Introduction of Introduction of KineticsKinetics

Rate of Chemical ReactionsRate of Chemical Reactions

What Factors Affect Reaction What Factors Affect Reaction Rates?Rates?

Your task is to discover a method to Your task is to discover a method to dissolve an effervescent tablet as dissolve an effervescent tablet as quickly as possible. You must report quickly as possible. You must report your results in g/s (yes, this means that your results in g/s (yes, this means that you must weigh your tablets before you must weigh your tablets before dissolving them). You have 15 minutes dissolving them). You have 15 minutes to to ““playplay”” and your best time may be and your best time may be submitted. Please keep track of the submitted. Please keep track of the variables you altered each trial. We will variables you altered each trial. We will discuss how you accomplished this task discuss how you accomplished this task at then end of the 15 minute period.at then end of the 15 minute period.

Chemical Kinetics and the Chemical Kinetics and the Human BodyHuman Body

How were brain functions affected by cold?How were brain functions affected by cold? How could you explain this?How could you explain this? Why do you suppose death could ensue?Why do you suppose death could ensue? Could chilling a live body ever be Could chilling a live body ever be

advantageous?advantageous? How can you use these ideas to explain why How can you use these ideas to explain why

the body becomes feverish when fighting the body becomes feverish when fighting infection?infection?

Considering that the body must maintain an Considering that the body must maintain an internal temperature of about 37internal temperature of about 37ºC, how can ºC, how can the body control reaction rates if it cannot the body control reaction rates if it cannot alter its temperature by more than a few alter its temperature by more than a few degrees?degrees?

CatalysisCatalysis

How are catalysts used at a crime How are catalysts used at a crime scene?scene?

Chemical KineticsChemical Kinetics

How are chemical reaction How are chemical reaction rates determined and what rates determined and what factors affect these rates?factors affect these rates?

ObjectivesObjectives

To understand rates of reaction and the To understand rates of reaction and the conditions affecting rates.conditions affecting rates.

To derive the rate equation, rate To derive the rate equation, rate constant, and reaction order from constant, and reaction order from experimental data.experimental data.

To use integrated rate laws.To use integrated rate laws. To understand collision theory of reaction To understand collision theory of reaction

rates and the role of activation energy.rates and the role of activation energy. To relate reaction mechanisms and rate To relate reaction mechanisms and rate

laws.laws.

Chemical KineticsChemical KineticsCh 14 hmwk problems:Ch 14 hmwk problems:

Sets: Sets: (how they are broken up in the (how they are broken up in the notes)notes)

11, 12, 11, 12, 13, 14, 15, 13, 14, 15, 20, 21, 25, 20, 21, 25, 29, 33, 38, 29, 33, 38, 45, 47, 50, 45, 47, 50, 55, 58, 59, 62, 55, 58, 59, 62, 66, 69, 87, 9866, 69, 87, 98

Why are Kinetics Why are Kinetics Important?Important? MedicinalMedicinal: How quickly or slowly : How quickly or slowly

will a medication work?will a medication work? EnvironmentalEnvironmental: Is the rate at : Is the rate at

which ozone depletion occurs which ozone depletion occurs equal to the rate at which it is equal to the rate at which it is formed?formed?

IndustrialIndustrial: How long will it take to : How long will it take to produce a chemical? Is there a produce a chemical? Is there a way to change this?way to change this?

What are Kinetics What are Kinetics Anyway?Anyway? Chemical Kinetics is the study of Chemical Kinetics is the study of

chemical reaction rates, or speeds.chemical reaction rates, or speeds. Can you think of an example of a Can you think of an example of a

chemical reaction with a fast rate?chemical reaction with a fast rate?combustion of gasolinecombustion of gasoline

Can you think of an example of a Can you think of an example of a chemical reaction with a slow rate?chemical reaction with a slow rate?

formation of a diamondformation of a diamond

What Makes a What Makes a Reactions Occur?Reactions Occur? If a reaction is to occur between two or If a reaction is to occur between two or

more reactants, the reactants must more reactants, the reactants must come in contact with each other, and come in contact with each other, and have enough energy to both beak have enough energy to both beak existing bonds and form new bonds.existing bonds and form new bonds.

The rate at which these reactants The rate at which these reactants come in contact with each other come in contact with each other affects the rate of the overall chemical affects the rate of the overall chemical reaction.reaction.

Factors Affecting Factors Affecting Reaction RatesReaction Rates The Physical State of the The Physical State of the

Reactants: Reactions are limited Reactants: Reactions are limited by the area of contact between by the area of contact between two reactants. Therefore, the two reactants. Therefore, the greater the surface area of a solid greater the surface area of a solid reactant, the faster a reaction can reactant, the faster a reaction can proceed.proceed.

Factors Affecting Factors Affecting Reaction RatesReaction Rates The Concentration of Reactants: The Concentration of Reactants:

As the concentration of one or As the concentration of one or more reactants increases, the more reactants increases, the more frequently the reactants can more frequently the reactants can collide, leading to increased collide, leading to increased reaction rates.reaction rates.

Factors Affecting Factors Affecting Reaction RatesReaction Rates The temperature at which the The temperature at which the

reaction occurs: The rate of a reaction occurs: The rate of a chemical reaction occurs increases chemical reaction occurs increases with increasing temperature. As with increasing temperature. As the molecules move more rapidly, the molecules move more rapidly, they collide more frequently and they collide more frequently and with greater energy, increasing the with greater energy, increasing the reaction rate.reaction rate.

Factors Affecting Factors Affecting Reaction RatesReaction Rates The presence of a catalyst: The presence of a catalyst:

Catalysts are reagents that Catalysts are reagents that increase the speed of a reaction increase the speed of a reaction without being consumed in the without being consumed in the reaction by affecting how the reaction by affecting how the reactants collide together. We reactants collide together. We will discuss this in greater depth will discuss this in greater depth later.later.

Reaction RatesReaction Rates

Since a rate is expressed as Since a rate is expressed as changechange over over timetime, a reaction rate is expressed , a reaction rate is expressed as the change in the concentration of as the change in the concentration of reactants over time.reactants over time.

For a simple reaction, A —› B, the rate For a simple reaction, A —› B, the rate would be can be described as the rate would be can be described as the rate of appearance of B or the rate of of appearance of B or the rate of disappearance of A, since one mole of disappearance of A, since one mole of A will produce to one mole of B.A will produce to one mole of B.

Reaction RatesReaction Rates

For a simple reaction, A —› B, the For a simple reaction, A —› B, the rate would be expressed as either rate would be expressed as either ΔΔ[B]/ [B]/ ΔΔtt or - or -ΔΔ[A]/ [A]/ ΔΔtt

Rates are always expressed as Rates are always expressed as positive quantities, therefore, a positive quantities, therefore, a negative sign must be used negative sign must be used because the [A] decreases as the because the [A] decreases as the reaction progressesreaction progresses

A reaction vessel initially contains A reaction vessel initially contains 1.00 M of reactant A. Twenty 1.00 M of reactant A. Twenty seconds later, the reaction vessel seconds later, the reaction vessel contains only 0.54 M of reactant A contains only 0.54 M of reactant A and 0.46 M of product B. Calculate and 0.46 M of product B. Calculate the average rate of the the average rate of the disappearance of reactant A.disappearance of reactant A. Recall, A —› B, and Recall, A —› B, and raterate = - = -ΔΔ[A]/ [A]/

ΔΔtt . . Rate = -[0.54M – 1.00M]/(20s – 0s)Rate = -[0.54M – 1.00M]/(20s – 0s) Rate = 0.46M/20sRate = 0.46M/20s Rate = 2.3 x 10Rate = 2.3 x 10-2-2 M/s M/s

Reaction Rates and Reaction Rates and StoichiometryStoichiometry What if the reaction stoichiometry What if the reaction stoichiometry

is more complicated than our is more complicated than our simple 1:1 ratio?simple 1:1 ratio?

LetLet’’s look at 2HI —› Hs look at 2HI —› H22 + I + I22

Rate = -½ Rate = -½ ΔΔ[HI]/[HI]/ΔΔtt = = ΔΔ[H[H22]/ ]/ ΔΔtt

= [I= [I22]/ ]/ ΔΔtt

Reaction Rates and Reaction Rates and StoichiometryStoichiometry In general, for the reactionIn general, for the reaction

aA + bB —› cC + dDaA + bB —› cC + dD

the rate is given bythe rate is given by

Rate= Rate=

-1/a-1/aΔΔ[A]/[A]/ΔΔtt = -1/b = -1/bΔΔ[B]/[B]/ΔΔtt

= 1/c= 1/cΔΔ[C]/ [C]/ ΔΔtt = 1/d = 1/dΔΔ[D]/ [D]/ ΔΔtt

How is the rate at which ozone disappears How is the rate at which ozone disappears related to the rate at which oxygen appears in related to the rate at which oxygen appears in the reaction the reaction 2O2O33 —› 3O—› 3O22??

Rate = -1/2Rate = -1/2ΔΔ[O[O33]/]/ΔΔtt = 1/3 = 1/3ΔΔ[[OO22]/]/ΔΔtt

If the rate at which OIf the rate at which O22 appears is 6.0 x 10 appears is 6.0 x 10-5-5 M/s M/s at a particular instant, at what rate is Oat a particular instant, at what rate is O33 disappearing at the same time?disappearing at the same time?

Rate = -1/2Rate = -1/2ΔΔ[O[O33]/]/ΔΔtt = 1/3 = 1/3ΔΔ[[OO22]/]/ΔΔtt

Rate = -1/2Rate = -1/2ΔΔ[O[O33]/]/ΔΔtt = 1/3(6.0 x 10 = 1/3(6.0 x 10-5-5 M/s) M/s)

Rate = -Rate = -ΔΔ[O[O33]/]/ΔΔtt = 2/3(6.0 x 10 = 2/3(6.0 x 10-5-5 M/s) M/s)

Rate = -Rate = -ΔΔ[O[O33]/]/ΔΔtt = 4.0 x 10 = 4.0 x 10-5-5 M/s M/s

You should now be able You should now be able to complete # 11 & 12to complete # 11 & 12

Rate LawsRate Laws

The The rate lawrate law is an equation that is an equation that relates the rate of a reaction to the relates the rate of a reaction to the concentration of reactants (and concentration of reactants (and catalysts) raised to various powerscatalysts) raised to various powers

The rate law for our last example, The rate law for our last example, aA + bB —› cC + dD, would be aA + bB —› cC + dD, would be expressed asexpressed as

Rate = Rate = kk[A][A]mm[B][B]nn, where , where kk is the is the rate rate constant and m and n are constant and m and n are typically typically small, whole numbers.small, whole numbers.

Rate LawsRate Laws

kk in the rate law is the in the rate law is the rate rate constantconstant defined as a defined as a proportionality constant in the proportionality constant in the relationship between rate and relationship between rate and concentrations.concentrations.

The rate constant is fixed at a The rate constant is fixed at a particular temperature (but will particular temperature (but will change as the temperature changes)change as the temperature changes)

Rate LawsRate Laws

We can classify reactions by their ordersWe can classify reactions by their orders

– The The reaction orderreaction order, with respect to a given , with respect to a given reactant species, equals the exponent of the reactant species, equals the exponent of the concentration of that species in the rate law, concentration of that species in the rate law, as determined experimentallyas determined experimentally

– The The overall orderoverall order of the reaction equals the of the reaction equals the sum of the orders of all reactant species in the sum of the orders of all reactant species in the rate lawrate law

– In the case Rate = In the case Rate = kk[A][A]mm[B][B]nn, the overall order , the overall order of this reaction is given by m+n.of this reaction is given by m+n.

Rate LawsRate Laws

For the following reaction, the rate law can be For the following reaction, the rate law can be written as is indicated below:written as is indicated below:

NH4+ + NO2

- —›—› N2 + 2H2O

Rate = k[NH4+][NO2

-]

Because the exponent ofBecause the exponent of [NH4+] is one, the rate

in NH4+ is first order. The rate in [NO2

-] is also first order. The overall reaction order is second order.

The exponents in a rate law indicate how the rate is affected by the concentration of each reactant.

For the following reaction, the rate law can be For the following reaction, the rate law can be written as is indicated below:written as is indicated below:

NH4+ + NO2

- —›—› N2 + 2H2ORate = k[NH4

+][NO2-],

What is the effect on the rate if What is the effect on the rate if [NO2-] doubles?

Because the exponent of [NO2-] in the rate law is

one, doubling its concentration will double the rate.

What is the effect on the rate if [NH4+] triples?

Because the exponent of [NH4+] in the rate law is

one, tripling its concentration will triple the rate.

For a given reaction, the rate law is For a given reaction, the rate law is given bygiven by

Rate = k[NORate = k[NO22]]22

What is the order of the reaction with respect What is the order of the reaction with respect to NOto NO22??

Because the exponent of NOBecause the exponent of NO22 in the equation is in the equation is 2, the reaction is second order with respect 2, the reaction is second order with respect to NOto NO22. Because the concentration of only . Because the concentration of only one reactant is expressed in the rate law, the one reactant is expressed in the rate law, the overall order is also second.overall order is also second.

What is the What is the effect on the rate if [NO2] triples?

If the concentration triples, then the rate increases by 9 since [3]2.

Rate Laws: Additional Rate Laws: Additional ExamplesExamples2N2N22OO55 —› —› 4NO2 + O2 Rate = k[NN22OO55]

CHCl3 + Cl2 —› CCl—› CCl44 + HCl + HCl Rate = Rate = kk[[CHCl3]][[Cl2]]½½

HH22 + I + I22 —› 2HI —› 2HI Rate = Rate = kk[H[H22][I][I22]]

Note: While exponents in rate laws are often Note: While exponents in rate laws are often the same as the coefficients in a balanced the same as the coefficients in a balanced equation, this is not a rule! equation, this is not a rule! Exponents in a Exponents in a rate law can only be determined rate law can only be determined experimentally!experimentally!

Determining a Rate Law Determining a Rate Law Experimentally – Initial Experimentally – Initial Rate MethodRate Method

Trial [NO] [H2] R0 (M/s)

1 0.10 0.10 1.23 x 10-3

2 0.10 0.20 2.46 x 10-3

3 0.20 0.10 4.92 x 10-3

Using the data above, determine the rate law, calculate the value of k and determine the rate when [NO] = 0.050M and [H2] = 0.150 M.

Determining a Rate Law Determining a Rate Law Experimentally – Initial Experimentally – Initial Rate MethodRate MethodA. Determine the rate lawA. Determine the rate law

Recall, the Rate = Recall, the Rate = kk[NO][NO]xx[H[H22]]yy. We . We can determine the values of x and can determine the values of x and y by using the experimental data y by using the experimental data and the equation and solving for and the equation and solving for the unknown variables.the unknown variables.

Determining a Rate Law Determining a Rate Law Experimentally – Initial Experimentally – Initial Rate MethodRate MethodLetLet’’s compare trial two to trial one:s compare trial two to trial one:

Rate 2Rate 2 = = [NO][NO]xx[H[H22]]yy = = [0.10][0.10]xx[.20][.20]yy

Rate 1 = [NO]Rate 1 = [NO]xx[H[H22]]yy = [0.10] = [0.10]xx[.10][.10]yy

We can simplify our expression and We can simplify our expression and substitute the values for Rate 1 and Rate substitute the values for Rate 1 and Rate 22

Rate 2Rate 2 = = 2.46 x 102.46 x 10-3-3 = = [.20][.20]yy

Rate 1 = 1.23 x 10Rate 1 = 1.23 x 10-3-3 = [.10] = [.10]yy

Notice, we are only left with one unknown Notice, we are only left with one unknown variable to solve for.variable to solve for.

Determining a Rate Law Determining a Rate Law Experimentally – Initial Experimentally – Initial Rate MethodRate MethodRate 2Rate 2 = = 2.46 x 102.46 x 10-3-3 = = [.20][.20]yy

Rate 1 = 1.23 x 10Rate 1 = 1.23 x 10-3-3 = [.10] = [.10]yy

22yy = 2 = 2

y must equal 1.y must equal 1.

So far we have determined that So far we have determined that

Rate = Rate = kk[NO][NO]xx[H[H22]]11. We must repeat . We must repeat the process to solve for x by the process to solve for x by comparing trial 3 to trial 1.comparing trial 3 to trial 1.

Determining a Rate Law Determining a Rate Law Experimentally – Initial Experimentally – Initial Rate MethodRate MethodRate 3Rate 3 = = [NO][NO]xx[H[H22]]yy = = [0.20][0.20]xx[.10][.10]yy

Rate 1 = [NO]Rate 1 = [NO]xx[H[H22]]yy = [0.10] = [0.10]xx[.10][.10]yy

Rate 3Rate 3 = = 4.92 x 104.92 x 10-3-3 = = [.20][.20]xx

Rate 1 = 1.23 x 10Rate 1 = 1.23 x 10-3-3 = [.10] = [.10]xx

22xx = 4 = 4

X must equal 2. Therefore, the X must equal 2. Therefore, the experimentally determined rate law is experimentally determined rate law is Rate = Rate = kk[NO][NO]22[H[H22]]11

Determining a Rate Law Determining a Rate Law Experimentally – Initial Experimentally – Initial Rate MethodRate MethodB. Calculate the value of B. Calculate the value of kk

If Rate = If Rate = kk[NO][NO]22[H[H22]]11, then simple , then simple substitution from the data table substitution from the data table will allow us to calculate will allow us to calculate kk..

Rate = Rate = kk[NO][NO]22[H[H22]]1 1 (use trial 1)(use trial 1)

1.23 x 101.23 x 10-3-3M/s = M/s = kk[0.10M][0.10M]22[0.10M][0.10M]11

1.23 x 101.23 x 10-3-3M/s = M/s = kk(0.0010M(0.0010M33))1.2 M1.2 M-2-2·s·s-1-1 = = kk

Determining a Rate Law Determining a Rate Law Experimentally – Initial Experimentally – Initial Rate MethodRate MethodC. C. Determine the rate when [NO] =

0.050M and [H2] = 0.150 MSince Rate = 1.2MRate = 1.2M-2-2·s·s-1-1[NO][NO]22[H[H22]]11

Rate = 1.2MRate = 1.2M-2-2·s·s--

11[0.050M][0.050M]22[0.150M][0.150M]11

Rate = 4.5 x 10Rate = 4.5 x 10-4-4 M/s M/s

You should now be able You should now be able to complete # 13, 14, 15, to complete # 13, 14, 15, 20, 21, 25 20, 21, 25

Instantaneous RatesInstantaneous Rates

Calculated from a Calculated from a graph of graph of experimental experimental data that data that represents represents concentration vs. concentration vs. time.time.

Instantaneous RatesInstantaneous Rates

Recall, the slope of a graph is Recall, the slope of a graph is ΔΔy/y/ΔΔx. In a concentration vs. x. In a concentration vs. time graph, the time graph, the ΔΔy = y = ΔΔ[reactant] and [reactant] and ΔΔx = x = ΔΔtime.time.

Therefore, the slope of the Therefore, the slope of the graph = graph = ΔΔ[reactant]/ [reactant]/ ΔΔtime time and will yield the and will yield the instantaneous rate of instantaneous rate of reaction progression at any reaction progression at any given point in time.given point in time.

Average reaction rates can Average reaction rates can be determined from the be determined from the average of several calculated average of several calculated instantaneous rates.instantaneous rates.

Rate OrdersRate Orders

Recall, the overall order of a reaction is Recall, the overall order of a reaction is equal to the sum of the exponents in a equal to the sum of the exponents in a rate lawrate law

Rate = Rate = kk[A][A]00 or Rate = or Rate = kk is zero order is zero order

Rate = Rate = kk[A][A]11 is first order is first order

Rate = Rate = kk[A][A]22 or Rate = or Rate = kk[A][B] is second [A][B] is second orderorder

Rate = Rate = kk[A][A]33 or Rate = or Rate = kk[A][A]22[B] is third order[B] is third order

A closer look at Zero-A closer look at Zero-order Reactionsorder Reactions

Rate = Rate = kk[A][A]00 or Rate = or Rate = kk is zero order is zero orderThis is referred to as the differential rate law.This is referred to as the differential rate law.

If we integrate the above equation, we find a If we integrate the above equation, we find a very useful equation:very useful equation:

[A] = -[A] = -kkt + [At + [A00] ]

This will allow us to solve for unknown This will allow us to solve for unknown concentration values of zero-order reactionsconcentration values of zero-order reactions

A closer look at Zero-A closer look at Zero-order Reactionsorder Reactions

A zero-order reaction begins with A zero-order reaction begins with [X] = 0.50 M. How long does it [X] = 0.50 M. How long does it take to react all but 0.10M if take to react all but 0.10M if kk=0.0410 M/min? What is the =0.0410 M/min? What is the half-life? What is the [X] after 5 half-life? What is the [X] after 5 min?min?

A closer look at Zero-A closer look at Zero-order Reactionsorder Reactions

A. [X] = 0.10 M, [XA. [X] = 0.10 M, [X00] = 0.50 M and ] = 0.50 M and kk=0.0410 M/min=0.0410 M/min

[A] = -[A] = -kkt + [At + [A00] ]

[0.10M] = -(0.0410M/min)t + [0.10M] = -(0.0410M/min)t + [0.50M] [0.50M]

-0.40M = -(0.0410M/min)t -0.40M = -(0.0410M/min)t

10. min = t10. min = t

A closer look at Zero-A closer look at Zero-order Reactionsorder ReactionsB. What is the half-life? B. What is the half-life? Recall, a half-life is the time it takes for Recall, a half-life is the time it takes for

half a reactant to be consumed.half a reactant to be consumed.

So, if [XSo, if [X00] = 0.50 M, then] = 0.50 M, then [X] must equal [X] must equal 0.25 M0.25 M

If [A] = -If [A] = -kkt + [At + [A00], then ], then

[0.25M] = -(0.0410M/min)t[0.25M] = -(0.0410M/min)t½½ + [0.50M] + [0.50M]

-0.25M = -(0.0410M/min)t-0.25M = -(0.0410M/min)t½½

6.1 min = t 6.1 min = t ½½

A closer look at Zero-A closer look at Zero-order Reactionsorder Reactions

C. What is the [X] after 5 min?C. What is the [X] after 5 min?

If [A] = -If [A] = -kkt + [At + [A00], then ], then

[X] = -(0.0410M/min)5min + [0.50M][X] = -(0.0410M/min)5min + [0.50M]

[X] = -.205M + 0.50M[X] = -.205M + 0.50M

[X] = 0.30 M[X] = 0.30 M

A closer look at Zero-A closer look at Zero-order Reactionsorder Reactions

Notice, [A] = -Notice, [A] = -kkt + [At + [A00] closely mimics] closely mimics

y = mx + by = mx + b

Therefore, a graph of change in Therefore, a graph of change in concentration vs. time will be linear for concentration vs. time will be linear for a zero-order reaction.a zero-order reaction.

A closer look at First-A closer look at First-order Reactionsorder Reactions

Rate = Rate = kk[A][A]11 is first order is first orderThis is referred to as the differential rate law.This is referred to as the differential rate law.

If we integrate the above equation, we find a the If we integrate the above equation, we find a the integrated rate law for first order equations:integrated rate law for first order equations:

ln[Aln[Att] - ln[A] - ln[A00] = -] = -kkt t OrOr

ln[Aln[Att] = -] = -kkt + ln[At + ln[A00] ] (y) = (mx) + b(y) = (mx) + b

This will allow us to solve for unknown This will allow us to solve for unknown concentration values of first-order reactionsconcentration values of first-order reactions

A closer look at A closer look at Second-order Second-order ReactionsReactions

Rate = Rate = kk[A][A]22 or is second order or is second order This is referred to as the differential rate law.This is referred to as the differential rate law.

If we integrate the above equation, we find a If we integrate the above equation, we find a the integrated rate law for second order the integrated rate law for second order equations:equations:

1/[A1/[Att] = ] = kkt + 1/[At + 1/[A00] ] (y) = (mx) + (b)(y) = (mx) + (b)

This will allow us to solve for unknown This will allow us to solve for unknown concentration values of second-order concentration values of second-order reactionsreactions

Determining Order Determining Order From Graphical DataFrom Graphical Data

Time (min) [X]

0 0.0200

200 0.0160

400 0.0131

600 0.0106

800 0.0086

1000 0.0069

1200 0.0056

1600 0.0037

An experiment was conducted were a reaction was allowed to proceed and the reactant concentration was measured over a period of time. What is the order of this reaction?

Determining Order Determining Order From Graphical DataFrom Graphical DataRecall, the integrated rate law of zero, Recall, the integrated rate law of zero,

first and second order reactions can be first and second order reactions can be rearranged in the form of y=mx + brearranged in the form of y=mx + b

Zero-order: [A] = -Zero-order: [A] = -kkt + [At + [A00]]First-order: ln[AFirst-order: ln[Att] = -] = -kkt + ln[At + ln[A00] ] Second-order: 1/[ASecond-order: 1/[Att] = ] = kkt + 1/[At + 1/[A00] ] If we create three plots of our data, we If we create three plots of our data, we

can determine order: [X] vs. t; ln[X] vs. can determine order: [X] vs. t; ln[X] vs. t; 1/[X] vs. t. The plot that yield a t; 1/[X] vs. t. The plot that yield a linear graph will determine its order.linear graph will determine its order.

Determining Order From Graphical Determining Order From Graphical DataData

Notice, only one graph yields a straight line: The plot of ln[X] vs. t. Therefore, the reaction must be first order.

You should now be able You should now be able to complete # 29, 33, 38to complete # 29, 33, 38

A Microscopic View of A Microscopic View of Reaction RatesReaction RatesConsider the reaction Consider the reaction

NO + ONO + O33 —› NO —› NO22 + O + O22

where Rate = where Rate = kk[NO][O[NO][O33]]Imagine these reactants in rapid, random Imagine these reactants in rapid, random

motion inside a flask. They strike the motion inside a flask. They strike the walls of the reaction vessel and collide walls of the reaction vessel and collide with other molecules.with other molecules.

Will all collisions result in a chemical Will all collisions result in a chemical reaction?reaction?

What conditions must be met for these What conditions must be met for these reactants to react together?reactants to react together?

How a Chemical How a Chemical Reaction Takes PlaceReaction Takes Place Only a tiny fraction of reactant Only a tiny fraction of reactant

collisions leads to reaction. Why?collisions leads to reaction. Why?– Reactant molecules must collide in Reactant molecules must collide in

the proper orientation.the proper orientation.– They must collide with the minimum They must collide with the minimum

energy required to initiate a energy required to initiate a chemical reaction. This is called chemical reaction. This is called activation energyactivation energy (E (Eaa) and the ) and the value varies from reaction to value varies from reaction to reaction.reaction.

Activation EnergyActivation Energy

The diagrams shows the change in potential energy of the molecules during the reaction. Energy must be supplied to the reactants to stretch and break reactant bonds into an intermediate form (represented at the peak of the graph) before proceeding to the final product. The energy difference between the starting molecule and the highest energy along the path represent the activation energy. The difference in the potential energies of the reactants and products indicate whether the reaction was endo- or exothermic.

The Arrhenius The Arrhenius EquationEquationMost reaction rate data is based on three factorsMost reaction rate data is based on three factors

1.1. The fraction of molecules possessing the minimum The fraction of molecules possessing the minimum EEaa or greater or greater

2.2. The number of collisions occurring per secondThe number of collisions occurring per second

3.3. The fraction of collisions that have the appropriate The fraction of collisions that have the appropriate orientationorientation

Arrhenius related these factors mathematically:Arrhenius related these factors mathematically:

kk = = AeAe-E-Eaa/RT/RT

The frequency factor, The frequency factor, AA, is considered constant., is considered constant.

The Arrhenius The Arrhenius EquationEquationThe Arrhenius can be written in The Arrhenius can be written in

several forms:several forms:kk = = AAee-Ea/RT-Ea/RT

lnlnkk = -E = -Eaa/RT + ln/RT + lnAA

ln(ln(kk11//kk22) = E) = Eaa/R (1/T/R (1/T22 – 1/T – 1/T11))(notice this last formula closely mimics y=mx + (notice this last formula closely mimics y=mx +

b)b)

Using the Arrhenius Using the Arrhenius EquationEquationUse the data to determine the activation Use the data to determine the activation

energy for the zero-order decomposition of HI energy for the zero-order decomposition of HI on a platinum surface.on a platinum surface.

Temp (K) Rate Constant (M/s)

573 2.91 x 10-6

673 8.38 x 10-4

773 7.65 x 10-2

We can determine the activation energy for a reaction from a plot of the lnk vs. 1/T. According to ln(ln(k1k1//k2k2) = E) = Eaa/R (1/T2 – 1/T1), /R (1/T2 – 1/T1), the slope of this graph will equal Ethe slope of this graph will equal Eaa/R./R.

Using the Arrhenius Using the Arrhenius EquationEquation

Temp (K)Rate Constant

(M/s) lnk 1/T (K-1)

573 2.91 x 10-6 -12.75 0.00175

673 8.38 x 10-4 -7.08 0.00149

773 7.65 x 10-2 -2.57 0.00129

ln k vs. 1/T

y = -22115x + 25.926

-14

-12

-10

-8

-6

-4

-2

0

0 0.0005 0.001 0.0015 0.002

1/T (K-1)

ln k

Using the Arrhenius Using the Arrhenius EquationEquation

The graph yields a straight line with The graph yields a straight line with a slope of -2.2x10a slope of -2.2x1044K. According K. According to the equation, ln(to the equation, ln(k1k1//k2k2) = E) = Eaa/R /R (1/T2 – 1/T1), this equals E(1/T2 – 1/T1), this equals Eaa/R./R.

So, -2.2x10So, -2.2x1044 K= E K= Eaa/R/R

-2.2x10-2.2x1044 K= E K= Eaa/8.314 J/K·mol/8.314 J/K·mol

EEaa= 1.8 x 10= 1.8 x 1022 kJ/mol kJ/mol

You should now be able You should now be able to complete # 45, 47, 50to complete # 45, 47, 50

Reaction MechanismsReaction Mechanisms

We know that reactants must change We know that reactants must change form (often forming an intermediate form (often forming an intermediate state) in order to proceed to becoming state) in order to proceed to becoming products.products.

This process can be described in much This process can be described in much greater detail if we look at a chemical greater detail if we look at a chemical reaction as occurring over a series of reaction as occurring over a series of steps.steps.

This is referred to as a reaction This is referred to as a reaction mechanism.mechanism.

Reaction MechanismsReaction Mechanisms

Reaction mechanisms can be very Reaction mechanisms can be very simple and involve only one step, simple and involve only one step, or very complicated, involving or very complicated, involving many steps.many steps.

Reaction mechanisms are often Reaction mechanisms are often speculated. It is very difficult to speculated. It is very difficult to scientifically prove that a chemical scientifically prove that a chemical reaction happens in a specific way.reaction happens in a specific way.

Reaction MechanismsReaction Mechanisms

The steps of a reaction mechanism The steps of a reaction mechanism can be described by the number of can be described by the number of

molecules that participate as molecules that participate as reactants in the step. This is called reactants in the step. This is called the the molecularitymolecularity of the reaction. of the reaction.HH33CNC —> HCNC —> H33CCNCCN unimolecularunimolecular

NO + ONO + O22 —> NO —> NO22 + O + O22 bimolecularbimolecular

And so onAnd so on

Reaction MechanismsReaction Mechanisms

Reactions involving only one step Reactions involving only one step are referred to as are referred to as elementaryelementary reactionsreactions

Reaction involving several steps Reaction involving several steps are referred to as are referred to as multistepmultistep reactions. This often involves the reactions. This often involves the formation and consumption of an formation and consumption of an intermediate.intermediate.

Multistep MechanismsMultistep Mechanisms

LetLet’’s consider the reaction s consider the reaction NONO22 + CO —> NO + CO + CO —> NO + CO22

Studies of the reaction have shown that it most Studies of the reaction have shown that it most likely occurs in two steps,likely occurs in two steps,

NONO22 + NO + NO22 —> NO —> NO33 + NO + NONONO33 + CO —> NO + CO —> NO22 + CO + CO22

Thus, we say the reaction occurs by a two-step Thus, we say the reaction occurs by a two-step mechanism.mechanism.

Each step is represented by an elementary Each step is represented by an elementary reaction and the rate law can be written reaction and the rate law can be written directly from its molecularity.directly from its molecularity.

Multistep MechanismsMultistep MechanismsThe chemical equations for the elementary The chemical equations for the elementary

reactions in a multistep mechanism must reactions in a multistep mechanism must always add to give the chemical equation of always add to give the chemical equation of

the overall process.the overall process.NONO22 + NO + NO22 —> NO —> NO33 + NO + NONONO33 + CO —> NO + CO —> NO22 + CO + CO22

2 NO2 NO22 + NO + NO33 + CO —> NO + CO —> NO22 + NO + NO33 + NO + CO + NO + CO22

This equation can be simplified by eliminating This equation can be simplified by eliminating substances that appear on both sides to yield:substances that appear on both sides to yield:

NONO22 + CO —> NO + CO + CO —> NO + CO22

Notice, NONotice, NO33 appears in the reaction mechanism, appears in the reaction mechanism, but it neither a reactant nor a product. It is but it neither a reactant nor a product. It is

formed in one elementary step and consumed formed in one elementary step and consumed in the next. NOin the next. NO33 is an intermediate. is an intermediate.

Multistep MechanismsMultistep Mechanisms

Each step of a mechanism has its own rate law.Each step of a mechanism has its own rate law. Each step is represented by an elementary Each step is represented by an elementary

reaction and the rate law can be written directly reaction and the rate law can be written directly from its molecularity.from its molecularity.

Step 1: NOStep 1: NO22 + NO + NO22 —> NO —> NO33 + NO Rate + NO Rate11 = = kk11[NO[NO22]]22

Step 2: NOStep 2: NO33 + CO —> NO + CO —> NO22 + CO + CO2 2 RateRate22 = = kk22[NO[NO33][CO]][CO]

It is important to remember that the overall rate It is important to remember that the overall rate law of a chemical reaction does not necessarily law of a chemical reaction does not necessarily follow the molecularity of the balanced chemical follow the molecularity of the balanced chemical equation! It must be determined equation! It must be determined experimentally.experimentally.

Multistep MechanismsMultistep Mechanisms

Most chemical reactions occur by Most chemical reactions occur by mechanisms involving two or more mechanisms involving two or more steps.steps.

These steps often occur at different These steps often occur at different rates.rates.

The overall rate of a chemical reaction The overall rate of a chemical reaction is determined by the slowest step of is determined by the slowest step of its mechanism. This is referred to as its mechanism. This is referred to as the rate-determining step.the rate-determining step.

Multistep Mechanisms Multistep Mechanisms – Slow Initial Step– Slow Initial Step

LetLet’’s take another look at our example.s take another look at our example.Step 1: NOStep 1: NO22 + NO + NO22 —> NO —> NO33 + NO + NO

Step 2: NOStep 2: NO33 + CO —> NO + CO —> NO22 + CO + CO22

Overall: NOOverall: NO22 + CO —> NO + CO + CO —> NO + CO22

If we know that step 1 is much, much If we know that step 1 is much, much slower than step two, step 1 must be slower than step two, step 1 must be rate-determining. Therefore, the rate-determining. Therefore, the overall rate law for this reaction must overall rate law for this reaction must be be Rate = Rate = kk[NO[NO22]]22

Multistep Mechanisms Multistep Mechanisms – Fast Initial Step– Fast Initial Step

If the initial step is fast, and therefore If the initial step is fast, and therefore not rate-determining, it is likely that not rate-determining, it is likely that

the rate-determining step involves an the rate-determining step involves an intermediate as a reactant.intermediate as a reactant.

LetLet’’s look at another example:s look at another example:Step 1: NO + BrStep 1: NO + Br22 <—> NOBr <—> NOBr22 (fast) (fast)

Step 2: NOBrStep 2: NOBr22 + NO —> 2NOBr (slow) + NO —> 2NOBr (slow)Overall: 2NO + BrOverall: 2NO + Br22 —> 2NOBr —> 2NOBr

How would you write a rate law for the How would you write a rate law for the overall reaction?overall reaction?

Multistep Mechanisms Multistep Mechanisms – Fast Initial Step– Fast Initial Step

You may have been tempted to say You may have been tempted to say

Rate = Rate = kk[NOBr[NOBr22][NO] ][NO] However, notice that an intermediate However, notice that an intermediate

appears in the rate law. The appears in the rate law. The concentrations of intermediates are concentrations of intermediates are usually unknown. Thus, our rate law usually unknown. Thus, our rate law depends on an unknown concentration. depends on an unknown concentration. We must rewrite our rate law to include We must rewrite our rate law to include only known quantities.only known quantities.

But how???But how???

Multistep Mechanisms Multistep Mechanisms – Fast Initial Step– Fast Initial Step

LetLet’’s go back to the mechanisms go back to the mechanismStep 1: NO + BrStep 1: NO + Br22 <—> NOBr <—> NOBr22 (fast) (fast)

Step 2: NOBrStep 2: NOBr22 + NO —> 2NOBr (slow) + NO —> 2NOBr (slow)

Overall: 2NO + BrOverall: 2NO + Br22 —> 2NOBr —> 2NOBr

Write a rate law the first stepWrite a rate law the first stepStep 1: NO + BrStep 1: NO + Br22 <—> NOBr <—> NOBr22 (fast) (fast)

RateRate11 = = kk11[NO][Br[NO][Br22]]

But notice the equilibrium? The rate above is for the But notice the equilibrium? The rate above is for the forward reaction, what about the reverse?forward reaction, what about the reverse?

RateRate-1-1 = = kk-1-1[NOBr[NOBr22]]

Multistep Mechanisms Multistep Mechanisms – Fast Initial Step– Fast Initial StepWrite a rate law the second stepWrite a rate law the second step

Step 2: NOBrStep 2: NOBr22 + NO —> 2NOBr (slow) + NO —> 2NOBr (slow)

RateRate22 = = kk22[NOBr[NOBr22][NO]][NO]

Recall, the original overall rate law was Recall, the original overall rate law was written as Rate = written as Rate = kk[NOBr[NOBr22][NO]. We ][NO]. We must rewrite the rate law without must rewrite the rate law without [NOBr[NOBr22].].

The three elementary rate laws written The three elementary rate laws written from the mechanism should give us from the mechanism should give us plenty of substitution to choose from.plenty of substitution to choose from.

Multistep Mechanisms Multistep Mechanisms – Fast Initial Step– Fast Initial StepWe must rewrite Rate = We must rewrite Rate = kk[[NOBrNOBr22][NO] using ][NO] using

the elementary rate laws shown below.the elementary rate laws shown below.RateRate11 = = kk11[NO][Br[NO][Br22]]

RateRate-1-1 = = kk-1-1[NOBr[NOBr22]]

RateRate22 = = kk22[NOBr[NOBr22][NO]][NO]

Recall that because of equilibrium, RateRecall that because of equilibrium, Rate11 = Rate = Rate-1-1. . So, So, kk11[NO][Br[NO][Br22] = ] = kk-1-1[NOBr[NOBr22]]

Notice, we can now solve for NOBrNotice, we can now solve for NOBr22..

((kk11//kk-1-1) [NO][Br) [NO][Br22] = [NOBr] = [NOBr22]]

Multistep Mechanisms Multistep Mechanisms – Fast Initial Step– Fast Initial StepLetLet’’s make the substitution.s make the substitution.

Rate = Rate = kk[[NOBrNOBr22][NO]][NO]

Rate = Rate = kk((kk11//kk-1-1) [NO][Br) [NO][Br22][NO]][NO]

Since Since kk ’’s are constant, lets are constant, let’’s rename s rename kk((kk11//kk-1-1) as K.) as K.

Rate = Rate = KK [NO][Br [NO][Br22][NO]][NO]

And now simplifyAnd now simplify

Rate = Rate = KK [NO] [NO]22[Br[Br22] where ] where KK = = kk((kk11//kk-1-1).).

This is a viable rate law since the concentrations This is a viable rate law since the concentrations of all species in the rate law are known.of all species in the rate law are known.

You should now be able You should now be able to complete # 55, 58, 59, to complete # 55, 58, 59, 6262

CatalysisCatalysis

A A catalyst catalyst is a substance that changes is a substance that changes the speed of a chemical reaction without the speed of a chemical reaction without undergoing a permanent chemical undergoing a permanent chemical change.change.

A catalyst can be homogeneous (in the A catalyst can be homogeneous (in the same phase as the reacting molecules) or same phase as the reacting molecules) or heterogeneous (in a different phase as heterogeneous (in a different phase as the reacting molecules)the reacting molecules)

Enzymes in the human body are Enzymes in the human body are examples of organic catalysts.examples of organic catalysts.

Homogeneous Homogeneous CatalysisCatalysis

HH22OO2(aq)2(aq) —> H —> H22OO(l)(l) + O + O2(g)2(g)

The decomposition of hydrogen peroxide can The decomposition of hydrogen peroxide can be catalyzed by Ibe catalyzed by I- - (given by the dissociation (given by the dissociation of KI) as shown.of KI) as shown.

Step 1: 2IStep 1: 2I--(aq)(aq) + 2K + 2K++

(aq)(aq) + H + H22OO2(aq)2(aq) —> I —> I2(aq)2(aq) + 2H + 2H22OO(g)(g)

Step 2: IStep 2: I2(aq)2(aq) + H + H22OO2(aq)2(aq) —> 2I —> 2I--(aq)(aq) + 2K + 2K++

(aq)(aq) + O + O2(g)2(g)

Notice, adding steps 1 & 2 yieldsNotice, adding steps 1 & 2 yields2H2H22OO2(aq)2(aq) —> 2H —> 2H22OO(l)(l) + O + O2(g)2(g)

How does this work?How does this work?

Homogeneous Homogeneous CatalysisCatalysis

As a general rule, a catalyst lowers the As a general rule, a catalyst lowers the overall activation energy for a overall activation energy for a chemical reactionchemical reaction

Heterogeneous Heterogeneous CatalysisCatalysis The initial set usually begins with The initial set usually begins with

adsorption of reactants onto the adsorption of reactants onto the catalyst surface.catalyst surface.

Atoms/Ions on the surface of a solid Atoms/Ions on the surface of a solid are very reactive because of the are very reactive because of the unused bonding capacity.unused bonding capacity.

Be sure to read this section of your Be sure to read this section of your text book carefully so you understand text book carefully so you understand how heterogeneous catalysis works!!!how heterogeneous catalysis works!!!

The decomposition reaction is determined to be first order. A graph of the partial pressure of HCOOH versus time for decomposition at 838 K is shown as the red curve in Figure 14.28. When a small amount of solid ZnO is added to the reaction chamber, the partial pressure of acid versus time varies as shown by the blue curve in Figure 14.28.

SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together

Formic acid (HCOOH) decomposes in the gas phase at elevated temperatures as follows:

Figure 14.28  Variation in pressure of HCOOH(g) as a function of time at 838 K. The red line corresponds to decomposition when only gaseous HCOOH is present. The blue line corresponds to decomposition in the presence of added ZnO(s).

SAMPLE INTEGRATIVE EXERCISE continued

(a) Estimate the half-life and first-order rate constant for formic acid decomposition.(b) What can you conclude from the effect of added ZnO on the decomposition of formic acid?(c) The progress of the reaction was followed by measuring the partial pressure of formic acid vapor at selected times. Suppose that, instead, we had plotted the concentration of formic acid in units of mol/L. What effect would this have had on the calculated value of k?(d) The pressure of formic acid vapor at the start of the reaction is 3.00 10 2 torr. Assuming constant temperature and ideal-gas behavior, what is the pressure in the system at the end of the reaction? If the volume of the reaction chamber is 436 cm3, how many moles of gas occupy the reaction chamber at the end of the reaction?(e) The standard heat of formation of formic acid vapor is ΔHf° = -378.6 kJ/mol. Calculate ΔHº for the overall reaction. Assuming that the activation energy (Ea) for the reaction is 184 kJ/mol, sketch an approximate energy profile for the reaction, and label Ea, ΔHº, and the transition state.

You should now be able You should now be able to complete # 66, 87, 88, to complete # 66, 87, 88, 9898

You Should Now Be Able You Should Now Be Able To…To…

Identify factors which affect reaction rates.Identify factors which affect reaction rates. Calculate the rate of production of a product Calculate the rate of production of a product

or consumption of a reactant using mole or consumption of a reactant using mole ratios and the given rate.ratios and the given rate.

Determine the rate law for a reaction from Determine the rate law for a reaction from given data, overall order and value of the given data, overall order and value of the rate constant (including units!)rate constant (including units!)

Determine the instantaneous rate of a Determine the instantaneous rate of a reaction.reaction.

You Should Now Be Able You Should Now Be Able To…To…

Use integrated rate laws to determine Use integrated rate laws to determine concentrations at a certain time, t, and create concentrations at a certain time, t, and create graphs to determine the order of a reaction. graphs to determine the order of a reaction. Also, determine the half-life reaction.Also, determine the half-life reaction.

Determine the activation energy for the Determine the activation energy for the reaction using the Arrhenius equation.reaction using the Arrhenius equation.

Graphically determine the activation energy Graphically determine the activation energy using the Arrhenius equation.using the Arrhenius equation.

Write the rate law from a given mechanism Write the rate law from a given mechanism and identify catalysts and intermediates and identify catalysts and intermediates present.present.