2. Chapter 1411 Copyright © by Houghton Mifflin Company. All rights reserved. Suggested Problems Ch 14 27, 28, 31, 33, 35, 41, 43, 45, 49, 53, 63, 67, 75, 77, 79, 91, 99, 101, 103, 109

3. Chapter 1422 Copyright © by Houghton Mifflin Company. All rights reserved. Figure 14.21: Enzyme action (lock-and-key model). Return to Slide 118

4. Chapter 1433 Copyright © by Houghton Mifflin Company. All rights reserved.

5. Chapter 1444 Copyright © by Houghton Mifflin Company. All rights reserved. Concentration-Time Equations First-Order Rate Law Using calculus, you get the following equation. ln 1/2 [A] o = kt 1/2 [A] o  [A] = 1/2 [A] o

6. Chapter 1455 Copyright © by Houghton Mifflin Company. All rights reserved. Half Life First-Order Rate Law You could write the rate law in the form  [A] = 1/2 {A] o 1/2 {A] o k[A] o =  t

7. Chapter 1466 Copyright © by Houghton Mifflin Company. All rights reserved. Half-life The half-life of a reaction is the time required for the reactant concentration to decrease to one-half of its initial value. For a first-order reaction, the half-life is independent of the initial concentration of reactant. In one half-life the amount of reactant decreases by one-half. Substituting into the first-order concentration-time equation, we get:

8. Chapter 1477 Copyright © by Houghton Mifflin Company. All rights reserved. Half-life The half-life of a reaction is the time required for the reactant concentration to decrease to one-half of its initial value. Solving for t 1/2 we obtain: Figure 14.8 illustrates the half-life of a first-order reaction.

9. Chapter 1488 Copyright © by Houghton Mifflin Company. All rights reserved. Figure 14.8: A graph illustrating that the half-life of a first-order reaction is independent of initial concentration. Half life, t 1/2, is the time it takes for the [R] to decrease by 1/2. This is exactly like radioactive decay.

10. Chapter 1499 Copyright © by Houghton Mifflin Company. All rights reserved. Half-life Sulfuryl chloride, SO 2 Cl 2, decomposes in a first- order reaction to SO 2 and Cl 2. At 320 o C, the rate constant is 2.2 x 10 -5 s -1. What is the half-life of SO 2 Cl 2 vapor at this temperature? Substitute the value of k into the relationship between k and t 1/2.

11. Chapter 1410 Copyright © by Houghton Mifflin Company. All rights reserved. A model of SO 2 Cl 2 (g) Hybridizations of S??

12. Chapter 1411 Copyright © by Houghton Mifflin Company. All rights reserved. Half-life Sulfuryl chloride, SO 2 Cl 2, decomposes in a first- order reaction to SO 2 and Cl 2. At 320 o C, the rate constant is 2.20 x 10 -5 s -1. What is the half-life of SO 2 Cl 2 vapor at this temperature? Substitute the value of k into the relationship between k and t 1/2.

13. Chapter 1412 Copyright © by Houghton Mifflin Company. All rights reserved. Half-life Sulfuryl chloride, SO 2 Cl 2, decomposes in a first- order reaction to SO 2 and Cl 2. At 320 o C, the rate constant is 2.20 x 10 -5 s -1. What is the half-life of SO 2 Cl 2 vapor at this temperature? Substitute the value of k into the relationship between k and t 1/2.

14. Chapter 1413 Copyright © by Houghton Mifflin Company. All rights reserved. Half-life For a second-order reaction, half-life depends on the initial concentration and becomes larger as time goes on. Again, assuming that [A] t = ½[A] o after one half- life, it can be shown that: Each succeeding half-life is twice the length of its predecessor.

15. Chapter 1414 Copyright © by Houghton Mifflin Company. All rights reserved. Graphing Kinetic Data In addition to the method of initial rates, rate laws can be deduced by graphical methods. If we rewrite the first-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line. y = mx + b

16. Chapter 1415 Copyright © by Houghton Mifflin Company. All rights reserved. Graphing Kinetic Data In addition to the method of initial rates, rate laws can be deduced by graphical methods. If we rewrite the first-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line. This means if you plot ln[A] versus time, you will get a straight line for a first-order reaction. (see Figure 14.9)(see Figure 14.9)

17. Chapter 1416 Copyright © by Houghton Mifflin Company. All rights reserved. Figure 14.9: A plot of log [R] versus time.

18. Chapter 1417 Copyright © by Houghton Mifflin Company. All rights reserved. Figure 14.9: A plot of log [R] versus time.

19. Chapter 1418 Copyright © by Houghton Mifflin Company. All rights reserved. Figure 14.9: A plot of log [R] versus time.

20. Chapter 1419 Copyright © by Houghton Mifflin Company. All rights reserved. Figure 14.9: A plot of log [R] versus time.

21. Chapter 1420 Copyright © by Houghton Mifflin Company. All rights reserved. Graphing Kinetic Data In addition to the method of initial rates, rate laws can be deduced by graphical methods. If we rewrite the second-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line. y = mx + b

22. Chapter 1421 Copyright © by Houghton Mifflin Company. All rights reserved. Graphing Kinetic Data In addition to the method of initial rates, rate laws can be deduced by graphical methods. If we rewrite the second-order concentration-time equation in a slightly different form, it can be identified as the equation of a straight line. This means if you plot 1/[A] versus time, you will get a straight line for a second-order reaction. Figure 14.10 illustrates the graphical method of deducing the order of a reaction.

23. Chapter 1422 Copyright © by Houghton Mifflin Company. All rights reserved. Collision Theory Rate constants vary with temperature. Consequently, the actual rate of a reaction is very temperature dependent. Why the rate depends on temperature can by explained by collision theory.

24. Chapter 1423 Copyright © by Houghton Mifflin Company. All rights reserved. Collision Theory Collision theory assumes that for a reaction to occur, reactant molecules must collide with sufficient energy and the proper orientation. The minimum energy of collision required for two molecules to react is called the activation energy, E a.

25. Chapter 1424 Copyright © by Houghton Mifflin Company. All rights reserved. Figure 14.11: Two test tubes containing the same solution, one in water. Photo courtesy of American Color. KMnO 4 + H 2 C 2 O 4 (oxalic acid)

26. Chapter 1425 Copyright © by Houghton Mifflin Company. All rights reserved. Figure 14.11: One test tube shows a reaction while the other does not. Photo courtesy of American Color. KMnO 4 + H 2 C 2 O 4 (oxalic acid)  CO 2 + H 2 O

27. Chapter 1426 Copyright © by Houghton Mifflin Company. All rights reserved. Transition-State Theory Transition-state theory explains the reaction resulting from the collision of two molecules in terms of an activated complex. An activated complex (transition state) is an unstable grouping of atoms that can break up to form products. A simple analogy would be the collision of three billiard balls on a billiard table.

28. Chapter 1427 Copyright © by Houghton Mifflin Company. All rights reserved. Transition-State Theory Transition-state theory explains the reaction resulting from the collision of two molecules in terms of an activated complex. Suppose two balls are coated with a slightly stick adhesive. We’ll take a third ball covered with an extremely sticky adhesive and collide it with our joined pair.

29. Chapter 1428 Copyright © by Houghton Mifflin Company. All rights reserved. Transition-State Theory Transition-state theory explains the reaction resulting from the collision of two molecules in terms of an activated complex. The “incoming” billiard ball would likely stick to one of the joined spheres and provide sufficient energy to dislodge the other, resulting in a new “pairing.” At the instant of impact, when all three spheres are joined, we have an unstable transition- state complex.

30. Chapter 1429 Copyright © by Houghton Mifflin Company. All rights reserved. Transition-State Theory Transition-state theory explains the reaction resulting from the collision of two molecules in terms of an activated complex. If we repeated this scenario several times, some collisions would be successful and others (because of either insufficient energy or improper orientation) would not be successful. We could compare the energy we provided to the billiard balls to the activation energy, E a.

31. Chapter 1430 Copyright © by Houghton Mifflin Company. All rights reserved. Figure 14.12: Importance of molecular orientation in the reaction of NO and CI 2.

32. Chapter 1431 Copyright © by Houghton Mifflin Company. All rights reserved. Potential-Energy Diagrams for Reactions To illustrate graphically the formation of a transition state, we can plot the potential energy of a reaction versus time. Figure 14.13 illustrates the endothermic reaction of nitric oxide and chlorine gas. Note that the forward activation energy is the energy necessary to form the activated complex. The  H of the reaction is the net change in energy between reactants and products.

33. Figure 14.13 Potential-energy curve for the endothermic reaction of nitric oxide and chlorine.

34. Chapter 1433 Copyright © by Houghton Mifflin Company. All rights reserved. Potential-Energy Diagrams for Reactions The potential-energy diagram for an exothermic reaction shows that the products are more stable than the reactants. Figure 14.14 illustrates the potential-energy diagram for an exothermic reaction. We see again that the forward activation energy is required to form the transition-state complex. In both of these graphs, the reverse reaction must still supply enough activation energy to form the activated complex.

35. Figure 14.14 Potential-energy curve for an exothermic reaction.

36. Chapter 1435 Copyright © by Houghton Mifflin Company. All rights reserved. Collision Theory and the Arrhenius Equation Collision theory maintains that the rate constant for a reaction is the product of three factors. 1.Z, the collision frequency 2.f, the fraction of collisions with sufficient energy to react 3.p, the fraction of collisions with the proper orientation to react

37. Chapter 1436 Copyright © by Houghton Mifflin Company. All rights reserved. Collision Theory and the Arrhenius Equation Z is only slightly temperature dependent. This is illustrated using the kinetic theory of gases, which shows the relationship between the velocity of gas molecules and their absolute temperature. or

38. Chapter 1437 Copyright © by Houghton Mifflin Company. All rights reserved.

39. Chapter 1438 Copyright © by Houghton Mifflin Company. All rights reserved. Collision Theory and the Arrhenius Equation Z is only slightly temperature dependent. This alone does not account for the observed increases in rates with only small increases in temperature. From kinetic theory, it can be shown that a 10 o C rise in temperature will produce only a 2% rise in collision frequency.

40. Chapter 1439 Copyright © by Houghton Mifflin Company. All rights reserved. Collision Theory and the Arrhenius Equation On the other hand, f, the fraction of molecules with sufficient activation energy, turns out to be very temperature dependent. It can be shown that f is related to E a by the following expression. Here e = 2.718…, and R is the ideal gas constant, 8.31 J/(mol. K).

41. Chapter 1440 Copyright © by Houghton Mifflin Company. All rights reserved. Collision Theory and the Arrhenius Equation On the other hand, f, the fraction of molecules with sufficient activation energy turns out to be very temperature dependent. From this relationship, as temperature increases, f increases. Also, a decrease in the activation energy, E a, increases the value of f.

42. Chapter 1441 Copyright © by Houghton Mifflin Company. All rights reserved. Collision Theory and the Arrhenius Equation On the other hand, f, the fraction of molecules with sufficient activation energy turns out to be very temperature dependent. This is the primary factor relating temperature increases to observed rate increases.

43. Chapter 1442 Copyright © by Houghton Mifflin Company. All rights reserved. Collision Theory and the Arrhenius Equation The reaction rate also depends on p, the fraction of collisions with the proper orientation. This factor is independent of temperature changes. So, with changes in temperature, Z and p remain fairly constant. We can use that fact to derive a mathematical relationship between the rate constant, k, and the absolute temperature.

44. Chapter 1443 Copyright © by Houghton Mifflin Company. All rights reserved. The Arrhenius Equation If we were to combine the relatively constant terms, Z and p, into one constant, let’s call it A. We obtain the Arrhenius equation: The Arrhenius equation expresses the dependence of the rate constant on absolute temperature and activation energy.

45. Chapter 1444 Copyright © by Houghton Mifflin Company. All rights reserved. The Arrhenius Equation It is useful to recast the Arrhenius equation in logarithmic form. Taking the natural logarithm of both sides of the equation, we get:

46. Chapter 1445 Copyright © by Houghton Mifflin Company. All rights reserved. The Arrhenius Equation It is useful to recast the Arrhenius equation in logarithmic form. We can relate this equation to the (somewhat rearranged) general formula for a straight line. y = b + m x A plot of ln k versus (1/T) should yield a straight line with a slope of (-E a /R) and an intercept of ln A. (see Figure 14.15)(see Figure 14.15)

47. Chapter 1446 Copyright © by Houghton Mifflin Company. All rights reserved. Figure 14.15: Plot of ln k versus 1/T.

48. Chapter 1447 Copyright © by Houghton Mifflin Company. All rights reserved. The Arrhenius Equation A more useful form of the equation emerges if we look at two points on the line this equation describes that is, (k 1, (1/T 1 )) and (k 2, (1/T 2 )). The two equations describing the relationship at each coordinate would be and

49. Chapter 1448 Copyright © by Houghton Mifflin Company. All rights reserved. The Arrhenius Equation A more useful form of the equation emerges if we look at two points on the line this equation describes that is, (k 1, (1/T 1 )) and (k 2, (1/T 2 )). We can eliminate ln A by subtracting the two equations to obtain