Factors That Affect Reaction Rates Presence of a Catalyst Surface - - PDF document

kinetics­presentation­Yoos.notebook 1 October 15, 2012

Chemical Kinetics

Kinetics

• In 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

• ccurs).

Factors That Affect Reaction Rates

• The Physical State of the Reactants
• The Concentrations of the Reactants
• Temperature
• The Presence of a Catalyst
• Surface area

Factors That Affect Reaction Rates

• Physical State of the Reactants
• In order to react, molecules must come in

contact with each other.

• The more homogeneous the mixture of

reactants, the faster the molecules can react.

Factors That Affect Reaction Rates

Air hole

The flame from the Bunsen burner shows the variation in color as more air is mixed with the natural gas. As more air is mixed with the natural gas, the rate of combustion reaction will increase and the flame gets more blue in color. The Concentrations of the Reactants As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.

Factors That Affect Reaction Rates

Temperature At higher temperatures, reactant molecules

• have more kinetic energy
• move faster
• collide more often and with greater energy.

Why do we keep food in fridge?

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Factors That Affect Reaction Rates

Presence of a Catalyst

• Catalysts speed up reactions by changing the

mechanism of the reaction.

• Catalysts are not consumed during the course
• f the reaction.

Surface Area

• When more surface area of a reactant is

exposed, a reaction will occur faster.

Factors That Affect Reaction Rates

• A steel bar will not ignite

if exposed to a flame. In contrast, a steel wool pad will burn and ignite quickly. 1 Which of the following does not play a part in determining the rate of a reaction?

A

Temperature

B

Concentration of the reactants

C

Catalyst

D

Surface area of the solid

E

Equilibrium constant

2 Food spoils at a faster rate at 25 degrees F compared to 5 degrees C.

Yes No

3 Why does a higher temperature cause a reaction to go faster?

A Only because there are more collisions per second. B Only because collisions occur with greater energy. C There are more frequent collisions and they are of greater energy.

4 Why does a higher reactant concentration make a reaction faster?

A Only because there are more collisions per second. B Only because collisions occur with greater energy. C There are more frequent collisions and they are of greater energy.

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5 Why does a greater surface area cause a reaction to proceed faster?

A Only because there are more collisions per second. B Only because collisions occur with greater energy. C There are more frequent collisions and they are of greater energy.

6 What happens to a catalyst in a reaction?

A It is unchanged.

B It is incorporated into the products. C It is incorporated into the reactants. D It evaporates away.

7 If a catalyst is used in a reaction_________.

A the energy of activation increases. B different reaction products are obtained. C the reaction rate increases. D it evaporates away.

Reaction Rates

Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

0 s 20 s 40 s 1 mol A 0.5 mol A 0.5 mol B 0.25 mol A 0.75 mol B

A B

Reaction Rates

C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq) In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times.

Time , t(s) [C4H9Cl](M) 0.0 0.1000 50 0.0905 00.0 0.0820 50.0 0.0741 200.0 0.0671 300.0 0.0549 400.0 0.0448 500.0 0.0368 800.0 0.0200 10000 0

Reaction Rates

C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)

The average rate of the reaction over each interval is the change in concentration divided by the change in time:

Time , t(s) [C4H9Cl](M) Average Rate (M/s) 0.0 0.1000 50 0.0905 1.9 x10­4 00.0 0.0820 1.7 x10­4

• 50. 0 0.0741 1.6 x10­4
• 200. 0 0.0671 1.4 x10­4

300.0 0.0549 1.22 x 10­4 400.0 0.0448 1.01 x10­4 500.0 0.0368 0.88 x10­4 800.0 0.0200 0.560 x10­4 10000 0

Average Rate = [C4H9Cl] t

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Reaction Rates

C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)

Time , t(s) [C4H9Cl](M) Average Rate (M/s) 0.0 0.1000 50 0.0905 1.9 x10­4 00.0 0.0820 1.7 x10­4

• 50. 0 0.0741 1.6 x10­4
• 200. 0 0.0671 1.4 x10­4

300.0 0.0549 1.22 x 10­4 400.0 0.0448 1.01 x10­4 500.0 0.0368 0.88 x10­4 800.0 0.0200 0.560 x10­4 10000 0

• Note that the average rate decreases as the reaction proceeds.

WHY?

Reaction Rates

• A plot of [C4H9Cl] 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.

C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)

Instantaneous rate at t=600s

[C4H9Cl] (M) x 10­2 Time (s)

Instantaneous rate at t=0s

• 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.

Reaction Rates

C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)

Instantaneous rate at t=600s

[C4H9Cl] (M) x 10­2 Time (s)

Instantaneous rate at t=0s

Reaction Rates and Stoichiometry

• In this reaction, the ratio
• f C4H9Cl to C4H9OH is

1:1.

• Thus, the rate of

disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH.

C4H9Cl(aq) + H2O(l) → C4H9OH(aq) + HCl(aq)

Instantaneous rate at t=600s

[C4H9Cl] (M) x 10­2 Time (s)

Instantaneous rate at t=0s

Rate [C4H9Cl] t ­ t [C4H9OH] = = 8 Which substance in the reaction below either appears or disappears the slowest? 4NH3 + 7O2 4NO2 + 6H2O

A

NH3

B

O2

C

H2O

D

NO2

E

The rates of appearance are the same as the rates of disappearance 9 Which of the following expressions could represent a reaction rate?

A

time/mass

B

concentration/time

C

energy/time

D

time/energy

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10 Which of the following set of units represents a reaction rate?

A

g/s

B

mol/s

C

s

D

j/s

E

s/j 11 A flask is charged with 0.124mol of A and allowed to react to form B according to the reaction A(g) B(g). the following data are obtained for [A] as the reaction proceeds. The average rate of disappearance of A between 10 s and 20 s is __________ mol/s.

A

2.2 10­3

B

1.1 10­3

C

4.4 10­4

D

454

E

9.9 10­3

12 A flask is charged with 0.124 mol of A and allowed to react to form B according to the reaction A(g) →B(g). The following data are obtained for [A] as the reaction proceeds:

A B C D E

590 The peroxydisulfate ion, S2O8

2­, reacts with the

iodide ion in aqueous solution via the reaction: S2O8

2­ (aq) + 3I­ ­­­> 2SO4 (aq) + I3 ­ (aq)

The average rate of disappearance of between 400.0 s and 800.0 s is __________ M/s. 13 A 2.8 x 10­5 B 1.4 x 10­5

C 5.8 x 10­5 D 3.6 x 10­4 E 2.6 x 10­4

The following data are collected for the concentration of iodide ion:

Reaction Rates and Stoichiometry

• What if the ratio is not 1:1?

For every 1 mole of I2 that is produced, 2 moles of HI must decompose. So the rate of disappearance of HI is twice as fast as the appearance of I2 or the appearance

• f H2.

2HI(g) H2(g) + I2(g)

Rate = ­ 1 2 [HI] t [I2] t =

Reaction Rates and Stoichiometry

• To generalize, then, for the reaction

aA + bB cC + dD

Rate = t [A] ­1 a [B] b ­1 t = [C] c ­1 t [D] d ­1 t = =

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At elevated temperatures, dinitrogen pentoxide decomposes to nitrogen dioxide and oxygen: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) When the rate of formation of O2 (g) is 5.5 10­4 M/s, the rate of decomposition of N2O5 is __________ M/s. 14

A 2.2 x 10­3 B 1.4 x 10­4 C 10.1 x 10­4 D 2.8 x 10­4 E 5.5 x 10­4

At elevated temperatures, dinitrogen pentoxide decomposes to nitrogen dioxide and oxygen: 2 N2O5 (g) → 4 NO2 (g) + O2 (g) When the rate of formation of NO2 (g) is 5.6 10­4 M/s, the rate of decomposition of N2O5 is __________ M/s. 15

A 2.2 x 10­3 B 1.4 x 10­4 C 10.1 x 10­4

D 2.8 x 10­4 E 5.5 x 10­4

Concentration and Rate

One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration.

Concentration and Rate

If we compare Experiments 1 and 2, we see that when [NH4

+] doubles, the initial rate doubles.

NH4

+(aq) + NO2−(aq)

N2(g) + 2 H2O(l)

Experiment Initial[ NH4

+] Initial [NO2 ­ ] Observed initial

number (M) [M] Rate (M/s) 1 0.0100 0.200 5.4 x10­7 2 0.0200 0.200 10.8 x10­7 3 0.0400 0.200 21.5 x10­7 4 0.200 0.0202 10.8 x10­7 5 0.200 0.0404 21.6 x10­7 6 0.200 0.0808 43.3 x10­7

Likewise, when we compare Experiments 5 and 6, we see that when [NO2

−] doubles, the initial rate doubles.

Concentration and Rate

NH4

+(aq) + NO2−(aq)

N2(g) + 2 H2O(l)

Experiment Initial[ NH4

+] Initial [NO2 ­ ] Observed initial

number (M) [M] Rate (M/s) 1 0.0100 0.200 5.4 x10­7 2 0.0200 0.200 10.8 x10­7 3 0.0400 0.200 21.5 x10­7 4 0.200 0.0202 10.8 x10­7 5 0.200 0.0404 21.6 x10­7 6 0.200 0.0808 43.3 x10­7

Concentration and Rate

• This means

Rate ∝ [NH4

+]

Rate ∝ [NO2

­]

Rate ∝ [NH4

+] [NO2 ­]

which, when written as an equation, becomes Rate = k [NH4

+] [NO2 −]

• This equation is called the rate law, and k is the

rate constant. Therefore,

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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 [NH4

+] [NO2 −]

the reaction is First­order in [NH4

+] and

First­order in [NO2

−].

Rate Laws

Rate = k [NH4

+] [NO2 −]

• The overall reaction order can be found by

adding the exponents on the reactants in the rate law.

• This reaction is second­order overall.

Order of the reaction can be determined only from experimental data. Concentration changes → Rate independent → zero order Concentration doubles → Rate doubles → first order Concentration doubles → Rate quadruples → second order

Order of a reaction

16 If the rate law for the reaction 2A + 3B products is first order in A and second order in B, then the rate law is:

A Rate = k [A][B] B Rate = k [A]2[B]3 C Rate = k [A][B]2 D Rate = k [A]2[B] E Rate = k [A]2[B]2

17 What will happen to the rate of a reaction if the molarity of a first­

• rder reactant is doubled?

A rate will double B rate will triple C rate will quadruple

D rate remains unchanged

18 What will happen to the rate of a reaction if the molarity of a second­order reactant is doubled?

A rate will double B rate will triple C rate will quadruple

D rate remains unchanged

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Practice Question

[NO] [O2] Initial rate M/s 1 0.0126 0.0125 1.41x10­2 2 0.0252 0.0250 1.13x10­1 3 0.0252 0.0125 5.64x10­2 R3 = K[NO]x [O2]y 5.64x10­2 R1 K[NO]x [O2]y 1.41x10­2 = k[0.0256]x [0.0125]y = 4 k[0.0126]x [0.0125]y 2x = 4; x = 2 R2/R3 = K[0.0252]x [0.025]y =1.13x10­1/5.64x10­2 k[0.0252]x [0.0125]y 2y = 2 y = 1 R = K[NO]2 [O2]

19

This reaction is _______order with respect to [OH­]. A

1st B 2nd C 3rd D 4th E zero 20 This reaction is _______order with respect to [ClO2]. A 1st B 2nd C 3rd D 4th E zero 21

The overall order of a reaction is 2. The units of the rate constant for the reaction are __________.

A M/s B M­1s­1 C

s­1

D M­1

E s / M2 Integrated rate law equations can be used to determine the concentration of either a reactant or a product at any time.

Integrated Rate Laws

In a typical problem, you may be asked to find either: 1) concentration of a reactant at a given time, t, after the start of a rxn 2) the time it takes for a certain amount of reactant to be consumed 3) the time it takes for the concentration of a reactant or product to reach a certain level

Integrated Rate Laws

Using calculus to integrate the rate law for a first­order process gives us R= k[A] time 0 to t Where

[A]0 is the initial concentration of A, and [A]t is the concentration of A at some time, t, during the course of the reaction.

ln [A]t [A]0 =­kt

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Integrated Rate Laws

Manipulating this equation produces… …which is in the form

y = mx + b

ln [A]t [A]0 =­kt ln[A]t ­ ln[A]0 = ­kt ln[A]t = ­kt + ln[A]0

First­Order Processes

Therefore, if a reaction is first­order, a plot

• f ln [A] vs. t will yield a straight line, and the

slope of the line will be ­k.

ln [A]t = ­kt + ln [A]0

First­Order Processes

Consider the process in which methyl isonitrile is converted to acetonitrile. CH3NC CH3CN CH3NC CH3CN

First­Order Processes

This data was collected for this reaction at 198.9 °C. CH3NC CH3CN

First­Order Processes

• When ln P is plotted as a function of time, a straight

line results.

• Therefore,
• The process is first­order.
• k is the negative of the slope: 5.1 × 10­5 s­1.

Second­Order Processes

Similarly, integrating the rate law for a process that is second­order in reactant A, R= k[A]2, we get 1 [A]t = kt +

1 [A]0

also in the form

y = mx + b

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Second­Order Processes

1 [A]t = kt + 1 [A]0

So if a process is second­order in A, a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k.

Second­Order Processes

The decomposition of NO2 at 300°C is described by the equation

and yields data comparable to this:

[NO2], M 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 00.0380

Time (s)

NO2 (g) NO (g) + O2 (g) 1 2

Second­Order Processes

Plotting ln [NO2] vs. t yields the graph below

The plot is not a straight line, so the process is not first­order in [A]. Time (s) [NO2], M ln [NO2] 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

Second­Order Processes

Graphing vs. t, however, gives this plot.

1 [NO2]

Because this is a straight line, the process is second order in [A] Time (s) [NO2], M 1 [NO2] 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

Half­Life

• Half­life is defined as the time required for
• ne­half of a reactant to react.
• Because [A] at t1/2 is one­half of the original

[A], [A]t = 0.5 [A]0.

Half­Life

For a first­order process, this becomes

ln 0.5 = −kt1/2 −0.693 = −kt1/2 = t1/2 0.693 k

NOTE: For a first­order process, then, the half­ life does not depend on [A]0.

ln 0.5[A]0 = ­kt1/2 [A]0

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Half­Life

For a second­order process, kt1/2 [A]0

1

0.5 = + [A]0

1 2 [A]0 kt1/2

= + [A]0

1

2­1 [A]0 [A]0

1 kt1/2

= = 1 [A]0 k = t1/2

22

Which of the following will yield a straight line for a first­order reaction? A [A] vs. time B log [A] vs. time C ln [A] vs. time D ln [A] vs. k E 1/[A] vs. k 23 Which of the following will yield a straight line for a second­order reaction? A [A] vs. time B log [A] vs. time C ln [A] vs. time D 1/[A] vs. time E 1/[A] vs. k 24 Which of the following will yield a straight line for a second­order reaction? A [A] vs. time B Pressure, P vs. time C 1/P vs. k D ln P vs. time E

1/P vs. time

25

The half­life for Substance X is 22 minutes. How much of a 10.0­gram sample will remain after 66 minutes? A 1.25 grams B 1.33 grams C 2.5 grams D 5.00 grams E 7.00 grams 26 A medical courier must deliver 8.0 mg of a substance in four hours. If the half­life of this substance is 2.0 hours, the original sample should be at least _____ mg.

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Temperature and Rate

• Generally, as temperature

increases, so does the reaction rate.

• This is because k is

temperature dependent.

High Temperature Low temperature

180 190 200 210 220 230 240 250 260 1 2 3 4 x y

temperature ( 0C)

x 10­3 k(s­1)

Maxwell–Boltzmann Distributions

Temperature is defined as a measure

• f the average kinetic

energy of the molecules in a sample.

Fraction of molecules Speed / KE Ea

At any temperature there is a wide distribution of kinetic energies.

Maxwell–Boltzmann Distributions

• As the temperature

increases, the curve flattens and broadens.

• Thus at higher

temperatures, a larger population of molecules has higher energy.

F r a c t i

• n
• f

m

• l

e c u l e s Speed / KE Ea

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.

Maxwell–Boltzmann Distributions

F r a c t i

• n
• f

m

• l

e c u l e s Speed / KE Ea

As a result, the reaction rate increases.

Maxwell–Boltzmann Distributions

This fraction of molecules can be found through the expression where Ea is the activation energy, R is the gas constant and T is the Kelvin temperature.

f = e

­Ea

RT F r a c t i

• n
• f

m

• l

e c u l e s Speed / KE Ea

The Collision Model

• In a chemical reaction, bonds are broken and new

bonds are formed.

• Molecules can only react if they collide with each
• ther.

Temperature and Rate

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The Collision Model

In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other. Furthermore, molecules must collide with the correct

• rientation and with enough energy to cause bond breakage

and formation.

Cl + ClON Cl2 + NO

Before collision Collision After collision

Temperature and Rate

Svante Arrhenius developed a mathematical relationship between the rate constant, 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.

­Ea RT

Temperature and Rate The Collision Model

Arrhenius Equation

y = m x + b Therefore, if k is determined experimentally at several temperatures, Ea can be calculated from the slope of a plot

• f ln k vs. 1/T.

Taking the natural logarithm of both sides, the Arrhenius equation becomes:

which is also in the form

Reaction Coordinate Diagrams

It is helpful to visualize energy changes throughout a process

• n a reaction coordinate diagram like this one for the

rearrangement of methyl isonitrile.

H3CNC H3CCN

Methyl isonitrile to Methyl nitrile

Reaction pathway Potential energy Ea

Δ E

Reaction Coordinate Diagrams

Reaction pathway P

• t

e n t i a l e n e r g y

Ea

ΔE

Reaction Coordinate Diagrams

Reaction pathway Potential energy Ea ΔE

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Reaction Coordinate Diagrams

Reaction pathway Potential energy Ea ΔE

Reaction Coordinate Diagrams

Reaction pathway Potential energy

Ea ΔE

Reaction Coordinate Diagrams

Reaction pathway P

• t

e n t i a l e n e r g y

Ea ΔE

• The diagram shows the

energy of the reactants and products (and, therefore, ∆E).

• The high point on the

diagram is the transition state.

• The energy gap between the reactants and the activated

complex is the activation energy barrier.

Reaction Coordinate Diagrams

Reaction pathway Potential energy

• The species present at the transition state is called the

activated complex.

• In order for a reaction to
• ccur, a minimum amount
• f energy must first be

applied to the reactants.

• This minimum energy

input is usually needed to break bonds.

Reaction Coordinate Diagrams Reaction Coordinate Diagrams

• This minimum energy input is

called the Activation Energy, or Ea, energy of activation.

• Activation energy is the energy

gap between the reactants and the activated complex.

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• In this example, the PE of

the products are lower than the PE of the reactants.

• Therefore, ∆H° is negative

and the reaction is exothermic.

Reaction Coordinate Diagrams

27 This potential energy diagram represents a reaction that is ____________ with a ∆H of _____.

A exothermic, ­10 kJ B exothermic, ­25 kJ C endothermic, +15 kJ D endothermic, +25 kJ

28 This potential energy diagram represents a reaction that is ____________ with a ∆H of _____.

A exothermic, ­5 kJ B exothermic, ­20 kJ

C endothermic, +5 kJ D endothermic, +25 kJ

29 The activation energy for this reaction is

A

+15 kJ

B

­15 kJ

C

+25 kJ

D

­25 kJ

30 The activation energy for the reverse reaction is

A +25 kJ B ­25 kJ C +40 kJ D ­40 kJ

31 What is another name for the activated complex?

A energy barrier B transition state C rate limiter D collision group E reactant or product

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32 At what stage of a reaction do atoms have the highest energy?

A reactant stage B product stage

C transition state stage

33 Activation energy is __________.

A the heat released in a reaction B the energy given off when reactants collide C generally very high for a reaction that takes place rapidly D an energy barrier between reactants and products

Catalysts

Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. This graph shows the decomposition of hydrogen peroxide both with and without a catalyst. Notice that the energies of reactants and products are unchanged by the catalyst.

Reaction Mechanisms

The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism.

Reaction Mechanisms

• Reactions may occur all at once or through several

discrete steps.

• Each of these processes is known as an elementary

reaction or elementary process.

Reaction Mechanisms

The molecularity of a process tells how many molecules are involved in the process.

Molecularity Elementary Reaction Rate Law Unimolecular A­­> products Rate =k[A] Bimolecular A+A ­­> products Rate =k[A]2 Bimolecular A+ B ­­> products Rate = k[A][B] Termolecular A+A+A­­> products Rate = k[A]3 Termolecular A+A+B­­> products Rate =k[A]2[B] Termolecular A+B+C­­> products Rate = k[A][B][C]

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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, rate­determining step. step 1 1→2 step 2 2→3

Toll A Toll B 1

2 3 slow

Toll A Toll B

1 2 3 slow

Multistep Mechanisms

In a multistep mechanism, the main idea about identifying the rate­determining step is this:

The rate law for the overall reaction is equivalent to the rate law for the slowest, or rate­determining step.

Slow Initial Step

• The rate law for this reaction is found experimentally

to be Rate = k [NO2]2

• CO is necessary for this reaction to occur, but the rate
• f the reaction does not depend on its concentration.
• This suggests the reaction occurs in two steps.

NO2(g) + CO (g) → NO (g) + CO2 (g)

Slow Initial Step

• A proposed mechanism for this reaction is

Step 1: NO2 + NO2 → NO3 + NO (slow) Step 2: NO3 + CO → NO2 + CO2 (fast)

• The NO3 intermediate is consumed in the second

step.

• As CO is not involved in the slow, rate­determining

step, it does not appear in the rate law.

Toll A Toll B 1

2 3 slow

Fast Initial Step

• The rate law for this reaction is found to be

Rate = k [NO]2 [Br2] This reaction is termolecular

• Because termolecular processes are rare∗, this

rate law suggests a two­step mechanism. 2 NO (g) + Br2 (g) → 2 NOBr (g)

∗ collision of three molecules in the correct orientation is very difficult

Fast Initial Step

• A proposed mechanism is

Step 1 includes the forward and reverse reactions.

K2

Step 2: NOBr2 + NO → 2NOBr (slow) Step 1: NO + Br2 ↔ NOBr2(fast)

K1 K­1

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Fast Initial Step

• The rate of the overall reaction depends upon the rate of

the slow step.

• The rate law for that step would be

Rate = K2[NOBr2] [NO]

• But how can we find [NOBr2]?

Fast Initial Step

• NOBr2 can react two ways:
• 1. With NO to form NOBr
• 2. By decomposition to reform NO and Br2
• The reactants and products of the first step are in

equilibrium with each other.

• Therefore,

Ratef = Rater

Fast Initial Step

• Because Ratef = Rater ,

k1 [NO] [Br2] = k−1 [NOBr2]

• Solving for [NOBr2] gives us

K1 K­1 [NO] [Br2]= [NOBr2]

Fast Initial Step

Substituting this expression for [NOBr2] in the rate law for the rate­determining step gives = k [NO]2 [Br2] K2K1 K­1

`

Rate = [NO][Br2][NO] 34 Which step is the rate determining step, in a multistep reaction mechanism?

A the first step B the last step C the slowest step D the fastest step

E

not enough information given.

35

What is the rate law for the

• verall reaction, given the

following mechanism: Step 1: A + B → C (slow) Step 2: C + A → 2D (fast)

A Rate = [A]2[B][C] B Rate = [A][B] C Rate = [A][C] D Rate = [A]2[B] / [D]2

kinetics­presentation­Yoos.notebook 19 October 15, 2012

36

What is the rate law for the overall reaction, given the following mechanism: Step 1: A + B ↔ C (fast) Step 2: C + A → 2D (slow)

A Rate = [A]2[B][C] B Rate = [A][B] C Rate = [A][C] D Rate = [A]2[B]

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.

Catalysts

One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.

Shown here is hydrogen reacting with ethylene. Both molecules are adsorbed onto the metal (blue) surface which acts as the catalyst.

• Automobiles are equipped

with catalytic converters, which are part of their exhaust systems.

• The exhaust gases

contain CO, NO, NO2, and unburned hydrocarbons that pass over surfaces impregnated with catalysts.

Catalytic Converters

• The catalysts promote the conversion of the exhaust gases into

CO2, H2O, and N2.

37 Why does a catalyst cause a reaction to proceed faster?

A Only because there are more collisions per second. B Only because collisions occur with greater energy. C Only because the activation energy is lowered. D There are more frequent collisions and they are of greater energy.

38 A catalyst can increase the rate of a

reaction __________.

A

By changing the value of the frequency factor (A)

B

By increasing the overall activation energy (Ea) of the reaction

C

By lowering the activation energy of the reverse reaction

D

By providing an alternative pathway with a lower activation energy

E

All of these are ways that a catalyst might act to increase the rate of reaction.

kinetics­presentation­Yoos.notebook 20 October 15, 2012

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. Liver contains an enzyme called catalase which speeds up the decomposition of H2O2.

Substrate entering active site of enzyme Enzyme/substrate complex Enzyme/products complex Products leaving active site of enzyme

Active site substrate

39 Which of the following acts as catalysts in the body?

A

Carbohydrates

B

Lipids

C

Nucleic Acids

D

Enzymes

E

Water