Chemical Kinetics CHAPTER 14 Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop

CHAPTER 14 Chemical Kinetics Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 2 Learning Objectives:  Factors Affecting Reaction Rate: o Concentration o State o Surface Area o Temperature o Catalyst  Collision Theory of Reactions and Effective Collisions  Determining Reaction Order and Rate Law from Data  Integrated Rate Laws  Rate Law  Concentration vs Rate  Integrated Rate Law  Concentration vs Time  Units of Rate Constant and Overall Reaction Order  Half Life vs Rate Constant (1 st Order)  Arrhenius Equation  Mechanisms and Rate Laws  Catalysts

CHAPTER 14 Chemical Kinetics Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 3 Lecture Road Map: ① Factors that affect reaction rates ② Measuring rates of reactions ③ Rate Laws ④ Collision Theory ⑤ Transition State Theory & Activation Energies ⑥ Mechanisms ⑦ Catalysts

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 4 Mechanisms of Reactions CHAPTER 14 Chemical Kinetics

Mechanisms Overall vs Individual Steps Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 5 Sometimes rate law has simple form – N 2 O 5  NO 2 + NO 3 – NO 2 + NO 3  N 2 O 5 But others are complex – H 2 + Br 2  2 HBr

Mechanisms Overall vs Individual Steps Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 6 Some reactions occur in a single step, as written Others involve a sequence of steps o Reaction Mechanism o Entire sequence of steps o Elementary Process o Each individual step in mechanism o Single step that occurs as written

Mechanisms Overall vs Individual Steps Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 7 o Exponents in rate law for elementary process are equal to coefficients of reactants in balanced chemical equation for that elementary process o Rate laws for elementary processes are directly related to stoichiometry o Number of molecules that participate in elementary process defines molecularity of step

Mechanisms Unimolecular Process Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 8 o Only one molecule as reactant o H 3 C—N  C  H 3 C—C  N o Rate = k[CH 3 NC] o 1 st order overall o As number of molecules increases, number that rearrange in given time interval increases proportionally

Mechanisms Bimolecular Process Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 9 o Elementary step with two reactants o NO(g) + O 3 (g)  NO 2 (g) + O 2 (g) o Rate = k[NO][O 3 ] o 2 nd order overall o From collision theory: o If [A] doubles, number of collisions between A and B will double o If [B] doubles, number of collisions between A and B will double o Thus, process is 1 st order in A, 1 st order in B, and 2 nd order overall

Mechanisms Termolecular Process Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 10 o Elementary reaction with three molecules o Extremely rare o Why? o Very low probability that three molecules will collide simultaneously o 3 rd order overall

Mechanisms Elementary Processes Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 11 MolecularityElementary StepRate Law Unimolecular A  products Rate = k[A] Bimolecular A + A  products Rate = k[A] 2 Bimolecular A + B  products Rate = k[A][B] Significance of elementary steps: o If we know that reaction is elementary step o Then we know its rate law

Mechanisms Multi-step Mechanisms Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 12 o Contains two or more steps to yield net reaction o Elementary processes in multi-step mechanism must always add up to give chemical equation of overall process o Any mechanism we propose must be consistent with experimentally observed rate law o Intermediate = species which are formed in one step and used up in subsequent steps o Species which are neither reactant nor product in overall reaction o Mechanisms may involve one or more intermediates

Mechanisms Example Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 13 The net reaction is: NO 2 (g) + CO(g)  NO(g) + CO 2 (g) The proposed mechanism is: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) 2NO 2 (g) + NO 3 (g) + CO(g)  NO 2 (g) + NO 3 (g) + NO(g) + CO 2 (g) or NO 2 (g) + CO(g)  NO(g) + CO 2 (g) 1

Mechanisms Rate Determining Step Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 14 o If process follows sequence of steps, slow step determines rate = rate determining step. o Think of an assembly line o Fast earlier steps may cause intermediates to pile up o Fast later steps may have to wait for slower initial steps o Rate-determining step governs rate law for overall reaction o Can only measure rate up to rate determining step

Mechanisms Example: Rate Determining Step Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 15 (CH 3 ) 3 CCl(aq) + OH – (aq)  (CH 3 ) 3 COH(aq) + Cl – (aq) chlorotrimethylmethane trimethylmethanol o Observed rate = k[(CH 3 ) 3 CCl] o If reaction was elementary o Rate would depend on both reactants o Frequency of collisions depends on both concentrations o  Mechanism is more complex than single step o What is mechanism? o Evidence that it is a two step process

Mechanisms Rate Determining Step as Initial Step Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 16 Step 1: (CH 3 ) 3 CCl(aq)  (CH 3 ) 3 C + (aq) + Cl – (aq) (slow) Step 2: (CH 3 ) 3 C + (aq) + OH – (aq)  (CH 3 ) 3 COH(aq) (fast) o Two steps each at different rates o Each step in multiple step mechanism is elementary process, so o Has its own rate constant and its own rate law o Hence only for each step can we write rate law directly o Observed rate law says that step 1 is very slow compared to step 2 o In this case step 1 is rate determining o Overall rate = k 1 [(CH 3 ) 3 CCl]

Mechanisms Mechanisms with Fast Initial Step Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 17 1 st step involves fast, reversible reaction Ex. Decomposition of ozone (No catalysts) Net reaction: 2O 3 (g)  3O 2 (g) Proposed mechanism: O 3 (g)  O 2 (g) + O(g) (fast) O(g) + O 3 (g)  2O 2 (g)(slow)

Mechanisms Is the Mechanism Rate Law Consistent? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 18 o Rate of formation of O 2 = Rate of reaction 2 = k 2 [O][O 3 ] o But O is intermediate o Need rate law in terms of reactants and products o and possibly catalysts o Rate (forward) = k f [O 3 ] o Rate (reverse) = k r [O 2 ][O] o When step 1 comes to equilibrium o Rate (forward) = Rate (reverse) o k f [O 3 ] = k r [O 2 ][O]

Mechanisms Is the Mechanism Rate Law Consistent? Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 19 o Solving this for intermediate O gives: o Substitution into rate law for step 2 gives: o Rate of reaction 2 = k 2 [O][O 3 ] = o where o This is observed rate law o Yes, mechanism consistent

Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 20 The reaction mechanism that has been proposed for the decomposition of H 2 O 2 is 1. H 2 O 2 + I – → H 2 O + IO – (slow) 2. H 2 O 2 + IO – → H 2 O + O 2 + I – (fast) What is the expected rate law? First step is slow so the rate determining step defines the rate law rate=k [H 2 O 2 ][I – ]

Group Problem Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 21 The reaction: A + 3B → D + F was studied and the following mechanism was finally determined: 1. A + B  C (fast) 2. C + B → D + E (slow) 3. E + B → F (very fast) What is the expected rate law? Rate Step 2=k 2 [C][B] Rate forward = k f [A][B] Rate reverse = k r [C] k f [A][B] = k r [C] [C]= k f [A][B]/k r Rate = k obs [A][B] 2

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 22 Catalysts CHAPTER 14 Chemical Kinetics

Catalyst Definition Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 23 o Substance that changes rate of chemical reaction without itself being used up o Speeds up reaction, but not consumed by reaction o Appears in mechanism, but not in overall reaction o Does not undergo permanent chemical change o Regenerated at end of reaction mechanism o May appear in rate law o May be heterogeneous or homogeneous

Catalyst Activation Energy Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 24 o By providing alternate mechanism o One with lower E a o Because E a lower, more reactants and collisions have minimum KE, so reaction proceeds faster

Catalyst Activation Energy Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 25

Catalyst Homogeneous Catalyst Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 26 Same phase as reactants Consider : S(g) + O 2 (g) + H 2 O(g)  H 2 SO 4 (g) S(g) + O 2 (g)  SO 2 (g) NO 2 (g) + SO 2 (g)  NO(g) + SO 3 (g) Catalytic pathway SO 3 (g) + H 2 O(g)  H 2 SO 4 (g) NO(g) + ½O 2 (g)  NO 2 (g) Regeneration of catalyst  Net: S(g) + O 2 (g) + H 2 O(g)  H 2 SO 4 (g) What is Catalyst? –Reactant (used up) in early step –Product (regenerated) in later step Which are Intermediates?

Catalyst Heterogeneous Catalyst Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 27 o Exists in separate phase from reactants o Usually a solid o Many industrial catalysts are heterogeneous o Reaction takes place on solid catalyst Ex. 3H 2 (g) + N 2 (g)  2NH 3 (g)

Catalyst Heterogeneous Catalyst Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 28 H 2 and N 2 approach Fe catalyst H 2 and N 2 bind to Fe & bonds break N—H bonds forming NH 3 formation complete NH 3 dissociates

Enzymes: Superoxide Dismutase 29 Miller, Anne-Frances. “Fe Superoxide Dismutase” Handbook of Metalloproteins. John Wiley & Sons, Ltd, Chinchester, 2001 Rodrigues, J. V; Abreu, I. A.; Cabelli, D; Teixeira, M. Biochemistry 2006, 45, Catalyst