the form of a linear equation

lnk=(EaR)(1T)+lnAy=mx+blnk=(EaR)(1T)+lnAy=mx+b

A plot of ln k versus 1T1T is linear with a slope equal to EaREaR and a y-intercept equal to ln A.

EXAMPLE 10.14

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 here. What is the activation energy for the reaction?

2HI(g)H2(g)+I2(g)2HI(g)H2(g)+I2(g)
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
Table 10.11

Solution Use the provided data to derive values of 1T1T and ln k:

1T(K−1)1T(K−1) 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
Table 10.12

Figure 10.14 is a graph of ln k versus 1T.1T. 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 negative 3, 1.6 times 10 superscript negative 3, and 1.8 times 10 superscript negative 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 10.14 This graph shows the linear relationship between ln k and 1T1T for the reaction 2HIH2+I22HIH2+I2 according to the Arrhenius equation.
Slope=Δ(lnk)Δ(1T)=(−14.860)(−3.231)(1.80×10−3K−1)(1.28×10−3K−1)=−11.6290.52×10−3K−1=2.2×104K=EaR Ea=−slope×R=(−2.2×104K×8.314 J mol−1K−1)1.8×105Jmol−1or180kJmol−1Slope=Δ(lnk)Δ(1T)=(−14.860)(−3.231)(1.80×10−3K−1)(1.28×10−3K−1)=−11.6290.52×10−3K−1=2.2×104K=EaR Ea=−slope×R=(−2.2×104K×8.314 J mol−1K−1)1.8×105Jmol−1or180kJmol−1

Alternative approach: A more expedient approach involves deriving activation energy from measurements of the rate constant at just two temperatures. In this approach, the Arrhenius equation is rearranged to a convenient two-point form:

lnk1k2=EaR(1T21T1)lnk1k2=EaR(1T21T1)

Rearranging this equation to isolate activation energy yields:

Ea=R(lnk2lnk1(1T2)(1T1))Ea=R(lnk2lnk1(1T2)(1T1))

Any two data pairs may be substituted into this equation—for example, the first and last entries from the above data table:

Ea=−8.314Jmol−1K−1(−3.231(−14.860)1.28×10−3K−11.80×10−3K−1)Ea=−8.314Jmol−1K−1(−3.231(−14.860)1.28×10−3K−11.80×10−3K−1)

and the result is Ea = 1.8 ×× 105 J mol−1 or 180 kJ mol−1

This approach yields the same result as the more rigorous graphical approach used above, as expected. In practice, the graphical approach typically provides more reliable results when working with actual experimental data.

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:

2N2O5(g)4NO(g)+3O2(g)2N2O5(g)4NO(g)+3O2(g)

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

Answer:

1.1 ×× 105 J mol−1 or 110 kJ mol−1

Catalysis

Among the factors affecting chemical reaction rates discussed earlier in this chapter was the presence of a catalyst, a substance that can increase the reaction rate without being consumed in the reaction. The concepts introduced in the previous section on reaction mechanisms provide the basis for understanding how catalysts are able to accomplish this very important function. Catalysts speed up the reaction by altering the reaction mechanism such that the new reaction mechanism has lower activation energy.

Figure 10.15 shows reaction diagrams for a chemical process in the absence and presence of a catalyst. Inspection of the diagrams reveals several traits of these reactions. Consistent with the fact that the two diagrams represent the same overall reaction, both curves begin and end at the same energies (in this case, because products are more energetic than reactants, the reaction is endothermic). The reaction mechanisms, however, are clearly different. The uncatalyzed reaction proceeds via a one-step mechanism (one transition state observed), whereas the catalyzed reaction follows a two-step mechanism (two transition states observed) with a notably lesser activation energy. This difference illustrates the means by which a catalyst functions to accelerate reactions, namely, by providing an alternative reaction mechanism with a lower activation energy. Although the catalyzed reaction mechanism for a reaction needn’t necessarily involve a different number of steps than the uncatalyzed mechanism, it must provide a reaction path whose rate determining step is faster (lower Ea).

Figure 10.15 Reaction diagrams for an endothermic process in the absence (red curve) and presence (blue curve) of a catalyst. The catalyzed pathway involves a two-step mechanism (note the presence of two transition states) and an intermediate species (represented by the valley between the two transitions states).

Key Concepts and Summary

Chemical reactions typically 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, activation energy, temperature, and dependence on collision orientation.

Key Equations

  • k=AeEa/RTk=AeEa/RT
  • lnk=(EaR)(1T)+lnAlnk=(EaR)(1T)+lnA
  • lnk1k2=EaR(1T21T1)lnk1k2=EaR(1T21T1)