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Chemical Kinetics

Chemical Kinetics. Chapter 14. How fast chemical reactions go. x 1 t 1. x 2 t 2. Chemical Kinetics. The speed of a car is its rate of change of position : D x/ D t = (x 2 – x 1 )/(t 2 – t 1 ).

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Chemical Kinetics

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  1. Chemical Kinetics Chapter 14 How fast chemical reactions go Chem 1422, Chapter 14

  2. x1 t1 x2 t2 Chemical Kinetics The speed of a car is its rate of change of position: Dx/Dt = (x2 – x1)/(t2 – t1) The speed of a chemical reaction is itsrate of change of concentration: D[conc]/DtNote that [ ] represents concentration (mol/L) Chemical Kinetics = Measurement and analysis of concentrations versus time. This analysis may also reveal the mechanism of a reaction (how it really works at the molecular level). Chem 1422, Chapter 14

  3. Chemical Kinetics • Chemical Kinetics is the study of how chemical reactions evolve in time and how fast they occur. • There are 4 important factors which affect rates of reactions: • reactant concentrations • temperature • presence of a catalyst • surface area of solid reactants • Goal: to understand chemical reactions at the molecular level. Chem 1422, Chapter 14

  4. Kinetic Plot conc time Chemical Kinetics AB One mole of A in a closed 1 liter reactor min [A]0 = 1.00 M[B]0 = 0.00 M [A]20 = 0.54 M[B]20 = 0.46 M [A]40 = 0.30 M[B]40 = 0.70 M Chem 1422, Chapter 14

  5. Kinetic Plot AB D[conc] vs. Dt (kinetic plot) Molarity (mol/L) Chem 1422, Chapter 14

  6. [ ] [ ] Δ B Δ A - = = 0.026 M/min initial reaction rate = Δt Δt Reaction Rates AB Calculate the initial reaction ratefrom t = 0 to t = 10 min The meaning of "+" and "-" rates [A] is decreasing as fast as [B] is increasing Chem 1422, Chapter 14

  7. Reaction Rates AB Calculate the reaction rate from t = 50 to t = 60 min The initial reaction rate is 0.026 M/min After 50 min, the reaction rate is 0.006 M/min Reaction rates decrease with time! Chem 1422, Chapter 14

  8. Reaction Rates AB When does the reaction stop (Rate = 0)? Rate00-10 = 0.026 M/min Rate10-20 = 0.020 M/min Rate20-30 = 0.014 M/min Rate30-40 = 0.010 M/min Rate40-50 = 0.008 M/min Rate50-60 = 0.006 M/min In an open system, when the limiting reagent is used up. (Chapters 3, 4) In a closed system, when equilibrium is reached. (Chapters 15, 16, 17) Chem 1422, Chapter 14

  9. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) Chem 1422, Chapter 14

  10. Reaction Rates C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq) D[conc]/ Dt is an averagerate. In the limit Dt  0, rate = {d[conc]/dt}t instantaneous rateat time t = slope of kinetic plot at time t. Chem 1422, Chapter 14

  11. aA + bB cC + dD Reaction Rates and Stoichiometry We can set up a reaction table for any chemical reaction: x/Dt is the rate of the chemical reaction (M/s) (treat stoichiometric coefficients as dimensionless numbers) Chem 1422, Chapter 14

  12. Reaction Rates and Stoichiometry rate the overall rate of the chemical reaction is always positive Each of these is positive because D[C] and D[D] are both positive but D[A] and D[B] are both negative (Dt is always positive) the rate of decrease of reactant A (M/s) the rate of decrease of reactant B (M/s) the rate of increase of product C (M/s) the rate of increase of product D (M/s) Chem 1422, Chapter 14

  13. = 0.56 M/s D[NH3] D[NH3] D[NH3] D[H2] D[H2] D[N2] + – = = – Dt Dt Dt Dt Dt Dt – = 3 1 1 1 3 2 1 2 Reaction Rates and Stoichiometry The kinetic behavior of the following reaction was studied: 3H2(g) + N2(g) → 2NH3(g) In one experiment, it was found that the rate of production of NH3 was 0.56 M/s. What was the rate of consumption of H2? = (–1.5)(0.56 M/s) =–0.84 M/s Answer: the rate of consumption of H2 was 0.84 M/s (the rate of change of H2 was –0.84 M/s) Chem 1422, Chapter 14

  14. = 0.56 M/s D[NH3] D[NH3] D[NH3] D[H2] D[H2] D[N2] + – = = – Dt Dt Dt Dt Dt Dt – = 3 1 1 1 3 2 1 2 Reaction Rates and Stoichiometry The kinetic behavior of the following reaction was studied: 3H2(g) + N2(g) → 2NH3(g) In one experiment, it was found that the rate of production of NH3 was 0.56 M/s. What was the rate of consumption of H2? = (–1.5)(0.56 M/s) =–0.84 M/s Another answer: the rate of the reaction was 0.28 M/s Chem 1422, Chapter 14

  15. Concentration Dependence Usually, theinitial rate changesif aninitial concentration changes NH4+(aq) + NO2–(aq) → N2(g) + 2H2O(l) As [NH4+] doubles, the rate doubles As [NO2–] doubles, the rate doubles Chem 1422, Chapter 14

  16. Concentration Dependence NH4+(aq) + NO2–(aq) N2(g) + 2H2O(l) As [NH4+] doubles, the rate doubles As [NO2–] doubles, the rate doubles We conclude that the rate of this reaction is proportional to [NH4+] and [NO2–] Rate law as a function of concentrations: rate = k[NH4+][NO2–] Constant k is the rate constant Chem 1422, Chapter 14

  17. General Rate Law aA + bB cC + dD The form of the General Rate Law is: rate = k[A]m[B]n • Only reactants appear in the rate law equation. • m is the order of the reaction with respect to A. • m is not necessarily equal to a. • n is the order of the reaction with respect to B. • n is not necessarily equal to b. • The overall order of the reaction is m + n. • k is the rate constant of the reaction. • m, n and k can only be determined by experiment Chem 1422, Chapter 14

  18. [ ] [ ] [ ] [ ] 1 1 Δ A Δ B 1 Δ C 1 Δ D - = + = = + = - rate =k[A]m[B]n b a Δt Δt c Δt d Δt General Rate Law aA + bB cC + dD The form of the General Rate Law is: rate = k[A]m[B]n • An order can be positive, zero, negative, an integer or a fraction. • Again, orders m & n must be determined by experiment; they are not generally related to stoichiometric coefficients a & b. • Rate constant k does not depend on concentration. Chem 1422, Chapter 14

  19. [ ] [ ] [ ] [ ] 1 1 Δ A Δ B 1 Δ C 1 Δ D - = + = = + = - rate =k[A][B] 3 2 Δt Δt 2 Δt 1 Δt Reaction Rates in Words Example: 2A + 3B → 2C + D • The rate of this reaction can be expressed in several different but equivalent ways. • The rate of the reaction is 1st order in A and 1st order in B. • The rate of the reaction is equal to 1/3 the rate of consumption of B. • The rate of the reaction is equal to the rate of production of D. • The rate of consumption of A is twice the rate of production of D. • The rate of production of C is 67% of the rate of consumption of B. Chem 1422, Chapter 14

  20. Exp [A]i [B]iRatei R2 k[A]2m[B]1n = R1 k[A]1m[B]1n m R2 [A]2 log log = m = R2 R2 [A]2 [A]2 R1 [A]1 = m log R1 R1 [A]1 [A]1 log Experimental Rate Law Stoichometry: aA + bB  products General Rate Law: Rate = k[A]m[B]n To determine m: Initial Rate Data 1 [A]1 [B]1 R1 (change the concentration of only reactant A) 2 [A]2 [B]1 R2 Equations: R1 = k[A]1m[B]1n R2 = k[A]2m[B]1n Ratio: Chem 1422, Chapter 14

  21. m [ ] [ ] [ ] [ ] 1 R2 [A]2 1 Δ A Δ B 1 Δ C 1 Δ D - = = + = = + = - rate =k[A]m[B]n R1 [A]1 b a Δt Δt c Δt d Δt Experimental Rate Law aA + bB cC + dD For simple doubling experiments, [A]2 = 2[A]1 so R2 = 2mR1. • Double [A]. If the rate does not change, m = 0. • Double [A]. If the rate doubles, m = 1. • Double [A]. If the rate quadruples, m = 2 (etc.) Chem 1422, Chapter 14

  22. m R2 [A]2 = R1 [A]1 Experimental Rate Law Rate = k [NO]m [H2]n Rate = k [NO]2 [H2] k = ? 2NO(g) + 2H2(g)  N2(g) + 2H2O(g) {3:1} R3/R1 = 4 = {2:1} R2/R1 = 2 = ([H2]2/[H2]1)n = 2n 2 = 2n n = 1 ([NO]3/[NO]1)m = 2m 4 = 2m m = 2 Chem 1422, Chapter 14

  23. m R2 [A]2 = R1 [A]1 Experimental Rate Law Rate = k [NO]m [H2]n Rate = k [NO]2 [H2] k = ? 2NO(g) + 2H2(g)  N2(g) + 2H2O(g) 1: 1.23E-3 Ms-1 = k (0.1 M)2(0.1 M) k = 1.23 M-2s-1 2: 2.46E-3 Ms-1 = k (0.1 M)2(0.2 M) k = 1.23 M-2s-1 3: 4.92E-3 Ms-1 = k (0.2 M)2(0.1 M) k = 1.23 M-2s-1 Rate = k [NO]2 [H2] k = 1.23 M-2s-1 Chem 1422, Chapter 14

  24. Experimental Rate Law Rate = k [NO]2 [H2] k = 1.23 M-2s-1 Note that the units of rate are Ms-1, but the units of k depend on the overall order: Ms-1 = k Mm+n k = M1–m–n s-1 1st order Rate = k[ ] k (s-1) 2nd order Rate = k[ ]2 k (M-1s-1) 3rd order Rate = k[ ]3 k (M-2s-1) 4th order Rate = k[ ]4 k (M-3s-1) etc. 2NO(g) + 2H2(g)  N2(g) + 2H2O(g) 1st order processes are very common in nature. Chem 1422, Chapter 14

  25. First Order Rate Laws Ratet = k[CH3NC]t CH3NC(g) → Products (methylisonitrile) The rate at time t, Ratet, is the slope of the line at time t. A plot of the logarithm of [CH3NC]t versus t produces a straight line (y = ax + b) with a negative slope. If natural logs are used, the slope is –k. This leads to the integrated first order rate law. [A]t = [A]oe-kt ln[A]t = -kt + ln[A]o Chem 1422, Chapter 14

  26. First Order Rate Laws Ratet = k’pt CH3NC(g) → Products (methylisonitrile) The kinetic data can also be expressed in terms of the partial pressure of CH3NC at time t. Kinetic data can be expressed in any units related to amount (moles, grams, etc.) Slope = -k’ s-1 Intercept = ln po pt = poe-k’t Chem 1422, Chapter 14

  27. ln[A]t = -kt + ln[A]o [A]t = [A]oe-kt First Order Rate Laws Half-life t1/2 (or t) is the time required for [A]t to decrease to half of its original value. At = A0e-kt At = A0e-kt = ½A0 ½ = e-kt ln(½) = -kt ln(2) = kt kt1/2 = ln 2 = 0.693... Chem 1422, Chapter 14

  28. (just for fun) First Order Rate Laws The general stoichiometry for a first order reaction: aA → bB Chem 1422, Chapter 14

  29. (just for fun) First Order Rate Laws The general stoichiometry for a first order reaction: aA → bB [A]0 [B]0 Chem 1422, Chapter 14

  30. First Order Rate Laws There are many first order ( exponential or logarithmic ) processes. For example: Integrated 1st-order rate law • Radioactive Decay • Ao = the initial amount (t=0; g, mol, etc.) of a radioactive isotope. • At = the amount of radioactive isotope after time t. • t = ln(2)/k = half-life of the isotope. • k = ln(2)/t = rate constant for the decay. Chem 1422, Chapter 14

  31. First Order Rate Laws There are many first order ( exponential or logarithmic ) processes. For example: Integrated 1st-order rate law • Bacterial Growth • No = the initial number of bacteria at t = 0. • Nt = the number of bacteria at time t. • k = growth-rate factor. • t2 = doubling time. • kt2 = ln(2) • Note • + sign for growth • – sign for decay Chem 1422, Chapter 14

  32. Temperature and Rate • Most chemical reactions speed up as temperature increases. • Food spoils faster when not refrigerated (bacterial growth) • Chemiluminescent "lite sticks" glow brighter at room temperature than in the refrigerator • The reason is that the rate “constant” for most chemical reactions increases with temperature. Chem 1422, Chapter 14

  33. Temperature and Rate H3CNC  H3CCN The intramolecular rearrangement of methylisonitrile to acetonitrile is 1st order rate = k[H3CNC] k = 0.03  10-3 s-1 at 190 Ck = 3.16  10-3 s-1 at 250 C k is a function of temperature Chem 1422, Chapter 14

  34. Reaction Rate Explained • Observation: reactions in and between solids are typically slow, whereas liquid and gas phase reactions can be very fast. • Observation: reaction rates are directly proportional to concentrations and temperature. • Goal: to develop a model that explains why solid state reactions are slow, and in general rates increase as concentrations and temperature increase. • The collision model: in order for two molecules to react, they must collide! Chem 1422, Chapter 14

  35. The Collision Model • Concentration: the more molecules there are, the greater the probability of collisions; • The greater the number of collisions, the faster the rate! • Temperature: chemical reactions always require bonds to be broken; • The higher the temperature, the more thermal (kinetic) energy available to break bonds. • Complication: not every collision leads to products. In fact, only a small fraction of collisions lead to product. Chem 1422, Chapter 14

  36. The Collision Model H2 + I2 → 2HI The molecules are headed for a collision! The result of most collisions is that the molecules just bounce apart with no change, so no products are formed. They collide! ... but then they just bounce apart, with no bonds broken. Chem 1422, Chapter 14

  37. The Collision Model Reactants H2 + I2 → 2HI For an intermolecular collision to be effective, the molecules must collide in the correct orientation and with sufficient energy to form a transition state (a highly energetic activated complex). Transition State Products Chem 1422, Chapter 14

  38. DE Activation Energy Activation Energy, Ea, is the minimum energy required to form the Transition State or activated complex energy of Reactants Ea energy of Products Chem 1422, Chapter 14

  39. H3CNC H3CCN Ea of Chemically Reversible Reactions (acetonitrile) (methylisonitrile) Forward Rate = kf[H3CNC] DEf = E(H3CCN) - E(H3CNC) exothermic Eaf = E(TS) - E(H3CNC) Reverse Rate = kr[H3CCN] DEr = E(H3CNC) - E(H3CCN) endothermic Ear = E(TS) - E(H3CCN) = Eaf + |DE| Chem 1422, Chapter 14

  40. The Arrhenius Equation Arrhenius discovered that k for any reaction: Chem 1422, Chapter 14

  41. The Arrhenius Equation Arrhenius discovered that k for any reaction: k increases when A increases k increases when T increases k decreases when Ea increases k is the rate “constant” Ea is the activation energy R = 8.314 J•mol-1•K-1 A is called the frequency factor, a number related to the probability of favorable collisions. Chem 1422, Chapter 14

  42. Temperature and Rate H3CNC  H3CCN y = m x + b k = 0.03  10-3 s-1 at 190 Ck = 3.16  10-3 s-1 at 250 C Ea = 157 kJ/mol A = 1.4  1013 s-1 Chem 1422, Chapter 14

  43. Reaction Mechanisms Many chemical reactions are complex, and actually occur in two or more elementary steps. For example: Do three reactant molecules collide, form an acitivated complex, and produce two molecules? NO! Balanced Chemical Reaction 2NO(g) + Br2(g)  2NOBr(g) Here is what actually happens! The actual mechanism of a chemical reaction is almost always a series of simple reactions called Elementary Steps Chem 1422, Chapter 14

  44. Reaction Mechanisms 2NO(g) + Br2(g)  2NOBr(g) Elementary Steps NO + Br2 NOBr2 NOBr2 + NO  N2O2Br2* N2O2Br2*  2NOBr 2NO + Br2  2NOBr The rate of the overall reaction is a combination of the step rates, but is often close to the rate of the slowest elementary step called the rate determining step or bottleneck. NOBr2 is a reactive intermediate N2O2Br2* is an activated complex Chem 1422, Chapter 14

  45. Reaction Mechanisms reactive intermediate reactive intermediate NO + Br2 NOBr2 NOBr2 + NO  N2O2Br2* N2O2Br2*  2NOBr 2NO + Br2  2NOBr transition state transition state A multi-step reaction mechanism provides detailed information about which bonds are broken and formed, and what reactive intermediates and transition states are formed. The balanced chemical equation provides information only about the initial reactants and final products of a reaction; it is always the sum of the elementary steps. Chem 1422, Chapter 14

  46. Reaction Mechanisms NO + Br2 NOBr2 NOBr2 + NO  N2O2Br2* N2O2Br2*  2NOBr 2NO + Br2  2NOBr R1 = k1[NO][Br2] R2 = k2[NOBr2][NO] R3 = k3[N2O2Br2] The number of molecules which react (collide) in an elementary step is called the molecularity of the step. Steps 1 and 2 are bimolecular, step 3 is unimolecular. In an elementary step rate equation, the overall order is equal to the molecularity, and the individual orders are equal to the stoichiometric coefficients. Chem 1422, Chapter 14

  47. Reaction Mechanisms NO + Br2 NOBr2 NOBr2 + NO  N2O2Br2* N2O2Br2*  2NOBr 2NO + Br2  2NOBr R1 = k1[NO][Br2] R2 = k2[NOBr2][NO] R3 = k3[N2O2Br2] rate = ? • The overall rate equation must be determined from experimental kinetic data, but we can be sure that ... • A three-molecule collision (termolecular step), as suggested by the overall reaction equation, is very unlikely; termolecular steps almost never happen. Chem 1422, Chapter 14

  48. Reaction Mechanisms NO + Br2 NOBr2 NOBr2 + NO  N2O2Br2* N2O2Br2*  2NOBr 2NO + Br2  2NOBr R1 = k1[NO][Br2] R2 = k2[NOBr2][NO] R3 = k3[N2O2Br2] rate = ? • The overall rate equation must be determined from experimental kinetic data. • The mechanism is a guess made after the overall rate equation is determined. • The overall rate equation derived from the mechanism (using differential calculus) must match the experimental equation. Chem 1422, Chapter 14

  49. Catalysis A catalyst is a chemical substance which, when added to a chemical reaction, increases the rate of that reaction. The catalyst does not appear in the reaction equation as a reactant or product because it is neither consumed nor produced in the reaction; thus, the balanced reaction equation is unchanged. The catalyst does take part in the reaction, however, by changing the actual mechanism of the reaction. Chem 1422, Chapter 14

  50. Catalysis For example, Cl or Br atoms in the upper atmosphere catalyze the decomposition of ozone: 2O3(g) → 3O2(g) (very slow) [Cl(g) or Br(g)] + 2O3(g) → 3O2(g) (very fast) There are two types of catalytic reaction: • homogeneous: reactants, products and catalyst are all in the same (usually fluid) phasee.g., Cl(g), O3(g) and O2(g) • heterogeneous: reactants and products are in a fluid phase (liquid or gas), the catalyst is a solid. Chem 1422, Chapter 14

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