Homework pages 463 – 466 1, 2, 5 – 11, 13, 15, 19, 21 - 24, 27, 30, 31, 39, 40, 41, 43 – 51,

1. Chapter 14 Homework pages 463 – 466 1, 2, 5 – 11, 13, 15, 19, 21 - 24, 27, 30, 31, 39, 40, 41, 43 – 51, 56, 64.

2. A B rate = D[A] D[B] rate = - Dt Dt Chapter 14 - Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). Ave. rates D[A] = change in concentration of A over time period Dt D[B] = change in concentration of B over time period Dt Because [A] decreases with time, D[A] is negative. 13.1

3. A B time rate = D[A] D[B] rate = - Dt Dt 13.1

4. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) time Br2(aq) 393 nm 393 nm Detector light [Br2] a Absorbance 13.1

5. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) [Br2]final – [Br2]initial D[Br2] average rate = - = - Dt tfinal - tinitial • Calculate the average rate for: • The first 200 s • From 300 s to 350 s 13.1

6. Br2(aq) + HCOOH (aq) 2Br-(aq) + 2H+(aq) + CO2(g) slope of tangent slope of tangent slope of tangent [Br2]final – [Br2]initial D[Br2] average rate = - = - Dt tfinal - tinitial instantaneous rate = rate for specific instance in time 13.1

7. rate k = [Br2] How does the rate depend on concentration? rate = k [Br2] = rate constant = 3.50 x 10-3 s-1 13.1

8. 2A B aA + bB cC + dD rate = - = = rate = - = - D[D] D[B] D[A] D[B] D[C] D[A] rate = 1 1 1 1 1 Dt Dt Dt Dt Dt Dt c d a 2 b Reaction Rates and Stoichiometry Two moles of A disappear for each mole of B that is formed. 13.1

9. D[CO2] = Dt D[CH4] rate = - Dt Write the rate expression for the following reaction: CH4(g) + 2O2(g) CO2(g) + 2H2O (g) D[H2O] = Dt D[O2] = - 1 1 Dt 2 2 If concentration of O2 decreases at a rate of 0.10 M/s, what is the reaction rate? What is the rate of appearance of CO2? 13.1

10. aA + bB cC + dD The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. Rate = k [A]x[B]y reaction is xth order in A reaction is yth order in B reaction is (x +y)th order overall 13.2

11. F2(g) + 2ClO2(g) 2FClO2(g) rate = k [F2]x[ClO2]y Double [F2] with [ClO2] constant Rate doubles x = 1 rate = k [F2][ClO2] Quadruple [ClO2] with [F2] constant Rate quadruples y = 1 13.2

12. F2(g) + 2ClO2(g) 2FClO2(g) 1 Rate Laws • Rate laws are always determined experimentally. • Reaction order is always defined in terms of reactant (not product) concentrations. • The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. rate = k [F2][ClO2] 13.2

13. Determine the rate law and calculate the rate constant for the following reaction from the following data: S2O82-(aq) + 3I-(aq) 2SO42-(aq) + I3-(aq) rate k = 2.2 x 10-4 M/s = [S2O82-][I-] (0.08 M)(0.034 M) rate = k [S2O82-]x[I-]y y = 1 x = 1 rate = k [S2O82-][I-] Double [I-], rate doubles (experiment 1 & 2) Double [S2O82-], rate doubles (experiment 2 & 3) = 0.08/M•s 13.2

14. Finding Reaction Orders by Experiment Given the reaction: O2 (g) + 2 NO(g) 2 NO2 (g) The rate law for this reaction in general terms is: Rate = k [O2]m[NO]n To find the reaction orders, we run a series of experiments, each with a different set of reactant concentrations, and determine the initial reaction rates. Experiment Initial Reactant Concentrations (mol/l) Initial Rate O2 NO (mol/L·s) 1 < 1.10 x 10-2* 1.30 x 10-2*>3.21 x 10-3 2 < 2.20 x 10-2 1.30 x 10-2>6.40 x 10-3 3 1.10 x 10-2* 2.60 x 10-2* 12.8 x 10-3 4 3.30 x 10-2 1.30 x 10-2 9.60 x 10-3 5 1.10 x 10-2 3.90 x 10-2 28.8 x 10-3

16. What are the units of k for a first order reaction? Zero order? Second order? The rate equation tells how the rate depends on the reactant concentrations. Can we derive an equation that tells how the concentration depends on time? These are the integrated rate laws. How can we determine the reaction order graphically?

17. A product rate = [A] M/s D[A] - M = k [A] Dt [A] = [A]0exp(-kt) ln[A] = ln[A]0 - kt D[A] rate = - Dt First-Order Reactions rate = k [A] = 1/s or s-1 k = [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 13.3

18. The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ? [A]0 = 0.88 M [A] = 0.14 M What is the concentration of A after 100 s?

19. The half-life of a reaction, t1/2, is the time required for the concentration of a reactant to decrease to 1/2 of its initial concentration. For a first order reaction only, the half-life is independent of the initial concentration. t1/2 = (1/k) ln 2

20. What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? 0.693 = = = k t½ ln2 ln2 0.693 = k k 5.7 x 10-4 s-1 First-Order Reactions The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t½ = t when [A] = [A]0/2 t½ = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s-1) 13.3

21. A product # of half-lives [A] = [A]0/n First-order reaction 1 2 2 4 3 8 4 16 13.3

23. k = (1/200s)ln2 = 0.0035 s-1

24. Concentration-Time Equation Order Rate Law Half-Life 1 1 = + kt [A] [A]0 = [A]0 t½ = t½ t½ = ln2 2k k 1 k[A]0 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions [A] = [A]0 - kt rate = k 0 ln[A] = ln[A]0 - kt 1 rate = k [A] 2 rate = k [A]2 13.3

25. Experiment 13 Rate = k[A]x Take ln of both sides. Ln(rate) = x ln[A] + ln k Ln(1/time) = x ln[A] + constant

26. COLLISION THEORY OF CHEMICAL KINETICS 1. Chemical reactions occur as a result of collisions between reacting molecules. 2. In order to react the colliding molecules must have a. the right orientation b. a total kinetic energy equal to or greater than the activation energy, Ea, which is the minimum energy required to initiate the chemical reaction.

28. Increasing the temperature usually has a dramatic effect on reaction rates. Chemists use a “rule of thumb” that a 10° rise in temperature doubles the rate of a reaction.

31. TRANSITION STATE THEORY The species temporarily formed by the reactant molecules before they form the product is called the transition state or activated complex. In the transition state old bonds are in the process of breaking and new bonds are beginning to form. The activation energy is the energy required to reach the transition state.

32. A + B C + D Endothermic Reaction Exothermic Reaction The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction. 13.4

34. Sample Problem For the reaction O3(g) + O(g) 2 O2(g) Ea(fwd) = 19 kJ and ΔH = -392 kJ. Draw a reaction energy diagram, show a possible transition state, and calculate Ea(rev).

35. lnk = - + lnA Ea 1 T R Temperature Dependence of the Rate Constant k = A •exp( -Ea/RT ) (Arrhenius equation) Eais the activation energy (J/mol) R is the gas constant (8.314 J/K•mol) T is the absolute temperature A is the frequency factor 13.4

36. lnk = - + lnA Ea 1 T R 13.4

38. An elementary reaction is a simple reaction that proceeds in one step. Most reactions are complex and require more than one step. A reaction mechanism is the sequence of elementary steps that leads to product formation. An intermediate is a species that appears in the mechanism, but not in the overall balanced rxn.

39. 2NO (g) + O2 (g) 2NO2 (g) Elementary step: NO + NO N2O2 Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 Reaction Mechanisms The sequence of elementary steps that leads to product formation is the reaction mechanism. 13.5

40. Elementary step: NO + NO N2O2 + Elementary step: N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. • The molecularity of a reaction is the number of molecules reacting in an elementary step. • Unimolecular reaction – elementary step with 1 molecule • Bimolecular reaction – elementary step with 2 molecules • Termolecular reaction – elementary step with 3 molecules 13.5

41. Unimolecular reaction Bimolecular reaction Bimolecular reaction A + B products A + A products A products The rate-determining step is the sloweststep in the sequence of steps leading to product formation. Rate Laws and Elementary Steps rate = k [A] rate = k [A][B] rate = k [A]2 • Writing plausible reaction mechanisms: • The sum of the elementary steps must give the overall balanced equation for the reaction. • The rate-determining step should predict the same rate law that is determined experimentally. 13.5

42. For elementary reactions only, the exponents in the rate law must correspond to the coefficients in the balanced equation. The elementary steps that make up a reaction mechanism must satisfy two requirements. 1. The sum of the elementary steps must give the overall balanced equation for the reaction. 2. The rate-determining step (the slow step) must predict the same rate law as is determined by experiment.

43. Step 1: Step 2: NO2 + NO2 NO + NO3 The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps: NO2+ CO NO + CO2 NO3 + CO NO2 + CO2 What is the equation for the overall reaction? What is the intermediate? NO3 What can you say about the relative rates of steps 1 and 2? rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2 13.5

44. For the reaction 2 X + Y Z rate = k[X][Y] Can this be an elementary reaction? For the reaction 2 A + B C rate = k[A]2 [B] Is this an elementary reaction?

45. Sample Problem Problem: The following two reactions are proposed as elementary steps in the mechanism for an overall reaction: (1) NO2Cl (g) NO2(g) + Cl(g) (2) NO2Cl (g + Cl(g) NO2(g) + Cl2(g) (a) Write the overall balanced equation. (b) Determine the molecularity of each step. (c) Write the rate law for each step. 2 C2H6 + 7 O2  4 CO2 + 6 H2O Is this likely to be an elementary process?

46. Mechanisms with a Slow Initial Step 2 NO2 + F2 2 NO2F The experimental rate law for this reaction is first order in both reactants. A proposed mechanism is: 1. NO2 + F2 NO2F + F slow 2. NO2 + F NO2F fast Show that this is a reasonable mechanism. What is the intermediate species?

47. A Catalyst is a substance that increases the rate of reaction without itself being consumed. A catalyst works by changing the mechanism. The new mechanism has a lower activation energy. It speeds up both forward and reverse reactions and cannot increase the final equilibrium yield, but it gets to the final equilibrium state faster.

48. uncatalyzed catalyzed Ea k ‘ Ea< Ea A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A •exp( -Ea/RT ) ratecatalyzed > rateuncatalyzed 13.6

49. Depletion of the Earth’s Ozone Layer - I The stratospheric ozone is vital to life on earth because ozone absorbs short-wavelength (~ 3x10-7m) ultraviolet (UV) radiation from the sun. UV photon O3 (g) O2 (g) + O(g) UVB Normally two reactions maintain the ozone in the stratosphere, and keep a level of ozone sufficient to absorb all of the UVB radiation that reaches the earth’s atmosphere: O2 (g) + O(g) O3 (g) [formation] O3 (g) + O(g) 2 O2 (g) [breakdown] Under normal conditions, the atomic oxygen is produced by the breakdown of molecular oxygen by it absorbing ultraviolet radiation UV photon O2 (g) 2 O(g) UVA

50. O3 (g) + Cl(g) ClO(g) + O2 (g) ClO(g) + O(g) Cl(g) + O2 (g) Depletion of the Earth’s Ozone Layer - II Chlorofluorocarbons (CFC’s) were introduced in the 1930’s as working gases to be used in refrigerators and air conditioners. They were found to breakdown in the stratosphere by UVAradiation and release atomic chlorine atoms to the atmosphere, which had a “catalytic” reaction with ozone that has been causing it’s concentration in the stratosphere to drop. . . Uv photon Freon -12 CF2Cl2 (g) CF2Cl(g) + Cl(g) UVA The • represents an unpaired electron, resulting from bond breaking, and the resultant molecule is called a “free radical” and is very reactive. . . . . O3 (g) + O(g) 2 O2 (g)