AP Chemistry Unit 8 - Kinetics

1. AP ChemistryUnit 8 - Kinetics Lesson 6 – Reaction Mechanisms Book Section: 14.6

2. Reaction Mechanisms • Reactions can occur in multiple steps. • The rate of reaction is determined by whichever step is the slowest… • just like traffic is determined by the slowest car in the line.

3. Multistep Mechanisms • The following reaction occurs in two steps: • NO2 + CO  NO + CO2 • Step 1: NO2 + NO2 NO3 + NO • Step 2: NO3 + CO  NO2 + CO2 • SUM: NO2 + CO  NO + CO2 • We can say that NO3is formed as an intermediate in the reaction.

4. Rate Laws for Elementary Reactions • If a reaction occurs in one step, the rate law is merely the product of all the reactant concentrations • A + B  C • Rate = k[A][B]

5. Rate Laws for Elementary Reactions • 2 A + B  C • A + A + B  C • Rate = k[A][A][B] = k[A]2[B] • For one-step reactions, • a A + b B  C • Rate = k[A]a[B]b

6. Rate Laws for Elementary Reactions • If a reaction occurs in one step, the rate law is merely the product of all the reactant concentrations • A + B  C • Rate = k[A][B]

7. Rate Laws for Elementary Reactions • What is the rate law for the following reaction? • H2 + Br2 2 HBr

8. Rate Laws for Elementary Reactions • What is the rate law for the following reaction? • H2 + Br2 2 HBr • Rate = k[H2][Br2]

9. Multistep Mechanisms • If a reaction has multiple steps, the slow one determines the reaction rate. • We will discuss two types – • Reactions with a slow initial step • Reactions with a fast initial step

10. Slow Initial Step • Step 1: NO2 + NO2 NO3 + NO k1 – slow • Step 2: NO3 + CO  NO2 + CO2k2 – fast • Overall: NO2 + CO  NO + CO2 • The rate law will be determined by the slow step – the first step slowly makes products that are immedately consumed in step 2. • Step 1 is the rate-determining step for this mechanism.

11. Slow Initial Step • The rate law for this reaction is based off of the slow step. • NO2 + NO2 NO3 + NO • Rate = k1[NO2][NO2] = k1[NO2]2

12. Try Yourself • What is the rate law for the following reaction? • N2O  N2 + O (slow) • N2O + O  N2 + O2 (fast)

13. Try Yourself • What is the rate law for the following reaction? • N2O  N2 + O (slow) • N2O + O  N2 + O2 (fast) • Rate = k[N2O]

14. Fast Initial Step • These become difficult because often times the rate is dependent on an intermediate, not a reactant. • Example: • Step 1: NO + Br2 NOBr2 k1 = fast • Step 2: NOBr2 + NO  2 NOBr k2 = slow • Overall: 2 NO + Br  2 NOBr

15. Fast Initial Step • Step 1: NO + Br2 NOBr2 k1 = fast • Step 2: NOBr2 + NO  2 NOBr k2 = slow • Overall: 2 NO + Br  2 NOBr • Step 2 is the rate determining step, and the rate law is • Rate = k2[NOBr2][NO] • But… NOBr2 is an intermediate, not a reactant, so our final answer can not use it.

16. Fast Initial Step • Example: • Step 1: NO + Br2 NOBr2 k1 = fast • Step 2: NOBr2 + NO  2 NOBr k2 = slow • Overall: 2 NO + Br  2 NOBr • We must solve for the NOBr2 in terms of the first step. • Since the reaction goes in both directions at the same rate, we can choose the direction at which it goes.

17. Fast Initial Step • If k1is the rate constant of the forward reaction, we can define k-1as the rate constant of the reverse reaction. • The forward reaction and the reverse reaction have the same rates. • k1[NO][Br2] = k-1[NOBr2] • [NOBr2] = k1/k-1[NO][Br2]

18. Fast Initial Step • Substituting into our rate, we get • Rate = k2k1/k-1[NO][Br2][NO] • Rate = k[NO]2[Br2] • Whew!

19. HW: 14.62, 14.64, 14.66, 14.68, 14.70 • This week: • Wednesday – Begin Unit 9 – Equilibrium : Equilibrium Concept (15.1) • Thursday – Equilibrium Constant (15.2-15.3) • Friday – Kinetics Exam • Kinetics Exam – Friday, Feb. 11th • Problem Set 7 – Tuesday, Feb 15th