1. AP CHEMISTRY Chapter 14 Chemical Kinetics

2. 14.1 Chemical Kinetics Study of how rapidly a reaction will occur. In addition to speed of reaction, kinetics also deals with the reaction mechanism (exactly how the reaction occurs) 1. Factors that affect reaction rate 1. Concentration of reactants – more particles means more frequent “effective” collisions occur 2. Temperature – higher temps, the molecules have more KE, are moving faster and are more likely to collide 3. Catalyst – chemical that speeds up a reaction by changing the reaction mechanism, not consumed in reaction 4. Physical state of reactants – (phase) 5. Surface area (for solids) – greater SA = faster reaction

3. 14.2 Reaction Rate The change in concentration of reactants or products over time.  The rate can be expressed as a rate of decomposition of reactants or rate of production of products.

5. Mathematical Expressions for Reactions Reaction: N 2(g) + 3 H 2(g) → 2NH 3(g) ***If you want to compare the rates using different substances, you use the mole ratios The rate of reaction is always positive, and is usually measured in M/s  Under certain conditions, the rate of formation of NH 3 is 0.28 M/s. What is the rate of change for N 2 ?  0.28 M NH 3 1 mol N 2 = 0.14 M/s s2 mol NH 3

7. Average Rate of Change Avg rate = - ∆[C 4 H 9 Cl] ∆t Avg rate = - 0.0905 – 0.1000 50 – 0 = 1.9 x 10 -4 M/s

8. Instantaneous Rate of Change The rate of reaction at a given moment in time The instantaneous rate can be determined by finding the slope of a tangent line at that point in time. All reactions slow down over time, therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction.

9. 14.3 Concentration and Rate Laws One can gain information about the rate of a rxn by seeing how the rate changes with changes in concentration The rate of rxn generally decreases as the concentration of reactant decreases

10. Determining Rate Laws Rate Law- shows how the rate depends on the concentration of reactants  Must be determined from experimental data  Rate depends only on the concentration of the reactants 2N 2 O 5 → 4 NO 2 + O 2 Rate = k[N 2 O 5 ] m  This is called the rate law  k is called the rate constant (depends on temperature and solvent used)  m is the order of the reactant – usually a positive integer (not necessarily the same as the coefficient)

11. General Terms For a general reaction, aA + bB → cC + dD The rate law has the form, Rate = k [A] m [B] n Example: NH 4 + + NO 2 - → N 2 + 2H 2 O Rate = k[NH 4 + ] m [NO 2 - ] n  m and n must be determined experimentally.

12. Reaction Orders The exponents m and n are called reaction orders. If the exponent is one, the rate is first order with respect to that reactant If the exponent is two, the rate is second order with respect to that reactant The overall reaction order is the sum of the orders with respect to each reactant Example: Rate = k[A] 2 [B] 1  The rate is 2 nd order with respect to [A], and first order with respect to [B]. The overall reaction order is 3.

13. Order versus Concentration The rate of a zero order is independent of [A].  Doubling the concentration multiplies the rate of the rxn by 1.  Tripling the concentration multiplies the rate of the rxn by 1. In first order the rate is directly proportional to the [A].  Doubling the concentration multiplies the rate of the rxn by 2.  Tripling the concentration multiplies the rate of the rxn by 3. In second order the rate is proportional to the square of the [A].  Doubling the concentration multiplies the rate of the rxn by 4.  Tripling the concentration multiplies the rate of the rxn by 9.

14. Units of Rate Constants

15. Consider the hypothetical equation and data: A + B → C + D Calculate the order of the reaction with respect to each reactant  Second order with respect to [A] and Zero order with respect to [B] What is the overall order of the reaction  2 nd order overall Write the rate law for the reaction.  Rate = k [A] 2 [B] 0 Calculate the value of k  K = 0.0625 M -1 s -1 #[A], M[B], MRate of formation of C (M/s) 10.200 0.0025 20.4000.2000.010 30.400 0.010

16. Example The following data were measured for the reaction of nitric oxide with hydrogen: 2NO (g) + 2H 2(g) → N 2(g) + 2H 2 O (g) Determine the rate law for this reaction.  rate = k[NO] 2 [H 2 ] What is the overall order of the reaction?  2 + 1 = 3 rd order overall Calculate the rate constant.  k = 1.2 M -2 s -1 Calculate the rate when [NO] = 0.050M and [H 2 ] = 0.150M  rate = 4.5 x 10 -4 M/s Exp #[NO] (M)[H 2 ] (M)Initial Rate 10.10 1.23 x 10-3 20.100.202.46 x 10-3 30.200.104.92 x 10-3

17. Example 2 Time (min)0123 [N2O5], M0.1600.1120.0800.056 rate (M/min) 0.0560.0390.0280.020

18. 14.4 Change of Concentration with Time A first-order reaction is one whose rate depends on the concentration of a single reactant raised to the first power.  Rate = k[A] Relating concentration of A at the start of the reaction, [A] 0, to its concentration at any other time, [A] t can be achieved using the following equation ln[A] t = -kt + ln[A] 0

19. Example The decomposition of dimethyl ether, (CH 3 ) 2 O, at 510 ⁰ C is a first-order process with a rate constant of 6.8 x 10 -4 s -1 : (CH 3 ) 2 O (g) → CH 4(g) + H 2(g) + CO (g) If the initial pressure of (CH 3 ) 2 O is 135 torr, what is its partial pressure after 1420 s  lnP t = -kt + lnP o  lnP t = -(6.8 x 10 -4 s -1 )(1420 s) + ln(135)  lnP t = -0.9656 + 4.91  lnP t = 3.94 = e (3.94)  P t = 51 sec

20. Second-Order Reactions

21. Example

22. 1 st Order Reaction Graphs Ln[A] t = -k t + ln[A] 0 y = m x + b For a first order reaction, a graph of ln[A] t vs time will give a straight line with a slope of –k.

23. 2 nd Order Reaction Graphs

24. Example Cyclopentadiene (C5H6) reacts with itself to form dicyclopentadiene (C10H12). A 0.0400 M solution of C5H6 was monitored as a function of time as the reaction 2C5H6 C10H12 proceeded. The following data was collected: What is the order of the reaction? What is the value of the rate constant? Time (s)[C5H6] (M) 0.00.0400 50.00.0300 100.00.0240 150.00.0200 200.00.0174

25. Time (s) [C5H6] (M) ln[C5H6]1/[C5H6] 0.00.0400-3.21925.0 500.0300-3.50733.3 1000.0240-3.73041.7 1500.0200-3.91250 2000.0174-4.05157.5

26. Half-Life

27. Example

28. 14.5 Temperature and Rate The rates of most chemical reactions increase as temperature increases. Why? Collision Model  Molecules must collide to react  The greater the number of collisions occurring per second, the greater the reaction rate.  Greater concentration and higher temperature both lead to an increased reaction rate according to the Collision Model.

29. Orientation Factor Molecules must be oriented in a certain way during collisions in order for a reaction to occur  Atoms must be suitably positioned for bonds to break and form new bonds

30. Activation Energy In order to react, colliding molecules must have a total kinetic energy equal to or greater than some minimum value Activation energy – the minimum energy needed to make a reaction happen. Activated Complex or Transition State – The arrangement of atoms at the top of the energy barrier.

31. Activation Energy

32. Fraction of molecules with Ea Said that reaction rate should increase with temperature. At high temperature more molecules have the energy required to get over the barrier The fraction of molecules that has an energy equal to or greater than the E a is given by the expression: f = e -Ea/Rt  E is Euler’s number (opposite of ln)  E a is activation energy  R is gas constant, 8.314 J/mol  T is temperature is Kelvin

33. Magnitude of f Suppose that Ea is 100 kJ/mol, a typical value of many reactions, and that T is 300K, around room temp.  Calculate the value of the fraction of molecules that has energy equal to or greater than Ea  f = e -Ea/Rt  -Ea/Rt = -100 x 103 J/mol/(8.314 J/mol)(300K)  = -40.1  e-40.1 = 3.87 x 10-18

34. Arrhenius Equation Arrhenius found that most reaction rate data obeyed an equation based on three factors:  The fraction of molecules possessing an energy of Ea or greater  The number of collisions occurring per second  The fraction of collisions that have the appropriate orientation

35. Arrhenius Equation He incorporated the 3 factors into his equation: K = Ae -Ea/RT  K is the rate constant  Ea is activation energy  R is the gas constant, 8.314 kJ/mol  T is absolute temperature  A = frequency factor, (is constant as temperature is varied)  As the magnitude of Ea increases, k decreases because the fraction of molecules that possess the required energy is smaller.

36. Determining Activation Energy

37. Example

38. Activation Energy from Graphs

40. 14.6 Reaction Mechanisms The process by which a reaction occurs is called a reaction mechanism There is activation energy for each elementary step. Slowest step (rate determining) must have the highest activation energy The reaction to the right occurs in 2 steps. The first step is the rate determining step.

41. Reaction Mechanisms 2NO 2 + F 2 → 2NO 2 F Rate = k[NO 2 ][F 2 ] The proposed mechanism is NO 2 + F 2 → NO 2 F + F (slow) F + NO 2 → NO 2 F (fast) F is called an intermediate. It is formed then consumed in the reaction

42. Molecularity Each of the 2 reactions is called an elementary step. The number of molecules that participate in an elementary step defines the molecularity.  If a single molecule, the reaction is unimolecular  If 2 reactant molecules, the reaction is bimolecular  If 3, the reaction is termolecular (very rare)

43. Rate Law for Elementary Reactions If we know that a reaction is an elementary step, then we know its rate law The rate of a unimolecular process will be first order. The rate of a bimolecular process will be second order

44. Multi-Step Mechanisms are most common Most chemical reactions occur by mechanisms that involve two or more elementary steps. Each step has its own rate constant and activation energy The overall rate of the reaction cannot exceed the rate of the slowest elementary step = rate determining step.

45. 14.7 Catalysts Speed up a reaction without being used up in the reaction Catalysts allow reactions to follow an alternate pathway that has a lower activation energy  Enzymes are biological catalysts  Homogeneous catalysts are in the same phase as the reactants  Heterogeneous catalysts are in a different phase as the reactants. Catalysts can speed up and slow down reactions

46. How Catalysts work Catalysts allow reactions to proceed by a different mechanisms – a new pathway New pathway has a lower activation energy

47. Examples of Catalysts Catalase is an enzyme in the livers of mammals that catalyzes the decomposition of hydrogen peroxide into water and oxygen. Nitrogenase is an enzyme in bacteria that live on the roots of certain plants. It converts N 2 into NH 3, which can then be used by the plants. Manganese dioxide will also act as a catalyst in the decomposition of hydrogen peroxide to water and oxygen gas.