Chemical Kinetics

1. Chemical Kinetics How are chemical reaction rates determined and what factors affect these rates?

2. Objectives • To understand rates of reaction and the conditions affecting rates. • To derive the rate equation, rate constant, and reaction order from experimental data. • To use integrated rate laws. • To understand collision theory of reaction rates and the role of activation energy. • To relate reaction mechanisms and rate laws.

3. Chemical Kinetics Ch 14 hmwk problems: 11 – 15, 20, 21, 25, 29, 33, 38, 45, 47, 50, 55, 58, 59, 62, 66, 69, 87, 98

4. Why are Kinetics Important? • Medicinal: How quickly or slowly will a medication work? • Environmental: Is the rate at which ozone depletion occurs equal to the rate at which it is formed? • Industrial: How long will it take to produce a chemical? Is there a way to change this?

5. What are Kinetics Anyway? • Chemical Kinetics is the study of chemical reaction rates, or speeds. • Can you think of an example of a chemical reaction with a fast rate? combustion of gasoline • Can you think of an example of a chemical reaction with a slow rate? formation of a diamond

6. What Makes a Reactions Occur? • If a reaction is to occur between two or more reactants, the reactants must come in contact with each other, and have enough energy to both beak existing bonds and form new bonds. • The rate at which these reactants come in contact with each other affects the rate of the overall chemical reaction.

7. Factors Affecting Reaction Rates • The Physical State of the Reactants: Reactions are limited by the area of contact between two reactants. Therefore, the greater the surface area of a solid reactant, the faster a reaction can proceed.

8. Factors Affecting Reaction Rates • The Concentration of Reactants: As the concentration of one or more reactants increases, the more frequently the reactants can collide, leading to increased reaction rates.

9. Factors Affecting Reaction Rates • The temperature at which the reaction occurs: The rate of a chemical reaction occurs increases with increasing temperature. As the molecules move more rapidly, they collide more frequently and with greater energy, increasing the reaction rate.

10. Factors Affecting Reaction Rates • The presence of a catalyst: Catalysts are reagents that increase the speed of a reaction without being consumed in the reaction by affecting why the reactants collide together. We will discuss this in greater depth later.

11. Reaction Rates • Since a rate is expressed as change over time, a reaction rate is expressed as the change in the concentration of reactants over time. • For a simple reaction, A —› B, the rate would be can be described as the rate of appearance of B or the rate of disappearance of A, since one mole of A will produce to one mole of B.

12. Reaction Rates • For a simple reaction, A —› B, the rate would be expressed as either Δ[B]/ Δt or -Δ[A]/ Δt • Rates are always expressed as positive quantities, therefore, a negative sign must be used because the [A] decreases as the reaction progresses

13. A reaction vessel initially contains 1.00 M of reactant A. Twenty seconds later, the reaction vessel contains only 0.54 M of reactant A and 0.46 M of product B. Calculate the average rate of the disappearance of reactant A. • Recall, A —› B, and rate = -Δ[A]/ Δt . • Rate = -[0.54M – 1.00M]/(20s – 0s) • Rate = 0.46M/20s • Rate = 2.3 x 10-2 M/s

14. Reaction Rates and Stoichiometry • What if the reaction stoichiometry is more complicated than our simple 1:1 ratio? • Let’s look at 2HI —› H2 + I2 • Rate = -½ Δ[HI]/Δt = Δ[H2]/ Δt = [I2]/ Δt

15. Reaction Rates and Stoichiometry • In general, for the reaction aA + bB —› cC + dD the rate is given by Rate= -1/aΔ[A]/Δt = -1/bΔ[B]/Δt = 1/cΔ[C]/ Δt = 1/dΔ[D]/ Δt

16. How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction 2O3—› 3O2? Rate = -1/2Δ[O3]/Δt = 1/3Δ[O2]/Δt If the rate at which O2 appears is 6.0 x 10-5 M/s at a particular instant, at what rate is O3 disappearing at the same time? Rate = -1/2Δ[O3]/Δt = 1/3Δ[O2]/Δt Rate = -1/2Δ[O3]/Δt = 1/3(6.0 x 10-5 M/s) Rate = -Δ[O3]/Δt = 2/3(6.0 x 10-5 M/s) Rate = -Δ[O3]/Δt = 4.0 x 10-5 M/s

17. You should now be able to complete # 11 & 12

18. Rate Laws • The rate law is an equation that relates the rate of a reaction to the concentration of reactants (and catalysts) raised to various powers • The rate law for our last example, aA + bB —› cC + dD, would be expressed as Rate = k[A]m[B]n, where k is the rate constant and m and n are typically small, whole numbers.

19. Rate Laws • k in the rate law is the rate constant defined as a proportionality constant in the relationship between rate and concentrations. • The rate constant is fixed at a particular temperature (but will change as the temperature changes)

20. Rate Laws We can classify reactions by their orders • The reaction order, with respect to a given reactant species, equals the exponent of the concentration of that species in the rate law, as determined experimentally • The overall order of the reaction equals the sum of the orders of all reactant species in the rate law • In the case Rate = k[A]m[B]n, the overall order of this reaction is given by m+n.

21. Rate Laws • For the following reaction, the rate law can be written as is indicated below: NH4+ + NO2-—› N2 + 2H2O Rate = k[NH4+][NO2-] • Because the exponent of[NH4+] is one, the rate in NH4+ is first order. The rate in [NO2-] is also first order. The overall reaction order is second order. • The exponents in a rate law indicate how the rate is affected by the concentration of each reactant.

22. For the following reaction, the rate law can be written as is indicated below:NH4+ + NO2-—› N2 + 2H2O Rate = k[NH4+][NO2-], What is the effect on the rate if [NO2-] doubles? Because the exponent of [NO2-] in the rate law is one, doubling its concentration will double the rate. What is the effect on the rate if [NH4+] triples? Because the exponent of [NH4+] in the rate law is one, tripling its concentration will triple the rate.

23. For a given reaction, the rate law is given by Rate = k[NO2]2 What is the order of the reaction with respect to NO2? Because the exponent of NO2 in the equation is 2, the reaction is second order with respect to NO2. Because the concentration of only one reactant is expressed in the rate law, the overall order is also second. What is the effect on the rate if [NO2] triples? If the concentration triples, then the rate increases by 9 since [3]2.

24. Rate Laws: Additional Examples 2N2O5 —›4NO2 + O2 Rate = k[N2O5] CHCl3 + Cl2—› CCl4 + HCl Rate = k[CHCl3][Cl2]½ H2 + I2 —› 2HI Rate = k[H2][I2] Note: While exponents in rate laws are often the same as the coefficients in a balanced equation, this is not a rule! Exponents in a rate law can only be determined experimentally!

25. Determining a Rate Law Experimentally – Initial Rate Method Using the data above, determine the rate law, calculate the value of k and determine the rate when [NO] = 0.050M and [H2] = 0.150 M.

26. Determining a Rate Law Experimentally – Initial Rate Method A. Determine the rate law Recall, the Rate = k[NO]x[H2]y. We can determine the values of x and y by using the experimental data and the equation and solving for the unknown variables.

27. Determining a Rate Law Experimentally – Initial Rate Method Let’s compare trial two to trial one: Rate 2 = [NO]x[H2]y = [0.10]x[.20]y Rate 1 = [NO]x[H2]y = [0.10]x[.10]y We can simplify our expression and substitute the values for Rate 1 and Rate 2 Rate 2 = 2.46 x 10-3 = [.20]y Rate 1 = 1.23 x 10-3 = [.10]y Notice, we are only left with one unknown variable to solve for.

28. Determining a Rate Law Experimentally – Initial Rate Method Rate 2 = 2.46 x 10-3 = [.20]y Rate 1 = 1.23 x 10-3 = [.10]y 2y = 2 y must equal 1. So far we have determined that Rate = k[NO]x[H2]1. We must repeat the process to solve for x by comparing trial 3 to trial 1.

29. Determining a Rate Law Experimentally – Initial Rate Method Rate 3 = [NO]x[H2]y = [0.20]x[.10]y Rate 1 = [NO]x[H2]y = [0.10]x[.10]y Rate 3 = 4.92 x 10-3 = [.20]x Rate 1 = 1.23 x 10-3 = [.10]x 2x = 4 X must equal 2. Therefore, the experimentally determined rate law is Rate = k[NO]2[H2]1

30. Determining a Rate Law Experimentally – Initial Rate Method B. Calculate the value of k If Rate = k[NO]2[H2]1, then simple substitution from the data table will allow us to calculate k. Rate = k[NO]2[H2]1 (use trial 1) 1.23 x 10-3M/s = k[0.10M]2[0.10M]1 1.23 x 10-3M/s = k(0.0010M3) 1.2 M-2·s-1 = k

31. Determining a Rate Law Experimentally – Initial Rate Method C. Determine the rate when [NO] = 0.050M and [H2] = 0.150 M SinceRate = 1.2M-2·s-1[NO]2[H2]1 Rate = 1.2M-2·s-1[0.050M]2[0.150M]1 Rate = 4.5 x 10-4 M/s

32. You should now be able to complete # 13, 14, 15, 20, 21, 25

33. Calculated from a graph of experimental data that represents concentration vs. time. Instantaneous Rates

34. Recall, the slope of a graph is Δy/Δx. In a concentration vs. time graph, the Δy = Δ[reactant] and Δx = Δtime. Therefore, the slope of the graph = Δ[reactant]/ Δtime and will yield the instantaneous rate of reaction progression at any given point in time. Average reaction rates can be determined from the average of several calculated instantaneous rates. Instantaneous Rates

35. Rate Orders Recall, the overall order of a reaction is equal to the sum of the exponents in a rate law Rate = k[A]0 or Rate = k is zero order Rate = k[A]1 is first order Rate = k[A]2 or Rate = k[A][B] is second order Rate = k[A]3 or Rate = k[A]2[B] is third order

36. A closer look at Zero-order Reactions Rate = k[A]0 or Rate = k is zero order This is referred to as the differential rate law. If we integrate the above equation, we find a very useful equation: [A] = -kt + [A0] This will allow us to solve for unknown concentration values of zero-order reactions

37. A closer look at Zero-order Reactions A zero-order reaction begins with [X] = 0.50 M. How long does it take to react all but 0.10M if k=0.0410 M/min? What is the half-life? What is the [X] after 5 min?

38. A closer look at Zero-order Reactions A. [X] = 0.10 M, [X0] = 0.50 M and k=0.0410 M/min [A] = -kt + [A0] [0.10M] = -(0.0410M/min)t + [0.50M] -0.40M = -(0.0410M/min)t 10 min = t

39. A closer look at Zero-order Reactions B. What is the half-life? Recall, a half-life is the time it takes for half a reactant to be consumed. So, if [X0] = 0.50 M, then[X] must equal 0.25 M If [A] = -kt + [A0], then [0.25M] = -(0.0410M/min)t½ + [0.50M] -0.25M = -(0.0410M/min)t½ 6.1 min = t ½

40. A closer look at Zero-order Reactions C. What is the [X] after 5 min? If [A] = -kt + [A0], then [X] = -(0.0410M/min)5min + [0.50M] [X] = -.205M + 0.50M [X] = 0.3 M

41. A closer look at Zero-order Reactions Notice, [A] = -kt + [A0] closely mimics y = mx + b Therefore, a graph of change in concentration vs. time will be linear for a zero-order reaction.

42. A closer look at First-order Reactions Rate = k[A]1 is first order This is referred to as the differential rate law. If we integrate the above equation, we find a the integrated rate law for first order equations: ln[At] - ln[A0] = -kt Or ln[At] = -kt + ln[A0] (y) = (mx) = b This will allow us to solve for unknown concentration values of first-order reactions

43. A closer look at Second-order Reactions Rate = k[A]2 or is second order This is referred to as the differential rate law. If we integrate the above equation, we find a the integrated rate law for second order equations: 1/[At] = kt + 1/[A0] (y) = (mx) + (b) This will allow us to solve for unknown concentration values of second-order reactions

44. Determining Order From Graphical Data An experiment was conducted were a reaction was allowed to proceed and the reactant concentration was measured over a period of time.What is the order of this reaction?

45. Determining Order From Graphical Data Recall, the integrated rate law of zero, first and second order reactions can be rearranged in the form of y=mx + b Zero-order: [A] = -kt + [A0] First-order: ln[At] = -kt + ln[A0] Second-order: 1/[At] = kt + 1/[A0] If we create three plots of our data, we can determine order: [X] vs. t; ln[X] vs. t; 1/[X] vs. t. The plot that yield a linear graph will determine its order.

46. Determining Order From Graphical Data Notice, only one graph yields a straight line: The plot of ln[X] vs. t. Therefore, the reaction must be first order.

47. You should now be able to complete # 29, 33, 38

48. A Microscopic View of Reaction Rates Consider the reaction NO + O3 —› NO2 + O2 where Rate = k[NO][O3] Imagine these reactants in rapid, random motion inside a flask. They strike the walls of the reaction vessel and collide with other molecules. Will all collisions result in a chemical reaction? What conditions must be met for these reactants to react together?

49. How a Chemical Reaction Takes Place • Only a tiny fraction of reactant collisions leads to reaction. Why? • Reactant molecules must collide in the proper orientation. • They must collide with the minimum energy required to initiate a chemical reaction. This is called activation energy (Ea) and the value varies from reaction to reaction.

50. Activation Energy The diagrams shows the change in potential energy of the molecules during the reaction. Energy must be supplied to the reactants to stretch and break reactant bonds into an intermediate form (represented at the peak of the graph) before proceeding to the final product. The energy difference between the starting molecule and the highest energy along the path represent the activation energy. The difference in the potential energies of the reactants and products indicate whether the reaction was endo- or exothermic.