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Chapter 14

Chapter 14. Chemical Kinetics. Kinetics. The study of reaction rates – the speed at which reactants are converted to products. Explosions happen in seconds- very fast. Coal (carbon) will turn into diamond  eventually.

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Chapter 14

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  1. Chapter 14 Chemical Kinetics

  2. Kinetics • The study of reaction rates – the speed at which reactants are converted to products. • Explosions happen in seconds- very fast. • Coal (carbon) will turn into diamond  eventually. • Spontaneous reactions are reactions that will happen  but we can’t tell how fast. • Kinetics is very important because it allows us to manipulate and control reactions.

  3. Reactions occur when . . . • Molecules find each other • Effective collisions happen between molecules • Frequency of collisions is high

  4. Factors that affect Reaction Rate: 1. Temperature • Hot = fast Cold = slow 2. Concentration of Reactants • more concentrated = faster 3. Physical conditions of reactants. • more surface area = faster 4. Presence of a Catalyst – • increases rate without being used up

  5. Reaction Rate • The change in concentration of a reactant or a product per unit of time. • Rates decrease with time • Typically rates are expressed as positive values • Instantaneous rate can be determined by finding the slop of a line tangent to a point representing a particular time

  6. Reaction Rate - “speed of the reaction” • Rate = Conc. of A at t2 Conc. of A at t1t2  t1 • Rate =D[A]Dt • Units: Change in concentration per time • For this reaction • N2 + 3H2 2NH3

  7. As the reaction progresses the concentration of H2 goes down N2 + 3H2 2NH3 Concentration [H2] Time

  8. As the reaction progresses the concentration N2 goes down 1/3 as fast. Why? N2 + 3H2 2NH3 Concentration [N2] [H2] Time

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

  10. Calculating Rates 1. Average rates are taken over long intervals 2. 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 (Finding the Derivative.)

  11. Average slope method Concentration D[H2] Dt Time

  12. Instantaneous slope method. Concentration D[H2] D t Time

  13. Find the Instantaneous rate of disappearance: • In order to find the instantaneous rate of disappearance at time 0, a tangent line was drawn and the following information was obtained: the concentration of the unknown changed from 0.100 M to 0.060 M from 0 seconds to 200 seconds. What is the instantaneous rate of disappearance at time 0? • 2.0 x 10-4 M/s

  14. Reaction Rates and Stoichiometry • Remember, the rate of disappearance is a  slope and the rate of appearance is a + slope • For general equation: aA + bBcC + dD General rate equation: Rate = -1D[A] = -1D[B] = 1D[C] = 1∆[D] a DtbDtcDtd∆t

  15. 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. • In our example N2 + 3H2 2NH3 -D[N2] = -D[H2] = D[NH3] Dt 3 Dt 2 Dt

  16. Practice • How is the rate of disappearance of ozone related to the rate of appearance of oxygen in the following equation: 2O3 3O2? • If the rate of appearance of O2, is 6.0x10-5 M/s at a particular instant, what is the value of the rate of disappearance of O3 at this same time? • 4.0x10-5 M/s

  17. The decomposition of N2O5 proceeds according to the following equation: • 2N2O5 4NO2 + O2 • If the rate of decomposition of N2O5 at a particular instant in a reaction vessel is 4.2x10-7 M/s, what is the rate of appearance of NO2? O2? • NO2 = 8.4x10-7 M/s • O2 = 2.1 x 10-7 M/s

  18. Rate Laws • To study the rate of a reaction and determine the rate law, a series of experiments must be done varying the initial concentrations of reactants. • This is called the Initial rate method. • The products do not appear in the rate law because we are only concerned with the very start of the experiment where conc. of reactants are carefully chosen.

  19. 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. • aA + bBcC + dD • Rate = k [A]x[B]y } The rate law expression • kis the rate constant(temperature dependent) • Unitsof k vary • x and y are integers, usually 0, 1 or 2 • reaction is xth order in A, reaction is yth order in B • reaction is (x +y)th order overall

  20. F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2]x [ClO2]y Initial [ ] (All readings recorded at same temp.) Double [F2] with [ClO2] constant: rate doubles Quadruple [ClO2] with [F2] constant: rate quadruples Therefore, rate law is: rate = k [F2]1 [ClO2]1 We say, “first order with respect to F2, first order with respect to ClO2, and 2nd order overall”

  21. Rate Laws - Key Points 1. Rate laws are always determined experimentally. 2. Reaction order is always defined in terms of reactant (not product) concentrations. 3. The order of a reactant is notrelated to the stoichiometric coefficient of the reactant in the balanced chemical equation. F2(g) + 2ClO2(g) 2FClO2(g) rate = k [F2]1 [ClO2]1 4. Once you obtain the Rate Law and the value of k, you can calculate the rate of reaction for any set of concentrations.

  22. Try this. A + B ---> C + D Run # Initial [A] Initial [B] Initial Rate (v0) ([A]0) ([B]0) 1 1.00 M 1.00 M 1.25 x 10-2 M/s 2 1.00 M 2.00 M 2.5 x 10-2 M/s 3 2.00 M 2.00 M 2.5 x 10-2 M/s What is the order with respect to A? What is the order with respect to B? What is the overall order of the reaction? Write the rate law expression 0 1 1 Rate = k [A]0 [B]1

  23. Try this. 2NO (g) + Cl2(g) ---> 2NOCl [NO(g)] (M) [Cl2(g)] (M) Initial Rate(Ms-1) 0.250 0.250 1.43 x 10-6 0.250 0.500 2.86 x 10-6 0.500 0.500 1.14 x 10-5 What is the order with respect to Cl2? What is the order with respect to NO? What is the overall order of the reaction? Write the rate law expression Calculate the value of k 1 2 3 Rate = k [NO]2 [Cl2]1 9.15 x 10-5 1/M2•s

  24. Units of k What the Order means: •Zero order means doubling or tripling the [reactant]o has NO effect on the rate. •First order means doubling the [reactant]o doubles the rate, tripling the [reactant]o triples the rate, etc… •Second order means doubling the [reactant]o quadruples the rate (raises to a power of 2), etc. . . • 1st Order overall units: 1/s • 2nd Order overall units: 1/M•s • 3rd Order overall units: 1/ M2•s

  25. Try this: BrO3- + 5 Br- + 6H+ ----> 3Br2 + 3 H2O Initial concentrations (M) Rate (M/s) Determine the rate law expression Calculate the value of k [BrO3-] [Br-] [H+] 0.10 0.10 0.10 8.0 x 10-4 0.20 0.10 0.10 1.6 x 10-3 0.20 0.20 0.10 3.2 x 10-3 0.10 0.10 0.20 3.2 x 10-3 Rate = k [BrO3]1 [Br-]1 [H+]2 fourth order overall 8.0 1/M3•s

  26. NH4+ + NO2- N2 + 2H2O • Determine the Rate Law Expression • Determine the value for k

  27. Types of Rate Laws • Differential Rate Law – describes how rate depends on concentration • Rate = k [A]x[B]y • Integrated Rate Law– describes how concentration depends on time • For each type of differential rate law, there is an integrated rate law and vise versa. • We can use these laws to help us better understand reaction mechanisms

  28. Integrated Rate Law Equations allow us to: 1. Determine the concentration of reactant remaining at any time after the reaction has started. 2. Determine the time required for a fraction of a sample to react. 3. Determine the time for a reactant concentration to fall to a certain level. 4. Verify whether a reaction is first order and determine the rate constant.

  29. Integrated Rate Law • Expresses the reaction concentration as a function of time. • Form of the equation depends on the order of the rate law (differential). • A Changing Rate = D[A]nDt • We will only work with n = 0, 1, and 2

  30. First Order • For the reaction 2N2O5 4NO2 + O2 • Rate = k[N2O5]1 • If concentration doubles the rate doubles. • If we integrate this equation with respect to time we get the Integrated Rate Law • ln[N2O5] = - kt + ln[N2O5]0 • ln is the natural log • [N2O5]0 is the initial concentration.

  31. First Order • General form-- Rate = D[A] / Dt = k[A] • ln[A] = - kt + ln[A]0 • In the form y = mx + b • y = ln[A] ; m = -k ; x = t ;b = ln[A]0 • A graph of ln[A] vs time is a straight line.

  32. First Order • A graph of ln[A] vs time is a straight line. • By getting the straight line you can prove it is first order

  33. First Order • For the reaction 2N2O5 4NO2 + O2 • Rate = k[N2O5]1 • ln[N2O5] = - kt + ln[N2O5]0

  34. Try this • 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 ? ln[A] = - kt + ln[A]0 ln[A] - ln[A]0 = - kt ln[0.14 M] - ln[0.88M]0 = - (2.8 x 10-2 s-1)t - 1.97  - 0.128 = - (2.8 x 10-2 s-1)t - 1.84 = - (2.8 x 10-2 s-1)t 66 s = t

  35. Second Order (Overall) • Rate = -D[A] / Dt = k[A]2 • integrated rate law: 1/[A] = kt + 1/[A]0 • y= 1/[A] m = k • x= t b = 1/[A]0 • A straight line if 1/[A] vs time is graphed • Knowing k and [A]0 you can calculate [A] at any time t

  36. Second Order • Looking at the data alone, you cannot tell if 1st or 2nd order. • To tell the difference you must graph the data, or look at the units of k.

  37. Zero Order Rate Law • Rate = k[A]0 = k • Rate does not change with concentration. • Integrated [A] = -kt + [A]0 • In the form y = mx + b • Most often when reaction happens on a surface because the surface area stays constant.

  38. Zero Order Concentration Time

  39. Zero Order Concentration k = D[A]/Dt D[A] Dt Time

  40. Zero Order

  41. Half Life • 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

  42. [A]0 ln t½ [A]0/2 = = = k [A]0 t½ = t½ = ln 2 .693 2k k k 1 k[A]0 Half Life • The time required to reach half the original concentration. • If the reaction is first order • t½ = t when [A] = [A]0/2 • If the reaction is 2nd order: • t½ = t when [A] = [A]0/2 • If the reaction is zero order: t½ = t when [A] = [A]0/2

  43. = t½ Ln 2 0.693 = k 5.7 x 10-4 s-1 Try This • What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? • Which order is it? How do you know? = 1200 s = 20 minutes

  44. Summary of Rate Laws

  45. Reaction Rates So, WHY do some reactions go fast and others are slow? Reaction Mechanisms Collision Theory and Activation Energy

  46. Reaction Mechanisms • The series of steps that actually occur in a chemical reaction. • Each step is called an elementary step. • Kinetics can tell us something about the mechanism • A balanced equation does not tell us how the reactants become products.

  47. Reaction Mechanisms -- Example • 2NO2 + F2 2NO2F • Rate = k[NO2] [F2] (determined experimentally) • The proposed mechanism is • NO2 + F2 NO2F + F (slow) • F + NO2 NO2F (fast) • F is called an intermediate It is formed then consumed in the reaction • The slow step is therate determiningstep!

  48. Reaction Mechanisms • Each of the two reactions is called an elementary step . • The rate for a reaction can be written from its molecularity • Molecularity is the number of pieces that must collide to produce the reaction indicated by the step. • Molecularity is the number of molecules which must collide effectively for reaction to occur.

  49. Unimolecular step involves one molecule - Rate is first order. • Bimolecular step - requires two molecules - Rate is second order • Termolecular step- requires three molecules - Rate is third order • Termolecular steps are almost never heard of because the chances of three molecules colliding effectively at the same time are miniscule.

  50. Rate Laws for Elementary Steps (these ARE based on the stoichiometry) • A products Rate = k[A] • A+A products Rate= k[A]2 • 2A products Rate= k[A]2 • A+B products Rate= k[A][B] • A+A+B Products Rate= k[A]2[B] • 2A+B Products Rate= k[A]2[B] • A+B+C Products; Rate= k[A][B][C]

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