Prentice Hall © 2003Chapter 14 Chapter 14 Chemical Kinetics CHEMISTRY The Central Science 9th Edition

Prentice Hall © 2003Chapter 14 Kinetics is the study of how fast chemical reactions occur There are 4 important factors which affect rates of reactions: –reactant concentration –temperature –action of catalysts –surface area (dust grain elevators/saw mills) 14.1: Factors that Affect Reaction Rates

Prentice Hall © 2003Chapter 14 The rate of a reaction is measured by the change in concentration with time For a reaction A  B 14.2: Reaction Rates

Prentice Hall © 2003Chapter 14 Suppose A reacts to form B. If we begin with 1.00 mol A At t = 0 (time zero) there is 1.00 mol A (100 red spheres) and no B present At t = 20 min, there is 0.54 mol A and 0.46 mol B At t = 40 min, there is 0.30 mol A and 0.70 mol B Picture & scenario from Text, P. 527

Prentice Hall © 2003Chapter 14 For the reaction A  B there are two ways of measuring rate: –the speed at which the products appear or –the speed at which the reactants disappear Note the negative sign!

Prentice Hall © 2003Chapter 14 Change of Rate with Time The most useful unit for rate is molarity volume is constant molarity and moles are directly proportional

C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Text, P. 529

The average rate of disappearance of reactant decreases with time Plot [C 4 H 9 Cl] versus time: The rate at any instant in time (instantaneous rate) is the slope of the tangent to the curve The units for average rate are mol/L·s or M/s Instantaneous rate is different from average rate We usually call the instantaneous rate the rate

Prentice Hall © 2003Chapter 14 Reaction Rate and Stoichiometry For the reaction C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) we know In general for aA + bB  cC + dD We want to describe the rate of change on a basis that is the same regardless of which reactant or product is chosen to be measured, so divide each change by its coefficient!

Prentice Hall © 2003Chapter 14 Think sandwiches: 2 bread slices + 3 sardines + 1 pickle  1 sandwich Write 4 rate expressions for this recipe

Prentice Hall © 2003Chapter 14 Spectrophotometer –Measures the amount of light absorbed by a sample Beer’s Law: Graph the change in concentration over a period of time Text, P. 532

Prentice Hall © 2003Chapter 14 Sample problems # 5, 7, 9, 11

Prentice Hall © 2003Chapter 14

Prentice Hall © 2003Chapter 14 Sample problems # 5, 7, 9 & 11

Prentice Hall © 2003Chapter 14 In general rates increase as concentrations increase NH 4 + (aq) + NO 2 - (aq)  N 2 (g) + 2H 2 O(l) 14.3: Concentration and Rate

Prentice Hall © 2003Chapter 14 For this reaction: –as [NH 4 + ] doubles with [NO 2 - ] constant, the rate doubles –as [NO 2 - ] doubles with [NH 4 + ] constant, the rate doubles –We conclude rate  [NH 4 + ][NO 2 - ] Rate law: The constant k is the rate constant

Prentice Hall © 2003Chapter 14 Exponents in the Rate Law For a general reaction with the rate law we say the reaction is mth order in reactant 1 and nth order in reactant 2 The overall order of reaction is m + n + …. A reaction can be zeroth order if m, n, … are zero The values of the exponents (orders) have to be determined experimentally They are not simply related to stoichiometry

Prentice Hall © 2003Chapter 14 Regarding k: –Its value is for a specific reaction (balanced equation) –Its units depend on the overall order of the reaction –Its value does not change with concentration of reactants or products –Its value does not change with time –Its value refers to a particular temperature (changes with temperature) –It depends on a catalyst presence –It must be determined experimentally

Prentice Hall © 2003Chapter 14 Using Initial Rates to Determines Rate Laws More about m and n: –When m or n = 1, the rate is directly proportional to the concentration of the reactant –When m or n = 2, the rate is directly proportional to the square of the concentration of that reactant A reaction is xth order if doubling the concentration causes a 2x increase in rate –When m or n = 0, the rate does not depend on the concentration of that reactant

Prentice Hall © 2003Chapter 14 Units for k For a reaction that is 1 st order overall, k = time -1 For a reaction that is 2 nd order overall, k = M -1 time -1 For a reaction that is 3 rd order overall, k = M -2 time -1 Example: M -1 s -1 is read as L/mol-s

Prentice Hall © 2003Chapter 14 Sample Problems #13, 17, 21, 23

Prentice Hall © 2003Chapter 14 The integrated rate equation relates time and concentration for chemical and nuclear reactions –Concentration of a reactant remaining –Time required for a fraction a reactant to be used up Half-life –Time required for a reactant concentration to fall to a certain level 14.4: The Change of Concentration with Time

Prentice Hall © 2003Chapter 14 An example of a reaction that is 1 st order in the reactant and 1 st order overall is: a A  products –Rate of reaction depends on the concentration of a single reactant –Common for many chemical reactions and all simple radioactive decays

Prentice Hall © 2003Chapter 14 where: [A] 0 = concentration of A at time t=0[A] = conc. of A at time t k = specific rate constant t = time elapsed since start a = stoichiometric coefficient of A in balanced overall equation The integrated rate equation for first order reactions is: or Relates concentration at the start to concentration at some time, t

Prentice Hall © 2003Chapter 14 Interpret the equation as: –y can be identified with ln[A] and plotted on the y-axis –m can be identified with –ak and is the slope of the line –x can be identified with t and plotted on the x-axis –b can be identified with ln[A] 0 and is the y-intercept

Prentice Hall © 2003Chapter 14 The order of the reaction can be determined from the graph that results in a straight line: Make three different graphs of sample data, look for the linear plot: 1Plot the [A] (y-axis) vs. time (x-axis) If the plot is linear then the reaction is zero order 2Plot the ln [A] (y-axis) vs. time (x-axis) If the plot is linear then the reaction is first order 3Plot 1/ [A] (y-axis) vs. time (x-axis) If the plot is linear then the reaction is second order

Prentice Hall © 2003Chapter 14

Prentice Hall © 2003Chapter 14 1 st order with respect to C 2 H 5 Br

Prentice Hall © 2003Chapter 14 Solve the first order integrated rate equation for t: The half-life, t 1/2, of a reactant is the time required for half of the reactant to be consumed, or the time at which [A]=1/2[A] 0 then The equation for half-life for a 1 st order reaction (doesn’t depend on initial concentration):

Prentice Hall © 2003Chapter 14 For reactions that are second order with respect to a particular reactant ( A  A + B) and second order overall, the rate equation is: Where: [A] 0 = concentration of A at time t=0 [A] = conc. of A at time t k = specific rate constant t = time elapsed since start a = stoichiometric coefficient of A

Prentice Hall © 2003Chapter 14 Compare the equation for a straight line and the second order rate-law expression:

Prentice Hall © 2003Chapter 14 Once again, make three different graphs of sample data, look for the linear plot 1. Plot [A] (y-axis) vs. time (x-axis) –Linear: the reaction is zero order 2. Plot ln [A] (y-axis) vs. time (x-axis) Linear: the reaction is first order 3. Plot 1/ [A] (y-axis) vs. time (x-axis) –Linear: the reaction is second order

Prentice Hall © 2003Chapter 14 Thus this is a second order reaction with respect to [NO 2 ]

Prentice Hall © 2003Chapter 14 For a second order reaction, half-life depends on the initial concentration:

Prentice Hall © 2003Chapter 14 Interesting website: ateLaws.html ateLaws.html Sample problems: #29, 31, 37

The Collision Model As temperature increases, the rate increases Since the rate law has no temperature term in it, the rate constant must depend on temperature 14.5: Temperature and Rate

Prentice Hall © 2003Chapter 14 The Collision Model The collision model: in order for molecules to react they must collide The rate of reaction is increased when: There are a greater the number of collisions There are more molecules present, (greater probability of collision) There is a higher temperature Complication: not all collisions lead to products only a small fraction of them do so

Prentice Hall © 2003Chapter 14 The Orientation Factor In order for reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products Cl + NOCl  NO + Cl 2 There are two possible ways that Cl atoms and NOCl molecules can collide; one is effective and one is not

Prentice Hall © 2003Chapter 14 Activation Energy Svante Arrhenius: molecules must posses a minimum amount of energy to react In order to form products, bonds must be broken in the reactants –Bond breakage requires energy Activation energy, E a, is the minimum energy required to initiate a chemical reaction

Prentice Hall © 2003Chapter 14 Cross section of mountain Boulder E activation hh h2h2 h1h1 E pot =mgh 2 E pot =mgh 1 Height  E pot = mg  h

Prentice Hall © 2003Chapter 14 Potential Energy Reaction Coordinate X 2 + Y 2 2 XY E activation - a kinetic quantity  E  H a thermodynamic quantity Representation of a chemical reaction

Prentice Hall © 2003Chapter 14 The relationship between the activation energy for forward and reverse reactions is –Forward reaction = E a (typically exothermic) The lower the E a, the faster the reaction –Reverse reaction = E a +  E (typically endothermic) –Difference between E forward and E reverse =  E Potential Energy Reaction Coordinate X 2 + Y 2 2 XY E activation - a kinetic quantity  E  H a thermodynamic quantity

Prentice Hall © 2003Chapter 14 More molecules have KE greater than E a, which leads to a greater rate of reaction Text, P. 546

Prentice Hall © 2003Chapter 14 The Arrhenius Equation Arrhenius: most reaction-rate data obeyed the Arrhenius equation: –k is the rate constant –E a is the activation energy –R is the gas constant (8.314 J/K-mol) –T is the temperature in K –A is called the frequency factor –It is a measure of the probability of a favorable collision –Both A and E a are specific to a given reaction

Prentice Hall © 2003Chapter 14 Determining the Activation Energy If we have a lot of data, E a and A can be determined graphically by rearranging the Arrhenius equation: A plot of ln k versus 1/T will have m = –E a /R and an intercept of ln A

Prentice Hall © 2003Chapter 14 If we do not have a lot of data, then we recognize From text, P. 547:

Prentice Hall © 2003Chapter 14 Sample Problems # 43, 47, 49, 51

Prentice Hall © 2003Chapter 14 Balanced chemical equation: information about the beginning and end of reaction Reaction mechanism: the path of the reaction detailed picture of the bonds are broken/formed during a reaction 14.6: Reaction Mechanisms

Prentice Hall © 2003Chapter 14 Elementary Steps Elementary step: any process that occurs in a single step Molecularity: the number of molecules present in an elementary step –Unimolecular: one molecule –Bimolecular: two molecules –Termolecular: three molecules –Uncommon

Prentice Hall © 2003Chapter 14 Multistep Mechanisms Some reactions occur in more than one step (P. 550): NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) Add the above steps to get the overall reaction: NO 2 (g) + CO(g)  NO(g) + CO 2 (g) Intermediate: a species which appears in an elementary step which is not a reactant or product

Prentice Hall © 2003Chapter 14 Rate Laws for Elementary Steps The rate law of an elementary step is determined by its molecularity: –Unimolecular = first order –Bimolecular = second order –Termolecular = third order Text, P. 551 Remember: You can’t tell by looking at a balanced equation whether 1 or more elementary steps are required!

Prentice Hall © 2003Chapter 14 Rate Laws for Multistep Mechanisms The rate-determining step governs the overall rate law for the reaction Rate-determining step: the slowest of the elementary steps A reaction can’t occur faster than its slowest step The reaction orders for an elementary step are equal to the coefficients for that step

Prentice Hall © 2003Chapter 14 Mechanisms with an Initial Fast Step It is possible for an intermediate to be a reactant in the rate-determining step (Text, P. 553): 2NO(g) + Br 2 (g)  2NOBr(g) The experimentally determined rate law is Rate = k[NO] 2 [Br 2 ] Consider the following mechanism (P. 554): The rate law is based on the slow step: Rate = k 2 [NOBr 2 ][NO]

Prentice Hall © 2003Chapter 14 –The rate law should not include concentrations of intermediates NOBr 2 is unstable, concentration difficult to determine Express the concentration of NOBr 2 in terms of NO and Br 2 assuming there is an equilibrium in step 1:

Prentice Hall © 2003Chapter 14 Substituting into the rate law, Rate = k 2 [NOBr 2 ][NO], (based on the slow step) The overall rate law becomes When simplified, this is consistent with the experimentally observed rate law: Rate = k[NO] 2 [Br 2 ]

Prentice Hall © 2003Chapter 14 Sample Problems # 59, 61

Prentice Hall © 2003Chapter 14 A catalyst changes the rate of a chemical reaction Two types: –Homogeneous: The catalyst and reaction is in one phase –Heterogeneous: The catalyst and reaction are in different phases 14.7: Catalysis

Prentice Hall © 2003Chapter 14

Prentice Hall © 2003Chapter 14 Homogeneous Catalysis Catalysts can operate by Increasing the number of effective collisions (from the Arrhenius equation): Increase k by increasing A or decreasing E a Add intermediates to the reaction

– Br - is a catalyst because it can be recovered at the end of the reaction Hydrogen peroxide decomposes very slowly: 2H 2 O 2(aq)  2H 2 O (l) + O 2(g) Generally, catalysts operate by lowering the E a for a reaction

Prentice Hall © 2003Chapter 14 Heterogeneous Catalysis Typical example: solid catalyst, gaseous reactants and products Catalytic converters in cars (Text, P. 559) Industrial catalysts

Prentice Hall © 2003Chapter 14 Adsorption: the binding of reactant molecules to the catalyst surface Adsorbed species are very reactive Molecules are adsorbed onto active sites on the catalyst surface

Prentice Hall © 2003Chapter 14 Enzymes Enzymes are biological catalysts Most enzymes are protein molecules with large molecular masses Very specific shapes Catalyze very specific reactions Substrates undergo reaction at the active site of an enzyme It locks into an enzyme and a fast reaction occurs If another substrate cannot displace the molecule, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors)

Prentice Hall © 2003Chapter 14 End of Chapter 14 Chemical Kinetics