Rate Mechanisms The Basics

Reaction Mechanisms O The series of steps that actually occur in a chemical reaction. O Kinetics can tell us something about the mechanism O A balanced equation does not tell us how the reactants become products.

Reaction Mechanisms O 2NO 2 + F 2 2NO 2 F O Rate = k[NO 2 ][F 2 ] through experimentation O The proposed mechanism is O NO 2 + F 2 NO 2 F + F (slow) O F + NO 2 NO 2 F(fast) O F is called an intermediate It is formed then consumed in the reaction

Reaction Mechanisms O Each of the two reactions is called an elementary step. O The rate for a reaction can be written from its molecularity. O Molecularity is the number of pieces that must come together. O Elementary steps add up to the balanced equation

O Unimolecular step involves one molecule - Rate is first order. O Bimolecular step - requires two molecules - Rate is second order O Termolecular step- requires three molecules - Rate is third order O Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.

Molecularity and Rate Laws O A products Rate = k[A] O A+A productsRate= k[A] 2 O 2A productsRate= k[A] 2 O A+B productsRate= k[A][B] O A+A+B products Rate= k[A] 2 [B] O 2A+B products Rate= k[A] 2 [B] O A+B+C products Rate= k[A][B][C]

How to get rid of intermediates O They can’t appear in the rate law. O Slow step determines the rate and the rate law O Use the reactions that form them O Use Hess’s Law to add the equations together. O The summary reaction shows the overall reaction that is taking place.

Example #1 3/2 O 2 O 3 O Proposed Mechanism A. NO + ½ O 2 NO 2 B. NO 2 NO + O C. O 2 + O O 3 O How can we determine if this is a feasible mechanism?

Example #1 Continued O Add the elementary reaction together: NO + ½ O 2 + NO 2 + O 2 + O NO 2 + NO + O + O 3 O Write the rate equation for each elementary step: A. rate = k[NO][O 2 ] 1/2 B. rate = k[NO 2 ] C. rate = k[O 2 ][O]

Example #1 Continued How 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.

Example #2 Write the overall reaction given the following mechanism: C 4 H 9 Br C 4 H Br - slow C 4 H H 2 O C 4 H 9 OH 2 + fast C 4 H 9 OH H 2 O C 4 H 9 OH + H 3 O + fast C 4 H 9 Br + 2H 2 O Br - + C 4 H 9 OH + H 3 O +

Example #2 Continued Write the rate expression for each elementary step. Rate = k[C 4 H 9 Br] Rate = k[C 4 H 9 + ] Rate = k[C 4 H 9 OH 2 + ] Because the first step in the rate determining step rate = k[C 4 H 9 Br]

Intermediate Step Sometimes 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?

NO + Cl 2 NOCl 2 fast k 1 [NO][Cl 2 ] k -1 [NOCl 2 ] NOCl 2 + NO2NOCl slow k 2 [NOCl 2 ][NO] 2 NO + Cl 2 2NOCl Set k 1 =k -1 and solve in term of [NOCl 2 ] k 1 [NO][Cl 2 ]/k -1 = [NOCl 2 ] Substitute this into the rate for the slow step: rate = k 2 (k 1 [NO][Cl 2 ]/k -1 ) [NO] = k[NO] 2 [Cl 2 ]

Try another one NO + O 2 NO 3 fast NO 3 + NO 2NO 2 slow Fast rate = k 1 [NO][O 2 ] = k -1 [NO 3 ] Slow rate = k 2 [NO 3 ][NO] Must get rid of the intermediate [NO 3 ] Rate = k[NO] 2 [O 2 ]

On Your Own! NO + NON 2 O 2 fast N 2 O 2 + O 2 2NO 2 slow Find the overall reaction and rate law 2NO + O 2 2NO 2 Rate = k[NO] 2 [O 2 ]