Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 1 of Theoretical Models for Chemical Kinetics  Kinetic-Molecular theory can be used to calculate the collision frequency.  In gases collisions per second.  If each collision produced a reaction, the rate would be about 10 6 M s -1.  Actual rates are on the order of 10 4 M s -1. ◦Still a very rapid rate.  Only a fraction of collisions yield a reaction. Collision Theory

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 2 of 61 Activation Energy  For a reaction to occur there must be a redistribution of energy sufficient to break certain bonds in the reacting molecule(s).  Activation Energy:  The minimum energy above the average kinetic energy that molecules must bring to their collisions for a chemical reaction to occur.

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 3 of 61 Activation Energy

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 4 of 61 Kinetic Energy

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 5 of 61 Collision Theory  If activation barrier is high, only a few molecules have sufficient kinetic energy and the reaction is slower.  As temperature increases, reaction rate increases.  Orientation of molecules may be important.

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 6 of 61 Collision Theory

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 7 of 61 Transition State Theory  The activated complex is a hypothetical species lying between reactants and products at a point on the reaction profile called the transition state.

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 8 of Effect of Temperature on Reaction Rates  Svante Arrhenius demonstrated that many rate constants vary with temperature according to the equation: k = Ae -E a /RT ln k = + lnA R -Ea T 1

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 9 of 61 Arrhenius Plot N 2 O 5 (CCl 4 ) → N 2 O 4 (CCl 4 ) + ½ O 2 (g) = -1.2  10 4 K R -Ea-Ea -E a = 1.0  10 2 kJ mol -1

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 10 of 61 Arrhenius Equation k = Ae -E a /RT ln k = + ln A R -Ea-Ea T 1 ln k 2 – ln k 1 = + ln A - - ln A R -Ea-Ea T2T2 1 R -Ea-Ea T1T1 1 ln = - R -Ea-Ea T2T2 1 k2k2 k1k1 T1T1 1

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 11 of Reaction Mechanisms  A step-by-step description of a chemical reaction.  Each step is called an elementary process.  Any molecular event that significantly alters a molecules energy of geometry or produces a new molecule.  Reaction mechanism must be consistent with:  Stoichiometry for the overall reaction.  The experimentally determined rate law.

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 12 of 61 Elementary Processes  Unimolecular or bimolecular.  Exponents for concentration terms are the same as the stoichiometric factors for the elementary process.  Elementary processes are reversible.  Intermediates are produced in one elementary process and consumed in another.  One elementary step is usually slower than all the others and is known as the rate determining step.

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 13 of 61 Slow Step Followed by a Fast Step H 2 (g) + 2 ICl(g) → I 2 (g) + 2 HCl(g) dtdt = k[H 2 ][ICl] d[P] Postulate a mechanism: H 2 (g) + 2 ICl(g) → I 2 (g) + 2 HCl(g) slow H 2 (g) + ICl(g) HI(g) + HCl(g) fast HI(g) + ICl(g) I 2 (g) + HCl(g) dtdt = k[H 2 ][ICl] d[HI] dtdt = k[HI][ICl] d[I 2 ] dtdt = k[H 2 ][ICl] d[P]

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 14 of 61 Slow Step Followed by a Fast Step

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 15 of 61 Fast Reversible Step Followed by a Slow Step 2NO(g) + O 2 (g) → 2 NO 2 (g) dt = -k obs [NO 2 ] 2 [O 2 ] d[P] Postulate a mechanism: dtdt = k 2 [N 2 O 2 ][O 2 ] d[NO 2 ] fast 2NO(g) N 2 O 2 (g) k1k1 k -1 slow N 2 O 2 (g) + O 2 (g) 2NO 2 (g) k2k2 dtdt = k 2 [NO] 2 [O 2 ] d[I 2 ] k -1 k1k1 2NO(g) + O 2 (g) → 2 NO 2 (g) K = k -1 k1k1 = [NO] [N 2 O 2 ] = K[NO] 2 k -1 k1k1 = [NO] 2 [N 2 O 2 ]

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 16 of 61 Catalytic Converters  Dual catalyst system for oxidation of CO and reduction of NO. CO 2 + N 2 CO + NO cat

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 17 of Catalysis

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 18 of 61  Worked Examples Follow:

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 19 of 61

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 20 of 61

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 21 of 61

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 22 of 61  CRS Questions Follow:

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 23 of 61 Energy reaction coordinate A+B X+Y Energy reaction coordinate A+B X+Y Energy reaction coordinate A+B X+Y Energy reaction coordinate A+B X+Y The reaction between A and B is determined to be a fairly fast reaction and slightly exothermic. Which of the following potential energy surfaces fit this description?

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 24 of 61 Energy reaction coordinate A+B X+Y Energy reaction coordinate A+B X+Y Energy reaction coordinate A+B X+Y The reaction between A and B is determined to be a fairly fast reaction and only slightly exothermic. Which of the following potential energy surfaces fit this description? Energy reaction coordinate A+B X+Y 1.

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 25 of 61 Energy reaction coordinate 1.  H = 60 kJ mol -1 R P 2.  H = -60 kJ mol  H = 140 kJ mol  H = -80 kJ mol  H = 80 kJ mol -1 A particular reaction was found to have forward and reverse activation energies of 60 and 140 kJ mol -1, respectively. The enthalpy change for the reaction is, (do not use a calculator)

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 26 of 61 Energy reaction coordinate 1.  H = 60 kJ mol -1 R P A particular reaction was found to have forward and reverse activation energies of 60 and 140 kJ mol -1, respectively. The enthalpy change for the reaction is, (do not use a calculator) 2.  H = -60 kJ mol  H = 140 kJ mol  H = -80 kJ mol  H = 80 kJ mol -1

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 27 of 61 Energy reaction coordinate A+B X+Y Energy reaction coordinate 1. A+B X+Y Energy reaction coordinate A+B X+Y Energy reaction coordinate A+B X+Y In which diagram to the right does the dashed line best represent the catalyzed version of the reaction’s potential energy profile?

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 28 of 61 In which diagram to the right does the dashed line best represent the catalyzed version of the reaction’s potential energy profile? Energy reaction coordinate A+B X+Y Energy reaction coordinate 1. A+B X+Y Energy reaction coordinate A+B X+Y Energy reaction coordinate A+B X+Y

Prentice-Hall © 2007 General Chemistry: Chapter 14 Slide 29 of 61  Textbook End of Chapter ?’s:  P.611- #1, 3, 11, 13, 17, 19,  21, 33, 47, 51, 55, 100, 101,  102, 103, 104, 105