Reaction Mechanisms The series of steps that actually occur in a chemical reaction Kinetics can tell us something about the mechanism A balanced equation does not tell us how the reactants become products ID: 1029832
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1. Rate MechanismsThe Basics
2. Reaction MechanismsThe series of steps that actually occur in a chemical reaction.Kinetics can tell us something about the mechanismA balanced equation does not tell us how the reactants become products.
3. Reaction Mechanisms 2NO2 + F2 2NO2F Rate = k[NO2][F2] through experimentationThe proposed mechanism isNO2 + F2 NO2F + F (slow)F + NO2 NO2F (fast) F is called an intermediate It is formed then consumed in the reaction
4. Reaction MechanismsEach of the two reactions is called an elementary step .The rate for a reaction can be written from its molecularity .Molecularity is the number of pieces that must come together.Elementary steps add up to the balanced equation
5. Unimolecular step involves one molecule - Rate is first order.Bimolecular step - requires two molecules - Rate is second orderTermolecular step- requires three molecules - Rate is third order Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.
6. Molecularity and Rate LawsA products Rate = k[A]A+A products Rate= k[A]22A products Rate= k[A]2A+B products Rate= k[A][B]A+A+B products Rate= k[A]2[B]2A+B products Rate= k[A]2[B]A+B+C products Rate= k[A][B][C]
7. How to get rid of intermediatesThey can’t appear in the rate law.Slow step determines the rate and the rate lawUse the reactions that form themUse Hess’s Law to add the equations together.The summary reaction shows the overall reaction that is taking place.
8. Example #13/2 O2 O3Proposed Mechanism NO + ½ O2 NO2NO2 NO + O O2 + O O3 How can we determine if this is a feasible mechanism?
9. Example #1 ContinuedAdd the elementary reaction together:NO + ½ O2 + NO2 + O2 + O NO2 + NO + O + O3Write the rate equation for each elementary step: A. rate = k[NO][O2]1/2 B. rate = k[NO2] C. rate = k[O2][O]
10. Example #1 ContinuedHow can we determine which step in the mechanism is the rate determining step? (The Slow one)The rate of the slow step will be equal to the experimentally determined rate law.
11. Example #2Write the overall reaction given the following mechanism:C4H9Br C4H9+ + Br- slowC4H9+ + H2O C4H9OH2+ fastC4H9OH2+ + H2O C4H9OH + H3O+ fastC4H9Br + 2H2O Br- + C4H9OH + H3O+
12. Example #2 ContinuedWrite the rate expression for each elementary step.Rate = k[C4H9Br]Rate = k[C4H9+]Rate = k[C4H9OH2+]Because the first step in the rate determining step rate = k[C4H9Br]
13. Intermediate StepSometimes the slow step includes an intermediate species that is produced in one step yet consumed in another step. NOTE: The final rate law CANNOT include an intermediate……So now what do we do?
14. NO + Cl2 NOCl2 fast k1[NO][Cl2] k-1[NOCl2] NOCl2 + NO 2NOCl slow k2[NOCl2][NO] 2 NO + Cl2 2NOCl Set k1=k-1 and solve in term of [NOCl2] k1[NO][Cl2]/k-1 = [NOCl2]Substitute this into the rate for the slow step: rate = k2 (k1[NO][Cl2]/k-1) [NO] = k[NO]2[Cl2]
15. Try another oneNO + O2 NO3 fastNO3 + NO 2NO2 slowFast rate = k1[NO][O2] = k-1[NO3]Slow rate = k2[NO3][NO]Must get rid of the intermediate [NO3]Rate = k[NO]2[O2]
16. On Your Own! NO + NO N2O2 fastN2O2 + O2 2NO2 slowFind the overall reaction and rate law2NO + O2 2NO2 Rate = k[NO]2[O2]