CHEMISTRY. Chapter 13. Chapter Outline. Factors Affecting Rate

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1 CHEMISTRY Fifth Edition Gilbert Kirss Foster Bretz Davies Chapter 3 Chemical Kinetics: Reactions in the Atmosphere Chemistry, 5 th Edition Copyright 207, W. W. Norton & Company Chapter Outline 3.4 Reaction Rates, Temperature, and the Arrhenius Equation 3.5 Reaction Mechanisms 3.6 Catalysis 2 2 Factors Affecting Rate Temperature: Increased temperature increases kinetic energy of molecules and molecular collisions. Activation energy (E a ): The minimum energy of molecular collisions required to break bonds in reactants, leading to formation of products These factors are incorporated into rate constant k by Arrhenius equation. 3 3

2 Arrhenius Equation k a / = Ae E RT R = universal gas constant [in J/(mol K)] T = temperature (in kelvin) A is a collisional frequency factor. Includes frequency of collisions and an orientation factor Log form: E a ln k= + l n A R T 4 4 Reaction Energy Profile High-energy transition state ( activated complex ) Activation energy (E a ) k a / = Ae E RT 5 5 Effect of Temperature k a / = Ae E RT As temperature increases, the fraction of collisions exceeding this minimum energy also increases, leading to an increase in k and an increase in the rate of the reaction

3 Molecular Orientation O 3 (g) + NO(g) O 2 (g) + NO 2 (g) 7 7 Energy Profiles O 3 (g) + NO(g) O 2 (g) + NO 2 (g) Low E a = large k, fast forward reaction 8 8 Energy Profiles O 3 (g) + NO(g) O 2 (g) + NO 2 (g) High E a = small k, slow reverse reaction 9 9 3

4 Graphical Determination of E a Ea ln k = - ln R T + A Slope = E a /R Intercept = ln A 0 0 Mathematical Determination of E a Ea ln k = - ln R T + A Consider determination of k at two different T: E a E a ln k = + ln A ln k2 = + ln A R T R T2 By subtraction: k E a ln = k2 R T2 T Reaction Mechanisms Reaction mechanism a set of steps that describe how a reaction occurs at the molecular level; must be consistent with the rate law for the reaction Elementary step molecular-level view of a single process taking place in a chemical reaction Intermediate a species produced in one step of a reaction and consumed in a subsequent step Molecularity the number of ions, atoms, or molecules involved in an elementary step in a reaction 2 2 4

5 Reaction Mechanisms Consider the reaction: 2 NO 2 (g) 2 NO(g) + O 2 (g) Observed rate law: Rate = k[no 2 ] 2 Proposed mechanism: Step : 2 NO 2 (g) NO(g) + NO 3 (g) Rate =? Step 2: NO 3 (g) NO(g) + O 2 (g) Rate =? Rate law depends on rate-determining step (RDS), the slowest step in the mechanism (may involve mathematical substitution!). 3 3 Reaction Mechanisms 4 4 Reaction Mechanisms 2 NO(g) + O 2 (g) 2 NO 2 (g) Observed rate law: Rate = k[no] 2 [O 2 ] Proposed mechanism: Step : NO + O 2 NO 3 Rate = k [NO][O 2 ] Step 2: NO 3 + NO 2 NO 2 Rate2 = k 2 [NO 3 ][NO] Step 2 is rate-determining step; step is fast and reversible: Rate(f) = rate(r); k f [NO][O 2 ] = k r [NO 3 ] By rearrangement: k f NO 3 = NOO2 kr 5 5 5

6 Mechanisms: Zero Order Reaction: NO 2 (g) + CO(g) NO(g) + CO 2 (g) Observed rate law: Rate = k[no 2 ] 2 Mechanism: 2 NO 2 (g) NO(g) + NO 3 (g) Rate = k[no 2 ] 2 NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) Rate = k2[no 3 ][CO] NO 2 (g) + CO(g) NO(g) + CO 2 (g) Sum of elementary steps = balanced reaction Rate law from RDS (step ): Rate = k[no 2 ] 2 Zeroth order in [CO], because not involved in RDS 6 6 Practice: Reaction Mechanisms For the reaction: 2 NO 2 + O 3 N 2 O 5 + O 2 The experimentally determined rate law is rate = k[no 2 ][O 3 ]. Identify the rate-determining step in the proposed two-step mechanism:. NO 2 + O 3 NO 3 + O 2 2. NO 3 + NO 2 N 2 O 5 Collect and Organize: Given a balanced chemical equation, an experimentally determined rate law, and a proposed two-step mechanism, we are asked to identify the rate-determining step based on the mechanism. 7 7 Practice: Reaction Mechanisms For the reaction: 2 NO 2 + O 3 N 2 O 5 + O 2 The experimentally determined rate law is rate = k[no 2 ][O 3 ]. Identify the rate-determining step in the proposed two-step mechanism:. NO 2 + O 3 NO 3 + O 2 2. NO 3 + NO 2 N 2 O 5 Analyze: The rate law based on the molecularity of the RDS must agree with the experimentally observed rate law. We can derive the rate laws for the elementary steps in the mechanism and compare them with the observed rate law

7 Practice: Reaction Mechanisms For the reaction: 2 NO 2 + O 3 N 2 O 5 + O 2 The experimentally determined rate law is rate = k[no 2 ][O 3 ]. Identify the rate-determining step in the proposed two-step mechanism:. NO 2 + O 3 NO 3 + O 2 2. NO 3 + NO 2 N 2 O 5 Solve: The rate laws obtained based on the molecularity of the elementary steps are Step : Rate = k[no 2 ][O 3 ] Step 2: Rate = k[no 3 ][NO 2 ] Step 2 matches the experimentally determined rate law. This shows that step 2 would be the rate determining step. 9 9 Practice: Reaction Mechanisms For the reaction: 2 NO 2 + O 3 N 2 O 5 + O 2 The experimentally determined rate law is rate = k[no 2 ][O 3 ]. Identify the rate-determining step in the proposed two-step mechanism:. NO 2 + O 3 NO 3 + O 2 2. NO 3 + NO 2 N 2 O 5 Think About It: The molecularity of the RDS matches the observed rate law. It is worth noting that this does NOT prove that the mechanism is correct, only that it is feasible Catalysts and the Ozone Layer 2 2 7

8 Ozone Layer Natural photodecomposition of ozone: Mechanism: () sunlight O ( g) O ( g) + O( g) O ( g) + O( g) 2 O ( g) (2) 3 2 Net: 2 O 3 (g) 3 O 2 (g) E a for step 2 = 7.7 kj/mol (slow step) CFC Emissions and Ozone Chlorofluorocarbons (CFC) CCl 2 F 2, CCl 3 F, CClF 3 stable! CFCs in stratosphere: hν () CCl F( g) CCl F( g) + Cl( g) (2) (3) Cl( g) + O ( g) ClO( g) + O ( g) ClO( g) + O ( g) Cl( g) + O ( g) Net: 2 O3(g) 3 O2(g) E a for Cl-catalyzed reaction = 2.2 kj/mol Energy Profiles for O 3 Decomposition hν 2 O 3 O Cl atom: Not consumed during the reaction Homogeneous catalyst

9 Catalytic Converters 2 NO N 2 + O 2 Metal surface speeds up conversion of NO to N 2 and O 2, reduces smog-producing emissions Enzymes and Biocatalysis Step : E + S ES Step 2: ES E + P k 2 k k - Rate = k[es]