 2009, Prentice-Hall, Inc. Chapter 14 Chemical Kinetics - PowerPoint Presentation

 2009, Prentice-Hall, Inc. Chapter 14Chemical Kinetics John D. BookstaverSt. Charles Community CollegeCottleville, MO Chemistry, The Central Science , 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

 2009, Prentice-Hall, Inc. KineticsIn kinetics we study the rate at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).

 2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates Physical State of the ReactantsIn order to react, molecules must come in contact with each other.The more homogeneous the mixture of reactants, the faster the molecules can react.

 2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates Concentration of ReactantsAs the concentration of reactants increases, so does the likelihood that reactant molecules will collide.

 2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates TemperatureAt higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy.

 2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates Presence of a CatalystCatalysts speed up reactions by changing the mechanism of the reaction.Catalysts are not consumed during the course of the reaction.

 2009, Prentice-Hall, Inc. Reaction Rates Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

 2009, Prentice-Hall, Inc. Reaction Rates In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times. C 4 H 9 Cl ( aq ) + H 2 O ( l )  C 4 H 9 OH ( aq ) + HCl (aq)

 2009, Prentice-Hall, Inc. Reaction Rates The average rate of the reaction over each interval is the change in concentration divided by the change in time: C 4 H 9 Cl ( aq ) + H 2 O ( l )  C 4 H 9 OH ( aq ) + HCl (aq) Average rate =  [C 4 H 9 Cl]  t

 2009, Prentice-Hall, Inc. Reaction Rates Note that the average rate decreases as the reaction proceeds.This is because as the reaction goes forward, there are fewer collisions between reactant molecules. C 4 H 9 Cl ( aq ) + H 2 O ( l )  C 4 H 9 OH ( aq ) + HCl (aq)

Suppose that at one point in the reaction, the concentration of A is 0.4658 M and 125s later the concentration has fallen to 0.4282M. During this time, what is the average rate of reaction?  2009, Prentice-Hall, Inc.

 2009, Prentice-Hall, Inc. Reaction Rates A plot of [C 4 H 9 Cl] vs. time for this reaction yields a curve like this. The slope of a line tangent to the curve at any point is the instantaneous rate at that time. C 4 H 9 Cl ( aq ) + H 2 O ( l )  C 4 H9OH ( aq ) + HCl ( aq )

 2009, Prentice-Hall, Inc. Reaction Rates All reactions slow down over time.Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction. C 4 H 9 Cl ( aq ) + H 2 O ( l )  C 4 H 9 OH ( aq ) + HCl ( aq )

 2009, Prentice-Hall, Inc. Reaction Rates and Stoichiometry In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. C 4 H 9 Cl ( aq ) + H2O( l )  C 4 H 9 OH(aq) + HCl(aq) Rate = -  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t

 2009, Prentice-Hall, Inc. Reaction Rates and Stoichiometry What if the ratio is not 1:1? 2 HI ( g )  H 2 ( g ) + I 2 ( g ) In such a case , Rate = − 1 2  [HI]  t =  [I 2 ]  t

 2009, Prentice-Hall, Inc. Reaction Rates and Stoichiometry To generalize, then, for the reaction a A + b B c C + d D Rate = − 1 a  [A]  t = − 1 b  [B]  t = 1 c  [C]  t 1 d  [D]  t =

 2009, Prentice-Hall, Inc. Concentration and Rate One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration.

Given the reaction, 2A + B  2C + DAt some point in the reaction, [C] = 0.2885M and 153s later, the concentration of C is 0.3546M. What is the average rate of reaction during this time period? What is the average rate of formation of D during this time?  2009, Prentice-Hall, Inc.

Given the reaction, 2A + B  3C + DIf –Δ[A]/Δt is 2.10 E-5 M/s, What is the rate of formation of C?   2009, Prentice-Hall, Inc.

Given A + 2B  C + 3DIf the rate of disappearance of B is -6.2E-4 M/s, what is the rate of reaction, the rate of disappearance of A and the rate of formation of D?  2009, Prentice-Hall, Inc.

If the initial concentration of hydrogen peroxide is 0.2546M and the initial rate of reaction is 9.32 E-4M/s. What will be the concentration of hydrogen peroxide remaining from its decomposition at t=35s. (reaction given for 1 mole of hydrogen peroxide decomposition)   2009, Prentice-Hall, Inc.

Given the same reaction, the initial concentration of hydrogen peroxide is 0.1108M and 12s later the concentration is 0.1060M. What is the initial rate of this reaction in both M/s and M/min.   2009, Prentice-Hall, Inc.

 2009, Prentice-Hall, Inc. Concentration and Rate If we compare Experiments 1 and 2, we see that when [NH4+] doubles, the initial rate doubles. NH 4 + ( aq ) + NO 2 − ( aq ) N 2 ( g ) + 2 H 2 O ( l )

 2009, Prentice-Hall, Inc. Concentration and Rate Likewise, when we compare Experiments 5 and 6, we see that when [NO2−] doubles, the initial rate doubles. NH 4 + ( aq ) + NO 2 − ( aq ) N 2 ( g ) + 2 H 2 O ( l )

 2009, Prentice-Hall, Inc. Concentration and Rate Likewise, when we compare Experiments 5 and 6, we see that when [NO2−] doubles, the initial rate doubles. NH 4 + ( aq ) + NO 2 − ( aq ) N 2 ( g ) + 2 H 2 O ( l )

 2009, Prentice-Hall, Inc. Concentration and Rate This meansRate  [NH4+ ] Rate  [NO 2 − ] Rate  [NH + ] [NO 2 − ] which, when written as an equation, becomes Rate = k [NH 4 + ] [NO 2 − ] This equation is called the rate law , and k is the rate constant. Therefore,

 2009, Prentice-Hall, Inc. Rate Laws A rate law shows the relationship between the reaction rate and the concentrations of reactants.The exponents tell the order of the reaction with respect to each reactant. Since the rate law is Rate = k [NH 4 + ] [NO 2 − ] the reaction is First-order in [NH 4 + ] and First-order in [NO 2 − ].

 2009, Prentice-Hall, Inc. Rate LawsRate = k [NH4+] [NO2−] The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second-order overall.

aA + bB  cC + dDThe rate law: Rate = k [A]m[B]nk = rate constantm = experimentally determined #, the value shows the order of An = experimentally determined #, the value shows the order of Bm + n = overall order of the reaction  2009, Prentice-Hall, Inc.

Rate = k[A]o[B]1m = 0; rxn is 0th order in An = 1; rxn is 1st order in Bm + n = 0 + 1; overall rxn is 1st orderSince anything [ ]0 = 1  simplifyRate = k[B]1 = k[B]  2009, Prentice-Hall, Inc.

H2SeO3 + 6I- + 4H+  Se + 2I3- + 3H2ORate = k[H2SeO3] [I3-]3 [H+]2Does the coefficient always correspond to the order of the reaction?  2009, Prentice-Hall, Inc.

Units of kRxn orderUnits of k0th order1st order2nd order3rd order nth order  2009, Prentice-Hall, Inc.

A + B  CGiven rate = k[A]0[B]What does this mean?  2009, Prentice-Hall, Inc.

A + B  CGiven rate = k[A]0[B]2What does this mean?Double ADouble BHalf BTriple B  2009, Prentice-Hall, Inc.

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A  BWe find that when [A] has fallen to half its initial value, the reaction proceeds at the same rate. What order is the reaction?  2009, Prentice-Hall, Inc.

H2O2  H2O + ½ O2For the decomposition of 1 mole of hydrogen peroxide, use k = 3.66 E-3 1/s to determine the instantaneous rate of the 1st order decomposition of 2.05M H2O2?  2009, Prentice-Hall, Inc.

Experiment [NO] [Cl 2 ] Initial Rate (M/s) 1 0.0125 0.0255 2.27E-5 2 0.0125 0.0510 4.55E-5 3 0.0250 0.0255 9.08E-5 Determine the reaction order through experimental data 2NO + Cl 2  2NOCl By comparison write the rate laws (differential rate law)

Experiment [NO] [Cl 2 ] Initial Rate (M/s) 1 0.0125 0.0255 2.27E-5 2 0.0125 0.0510 4.55E-5 3 0.0250 0.0255 9.08E-5 Determine the reaction order through experimental data 2NO + Cl 2  2NOCl By comparison write the rate laws (differential rate law) Suppose there is a hypothetical experiment 4, which has [NO]=0.0500M and [Cl 2 ]=0.0255M what is the rate in this experiment?

Determine the reaction order through experimental data S 2 O 8 -2 + 3I -  2SO 4 -2 + I 3 - By comparison write the rate laws. What is the order for each reactant and overall? Solve for the rate constant. What would be the initial rate of reaction if [S 2 O 8 -2 ] = 0.083M and [I - ] =0.115M Experiment [S 2 O 8 -2 ] [l - ] Initial Rate (M/s) 1 0.038 0.060 1.4E-5 2 0.076 0.060 2.8E-5 3 0.076 0.120 5.6E-5  2009, Prentice-Hall, Inc.

Concentration and Reaction Rate   2009, Prentice-Hall, Inc.

 The rate expression will always have the formInitial rate = k [A]m[B]n[C]pk = rate constant[A] = conc. of reactant A[B] = conc. of reactant B[C] = conc. of catalyst (rare on AP/IB)m = order of rxn for reactant An = order of rxn for reactant Bp = order of rxn for catalyst  2009, Prentice-Hall, Inc.

 Initial rate = k [A]m[B]n[C]pExponents can be zero, whole numbers, or fractions (rare)Must be determined by experimentationThe rate constant, kTemperature dependentDetermined experimentally  2009, Prentice-Hall, Inc.

 2009, Prentice-Hall, Inc. Half-Life Half-life is defined as the time required for one-half of a reactant to react.Because [A] at t1/2 is one-half of the original [A], [A] t = 0.5 [A] 0 .

Zeroth order half-life   2009, Prentice-Hall, Inc.

Zeroth order half-life [A]t = -kt +[A]o (not in calculus – trust me)  2009, Prentice-Hall, Inc.

Zeroth order half-life [A]t = -kt +[A]o This looks like the equation for what??? - lineGraph [A] vs. t will produce a straight line, with slope = -k and y-intercept of [A]0  2009, Prentice-Hall, Inc.

 2009, Prentice-Hall, Inc. First-Order Processes So how do we straighten the line???

First order half-life   2009, Prentice-Hall, Inc.

 2009, Prentice-Hall, Inc. First-Order Processes When ln P is plotted as a function of time, a straight line results.Therefore,The process is first-order.k is the negative of the slope: 5.1  10 -5 s − 1 .

 2009, Prentice-Hall, Inc. First-Order Processes   ln [A] t = - k t + ln [A] 0  

In the first order decomposition of hydrogen peroxide, k = 3.66E-3 1/s and [H2O2]0 = 0.882M Determine the time at which [H2O2] = 0.600MDetermine the [H2O2] after 225s  2009, Prentice-Hall, Inc.

The decomposition of NH2NO2 has a rate law of rate = k[NH2NO2]. k = 5.62E-3 1/min at 15oC and [NH2NO2]o = 0.105 At what time is [NH2NO2] = 0.0250MWhat is [NH2NO2] after 6.00hrsWhat is the rate of reaction after 35 mins, if the [NH2NO2]0=0.0750M  2009, Prentice-Hall, Inc.

 2009, Prentice-Hall, Inc. Half-Life Half-life is defined as the time required for one-half of a reactant to react.Because [A] at t1/2 is one-half of the original [A], [A] t = 0.5 [A] 0 .

 2009, Prentice-Hall, Inc. Half-Life For a first-order process, this becomes 0.5 [A] 0 [A] 0 ln = − kt 1/2 ln 0.5 = − kt 1/2 − 0.693 = − kt 1/2 = t 1/2 0.693 k NOTE: For a first-order process, then, the half-life does not depend on [A] 0 .

What is the rate constant of the first order decomposition of N2O5 if the t1/2 = 120s  2009, Prentice-Hall, Inc.

How long will it take for an initial [N2O5] to reach 1/16 of its value?Find the mass of dinitrogen pentoxide remaining after a 4.80g sample has decomposed for 10mins.Determine the time for 90% of a N2O5 sample to undergo decomposition.  2009, Prentice-Hall, Inc.

The half-life of the first order decomposition of SO2Cl2 at 320oC is 8.75 hr. What is the value of the rate constant? What is the pressure of sulfuryl chloride 3.00 hr after the start of the reaction, if the initial pressure of sulfuryl chloride is 722mmHg?  2009, Prentice-Hall, Inc.

 2009, Prentice-Hall, Inc. Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, we get (the math is a little more complicated) 1 [A] t = kt + 1 [A] 0 also in the form y = mx + b

 2009, Prentice-Hall, Inc. Second-Order Processes So if a process is second-order in A, a plot of vs. t will yield a straight line, and the slope of that line is k. 1 [A] t = kt + 1 [A] 0 1 [A]

 2009, Prentice-Hall, Inc. Second-Order Processes The decomposition of NO 2 at 300 ° C is described by the equation NO 2 ( g ) NO ( g ) + O 2 ( g ) and yields data comparable to this: Time ( s ) [NO 2 ], M 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380 1 2

 2009, Prentice-Hall, Inc. Second-Order Processes Plotting ln [NO 2 ] vs . t yields the graph at the right. Time ( s ) [NO 2 ], M ln [NO 2 ] 0.0 0.01000 − 4.610 50.0 0.00787 − 4.845 100.0 0.00649 − 5.038 200.0 0.00481 − 5.337 300.0 0.00380 − 5.573 The plot is not a straight line, so the process is not first-order in [A].

 2009, Prentice-Hall, Inc. Second-Order Processes Graphing ln vs. t , however, gives this plot. Time ( s ) [NO 2 ], M 1/[NO 2 ] 0.0 0.01000 100 50.0 0.00787 127 100.0 0.00649 154 200.0 0.00481 208 300.0 0.00380 263 Because this is a straight line, the process is second-order in [A]. 1 [NO 2 ]

 2009, Prentice-Hall, Inc. Half-Life For a second-order process, 1 0.5 [A] 0 = kt 1/2 + 1 [A] 0 2 [A] 0 = kt 1/2 + 1 [A] 0 2 − 1 [A] 0 = kt 1/2 1 [A] 0 = = t 1/2 1 k [A] 0

Find the rate constant, and the half-life of the decomposition of 1.00M NO2  2009, Prentice-Hall, Inc.

(2nd order) A  B, takes 55s for the [A] to fall to 0.40 from [A]o = 0.80M What is the k for this reaction?How long would it take for [A] to fall to 0.20M? to 0.10M?  2009, Prentice-Hall, Inc.

In a process where Cl atoms combine to form Cl2­, the reaction follows a second order rate law. The first half-life is 12ms (initial concentration is .500 M). How long will it take for the [Cl] to drop to 1/8 of its initial value?  2009, Prentice-Hall, Inc.

Integrated Rate LawsUse graphical methods to determine the order of a given reactant. The value of the rate constant k is equal to the absolute value of the slope of the best fit line which was decided by performing 3 linear regressions and analyzing the regression correlation coefficient r.  2009, Prentice-Hall, Inc.

Integrated Rate Law: [ ] / timeCreate 3 graphsTime always goes on the x-axisGraph 1: [A] on y-axisGraph 2: ln[A] on y-axisGraph 3: 1/[A] on y-axis [A]-1Alphabetic order: Concentration = 0 orderNatural log = 1st orderReciprocal = 2nd order  2009, Prentice-Hall, Inc.

Integrated Rate Law: [ ] / timeAlphabetic order: Concentration = 0 orderNatural log = 1st orderReciprocal = 2nd orderChoose the graph that forms a straight line.(r value closest to ±1)The equation formed by line allows you to predict either [ ] at a time or a time for a specific [ ].  2009, Prentice-Hall, Inc.

Integrated Rate Law: [ ] / time y = mx + bZero order: [A] = -kt + [A]oFirst order: ln[A] = -kt + ln[Ao]Second order 1/[A] = kt + 1/[Ao]*** |slope| = k ***Rate expression =Rate law Rate k[A]order determined from graph  2009, Prentice-Hall, Inc.

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[A] Time(min) 0.80 0 0.60 8 0.35 24 0.20 40  2009, Prentice-Hall, Inc. Determine the order of this reaction using the given experimental data . Determine k.

Given the following information, determine the order of the reaction and the value of k, the reaction constant.Concentration (M)Time (s)1.0 100.5020 0.3330  2009, Prentice-Hall, Inc.

 2009, Prentice-Hall, Inc. Temperature and Rate Generally, as temperature increases, so does the reaction rate.This is because k is temperature dependent.

 2009, Prentice-Hall, Inc. The Collision ModelIn a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other.

 2009, Prentice-Hall, Inc. The Collision Model Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

 2009, Prentice-Hall, Inc. Activation Energy In other words, there is a minimum amount of energy required for reaction: the activation energy, Ea. Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

http://phet.colorado.edu/en/simulation/reactions-and-rates  2009, Prentice-Hall, Inc.

 2009, Prentice-Hall, Inc. Reaction Coordinate Diagrams It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile.

 2009, Prentice-Hall, Inc. Reaction Coordinate Diagrams The diagram shows the energy of the reactants and products (and, therefore, E). The high point on the diagram is the transition state . The species present at the transition state is called the activated complex . The energy gap between the reactants and the activated complex is the activation energy barrier .

 2009, Prentice-Hall, Inc. Maxwell–Boltzmann Distributions Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. At any temperature there is a wide distribution of kinetic energies.

 2009, Prentice-Hall, Inc. Maxwell–Boltzmann Distributions As the temperature increases, the curve flattens and broadens.Thus at higher temperatures, a larger population of molecules has higher energy.

 2009, Prentice-Hall, Inc. Maxwell–Boltzmann Distributions If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. As a result, the reaction rate increases.

 2009, Prentice-Hall, Inc. Maxwell–Boltzmann Distributions This fraction of molecules can be found through the expression where R is the gas constant and T is the Kelvin temperature. f = e - E a RT

 2009, Prentice-Hall, Inc. Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and Ea: k = A e where A is the frequency factor , a number that represents the likelihood that collisions would occur with the proper orientation for reaction. - E a RT

 2009, Prentice-Hall, Inc. Arrhenius Equation Taking the natural logarithm of both sides, the equation becomesln k = - ( ) + ln A 1 T y = m x + b Therefore, if k is determined experimentally at several temperatures, E a can be calculated from the slope of a plot of ln k vs. . E a R 1 T

 2009, Prentice-Hall, Inc. Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism.

 2009, Prentice-Hall, Inc. Reaction MechanismsReactions may occur all at once or through several discrete steps. Each of these processes is known as an elementary reaction or elementary process.

 2009, Prentice-Hall, Inc. Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process.

 2009, Prentice-Hall, Inc. Multistep Mechanisms In a multistep process, one of the steps will be slower than all others.The overall reaction cannot occur faster than this slowest, rate-determining step.

 2009, Prentice-Hall, Inc. Slow Initial Step The rate law for this reaction is found experimentally to beRate = k [NO2]2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps. NO 2 ( g ) + CO ( g )  NO ( g ) + CO 2 ( g )

 2009, Prentice-Hall, Inc. Slow Initial Step A proposed mechanism for this reaction isStep 1: NO2 + NO2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast) The NO 3 intermediate is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

 2009, Prentice-Hall, Inc. Fast Initial Step The rate law for this reaction is found to beRate = k [NO]2 [Br2] Because termolecular processes are rare, this rate law suggests a two-step mechanism. 2 NO ( g ) + Br 2 ( g )  2 NOBr ( g )

 2009, Prentice-Hall, Inc. Fast Initial Step A proposed mechanism is Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1 includes the forward and reverse reactions. Step 1: NO + Br 2 NOBr 2 (fast)

 2009, Prentice-Hall, Inc. Fast Initial StepThe rate of the overall reaction depends upon the rate of the slow step. The rate law for that step would beRate = k2 [NOBr 2 ] [NO] But how can we find [NOBr 2 ]?

 2009, Prentice-Hall, Inc. Fast Initial StepNOBr 2 can react two ways:With NO to form NOBrBy decomposition to reform NO and Br2 The reactants and products of the first step are in equilibrium with each other. Therefore, Rate f = Rate r

 2009, Prentice-Hall, Inc. Fast Initial Step Because Ratef = Rater , k 1 [NO] [Br 2 ] = k − 1 [NOBr 2 ] Solving for [NOBr 2 ] gives us k 1 k − 1 [NO] [Br 2 ] = [NOBr 2 ]

 2009, Prentice-Hall, Inc. Fast Initial Step Substituting this expression for [NOBr2] in the rate law for the rate-determining step gives k 2 k 1 k − 1 Rate = [NO] [Br 2 ] [NO] = k [NO] 2 [Br 2 ]

 2009, Prentice-Hall, Inc. Catalysts Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction.Catalysts change the mechanism by which the process occurs.

 2009, Prentice-Hall, Inc. Catalysts One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.

 2009, Prentice-Hall, Inc. Enzymes Enzymes are catalysts in biological systems.The substrate fits into the active site of the enzyme much like a key fits into a lock.