1. Chapter 14 Chemical Kinetics (part 2)

2. The Collision Model Goal: develop a model that explains why rates of reactions increase as concentration and temperature increases. The collision model: in order for molecules to react they must collide. The greater the number of collisions the faster the rate. The more molecules present, the greater the probability of collision and the faster the rate. Temperature and Rate

3. The Collision Model The higher the temperature, the more energy available to the molecules and the faster the rate. Complication: not all collisions lead to products. In fact, only a small fraction of collisions lead to product. The Orientation Factor In order for reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products. Temperature and Rate

4. The Orientation Factor Consider: Cl + NOCl  NO + Cl 2 There are two possible ways that Cl atoms and NOCl molecules can collide; one is effective and one is not. Temperature and Rate

5. The Orientation Factor Temperature and Rate

6. Activation Energy Arrhenius: molecules must posses a minimum amount of energy to react. Why? –In order to form products, bonds must be broken in the reactants. –Bond breakage requires energy. Activation energy, E a, is the minimum energy required to initiate a chemical reaction. Temperature and Rate

7. Activation Energy Consider the rearrangement of methyl isonitrile: –In H 3 C-N  C, the C-N  C bond bends until the C-N bond breaks and the N  C portion is perpendicular to the H 3 C portion. This structure is called the activated complex or transition state. –The energy required for the above twist and break is the activation energy, E a. –Once the C-N bond is broken, the N  C portion can continue to rotate forming a C-C  N bond. Temperature and Rate

10. Activation Energy The change in energy for the reaction is the difference in energy between CH 3 NC and CH 3 CN. The activation energy is the difference in energy between reactants, CH 3 NC and transition state. The rate depends on E a. Notice that if a forward reaction is exothermic (CH 3 NC  CH 3 CN), then the reverse reaction is endothermic (CH 3 CN  CH 3 NC). Temperature and Rate

11. Activation Energy How does a methyl isonitrile molecule gain enough energy to overcome the activation energy barrier? From kinetic molecular theory, we know that as temperature increases, the total kinetic energy increases. We can show the fraction of molecules, f, with energy equal to or greater than E a is where R is the gas constant (8.314 J/mol·K). Temperature and Rate

12. Activation Energy Temperature and Rate

13. The Arrhenius Equation Arrhenius discovered most reaction-rate data obeyed the Arrhenius equation: –k is the rate constant, E a is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K. –A is called the frequency factor. –A is a measure of the probability of a favorable collision. –Both A and E a are specific to a given reaction. Temperature and Rate

14. Determining the Activation Energy If we have a lot of data, we can determine E a and A graphically by rearranging the Arrhenius equation: From the above equation, a plot of ln k versus 1/T will have slope of –E a /R and intercept of ln A. Temperature and Rate

16. Determining the Activation Energy If we do not have a lot of data, then we recognize Temperature and Rate

17. 1.For the following reaction producing 1 mol of oxygen gas at a particular temperature, ∆H = –200 kJ. NO(g) + O 3 (g) → NO 2 (g) + O 2 (g) The activation energy is 11 kJ/mol. What is the activation energy for the reverse reaction? a)11 kJ/mol b)188 kJ/mol c)200 kJ/mol d)211 kJ/mol e)222 kJ/mol

18. 2.The rate constants for the first-order decomposition of a compound are 5.50 × 10 –4 s -1 at 39°C and 2.32 × 10 –3 s -1 at 56°C. What is the value of the activation energy for this reaction? (R = 8.31 J/(mol · K)) a)0.704 kJ/mol b)72.2 kJ/mol c)1.54 kJ/mol d)0.667 kJ/mol e)31.4 kJ/mol

19. Kinetics lab discussion

21. The balanced chemical equation provides information about the beginning and end of reaction. The reaction mechanism gives the path of the reaction. Mechanisms provide a very detailed picture of which bonds are broken and formed during the course of a reaction. Elementary Steps Elementary step: any process that occurs in a single step. Reaction Mechanisms

22. Elementary Steps Molecularity: the number of molecules/atoms/ions present in an elementary step. –Unimolecular: one molecule in the elementary step, –Bimolecular: two molecules in the elementary step, and –Termolecular: three molecules in the elementary step. It is not common to see termolecular processes (statistically improbable). Reaction Mechanisms

23. Multistep Mechanisms Some reaction proceed through more than one step: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) Notice that if we add the above steps, we get the overall reaction: NO 2 (g) + CO(g)  NO(g) + CO 2 (g) Reaction Mechanisms

24. Multistep Mechanisms If a reaction proceeds via several elementary steps, then the elementary steps must add to give the balanced chemical equation. Intermediate: a species which appears in an elementary step which is not a reactant or product. Reaction Mechanisms

25. Rate Laws for Elementary Steps The rate law of an elementary step is determined by its molecularity: –Unimolecular processes are first order, –Bimolecular processes are second order, and –Termolecular processes are third order. Rate Laws for Multistep Mechanisms Rate-determining step: is the slowest of the elementary steps. Reaction Mechanisms

26. Rate Laws for Elementary Steps Reaction Mechanisms

27. Rate Laws for Multistep Mechanisms Therefore, the rate-determining step governs the overall rate law for the reaction. Mechanisms with an Initial Fast Step It is possible for an intermediate to be a reactant. Consider 2NO(g) + Br 2 (g)  2NOBr(g) Reaction Mechanisms

28. Mechanisms with an Initial Fast Step 2NO(g) + Br 2 (g)  2NOBr(g) The experimentally determined rate law is Rate = k[NO] 2 [Br 2 ] Consider the following mechanism Reaction Mechanisms

29. Mechanisms with an Initial Fast Step The rate law is (based on Step 2): Rate = k 2 [NOBr 2 ][NO] The rate law should not depend on the concentration of an intermediate (intermediates are usually unstable). Assume NOBr 2 is unstable, so we express the concentration of NOBr 2 in terms of NOBr and Br 2 assuming there is an equilibrium in step 1 we have Reaction Mechanisms

30. Mechanisms with an Initial Fast Step By definition of equilibrium: Therefore, the overall rate law becomes Note the final rate law is consistent with the experimentally observed rate law. Reaction Mechanisms

31. A catalyst changes the rate of a chemical reaction. There are two types of catalyst: –homogeneous, and –heterogeneous. Chlorine atoms are catalysts for the destruction of ozone. Homogeneous Catalysis The catalyst and reaction is in one phase. Reaction Mechanisms

32. Homogeneous Catalysis Hydrogen peroxide decomposes very slowly: 2H 2 O 2 (aq)  2H 2 O(l) + O 2 (g) In the presence of the bromide ion, the decomposition occurs rapidly: –2Br - (aq) + H 2 O 2 (aq) + 2H + (aq)  Br 2 (aq) + 2H 2 O(l). –Br 2 (aq) is brown. –Br 2 (aq) + H 2 O 2 (aq)  2Br - (aq) + 2H + (aq) + O 2 (g). –Br - is a catalyst because it can be recovered at the end of the reaction. Catalysis

33. Homogeneous Catalysis Generally, catalysts operate by lowering the activation energy for a reaction. Catalysis

34. Catalysis

35. Homogeneous Catalysis Catalysts can operate by increasing the number of effective collisions. That is, from the Arrhenius equation: catalysts increase k be increasing A or decreasing E a. A catalyst may add intermediates to the reaction. Example: In the presence of Br -, Br 2 (aq) is generated as an intermediate in the decomposition of H 2 O 2. Catalysis

36. Homogeneous Catalysis When a catalyst adds an intermediate, the activation energies for both steps must be lower than the activation energy for the uncatalyzed reaction. The catalyst is in a different phase than the reactants and products. Heterogeneous Catalysis Typical example: solid catalyst, gaseous reactants and products (catalytic converters in cars). Most industrial catalysts are heterogeneous. Catalysis

37. Heterogeneous Catalysis First step is adsorption (the binding of reactant molecules to the catalyst surface). Adsorbed species (atoms or ions) are very reactive. Molecules are adsorbed onto active sites on the catalyst surface. Catalysis

38. Catalysis

39. Heterogeneous Catalysis Consider the hydrogenation of ethylene: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g),  H = -136 kJ/mol. –The reaction is slow in the absence of a catalyst. –In the presence of a metal catalyst (Ni, Pt or Pd) the reaction occurs quickly at room temperature. –First the ethylene and hydrogen molecules are adsorbed onto active sites on the metal surface. –The H-H bond breaks and the H atoms migrate about the metal surface. Catalysis

40. Heterogeneous Catalysis –When an H atom collides with an ethylene molecule on the surface, the C-C  bond breaks and a C-H  bond forms. –When C 2 H 6 forms it desorbs from the surface. –When ethylene and hydrogen are adsorbed onto a surface, less energy is required to break the bonds and the activation energy for the reaction is lowered. Enzymes Enzymes are biological catalysts. Most enzymes are protein molecules with large molecular masses (10,000 to 10 6 amu). Catalysis

41. Enzymes Enzymes have very specific shapes. Most enzymes catalyze very specific reactions. Substrates undergo reaction at the active site of an enzyme. A substrate locks into an enzyme and a fast reaction occurs. The products then move away from the enzyme. Catalysis

42. Enzymes Only substrates that fit into the enzyme lock can be involved in the reaction. If a molecule binds tightly to an enzyme so that another substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors). The number of events (turnover number) catalyzed is large for enzymes (10 3 - 10 7 per second). Catalysis

43. Enzymes Catalysis