1. Activation Energy Section 16.3 (AHL)

2. Introduction All chemical reactions require minimum energy (activation energy, E a ) to occur At higher temperatures, a greater proportion of the molecules have an energy greater than the activation energy This is the major reason why reactions occur more rapidly at higher temperatures

3. Don’t forget! The Maxwell-Boltzmann Distribution

4. Arrhenius Equation The effect of temperature on the rate constant for many reactions is found by the expression: This is called the Arrhenius equation

5. Meaning E a is the activation energy T is the absolute temperature in Kelvin R is the gas constant (8.314 J K -1 mol -1 ) A is called the Arrhenius constant The Arrhenius constant is dependent on the collision rate and steric factors

6. Simplified Arrhenius Equation If you take logarithms to the base e and simplify you get: This is the equation of a straight line The rate can be found by calculating the gradient of the line

8. Example Problem The rate constant, k, was determined for a reaction at various temperatures. Here are the results: k in mol -1 dm 3 s -1 Temperature in °C 6.81 x 10 -6 5.00 1.40 x 10 -5 15.00 2.93 x 10 -5 25.00 6.11 x 10 -5 35.00

9. Problem Continued First draw up a table showing T -1 and ln k values, T must be in absolute temp. (K) Plot a graph of ln k against T -1 Take the ln k axis from -12 to -8 and the T -1 axis from 0.0030 to 0.0038 (Use graph paper provided)

10. ln k mol -1 dm 3 s -1 T -1 (K-1) -11.903.597x10 -3 -11.183.472x10 -3 -10.443.356x10 -3 -9.703.247x10 -3 Table of lnk and T -1 to graph

11. Problem Continued Calculate the gradient of your Arrhenius plot and use it to determine a value for E a in kJ mol -1 y-intercept = about -8.28 Gradient is about -6098 So y = -6098x - 8.28

12. Problem Continued Gradient = -E a /R = -6098 -E a = (-6098 K) (8.314 J mol -1 K -1 ) E a = 50698 J mol -1 E a = 50.7 kJ mol -1

13. Problem Continued Calculate an approximate value for the Arrhenius constant, A, using the Arrhenius plot ln A is the y-intercept, so ln A = -8.28 A = 2.54 x 10 -4 mol -1 dm 3 s -1 A reaction with a steeper gradient will have a higher E a

14. Checking with Excel

16. Catalysts There are two types of catalysts: Heterogeneous catalyst Homogeneous catalyst

17. Heterogeneous Catalyst This is when the catalyst and the reactants are in different phases Often this involves the reaction of gas molecules on the surface of a solid

18. Homogeneous Catalyst This is when the catalyst and the reactants are in the same phase Usually involves the formation of an intermediate during the reaction