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Chapter 14 Chemical Kinetics

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  1. Chapter 14Chemical Kinetics

  2. Chemical kinetics is the study of: - The rates of chemical reactions - The factors (variables)that affect rates - The reaction mechanisms • Variable in Reaction Rates: - Concentrations of reactants - Temperature - Surface area - Catalysts - Inhibitors

  3. Contents in Chapter 14 14-1 Rate of a Chemical Reaction 14-2 Measuring Reaction Rates 14-3 Effect of Concentration on Reaction Rates: The Rate Law 14-4 Zero-Order Reactions 14-5 First-Order Reactions 14-6 Second-Order Reactions 14-7 Reaction Kinetics: A Summary 14-8 Theoretical Models for Chemical Kinetics 14-9 The Effect of Temperature on Reaction Rates 14-10 Reaction Mechanisms 14-11 Catalysis

  4. 14-1 Rate of a Chemical Reaction • Rate of a chemical reaction: the change in concentration for formation of product or disappearance of reactant per unit of time. Unit in general: mol L–1 s–1 (M s–1) Example: 2 Fe3+(aq) + Sn2+(aq)→ 2 Fe2+(aq) + Sn4+(aq) and, 38.5 s after reaction starts, [Fe2+] = 0.0010 M

  5. General rate of reaction*** (average rate of reaction) Example: 2 Fe3+(aq) + Sn2+(aq)→ 2 Fe2+(aq) + Sn4+(aq) and, 38.5 s after reaction starts, [Fe2+] = 0.0010 M

  6. 14-2 Measuring Reaction Rates • Decomposition of H2O2 For Example H2O2(aq) → H2O(l) + ½ O2(g) Monitoring H2O2 disappearance for example: remove small samples of the reaction mixture then titration with KMnO4: 2MnO4-(aq) + 5H2O2(aq) + 6H+(aq) → 2Mn2+(aq) + 8H2O(l) + 5O2(g)

  7. Example of experimental data and derived rate data

  8. = • Graphical representation and explanation*** Average rate: The concentration change in a given time interval (dash) Initial rate: The instantaneous rate at t= 0. (black) Instantaneous rate: The slope of the tangent to the curve at a precise point in the reaction. (blue) Raw data (red)

  9. 14-3 Effect of Concentration on Reaction Rates: The Rate Law*** • Rate law (rate equation): An expression for a reaction relates the reaction rate to the concentrations of the reactants. For a hypothetical reaction: aA + bB gG + hH Rate law for this reaction can be expressed as: Rate of reaction = k[A]m[B]n [A] and [B]: The molarities of the reactants. m and n: Reaction order of A and B, respectively. m + n : Overall order of reaction k: Rate constant • Exponents m and n are determined by experiment, irrelevant tothe stoichiometric coefficients a and b.

  10. Units for rate constants: M(1– overall reaction order) s–1

  11. Distinction between rate and rate constant • Rateof a reaction: The change in concentration with time. Except for zero-order reaction, the rate of a reaction varies as concentration vary. • Rate constant: Remains constant throughout a reaction, regardless of the initial concentrations of the reactants • Value of the rate constant (k) depends on • The specific reaction • The temperature • The presence or absence of a catalyst.

  12. Method of initial rates A method of establishing the rate law for a reaction, i.e., finding the values of the exponents in the rate law, and the value of k. (It is performed by a series of experiments in which the initial concentration of one reactant is varied, other reactants are held constant.) Double the concentration of a reactant A, if reaction is zero-order in A no effect on rate. (20) first-order in A the ratedoubles (21). second-order in A the ratequadruples (22). third-order in A the rateeightfold (23)

  13. Example of Establishing a Rate Law For the reaction: 2 HgCl2(aq) + C2O42–(aq) → 2 Cl–(aq) + 2 CO2(g) + Hg2Cl2(s) Experimental results: Solution: rate of reaction = k[HgCl2]m[C2O42–]n

  14. (Continuous) • Comparing Exp. 2 and Exp.3: • Rate2/Rate3 = 2 = 21 = {[HgCl2]Exp2/[HgCl2]Exp3}m= 2m, m = 1 • Order in HgCl2 is 1. • Comparing Exp. 1 and Exp.3: • Rate2/Rate1 = 4 = 22 = {[C2O42–]Exp2/[C2O42–]Exp1}n= 2n, n = 2 • Order in C2O42– is 2. • The rate law of this reaction is: • Rate = k [HgCl2][C2O42–]2 • Overall order of this reaction is 3. • Calculate the rate constant (k), e.g., from Exp.1:

  15. 14-4 Zero-Order Reactions • Rate of reaction remains constant (independent of the concentration of the reactant) • Concentration versus time is a straight line with a negative slope. [A]=0 at tf • Integrated rate law equation of zero order reaction: • [A]t = –kt + [A]0 • Half-life of zero order reaction: t½= [A]0/2k • Half-life ( t½ ): The time in which one-half of the reactant originally present is consumed.

  16. (Continuous) • Integrated rate law (equation): Derived from a rate law (equation) by the calculus technique of integration. It relates the concentration of a reactant to elapsed time from the start of a reaction. • Zero-order integrated rate law derivation: A → products, the rate can be expressed as: y = mx + b

  17. 14-5 First-Order Reactions First- order integrated rate law derivation: [A]o: initial conc. of A [A]t: conc. of A at time "t" k: rate constant or Finally, • Integrated rate law equation of first-order reaction: • ln[A]t = –kt + ln[A]0 • y = mx + b • Half-life of first-order reaction: t½ = 0.693/k

  18. Example: Decomposition of H2O2 2H2O2(aq) → 2H2O(l) + O2(g) ln[A]t = –kt + ln[A]0 y = mx + b It is a first-order reaction!!

  19. Skilled your calculator!!

  20. 14-5 (Continuous) • Reaction involving gases: Example

  21. 14-5 (Continuous) The constancy of the half-life is proof that the reaction is first order.

  22. 14-5 (Continuous) • Examples of first-order reactions

  23. 14-6 Second-Order Reactions Second- order integrated rate law derivation: • Integrated rate law equation of second-order reaction: • Half-life of second-order reaction: A + B → C, i.e., Rate = k[A][B], tedious work, no further discussion in this course

  24. 14-6 (Continuous) • Pseudo First-Order Reactions For example: CH3COOCH2CH3 + H2O → CH3COOH + C2H5OH The molarity of the water remains essentially constant throughout the reaction (55.5 M). The reaction appears to be zero order in H2O, first order in CH3COOCH2CH3. First order overall.

  25. 14-7 Reaction Kinetics: A Summary****** • Determine the order by straight line • [A] vs time: zero order • ln[A] vs. time first order • 1/[A] vs. time second order

  26. 14-8 Theoretical Models for Chemical Kinetics • Collision Theory: The rate of a reaction is proportional to the number of effective collision per unit of volume per unit of time.

  27. (Continuous) • Variables for effective collisions • Activation energy (Ea): the minimum energy that must be supplied for an effective collisions, i.e., for a reaction to occur. • Temperature effect: increasing temperature • increases molecules with enough KE to react • increases the collision frequency. • Spatial orientations (steric factor).

  28. (Continuous) • Temperature effect: Distribution of Kinetic Energies of Molecules At the higher temperature T2(red), the fraction is considerably larger than at the lower temperature T1(blue).

  29. (Continuous) • Spatial orientation effect Orientation is not a factor Orientation is an important factor

  30. (Continuous) • Steric factor: The probability of favorable orientations of colliding molecules.

  31. 14-8 (Continuous) • Transition state theory: The theory assumes a chemical equilibrium between reactants and activated complex in transition state, and the theory can explains the reaction rates as well as the elementary step of a chemical reactions. • Transition state: An intermediate state between the reactants and products. • Activated complex: An intermediate in a chemical reaction formed through collisions. • Reaction profile: A graphical representation of a chemical reaction in terms of the energies of the reactants, activated complex(es), and products.

  32. (Continuous) Activated complex Partial bond Heat of reaction, ΔH = Ea(forward) – Ea(reverse)

  33. 14-9 The Effect of Temperature on Reaction Rates • Arrhenius equation for the rate constant (k):*** k= Zo•p•e−Ea/RT = Ae−Ea/RT • A: frequency factor, expression of collision frequency (Z0) and steric factor (p) , i.e., the number of collisions per unit volume per unit time that are capable of leading to reaction. • e−Ea/RT: fraction of collisions with sufficiently energetic to produce a reaction. • Ea: activation energy (kJ/mol) • R: gas constant (8.3145 J/K‧mol) • T: absolute temperature (K)

  34. Extending of the Arrhenius equation • When plotting ln k (y) vs. 1/T (x): k = Ae–Ea/RT y = mx + b where m = –Ea/R Ea = –(m×R) = –(slope)×R • Rates at different temperature: or

  35. *

  36. 14-10 Reaction Mechanisms • Reaction mechanism: A set of elementary processes by which a reaction is proposed to occur. • A proposed mechanism must: • consistent with the experimentally determined rate law. • consistent with the stoichiometry of the overall reaction. • Steps in the study of a reaction mechanism: 1. Measuring the reaction rate 2. Formulating the rate law 3. Postulating plausible reaction mechanism

  37. (Continuous) • Elementary process: A single step at the molecular level in the progress of the overall reaction. • Molecularity: The number of free atoms, ions, or molecules that collide or dissociate in an elementary reaction. unfavorable

  38. Rate-determining step: The elementary process in establishing the rate of the overall reaction. It is usually the slowest step. • Intermediate: A substance that is produced in one elementary process in a reaction mechanism and consumed in another elementary process. • The intermediate does not appear in the overall reaction. • Catalyst: A substance that increase the rate of a reaction without itself being consumed in the overall reaction. • A catalyst changes a reaction mechanism with a lower activation energy. • The catalyst does not appear in the overall reaction.

  39. Mechanism-with a slow step followed by a fast step Q: For the reaction: Experiment resulted in rate law: Rate = k[H2][ICl] Postulating a possible mechanism. • Possible mechanism: Step 1: (slow, rate determine step) Step 2: (fast) Overall: • Step 1 consist with the rate law: rate= k1[H2][ICl] =k[H2][ICl]. • HIis the intermediate in this reaction.

  40. (Continuous) Reaction profile

  41. Mechanism-with a fast reversible step followed by a slow step Q: For the reaction: 2NO(g) + O2(g)→ 2NO2(g) Experiment resulted in rate law: Rate = k[NO]2[O2] Intermediate N2O2 is detected during the experiment. Postulating a possible mechanism. • Possible mechanism: (fast) Step 1: 2NO N2O2 Step 2: N2O2 + O2 2NO2 (slow, rate determine step) Overall: 2NO(g) + O2(g) → 2NO2(g) Step1 represent a fast reaching dynamic equilibrium: k1[NO]2 = k–1[N2O2] or [N2O2] = (k1/k–1)[NO]2 Substitute into step 2:

  42. Smog-An Environmental Problem with Roots in Chemical Kinetics A photochemical smog for example: PAN: peroxyacetyl nitrate, a powerful lacrimator

  43. Catalytic Converters • A dual catalyst system for example • 2CO(g) + unburned hydrocarbons + 2NO(g) 2CO2(g) + 2N2(g) • CO(g) + unburned hydrocarbons + O2(g) CO2(g) + H2O(g)

  44. 14-11 Catalysis • The role of catalyst • Catalyst increases the reaction ratewithout itself being used up in a chemical reaction. • A catalyst works by changing the mechanism of a chemical reaction. • The catalyst is consumed in one step of the mechanism, but is regenerated in another step. • The pathway of a catalyzed reaction has a lower activation energy than that of an uncatalyzed reaction, i.e., more molecules at a fixed temperature have the necessary activation energy.