KINETICS

1. KINETICS • The study of reaction rates. • Spontaneous reactions are reactions that will happen - but we can’t tell how fast. • Diamond will spontaneously turn to graphite – eventually. • Reaction mechanism- the steps by which a reaction takes place.

2. The part of chemistry that is thermodynamics and the part that is kinetics.

3. FACTORS AFFECTING RATE • 1. Nature of Reactants • 2. Concentration of Reactants • 3. Temperature • 4. Catalysts • 5. Surface Area of Reactants • 6. Adding an Inert Gas (NO EFFECT).

4. COLLISION THEORY • Molecules must collide to react. • Only two particles collide at a time. • Must have proper orientation.

5. COLLISION THEORY • Must collide with enough energy. • Only a small number of collisions produce reactions. • Concentration affects rates because collisions are more likely. • Temperature and rate are related.

6. ENERGY PLOT • Know Ea, transition state, and activated complex. Know exothermic versus endothermic.

7. Potential Energy Reactants Products Reaction Coordinate

8. Potential Energy Activation Energy Ea Reactants Products Reaction Coordinate

9. Activated complex Potential Energy Reactants Products Reaction Coordinate

10. Potential Energy } Reactants DE Products Reaction Coordinate

11. Br---NO Potential Energy Br---NO Transition State 2BrNO 2NO + Br 2 Reaction Coordinate

12. MAXWELL BOLTZMANN DISTRIBUTION

13. MAXWELL BOLTZMANN DISTRIBUTION

14. REACTION RATE • Rate = [A] at t2 – [A] at t1 t2- t1 • Rate =D[A]Dt • Change in concentration per unit time. • [reactants] decreases with time. • [products] increases with time

15. Concentration • N2 + 3H2→ 2NH3 • As the reaction progresses the concentration H2 goes down [H2] Time

16. Concentration • N2 + 3H2→ 2NH3 • As the reaction progresses the concentration N2 goes down 1/3 as fast. [N2] [H2] Time

17. Concentration • N2 + 3H2→ 2NH3 • As the reaction progresses the concentration NH3 goes up. [N2] [H2] [NH3] Time

18. CALCULATING RATES • Average rates are taken over a range of time. Worry about coefficients. • Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time • Derivative. • RATES ARE ALWAYS POSITIVE!

19. Concentration • AVERAGE SLOPE METHOD D[H2] Dt Time

20. Concentration • INSTANTANEOUS SLOPE METHOD D[H2] D t Time

21. INSTANTANEOUS RATE

22. DEFINING RATE • We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product. • example N2 + 3H2→ 2NH3 • -D[N2] = -3D[H2] = 2D[NH3] • Dt Dt Dt Negative because [reactant] goes down.

23. RATE LAWS • Reactions are reversible. • As products accumulate they can begin to turn back into reactants. • Early on the rate will depend on only the amount of reactants present. • We want to measure the reactants as soon as they are mixed. • This is called the Initial rate method.

24. RATE LAWS • Two key points: • The concentration of the products do not appear in the rate law because this is an initial rate. • The order must be determined experimentally; they can’t be obtained from the equation.

25. 2 NO2 2NO + O2 • You will find that the rate will only depend on the concentration of the reactants. • Rate = k[NO2]n • This is called a rate law expression. • k is called the rate constant. • n is the order of the reactant -usually a positive integer.

26. 2 NO2 2NO + O2 • Oxygen can appear only half as rapidly as the nitrogen dioxide disappears while NO appears twice as fast as oxygen appears. • Calculate the AVERAGE rate at which [NO2] changes in the first 50.0 seconds given the [NO2] at 0.0s = 0.0100M and [NO2] at 50s = 0.0079M . • RATE = −Δ [NO] = −[.0079]−[0.0100] • Δt50.0 s • = −[−4.2 × 10−5M / sec] = 4.2 × 10−5M /sec

27. 2 NO2 →2NO + O2 • In terms of NO2: • Rate = -Δ[NO2] = k[NO2]n Δt In terms of O2: • Rate’ = -Δ[O2] = k’[O2]n Δt So: Rate = 2 x Rate’ OR k[NO2]n = 2k’[NO2]n

28. RELATIVE RATES • We can consider the appearance of products along with the disappearance of reactants. • The reactant’s concentration is declining, the products is increasing. • Respect the algebraic sign AND respect the stoichiometry. • Divide the rate of change in concentration of each reactant by its stoichiometric coefficient in the balanced chemical equation.

29. RELATIVE RATES • 2 NO2 →2NO + O2 • Thus..... • Rate of reaction = - 1Δ[NO2] = 1 Δ[NO] = Δ [ O2] • 2 Δtime2 ΔtimeΔtime • There are 2NO2 molecules consumed for every O2 molecule produced.

30. ANOTHER EXAMPLE • Using the coefficients from the balanced equation, you should be able to give relative rates. For example: 4 PH3(g) → P4(g) + 6 H2(g) • Initial rate rxn = = + = +

31. PRACTICE ONE • What are the relative rates of change in concentration of the products and reactant in the decomposition of nitrosyl chloride, NOCl? 2 NOCl(g) → 2 NO(g) + Cl2(g)

32. TYPES OF RATE LAWS • Differential Rate law - describes how rate depends on concentration. • Integrated Rate Law - Describes how concentration depends on time. • For each type of differential rate law, there is an integrated rate law and vice versa. • Rate laws can help us better understand reaction mechanisms.

33. TYPES OF RATE LAWS • Reactions are reversible. • When the rate of the forward = the rate of the reverse we have EQUILIBRIUM! To avoid this complication we will discuss reactions soon after mixing--initial reactions rates, and not worry about the buildup of products and how that starts up the reverse reaction. • So reaction rate will only depend on [reactants] and initial [reactants] right after mixing.

34. DETERMINING RATE LAWS • The first step is to determine the form of the rate law (especially its order). • Must be determined from experimental data. • For this reaction • 2 N2O5(aq) → 4NO2(aq) + O2(g) • The reverse reaction won’t play a role

35. DIFFERENTIAL RATE LAW • Describes how the reaction rate varies with the concentration of various species in a system. • Rates generally depend on reactant concentrations. • To find the exact relation between rate and concentration, we must conductexperiments and collect information.

36. DIFFERENTIAL RATE LAW • C • aA + bB → xX • Becomes Initial rxn rate = k[A]m[B]n[C]p • NOT COEFFICIENTS • K = temperature dependent

37. DIFFERENTIAL RATE LAW • Exponents can be zero, whole numbers or fractions AND MUST BE DETERMINED BYEXPERIMENTATION!! • THE RATE CONSTANT, k • temperature dependent and must be evaluated by experiment. • Example: rate = k[A]. If k is 0.090/hr when [A] = 0.018 mol/L • rate = (.0090/hr)(0.018 mol/L) = 0.00016 mol/(L• hr)

38. ORDER OF A REACTION • order with respect to a certain reactant is the exponent on its concentration term in the rate expression. • order of the reaction is the sum of all the exponents on all the concentration terms in the expression. • DETERMINATION OF THE RATE EXPRESSION • aA + bB → xX • initial rate = k[A]om[B]on - the little subscript “o” means “original” or at “time zero”.

39. ORDER OF A REACTION 1. Zero order: The change in concentration of reactant has no effect on the rate. These are not very common. General form of rate equation: Rate = k 2. First order: Rate is directly proportional to the reactants concentration; doubling [rxt], doubles rate. These are very common! Nuclear decay reactions usually fit into this category. General form of rate equation: Rate = k [A]1 = k[A] 3. Second order: Rate is quadrupled when [rxt] is doubled and increases by a factor of 9 when [rxt] is tripled etc. These are common, particularly in gas-phase reactions. General form of rate equation: Rate = k [A]2 or Rate = k[A]1[B]1which has an overall order of two (second order).

40. ORDER OF A REACTION • To determine the order of the reaction: • 1. Determine skeleton rate law • 2. Determine exponents using data from trials. • 3. Determine k with its units. • 4. Write the rate law. • Adding the orders of each reactant gives the overall order of the reaction.

41. EXAMPLE • NO(g) + Cl2(g) → NOCl(g) at 295K • Rate = k[NO]m[Cl2]n

42. HOW TO SOLVE • 1. Skeleton: Rate = k[NO]m[Cl2]n • 2. Determine order: Look for two trials where the concentration of a reactant was held constant. Next, focus on the other reactant. Ask yourself how it’s concentration changed for the same two trials. Was it doubled? Was it tripled? Was it halved? Once you have determined the factor by which the concentration of the other reactant was changed, determine how that affected the rate for those same two trials.

43. HOW TO SOLVE • To find the rate order of NO, find two trials where [Cl2] is constant and [NO] changes. From experiment 1 to experiment 3, [NO] triples while [Cl2] is constant. Since the initial rate increases by a factor of 9x, the rate is directly proportional to [NO]2 and the RATE ORDER FOR [NO] IS 2. • Experiment 39.0 x 10-3k[0.150]n [0.050]m 9 = 3n • Experiment 1 1.0 x 10-3k[0.050]n [0.050]m n = 2

44. HOW TO SOLVE • To find the rate order of Cl2, find two trials where [NO] is constant and [Cl2] changes. From experiment 1 to experiment 2, [Cl2] triples while [NO] is constant. Since the initial rate triples, the rate is directly proportional to [Cl2]1 and the RATE ORDER FOR Cl2 IS 1. • Experiment 23.0 x 10-3k[0.050]n[0.150]m3 = 3m • Experiment 1 1.0 x 10-3k[0.050]n[0.050]mm = 1

45. HOW TO SOLVE • THE RATE LAW EXPRESSION IS RATE = k[NO]2[Cl2]1 • OVERAL ORDER: 3 • NOTE: The rate orders aren’t always the same as the coefficients.

46. HOW TO SOLVE • 3. Determine k and units: Use any experimental data to find the value of k and its units. The units for K are one less that the overall order. Using experiment 1 data, plug the values into the rate law and solve for k • Rate = k[NO]2[Cl2]1 = 1.0 x 10-3M = k (0.050M)2 (0.050M)1 • k = (1.0 x 10-3M/s) / [(0.050M)2 (0.050M)1] • k = 8.0M-2x s-1 or 8.0L2 / (mol2 x s) • The larger the k, the faster the reaction will be. k only changes with changes in TEMPERATURE. Increasing the temperature will increase k and increase the reaction rate.

47. PRACTICE TWO • In the following reaction, a Co-Cl bond is replaced by a Co-OH2 bond. • [Co(NH3)5Cl]+2 + H2O → [Co(NH3)5H2O]+3 + Cl • Initial rate = k{[Co(NH3)5Cl]+2}m • Using the data below, find the value of m in the rate expression and calculate the value of k. • Exp. Initial Concentration Initial rate of [Co(NH3)5Cl]+2(M) mol/(L• min) • 1 1.0 × 10-3 1.3 × 10-7 • 2 2.0 × 10-3 2.6 × 10-7 • 3 3.0 × 10-3 3.9 × 10-7 • 4 1.0 × 10-3 1.3 × 10-7

48. PRACTICE THREE • The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation: BrO3-(aq) + 5 Br –(aq) + 6 H+(aq) → 3 Br2(l) + 3 H2O(l) • The table below gives the results of four experiments. Using these data, determine the orders for all three reactants, the overall reaction order, and the value of the rate constant. What is the value of k? • What are the units of k? • Experiment Initial [BrO3-] Initial [Br–] Initial [H+] Measured initial • rate (mol/L•s) • 1 0.10 0.10 0.10 8.0 × 10-4 • 2 0.20 0.10 0.10 1.6 × 10-3 • 3 0.20 0.20 0.10 3.2 × 10-3 • 4 0.10 0.10 0.20 3.2 × 10-3

49. PRACTICE THREE • The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation: BrO3-(aq) + 5 Br –(aq) + 6 H+(aq) → 3 Br2(l) + 3 H2O(l) • The table below gives the results of four experiments. Using these data, determine the orders for all three reactants, the overall reaction order, and the value of the rate constant. What is the value of k? • What are the units of k? • Experiment Initial [BrO3-] Initial [Br–] Initial [H+] Measured initial • rate (mol/L•s) • 1 0.10 0.10 0.10 8.0 × 10-4 • 2 0.20 0.10 0.10 1.6 × 10-3 • 3 0.20 0.20 0.10 3.2 × 10-3 • 4 0.10 0.10 0.20 3.2 × 10-3

50. TYPES OF RATE LAWS • Differential Rate law – data table contains concentration and rate data • Integrated Rate Law – data table contains concentration and time data. Only one reactant.