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Rate Laws

Rate Laws. Example: Determine the rate law for the following reaction given the data below. H 2 O 2 (aq) + 3 I - (aq) + 2H + (aq) I 3 - (aq) + H 2 O (l). [H 2 O 2 ] [I - ] [H + ] Initial Expt # (M) (M) (M) Rate (M /s) 1 0.010 0.010 0.00050 1.15 x 10 -6

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Rate Laws

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  1. Rate Laws Example: Determine the rate law for the following reaction given the data below. H2O2(aq) + 3 I-(aq) + 2H+(aq) I3-(aq) + H2O (l) [H2O2] [I-] [H+] Initial Expt # (M) (M) (M) Rate (M /s) 1 0.010 0.010 0.00050 1.15 x 10-6 2 0.020 0.010 0.00050 2.30 x 10-6 3 0.010 0.020 0.00050 2.30 x 10-6 4 0.010 0.010 0.00100 1.15 x 10-6

  2. Rate Laws Rate = k[H2O2]m[I-]n[H+]p Compare Expts. 1 and 2: Rate 2 = 2.30 x 10-6 M/s = 2.00 Rate 1 = 1.15 x 10-6 M/s Rate 2 = k (0.020 M)m(0.010 M)n(.00050 M)p= 2.00 Rate 1 k (0.010 M)m(0.010 M)n(.00050 M)p (0.020)m = 2.0m = 2.00 (0.010)m 2.0m = 2.00 only if m = 1 Rate = k[H2O2]1[I-]n[H+]p

  3. Rate Laws Rate = k[H2O2]1[I-]n[H+]p Compare Expts. 1 and 3: Rate 3 = 2.30 x 10-6 M/s = 2.00 Rate 1 = 1.15 x 10-6 M/s Rate 3 = k (0.010 M)1(0.020 M)n(.00050 M)p= 2.00 Rate 1 k (0.010 M)1(0.010 M)n(.00050 M)p (0.020)n = 2.0n = 2.00 (0.010)n 2.0n = 2.00 only if n = 1 Rate = k[H2O2]1[I-]1[H+]p

  4. Rate Laws Rate = k[H2O2]1[I-]1[H+]p Compare Expts. 1 and 4: Rate 4 = 1.15 x 10-6 M/s = 1.00 Rate 1 = 1.15 x 10-6 M/s Rate 4 = k (0.010 M)1(0.010 M)1(.00100 M)p= 1.00 Rate 1 k (0.010 M)1(0.010 M)1(.00050 M)p (0.00100)p = 2.0p = 1.00 (0.00050)p 2p = 1.00 only if p = 0 Rate = k[H2O2]1[I-]1[H+]0 Rate = k[H2O2] [I-]

  5. Rate Laws For the reaction: 5Br-(aq) + BrO3-(aq) + 6 H+(aq) 3 Br2(aq) + 3 H20 (l) the rate law was determined experimentally to be: Rate = k[Br-] [BrO3-] [H+]2 • The reaction is first order with respect to Br-, first order with respect to BrO3-, and second order with respect to H+.

  6. Rate Laws • The previous reaction is fourth order overall. • The overall reaction order is the sum of all the exponents in the rate law. • Note: In most rate laws, the reaction orders are 0, 1, or 2. • Reaction orders can be fractional or negative.

  7. Rate Laws • The value of the rate constant can be determined from the initial rate data that is used to determine the rate law. • Select one set of conditions. • Substitute the initial rate and the concentrations into the rate law. • Solve the rate law for the rate constant, k.

  8. Rate Laws • The value of the rate constant, k, depends only on temperature. • It does not depend on the concentration of reactants. • Consequently, all sets of data should give the same rate constant.

  9. Rate Laws Example: The following data was used to determine the rate law for the reaction: A + B C Calculate the value of the rate constant if the rate law is: Rate = k [A]2[B] Expt # [A] (M) [B] (M) Initial rate (M /s) 1 0.100 0.100 4.0 x 10-5 2 0.100 0.200 8.0 x 10-5 3 0.200 0.100 16.0 x 10-5

  10. Rate Laws • Select a set of conditions: Experiment 1: Rate = 4.0 x 10-5 M /s [A] = 0.100 M [B] = 0.100 M • Substitute data into the rate law: 4.0 x 10-5M = k (0.100 M)2 (0.100 M) s • Solve for k k = 4.0 x 10-5 M/s = 0.040 = 0.040 M-2s-1 (0.100 M)2 (0.100M) M2 . s Rate = k [A]2[B]

  11. Rate Laws • Important: • The units of the rate constant will depend on the overall order of the reaction. • You must be able to report your calculated rate constant using the correct units.

  12. First Order Reactions • A first orderreaction is one whose rate depends on the concentration of a single reactant raised to the first power: Rate = k[A] • Using calculus, it is possible to derive an equation that can be used to predict the concentration of reactant A after a given amount of time has elapsed.

  13. First Order Reactions • To predict the concentration of a reactant at a given time during the reaction: ln[A]t = -kt + ln[A]0 where ln = natural logarithm (not log) t = time (units depend on k) [A]t = conc. of A at time t [A]0 = initial concentration of A k = rate constant

  14. First Order Reactions Example: A certain pesticide decomposes in water via a first order reaction with a rate constant of 1.45 yr-1. What will the concentration of the pesticide be after 0.50 years for a solution whose initial concentration was 5.0 x 10-4 g/mL? Given: k = 1.45 yr-1 t = 0.50 yr [A]o = 5.0 x 10-4 g/mL Find: [A]t=0.5 yr ln[A]t = -kt + ln[A]0

  15. First Order Reactions ln[A]0.5 yr = -kt + ln[A]0 ln[A]0.5 yr = - (1.45 x 0.50 yr) + ln (5.0 x 10-4) yr ln[A]0.5 yr = - 0.725 + -7.601 = - 8.326 To find the value of [A]0.5 yr, use the inverse natural logarithm, ex. [A]0.5 yr = e-8.326 = 2.42 x 10-4 = 2.4 x 10-4 g/mL

  16. First Order Reactions • The time required for the concentration of a reactant to drop to one half of its original value is called the half-life of the reaction. • t1/2 • After one half life has elapsed, the concentration of the reactant will be: • [A]t = ½ [A]0 • For a first order reaction: t1/2 = 0.693 k ½

  17. First Order Reactions • Given the half life of a first order reaction and the initial concentration of the reactant, you can calculate the concentration of the reactant at any time in the reaction. • Use t1/2 = 0.693/k to find the rate constant • Substitute in the value for k and the initial concentration to find the value for [A]t: ln[A]t = -kt + ln[A]0

  18. First Order Reactions Example: A certain pesticide has a half life of 0.500 yr. If the initial concentration of a solution of the pesticide is 2.5 x 10-4 g/mL, what will its concentration be after 1.5 years? Given: [A]0 = 2.5 x 10-4 g/mL t1/2 = 0.500 yr t = 1.5 yr Find: [A]1. 5 yr First, find the value for k.

  19. First Order Reactions • Find the rate constant, k: k = 0.693/t1/2 k = 0.693 = 1.386 yr-1 0.500 yr • Substitute data into ln[A]1.5 yr = - (1.386 x 1.5 yr) + ln(2.5 x 10-4) yr = - 10.373 ln[A]t = -kt + ln[A]0

  20. First Order Reactions • Solve for [A]1. 5 yr using the inverse natural logarithm: [A]1. 5 yr = e-10.373 = 3.1 x 10-5 g/mL

  21. First Order Reactions • There are two other (simpler!!) ways to solve this problem: Draw a picture: 1.5 yr = 3 half lives 2.5 x 10-4 g/mL 0.5 yr 1.25 x 10-4 g/mL 0.5 yr 0.625 x 10-4 g/mL 0.5 yr 0.31 x 10-4 g/mL

  22. First Order Reactions • For a first order reaction, you can also determine the concentration (or mass or moles) of a material left at a given amount of time by: [A]t = 1x [A]0 2 Elapsed time Half life

  23. First Order Reactions [A]t = 1x[A]0 2 For the previous example: [A]1.5 yr = 1(1.5 yr/0.5 yr) x (2.5 x 10-4 g/mL) 2 [A]1.5 yr = 3.1 x 10-5 g/mL Elapsed time Half life

  24. First Order Reactions Example: Cobalt-57 has a half life of 270. days. How much of an 80.0 g sample will remain after 4.44 years if it decomposes via a first order reaction? Given: t1/2 = 270. days =0.740 yr A0 = 80.0 g t = 4.44 yr Find: A4.44 yr We can use mass (g) instead of the concentration.

  25. First Order Reactions • Calculate the rate constant, k. k = 0.693 = 0.693 = 0.937 yr -1 t1/2 0.740 yr • Substitute data into lnAt = -kt + lnA0 lnA4.44 yr = - (0.937 yr-1 x 4.44 yr) + ln (80.0) lnA4.44 yr = 0.222

  26. First Order Reactions • Solve for A4.44 yr using the inverse natural logarithm: A4.44 yr = e0.222 = 1.25 g

  27. First Order Reactions • Since 4.44 years is exactly 6 half lives, we can also draw a picture to solve this problem. • 4.44 yr = 6.00 0.740 yr • When one half life (270 days or 0.740 yr) has elapsed, half of the original 80.0 g will have decomposed.

  28. First Order Reactions 80.0 g 270 days 40.0 g 270 days After 6 half-lives have elapsed, 1.25 g of Co-57 are left. 20.0 g 270 days 10.0 g 270 days 5.00 g 270 days 2.50 g 270 days 1.25 g

  29. First Order Reactions • You can also solve this problem using the third approach: [A]4.44 yr = 1(4.44 yr/0.74 yr) x (80.0 g) 2 [A]4.44yr = 1.25 g

  30. Chemical Equilibrium • One of the challenges that industrial chemists face is to maximize the yield of product obtained in a reaction. • Many reactions do not go to completion. • The reaction stops short of the theoretical yield. • Unreacted starting materials are still present.

  31. Chemical Equilibrium Consider the reaction to produce ammonia: N2 (g) + 3 H2 (g) 2 NH3 (g)

  32. Chemical Equilibrium • After a period of time, the composition of the reaction mixture stays the same even though most of the reactants are still present. • Although it is not apparent, chemical reactions are still occurring within the reaction mixture. • N2 (g) + 3 H2 (g) 2 NH3 (g) • 2 NH3 (g) N2 (g) + 3 H2 (g)

  33. N2 (g) + 3 H2 (g) 2 NH3 (g) Chemical Equilibrium • The reaction has reached chemical equilibrium and is best represented by the equation: • The double arrow is used to indicate that the reaction is an equilibrium reaction. • It indicates that the reaction occurs in both directions simultaneously.

  34. Chemical Equilibrium • Chemical equilibrium: • A state of dynamic balance in which the opposing reactions are occurring at equal rates • Rate of forward reaction (reactants to products) = rate of reverse reaction (products decomposing to reactants)

  35. Chemical Equilibrium • At chemical equilibrium, the concentrations of the reactants and products do not change. • Note: This does not mean that the concentrations of the reactants and products are identical to each other.

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