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# Chemical Kinetics

Chemical Kinetics. The area of chemistry that concerns reaction rates and reaction mechanisms . Reaction Rate. The change in concentration of a reactant or product per unit of time. 2NO 2 (g)  2NO(g) + O 2 (g) . Reaction Rates:. 1. Can measure disappearance of reactants. Download Presentation ## Chemical Kinetics

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1. Chemical Kinetics The area of chemistry that concernsreaction rates and reaction mechanisms.

2. Reaction Rate The change in concentration of a reactant or product per unit of time

3. 2NO2(g)  2NO(g) + O2(g) Reaction Rates: 1. Can measure disappearance of reactants 2. Can measure appearance of products 3. Are proportional stoichiometrically

4. 2NO2(g)  2NO(g) + O2(g) Reaction Rates: 4. Are equal to the slope tangent to that point 5. Change as the reaction proceeds, if the rate is dependent upon concentration [NO2] t

5. Rate Laws Differential rate lawsexpress (reveal) the relationship between the concentration of reactants and the rate of the reaction. The differential rate law is usually just called “the rate law.” Integrated rate lawsexpress (reveal) the relationship between concentration of reactants and time

6. Writing a (differential) Rate Law Problem- Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction: 2 NO(g) + Cl2(g)  2 NOCl(g)

7. Writing a Rate Law Part 1– Determine the values for the exponents in the rate law: R = k[NO]x[Cl2]y In experiment 1 and 2, [Cl2] is constant while [NO] doubles. The rate quadruples, so the reaction is second order with respect to [NO]  R = k[NO]2[Cl2]y

8. Writing a Rate Law Part 1– Determine the values for the exponents in the rate law: R = k[NO]2[Cl2]y In experiment 2 and 4, [NO] is constant while [Cl2] doubles. The rate doubles, so the reaction is first order with respect to [Cl2]  R = k[NO]2[Cl2]

9. Writing a Rate Law Part 2– Determine the value for k, the rate constant, by using any set of experimental data: R = k[NO]2[Cl2]

10. Writing a Rate Law Part 3– Determine the overall order for the reaction. R = k[NO]2[Cl2] = 3 2 + 1  The reaction is 3rd order Overall order is the sum of the exponents, or orders, of the reactants

11. Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why?

12. Collision Model Collisions must havesufficient energyto produce the reaction (must equal or exceed the activation energy). 1. Colliding particles must becorrectly orientedto one another in order to produce a reaction. 2.

13. Factors Affecting Rate Increasing temperature always increases the rate of a reaction. • Particles collide more frequently • Particles collide more energetically Increasing surface areaincreases the rate of a reaction Increasing ConcentrationUSUALLY increases the rate of a reaction Presence of Catalysts, which lower the activation energy by providing alternate pathways

14. Endothermic Reactions

15. Exothermic Reactions

16. Catalysis • Catalyst: A substance that speeds up a reaction without being consumed • Enzyme: A large molecule (usually a protein) that catalyzes biological reactions. • Homogeneous catalyst: Present in the same phase as the reacting molecules. • Heterogeneous catalyst: Present in a different phase than the reacting molecules.

17. Lowering of Activation Energy by a Catalyst

18. Catalysts Increase the Number of Effective Collisions

19. Reaction Mechanism The reaction mechanism is the series of elementary stepsby which a chemical reaction occurs. • The sum of the elementary steps must give the overall balanced equation for the reaction • The mechanism must agree with the experimentally determined rate law

20. Rate-Determining Step In a multi-step reaction, the slowest stepis the rate-determining step. It therefore determines the rate of the reaction. The experimental rate law must agree with the rate-determining step

21. Identifying the Rate-Determining Step For the reaction: 2H2(g) + 2NO(g)  N2(g) + 2H2O(g) The experimental rate law is: R = k[NO]2[H2] Which step in the reaction mechanism is the rate-determining (slowest) step? Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g) Step #2 N2O(g) + H2(g)  N2(g) + H2O(g) Step #1 agrees with the experimental rate law

22. Identifying Intermediates For the reaction: 2H2(g) + 2NO(g)  N2(g) + 2H2O(g) Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?) Step #1 H2(g) + 2NO(g)  N2O(g) + H2O(g) Step #2 N2O(g) + H2(g)  N2(g) + H2O(g) 2H2(g) + 2NO(g)  N2(g) + 2H2O(g)  N2O(g) is an intermediate

23. Consider the hypothetical reaction shown below. Overall reaction: 2A + C2 A2C + C Reaction mechanism: Elementary step 1 A + C2 AC + C slow Elementary step 2 AC + A A2C fast What is the rate Law? Rate = k [A]1[C2]1

24. Consider the mechanism Step 1 ClO-(aq) + H2O(l) <-> HOCl(aq) + OH-(aq) K, fast Step 2 I-(aq) + HOCl(aq)  HOI(aq) + Cl-(aq) k1, slow Step 3 HOI(aq) + OH-(aq)  OI-(aq) + H2O(l) k2, fast Overall ClO- + I- OI- + Cl- What is the rate law? You might think it is rate =k[I-] [HOCl] [HOCl] is an intermediate though and therefore cannot be in the rate law as we do not know its concentration. We therefore need to substitute for it. rate = K [ClO-] [I-] [OH-] -1

25. Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why?

26. Collision Model Collisions must havesufficient energyto produce the reaction (must equal or exceed the activation energy). 1. Colliding particles must becorrectly orientedto one another in order to produce a reaction. 2.

27. Factors Affecting Rate Increasing temperature always increases the rate of a reaction. • Particles collide more frequently • Particles collide more energetically Increasing surface areaincreases the rate of a reaction Increasing ConcentrationUSUALLY increases the rate of a reaction Presence of Catalysts, which lower the activation energy by providing alternate pathways

28. Endothermic Reactions

29. Exothermic Reactions

30. Catalysis • Catalyst: A substance that speeds up a reaction without being consumed • Enzyme: A large molecule (usually a protein) that catalyzes biological reactions. • Homogeneous catalyst: Present in the same phase as the reacting molecules. • Heterogeneous catalyst: Present in a different phase than the reacting molecules.

31. Lowering of Activation Energy by a Catalyst

32. Catalysts Increase the Number of Effective Collisions

33. Not Tested or taught, but could show up on AP Test Determining Order withConcentration vs. Time data (the Integrated Rate Law) Zero Order: First Order: Second Order:

34. Not Tested or taught, but could show up on AP Test Problem: Find the integrated rate law and the value for the rate constant, k A graphing calculator with linear regression analysis greatly simplifies this process!! (Click here to download my Rate Laws program for theTi-83 and Ti-84)

35. Not Tested or taught, but could show up on AP Test Time vs. [H2O2] Regression results: y = ax + b a = -2.64 x 10-4 b = 0.841 r2 = 0.8891 r = -0.9429

36. Not Tested or taught, but could show up on AP Test Time vs. ln[H2O2] Regression results: y = ax + b a = -8.35 x 10-4 b = -.005 r2 = 0.99978 r = -0.9999

37. Not Tested or taught, but could show up on AP Test Regression results: y = ax + b a = 0.00460 b = -0.847 r2 = 0.8723 r = 0.9340

38. Not Tested or taught, but could show up on AP Test And the winner is… Time vs. ln[H2O2] 1. As a result, the reaction is 1st order 2. The (differential) rate law is: 3. The integrated rate law is: 4. But…what is the rate constant, k ?

39. Not Tested or taught, but could show up on AP Test Method #1: Calculate the slope from the Time vs. ln[H2O2] table. Now remember:  k = -slope k = 8.32 x 10-4s-1

40. Not Tested or taught, but could show up on AP Test Finding the Rate Constant, k Method #2: Obtain k from the linear regresssion analysis. Regression results: y = ax + b a = -8.35 x 10-4 b = -.005 r2 = 0.99978 r = -0.9999 Now remember:  k = -slope k = 8.35 x 10-4s-1

41. Not Tested or taught, but could show up on AP Test Rate Laws Summary

42. Not Tested Heterogeneous Catalysis Carbon monoxide and nitrogen monoxide adsorbed on a platinum surface Step #1: Adsorption and activation of the reactants.

43. Not Tested Heterogeneous Catalysis Carbon monoxide and nitrogen monoxide arranged prior to reacting Step #2: Migration of the adsorbed reactants on the surface.

44. Not Tested Heterogeneous Catalysis Carbon dioxide and nitrogen form from previous molecules Step #3: Reaction of the adsorbed substances.

45. Not Tested Heterogeneous Catalysis Carbon dioxide and nitrogen gases escape (desorb) from the platinum surface Step #4: Escape, or desorption, of the products.

46. Not Tested, or taught, but could show up on AP Test The Arrhenius Equation • k = rate constant at temperature T • A = frequency factor • Ea = activation energy • R = Gas constant, 8.31451 J/K·mol

47. Not Tested, or taught, but could show up on AP Test The Arrhenius Equation, Rearranged • Simplifies solving for Ea • -Ea / R is the slope when (1/T) is plotted against ln(k) • ln(A) is the y-intercept • Linear regression analysis of a table of (1/T) vs. ln(k) can quickly yield a slope • Ea = -R(slope)

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