1. 1 REACTION KINETICS Reaction rates Reaction order Reaction mechanisms Collision frequency Energy profile diagrams Arrhenius equation Catalysts

2. 2 AIMS To explain the factors which affect the rates of chemical reactions. To develop an ability to analyse experimental data to calculate reaction rates and rate constants. To explain how the temperature affects the rate of a reaction and to show how the effect can be predicted quantitatively.

3. 3 REACTION RATES Affected by: – chemical nature of the reactants – ability of the reactants to come in contact with each other – concentrations of the reactants – temperature of the reacting system – availability of catalysts

4. 4 Rate of reaction = = Ms -1

5. 5 Because the amounts of products and reactants are related by stoichiometry, any substance in the reaction can be used to express the reaction rate. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) propane oxygen carbon dioxide water – O 2 is reacting 5x faster than propane. – CO 2 is formed 3x faster than propane is reacting. – H 2 O is formed 4x faster than propane is reacting.

6. 6 C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 O (g) concentration of reactants decrease  ve concentration of products increase+ve rate of decrease of concentration with time rate of increase of concentration with time

7. 7 (a)N 2 + 3H 2  2NH 3 Rate = (b) H 2 O 2 + 2H + + 3I¯  I 3 ¯ + 2H 2 O Rate = PRACTICE EXAMPLES

8. 8 The Rate Law allows us to calculate the rate of the reaction if the concentrations of reactants are known. A + B  products Rate = k[A] m [B] n

9. 9 k = rate constant m = order with respect to reactant A n = order with respect to reactant B (m + n) = total order The exponents n and m – are determined by experiment by studying how changes in concentration affect the rate of reaction – often unrelated to stoichiometric coefficients – can be +ve,  ve or fractional depending on the mechanism.

10. 10 k depends on: the specific reaction the presence of a catalyst (if any) the temperature -larger the value of k, the faster the reaction goes. -units depend on the form of the rate law Rate = k[A] m [B] n

11. 11 PRACTICE EXAMPLE A + B  products rate = k[A] m [B] n CALCULATE m, n and k 1234512345

12. 12 PRACTICE EXAMPLE For the reaction: NO 2 (g) + CO(g)  NO(g) + CO 2 (g) the variation of the initial rate with the initial concentration of NO 2 (the initial concentration of CO kept constant), was found to be as follows: Experiment[NO 2 ]/mol dm -3 Initial rate/mol dm -3 s -1 10.1500.010 20.3000.040 30.6000.160 40.9000.360 Deduce the order of the reaction with respect to NO 2

13. 13 Concentration and time We often want to know the concentrations of reactants and products at some specified time after the reaction has started. By means of calculus, we can transform a rate law into a mathematical relationship between concentration and time called an integrated rate equation. A → products

14. 14 REACTION ORDER Zero Order Rate = – = k[A] 0 = k [A] t = -kt + [A] 0 [A] t /mol dm -3 t/s slope = -k k: mol dm –3 s –1 m + n + … =0

15. 15 First–order reactions Rate = – = k[A] 1 k = – ln = ktorln[A] t = -kt + ln[A] 0 ln[A] t t/s slope = -k k: s –1 m + n + … =1

16. 16 Second  order reactions Rate = – = k[A] 2 k = = kt t/s slope = k /mol -1 dm 3 k: mol -1 dm 3 s –1 m + n + … =2

17. 17 PRACTICE EXAMPLE t[A] 00.05 5000.0167 8000.0119 15000.0071 20000.0056 tln[A] 0-2.99573 500-4.09235 800-4.43122 1500-4.94766 2000-5.18499 For the alkaline hydrolysis of ethyl nitrobenzoate (A) the following data were obtained at 25  C. Time/s050080015002000 [A]/mol dm -3 0.05000.01670.01190.00710.0056 Evaluate the order of the reaction and calculate k t1/[A] 020 50059.88024 80084.03361 1500140.8451 2000178.5714

18. 18

19. 19 For the reaction A  products, the following data is given in the table below: [A]/MTime/s 0.6000 0.497100 0.413200 0.344300 0.285400 0.198600 0.0941000 A) Show that the reaction is first order by plotting the appropriate graph for first order reactions. B) What is the value of the rate constant, k? C) What is [A] at t = 750 s? PRACTICE EXAMPLE

20. time/sln[A] 0-0.51 100-0.699 200-0.884 300-1.07 400-1.26 600-1.62 1000-2.36  y/  x = (-2.36-(-0.51))/(1000-0) s= -1.85 x 10 -3 s -1

21. REACTION MECHANISMS elementary process : one-step reaction mechanism : entire series of elementary processes The number of reactant particles involved in an elementary reaction: – unimolecular – bimolecular – termolecular -orders wrt reactants in rate law for ELEMENTARY step is the same as stoichiometric coefficient of reactants -if species consumed in one elementary step and produced in another  not included in overall reaction molecularity

22. 22 1.N 2 O 5(g)  NO 2(g) + NO 3(g) 2. NO 2(g) + NO 3(g)  NO 2(g) + O 2(g) + NO (g) 3. NO (g) + NO 3(g)  2NO 2(g) 2N 2 O 5(g)  4NO 2(g) + O 2(g) unimolecular → bimolecular ← bimolecular Elementary step mechanism

23. COLLISION FREQUENCY Number of collisions between gas molecules Depends on: – Temperature – Concentration – Steric factors (orientation) – Energy (to break bonds)

24. N 2 O (g) + NO (g) ⇌ N 2(g) + NO 2(g)

25. Activation energy Minimum energy above the average kinetic energy that molecules must bring to their collisions for a reaction to occur

26. N 2 O (g) + NO (g) ⇌ N 2(g) + O 2(g) ENERGY PROFILES

27. PRACTICE EXAMPLE A single step reversible reaction has an activation energy for the forward reaction (E a,f ) of 28.9 kJ and 41.8 kJ for the reverse reaction (E a,r ) Draw a potential energy level diagram and indicate  H

28. ARRHENIUS EQUATION describes the way in which a reaction rate changes with temperature.

29. A  pre-exponential factor or frequency factor -collision rate and is effectively temperature independent. E a  activation energy -energy the molecules require to make a collision reactive. k  rate constant -characteristic of a specific reaction -indicates the fraction of successful collisions at a given temperature -varies with temperature -for fast reactions k is large -for slow reactions k is small.

30. Hence the rate constant ( ) is given by the total number of collisions per second (A) multiplied by the fraction of those collisions which involve molecules with sufficient energy to react( ). R = 8.314 J mol -1 K -1

31. Plotting the Arrhenius Equation If ln k for a reaction is plotted against 1/T, E a and A are obtained from the slope and intercept respectively.

32. Arrhenius Plot ln k 1/T (K -1 ) intercept = ln A slope = -E a /R

33. At TWO temperatures (for a single reaction): ln k 2 = ln A – E a /RT 2 ln k 1 = ln A – E a /RT 1  ln k 2 – ln k 1 = – E a /R(1/T 2 – 1/T 1 ) ln(k 2 /k 1 ) = – E a /R(1/T 2 – 1/T 1 ) ln(k 2 /k 1 ) = + E a /R(1/T 1 – 1/T 2 )

34. PRACTICE EXAMPLE The rate constant for the reaction: H 2(g) + I 2(g)  2HI (g) has been determined at the following temperatures: k/M -1 s -1 T/K 5.4 x 10 -4 599 2.8 x 10 -2 683 Calculate the activation energy, E a, for the reaction in J mol -1.

35. CATALYSTS This is a substance that increases the rate of a chemical reaction without itself being used up. It participates by changing the mechanism of the reaction. It provides a path with lower activation energy than the uncatalysed reaction. If the activation energy is smaller, a greater fraction of molecules will have the minimum energy to react.

36. Potential Energy Diagram Potential Energy reactants products catalysed reaction uncatalysed reaction EaEa E a, cat ΔHΔH

37. Types of catalysts homogeneous catalysts – exist in same phase as reactants heterogeneous catalysts – exist in a separate phase

38. Heterogeneous Catalysis Catalyst promotes a reaction on its surface. The reactant molecule is adsorbed on the surface of the catalyst where interaction with the surface increases reactivity (ACTIVE SITE) followed by product(s) desorption. If something blocks the active sites it poisons the surface by destroying the catalytic properties.