Chemistry: The Molecular Science Moore, Stanitski and Jurs

1. Chemistry: The Molecular Science Moore, Stanitski and Jurs Chapter 13: Chemical Kinetics: Rates of Reactions

2. Inquiring Minds Want to Know…. • Will a reaction happen? (Ch. 18) • How fast or slow does a reaction happen? (Ch. 13) • When will the reaction “stop” or reach equilibrium? (Ch. 14 and 17) • Drugs in the body. • Ozone in the atmosphere. • Graphite to diamond.

3. Chemical Kinetics Chemical kinetics studies the rate at which a chemical reaction occurs and the pathway taken. Rate is the change of something per unit time. Examples:distance/time $/time concentration/time C (graphite) C (diamond) excruciatingly slow rate = -[C(graphite)]/ time CH4(g) + 2 O2(g) CO2 (g) + 2 H2O (g) rapid rate = -[CH4(g)] /time

4. Chemical Kinetics Homogeneous- reactants & products in one phase. ex. molecules more likely to come together in gas or liquid phase Heterogeneous- species in multiple phases. ex. solid catalyst and air (catalytic converter)

5. Reaction Rate Factors affecting the speed of a reaction: (Homogeneous) [X], M or P • Concentration of reactants k, units include s-1 • Properties (phase, structure…) • Temperature • Catalysts Rate = -∆[Cv+]/∆time = k [Cv+ ]n later

6. change in concentration of Cv+ elapsed time -Δ [Cv+ ] Δt rate = = Reaction Rate Change in [reactant] (or [product]) per unit time. Cresol violet (Cv+; a dye) decomposes in NaOH(aq): Cv+(aq) + OH- (aq)  CvOH(aq)

7. 1 a 1 b 1 c 1 d Δ[D] Δt Δ[C] Δt Δ[B] Δt Δ[A] Δt + + − − Rate = = = = Reaction Rates and Stoichiometry For any general reaction: a A + b B c C + d D The overall rate of reaction is: Cv+(aq) + OH-(aq)  CvOH(aq) Loss of 1 Cv+ → Gain of 1 CvOH Rate of Cv+ loss= = Rate of CvOH gain Rate = -∆[Cv+]/∆time = ∆[CvOH]/∆time

8. 1 2 1 2 Δ[H2] Δt Δ[HI] Δt Δ[HI] Δt Δ[HI] Δt Δ[I2] Δt Rate = = = − − + + − (-0.0040) = So = +0.0080 mol L-1 s-1 Reaction Rates and Stoichiometry H2 (g) + I2 (g) → 2 HI (g) The rate of loss of I2 is 0.0040 mol L-1 s-1. What is the rate of formation of HI ?

9. Sample calculation Avg. rate = Δ[Cv+] Δt = -(3.68x10-5 - 5.00x10-5 ) mol/L (10.0 – 0.0) s = 1.32 x 10-6mol L-1s-1 (M/s) Rates are always positive. Reaction Rate The rate of the Cv+ reaction can be calculated: Time, t [Cv+] Average rate (s) (mol / L) (mol L-1 s-1) 0.0 5.000 x 10-5 10.0 3.680 x 10-5 20.0 2.710 x 10-5 30.0 1.990 x 10-5 40.0 1.460 x 10-5 50.0 1.078 x 10-5 60.0 0.793 x 10-5 80.0 0.429 x 10-5 100.0 0.232 x 10-5 13.2 x 10-7 9.70 x 10-7 7.20 x 10-7 5.30 x 10-7 3.82 x 10-7 2.85 x 10-7 1.82 x 10-7 0.99 x 10-7 trend:? Rate depends on conc. Rate decrease with conc. React.

10. Average Rate and Instantaneous Rate Graphical view of Cv+ reaction: 5.0E-5 4.0E-5 3.0E-5 2.0E-5 1.0E-5 0 Rate = -∆[Cv+]/∆time [Cv+] (mol/L) t (s) 0 20 40 60 80 100 Average rate = slope of the blue or grey triangle …but the avg. rate depends on interval chosen.

11. Average Rate and Instantaneous Rate 5.0E-5 4.0E-5 3.0E-5 2.0E-5 1.0E-5 0 Rate = -∆[Cv+]/∆time [Cv+] (mol/L) t (s) 0 20 40 60 80 100 • Instantaneous rate = slope of a line tangent to the curve. • t = 5 s and t = 75 s have different instantaneous rates • Cannot predict [Cv+] at a time too far away since the rate changes.

12. t [Cv+] Rate of Cv+ Rate/[Cv+] (s) (M) loss (M / s) (s-1) 0 5.00 x 10-5 13.2 x 10-7 0.0264 40 1.46 x 10-5 3.82 x 10-7 0.0262 The Rate Law Rate may change when [reactant] changes. • Cv+ example shows this. • For Cv+ the rate is proportional to concentration. k (~10 second interval) rate = k [Cv+]n y = m x

13. The Rate Law 5.0E-5 4.0E-5 3.0E-5 2.0E-5 1.0E-5 0 The rate law. rate = k [Cv+]1 [Cv+] (mol/L) Rate rate t (s) 0 20 40 60 80 100 [Cv+] (mol/L)

14. Rate Law and Order of Reaction A general reaction will usually have a rate law: rate = k [A]m [B]n . . . where krate constant (T, properties, catalyst) m, norder for A & B, respectively m + n + … overall order of the reaction • The orders are usually integers (-2, -1, 0, 1, 2…), but may also be fractions (½, ⅓…) • The concentration of a reactant with a zero-order dependence has no effect on the rate of the reaction. (There just needs to be some reactant around.

15. The experimentally determined rate law is The Rate Law – Reaction Order • The reaction is: • second order in NO • first order in H2 • third order overall.

16. The Rate Law – Reaction Order • #22. • 2 NO (g) + Br2(g) 2 NOBr (g) • Experiment shows that the reaction is first-order in Br2 and second-order in NO. • Write the rate law for the reaction. • Rate = k[Br2][NO]2 • If the concentration of Br2 is tripled, how will the reaction rate change? • 3X • What happens to the reaction rate when the concentration of NO is doubled? • 4X

17. Determining Rate Laws from Initial Rates Rate laws must be measured experimentally. They cannotbe predicted from reaction stoichiometry. Initial Rate Method To find the order for a reactant: • Run the experiment with known [reactant]0. • Measure the initial rate of reaction (slope at t = 0, <2% consumed.) • Change [reactant]0 of 1 reactant; keep all others constant. • Re-measure the initial rate. • The ratio of the two rates gives the order for the chosen reactant.

18. CH3COOCH3 + OH- CH3COO- + CH3OH Determining Rate Laws from Initial Rates Data for the reaction of methyl acetate with base: change concentration of each reactant in turn Initial concentration (M) Expt. [CH3COOCH3]0 [OH-]0 Initial rate (M/s) 1 0.040 0.040 2.2 x 10-4 2 0.040 0.080 4.5 x 10-4 3 0.0800.080 9.0 x 10-4 > x2 > x2 Start with generic rate law: rate = k [CH3COOCH3]m [OH-]n

19. Determining Rate Laws from Initial Rates Initial concentration (M) Expt. [CH3COOCH3] [OH-] Initial rate (M/s) 1 0.040 0.040 2.2 x 10-4 2 0.040 0.080 4.5 x 10-4 3 0.0800.080 9.0 x 10-4 Dividing the first two data sets ([OH] changing): 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 Thus: 2.05 = (2.00)n 2.05 = (2.00)n and n = 1 rate = k [CH3COOCH3]m[OH-]1

20. Determining Rate Laws from Initial Rates Use experiments 2 & 3 to find m (dependance on CH3COOCH3]): 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 So: 2.00 = (2.00)m (1)n 2.00 = (2.00)m and m = 1 Also 1st order with respect to CH3COOCH3. rate = k [CH3COOCH3]1[OH-]1

21. Determining Rate Laws from Initial Rates The rate law is: rate = k [CH3COOCH3][OH-] Overall order for the reaction is: m + n = 1 + 1 = 2 The reaction is: 2nd order overall. 1st orderinOH- 1st orderinCH3COOCH3

22. rate [CH3COOCH3 ][OH-] k = 2.2 x 10-4 M/s (0.040 M)(0.040 M) k = Determining Rate Laws from Initial Rates If a rate law is known, k can be determined: rate = k [CH3COOCH3][OH-] Using Exp. 1: k = 0.1375 M-1 s-1 = 0.1375 L mol-1 s-1 Could repeat for each run, take an average… But a graphical method is better.

23. Practice Rate = k[NO]n[O2]m 2NO(g) + O2(g)  2NO2 Initial rates were measured at 25 ºC starting with various concentrations of reactants. What is the rate law? (m,n) Calculate the rate constant (k).

24. Practice – What is n? Rate = k[NO]n[O2]m 2x 4x The reaction is second order with respect to NO. Why can’t we compare experiment 1 and 3 ??? n = 2

25. Practice – What is m? Rate = k[NO]2[O2]m 2x 2x The reaction is first order with respect to oxygen. m = 1

26. Practice – What is k? Rate = k[NO]2[O2] Using Exp. 1… Units?... rate always has units of M/s.

27. Practice – What is k? Iodide ion is oxidized in acidic solution to the triiodide ion, I3- , by hydrogen peroxide. A series of four experiments were run at different concentrations, and the initial rates of I3- formation were determined. From the following data, obtain the reaction orders with respect to H2O2, I-, and H+. Calculate the numerical value of the rate constant.

28. Practice – What is m,n,p? Rate = k[H2O2]m[I-]n[H+]p 2x 2x 3x 9x 2x 1x Rate = k[H2O2]1[I-]2[H+]0

29. Because [H ]0 = 1, the rate law is: Rate = k[H2O2][I-]2 We can now calculate the rate constant by substituting values from any of the experiments. Using Experiment 1:

30. Change of Concentration with Time A rate law simply tells you how the rate of reaction changes as reactant concentrations change. • The integrated rate law shows how a reactant concentration changes over a period of time.

31. Δ[A] Δt rate = – = k [A] d [A] dt = - k [A] The Integrated Rate Law (Calculus) Consider a 1st-order reaction: A products A big change A little change (differential equation) Integrates to: ln [A]t= −k t + ln [A]0 y = m x + b (straight line) Ifa reaction is 1st-order,a plot of ln [A] vs. t will be linear.

32. Graphing Kinetic Data In addition to the method of initial rates, rate laws and k can be deduced by graphical methods. For first-order reaction: y = mx + b This means if we plot ln[A]t versus t, we will get a straight line for a first-order reaction.

33. You could write the rate law in the form Concentration-Time Equations Second-Order Rate Law Using calculus, you get the following equation. Here [A]t is the concentration of reactant A at time t, and [A]o is the initial concentration.

34. Graphing Kinetic Data • For second-order reaction: y = mx + b • This means if we plot 1/[A]t versus time, • we will get a straight line for a second-order reaction.

35. Concentration-Time Equations For a zero-order reaction we could write the rate law in the form Rearrange get the following equation. Here [A]t is the concentration of reactant A at time t, and [A]o is the initial concentration.

36. Graphing Kinetic Data • For zero-order reaction: y = mx + b • This means if we plot [A]t versus time, • we will get a straight line for a second-order reaction.

37. First-order reaction slope = -k time t ln[A] A products Second-order reaction slope = k 1/[A] Order Rate law Integrated rate law Slope time t Zeroth-order reaction 1 [A]t 1 [A]0 2 rate = k[A]2 = kt + +k slope = -k [A] time t The Integrated Rate Law 0 rate = k [A]t = -kt + [A]0 -k 1 rate = k[A] ln[A]t = -kt + ln[A]0 -k y = mx + b The most accurate k is obtained from the slope of a plot.

38. The Integrated Rate Law Rate data for the decomposition of cyclopentene C5H8(g)  C5H6(g) + H2(g) were measured at 850°C. Determine the order of the reaction from the following plots of those data: NOT a line line! NOT a line • The reaction is first order (the only linear plot) • k = -1 x (slope) of this plot.

39. Example 1 The decomposition of N2O5 to NO2 and O2 is first order with a rate constant of 4.800 x 10-4 s-1. If the initial concentration of N2O5 is 1.65 x 10-2 mol/L, what is the concentration of N2O5 after 825 seconds? The first-order time-concentration equation for this reaction would be:

40. Practice- using integrated rate laws The decomposition of N2O5 to NO2 and O2 is first order with a rate constant of 4.800 x 10-4s-1. If the initial concentration of N2O5 is 1.65x10-2 mol/L, what is the concentration of N2O5 after 825 seconds? eln(x) = x

41. Example 2 Consider the following first-order reaction: CH3CH2Cl(g) C2H4(g) + HCl(g) The initial concentration of ethyl chloride was 0.00100M. After heating at 500.0ºC for 155.00 s the concentration was reduced to 0.000670 M. What was the concentration of ethyl chloride after a total of 256.00 s?

42. The initial concentration of ethyl chloride was 0.00100M. After heating at 500.0ºCfor 155.00 s the concentration was reduced to 0.000670 M. What was the concentration of ethyl chloride after a total of 256.00 s? Example 1 [C2H5Cl]0 = 0.00100M [C2H5Cl]t1 = 0.000670M t1 = 155.00s t2 = 256.00s, [C2H5Cl]t2=? = 2.58 × 10-3 s-1

43. Example 1 The initial concentration of ethyl chloride was 0.00100M. After heating at 500.0ºCfor 155.00 s the concentration was reduced to 0.000670 M. What was the concentration of ethyl chloride after a total of 256.00 s?

44. Half-Life t1/2 =time for [reactant] to fall to ½[reactant]0. • For 1st -order reactions • t1/2is independent of the starting concentration. • only true for 1st order reactions (not 0th, 2nd…) • t1/2 is constant for a given 1st -order reaction.

45. ln 2 k 0.693 k t1/2 = = Half-Life For a 1st-order reaction: ln[A]t = -kt + ln[A]0 When t = t1/2,[A]t = ½[A]0 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 }

46. 0.010 0.008 0.006 0.004 0.002 0 [cisplatin] (mol/L) k = = 0.693 475 min ln 2 t1/2 t (min) 0 400 800 1200 1600 2000 = 1.46 x 10-3 min-1 Half Life t1/2 of a 1st-order reaction can be used to find k. • For cisplatin (a chemotherapy agent): • 0.0100 M  0.0050 M, after 475 min • (t½=475minutes) graphically t½ t½ t½

47. Calculating [ ] or t Use an integrated rate equation or half-life equation. Example In a 1st-order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s. Calculate: • [reactant],1600.s after initiation. • t for [reactant] to drop to 1/16th of its initial value. • t for [reactant] to drop to 0.0500 mol/L.

48. Calculating [ ] or t In a 1st-order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s (a) Calculate [reactant] ,1600.s after initiation. 1st order: k = ln 2/ t½ = 0.6931/(400. s) = 1.733x10-3 s-1 ln[A]t = -(0.001733 s-1)(1600 s) +ln(0.500) ln[A]t = -2.773 + -0.693 = -3.466 [A]t= e-3.466 = 0.0312 mol/L ln [A]t = -kt + ln [A]0 OR…1600 s = 4 t1/2 so 0.500 → 0.250 → 0.125 → 0.0625 → 0.0313 M.

49. [reactant]0 [reactant]0t1/2 [reactant]0 [reactant]0t1/2 [reactant]0 [reactant]0t1/2 1 2 1 4 1 2 1 8 1 4 1 8 1 16 [reactant]0 [reactant]0t1/2 4 t1/2 Calculating [ ] or t In a 1st-order reaction, [reactant]0 = 0.500 mol/L and t1/2 = 400.s (b) Calculate t for [reactant] to drop to 1/16th of its initial value. = 4 (400 s) = 1600 s

50. -2.303 -0.001733 s-1 t = Calculating [ ] or t In a 1st-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 = 1.33 x 103 s