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- 1. Chapter 14 Chemical Kinetics CHEMISTRY The Central Science 9th Edition
- 2. <ul><li>Kinetics is the study of how fast chemical reactions occur. </li></ul><ul><li>There are 4 important factors which affect the rates of chemical reactions: </li></ul><ul><ul><li>reactant concentration, </li></ul></ul><ul><ul><li>temperature, </li></ul></ul><ul><ul><li>action of catalysts, and </li></ul></ul><ul><ul><li>surface area. </li></ul></ul>Kinetics
- 3. <ul><li>The speed of a reaction is defined as the change that occurs per unit time. </li></ul><ul><ul><li>It is determined by measuring the change in concentration of a reactant or product with time. </li></ul></ul><ul><ul><li>The speed the of the reaction is called the reaction rate . </li></ul></ul><ul><li>For a reaction A B </li></ul><ul><li>Suppose A reacts to form B. Let us begin with 1.00 mol A. </li></ul>Reaction Rates
- 4. Change in Concentration of Reactions 10 20
- 5. <ul><ul><li>At t = 0 (time zero) there is 1.00 mol A (100 red spheres) and no B present. </li></ul></ul><ul><ul><li>At t = 10 min, there is 0.54 mol A and 0.26 mol B. </li></ul></ul><ul><ul><li>At t = 20 min, there is 0.30 mol A and 0.70 mol B. </li></ul></ul><ul><ul><li>Calculating, </li></ul></ul>Calculating Reaction Rates Using Products
- 6. <ul><li>For the reaction A B there are two ways of measuring rate: </li></ul><ul><ul><li>the speed at which the products appear (i.e. change in moles of B per unit time), or </li></ul></ul><ul><ul><li>the speed at which the reactants disappear (i.e. the change in moles of A per unit time). </li></ul></ul><ul><ul><li>The equation, when calculating rates of reactants, is multiplied by -1 to compensate for the negative concentration. </li></ul></ul><ul><ul><ul><li>By convention rates are expressed as positive numbers. </li></ul></ul></ul>Calculating Reaction Rates Using Reactants
- 7. <ul><li>Most useful units for rates are to look at molarity. Since volume is constant, molarity and moles are directly proportional. </li></ul><ul><li>Consider: </li></ul><ul><li>C 4 H 9 Cl( aq ) + H 2 O( l ) C 4 H 9 OH( aq ) + HCl( aq ) </li></ul>Reaction Rates Reactants Products
- 8. Reaction Rates for C 4 H 9 Cl
- 9. <ul><li>C 4 H 9 Cl( aq ) + H 2 O( l ) C 4 H 9 OH( aq ) + HCl( aq ) </li></ul><ul><ul><li>We can calculate the average rate in terms of the disappearance of C 4 H 9 Cl. </li></ul></ul><ul><ul><li>The units for average rate are mol/L·s or M/s . </li></ul></ul><ul><ul><li>The average rate decreases with time. </li></ul></ul><ul><ul><li>We plot [C 4 H 9 Cl] versus time. </li></ul></ul><ul><ul><li>The rate at any instant in time (instantaneous rate) is the slope of the tangent to the curve. </li></ul></ul><ul><ul><li>Instantaneous rate is different from average rate. </li></ul></ul><ul><ul><li>We usually call the instantaneous rate the rate. </li></ul></ul>Properties of C 4 H 9 Cl Reaction
- 10. Instantaneous Reaction Rates for C 4 H 9 Cl
- 11. <ul><li>For the reaction </li></ul><ul><li>C 4 H 9 Cl( aq ) + H 2 O( l ) C 4 H 9 OH( aq ) + HCl( aq ) </li></ul><ul><li>we know </li></ul><ul><li>In general for </li></ul><ul><li>a A + b B c C + d D </li></ul>Reaction Rates and Stoichiometry
- 12. <ul><li>In general rates increase as concentrations increase. </li></ul><ul><li>NH 4 + ( aq ) + NO 2 - ( aq ) N 2 ( g ) + 2H 2 O( l ) </li></ul>Concentration and Rate Table
- 13. <ul><li>For the reaction </li></ul><ul><li>NH 4 + ( aq ) + NO 2 - ( aq ) N 2 ( g ) + 2H 2 O( l ) </li></ul><ul><li>we note </li></ul><ul><ul><li>as [NH 4 + ] doubles with [NO 2 - ] constant the rate doubles, </li></ul></ul><ul><ul><li>as [NO 2 - ] doubles with [NH 4 + ] constant, the rate doubles, </li></ul></ul><ul><ul><li>We conclude rate [NH 4 + ][NO 2 - ]. </li></ul></ul><ul><li>Rate law: </li></ul><ul><li>The constant k is the rate constant. </li></ul>Concentration and Rate Equation
- 14. <ul><li>For a general reaction with rate law </li></ul><ul><li>we say the reaction is m th order in reactant 1 and n th order in reactant 2. </li></ul><ul><li>The overall order of reaction is m + n + …. </li></ul><ul><li>A reaction can be zeroth order if m , n , … are zero. </li></ul><ul><li>Note the values of the exponents (orders) have to be determined experimentally. They are not simply related to stoichiometry. </li></ul>Exponents in the Rate Law
- 15. <ul><li>A reaction is zero order in a reactant if the change in concentration of that reactant produces no effect. </li></ul><ul><li>A reaction is first order if doubling the concentration causes the rate to double. </li></ul><ul><li>A reaction is n th order if doubling the concentration causes an 2 n increase in rate. </li></ul><ul><li>Note that the rate constant does not depend on concentration. </li></ul>Determining Order of Reactions
- 16. <ul><li>First Order Reactions </li></ul><ul><li>Goal: convert rate law into a convenient equation to give concentrations as a function of time. </li></ul><ul><li>For a first order reaction, the rate doubles as the concentration of a reactant doubles. </li></ul>Concentration Change with Time
- 17. <ul><li>A plot of ln[A] t versus t is a straight line with slope - k and intercept ln[A] 0 . </li></ul><ul><ul><li>Put into simple math language : y = m x + b </li></ul></ul><ul><li>Plotting of this equation for given reaction should yield a straight line if the reaction is first order. </li></ul><ul><li>In the above we use the natural logarithm, ln, which is log to the base e . </li></ul>Plotting First Order Reactions
- 18. <ul><li>Left graph plotted with pressure vs. t, and right graph plotted with ln(pressure) vs. t. </li></ul>Example Plots of a 1 st Order Reaction
- 19. <ul><li>Second Order Reactions </li></ul><ul><li>For a second order reaction with just one reactant </li></ul><ul><li>A plot of 1/[A] t versus t is a straight line with slope k and intercept 1/[A] 0 </li></ul><ul><ul><li>Put into simple math language: y = m x + b </li></ul></ul><ul><li>For a second order reaction, a plot of ln[A] t vs. t is not linear. </li></ul><ul><ul><li>However, a plot of 1/[A] t versus t is a straight line </li></ul></ul>Concentration Change with Time
- 20. <ul><li>Left is Ln[NO 2 ] vs t, and right is 1/[NO 2 ] vs. t. </li></ul>Example plots of a 2 nd Order Reaction
- 21. <ul><li>First Order Reactions </li></ul><ul><li>Half-life is the time taken for the concentration of a reactant to drop to half its original value. </li></ul><ul><li>For a first order process, half life, t ½ is the time taken for [A] 0 to reach ½[A] 0 . </li></ul><ul><li>Mathematically defined by: </li></ul><ul><li>The half-life for a 1 st order reactions depends only on k </li></ul>Half-Life Reactions
- 22. <ul><li>Second Order </li></ul><ul><li>Mathematically defined by: </li></ul><ul><li>A second order reaction’s half-life depends on the initial concentration of the reactants </li></ul>Half-Life Reaction
- 23. <ul><li>The Collision Model </li></ul><ul><li>Most reactions speed up as temperature increases. (E.g. food spoils when not refrigerated.) </li></ul><ul><li>When two light sticks are placed in water: one at room temperature and one in ice, the one at room temperature is brighter than the one in ice. </li></ul><ul><li>The chemical reaction responsible for chemiluminescence is dependent on temperature: the higher the temperature, the faster the reaction and the brighter the light. </li></ul>Temperature and Rate
- 24. The Collision Model <ul><li>As temperature increases, the rate increases. </li></ul>
- 25. <ul><li>Goal: develop a model that explains why rates of reactions increase as concentration and temperature increases. </li></ul><ul><li>The collision model: in order for molecules to react they must collide. </li></ul><ul><li>The greater the number of collisions the faster the rate. </li></ul><ul><li>The more molecules present, the greater the probability of collision and the faster the rate. </li></ul><ul><li>Faster moving molecule collide with greater energy and more frequently, increasing reaction rates. </li></ul>Collision Model: The Central Idea
- 26. <ul><li>The Collision Model </li></ul><ul><li>The higher the temperature, the more energy available to the molecules and the faster the rate. </li></ul><ul><li>Complication: not all collisions lead to products. In fact, only a small fraction of collisions lead to product. </li></ul><ul><ul><li>Why is this? </li></ul></ul><ul><li>The Orientation Factor </li></ul><ul><li>In order for reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products. </li></ul>The Speed of a Reaction
- 27. <ul><li>Consider: </li></ul><ul><li>Cl + NOCl NO + Cl 2 </li></ul><ul><li>There are two possible ways that Cl atoms and NOCl molecules can collide; one is effective and one is not. </li></ul>The Orientation Factor
- 28. <ul><li>Arrhenius: molecules must posses a minimum amount of energy to react. Why? </li></ul><ul><ul><li>In order to form products, bonds must be broken in the reactants. </li></ul></ul><ul><ul><li>Bond breakage requires energy. </li></ul></ul><ul><li>Activation energy, E a , is the minimum energy required to initiate a chemical reaction. </li></ul>Activation Energy
- 29. Energy Profile for Methly Isonitrile
- 30. <ul><li>How does a methyl isonitrile molecule gain enough energy to overcome the activation energy barrier? </li></ul><ul><li>From kinetic molecular theory, we know that as temperature increases, the total kinetic energy increases. </li></ul><ul><li>We can show the fraction of molecules, f , with energy equal to or greater than E a is </li></ul><ul><ul><li>where R is the gas constant (8.314 J/mol·K). </li></ul></ul>Fraction of Molecules Possessing E a
- 31. Activation Energy, E a , Plot
- 32. <ul><li>Arrhenius discovered most reaction-rate data obeyed the Arrhenius equation: </li></ul><ul><ul><li>k is the rate constant, E a is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K. </li></ul></ul><ul><ul><li>A is called the frequency factor. </li></ul></ul><ul><ul><ul><li>A is a measure of the probability of a favorable collision. </li></ul></ul></ul><ul><ul><li>Both A and E a are specific to a given reaction. </li></ul></ul>The Arrhenius Equation
- 33. <ul><li>If we have a lot of data, we can determine E a and A graphically by rearranging the Arrhenius equation: </li></ul><ul><li>From the above equation, a plot of ln k versus 1/ T will have slope of – E a /R and intercept of ln A . </li></ul>Determing Activation Energy
- 34. Ln k versus 1/T <ul><li>E a can be determined by finding the slope of the line. </li></ul>
- 35. <ul><li>The balanced chemical equation provides information about the beginning and end of reaction. </li></ul><ul><li>The reaction mechanism gives the path of the reaction (i.e., process by which a reaction occurs). </li></ul><ul><ul><li>Mechanisms provide a very detailed picture of which bonds are broken and formed during the course of a reaction. </li></ul></ul><ul><li>Elementary Steps </li></ul><ul><li>Elementary step: any process that occurs in a single step. </li></ul>Reaction Mechanisms
- 36. <ul><li>Molecularity: the number of molecules present in an elementary step. </li></ul><ul><ul><li>Unimolecular: one molecule in the elementary step, </li></ul></ul><ul><ul><li>Bimolecular: two molecules in the elementary step, and </li></ul></ul><ul><ul><li>Termolecular: three molecules in the elementary step. </li></ul></ul><ul><li>It is not common to see termolecular processes (statistically improbable). </li></ul>Properties of the Elementary Step Process
- 37. <ul><li>Some reaction proceed through more than one step: </li></ul><ul><ul><li>Consider the reaction of NO 2 and CO </li></ul></ul><ul><li>NO 2 ( g ) + NO 2 ( g ) NO 3 ( g ) + NO( g ) </li></ul><ul><li>NO 3 ( g ) + CO( g ) NO 2 ( g ) + CO 2 ( g ) </li></ul><ul><ul><li>Notice that if we add the above steps, we get the overall reaction: </li></ul></ul><ul><li>NO 2 ( g ) + CO( g ) NO( g ) + CO 2 ( g ) </li></ul><ul><li>The elementary steps must add to give the balanced chemical equation. </li></ul><ul><li>Intermediate : a species which appears in an elementary step which is not a reactant or product </li></ul>Multistep Mechanisms
- 38. <ul><li>Rate Laws for Elementary Steps </li></ul><ul><li>The rate law of an elementary step is determined by its molecularity: </li></ul><ul><ul><li>Unimolecular processes are first order, </li></ul></ul><ul><ul><li>Bimolecular processes are second order, and </li></ul></ul><ul><ul><li>Termolecular processes are third order. </li></ul></ul><ul><li>Rate Laws for Multistep Mechanisms </li></ul><ul><li>Rate-determining step: is the slowest of the elementary steps. </li></ul>Reaction Mechanisms
- 39. <ul><li>Rate Laws for Elementary Steps </li></ul>Table 14.3, Page 551
- 40. <ul><li>It is possible for an intermediate to be a reactant. </li></ul><ul><li>Consider </li></ul><ul><li>2NO( g ) + Br2( g ) 2NOBr( g ) </li></ul><ul><li>The experimentally determined rate law is </li></ul><ul><ul><ul><ul><ul><li>Rate = k[NO] 2 [Br2] </li></ul></ul></ul></ul></ul><ul><li>Consider the following mechanism </li></ul>Initial Fast Step of a Mechanisms
- 41. <ul><li>The rate law is (based on Step 2 ): </li></ul><ul><li>Rate = k 2 [NOBr 2 ][NO] </li></ul><ul><li>The rate law should not depend on the concentration of an intermediate (intermediates are usually unstable). </li></ul><ul><li>Assume NOBr 2 is unstable, so we express the concentration of NOBr 2 in terms of NOBr and Br 2 assuming there is an equilibrium in step 1 we have </li></ul>Intermediate as a Reactant
- 42. <ul><li>By definition of equilibrium: </li></ul><ul><li>Therefore, the overall rate law becomes </li></ul><ul><li>Note the final rate law is consistent with the experimentally observed rate law. </li></ul>Intermediate as a Reactant Conts.
- 43. <ul><li>A catalyst changes the rate of a chemical reaction. </li></ul><ul><ul><li>Catalyst lower the overall E a for a chemical reaction. </li></ul></ul><ul><li>There are two types of catalyst: </li></ul><ul><ul><li>Homogeneous and heterogeneous </li></ul></ul><ul><ul><li>Example: Cl atoms are catalysts for the destruction of ozone. </li></ul></ul><ul><ul><li>Homogeneous Catalysis </li></ul></ul><ul><li>The catalyst and reaction is in one phase. </li></ul><ul><li>Heterogeneous Catalysis </li></ul><ul><li>The catalyst and reaction exists in a different phase. </li></ul>Catalysis
- 44. The Effects of a Catalyst
- 45. <ul><li>Catalysts can operate by increasing the number of effective collisions (i.e., from the Arrhenius equation: catalysts increase k which results in increasing A or decreasing E a . </li></ul><ul><li>A catalyst may add intermediates to the reaction. </li></ul><ul><ul><li>Example: In the presence of Br - , Br 2 ( aq ) is generated as an intermediate in the decomposition of H 2 O 2 . </li></ul></ul>Functions of the Catalysis
- 46. End of Chapter 14 Chemical Kinetics

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