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Figure 1. A plot of the natural log of the rate constant for the reaction at different temperatures versus the inverse of the temperature in kelvin is a straight line with a slope equal to -Ea/R. |
Practice Problem 11
Use the following data to determine the activation energy for the decomposition of HI:
Temperature (K) Rate Constant (M/s)
573 2.91 x 10-6
673 8.38 x 10-4
773 7.65 x 10-2
Solution
We can determine the activation energy for a reaction from a plot of the natural log of the rate constants versus the reciprocal of the absolute temperature. We therefore start by calculating 1/T and the natural logarithm of the rate constants:
| ln k | 1/T (K-1) |
| -12.75 | 0.00175 |
| -7.08 | 0.00149 |
| -2.57 | 0.00129 |
When we construct a graph of these data, we get a straight line with a slope of -22,200 K, as seen in Figure 1.
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Figure 1. A plot of the natural log of the rate constant for the reaction at different temperatures versus the inverse of the temperature in kelvin is a straight line with a slope equal to -Ea/R. |
According to the Arrhenius equation, the slope of this line is equal to -Ea/R:
When this equation is solved, we get the follwoing value for the activation energy for this reaction:
Ea = 183 kJ/mol
