1. 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) 10.0100.0100.000501.15 x 10 -6 20.0200.0100.000502.30 x 10 -6 30.0100.0200.00050 2.30 x 10 -6 40.0100.0100.001001.15 x 10 -6

2. Rate Laws Rate = k[H 2 O 2 ] 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 = 2.0 m = 2.00 (0.010) m 2.0 m = 2.00 only if m = 1 Rate = k[H 2 O 2 ] 1 [I - ] n [H + ] p

3. Rate Laws Rate = k[H 2 O 2 ] 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 = 2.0 n = 2.00 (0.010) n 2.0 n = 2.00 only if n = 1 Rate = k[H 2 O 2 ] 1 [I - ] 1 [H + ] p

4. Rate Laws Rate = k[H 2 O 2 ] 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 = 2.0 p = 1.00 (0.00050) p 2 p = 1.00 only if p = 0 Rate = k[H 2 O 2 ] 1 [I - ] 1 [H + ] 0 Rate = k[H 2 O 2 ] [I - ]

5. Rate Laws For the reaction: 5Br - (aq) + BrO 3 - (aq) + 6 H + (aq) 3 Br 2 (aq) + 3 H 2 0 (l) the rate law was determined experimentally to be: Rate = k[Br - ] [BrO 3 - ] [H + ] 2 The reaction is first order with respect to Br -, first order with respect to BrO 3 -, 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: Expt #[A] (M)[B] (M)Initial rate (M /s) 10.1000.1004.0 x 10 -5 20.1000.2008.0 x 10 -5 30.2000.100 16.0 x 10 -5 Rate = k [A] 2 [B]

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 -5 M = 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 -2 s -1 (0.100 M) 2 (0.100M) M 2. 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 order reaction 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? ln[A] t = -kt + ln[A] 0 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

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, e x. [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. t 1/2 After one half life has elapsed, the concentration of the reactant will be: [A] t = ½ [A] 0 For a first order reaction: t 1/2 = 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 t 1/2 = 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 t 1/2 = 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/t 1/2 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: 2.5 x 10 -4 g/mL 1.25 x 10 -4 g/mL 0.625 x 10 -4 g/mL 0.31 x 10 -4 g/mL 1.5 yr = 3 half lives 0.5 yr

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 = 1 x [A] 0 2 Elapsed time Half life

23. First Order Reactions [A] t = 1 x[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:t 1/2 = 270. days =0.740 yr A 0 = 80.0 g t = 4.44 yr Find:A 4.44 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 t 1/2 0.740 yr Substitute data into lnA t = -kt + lnA 0 lnA 4.44 yr = - (0.937 yr -1 x 4.44 yr) + ln (80.0) lnA 4.44 yr = 0.222

26. First Order Reactions Solve for A 4.44 yr using the inverse natural logarithm: A 4.44 yr = e 0.222 = 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 40.0 g 20.0 g 10.0 g 5.00 g 2.50 g 1.25 g 270 days After 6 half-lives have elapsed, 1.25 g of Co-57 are left.

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: N 2 (g) + 3 H 2 (g) 2 NH 3 (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. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 2 NH 3 (g) N 2 (g) + 3 H 2 (g)

33. Chemical Equilibrium The reaction has reached chemical equilibrium and is best represented by the equation: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 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.