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12.5 Collision theory  (Page 4/11)

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A convenient approach to determining E a 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:

ln k = ( − E a R ) ( 1 T ) + ln A y = m x + b

Thus, a plot of ln k versus 1 T gives a straight line with the slope − E a R , from which E a may be determined. The intercept gives the value of ln A .

Determination of E a

The variation of the rate constant with temperature for the decomposition of HI( g ) to H 2 ( g ) and I 2 ( g ) is given here. What is the activation energy for the reaction?

2HI( g ) ⟶ H 2 ( g ) + I 2 ( 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

Solution

Values of 1 T and ln k are:

1 T ( 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

[link] is a graph of ln k versus 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 1 T (one near each end of the line is preferable). For example, the value of ln k determined from the line when 1 T = 1.25 × 10 −3 is −2.593; the value when 1 T = 1.78 × 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 1 T for the reaction 2HI ⟶ H 2 + I 2 according to the Arrhenius equation.

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

Slope = Δ ( ln k ) Δ ( 1 T ) = ( −14.447 ) − ( −2.593 ) ( 1.78 × 10 −3 K −1 ) − ( 1.25 × 10 −3 K −1 ) = −11.854 0.53 × 10 −3 K −1 = 2.2 × 10 4 K = − E a R

Thus:

E a = −slope × R = − ( −2.2 × 10 4 K × 8.314 J mol −1 K −1 )
E a = 1.8 × 10 5 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:

ln k = ( − E a R ) ( 1 T ) + ln A

can be rearranged as shown to give:

Δ ( ln k ) Δ ( 1 T ) = − E a R

or

ln k 1 k 2 = E a R ( 1 T 2 − 1 T 1 )

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

E a = − R ( ln k 2 − ln k 1 ( 1 T 2 ) − ( 1 T 1 ) )

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) 1 T ( K −1 ) 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 1 T and ln k , we can substitute into the equation:

E a = −8.314 J mol −1 K −1 ( −3.231 − ( −14.860 ) 1.28 × 10 −3 K −1 − 1.80 × 10 −3 K −1 )

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.

Check your learning

The rate constant for the rate of decomposition of N 2 O 5 to NO and O 2 in the gas phase is 1.66 L/mol/s at 650 K and 7.39 L/mol/s at 700 K:

2N 2 O 5 ( g ) ⟶ 4 NO ( g ) + 3O 2 ( g )

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.

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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