Lecture Slides on Chemical Kinetics - General Chemistry | CHEM 162, Study notes of Chemistry

Download Lecture Slides on Chemical Kinetics - General Chemistry | CHEM 162 and more Study notes Chemistry in PDF only on Docsity! 1 Chapter 15. Chemical Kinetics ― reaction rates ― reaction mechanisms ALL reactions (biochemical, chemical, inorganic, organic, gas phase, solution, solid, etc.) aA + bB products 15.1 Reaction Rates 2 NO2 (g) 2 NO (g) + O2 (g) Rate of consumption of NO2 Rate of production of NO (Rate of production of O2) → Stoichiometry matters → Rate = – ∆ [NO2] ∆t e.g. mol/L e.g. seconds (s) +ve quantity by convention = = 2 Copyright © Houghton Mifflin Company. All rights reserved. Fig. 15.1: Starting with pure nitrogen dioxide at 300°C, conc. of reactant and products with time 2 15.2 Rate Laws 2 NO2 (g) 2 NO (g) + O2 (g) If we can assume that the overall reaction as written is irreversible Rate = k [NO2]n k = rate constant n = order of the reactant (or order of reaction) Rate law (for the given rxn) empirical constants (must be found experimentally) Rate = – ∆ [NO2] ∆t = – d [NO2] = dt k [NO2]n Rate = – d [NO2] = dt k [NO2]n differential rate law ≡ rate law → gives information on how rate depends on concentration 1 1 * ∫ d [NO2] =[NO2] – k dt [NO2] [NO2]0 ∫ 0 t ln [NO2] = ln [NO2]0 – kt Integrated rate law → gives information on how concentration depends on time alternatively : [NO2] = [NO2]0 exp (– kt) concentration time → You can get the integrated rate law from the differential rate law and vice versa. 2 Integrated rate law (Integral rate law) Suppose n = 1 – d [NO2] = dt k [NO2] 2 * 5 → Summary of the concepts discussed in 15.1 to 15.3 Chemical Kinetics : ― reaction rates ― reaction mechanisms aA + bB products Rate = – d [A] = dt k [A]n Reaction rate: (differential rate law) Assumes reverse reaction is negligible, i.e. forward reaction is irreversible. • k = rate constant • n = order of the reaction w.r.t. reactant A • rate law relates reaction rate to concentration(s) of reactant(s) • rate laws are empirical (k and n are experimentally determined) • rate laws differential ― method of initial rates integrated 15.4 The Integrated Rate Law → Rate laws of the form [A] vs t • Exact form depends on the order of the reaction a A products Rate = – d [A] = dt k [A]n (differential rate law) 1 First – Order Rate Law (n = 1) Rate = – d [A] = dt k [A] ∫ – d [A] =[A] k dt [A] [A]0 ∫ 0 t ln [A] = – kt + ln [A]0 or [A] = [A]0 exp (– kt) or ln ([A]/[A]0) = – kt • To test if a reaction is first-order, plot ln [A] vs t, it should be linear. Slope = – k and intercept = ln [A]0 6 Half – Life of a 1st – Order Reaction Half – life of a reaction ≡ t1/2 ― time for a reactant to reach half its initial concentration ― i.e. t = t1/2 at [A]0/2 = [A] Since ln ([A]/[A]0) = – kt t1/2 = – 1 ln [A]0/2 k [A]0 = ln 2 k t1/2 = 0.693 k Note: ― t1/2 is obtained if k is known and k is obtained if t1/2 is known. ― t1/2 is independent of concentration. Example of 1st – Order Reaction 2 N2O5 (soln) 4 NO2 (soln) + O2 (g) Rate = – d [N2O5] = dt k [N2O5] (differential rate law) ln [N2O5] = – kt + ln [N2O5]0 (integrated rate law) → Example 15.2 has data for the gas phase reaction : 2 N2O5 (g) 4 NO2 (g) + O2 (g) 500.0707 00.1000 Time (s)[N2O5] (mol/L) • To verify that the rate law is first-order in N2O5 concentration, plot ln [N2O5] vs time If linear, rate law is first-order and slope gives the rate constant k. Slope = – k k = 6.93 x 10–3 s–1 Figure 15.3: A plot of ln[N205] versus time. 7 Since [N2O5]0 = 0.100 mol/L ln [N2O5]0 = –2.303 Also k = 6.93 x 10–3 s–1 ln [N2O5] = – kt + ln [N2O5]0 Complete integrated rate law for the N2O5 decomposition reaction. → Half – life of this reaction t1/2 = 0.693 k = 0.693 6.93 x 10–3 s–1 = 100 s Note that t1/2 is independent of initial concentration as illustrated in Fig. 15.4. Copyright © Houghton Mifflin Company. All rights reserved. Figure 15.4: Plot of [N2O5] versus time for the decomposition reaction of N2O5. 10 → According to the plots, the dimerization reaction is 2nd order. So Rate = – d [C4H6] = dt k [C4H6]2 k = slope = 6.14 x 10–2 L.mol–1s–1 t1/2 = 1 k [A]0 = 1 (6.14 x 10–2 L/mol.s) (1.0 x 10–2 mol/L) = 1.63 x 103 s Zero – Order Rate Laws a A products Rate = – d [A] = dt k [A]0 = k (differential rate law) ∫ – d [A] = k dt [A] [A]0 ∫ 0 t [A] = – kt + [A]0 integrated rate law → Half – life of a zero – order reaction : t1/2 = [A]0 2k Example of a zero – order reaction 2 N2O (g) 2 N2 (g) + O2 (g) Decomposition on a Pt (catalyst) surface 11 Copyright © Houghton Mifflin Company. All rights reserved. Figure 15.6: Plot of [A] versus t for a zero-order reaction