1. Activation Energy E a : is the minimum energy that reactants must have to form products. the height of the potential barrier (sometimes called the energy barrier 1 Activation Energy Curve called : 1. Potential Energy Hill Or 2. Potential Energy Barrier

2. 2 Activation energy, E a The shaded part of the Maxwell-Boltzmann distribution curve represents number of particles that have enough collision energy for a reaction (i.e. the energy is ≥ E a ). Suppose: Number of particles 100 Ea= 70 j/mole 25 particles have 20 j/mole 40 particles have 45 j/mole 20 particles have 50 j/mole 10 particles have 60 j/mole 5 particles have 70 j/mole

3. Maxwell–Boltzmann Distributions Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. At any temperature there is a wide distribution of kinetic energies. 3

4. Maxwell–Boltzmann Distributions As the temperature increases, the curve flattens and broadens. Thus at higher temperatures, a larger population of molecules has higher energy. 4

5. Maxwell–Boltzmann Distributions If the dotted line represents the activation energy, as the temperature increases, so does the number of molecules that can overcome the activation energy barrier. As a result, the reaction rate increases. 5

6. Maxwell–Boltzmann Distributions This fraction of molecules can be found through the expression: where R is the gas constant and T is the temperature in Kelvin. 6

7. Arrhenius Equation A mathematical relationship between k : ( rate constant of chemical reaction) and Ea: activation energy. where A : “Frequency Factor”-- a constant indicating how many collisions have the correct orientation to form products. 7

8. Arrhenius Equation: Temperature Dependence of the Rate Constant E a = the activation energy (J/mol) R = the gas constant (8.314 J/Kmol) T = is the absolute temperature ( in Kelvin) A = is the frequency factor 8

9. Arrhenius Equation Taking the natural logarithm (ln) of both sides, the equation becomes, Ln (natural logarithm): is inverse function of exponential function. 9 e ln(x) = x ln(e x ) = x

10. Arrhenius Equation 10 When k is determined experimentally at several temperatures, E a can be calculated from the slope of a plot, Slope= -E a /R Straight Line Equation y = mx + b ln(k) = - E a /R(1/T) + ln(A)

11. 11 Application of Arrhenius Equation From k = A e – Ea / R T, calculate A, E a, k at a specific temperature and T. The reaction: 2 NO 2 (g) -----> 2NO(g) + O 2 (g) The rate constant k = 1.0e-10 s -1 at 300 K and the activation energy E a = 111 kJ mol -1. What are A, k at 273 K and T when k = 1e-11? Method: derive various versions of the same formula k = A e – Ea / R T A = k e Ea / R T A / k = e Ea / R T ln (A / k) = E a / R T Make sure you know how to transform the formula into these forms.

12. Exothermic Reaction Potential Energy of reactant = Energy of chemical bond = Heat content = H H product < H reactant Enthalpy ( ∆ H) ∆ H = H product - H reactant < 0 ∆ H = negative value (-) Activation Energy (E a ) = H transition state - H reactant Exothermic reaction has Low E a 12 A → B + Heat Exothermic

13. Endothermic Reaction Potential Energy of reactant = Energy of chemical bond = Heat content = H H product > H reactant Enthalpy ( ∆ H) ∆ H = H product - H reactant > 0 ∆ H = positive value (+) Activation Energy (E a ) = H transition state - H reactant Endothermic reaction has High E a 13 A + Heat → B Endothermic

14. 14 Activation Energy and Enthalpy The E a for a reaction cannot be predicted from ∆H. ∆H is determined only by the difference in potential energy between reactants and products. △ H has no effect on the rate of reaction. The rate depends on the size of the activation energy Ea Reactions with low E a occur quickly. Reactions with high E a occur slowly. Potential energy diagram for the combustion of octane.

15. 15 Activation Energy for Reversible Reactions Potential energy diagrams  both forward and reverse reactions. follow left to right for the forward reaction follow right to left for the reverse reaction

16. Activated Complex (Transition State) 16 Activated complex is unstable compound and can break to form product. Activated complex: The arrangement of atoms found at the top of potential energy hill or barrier.

17. Activated Complex (Transition State) 1. The collision must provide at least the minimum energy necessary to produce the activated complex. 2. It takes energy to initiate the reaction by converting the reactants into the activated complex. 3.If the collision does not provide this energy, products cannot form. 17

18. 18 Analyzing Reactions Using Potential Energy Diagrams E a(rev) is greater than E a(fwd) Forward Reaction is Exothermic Reaction Reversible Reaction is Endothermic Reaction 1.BrCH 3 molecule and OH - must collide with the correct orientation and sufficient energy and an activated complex forms. 2. When chemical bonds reform, potential energy decreases and kinetic energy increases as the particles move apart.

19. 19 Reaction Rates and Reaction Mechanisms Initial rate is found by determining the slope of a line tangent to the curve at time zero. Initial rate is the rate of a chemical reaction at time zero. products of the reaction are not present, so the reverse reaction cannot occur it is a more accurate method for studying the relationship between concentration of reactant and reaction rate

20. 20 The Effect of Temperature on Reaction Rates Reaction rate = k [A} x [B] y [C] z (Concentration effect at constant T) k = A exp ( – E a / RT)(Temperature effect) Use graphic method to discuss the variation of k vs. T variation of k vs 1 / T variation of ln(k) vs T variation of ln(k) vs 1 / T See a potential multiple choice question in an exam?

21. 21 Quiz # 1 Write down the rate laws and describe them as uni- bi- or ter-molecular steps? I.1. Cl 2  2 Cl 2. 2 Cl  Cl 2 3. 2Cl + CH 4  Cl 2 + CH 4 * 4. Cl + CH 4  HCl + CH 3 5. CH 3 + Cl  CH 3 Cl II. In Arhenius Equation, what is A, Ea, R= ………., and T III. In the next potential diagram, Define, intermediates, reactant, product, Ea, ∆ H 21 ………. reaction …………….

22. Reaction Mechanism Most chemical reactions occur by a series of steps called the reaction mechanism. The sum of the elementary steps must give the overall balanced equation for the reaction. The mechanism must agree with the experimentally determined rate law. 22

23. Multistep Mechanisms In a multistep process, one of the steps will be slower than all others. The overall reaction cannot occur faster than this slowest, rate-determining step. 23

24. Rate Laws and Rate Determining Steps The rate-determining step is the slowest step in the sequence of steps leading to product formation. 24

25. Slow Initial Step The rate law for this reaction is found experimentally to be Rate = k [NO 2 ] 2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps. NO 2 (g) + CO (g)  NO (g) + CO 2 (g) 25

26. Slow Initial Step A proposed mechanism for this reaction is Step 1: NO 2 + NO 2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast) The NO 3 intermediate is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law. 26

27. Fast Initial Step The rate law for this reaction is found (experimentally) to be Because termolecular processes are rare, this rate law suggests a two-step mechanism. 27

28. Fast Initial Step A proposed mechanism is Step 1 is an equilibrium: it includes the forward and reverse reactions. 28

29. Fast Initial Step The rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would be But how can we find [NOBr 2 ]? 29

30. Fast Initial Step NOBr 2 can react two ways: With NO to form NOBr By decomposition to reform NO and Br 2 The reactants and products of the first step are in equilibrium with each other. Therefore, Rate f = Rate r 30

31. Fast Initial Step Because Rate f = Rate r, k 1 [NO] [Br 2 ] = k −1 [NOBr 2 ] Solving for [NOBr 2 ] gives us k1k−1k1k−1 [NO] [Br 2 ] = [NOBr 2 ] 31

32. Fast Initial Step Substituting this expression for [NOBr 2 ] in the rate law for the rate-determining step gives 32

33. 33 Rate Laws and Mechanisms A mechanism is a collection of elementary steps devise to explain the reaction in view of the observed rate law. For Example: 2 NO 2 (g) + F 2 (g)  2 NO 2 F (g), overall reaction the rate law is,rate = k [NO 2 ] [F 2 ]. Can the elementary reaction be the same as the overall reaction? If they were the same the rate law would have been rate = k [NO 2 ] 2 [F 2 ], Therefore, they the overall reaction is not an elementary reaction. Its mechanism is proposed next.

34. 34 Rate-determining Step in a Mechanism The rate determining step is the slowest elementary step in a mechanism, and the rate law for this step is the rate law for the overall reaction. The ( determined ) rate law is,rate = k [NO 2 ] [F 2 ], for the reaction, 2 NO 2 (g) + F 2 (g)  2 NO 2 F (g), and a two-step mechanism is proposed: i NO 2 (g) + F 2 (g)  NO 2 F (g) + F (g) ii NO 2 (g) + F (g)  NO 2 F (g) Which is the rate determining step? Answer: The rate for step i is rate = k [NO 2 ] [F 2 ], which is the rate law, this suggests that step i is the rate-determining or the s-l-o-w step.

35. 35 Deriving a Rate Law From a Mechanism The decomposition of H 2 O 2 in the presence of I – follow this mechanism, iH 2 O 2 + I –  k 1  H 2 O + IO – slow ii H 2 O 2 + IO –  k 2  H 2 O + O 2 + I – fast What is the rate law? Energy E ai E aii reaction

36. 36 Deriving a rate law from a mechanism The decomposition of H 2 O 2 in the presence of I – follow this mechanism, iH 2 O 2 + I –  k 1  H 2 O + IO – slow ii H 2 O 2 + IO –  k 2  H 2 O + O 2 + I – fast What is the rate law? Solution The slow step determines the rate, and the rate law is: rate = k 1 [H 2 O 2 ] [I – ]

37. 37 Deriving a rate law from a mechanism Derive the rate law for the reaction, H 2 + Br 2 = 2 HBr, from the proposed mechanism: i Br 2  2 Br fast equilibrium ( k 1, k -1 ) iiH 2 + Br  k 2  HBr + H slow iii H + Br  k 3  HBr fast Solution: The fast equilibrium condition simply says that k 1 [Br 2 ] = k -1 [Br] 2 and[Br] = ( k 1 / k -1 [Br 2 ]) ½ The slow step determines the rate law, rate = k 2 [H 2 ] [Br] Br is an intermediate = k 2 [H 2 ] ( k 1 / k -1 [Br 2 ]) ½ = k [H 2 ] [Br 2 ] ½; k = k 2 ( k 1 / k -1 ) ½ M -½ s - 1 total order 1.5

38. Catalysis A substance can speed up a reaction without being consumed itself. The catalyst is to provide a new pathway for the reaction and to decrease activation energy. 38

39. 39 Catalysis A catalyst is a substance that changes the rate of a reaction by lowing the activation energy, E a. It participates a reaction in forming an intermediate, but is regenerated. Enzymes are selective catalysts. A catalyzed reaction, NO (catalyst) 2 SO 2 (g) + O 2 —  2 SO 3 (g) via the mechanism i2 NO + O 2  2 NO 2 (3 rd order) iiNO 2 + SO 2  SO 3 + NO

40. 40 Homogenous and heterogeneous catalysts A catalyst in the same phase (gases and solutions) as the reactants is a homogeneous catalyst. It effective, but recovery is difficult. When the catalyst is in a different phase than reactants (and products), the process involve heterogeneous catalysis. Example: Platinum is often used to catalyze hydrogenation