1. CATALYST 1. 2. 3. You are in a car, how would you calculate the speed of the car? No, you cannot look at the speedometer.

2. HOW DO YOU CALCULATE THE SPEED OF A CHEMICAL REACTION?

3. LECTURE 6.3 – RATES OF REACTIONS AND RATE LAWS

4. TODAY’S LEARNING TARGETS LT 6.6 – For a given chemical reaction, I can describe the average rate of reaction using both chemical formulas and given molarities. LT 6.7 – I can explain how to relate the average rate of a reaction to the instantaneous rate of reaction using a plot of concentration vs. time. LT 6.8 – I can hypothesize the rate law for a chemical reaction using experimental data. LT 6.9 – I can calculate the reaction order, rate constant, and instantaneous rate of a chemical reaction using experimental data and a rate law.

5. RATES When we describe the rate of any process, we mean: When examining the rate of chemical reaction, we examine change in concentration over a period of time:

6. RATES OF REACTIONS When we describe the rate of any process, we can think of the reaction: aA  bB We can describe it as the rate of disappearance of A or the appearance of B:

7. CLASS EXAMPLE You run a chemical reaction whereby C 4 H 9 Cl is formed. At t = 0, the concentration is 0.1000 M. After 50.0 s, the new concentration is determined to be 0.0905 M. What is the average rate of reaction?

8. TABLE TALK Complete the average rate in the table below: Time (s)[C 4 H 9 Cl] (M)Average Rate (M/s) 0.00.1000 1.9 x 10 -4 50.00.0905 100.00.0820 150.00.0741 200.00.0671 300.00.0549 400.00.0448

9. A NOTE ON COEFFICIENTS Let us assume that we need to react 2 A’s to make one B: 2 A  B This means 2 moles of A disappear for every 1 mole of B formed Therefore for the reaction aA + bB  cC + dD has the rate:

10. GRAPHING AVERAGE RATE

11. INSTANTANEOUS RATE While the average rate is useful, it has limitations because we cannot identify the rate at a given period of time. The graph on the last slide allows us to examine the instantaneous rate. The slope of the tangent line at a given point in time gives us the instantaneous rate.

12. CLASS EXAMPLE Using the graph below, calculate the instantaneous rate of reaction at t = 20 s

13. TABLE TALK Using the graph below, calculate the instantaneous rate of reaction at t = 50 s

14. WHITE BOARD PROBLEMS

15. 1.Write ALL the possible average rate equations for the following chemical reaction: 2 HgO  2 Hg + O 2 2. For the following chemical reaction: 2 NO 2 (g) + O 3 (g)  N 2 O 5 (s) + O 2 (g) Complete the following table Time (s)[NO 2 ] (M)Average Rate (M/s) 0.01.000 250.00.905 1000.00.820

16. WHITE BOARD PROBLEMS 3.Calculate the instantaneous rate at t = 0 s for the graph below. 4. Calculate the instantaneous rate at t = 4 s for the graph below.

17. DIET COKE AND MENTOS

18. WHAT DOES THE DATA MEAN? With your table, analyze the data for the Diet Coke and Mentos experiment run at different concentrations.

19. RATE LAW The initial concentrations of the reactants allow for us to determine the rate of a reaction. The rate law shows the relationship between the reactant concentrations and the rate Must be experimentally determined For the general reaction: a A + b B  c C + d D The rate law is: k = rate constant (temperature dependent) m and n = whole numbers greater than 0 Allows us to calculate the rate for any concentration!

20. REACTION ORDER We classify the reaction order in order to examine the mechanism of the reaction. For the rate law: The reaction order is:

21. CLASS EXAMPLE For the reaction A + B  AB, the following data were obtained. a) Determine the rate law for this reaction b) Determine the overall order of the reaction Experiment[A] (M)[B] (M)Initial Rate (M/s) 10.7200.1800.470 20.720 1.880 30.360 0.117

22. TABLE TALK The following data were calculated for the following reaction: 2 NO (g) + O 2 (g)  2 NO 2 (g) What is the rate law for this reaction? Experiment[NO] (M)[O 2 ] (M)Initial Rate (M/s) 10.01260.01251.41 x 10 -2 20.02520.01255.64 x 10 -2 30.02520.02501.13 x 10 -1

23. DETERMINING RATE CONSTANTS Because rate laws are experimentally determined, rate constants must also be determined experimentally. Once the exponents are known, one of the sets of data can be put in the rate law and k can be determined

24. UNITS OF RATE CONSTANTS Units of k depend on the reaction order The units of k must cancel out concentration units so that the rate can be in units of M x s -1

25. CLASS EXAMPLE For the reaction A + B  AB, the following data were obtained. a)Using the rate law previously completed, determine the value of k. b)What are the units of k? Experiment[A] (M)[B] (M)Initial Rate (M/s) 10.7200.1800.470 20.720 1.880 30.360 0.117

26. TABLE TALK The following data were calculated for the following reaction: 2 ClO 2 (aq) + 2 OH - (aq)  ClO 3 - (aq) + ClO 2 - (aq) What is the rate law for this reaction? What is the value of k (with units)? What is the reaction order? Experiment[ClO 2 - ] (M)[OH - ] (M)Initial Rate (M/s) 10.0600.0300.0248 20.0200.0300.00276 30.0200.0900.00828

27. COLLABORATIVE PROBLEMS In groups of 4, assign each person as 1, 2, 3, or 4. You will begin working on the problem for the number you have been assigned. Pass to the next person when you finish a. Begin b for the problem that you have been passed. Do until question a – d for all 4 problems are completed.

28. WORK TIME Begin working on book problems: 14.2, 14.3, 14.4, 14.21, 14.22, 14.33, 14.34, 14.35, 14.36, and 14.37 These will be due next class period.

29. EXIT TICKET 1.Calculate the average rate of reaction for the following reaction: CH 3 OH (aq) + HCl (aq)  CH 3 Cl (aq) + H 2 O (l) 2. Determine the rate law and the value of k for the following reaction: S 2 O 3 2- (aq) + 3 I - (aq)  2 SO 4 2- (aq) + I 3 - (aq) Experiment[S 2 O 3 2- ] (M)[I - ] (M)Initial Rate (M/s) 10.0180.0362.6 x 10 -6 20.0270.0363.9 x 10 -6 30.0360.0547.8 x 10 -6 40.0500.0721.4 x 10 -5 Time (s)[HCl] (M) 01.85 541.58

30. RATE YOURSELF Using your learning target log, rate yourself 1 – 4 on 6.6 to 6.9

32. CLOSING TIME Read 14.1 to 14.3 Book problems 14.2, 14.3, 14.4, 14.21, 14.22, 14.33, 14.34, 14.35, 14.36, and 14.37