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AP Chemistry Unit 8 - Kinetics

AP Chemistry Unit 8 - Kinetics. Lesson 6 – Reaction Mechanisms Book Section: 14.6. 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.

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AP Chemistry Unit 8 - Kinetics

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  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

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