CHEM Chemical Kinetics. Reaction Mechanisms

Transcription

1 Chemical Kinetics Deri ed Rate La s from Derived Rate Laws from Reaction Mechanisms

2 Reaction Mechanism Determine the rate law by experiment Devise a reaction mechanism If the predicted and experimental rate laws do not agree Predict the rate law for the mechanism If the predicted and experimental rate laws agree Loo for additional supporting evidence 2

3 Reaction Mechanism A sequence of one or more elementary reaction steps together forms a reaction mechanism. In a mechanism, elementary steps proceed at various speeds (governed by different rate constants, ). Elementary reaction steps must be balanced (as do all chemical reactions). The slowest step is the rate-determining step. It is the bottlenec in the formation of products. 3

4 Reaction Mechanism A rate law derived from a set of mechanisms should only consist of concentrations of reactants and/or products, no intermediates. In predicting the rate law for an elementary step, the exponents for the concentration terms are the same as the stoichiometric coefficients. To propose a mechanism requires the nowledge of chemistry to give plausible elementary processes. In this course, you will not be ased to propose mechanisms, but you will be ased to derive the rate laws from given mechanisms. 4

5 Reaction Mechanism - Terminologies Molecularity is the number of reacting species (i.e. atoms, ions or molecules) in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction. Uni-, Bi-, Termolecular Involving 1 species Involving 2 species Involving 3 species 5

6 Reaction Mechanism 1. Unimolecular Elementary Step A 1B or A 1 B C There is only one molecule reacting, namely species "A" is reacting. This unimolecular reaction step implies the rate law, [A] d[b] - d 1[A] Involving a single species. Examples: Decomposition of hydrogen peroxide Decomposition of ammonium nitrite H 2 O 2 (aq) H 2 O(l)+½ O 2 (g) NH 4 NO 2 (g) N 2 (g) + 2 H 2 O(g) Recall, this is a 1 st order reaction. 6

7 Reaction Mechanism 2. Reversible Unimolecular Elementary Step A B At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. R forward = R reverse 1 [A] = -1 [B] Rearrange, 1 1 [B] [A] This is the definition of K eq. 7

8 Reaction Mechanism 2. Reversible Unimolecular Elementary Step (cont d) A B Species "A" is in equilibrium with species B". The forward reaction is governed by 1. The reverse reaction is governed by -1. The rate, R, is equal to the rate of the forward step minus the rate of the reverse step. This implies the rate law, R [A] d 1[A] 1d[B] [B] [A ] decreasing in forward reaction [A ] increasing in reverse reaction 8

9 Reaction Mechanism 3. Bimolecular Elementary Step requires two molecules coming together at the same time. A or A B 2C Implies this rate law. [A] d[b] d[c] d 2 B 2Implies this C D rate law. [A][B] [A] d[b] d[c] d[d] d 2 [A][B] 9

10 Reaction Mechanism 3. Bimolecular Elementary Step or A A C D Implies this rate law. 2 2 rate law. 21d[A] d[c] d[d] 2 [A] Examples of reactions involving a bimolecular elementary step. NO + O 3 NO 2 +O 2 Rate = [NO][O 3 ] Involves 2 HI H 2 + I 2 Rate = [HI] 2 collisions of two species. 10

11 Reaction Mechanism 4. Reversible Bimolecular Elementary Step A + B C Implies this rate law. or R- d[a] d[b] d 2d[C] 2[A][B] [C] A + B C + D Implies this rate law. R- d 2[A][B] 2 [A] d[b] d[c] d[d] [C][D] 11

12 Reaction Mechanism 4. Reversible Bimolecular Elementary Step (cont d) or Implies this rate law. 2 A C + D rate law. R- 21d 2[A] d[c] d[d] 2[A] 2 [C][D] At equilibrium, R forward = R reverse where R forward = 2 [A] 2 2 [A] 2 = -2 [C][D] R reverse [C][D]2= -2 Equilibrium lb constant [C][D ] K 2eq 2 [A] 12

13 Reaction Mechanism 2 HI H 2 (g) + I 2 (g) At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. R f = 1 [HI] 2 R forward = R reverse 1 [HI] 2 = -1 [H 2 ] [I 2 ] R r = 1 [H 2 ] [I 2 ] r -1[ 2] [ 2 ] At the start 1[H 2][I 2] K 2 eq 1 [HI] At equilibrium 13

14 Reaction Mechanism 5. Termolecular Elementary Step Termolecular reaction steps require three molecules coming together at the same time. A A A 3Implies this B products rate law. or 3rate law. 31d[A] d[b] 3[A] A 3 A B C products Implies this rate law. 12d[A] d[b] d[c] 3[A] 2 [B] 14

15 Reaction Mechanism 5. Termolecular Elementary Step or A 3 B C D products Implies this rate law. d[a] d[b] d[c] d[d] 3 [A][B][C] Example: The reaction mechanism for 2NO(g)+O O 2 (g) 2 NO 2 (g) involves a 3-body collision one-step mechanism. Rate = [NO] 2 [O 2 ] 15

16 Reaction Mechanism Recall, the equation in an elementary step represents the reaction at the molecular level, not the overall reaction. What about higher orders such as 4 th or 5 th orders? Simultaneous collision of 3 molecules is rare. In nature, we observe lots of 2- body collisions, very few 3-body collisions and not much else. 16

17 Most balanced equations do not literally describe how a reaction occurs in terms of the collisions made or the actual sequence of events. The combustion of hexane illustrates this point: 2 C 6 H 14 (g) + 19 O 2 (g) 12 CO 2 (g) + 14 H 2 O (g) If it were to tae things literally as written, the reaction is saying 2 hexane molecules and 19 oxygen molecules somehow collide simultaneously and fight among themselves until 12 molecules of carbon dioxides and 14 molecules of waters form. In nature, we observe lots of 2-body collisions, very few 3-body collisions i and not much else. 17

18 A one-step mechanism involving i the collision i of NO and O 3. 1NO + 1O 3 NO 2 + O 2 Since this is the only step in the reaction mechanism, then the rate law can be written directly from the stoichiometry of the step. d[no] d[o 3] d[no 2] d[o 2Rate - - ] d[no] d[o 3] d[no 2] d[o 2] Rate - - [NO] [O ] 18

19 Example: The overall reaction of N 2 O 5 (g) + NO(g) 3 NO 2 (g) occurs in a one-step mechanism where the two reactants collide to form the product. The derived rate law can be determined to be Rate = [N 2 O 5 ][NO] Suggestion of a bimolecular single step mechanism. 19

20 Mechanism: A sequence of one or more elementary reaction steps that proceed at various speeds. Step Step 1 : A 12 B : B 2 C Each step is governed ed by its rate constant. What would the energy profile diagram loo lie? If 1 >> 2, If 1 << 2, E a1 E a2 E a1 E a2 Step 2 is the rate determining step (RDS). This implies that isolation of B is good. Step1 is the rate determining step (RDS). This implies that isolation of B is not easy. 20

21 Mechanism: A sequence of one or more elementary reaction steps that proceed at various speeds. for an overall exothermic reaction: 21. Identify A, B, C, D, and E. the reaction mechanism Step 1Consider Step 1 : A B C : C 2 D E (slow) (fast) 2. What is the overall reaction? 3. What is the rate law? A and B are reactants C is the intermediate D and E are the products A + B D + E In this mechanism, the rate law can be written directly from the slowest step. Rate = 1 [A] [B] 21

22 Mechanism: A sequence of one or more elementary reaction steps that proceed at various speeds. Consider the reaction mechanism for an overall exothermic reaction: 24 the reaction mechanism Step 1Consider Step 1 : A 2B C : C D E (slow) (fast) 4. Setch the reaction coordinate of the reaction. Since step 1 is the rate determining step, 1 << 2. E a1 E a2 E a1 > E a2 22

23 Example: CO (g) + NO 2 (g) NO (g) + CO 2 (g) This reaction proceeds via two reaction mechanisms. Above 600K, a one-step mechanism. Below 600K, a two-step mechanism. 23

24 Example: Experimentally determined Rate = [CO][NO 2 ] CO (g) + NO 2 (g) NO (g) + CO 2 (g) (balanced reaction) Above 600K, the reaction mechanism involves the collision between CO and NO 2. A one-step elementary step which describes the collision 1 CO + 1NO 2 NO + CO 2 of CO and NO 2. Reasonable - a single collision i of two molecules l (with correct orientation and minimum energy) would lead to the exchange of an oxygen between CO and NO 2. 22d[NO ] d[co] d[no] d[co - - [ CO] [NO 2 ] ] 1 1 Derived rate law is consistent with the experimental rate law. 24

25 Example: CO (g) + NO 2 (g) NO (g) + CO 2 (g) (balanced reaction) Below 600K, the experimentally determined rate law is Rate = [NO 2 ] 2 Proposed mechanism is a 2-step mechanism: 25

26 Example: Below 600K, experimentally determined rate law is Rate = [NO 2 ] 2 CO (g) + NO 2 (g) NO (g) + CO 2 (g) Step Step 211: NO 2 NO 2 NO NO 32 : NO 3 CO NO 2 CO 2(slow) (fast) 1. Identify the intermediate, if any. NO 3 is the intermediate. 2. What is the overall reaction? NO 2 + NO 2 + NO 3 + CO NO + + NO 2 + NO 3 +CO 2 NO 2 + CO CO 2 + NO Yes, the elementary steps add to give the overall reaction. 3. Which is the rate determining step? Step 1 26

27 Example: Below 600K, experimentally determined rate law is Rate = [NO 2 ] 2 CO (g) + NO 2 (g) NO (g) + CO 2 (g) Proposed mechanism: Step : NO NO 1 NO NO Step : NO CO NO 2CO 32(slow) (fast) NO 3 is very reactive; consistent with step 1 being the RDS. 4. What is the rate law of the proposed mechanism? Is it consistent with the experimentally determined rate law? Since step 1 is the RDS, use step 1, the bimolecular step involving the collision of two NO 2, to determine the rate law. Rate = rate of the slow step = 1 [NO 2 ] 2 The mechanism ss rate law is consistent with the experimentally determined rate law. Confirmed by reacting two NO 2 molecules and loo for yielding NO 3 as a product. NO 3 is highly reactive and is capable of transferring an oxygen atom to CO to give CO 2. 27

28 Consider the following reaction mechanisms proposed for the thermal decomposition of NO 2. 2 NO 2 2 NO + O 2 Experimental rate law is Two possible mechanisms: Rate = [NO 2 ] 2 Mechanism 2 Step 2Step Step 2 2Mechanism Step 1Mechanism 1 1 : NO 2 NO O 2 2: O NO O NO 21: NO 1NO NO 223 2: NO NO O (slow) (fast) (slow) (fast) Which mechanism is consistent with the observed rate law? 28

29 Consider the following reaction mechanism proposed for the overall reaction Experimental rate law is 2 NO 2 2 NO + O 2 Rate = [NO 2 ] 2 Step 2Step 1Mechanism 1 1 : NO 2 NO O 2 2: O NO O 2Derived rate law from Mechanism 1 is NO (slow) (fast) Derived rate law from Mechanism 1 is Rate = 1 [NO 2 ] 29

30 Consider the following reaction mechanism proposed for the overall reaction Experimental rate law is 2 NO 2 2 NO + O 2 Mechanism 2 Step 2: Rate = [NO 2 ] NO NO NO (slow) Step : NO NO O (fast) Derived rate law from Mechanism 2 is Rate = 1 [NO 2 ] 2 30

31 Consider the following reaction mechanism proposed for the overall reaction Experimentally determined rate law is 2 NO + O 2 2 NO 2 Rate = [NO] 2 [O 2 ] Consider a single step mechanism: 222 2Derived rate law is NO O 1 NO Derived rate law appears to be consistent with the experimental rate law. Rate = 1 [NO] 2 [O 2 ] The mechanism invoes a termolecular step, which is very unliely since three way collisions are less liely to tae place. A mechanism Involving two species collision would be more probable. 31

32 Consider the following reaction mechanism proposed for the overall reaction 2 NO + O 2 2 NO 2 Consider a two-step mechanism: 1Step 2Step : : NO NO N O 22 O 2 N 2O NO 22 Derived rate law is Rate = rate of the slow step = 2 [N 2 O 2 ][O 2 ] 22(fast) (slow) Need to remove N 2 O 2 from the rate law. N 2 O 2 is an intermediate. The rate law cannot contain intermediates. Use the fast equilibrium step to find an expression for N 2 O 2. 1 [NO] 2 = -1 [N 2 O 2 1]2 [N O 2] [NO 2] 1 Substitute this bac into the above rate equation to remove N 2 O

33 Consider the following reaction mechanism proposed for the overall reaction 2 NO + O 2 2 NO 2 Consider a two-step mechanism: 1Step 2Step Derived rate law is : : NO N 2 O 2NO O 2 Rate = rate of the slow step = 2 [N 2 O 2 ][O 2 ] 2N 2O NO 22 22(fast) Experimentally determined rate law Rate = [NO] 2 [O 2 ] (slow) Since [N 12O 2] [NO ] 1 Need to remove dinitrogen dioxide, N 2 O 2, from the rate law. Derived rate law is consistent with the experimental rate law. This two step mechanism each involving a bimolecular step is more plausible. Rate 2 1 [NO] [O 2 1 Rate 2' [NO] 2[O 2 ] ] ; Let '

34 Ozone depletion Experimental rate law: Example: 221[O ] Rt 32 O 3 (g) 3 O 2 (g) Rate [ O ] [ O ] [O 22 ] Proposed mechanism: Derived rate law: Step 1: O O O 232 2Step : O O 2O d[o 2] 2[O 3 ][O] Need to remove [O ] O from the rate From step 1, 1 [O 3 ] = -1 [O 2 ][O] Rate [O 2law as O is an intermediate. [O 3[ O] 1 [O 21 ] ] (fast) 3 (slow) 1[O 3] Rate [O 3] ; Let ' 1[O 2 ] 23' ] Derived rate law is consistent with the experimental rate law. 34

35 Steady State Approximation The steady-state approximation is a method used to derive a rate law. The method is based on the assumption that one intermediate in the reaction mechanism is consumed as quicly as it is generated. Its concentration remains unchanged for most of the reaction. The system reaches a steady-state, hence the name of the technique is called steady state approximation. When steady-state is reached, there is no change observed in the concentration of the intermediate. 35

36 Steady State Approximation Example: H 2 (g) + 2 ICl (g) I 2 (g) + 2 HCl (g) Experimental rate law: Rate = [H 2 ][ICl] Proposed mechanism: Step 2Step 11 : H 2(g) ICl (g) HI (g) HCl (g) 2 : HI (g) ICl (g) I 2(g) HCl (g) Since step 1 is the slow step, the derived rate law (slow) (fast) is consistent with the experimental rate law. Rate = [H 2 ][ICl] 36

37 Steady State Approximation HI is being produced in step 1 and quicly removed in step 2 [HI] is pretty much constant throughout the reaction (with the exception of the beginning and the end of the reaction). H 2 (g) + 2 ICl (g) I 2 (g) + 2 HCl (g) Experimental rate law: Rate = [H Proposed mechanism: 2 ][ICl] 1 Step : H (g) 2 ICl (g) HI (g) HCl (g) (fast) 22 2Step : HI (g) ICl (g) I (g) HCl (g) (slow) 1Derived rate law: Need to remove 2HI from the rate d[i 2] ] 2law as HI is an [HI][ICl] intermediate. d[hi] 0[HI] is pretty much constant throughout the reaction (with the Steady State Approximation d 2[HI] 0 1[H 2][ICl] Production of HI (step 1) 1[ H 2][ICl] 2[HI][ICl] [H 2][ICl] [ HI] 1 2[ICl] [HI][ICl] Removal of HI (step 2) Substitute [HI] Into into the rate t law. l d 212 2[I ] [H 2][ICl] [ICl] [ICl] Derived rate law is consistent with the experimental rate law. d[i 2] 1[H 2 ][ICl] 37

38 Chain Reactions Example: Chain reactions are complex reactions that involve chain carriers, reactive intermediates which react to produce more intermediates. The elementary steps in a chain reaction may be classified into: initiation, propagation, inhibition, and termination steps. Chlorofluorocarbons (CFCs) destruction of the ozone layer Initiation: Thermally or photochemically produces Cl radicals Propagation: Regenerates more Cl radicals Inhibition: A step involving product molecules being destroyed. Termination: Cl radicals deactivates by reacting to form an inactive product. 38

39 Reaction Mechanism What about this reaction in Experiment 4? Experiment 4: Does this mean that 5+1+6=12 particles must come together and collide? 5 Br - (aq) + BrO - 3 (aq) + 6 H + (aq) 3 Br 2 (aq) + 3 H 2 O (l) Experimentally determined rate law is Rate = [Br - ][ BrO ][H ] First order with respect to Br - and BrO 3- ions Second order with respect to H + ions Overall reaction is of = 4. We observe that the reaction is found to be quite fast. It means that even though the balanced equation involves a large number of molecules, the reaction does not proceed by simultaneous collision of all these reacting particles. The mechanism can involve two or maximum three collisions simultaneous. 39

40 Reaction Mechanism What about this reaction in Experiment 4? 5 Br - (aq) + BrO - 3 (aq) + 6 H + (aq) 3 Br 2 (aq) + 3 H 2 O (l) This is a rather complex mechanism. According to Field et al., 1972; Pelle et al., 2004, the reaction is thought to occur by the following collection of bimolecular elementary reactions. The derived rate law for this mechanism is outside the scope of this course. It is found to be consistent with the experimental rate law. First order with respect to Br - and BrO 3- ions 3 Second order with respect to H + ions 40

41 Reaction Mechanism Determine the rate law by experiment Devise a reaction mechanism If the predicted and experimental rate laws do not agree Predict the rate law for the mechanism If the predicted and experimental rate laws agree Loo for additional supporting evidence Rate laws can prove a mechanism is wrong, but can t prove one right! 41