1. Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Rate of Reaction TEXT REFERENCE Masterton and Hurley Chapter 11

2. Chemistry 1011 Slot 52 11.7 Reaction Mechanisms YOU ARE EXPECTED TO BE ABLE TO: Define reaction mechanism and show how the reaction order is dependent upon the mechanism by which a reaction takes place. For a reaction taking place in more than one step, identify the rate determining step and identify reaction intermediates. Determine if a proposed reaction mechanism is consistent with experimental rate data.

3. Chemistry 1011 Slot 53 Reaction Mechanism Description of path or sequence of steps by which a reaction occurs Simplest case: single collision Frequently more than one step Rate expression and order of reaction depend on mechanism This is why rate law and order must be determined experimentally

4. Chemistry 1011 Slot 54 Reaction of CO (g) with NO 2(g) CO (g) + NO 2(g)  NO (g) + CO 2(g) At high temperatures, reaction is single step Rate = k[CO] x [NO 2 ] At low temperatures, reaction is two step process NO 2(g) + NO 2(g)  NO 3(g) + NO (g) CO (g) + NO 3(g)  CO 2(g) + NO 2(g) CO (g) + NO 2(g)  NO (g) + CO 2(g) Rate = k [NO 2 ] 2

5. Chemistry 1011 Slot 55 Elementary Steps The individual steps that make up a reaction pathway are called elementary steps Steps may be unimolecular, bimolecular, etc The molecularity is the number of molecules that are involved in an elementary process

6. Chemistry 1011 Slot 56 Rate Law for Elementary Steps For a complete reaction, the rate law and the order must be determined experimentally. This is because the mechanism may be simple or complex, and this determines the rate law and order For elementary reactions, however, the rate law can be determined from the equation

7. Chemistry 1011 Slot 57 Rate Law for Elementary Steps Elementary Step Molecularity Rate Law A  product unimolecular Rate = k[A] A + B  product bimolecular Rate = k[A][B] A + A  product bimolecular Rate = k[A] 2 2A + B  product termolecular Rate = k[A] 2 [B] For the elementary step: NO 2(g) + NO 2(g)  NO 3(g) + NO (g) The rate law is: Rate = k[ NO 2 ] 2

8. Chemistry 1011 Slot 58 Slow and Fast Steps Frequently, one step in a mechanism will be slower than the others This is the rate determining step Imagine a two step process: Step 1: A + B  X + I SLOW Step 2: A + I  Y FAST Total: 2A + B  X + Y The first step is the rate determining step

9. Chemistry 1011 Slot 59 Overall Rate Law and Order Rate (First Step) = k 1 [A][B] This slow first step determines the overall rate This is the rate law for the overall reaction: 2A + B  X + Y which is therefore second order

10. Chemistry 1011 Slot 510 Mechanisms and Rate Laws The rate law expression is determined by experiment The mechanism is a hypothesis about the way the reaction occurs The proposed mechanism must yield a rate law expression that is consistent with experiment

11. Chemistry 1011 Slot 511 Mechanism with a Fast Initial Step Sometimes the first step in a reaction mechanism, which results in the creation of a reaction intermediate, will be FAST The second step, where the reaction intermediate is a reactant, may be SLOW When this happens, the rate determining step will be the second step The rate law expression should then include the concentration of the reaction intermediate

12. Chemistry 1011 Slot 512 Reaction Intermediates and Rate Law The concentration of an intermediate cannot be measured The concentration of an intermediate cannot be included in a rate law expression if this is to be compared to experiment The final rate law expression can only include species occurring in the balanced equation

13. Chemistry 1011 Slot 513 Reaction Intermediates and Rate Law - an Example Step 1: NO (g) + Cl 2(g) NOCl 2(g) FAST Step 2: NOCl 2(g) + NO (g)  2NOCl (g) SLOW Overall: 2 NO (g) + Cl 2(g)  2NOCl (g) Rate of overall reaction = rate of step 2 Rate = k 2 [NOCl 2 ][NO] This is unsatisfactory; [NOCl 2 ] cannot be measured

14. Chemistry 1011 Slot 514 Eliminating Intermediates from the Rate Law Expression The first (fast) step in the reaction NO (g) + Cl 2(g) NOCl 2(g) FAST is reversible The reactants and products are in equilibrium rate forward = rate reverse k 1 [NO][Cl 2 ] =k  [NOCl 2 ]

15. Chemistry 1011 Slot 515 Eliminating Intermediates from the Rate Law Expression [NOCl 2 ] = k 1 [NO][Cl 2 ] k  Substitute in overall rate law expression Rate of reaction = rate of step 2 Rate = k 2 k 1 [NO] 2 [Cl 2 ] k 

16. Chemistry 1011 Slot 516 Limitations of Mechanism Studies Mechanisms are suggested in order to explain observed rate laws and orders of reaction Often more than one mechanism can explain experimental results