Chapter 12 Chemical Kinetics.

Chapter 12 Chemical Kinetics

General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
ASSIGNMENTS AP Chem Stop here Thursday Work on problem sets from handout AND the separate page... HW: ch to 12.4 #1-13 due Monday. HW: Read ch. 12 by Mon. HW: Read lab report TEST next WED. ch. 12 TURN in last week’s lab report. Yesterday’s lab report due next Thursday. Kinetics LAB tomorrow - switch rooms Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Chapter 12 Chemical Kinetics: Rates and Mechanisms of Chemical Reactions Chapter Twelve

12.2 Rate Laws: An Introduction
Chapter 12 Table of Contents 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

TURN in Clue Mystery Lab Report
Chapter 12 Table of Contents AP CHEM - WED - 1/29/14 CW: Notes 12.1 to 12.3 TURN in Clue Mystery Lab Report CW/HW: Assignment - 2 handouts #1-6 problems & (but only have to complete 1-6 for sect to 12.3 by Tuesday 2-4 HW: Read over Kinetics Lab for Friday On Friday, you will receive the ch take home test to complete by Monday, Feb. 10th Copyright © Cengage Learning. All rights reserved

Chemical Kinetics: A Preview
Chemical kinetics is the study of: the rates of chemical reactions factors that affect these rates the mechanisms by which reactions occur Reaction rates vary greatly – some are very fast (burning, precipitation) and some are very slow (rusting, disintegration of a plastic bottle in sunlight, production of a diamond). Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Variables in Reaction Rates
Concentrations of reactants: Reaction rates generally increase as the concentrations of the reactants are increased. Temperature: Reaction rates generally increase rapidly as the temperature is increased. Surface area: For reactions that occur on a surface rather than in solution, the rate increases as the surface area is increased. Catalysts: Catalysts speed up reactions and inhibitors slow them down. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

The Meaning of Rate The rate of a reaction is the change in concentration of a product per unit of time (rate of formation of product). Rate is also viewed as the negative of the change in concentration of a reactant per unit of time (rate of disappearance of reactant). The rate of reaction often has the units of moles per liter per unit time (mol L–1 s–1 or M s–1) rate of disappearance of reactant or of formation of product General rate of reaction = stoichiometric coefficient of that reactant or product in the balanced equation Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Change in concentration of a reactant or product per unit time.
Section 12.1 Reaction Rates Reaction Rate Change in concentration of a reactant or product per unit time. [A] means concentration of A in mol/L; A is the reactant or product being considered. Rate Return to TOC Copyright © Cengage Learning. All rights reserved

If the rate of consumption of H2O2 is 4.6 M/h, then … … the rate of formation of H2O must also be 4.6 M/h, and … … the rate of formation of O2 is 2.3 M/h Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

2 H2O2  2 H2O + O2 1 L 2.960 g O2 ( mole) produced in 60 s means … Rate = mol H2O2/L 60 s … mol H2O2 reacted in 60 s. = M H2O2 s–1 Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

The Decomposition of Nitrogen Dioxide
Section 12.1 Reaction Rates The Decomposition of Nitrogen Dioxide Return to TOC Copyright © Cengage Learning. All rights reserved

Value of the rate at a particular time.
Section 12.1 Reaction Rates Instantaneous Rate Value of the rate at a particular time. Can be obtained by computing the slope of a line tangent to the curve at that point. Return to TOC Copyright © Cengage Learning. All rights reserved

Average vs. Instantaneous Rate
Instantaneous rate is the slope of the tangent to the curve at a particular time. We often are interested in the initial instantaneous rate; for the initial concentrations of reactants and products are known at this time. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Shows how the rate depends on the concentrations of reactants.
Section 12.3 Rate Laws: An Introduction Atomic Masses Rate Law Shows how the rate depends on the concentrations of reactants. For the decomposition of nitrogen dioxide: 2NO2(g) → 2NO(g) + O2(g) Rate = k[NO2]n: k = rate constant n = order of the reactant Return to TOC Copyright © Cengage Learning. All rights reserved

Section 12.3 Rate Laws: An Introduction Atomic Masses Rate Law Rate = k[NO2]n The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate. Return to TOC Copyright © Cengage Learning. All rights reserved

Section 12.3 Rate Laws: An Introduction Atomic Masses Rate Law Rate = k[NO2]n The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation. Return to TOC Copyright © Cengage Learning. All rights reserved

Section 12.3 Rate Laws: An Introduction Atomic Masses Types of Rate Laws Differential Rate Law (rate law) – shows how the rate of a reaction depends on concentrations. Integrated Rate Law – shows how the concentrations of species in the reaction depend on time. Return to TOC Copyright © Cengage Learning. All rights reserved

Section 12.3 Rate Laws: An Introduction Atomic Masses Rate Laws: A Summary Because we typically consider reactions only under conditions where the reverse reaction is unimportant, our rate laws will involve only concentrations of reactants. Because the differential and integrated rate laws for a given reaction are related in a well– defined way, the experimental determination of either of the rate laws is sufficient. Return to TOC Copyright © Cengage Learning. All rights reserved

Section 12.3 Rate Laws: An Introduction Atomic Masses Rate Laws: A Summary Experimental convenience usually dictates which type of rate law is determined experimentally. Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law. Return to TOC Copyright © Cengage Learning. All rights reserved

Videoclip Summary Kinetic #1 - 7 minutes
Section 12.3 Rate Laws: An Introduction Atomic Masses Videoclip Summary Kinetic #1 - 7 minutes Click below to watch videoclip. Link for Lesson minute video CosmoLearning Videoclips Return to TOC Copyright © Cengage Learning. All rights reserved

Section 12.3 Determining the Form of the Rate Law The Mole Determine experimentally the power to which each reactant concentration must be raised in the rate law. Return to TOC Copyright © Cengage Learning. All rights reserved

Method of Initial Rates
Section 12.3 Determining the Form of the Rate Law The Mole Method of Initial Rates The value of the initial rate is determined for each experiment at the same value of t as close to t = 0 as possible. Several experiments are carried out using different initial concentrations of each of the reactants, and the initial rate is determined for each run. The results are then compared to see how the initial rate depends on the initial concentrations of each of the reactants. Return to TOC Copyright © Cengage Learning. All rights reserved

Overall Reaction Order
Section 12.3 Determining the Form of the Rate Law The Mole Overall Reaction Order The sum of the exponents in the reaction rate equation. Rate = k[A]n[B]m Overall reaction order = n + m k = rate constant [A] = concentration of reactant A [B] = concentration of reactant B Return to TOC Copyright © Cengage Learning. All rights reserved

The Rate Law of a Chemical Reaction
The rate law for a chemical reaction relates the rate of reaction to the concentrations of reactants. aA + bB + cC … products rate = k[A]n[B]m[C]p … The exponents (m, n, p…) are determined by experiment. Exponents are not derived from the coefficients in the balanced chemical equation, though in some instances the exponents and the coefficients may be the same. The value of an exponent in a rate law is the order of the reaction with respect to the reactant in question. The proportionality constant, k, is the rate constant. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

The Rate Law Rate = k[A]1 = k[A] Reaction is first order in A Rate = k[A]2 Reaction is second order in A If we triple the concentration of A in a second-order reaction, the rate increases by a factor of ________. Rate = k[A]3 Reaction is third order in A 9 Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

More About the Rate Constant k
The rate of a reaction is the change in concentration with time, whereas the rate constant is the proportionality constant relating reaction rate to the concentrations of reactants. The rate constant remains constant throughout a reaction, regardless of the initial concentrations of the reactants. The rate and the rate constant have the same numerical values and units only in zero-order reactions. For reaction orders other than zero, the rate and rate constant are numerically equal only when the concentrations of all reactants are 1 M. Even then, their units are different. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Method of Initial Rates
The method of initial rates is a method of establishing the rate law for a reaction— finding the values of the exponents in the rate law, and the value of k. A series of experiments is performed in which the initial concentration of one reactant is varied. Concentrations of the other reactants are held constant. When we double the concentration of a reactant A, if: there is no effect on the rate, the reaction is zero- order in A. the rate doubles, the reaction is first-order in A. the rate quadruples, the reaction is second-order in A. the rate increases eight times, the reaction is third-order in A. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

The concentration of NO was held the same in Experiments 1 and 2 …
… while the concentration of Cl2 in Experiment 2 is twice that of Experiment 1. The rate in Experiment 2 is twice that in Experiment 1, so the reaction must be first order in Cl2. Which two experiments are used to find the order of the reaction in NO? How do we find the value of k after obtaining the order of the reaction in NO and in Cl2? Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Videoclip Finding Rate Laws from data - 3.5 minutes
Section 12.3 Rate Laws: An Introduction Atomic Masses Videoclip Finding Rate Laws from data minutes Click below to watch videoclip. Link for Lesson 12.3 Rate Law 3.5 minutes video CosmoLearning Videoclips Return to TOC Copyright © Cengage Learning. All rights reserved

Rate Laws: An Introduction Atomic Masses
Section 12.3 Rate Laws: An Introduction Atomic Masses Problems Day 1 - Stop here and work practice problems assigned. 2 handouts - #1-6 practice problems (Dr. paper) # to 12.4 assigned problems (but only have to do section 12.1 to 12.3 for now but should read ahead for section 12.4 in textbook reading assignment due Tues Feb. 4th - Friday - Kinetics Lab (Read and look over pre-lab) Turn in Clue Lab report today Return to TOC Copyright © Cengage Learning. All rights reserved

Assignments - Notes 12.3 revisited & 12.4 problems - handouts CW/HW: #7-12 Bradley worksheet Chem 201B CW/HW: #7-13 worksheet chemical “Kinetics p.279” CW/HW: problems D & E new handout with notes HW: Ch Take Home Test Get help on any other problems previously assigned Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Videoclip introduction similar to our Kinetic Lab data
Section 12.3 Rate Laws: An Introduction Atomic Masses Videoclip introduction similar to our Kinetic Lab data integrated-rate-law-in-0th-1st-and-2nd-order/ Return to TOC Copyright © Cengage Learning. All rights reserved

Kinetics III Integrated Rate Law - 0,1,2 order
Section 12.3 Rate Laws: An Introduction Atomic Masses Example from videoclip 2CO2 --> 2CO + O2 t (time) vs concentration M M M M Return to TOC Copyright © Cengage Learning. All rights reserved

Rate = k[A] Integrated: ln[A] = –kt + ln[A]o The Integrated Rate Law
Section 12.4 The Integrated Rate Law First-Order Rate = k[A] Integrated: ln[A] = –kt + ln[A]o [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Return to TOC Copyright © Cengage Learning. All rights reserved

First-Order Reactions
In a first-order reaction, the exponent in the rate law is 1. Rate = k[A]1 = k[A] The integrated rate law describes the concentration of a reactant as a function of time. For a first-order process: ln [A]t [A]0 = –kt Look! It’s an equation for a straight line! ln [A]t – ln [A]0 = –kt ln [A]t = –kt + ln [A]0 y = mx+b At times, it is convenient to replace molarities in an integrated rate law by quantities that are proportional to concentration. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Half-Life of Reactions
Section 12.4 The Integrated Rate Law Half-Life of Reactions Click above when hand appears to see visual. Be in play mode. Use my log-in as needed to access. “Bryant123” Return to TOC Copyright © Cengage Learning. All rights reserved

The Integrated Rate Law
Section 12.4 The Integrated Rate Law Plot of ln[N2O5] vs Time Return to TOC Copyright © Cengage Learning. All rights reserved

Half-life of a Reaction
The half-life (t½) of a reaction is the time required for one-half of the reactant originally present to be consumed. At t½, [A]t = ½[A]0, and for a first order reaction: ln ½[A]0 [A]0 = –kt½ ln (½) = –kt½ – = –kt½ t½ = /k Thus, for a first-order reaction, the half-life is a constant; it depends only on the rate constant, k, and not on the concentration of reactant. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Half–life does not depend on the concentration of reactants.
Section 12.4 The Integrated Rate Law First-Order Half–Life: k = rate constant Half–life does not depend on the concentration of reactants. Return to TOC Copyright © Cengage Learning. All rights reserved

The Integrated Rate Law Exercise
Section 12.4 The Integrated Rate Law Exercise A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? ln(0.65) = –k(55) + ln(1) k = 7.8 x 10-3 min-1. If students use [A] = 35 in the integrated rate law (instead of 65), they will get k = 1.9 x 10-2 min-1. Note: Use the red box animation to assist in explaining how to solve the problem. Return to TOC Copyright © Cengage Learning. All rights reserved

k = 7.8 x 10–3 min–1 The Integrated Rate Law Exercise
Section 12.4 The Integrated Rate Law Exercise A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? 1st order so look at ln (xo/x) = kt If 35% is completed then 65% remains - original would 100% or 1 so ln (1/.65) = kt where t is 55 minutes then solve for k k = 7.8 x 10–3 min–1 ln(0.65) = –k(55) + ln(1) k = 7.8 x 10-3 min-1. If students use [A] = 35 in the integrated rate law (instead of 65), they will get k = 1.9 x 10-2 min-1. Note: Use the red box animation to assist in explaining how to solve the problem. Return to TOC Copyright © Cengage Learning. All rights reserved

Second-Order Reactions
A reaction that is second order in a reactant has a rate law in which the exponent for that reactant is 2. Rate = k[A]2 The integrated rate law has the form: What do we plot vs. time to get a straight line? –––– = kt + –––– [A]t [A]0 The half-life of a second-order reaction depends on the initial concentration as well as on the rate constant k: 1 t½ = ––––– k[A]0 Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Rate = k[A]2 Integrated: The Integrated Rate Law Second-Order
Section 12.4 The Integrated Rate Law Second-Order Rate = k[A]2 Integrated: [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Return to TOC Copyright © Cengage Learning. All rights reserved

Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time
Section 12.4 The Integrated Rate Law Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time Return to TOC Copyright © Cengage Learning. All rights reserved

Each successive half–life is double the preceding one.
Section 12.4 The Integrated Rate Law Second-Order Half–Life: k = rate constant [A]o = initial concentration of A Half–life gets longer as the reaction progresses and the concentration of reactants decrease. Each successive half–life is double the preceding one. Return to TOC Copyright © Cengage Learning. All rights reserved

Write the rate law for this reaction. ? rate = k[A]2 b) Calculate k.
Section 12.4 The Integrated Rate Law Exercise For a reaction aA  Products, [A]0 = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively. Write the rate law for this reaction. ? rate = k[A]2 b) Calculate k. k = 8.0 x 10-3 M–1min–1 Calculate [A] at t = 525 minutes. [A] = 0.23 M a) rate = k[A]2 We know this is second order because the second half–life is double the preceding one. b) k = 8.0 x 10-3 M–1min–1 25 min = 1 / k(5.0 M) c) [A] = 0.23 M (1 / [A]) = (8.0 x 10-3 M–1min–1)(525 min) + (1 / 5.0 M) Return to TOC Copyright © Cengage Learning. All rights reserved

Rate = k[A]0 = k Integrated: [A] = –kt + [A]o The Integrated Rate Law
Section 12.4 The Integrated Rate Law Zero-Order Rate = k[A]0 = k Integrated: [A] = –kt + [A]o [A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Return to TOC Copyright © Cengage Learning. All rights reserved

A Zero-Order Reaction rate = k[A]0 = k
Rate is independent of initial concentration Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Section 12.4 The Integrated Rate Law Zero-Order Half–Life: k = rate constant [A]o = initial concentration of A Half–life gets shorter as the reaction progresses and the concentration of reactants decrease. Return to TOC Copyright © Cengage Learning. All rights reserved

Section 12.4 The Integrated Rate Law Concept Check How can you tell the difference among 0th, 1st, and 2nd order rate laws from their graphs? For the zero-order reaction, the graph of concentration versus time is a straight line with a negative slope. For a first-order graph, the graph is a natural log function. The second-order graph looks similar to the first-order, but with a greater initial slope. Students should be able to write a conceptual explanation of how the half-life is dependent on concentration (or in the case of first-order reactions, not dependent). Return to TOC Copyright © Cengage Learning. All rights reserved

Summary of Kinetic Data
Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Assignments Thurs 2-6 CW/HW: #7-12 Bradley worksheet Chem 201B CW/HW: #7-13 worksheet chemical “Kinetics p.279” CW/HW: problems D & E new handout with notes HW: Ch Take Home Test Get help on any other problems previously assigned Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

ASSIGNMENTS AP Chem Stop here Thursday Work on problem sets from handout AND the separate page... HW: ch to 12.4 #1-13 due Monday. HW: Read ch. 12 by Mon. HW: Read lab report TEST next WED. ch. 12 TURN in last week’s lab report. Yesterday’s lab report due next Thursday. Kinetics LAB tomorrow - switch rooms Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

ASSIGNMENTS AP Chem HW ch. 12 handout section due - check answers; make corrections with WORK shown to get credit. HW: ch to 12.4 #1-13 due today. Notes TEST ch. 12 WEDNESDAY. CW: Quiz ch. 12 (12 problems - 20 min.) If you haven’t read ch. 12, you need to along with packet handout. HW: Lab report due FRIDAY. The other lab report not formal lab report from EOC day. (11th will do Thurs.) Turn in previous labs, etc. this WEEK! Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Assignments Mon Notes - Kinetics starting with 12.6 packet with reaction mechanisms then the math/science packet. Turn in ch Test ANSWERS - We will put in the computer later. HW: Kinetics I tutorial packet for grade. HW: Kinetics checklist packet with provided answers for practice and test preparation - will not count as grade Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Integrated Rate Law Summary

The Integrated Rate Law Exercise
Section 12.4 The Integrated Rate Law Exercise A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? ln(0.65) = –k(55) + ln(1) k = 7.8 x 10-3 min-1. If students use [A] = 35 in the integrated rate law (instead of 65), they will get k = 1.9 x 10-2 min-1. Note: Use the red box animation to assist in explaining how to solve the problem. Return to TOC Copyright © Cengage Learning. All rights reserved

k = 7.8 x 10–3 min–1 The Integrated Rate Law Exercise
Section 12.4 The Integrated Rate Law Exercise A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? 1st order so look at ln (xo/x) = kt If 35% is completed then 65% remains - original would 100% or 1 so ln (1/.65) = kt where t is 55 minutes then solve for k k = 7.8 x 10–3 min–1 ln(0.65) = –k(55) + ln(1) k = 7.8 x 10-3 min-1. If students use [A] = 35 in the integrated rate law (instead of 65), they will get k = 1.9 x 10-2 min-1. Note: Use the red box animation to assist in explaining how to solve the problem. Return to TOC Copyright © Cengage Learning. All rights reserved

Zero order First order Second order 4.7 M 3.7 M 2.0 M
Section 12.4 Exercise The Integrated Rate Law 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 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 Return to TOC Copyright © Cengage Learning. All rights reserved

Most chemical reactions occur by a series of elementary steps.
Section 12.5 Reaction Mechanisms 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. Return to TOC Copyright © Cengage Learning. All rights reserved

Reaction Mechanisms Analogy: a banana split is made by steps in sequence: slice banana; three scoops ice cream; chocolate sauce; strawberries; pineapple; whipped cream; end with cherry. A chemical reaction occurs according to a reaction mechanism—a series of collisions or dissociations—that lead from initial reactants to the final products. An elementary reaction represents, at the molecular level, a single step in the progress of the overall reaction. A proposed mechanism must: account for the experimentally determined rate law. be consistent with the stoichiometry of the overall or net reaction. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Molecularity The molecularity of an elementary reaction refers to the number of free atoms, ions, or molecules that collide or dissociate in that step. Termolecular processes are unusual, for the same reason that three basketballs shot at the same time are unlikely to collide at the same instant … Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Elementary Steps (Molecularity)
Section 12.5 Reaction Mechanisms 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. Return to TOC Copyright © Cengage Learning. All rights reserved

NO2(g) + CO(g) → NO(g) + CO2(g)
Section 12.5 Reaction Mechanisms A Molecular Representation of the Elementary Steps in the Reaction of NO2 and CO NO2(g) + CO(g) → NO(g) + CO2(g) Return to TOC Copyright © Cengage Learning. All rights reserved

The Rate-Determining Step
The rate-determining step is the crucial step in establishing the rate of the overall reaction. It is usually the slowest step. A reaction is only as fast as its slowest step. Some two-step mechanisms have a slow first step followed by a fast second step, while others have a fast reversible first step followed by a slow second step. Fast Mechanism for 2 NO + O2  2 NO2 Slow The rate-determining step (slowest step) determines the rate law and the molecularity of the overall reaction. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Reaction Mechanism Requirements
Section 12.5 Reaction Mechanisms 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. Return to TOC Copyright © Cengage Learning. All rights reserved

Decomposition of N2O5 at 67 °C
Initial amount After one half-life, half the N2O5 has reacted. After two half-lives, half of the remaining N2O5 has reacted—three-fourths has been consumed. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Step 2: NO2 + NO3 → NO + O2 + NO2 (slow)
Section 12.5 Reaction Mechanisms 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( ) Return to TOC Copyright © Cengage Learning. All rights reserved

The reaction A + 2B  C has the following proposed mechanism:
Section 12.5 Reaction Mechanisms 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. Return to TOC Copyright © Cengage Learning. All rights reserved

Theories of Chemical Kinetics: Collision Theory
Before atoms, molecules, or ions can react, they must first collide. An effective collision between two molecules puts enough energy into key bonds to break them. The activation energy (Ea) is the minimum energy that must be supplied by collisions for a reaction to occur. A certain fraction of all molecules in a sample will have the necessary activation energy to react; that fraction increases with increasing temperature. (general rule: 10 degree increase in temp. (Celsius/Kelvin) = doubles rate) The spatial orientations of the colliding species may also determine whether a collision is effective. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Importance of Orientation
One hydrogen atom can approach another from any direction … Effective collision; the I atom can bond to the C atom to form CH3I Ineffective collision; orientation is important in this reaction. … and reaction will still occur; the spherical symmetry of the atoms means that orientation does not matter. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Molecules must collide to react. Main Factors: Activation energy, Ea
Section 12.6 A Model for Chemical Kinetics Collision Model Molecules must collide to react. Main Factors: Activation energy, Ea Temperature Molecular orientations Return to TOC Copyright © Cengage Learning. All rights reserved

Distribution of Kinetic Energies
At higher temperature (red), more molecules have the necessary activation energy. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Energy that must be overcome to produce a chemical reaction.
Section 12.6 A Model for Chemical Kinetics Activation Energy, Ea Energy that must be overcome to produce a chemical reaction. Return to TOC Copyright © Cengage Learning. All rights reserved

Transition State Theory
The configuration of the atoms of the colliding species at the time of the collision is called the transition state. The transitory species having this configuration is called the activated complex. A reaction profile shows potential energy plotted as a function of a parameter called the progress of the reaction. Reactant molecules must have enough energy to surmount the energy “hill” separating products from reactants. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

A Model for Chemical Kinetics
Section 12.6 A Model for Chemical Kinetics Transition States and Activation Energy Click below to watch visual animation. Return to TOC Copyright © Cengage Learning. All rights reserved

An Analogy for Reaction Profiles and Activation Energy

For Reactants to Form Products
Section 12.6 A Model for Chemical Kinetics 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. Return to TOC Copyright © Cengage Learning. All rights reserved

A Model for Chemical Kinetics
Section 12.6 A Model for Chemical Kinetics The Gas Phase Reaction of NO and Cl2 Click below to watch visual animation. Return to TOC Copyright © Cengage Learning. All rights reserved

Effect of Temperature on the Rates of Reactions
In 1889, Svante Arrhenius proposed the following expression for the effect of temperature on the rate constant, k: k = Ae–Ea/RT The constant A, called the frequency factor, is an expression of collision frequency and orientation; it represents the number of collisions per unit time that are capable of leading to reaction. The term e–Ea/RT represents the fraction of molecular collisions sufficiently energetic to produce a reaction. Prentice Hall © 2005 Chapter Thirteen General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

R = gas constant (8.3145 J/K·mol) T = temperature (in K)
Section 12.6 Arrhenius Equation A Model for Chemical Kinetics Note the division sign “/”. A = frequency factor Ea = activation energy R = gas constant ( J/K·mol) T = temperature (in K) Return to TOC Copyright © Cengage Learning. All rights reserved

Ea = 53 kJ A Model for Chemical Kinetics Exercise
Section 12.6 A Model for Chemical Kinetics 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)] Return to TOC Copyright © Cengage Learning. All rights reserved

A substance that speeds up a reaction without being consumed itself.
Section 12.7 Catalysis Catalyst A substance that speeds up a reaction without being consumed itself. Provides a new pathway for the reaction with a lower activation energy. Return to TOC Copyright © Cengage Learning. All rights reserved

Catalysis Teaching Videoclip & Example
Section 12.7 Catalysis Catalysis Teaching Videoclip & Example Optional Click below. Must be logged in cengage to watch. /2944.html Good example to show Elephant Toothpaste experiment demonstration. Return to TOC Copyright © Cengage Learning. All rights reserved

Effect of a Catalyst on the Number of Reaction-Producing Collisions
Section 12.7 Catalysis Effect of a Catalyst on the Number of Reaction-Producing Collisions Return to TOC Copyright © Cengage Learning. All rights reserved

Heterogeneous Catalyst
Section 12.7 Catalysis 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. Return to TOC Copyright © Cengage Learning. All rights reserved

Heterogeneous Catalysis Click picture below to watch animation.
Section 12.7 Catalysis Heterogeneous Catalysis Click picture below to watch animation. Return to TOC Copyright © Cengage Learning. All rights reserved

Heterogeneous Catalyst
Section 12.7 Catalysis 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. Return to TOC Copyright © Cengage Learning. All rights reserved

Exists in the same phase as the reacting molecules.
Section 12.7 Catalysis Homogeneous Catalyst Exists in the same phase as the reacting molecules. Enzymes are nature’s catalysts. Return to TOC Copyright © Cengage Learning. All rights reserved

Homogeneous Catalysis - Click below for visual. Requires log-in
Section 12.7 Catalysis Homogeneous Catalysis - Click below for visual. Requires log-in Return to TOC Copyright © Cengage Learning. All rights reserved

Chapter 12 Chemical Kinetics.

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Chapter 12 Chemical Kinetics.

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