1. Brown and Lemay Pages

2.  The term reaction rate refers to how fast a  product is formed or a reactant is consumed.  Consider the following generic reaction:  aA(g) + bB(g) cC(g) + dD(g)  Rate ave = - 1 a Δ[A] Δt = - 1 b Δ[B] Δt = 1 c Δ[C] Δt = 1 d Δ[D] Δt

3.  Rates are always positive quantities.  The concentration of reactants do not change  at the same rate when coefficients are  different.  Consider the following specific example:  N 2 (g) + 2O 2 (g) 2NO 2 (g)  Rate ave = - Δ[N 2 ] Δt = - 1 2 Δ[O 2 ] Δt = 1 2 Δ[NO 2 ] Δt = M s

4.  The following graph shows the decomposition  of N 2 as a function of time.  There are two different rates shown in the  graph.  The instantaneous rate is the rate of ◦ change (Δ) at a specific instant of time, ◦ i.e. 40 s.

5.  In a non-calculus environment, this is ◦ determined by calculating the slope of a  tangent line containing that point.   The average rate is computed over a ◦ time interval necessitating using two ◦ different times. Rate inst = Δ[N 2 ] Δt - = - 0.30 M – 0.55 M 60. s – 30. s Rate inst = 8.3 x 10 -3 M s -1 Rate inst = Δ[N 2 ] Δt - =

6.  This is the average rate from 20. s to 60. s. - 0.30 M – 0.67 M 60. s – 20. s Rate avs = Δ[N 2 ] Δt - = Rate ave = 9.2 x 10 -3 M s -1

7. Time (s) [N2][N2] [N 2 ] vs Time ΔtΔt Δ [N2] instantaneous rate average rate

8.  Chemical reactions involve the breaking of  bonds (endothermic) and the forming of  bonds (exothermic).  The factors affecting reaction rates:  The physical state of reactant rates. ◦ When reactants are in different phases ◦ (states), the reaction is limited to their ◦ area of contact.

9.  The concentration of reactants. ◦ As the concentration of reactants ◦ increase, the rate of reaction increases.  Temperature ◦ As the temperature is increased, ◦ molecules have more kinetic energy ◦ resulting in higher reaction rates.  Presence of a catalyst. ◦ Increases the rate of reaction without ◦ being consumed.

10.  Consider the following generic reaction:  aA(g) + bB(g) cC(g) + dD(g)  The rate law is written:  Rate = k[A] x [B] y  k is called the rate constant and is ◦ temperature dependent.  The exponents x and y are called ◦ reaction orders.

11.  The overall reaction order is the sum of the  orders with respect to each reactant.  The reaction orders in a rate law indicate how  the rate is affected by the concentration of  the reactants.  The effect of reactant concentration can ◦ not be predicted from the balanced ◦ equation.  The effect of reactant concentration can ◦ only be determined empirically!

12.  The rate of a reaction will decrease as the  reaction proceeds because the reactants are  being consumed.  A very important distinction needs to be  made at this point:  The rate of a reaction depends on  concentration but k, the rate constant,  does not!  The rate constant and the rate of ◦ reaction are affected by temperature ◦ and the presence of a catalyst.

13. (a) How is the rate of disappearance of N 2  and O 2 related to the rate of appearance of NO 2 in the reaction shown below.  N 2 (g) + 2O 2 (g) 2NO 2 (g) Rate = - Δ[N 2 ] Δt = - 1 2 Δ[O 2 ] Δt = = 1 2 Δ[NO 2 ] Δt

14.  (b) If the rate of decomposition of N 2 at an  instant is 3.7 x 10 -6 Ms -1, what is the rate  of disappearance of O 2 and the rate of  appearance of NO 2 ? Rate = - Δ[N 2 ] Δt = 3.7 x 10 -6 Ms -1 Δ[O 2 ] Δt = 2 × 3.7 x 10 -6 Ms -1 = 7.4 x 10 -6 M/s Δ[NO 2 ] Δt 2 × 3.7 x 10 -6 Ms -1 = 7.4 x 10 -6 M/s =

15.  The following data was measured for the  reaction of nitrogen and oxygen to form  nitrogen(IV) oxide.  N 2 (g) + 2O 2 (g) 2NO 2 (g)

16. .. Trial[N 2 ] (M) [O 2 ] (M) Init Rate (M/s) 10.12 1.19×10 -3 20.120.232.44×10 -3 30.220.124.92×10 -3

17. (a) Determine the rate law for this reaction. Rate 2 Rate 1 = k[N 2 ] x [O 2 ] y 2 = 2y2y y = 1 2.44 × 10 -3 M/s 1.19 × 10 -3 M/s = k(0.12) x (0.23) y k(0.12) x (0.12) y

18. Rate 3 Rate 1 = k[N 2 ] x [O 2 ] y 4.92 × 10 -3 M/s 1.19 × 10 -3 M/s = k(0.22) x (0.12 ) k(0.12) x (0.12) 4 = 2x2x x = 2 Rate = k[N 2 ] 2 [O 2 ]

19. (b) Calculate the rate constant. Rate = k[N 2 ] 2 [O 2 ] 1.19 × 10 -3 Ms -1 = k(0.12 M) 2 (0.12 M) k = 0.69 M -2 s -1 (c) Calculate the rate when [N 2 ] = 0.047 M and [O 2 ] = 0.16 M Rate = 0.69 M -2 s -1 × (0.047 M) 2 × 0.16 M Rate = 2.4 x 10 -4 M/s

20. The second type of rate law is called the integrated rate law which shows the concentration as a function of time. There are three possible orders for a reaction, zero, first, or second.

21. The integrated rate law is given by: [A] = -kt + [A] 0 A plot of [A] vs Time produces a straight line with a slope equal to the negative rate contant, -k.

22. The reaction rate is independent of concentration. The reaction rate is independent of time.

23. For 0 th order kinetics:  Rate Law: Rate = k  Integrated Rate Law: [A] = -kt + [A] 0  Plot needed for straight line: [A] vs T  Slope: -k  Half-Life:t 1/2 = [A] 0 2k

24. The integrated rate law for 1 st order kinetics is given by:  ln[A] = -kt + ln[A] 0 A plot of ln[A] vs Time produces a straight line with a slope equal to the negative rate contant, -k.

25. The reaction rate is directly proportional to the concentration. The reaction rate decreases but not linearly with time.

26. A first-order reaction is one in which the rate depends on the concentration of a single reactant to the first power. Rate = - Δ [A] ΔtΔt = k[A] aA(g) Product Δ [A] ΔtΔt = - k [A] ln[A] t – ln[A] 0 = -kt

27. ln[A] t = -kt + ln[A] 0 y = mx + b The slope of the straight line gives -k, the rate constant, and ln[A] 0 is the y-intercept. For a first-order reaction, the half-life (the time required for one-half of a reactant to decompose) has a constant value. t 1/2 = ln 2 k = 0.693 k

28. For a first-order reaction, the half-life is only dependent on k, the rate constant and remains the same throughout the reaction. The half-life is not affected by the initial concentration of the reactant.

29. For 1 st order kinetics:  Rate Law: Rate = k[A]  Integrated Rate Law: ln[A] = -kt + ln[A] 0  Plot needed for straight line: ln[A] vs T  Slope: -k  Half-Life:t 1/2 = 0.693 k

30. The integrated rate law for 2 nd order kinetics is given by:  1/[A] = kt + 1/[A] 0 A plot of 1/[A] vs Time produces a straight line with a slope equal to the rate contant, k.

31. The reaction rate is directly proportional to the concentration. The reaction rate decreases but not linearly with time.

32. A second-order reaction is one in which the rate depends on the concentration of a single reactant concentration raised to the second power or concentrations of two different reactants, each raised to the first power. Rate = - Δ [A] ΔtΔt = k[A][B] aA(g) Product aA(g) + bB(g) Product - Δ [B] ΔtΔt = or Rate = - Δ [A] 2 ΔtΔt = k[A] 2

33. The integrated rate law for a second-order reaction is given by: t = k 1 [A] t + 1 [A] 0 y = mx + b The slope of the straight line gives k, the rate constant, and is the y-intercept. 1 [A] 0

34. For a second-order reaction, the half-life (the time required for one-half of a reactant to decompose) is double the preceding one. t 1/2 = 1 k[A] 0

35. For 2 nd order kinetics:  Rate Law: Rate = k[A] 2  Integrated Rate Law:  Plot needed for straight line:   Slope: k Half-Life:t 1/2 = 1 1 [A] = kt + 11 [A] 0 [A] 1 1 1 [A] 0 vs T k [A] 0

36.  Particles must collide for chemical reactions to occur.  Not every collision leads to a reaction.  For a reaction to occur, an “effective collision” must take place.  An “effective collision” consists of two conditions.

37. The colliding particles must approach each other at the proper angle. H2H2 Cl 2 v1v1 v2v2 v4v4 v3v3 HCl

38.  In addition, the colliding particles must have sufficient energy.  At the point of impact, ΔKE = ΔPE.  An elastic collision is assumed in which the law of conservation of mass-energy applies.  The KE/molecule must be sufficient to break the H-H and the CI-CI in both hydrogen and chlorine.

39.  Keep in mind that temperature is a measure of the KE/molecule.  Some molecules will be traveling faster than others while the slower molecules will bounce off each other without reacting.

40. A less than ideal set of conditions for a collision. H2H2 Cl 2 v1v1 v2v2 v4v4 v3v3 H2H2

41.  The bonding found in the reactants. ◦ triple bonds are stronger than double bonds which are stronger than single bonds.  The temperature of the system (reactants and products).  The initial concentration of the reactants.  The amount of surface area when reactants are in more than one phase.  The use of a catalyst to provide an alternate pathway.

42.  In order to react, colliding molecules must have KE equal to or greater than a minimum value.  This energy is called the activation energy (E a ) which varies from reaction to reaction. ◦ E a depends on the nature of the reaction and is independent of temperature and concentration.

43.  The rate of a reaction depends on E a.  At a sufficiently high temperature, a greater number of molecules have a KE > E a.

44. lower temperature higher temperature EaEa

45.  For the Maxwell-Boltzmann Distribution, only those molecules with energies in excess of E a react.  As shown on the graph, as the temperature is increased:  The curve shifts to higher energies meaning more molecules to have energies greater than E a.  The curve gets broader and flatter but the area remains the same under the curve.

46. Minimum Energy for Reaction Energy Content of Reactants Energy Content of Products EaEa Δ H = -

47. Energy Content of Products Energy Content of Reactants Minimum Energy for Reaction EaEa Δ H = +

48.  The Arrhenius equation illustrates the dependence of the rate constant, k, on the frequency factor, A, the activation energy, E a, the gas constant, R, and the absolute temperature, T. k = Ae -Ea/RT The frequency factor, A, is related to the frequency of collisions and the probability of these collisions being favorably oriented.

49. Taking the ln of both sides gives: ln k = ln A – E a /RT Rearranging the equation above shows that the ln k is directly proportional to 1/T. ln k = -E a /R 1T1T + ln A y = m x + b The equation also shows that the reaction rate decreases as the activation energy increases.

50. An unknown gas at 300.°C has a rate constant equal to 2.3 x 10 -10 s -1 and at 350.°C the rate constant is determined to be 2.4 x 10 -8 s -1. (a) Calculate E a for this reaction. T 1 = 300.°C = 573 KT 2 = 350.°C = 623 K k 1 = 2.3 x 10 -10 s -1 k 2 = 2.4 x 10 -8 s -1 R = 8.31 J mol -1 K -1 1 T2T2 ln k2k2 k1k1 = EaEa R [ 1 T1T1 - ] 1 T2T2

51. .. 2.4 × 10 -8 s -1 2.3 × 10 -10 s -1 = EaEa 8.31 J mol -1 K -1 [ 1 573 K 1 623 K - ] EaEa = 2.7 × 10 5 J/mol ln k2k2 k1k1 = EaEa R [ 1 T1T1 - ] 1 T2T2

52. (b) Determine the rate constant at 425°C. ln k2k2 2.4 x 10 -10 s -1 2.7 x 10 5 J mol -1 8.31 J mol -1 K -1 = [ 1 573 K 1 698 K - ] k2k2 = 6.2 x 10 -6 s -1

53.  A reaction mechanism consists of a series of elementary steps indicating how reacting particles rearrange themselves to form products.  Elementary steps are individual steps showing what molecules must collide with each other and the proper sequence of collisions.  Most reactions occur in more than one step.

54. A reaction mechanism provides much more information than a balanced chemical equation.  Intermediates are species that are formed and consumed between elementary steps which never appear in a balanced chemical equation.  Not all elementary steps occur at the same rate and consequently are designated as fast or slow steps.

55. Each elementary step has a molecularity associated with it.  An elementary step is referred to as a unimolecular step when one molecule decomposes or rearranges to form a product as shown below: A B  A unimolecular reaction follows a first-order rate law, i.e.  Rate = k[A]

56.  An elementary step is referred to as a bimolecular step when two molecules collide to form a product as shown below: A + BC  A bimolecular reaction follows a second-order rate law, i.e.  Rate = k[A][B]

57.  An elementary step is referred to as a termolecular step when three molecules collide to form a product as shown below: A + B+ C D  A termolecular reaction follows a third-order rate law, i.e.  Rate = k[A][B][C]  Termolecular reactions are very rare.

58.  The slowest step in a mechanism is called the rate determining step and determines the rate of the overall reaction.  Keep in mind:

59.  The rate of the overall reaction is the same as the rate of the rate-determining step.  Write the rate expression for the slowest step.  The rate expression must only include those species that appear in the balanced equation.  Intermediates are species that are produced in one step and consumed in the next step.  Intermediates can not appear in the rate expression because their concentrations are always small and undetectable.

60. Determine the overall reaction and the rate expression for the suggested mechanism: NO 2 (g) + NO 2 (g) NO 3 (g) + NO(g) (slow) NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) (fast) NO 2 (g) + NO 2 (g) NO 3 (g) + NO(g) (slow) NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) (fast) NO 2 (g) + CO(g) NO(g) + CO 2 (g)

61. Note that the NO 3 is an intermediate because it is produced in one step of the reaction and consumed in the next. Because the rate-determining step comes first, the intermediate NO 3 cancels out. This is good because intermediates cannot appear in the overall reaction. The rate expression is determined by using the rate-determining step and is given by: Rate = k[NO 2 ][NO 2 ] = k[NO 2 ] 2

62. The following reaction 2NO 2 (g) + O 3 (g) N 2 O 5 (g) + O 2 (g) Rate = k[NO 2 ][O 3 ] Which of the following mechanisms predicts a satisfactory rate law? Mechanism 1: NO 2 (g) + NO 2 (g) N 2 O 2 (g) + O 2 (g) (slow) N 2 O 2 (g) + O 3 (g) N 2 O 5 (g) (fast) Mechanism 2: NO 2 (g) + O 3 (g) NO 3 (g) + O 2 (g)(slow) NO 3 (g) + NO 2 (g) N 2 O 5 (g)(fast)

63. For Mechanism 1: NO 2 (g) + NO 2 (g) N 2 O 2 (g) + O 2 (g) (slow) N 2 O 2 (g) + O 3 (g) N 2 O 5 (g) (fast) The first criteria is satisfied because after adding the elementary steps, the overall equation for the reaction is correct. However, the predicted rate law determined from the rate-determining step is: which does not agree with given rate law Rate = k[NO 2 ] 2 which does not agree with given rate law 2NO 2 (g) + O 3 (g) N 2 O 5 (g) + O 2 (g)

64. For Mechanism 2: NO 2 (g) + O 3 (g) NO 3 (g) + O 2 (g)(slow) NO 3 (g) + NO 2 (g) N 2 O 5 (g)(fast) 2NO 2 (g) + O 3 (g) N 2 O 5 (g) + O 2 (g) The first criteria is satisfied because after adding the elementary steps, the overall equation for the reaction is correct. The predicted rate law determined from the rate- determining step is: which does agrees with given rate law! Rate = k[NO 2 ][O 3 ] which does agrees with given rate law! intermediate is not a reactant in the rate-determining step

65. When the intermediate is a reactant in the rate-determining step, mechanisms are more involved. Determine the overall reaction and the rate expression for the suggested mechanism: NO(g) + Br 2 (g) NOBr 2 (g)(fast) NOBr 2 (g) + NO(g) 2NOBr(g)(slow)

66. NO(g) + Br 2 (g) NOBr 2 (g)(fast) NOBr 2 (g) + NO(g) 2NOBr(g)(slow) The predicted rate law determined from the rate-determining step is: Rate = k[NOBr 2 ][NO] which is incorrect because intermediates can not appear in the rate law 2NO(g) + Br 2 (g) 2NOBr(g)

67. When this situation occurs, you must equate the forward and reverse rates obtained from the fast step as shown below: NO(g) + Br 2 (g) NOBr 2 (g) (fast) Rate f = Rate r k f [NO][Br 2 ] = k r [NOBr 2 ] [NOBr 2 ] = kfkf krkr [NO][Br 2 ] = k[NO][Br 2 ] [NOBr 2 ]

68. Now you can use the rate law obtained originally but substitute for [NOBr 2 ]. Rate = k[NOBr 2 ][NO] [NOBr 2 ] = k[NO][Br 2 ] Rate = k[NO] 2 [Br 2 ] (There is no need to keep track of the individual rate constants because they would be incorporated into a single constant.)

69. A catalyst is a substance that increases the speed of a reaction without being consumed. Catalysts can be either homogeneous or heterogeneous depending if they are in the same phase as the reactants. The function of a catalyst is to provide an alternate pathway (mechanism) for a reaction which lowers the activation energy, E a.

70. Arrhenius equation shows that as activation energy, E a, is made smaller, the rate constant, k, increases. k = Ae -Ea/RT

72. For the given mechanism, identify the catalyst and the intermediate. O 3 + Br O 2 + BrO BrO + O Br + O 2 Remember an intermediate is produced in one step and consumed in the other making BrO the intermediate. Remember a catalyst remains chemically unaltered during the reaction making the Br the catalyst.