Chapter 13 Chemical Kinetics

1. Chapter 13Chemical Kinetics Read & Study: Chapter 13 View the videos in the Science Learning Center (SLC), N-604. One is entitled, “Catalysis: Technology for a Clean Environment” and the other is “Reaction Rates”.

2. Chapter Overview 1. Introduction - Background and Key Definitions 2. Factors Influencing the Rates of Reactions 3. Rate Laws 4. Activation Energies 5. Reaction Mechanisms - The “Paragraphs” of the Chemical Language 6. Effects of Catalysts

3. Introduction 1. Chemical Kinetics - The study of how fast chemi- cal reactions take place; a study of the “RATE” of a reaction. 2. Chemical Reaction Rate - A change in the amount or concentration of a reactant or product in a given period of time. Rate = - D[Reactant] = + D[Product] D time D time

4. 3. Kinetics vs. thermodynamics - Factors to be con- sidered when predicting whether or not a change will take place: 1) Energy (Enthalpy) Change 2) The Change in disorder (Entropy) 3) The RATE of the change Thermodynamics Kinetics 4. Reaction Mechanism - A detailed molecular-, atomic-, or ionic-level “picture” or model of how a reaction takes place.

5. How does NO (g) react with O2 (g) in the atmosphere to form NO2 (g), a major contributor to smog? 2 NO (g) + O2 (g) 2 NO2 (g) R1 Step 1: NO (g) + O2 (g) NO3 (g) Fast R2 Step 2: NO3 (g) NO (g) + O2 (g) Fast R3 Step 3: NO3 (g) + NO (g) NO2 (g) Slow

6. By measuring the rates of reactions under various conditions, rate laws can be deduced. The chemist, then, designs reaction mechanisms to “explain” the observed rate laws. • 5. Factors affecting the rates of reactions - • 1) The Nature of the Reactants • 2) The Reactants’ Concentrations • 3) The Reactants Total Surface Area • 4) The Temperature • 5) The presence or absence of catalysts Memorize!! Memorize!! Memorize!! Memorize!! Memorize!!

7. FACTORS AFFECTING REACTION RATES • 1) The Nature of the Reactants - • Ions in solution tend to react very quickly. • Covalent molecules tend to react more slowly. • Large covalent molecules tend to react more • slowly than small covalent molecules. • Molecules with strong covalent bonds tend to • react more slowly than those with weak bonds.

8. 2) The Concentrations of the Reactants - • For a reaction to occur, three things must happen - • The reacting molecules, atoms, or ions, must • collide. • They must be properly oriented with respect to one • another. • The must have sufficient energy. • The probability of these three factors being present is • increased as the concentration of the reactants is • increased.

9. Proper Orientation is Necessary

10. Suffcient Energy is Necessary Bounces Off! Reacts!

11. Homogeneous Reactions - Reactions that take place in a single solution (phase). Cu2+ + 4 NH3 (aq) Cu(NH3)42+ NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l) • 3) The Reactants’ State of Subdivision - • Heterogeneous Reactions - Reactions in which • reactants are in different phases. The reactions • take place at the interface between the phases. • The larger the surface area of the interface, the • higher the rate of reaction.

12. V = 1 m3 A = 6 m2 V = 1 m3 A = 8 m2 V = 1 m3 A = 12 m2 V = 1 m3 A = 36 m2 V = 1 m3 A = 20 m2 V = 1 m3 A = ?? m2 Powder

13. http://blog.wired.com/wiredscience/2008/03/top-10-amazing.htmlhttp://blog.wired.com/wiredscience/2008/03/top-10-amazing.html http://pubs.acs.org/cen/government/86/8608gov1.html

14. Examples of Heterogenous Reactions - C (s) + O2 (g) CO2 (g) Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g) AgCl (s) + 2 NH3 (aq) Ag(NH3)2+ + Cl - • 4) The Temperature - Reaction rates are increased • by increases in temperature. • Rule of Thumb:A 10oC rise in temperature will • doublethe rate of the reaction.

15. Practice Exercise: The temperature at which a reaction is run is raised from 0oC to 40oC. How much is the rate of the reaction increased? toC Rate 0 R1 10 2R1 20 4R1 30 8R1 40 16R1 = R2 R2 = 2(t2 - t1)/10R1 If you want to slow down the reaction that causes your milk to sour, you put the milk in the refrigerator to lower its temperature.

16. What about the Thermite Reaction and temperature? http://blog.wired.com/wiredscience/2008/03/top-10-amazing.html

17. 5) Catalysis - How can a catalyst change the rate of a • reaction and yet NOT be consumed? • Thermodynamic Answer: By changing the energy • pathway. Activation Energy Reactants Energy Products With a catalyst, the “energy hump” is lowered -- the energy pathway is changed.

18. Kinetic Answer: The catalyst is a reactantin one step of the reaction mechanism and a product in a subsequent step. Homogeneous Catalysis - Reactions wherein the catalyst is present in the same phase as the reactants. 2 SO2 (g) + O2 (g) 2 SO3 (g) NO(g) Catalyst Heterogeneous Catalysis - Reactions wherein the cat- alyst is present in a separate phase from the reactants. Pt (s) 2 H2 (g) + O2 (g) 2 H2O (g) Catalyst

19. Heterogeneous Catalysis

20. Hydrogen reacts very exothermically with oxygen. Even so, they can be held in the same container together for years without reacting unless a catalyst or source of energy, such as a spark, are introduced into the mixture. Rate Laws Rate Law - A mathematical equation that shows how the rate of a reaction is affected by the concentration of re- actants or products in a reaction. Measuring the Rate of a Reaction - Follow the disap- pearance of a reactant or the appearance of a product.

21. 2 N2O5 (g)  2 N2O4 (g) + O2 (g) Rate = +D[O2]/Dt = +(1/2)(D[N2O4]/Dt) = - (1/2)(D[N2O5]/Dt)

22. Decomposition of N2O5 (g)

23. Reaction of Ethylene w/ O3 (g) Another Case!

24. Measuring the Rate of a Reaction • Method of Initial Rates – A method that involves running • A reaction several times starting with different initial • Concentrations of reactants. • Hold the temperature constant. • If Present, keep the solvent constant. • Vary (change) only one concentration at a • time.

25. Method of Initial Rates 2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g) 866oC Run Reactant Rate of Number Concentration decrease in [NO] [H2] [NO] 1 0.10 0.010 0.062 2 0.10 0.040 0.246 3 0.30 0.010 0.558

26. Method of Initial Rates Effect of [H2] on the Rate: 0.040 M0.246 M/s 0.010 M 0.062 M/s = 4.0 = 4.0 As the [H2] is raised by a factor of 4.0, the rate of decrease in [NO] (the rate of the reaction) is raised by a factor of 4.0.  Rate  [H2] Effect of [NO] on the Rate: 0.30 M0.558 M/s 0.10 M 0.062 M/s  Rate  [NO]2 = 3.0 = 9.0

27. Method of Initial Rates Rate  [H2][NO]2 Rate = k [H2][NO]2This is theRATE LAW! where k is called the “Reaction Rate Constant”! k is independent of reactant concentrations but IS dependent on temperature. The Rate Law exponents are usually NOT the same as the coefficients in the balanced chemical equation. Therefore, the rate law cannot be predicted. It MUST be experimentally determined.

28. k aA + bB  cC + dD Rate = k [A]m[B]n Run Reactant Initial Number Concentration Rate [A] [B] 1 0.030 0.010 1.7 x 10-8 2 0.060 0.010 6.8 x 10-8 3 0.030 0.020 4.9 x 10-8

29. rate 1 = k(0.030)m(0.010)n = 1.7 x 10-8 M/s rate 2 = k(0.060)m(0.010)n = 6.8 x 10-8 M/s rate 11.7 x 10-8(0.030)m0.030m rate 2 6.8 x 10-8 (0.060)m 0.060 = = = 0.25 = (0.50)m Log (0.25) = Log (0.50)m = m Log (0.50) Log (0.25) = m = 2.0 Log (0.50)

30. rate 1 = k(0.030)m(0.010)n = 1.7 x 10-8 M/s rate 3 = k(0.030)m(0.020)n = 4.9 x 10-8 M/s rate 11.7 x 10-8(0.010)n0.010n rate 3 4.9 x 10-8 (0.020)n 0.020 = = = 0.35 = (0.50)m Log (0.35) = Log (0.50)n = n Log (0.50) Log (0.35) = n = 1.5 Log (0.50) Rate = k [A]2[B]3/2

31. 2 I- + S2O82- I2 (aq) + 2 SO42- Run [I-] [S2O82-] Initial Rate 1 0.15 0.45 2.6 x 10 - 4 2 0.15 0.25 1.4 x 10 - 4 3 0.50 0.45 8.6 x 10 - 4 Find the Rate Law: Rate = k [I-]m[S2O82- ]n

32. rate 1 = k(0.15)m(0.45)n = 2.6 x 10 -4 M/s rate 3 = k(0.50)m(0.45)n = 8.6 x 10 -4 M/s rate 12.6 x 10-4(0.15)m0.15m rate 3 8.6 x 10-4 (0.50)m 0.50 = = = = 0.30 0.30 = (0.30)m Log (0.30) = Log (0.30)m = m Log (0.30) Log (0.30) = m = 1.0 Log (0.30)

33. rate 1 = k(0.15)m(0.45)n = 2.6 x 10 -4 M/s rate 2 = k(0.15)m(0.25)n = 1.4 x 10 -4 M/s rate 12.6 x 10-4(0.45)n0.45n rate 2 1.4 x 10-4 (0.25)n 0.25 = = = = 1.86 1.86= (1.80)n Log (1.86) = Log (1.80)n = n Log (1.80) Log (1.86) = n = 1.0 Log (1.80) ∴ Rate = k [I-][S2O82- ]

34. Rate Laws The Order of a Reaction – The sum of the exponents in theRate Lawfor the reaction 866oC 2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g) k1 Rate = k1 [H2][NO]2 Determined by Experiment First Order with respect to [H2] Second Order with respect to [NO] Third order OVERALL

35. Rate Laws a A + b B  c C + d D k2 Determined by Experiment Rate = k2 [A]2[B]3/2 Second Order with respect to [A] 1.5 Order with respect to [B] 3.5 order OVERALL 2 I- + S2O82- I2 (aq) + 2 SO42- Rate = k [I-][S2O82- ] First Order with respect to both [I-] and [S2O82-] and second order OVERALL

36. Rate Laws a A + b B + c C  d D + e E k4 Rate = k4 [A]m[B]n Subtances involved in a reaction that have NO concentration effect on the rate of the reaction are said to have a “zeroth order” effect. The reaction above is zeroth order with respect to [C]. REMEMBER: The rate law for a reaction CANNOT be predicted from the chemical equation! It MUST be experimentally determined!

37. Rate Laws Class Exercise: The following reaction is second order with respect to acetaldehyde. Write the Rate Law for this reaction. CH3-C –H (g)  CH4 (g) + CO (g) || k O Rate = k [CH3-C –H]2 || O Rate Constants: The value of the rate constant, k, for a given reaction depends on the NATURE of the reactants and the TEMPERATURE, but NOT on the concentration of reactants.

38. Rate Constants Class Exercise: A + 2 B  C Rate = k [B] = 0.012 mol·L-1·s-1 When [B]0 = 0.60 M and [A]0 = 0.020 M, Find k. k = rate/[B]0 = (0.012 mol·L-1·s-1)/(0.60 mol·L-1) = 0.020 s-1 Find the rate of the reaction when [A]0 = [B]0 = 0.010 M rate = k [B] = (0.020 s-1)(0.010 M) = 2.0 x 10-4 M·s-1

39. Rate Law Characteristics Zero Order Reactions: Reactions with rates that are not dependent on the concentration of reactants. [Reactant] Reaction Rate Time Time The rate of metabolism of alcohol in the body is indepen- dent of [alcohol]. The RATE is constant!

40. Rate Law Characteristics First Order Reactions: 2 N2O5 (g)  4 NO (g) + O2 (g) Rate = k[N2O5] In General: Rate = k[A] [A] ln [A] Time Time

41. Rate Law Characteristics [A] = [A]o e-ktIntegrated Rate Law (1st Order) ln [A] = ln [A]o + (-kt) ln [A] = - kt + ln [A]o y = mx + b Half-Life: The time required for one-half of a reactant to disappear in a reaction. For FIRST ORDER reactions, The half-life is INDEPENDENTof reactant concentration. O.K. Let’s Prove That!

42. Rate Law Characteristics ln [A] = ln [A]o + (-kt) [A] [A]o ln = -kt But…[A] = ½ [A]o at time t1/2 [A]o 2[A]o ln = ln (1/2) = -kt1/2 kt1/2 = - ln (1/2) = ln 2 = 0.693 t1/2 = 0.693/k = CONSTANT

43. The Half-Life (t1/2) of a First Order process is inversely proportional to the rate constant, k, and is independent of the reactant concentration. 2nd Order Reactions: Rate = k[A]2Simplest Type Rate = k[A][B] Also 2nd Order k 2 NO2 (g)  2 NO (g) + O2 (g) Rate = k[NO2]2 Integrated Rate Law (2nd Order) 1/[A] = kt + 1/[A]o y = mx + b

44. At time t1/2, [A] = 1/2[A]o . Therefore 2/[A]o = kt1/2 + 1/[A]o 2/[A]o - 1/[A]o = kt1/2 1 [A]o = kt1/2∴ t1/2 = 1/k[A]o How can you tell 1st order reactions from 2nd order? 1st Order 2nd Order ln [A] 1/[A] time time

45. Temperature Effects on Reaction Rates k = Ae-Ea/RT Arrhenius Plot Take log of both sides: ln k = ln A + ln e -Ea/RT ln k = ln A - Ea/RT ln k = -(Ea/R)(1/T) + ln A y = mx + b Slope = - Ea/R Ea = -R(Slope)

46. Temperature Effects on Reaction Rates If you know the k at one temperature, you can find k for a different temperature with the Arrhenius Equation. ln k1 = -(Ea/R)(1/T1) + ln A ln k2 = -(Ea/R)(1/T2) + ln A ln k1 + Ea/RT1 = ln A ln k2 + Ea/RT2 = ln A ln k1 + Ea/RT1 = ln k2 + Ea/RT2

47. Temperature Effects on Reaction Rates ln k1 + Ea/RT1 = ln k2 + Ea/RT2 ln k1 - ln k2 = [Ea/R][(1/T2)-(1/T1)] = ln (k1/k2) Class Exercise: Ea = 18.7 kJ/mol k1 = 0.0400 s-1 T1 = 273 K T2 = 298 K k2 = ?? ln k2 = ln k1 - [Ea/R][(1/T2)-(1/T1)] ln k2 = -2.538 k2 = 0.0798 s-1

48. Activation Energy: The minimum energy necessary for a reaction to occur.

49. Temperature Effects on Reaction Rates Arrhenius Plot k = Ae-Ea/RT

50. Reaction Mechanisms Reaction Mechanism: A series of consecutive molecular- level events or steps that lead from reactants to products. They can be disproven but never proven. For a mechanism to be valid, it must account for every- thing that is known about the reaction, including the rate law for the reaction and its stoichiometry. Elementary Process: A single reaction step in a reaction mechanism.