1. AP CHEMISTRY CHAPTER 12 KINETICS

2. Chemical Kinetics Thermodynamics tells us if a reaction can occur
Kinetics tells us how quickly the reaction occurs
some reactions that are thermodynamically feasible are kinetically so slow as to be imperceptible

3. Reaction Rate- change in concentration of a reactant or product per unit time. [A] = concentration in mol/L Rate = [A] t

4. If the rate expression involves a reactant: Rate = − [A] t (negative because [ ] decreases) The above gives the average rate.

5. 2NO2 2NO + O2

6. To get an instantaneous rate, we can compute the slope of a line tangent to the curve at that point. Rate = −(slope of the tangent line)

8. The rate of a reaction is not constant but changes with time because concentrations change with time. We will only work with reaction rates that are “initial rates” (reverse reaction is negligible)

9. Differential Rate Law or Rate Equation For the reaction aA+bB + … gG +hH+ … Rate = k[A]m[B]n… [A] & [B] represent molarities

10. The exponents are positive or negative, integers or fractions
The exponents are positive or negative, integers or fractions. usually positive integers (small whole numbers) k = rate constant *value depends on reaction, temperature and presence of a catalyst *faster the reaction, larger the k value

11. The exponents determine the order of the reactants
The exponents determine the order of the reactants. The sum of the exponents is the order of the reaction R = k[A][B]2 is first order in A, second order in B and third order overall.

12. The units of k can be calculated by reaction orders and units of concentration and rate. For example, if rate is in mol/Ls in the above rate law, we can find the units for k as follows: Rate = k so k = mol [A][B] L  s mol mol L L2 This simplifies to: k = L mol2.s

13. The “shortcut” to determining units of k is as follows: k = Lx/(molx
The “shortcut” to determining units of k is as follows: k = Lx/(molx.time). The value of x will be one less than the order of the reaction. If the reaction is 3rd order, k = L2/mol2.time. If the reaction is 2nd order, k = L/mol.time If the reaction is 1st order, k = 1/time

14. Determining Differential Rate Laws from Experimental Data If doubling the initial [ ] of a reactant causes the initial rate to double, the reaction is first order in that reactant.

15. If doubling the initial [ ] of a reactant causes the initial rate to quadruple, the reaction is second order in that reactant.

16. If doubling the initial [ ] of a reactant does not change the initial rate, the reaction is zero order in that reactant and that reactant is removed from the rate law.

17. 2A + B 2C Ex. [A] [B] Rate 1 0.1 0.2 0.10 2 0.1 0.4 0.20 3 0.2 0.4 0.80 Determine the rate law:
Rate = k[A]x[B]y
Rate = k[A]x[B]1
Rate = k[A]2[B]
Because 21=2, B is 1st order
Because 22 =4, A is 2nd order

18. A  B [A] Rate 0.05 3 x 10-4 0.10 1.2 x 10-3 0.20 4.8 x 10-3 Determine the rate law:
Rate = k[A]x
Rate = k[A]2
Because 22=4, A is 2nd order

19. 1 2 1
A + B  C Exp [A] [B] Rate ×10 ×10 ×105 Determine the rate law: 2 1 2
Rate = k[A]x[B]y
Rate = k[A]0[B]y
Rate = k [B]
Because 20=1, A is 0 order
Because 21=2, B is 1st order

20. 2 1 2
A + 2B 2C Ex [A] [B] Rate Determine the rate law: 2 2 2
Rate = k[A]x[B]y
Rate = k[A]1[B]y
Rate = k[A]1[B]0
Rate = k[A]
Because 21=2, A is 1st order
No control in this example. Because doubling A causes the rate to double, doubling B must have had no effect. B is zero order.

21. 1 2 2
NH4+ + NO2  N2 + 2H2 Exp [NH4+] [NO2] Rate(mol/Lmin) × 10 × 10 × 10 Determine the rate law: Determine the value of k and its units: 2 1 2
Rate = k[NH4+]x[NO2]y
Rate = k[NH4+]x[NO2]1
Rate = k[NH4+] [NO2]
1.35×107mol/Lmin = k(0.100mol/L)(0.005mol/L)
k = 2.7 ×104 L/mol.min
Because 21=2, NH4+ is 1st order orer
Because 21=2, NO2- is 1st order

22. Determine the rate law: Rate = k[A]x[B]y 42=16 so [A] is 2nd order
Experiment
[A]
[B]
Initial Rate
mol L-1s-1 1
10.1
2.01
5.00 2
19.8
3.99
80.0 3
40.2
2.00 2 16 2 4 1 16
Determine the rate law:
Rate = k[A]x[B]y
42=16 so [A] is 2nd order
Rate = k[A]2[B]y
Because A is 2nd order, doubling A will cause the rate to quadruple. When both A and B were doubled, the rate was 16 times greater. This means that B is also 2nd order.
Rate = k[A]2[B]2
Determine the value of k and its units.
5.00 mol L-1s-1 = k(10.1mol/L)2(2.01mol/L)2
k = L3mol-3s-1

23. Integrated Rate Law -expresses how the concentration of the reactant depends on time -instead of changing initial concentrations and using multiple experiments, one experiment is done and concentration changes over time are measured.

24. 1st order integrated rate law

25. A plot of ln[A] vs t always gives a straight line for a 1st order reaction. The slope = -k

27. Ex. At 400oC, the 1st order conversion of cyclopropane into propylene has a rate constant of 1.16 × 10-6s-1. If the initial concentration of cyclopropane is 1.00 × 10-2 mol/L at 400oC, what will its concentration be 24.0 hrs after the reaction begins?
24 hrs 3600s = 86400s
1 hr
ln[A]t = −kt ln [A]t = −1.16×10−6(86400)
[A] ×10−2
[A]t = 9.05×10−3

28. Radioactive decay is first order
Radioactive decay is first order. Half-life (t1/2) is the length of time required for the concentration of a reactant to decrease to half of its initial value t1/2 = 0.693/k
Where does come from?
It is the natural log of 2.

29. A fast reaction with a short t1/2 has a large k
A fast reaction with a short t1/2 has a large k. A slow reaction with a long t1/2 has a small k.

30. A Plot of (N2O5) Versus Time for the Decomposition Reaction of N2O5

31. Example: The decomposition of SO2Cl and Cl2 is a first order reaction with k = 2.2×10−5s−1 at 320oC. Determine the half-life of this reaction.
t1/2 = 0.693/k
t1/2 = 0.693/(2.2×10−5s−1)
t1/2 = s or 525 min or 8.75 hours

32. ln [A]t = −kt [A]0 ln 1.21 × 103 = −k(180 min) 1.00
Example: The decomposition of N2O5 dissolved in CCl4 is a first order reaction. The chemical change is: N2O5  4NO2 + O2 At 45oC the reaction was begun with an initial N2O5 concentration of 1.00 mol/L. After 3.00 hours the N2O5 concentration had decreased to 1.21×10−3 mol/L. What is the half-life of N2O5 expressed in minutes at 45oC?
ln [A]t = −kt
[A]0
ln 1.21 × 103 = −k(180 min)
k = min t1/2 = 0.693/k
t1/2 = 0.693/ min1 = 18.6 min

33. Differential rate law
Second order integrated rate law Rate = k[A] = kt [A]t [A]0 A plot of 1/[A]t versus t produces a straight line with slope k.

34. (a) A Plot of In(C4H6) Versus t (b) A Plot of 1/(C4H6) Versus t

35. Differential rate law
This equation is not found on the AP formula sheet and will not have to be calculated. Make sure that you understand the graphing.
Zero Order Rate = k [A]t  [A]o = kt A plot of [A] versus t produces a straight line with slope -k.

36. A Plot of (A) Versus t for a Zero-Order Reaction

37. An easy way to keep the graphs straight is to remember CLR=0,1,2
C=concentration, L = ln concentration, R=reciprical of concentration
If the concentration vs time is straight, it is C (0 order).
If natural log of concentration vs time is straight it is L=1st order.
If reciprical of concentration vs time is straight it is R=2nd order.
CLR
0 1 2

38. What order is this reaction?

39. What order is this reaction?

40. Reaction Mechanisms intermediate - a species that is neither a product nor a reactant in the overall equation -is used up in a subsequent step elementary step- a reaction whose rate law can be written from its molecularity (balanced equation)

41. molecularity- the number of species that must collide to produce the reaction represented by the elementary step. Unimolecular step- a reaction step involving only one molecule Bimolecular step- a reaction step involving the collision of two molecules (Rate law always 2nd order)

42. Rate-determining step -slowest step Lucy clip Elementary Reaction -agrees with the balanced equation

43. Reaction mechanisms must: 1. Add up to the overall balanced equation. 2. Agree with the rate law

44. We can’t prove a mechanism absolutely
We can’t prove a mechanism absolutely. We can only come up with a possible mechanism. Look at this video where we can see that the reaction is occuring through a series of steps (mechanism). This is called a “clock reaction”. clock reaction

45. The iodine clock reaction uses a solution of iodate ion to which an acidified solution of sodium bisulfite is added. Iodide ion is generated by the following slow reaction between the iodate and bisulfite: IO3− (aq) + 3HSO3− (aq) → I− (aq) + 3HSO4−(aq) This is the rate determining step. The iodate in excess will oxidize the iodide generated above to form iodine: IO3− (aq) + 5I− (aq) + 6H+ (aq) → 3I2 + 3H2O (l) However, the iodine is reduced immediately back to iodide by the bisulfite: I2 (aq) + HSO3− (aq) + H2O (l) → 2I− (aq) + HSO4−(aq) + 2H+ (aq) When the bisulfite is fully consumed, the iodine will survive (i.e., no reduction by the bisulfite) to form the dark blue complex with starch.

46. Ex. Elementary Rxn : NO + N2O  NO2+ N2
Rate Law: R = k[NO][N2O]

47. Rate law for the reaction: Rate = k[NO2][F2]
Ex. The reaction 2NO2(g) + F2(g)  2NO2F is thought to proceed via the following two-step mechanism: NO2 + F2  NO2F + F slow F + NO2  NO2F fast
Rate law for the reaction:
Rate = k[NO2][F2]

48. When an intermediate is a reactant in the rate-determining step, the derivation of the rate law is more difficult.

49. Ex. NO2 + CO  NO + CO k1 Mechanism: NO2 + NO2  NO3 + NO Both fast w/ k equal rates NO3 + CO  CO2 + NO slow

50. Ex. NO2 + CO  NO + CO k1 Mechanism: NO2 + NO2  NO3 + NO Both fast w/ k equal rates NO3 + CO  CO2 + NO slow Slow step determines rate law (rate-determining step) Rate law: R = k[NO3][CO] But, NO3 was an intermediate. We must come up with something equal to NO3 to substitute k[NO2]2 = k[NO3][NO] [NO3] = [NO2] [NO] R = k [NO2]2[CO] [NO]

51. Ex. Cl2 + CHCl3  HCl + CCl4 Mechanism k Cl2  2Cl fast k Cl + CHCl3  HCl + CCl3 slow Cl + CCl3  CCl fast Rate Law:

52. Ex. Cl2 + CHCl3  HCl + CCl4 Mechanism k Cl2  2Cl fast k Cl + CHCl3  HCl + CCl3 slow Cl + CCl3  CCl fast Rate Law: R = k[Cl][CHCl3] Cl is an intermediate k[Cl2] = k[Cl] [Cl2]1/2 = [Cl] R = k[Cl2]1/2[CHCl3]

53. Increasing temperature increases reaction speed
Increasing temperature increases reaction speed. Rate and rate constants often double for every 10o increase in temperature. (Rule of thumb)

54. Plot Showing the Number of Collisions with a Particular Energy at T1 and T2, where T2>T1

55. Molecules must collide to react
Molecules must collide to react. Only a small portion of collisions produce a reaction.

56. Several Possible Orientations for a Collision Between Two BrNO Molecules

57. Activation energy (EA) - energy that must be overcome to produce a chemical reaction.

58. Rate of reaction depends on EA, not E
Rate of reaction depends on EA, not E. E has no effect on rate of reaction. The higher the EA, the slower the reaction at a given temperature.

59. Molecules and collisions have varying energies
Molecules and collisions have varying energies. As temperature increases, more collisions will have sufficient energy to overcome the activation energy. As temperature doubles, the fraction of effective collisions increases exponentially. Reaction rate is smaller than would be predicted from the number of collisions having sufficient energy to react. This is because of molecular orientations.

60. 2 factors: 1. sufficient energy 2. proper orientation

61. This calculation is not required on the AP exam.
Arrhenius Equation- -can be used to calculate EA ln k = ln A - EA k = rate constant RT R = J/K mol T = Kelvin temp A = frequency factor (constant as temperature changes) EA = activation energy As EA increases, k decreases.

62. The Arrhenius equation describes a line
The Arrhenius equation describes a line. We can plot 1/T vs ln k and get a straight line whose slope is equal to -EA/R.

63. Plot of In(k) Versus 1/T for the Reaction 2N2O5(g) ® 4NO2(g) + O2(g)
Slope = -EA/R
Plot of In(k) Versus 1/T
for the Reaction
2N2O5(g) ® 4NO2(g) + O2(g)

64. A variation of the Arrhenius equation can be used to calculate EA or to find k at another temperature if EA is known: ln k1 = EA k2 R T2 T1
This calculation is not required on the AP exam.

65. Catalysis
Catalysts, such as enzymes in a biological system or the surfaces in an automobile's catalytic converter, increase the rate of a chemical reaction.

66. catalyst- substance that speeds up a reaction without being consumed
catalyst- substance that speeds up a reaction without being consumed. -produces a new reaction pathway with a lower activation energy -A catalyst lowers the EA for both the forward and reverse reaction. Some catalysts change the mechanism of the reaction greatly and others simply change the rate-determining step.

69. enzymes- biological catalysts

70. homogeneous catalyst- present in the same phase as the reacting molecules (usually liquid phase)

71. heterogeneous catalyst- exists in a different phase -usually involves gaseous reactants being adsorbed on the surface of a solid catalyst (such as a car’s catalytic converter) Either a new reaction intermediate is formed or the probability of successful collisions is increased. This is sometimes called a surface catalyst. Absorption involves penetration.

72. Heterogeneous Catalysis of the Hydrogenation of Ethylene

74. Catalytic Converter

75. How does the mechanism for the catalyzed reaction differ from the mechanism of the uncatalyzed reaction?