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Chemical Kinetics CHAPTER 14 Part B Chemistry: The Molecular Nature of Matter, 6 th edition

Chemical Kinetics CHAPTER 14 Part B Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson , Brady, & Hyslop. CHAPTER 14 Chemical Kinetics. Learning Objectives: Factors Affecting Reaction Rate : Concentration State Surface Area Temperature Catalyst

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Chemical Kinetics CHAPTER 14 Part B Chemistry: The Molecular Nature of Matter, 6 th edition

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  1. Chemical Kinetics CHAPTER 14 Part B Chemistry: The Molecular Nature of Matter, 6th edition By Jesperson, Brady, & Hyslop

  2. CHAPTER 14 Chemical Kinetics • Learning Objectives: • Factors Affecting Reaction Rate: • Concentration • State • Surface Area • Temperature • Catalyst • Collision Theory of Reactions and Effective Collisions • Determining Reaction Order and Rate Law from Data • Integrated Rate Laws • Rate Law  Concentration vs Rate • Integrated Rate Law  Concentration vs Time • Units of Rate Constant and Overall Reaction Order • Half Life vs Rate Constant (1st Order) • Arrhenius Equation • Mechanisms and Rate Laws • Catalysts Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  3. CHAPTER 14 Chemical Kinetics Lecture Road Map: Factors that affect reaction rates Measuring rates of reactions Rate Laws Collision Theory Transition State Theory& Activation Energies Mechanisms Catalysts Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  4. CHAPTER 14 Chemical Kinetics Integrated Rate Laws Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  5. Integrated Rate Laws Concentration & Time Rate law tells us how speed of reaction varies with concentrations. Sometimes want to know • Concentrations of reactants and products at given time during reaction • How long for the concentration of reactants to drop below some minimum optimal value  Need dependence of rate on time Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  6. Integrated Rate Laws First Order Integrated Rate Law • Corresponding to reactions • A  products • Integrating we get • Rearranging gives • Equation of line y = mx + b Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  7. Integrated Rate Laws First Order Integrated Rate Law Slope = –k Yields straight line • Indicative of first order kinetics • Slope = –k • Intercept = ln [A]0 • If we don't know already Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  8. Integrated Rate Laws 2nd Order Integrated Rate Law • Corresponding to special second order reaction • 2B  products • Integrating we get • Rearranging gives • Equation of line y = mx + b Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  9. Integrated Rate Laws 2nd Order Integrated Rate Law Slope = +k Yields straight line • Indicative of 2nd order kinetics • Slope = +k • Intercept = 1/[B]0 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  10. Integrated Rate Laws Graphically determining Order Make two plots: • ln [A] vs. time • 1/[A] vs. time • If ln [A] is linear and 1/[A] is curved, then reaction is 1st order in [A] • If 1/[A] plot is linear and ln [A] is curved, then reaction is 2nd order in [A] • If both plots give horizontal lines, then 0th order in [A] Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  11. Integrated Rate Laws Graphically determining Order Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  12. Integrated Rate Laws Example: SO2Cl2 SO2 + Cl2 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  13. Integrated Rate Laws Example: SO2Cl2 SO2 + Cl2 Reaction is 1st order in SO2Cl2 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  14. Integrated Rate Laws Example: HI(g)  H2(g) + I2(g) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  15. Integrated Rate Laws Example: HI(g)  H2(g) + I2(g) Reaction is second order in HI. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  16. Group Problem A plot for a zeroth order reaction is shown. What is the proper label for the y-axis in the plot ? A. Concentration B. ln of Concentration C. 1/Concentration D. 1/ ln Concentration Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  17. Integrated Rate Laws Half Life (t1/2) for first order reactions Half-life = t½ We often use the half life to describe how fast a reaction takes place First Order Reactions • Set • Substituting into • Gives • Canceling gives ln 2 = kt½ • Rearranging gives Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  18. Integrated Rate Laws Half Life (t1/2) for First Order Reactions Observe: • t½ is independent of [A]o • For given reaction (and T) • Takes same time for concentration to fall from • 2 M to 1 M as from • 5.0  10–3M to 2.5  10–3M • k1 has units (time)–1, so t½ has units (time) • t½ called half-life • Time for ½ of sample to decay Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  19. Integrated Rate Laws Half Life (t1/2) Does this mean that all of sample is gone in two half-lives (2 × t½)? No! • In 1stt½, it goes to ½[A]o • In 2ndt½, it goes to ½(½[A]o) = ¼[A]o • In 3rdt½, it goes to ½(¼[A]o) = ⅛[A]o • In ntht½, it goes to [A]o/2n Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  20. Integrated Rate Laws Half Life (t1/2) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  21. Integrated Rate Laws Half Life (t1/2): First Order Example 131I is used as a metabolic tracer in hospitals. It has a half-life, t½ = 8.07 days. How long before the activity falls to 1% of the initial value? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  22. Group Problem The radioactive decay of a new atom occurs so that after 21 days, the original amount is reduced to 33%. What is the rate constant for the reaction in s–1? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  23. Group Problem The half-life of I-132 is 2.295 h. What percentage remains after 24 hours? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  24. Integrated Rate Laws Half Life (t1/2): Carbon-14 Dating Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  25. Integrated Rate Laws Half Life (t1/2): Second Order Reactions How long before [A] = ½[A]o? • t½, depends on [A]o • t½, not useful quantity for a second order reaction Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  26. Group Problem The rate constant for the second order reaction 2A → B is 5.3 × 10–5M–1 s–1. What is the original amount present if, after 2 hours, there is 0.35 M available? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  27. CHAPTER 14 Chemical Kinetics Collision Theory Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  28. Collision Theory Reaction Rates Collision Theory As the concentration of reactants increase • The number of collisions increases • Reaction rate increases As temperature increases • Molecular speed increases • Higher proportion of collisions with enough force (energy) • There are more collisions per second • Reaction rate increases Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  29. Collision Theory Reaction Rates Rate of reaction proportional to number of effective collisions/sec among reactant molecules Effective collision • One that gives rise to product e.g. At room temperature and pressure • H2 and I2 molecules undergoing 1010 collisions/sec • Yet reaction takes a long time • Not all collisions lead to reaction Only very small percentage of all collisions lead to net change Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  30. Collision Theory Molecular Orientation Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  31. Collision Theory Temperature • As T increases • More molecules have Ea • So more molecules undergo reaction Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  32. Collision Theory Activation Energy (Ea) Molecules must possess certain amount of kinetic energy (KE) in order to react Activation Energy, Ea = Minimum KE needed for reaction to occur • Get energy from collision with other molecules • If molecules move too slowly, too little KE, they just bounce off each other • Without this minimum amount, reaction will not occur even when correctly oriented Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  33. CHAPTER 14 Chemical Kinetics Transition State Theory Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  34. Transition State Molecular Basis of Transition State Theory KE KE KE decreasing as PE increases Is the combined KE of both molecules enough to overcome Activation Energy PE KE KE Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  35. Transition State Molecular Basis of Transition State Theory Activation energy (Ea) = hill or barrier between reactants and products Potential Energy Heat of reaction (H) = difference in PE between products and reactants Products Hreaction = Hproducts – Hreactants Reaction Coordinate (progress of reaction) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  36. Transition State Exothermic Reactions • Exothermic reaction • Products lower PE than reactants Potential Energy Exothermic Reaction H = – Products Reaction Coordinate (progress of reaction) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  37. Transition State Exothermic Reactions • Hreaction < 0 (negative) • Decrease in PE of system • Appears as increase in KE • So the temperature of the system increases • Reaction gives off heat • Can’t say anything about Ea from size of H • Ea could be high and reaction slow even if Hrxn large and negative • Ea could be low and reaction rapid Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  38. Transition State Endothermic Reactions Endothermic Reaction H = + Hreaction = Hproducts – Hreactants Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  39. Transition State Endothermic Reactions • Hreaction > 0 (positive) • Increase in PE • Appears as decrease in KE • So temperature of the system decreases • Have to add E to get reaction to go • EaHrxnasEaincludesHrxn • If Hrxn large and positive • Ea must be high • Reaction very slow Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  40. Transition State Activated Complex • Arrangement of atoms at top of activation barrier • Brief moment during successful collision when • bond to be broken is partially broken and • bond to be formed is partially formed Example Transition State Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  41. Group Problem Draw the transition state complex, or the activated complex for the following reaction: CH3CH2O- + H3O+ CH3CH2OH + H2O Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  42. CHAPTER 14 Chemical Kinetics Activation Energies Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  43. Ea Arrhenius Equation The rate constant is dependent on Temperature, which allows us to calculate Activation Energy, Ea Arrhenius Equation: Equation expressing temperature-dependence of k • A = Frequency factor has same units as k • R = gas constant in energy units = 8.314 J mol–1 K–1 • Ea = Activation Energy—has units ofJ/mol • T = Temperature inK Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  44. Ea Calculating Activation Energy • Method 1. Graphically • Take natural logarithm of both sides • Rearranging • Equation for a line • y = b + mx Arrhenius Plot • Plot lnk(y axis) vs. 1/T (x axis)  yield a straight line • Slope = -Ea/R • Intercept = A Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  45. Ea Arrhenius Equation: Graphing Example Given the following data, predict k at 75 ˚C using the graphical approach ln (k) = –36.025/T – 6.908 ln (k) = –36.025/(348) – 6.908 = – 7.011 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  46. Ea Arrhenius Equation: Graphing Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  47. Ea Arrhenius Equation Sometimes a graph is not needed • Only have two k s at two Ts Here use van't Hoff Equation derived from Arrhenius equation: Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  48. Ea Arrhenius Equation: Ex Vant Hoff Equation CH4 + 2 S2 CS2 + 2 H2S Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  49. Group Problem Given that k at 25 ˚C is 4.61 × 10–1M/s and that at 50 ˚C it is 4.64 × 10–1 M/s, what is the activation energy for the reaction? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

  50. Group Problem A reaction has an activation energy of 40 kJ/mol. What happens to the rate if you increase the temperature from 70˚C to 80 ˚C? A. Rate increases approximately 1.5 times B. Rate increases approximately 5000 times C. Rate does not increase D. Rate increases approximately 3 times Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

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