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Chemical Kinetics Lesson 2
Chapter 12 Chemical Kinetics Lesson 2
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12.2 Rate Laws: An Introduction
Reaction Rates Rate Laws: An Introduction Determining the Form of the Rate Law The Integrated Rate Law Reaction Mechanisms 12.6 A Model for Chemical Kinetics 12.7 Catalysis Copyright © Cengage Learning. All rights reserved
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Summary of the Rate Laws
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Zero order First order Second order 4.7 M 3.7 M 2.0 M Exercise
Consider the reaction aA Products. [A]0 = 5.0 M and k = 1.0 x 10–2 (assume the units are appropriate for each case). Calculate [A] after 30.0 seconds have passed, assuming the reaction is: Zero order First order Second order a) 4.7 M [A] = –(1.0×10–2)(30.0) + 5.0 b) 3.7 M ln[A] = –(1.0×10–2)(30.0) + ln(5.0) c) 2.0 M (1 / [A]) = (1.0×10–2)(30.0) + (1 / 5.0) 4.7 M 3.7 M 2.0 M Copyright © Cengage Learning. All rights reserved
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Most chemical reactions occur by a series of elementary steps.
Reaction Mechanism Most chemical reactions occur by a series of elementary steps. An intermediate is formed in one step and used up in a subsequent step and thus is never seen as a product in the overall balanced reaction. Copyright © Cengage Learning. All rights reserved
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NO2(g) + CO(g) → NO(g) + CO2(g)
A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO NO2(g) + CO(g) → NO(g) + CO2(g) Copyright © Cengage Learning. All rights reserved
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Elementary Steps (Molecularity)
Unimolecular – reaction involving one molecule; first order. Bimolecular – reaction involving the collision of two species; second order. Termolecular – reaction involving the collision of three species; third order. Copyright © Cengage Learning. All rights reserved
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Rate-Determining Step
A reaction is only as fast as its slowest step. The rate-determining step (slowest step) determines the rate law and the molecularity of the overall reaction. Copyright © Cengage Learning. All rights reserved
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Reaction Mechanism Requirements
The sum of the elementary steps must give the overall balanced equation for the reaction. The mechanism must agree with the experimentally determined rate law. Copyright © Cengage Learning. All rights reserved
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Decomposition of N2O5 Copyright © Cengage Learning. All rights reserved
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Step 2: NO2 + NO3 → NO + O2 + NO2 (slow)
Decomposition of N2O5 2N2O5(g) 4NO2(g) + O2(g) Step 1: N2O NO2 + NO3 (fast) Step 2: NO2 + NO3 → NO + O2 + NO2 (slow) Step 3: NO3 + NO → 2NO2 (fast) 2( ) Copyright © Cengage Learning. All rights reserved
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The reaction A + 2B C has the following proposed mechanism:
Concept Check The reaction A + 2B C has the following proposed mechanism: A + B D (fast equilibrium) D + B C (slow) Write the rate law for this mechanism. rate = k[A][B]2 rate = k[A][B]2 The sum of the elementary steps give the overall balanced equation for the reaction. Copyright © Cengage Learning. All rights reserved
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Molecules must collide to react. Main Factors: Activation energy, Ea
Collision Model Molecules must collide to react. Main Factors: Activation energy, Ea Temperature Molecular orientations Copyright © Cengage Learning. All rights reserved
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Energy that must be overcome to produce a chemical reaction.
Activation Energy, Ea Energy that must be overcome to produce a chemical reaction. Copyright © Cengage Learning. All rights reserved
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Transition States and Activation Energy
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Change in Potential Energy
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For Reactants to Form Products
Collision must involve enough energy to produce the reaction (must equal or exceed the activation energy). Relative orientation of the reactants must allow formation of any new bonds necessary to produce products. Copyright © Cengage Learning. All rights reserved
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The Gas Phase Reaction of NO and Cl2
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R = gas constant (8.3145 J/K·mol) T = temperature (in K)
Arrhenius Equation A = frequency factor Ea = activation energy R = gas constant ( J/K·mol) T = temperature (in K) Copyright © Cengage Learning. All rights reserved
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Linear Form of Arrhenius Equation
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Linear Form of Arrhenius Equation
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Exercise Chemists commonly use a rule of thumb that an increase of 10 K in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25°C to 35°C? Ea = 53 kJ Ea = 53 kJ ln(2) = (Ea / J/K·mol)[(1/298 K) – (1/308 K)] Copyright © Cengage Learning. All rights reserved
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A substance that speeds up a reaction without being consumed itself.
Catalyst A substance that speeds up a reaction without being consumed itself. Provides a new pathway for the reaction with a lower activation energy. Copyright © Cengage Learning. All rights reserved
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Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction
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Effect of a Catalyst on the Number of Reaction-Producing Collisions
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Heterogeneous Catalyst
Most often involves gaseous reactants being adsorbed on the surface of a solid catalyst. Adsorption – collection of one substance on the surface of another substance. Copyright © Cengage Learning. All rights reserved
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Heterogeneous Catalysis
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Heterogeneous Catalyst
Adsorption and activation of the reactants. Migration of the adsorbed reactants on the surface. Reaction of the adsorbed substances. Escape, or desorption, of the products. Copyright © Cengage Learning. All rights reserved
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Exists in the same phase as the reacting molecules.
Homogeneous Catalyst Exists in the same phase as the reacting molecules. Enzymes are nature’s catalysts. Copyright © Cengage Learning. All rights reserved
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Homogeneous Catalysis
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