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Chapter 12 Lecture Notes Chemical Kinetics “The study of speeds of reactions and the nanoscale pathways or rearrangements by which atoms and molecules are transformed to products” Chapter 13: Chemical Kinetics: Rates of Reactions © 2008 Brooks/Cole 1 © 2008 Brooks/Cole 2 Reaction Rate Reaction Rate Combustion of Fe(s) powder: © 2008 Brooks/Cole 3 © 2008 Brooks/Cole 4 Reaction Rate Reaction Rate + Change in [reactant] or [product] per unit time. Average rate of the Cv reaction can be calculated: Time, t [Cv+] Average rate + Cresol violet (Cv ; a dye) decomposes in NaOH(aq): (s) (mol / L) (mol L-1 s-1) 0.0 5.000 x 10-5 13.2 x 10-7 10.0 3.680 x 10-5 9.70 x 10-7 20.0 2.710 x 10-5 7.20 x 10-7 30.0 1.990 x 10-5 5.30 x 10-7 40.0 1.460 x 10-5 3.82 x 10-7 + - 50.0 1.078 x 10-5 Cv (aq) + OH (aq) → CvOH(aq) 2.85 x 10-7 60.0 0.793 x 10-5 -7 + + -5 1.82 x 10 change in concentration of Cv Δ [Cv ] 80.0 0.429 x 10 rate = = 0.99 x 10-7 elapsed time Δt 100.0 0.232 x 10-5 © 2008 Brooks/Cole 5 © 2008 Brooks/Cole 6 1 Reaction Rates and Stoichiometry Reaction Rates and Stoichiometry Cv+(aq) + OH-(aq) → CvOH(aq) For any general reaction: a A + b B c C + d D Stoichiometry: The overall rate of reaction is: Loss of 1 Cv+ → Gain of 1 CvOH Rate of Cv+ loss = Rate of CvOH gain 1 Δ[A] 1 Δ[B] 1 Δ[C] 1 Δ[D] Rate = − = − = + = + Another example: a Δt b Δt c Δt d Δt 2 N2O5(g) 4 NO2(g) + O2(g) Reactants decrease with time. Products increase with time. Negative sign. Positive sign Negative rate Positive rate Rate of loss of N2O5 divided by -2, equals rate of gain of O2 © 2008 Brooks/Cole 7 © 2008 Brooks/Cole 8 Reaction Rates and Stoichiometry Average Rate and Instantaneous Rate For: Graphical view of Cv+ reaction: 5.0E-5 H2 (g) + I2 (g) → 2 HI (g) -1 -1 the rate of loss of I2 is 0.0040 mol L s . What is 4.0E-5 the rate of formation of HI ? 3.0E-5 Δ[H ] Δ[I ] Δ[HI] ] (mol/L) 1 + Rate = − 2 = − 2 = + 2.0E-5 Δt Δt 2 Δt [Cv 1.0E-5 Δ[H2] 1 Δ[HI] Rate = − = − (-0.0040) = + 0 Δt 2 Δt 0 20 40 60 80 100 t (s) Δ[HI] So = +0.0080 mol L-1 s-1 Δt © 2008 Brooks/Cole 9 © 2008 Brooks/Cole 10 Average Rate and Instantaneous Rate Effect of Concentration on Reaction Rate Rate may change when [reactant] changes. • Cv+ example shows this. • For Cv+ the rate is proportional to concentration. t [Cv+] Rate of Cv+ Rate/[Cv+] (s) (M) loss (M / s) (s-1) 0 5.00 x 10-5 1.54 x 10-6 0.0308 80 4.29 x 10-6 1.32 x 10-7 0.0308 © 2008 Brooks/Cole 11 © 2008 Brooks/Cole 12 2 Rate Law and Order of Reaction Determining Rate Laws from Initial Rates A general reaction will usually have a rate law: rate = k [A]m [B]n . The orders are usually integers (-2, -1, 0, 1, 2…), but may also be fractions (½, ⅓…) © 2008 Brooks/Cole 13 © 2008 Brooks/Cole 14 Determining Rate Laws from Initial Rates Determining Rate Laws from Initial Rates Data for the reaction of methyl acetate with base: - - CH3COOCH3 + OH CH3COO + CH3OH Dividing the first two data sets: 4.5 x 10-4 M/s = k (0.040 M)m(0.080 M)n 2.2 x 10-4 M/s = k (0.040 M)m(0.040 M)n 1 raised to any Rate law: power = 1 m - n rate = k [CH3COOCH3] [OH ] © 2008 Brooks/Cole 15 © 2008 Brooks/Cole 16 Determining Rate Laws from Initial Rates Determining Rate Laws from Initial Rates Use experiments 2 & 3 to find m: 9.0 x 10-4 M/s = k (0.080 M)m(0.080 M)n 4.5 x 10-4 M/s = k (0.040 M)m(0.080 M)n © 2008 Brooks/Cole 17 © 2008 Brooks/Cole 18 3 Determining Rate Laws from Initial Rates The Integrated Rate Law If a rate law is known, k can be determined: Calculus is used to integrate a rate law. st rate Consider a 1 -order reaction: A products k = [CH COOCH ][OH-] Δ[A] 3 3 rate = – = k [A] Δt 2.2 x 10-4 M/s Using run 1: k = d [A] (0.040 M)(0.040 M) = – = k [A] (as a differential equation) dt -1 -1 -1 -1 k = 0.1375 M s = 0.1375 L mol s Integrates to: ln [A]t = −k t + ln [A]0 y = m x + b (straight line) Could repeat for each run, take an average… But a graphical method is better. © 2008 Brooks/Cole 19 © 2008 Brooks/Cole 20 The Integrated Rate Law Zeroth-order reaction The Integrated Rate Law slope = - k Rate data for the decomposition of cyclopentene The reaction: A products [A] C H (g) C H (g) + H (g) doesn’t have to be 1st order. 5 8 → 5 6 2 time t were measured at 850°C. Determine the order of the Some common integrated rate laws: First-order reaction reaction from the following plots of those data: slope = - Order Rate law Integrated rate law Slope k time t ln[A] 0 rate = k [A]t = -kt + [A]0 -k 1 rate = k[A] ln[A] = -kt + ln[A] -k t 0 Second-order reaction 2 1 1 2 rate = k[A] = kt + +k k [A]t [A]0 slope = 1/[A] y x time t • The reaction is first order (the only linear plot) • k = -1 x (slope) of this plot. The most accurate k is obtained from the slope of a plot. © 2008 Brooks/Cole 21 © 2008 Brooks/Cole 22 Half-Life Half-Life For a 1st-order reaction: ln[A]t = -kt + ln[A]0 st Half-lives are only useful for 1 -order reactions. When t = t1/2 [A]t = ½[A]0 Why? Then: ln(½[A]0) = -kt1/2 + ln[A]0 ln(½[A]0/[A]0) = -kt1/2 {note: ln x – ln y = ln(x/y)} ln(½) = -ln(2) = -kt1/2 {note: ln(1/y) = –ln y } © 2008 Brooks/Cole 23 © 2008 Brooks/Cole 24 4 Half Life Calculating [ ] or t from a Rate Law st t1/2 of a 1 -order reaction can be used to find k. Use an integrated rate equation. For cisplatin (a chemotherapy agent): • ½ the cisplatin lost after 475 min. • (0.0100 M → 0.0050 M) 0.010 • [cisplatin] halves every 475 min 0.008 k = ln 2 = 0.693 a) [reactant],1600.s after initiation. 0.006 t 475 min 1/2 b) t for [reactant] to drop to 1/16th of its initial value. -3 -1 0.004 = 1.46 x 10 min c) t for [reactant] to drop to 0.0500 mol/L. [cisplatin] (mol/L) 0.002 0 0 400 800 1200 1600 2000 t (min) © 2008 Brooks/Cole 25 © 2008 Brooks/Cole 26 Calculating [ ] or t from a Rate Law Calculating [ ] or t from a Rate Law st st In a 1 -order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s In a 1 -order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s (a) Calculate [reactant] ,1600.s after initiation. (b) Calculate t for [reactant] to drop to 1/16th of its initial value. st 1 order: [reactant] 1 [reactant] t 0 2 0 1/2 -3 -1 1 k = ln 2/ t½ = 0.6931/(400. s) = 1.733x10 s [reactant] 1 [reactant] t 2 0 4 0 1/2 and ln [A]t = -kt + ln [A]0 1 1 [reactant] [reactant] t 4 0 8 0 1/2 -1 so ln[A]t = -(0.001733 s )(1600 s) +ln(0.500) 1 1 [reactant] [reactant] t 8 0 16 0 1/2 ln[A]t = -2.773 + -0.693 = -3.466 4 t1/2 = 4 (400 s) = 1600 s -3.466 [A]t = e = 0.0312 mol/L Note: part (a) could be solved in a similar way. 1600 s = 4 t1/2 so 0.500 → 0.250 → 0.125 → 0.0625 → 0.0313 M. © 2008 Brooks/Cole 27 © 2008 Brooks/Cole 28 Calculating [ ] or t from a Rate Law Nanoscale View: Elementary Reactions st In a 1 -order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s (c) Calculate t for [reactant] to drop to 0.0500 mol/L ? From part (a): k = 1.733 x 10-3 s-1 ln [A]t = -kt + ln [A]0 then ln (0.0500) = -(0.001733 s-1) t + ln(0.500) -2.996 = -(0.001733 s-1) t – 0.693 t = -2.303 -0.001733 s-1 3 t = 1.33 x 10 s © 2008 Brooks/Cole 29 © 2008 Brooks/Cole 30 5 Elementary reactions Unimolecular Reactions 2-butene isomerization is unimolecular: H3C CH3 H3C H unimolecular C=C (g) C=C (g) H H H CH3 bimolecular unimolecular unimolecular © 2008 Brooks/Cole 31 © 2008 Brooks/Cole 32 Unimolecular Reactions Transition State transition state or activated complex ) J 500 -21 Ea is the activation 400 energy, the minimum E to J 300 -21 go over the barrier.
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