reaction rates 12.1 reaction rates - welcome to...

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1Chemistry: OpenStax

Ch. 12: Kinetics

An agama lizard basks in the sun. As its body warms, thechemical reactions of its metabolism speed up.

12.1 Reaction Rates� Chemical Kinetics is the study of how fast reactions

take place.

� Reactions can be fast (reaction of sodium metal with

water) or very slow (rusting of iron).

� Why do you need to study kinetics?

� Predict rate/speed of reactions

� Drug delivery: time-release capsules

� Understanding reaction mechanisms

� Understanding metabolism regulation

2

� Reaction rate (or speed) = change in amount of

substance per unit of time.

� What factors can we change in lab to alter the rate of

a reaction?

� Concentration

� Temperature

� Catalysts

� Surface area of solids 3

Reaction Rates Calculating Rates of Substances

� For the reaction: A � B, we can measure the rate of

disappearance of reactants:

� (assigned negative so rate is a positive value)

and appearance of products:

4

Image from Principles of General Chemistry http://2012books.lardbucket.org/books/principles-of-general-chemistry

-v1.0/s18-chemical-kinetics.html, CC-BY-NC-SA 3.0 license

Rates of Substances

� Reaction: A � B

5Image from Principles of General Chemistry http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s18-chemical-kinetics.html, CC-BY-NC-SA 3.0 license

Rates of Substances

� 2H2O2(aq) � 2H2O(l) + O2(g)

� What is the rate of disappearance of H2O2 for each

time interval?

6

Time (hr) [H2O2], M ∆∆∆∆[H2O2], M ∆∆∆∆t (hr) Rate of H2O2, M/hr

0 1.000

6.00 0.500

12.00 0.250

18.00 0.125

24.00 0.0625

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Rates of Substances

� 2H2O2(aq) � 2H2O(l) + O2(g)

� What is the rate of disappearance of H2O2 for each

time interval? Units: M = mol/L = mol�L-1

7


0 1.000

-0.500 6.00 0.08336.00 0.500

12.00 0.250

18.00 0.125

24.00 0.0625


0 1.000

-0.500 6.00 0.08336.00 0.500

-0.250 6.00 0.041712.00 0.250

-0.125 6.00 0.020818.00 0.125

-0.0625 6.00 0.010424.00 0.0625

2 H2O2(aq) � 2 H2O(l) + O2(g)

8

Rates of Substances

Rates are related by coefficients. Which curve is the reactant? How

can you tell the difference between the curves of the products?

0

0.2

0.4

0.6

0.8

1

1.2

0 5 10 15 20 25

Concentr

ation (

M)

Time (hr)

2 H2O2 --> 2 H2O + O2

0

0.2

0.4

0.6

0.8

1

1.2

0 5 10 15 20 25 30 35 40 45 50

Concentr

ation (

M)

Time (hr)

2 H2O2 �� 2 H2O + O2

� 2 H2O2(aq) � 2 H2O(l) + O2(g)

� This is called the “rate expression”. We can use this expression (or equation) to determine the rate of each substance and the rate of reaction over a given time period.

Rate of reaction = rate of substance / its coefficient

Notes about rate expressions:

1) Reactants (-); Products (+);

2) Coefficients are in the denominator;

3) Every component in equation is equal to all others (each one

calculates the rate of reaction).

9

Rates of Substances, Reaction

� 2 H2O2(aq) � 2 H2O(l) + O2(g)

Notes about rate expressions:

1) Reactants (-); Products (+);

2) Coefficients are in the denominator;

3) Every component in equation is equal to all others (each one

calculates the rate of reaction).

Write the balanced equation for the rate expression below:

10

Rates of Substances, Reaction

2 H2O2(aq) � 2 H2O(l) + O2(g)If H2O2 disappears at 0.20 M/s, 1) What is the rate of appearance of H2O? 2) What is the rate of appearance of O2? 3) What is the rate of reaction?

Two methods: a) rate expression and b) stoichiometrya)

b) 0.20 mol H2O2/L/s x (2 mol H2O / 2 mol H2O2)

11

Rates of Substances, Reaction Rates of Substances – Practice

� 2H2O2(aq) � 2H2O(l) + O2(g)

� What are the rates of appearance of H2O and O2

from 0 – 6 hours?

� From 6-12 hours?

12

0.0833 M/hr, 0.0417 M/hr (mol�L-1�hr-1)

0.0417 M/hr, 0.0208 M/hr

Time (hr) Rate of H2O2,M/hr

Rate of H2O, M/hr

Rate of O2, M/hr

0 – 6.00 0.0833 0.0833 0.0417

6.00 – 12.00 0.0417 0.0417 0.0208

12.00 – 18.00 0.0208 0.0208 0.0104

18.00 – 24.00 0.0104 0.0104 0.00520

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Rates of Substances, ReactionWrite the rate expression and calculate rates for the

reaction: SiO3 + 4C � SiC + 3CO

1) If the rate of disappearance of SiO3 is 0.25 M/s, what are

the rates of disappearance of C, appearance of SiC and

CO?

C = 1.0 M/s; SiC = 0.25 M/s; CO = 0.75 M/s

2) What is the rate of the reaction?

Reaction = 0.25 M/s

Examples: 12.1, 12.213

Calculating Rates – Practice

� 5Br - (aq) + BrO3- (aq) + 6H+ (aq) � 3Br2(aq) + 3H2O(l)

� If the rate of appearance of Br2 is 1.2 x 10-3 M/s, what is the rate of

disappearance of Br -?

� 1.2 x 10-3 M/s * (5 mol Br -/3 mol Br2) = 2.0 x 10-3 M/s

� C6H12O6(aq) � 2C2H5OH(aq) + 2CO2(g)

� If rate of appearance of ethanol is 4.2 x 10-2 M/s, what is rate of

disappearance of glucose? 2.1 x 10-2 M/s

� SiO3 + 4C � SiC + 3CO� If the rate of appearance of CO is 0.78 M/s, what is the rate with

respect to C? 1.0 M/s

14

� Average Rate: rate of substance over a given time interval

� Rate = (Mf – Mi) / (tf – ti)

� Instantaneous rate: rate at a specific time

� Determined by measuring slope of tangent at a point

� Initial rate: rate at time = 0

� Determined by measuring slope of tangent at t = 0

� These rates will all have different values. Rates are not

constant; they change over time. If they were constant,

graphs would be straight lines.

15

12.3 Rate Laws; Reaction Orders Reactant Concentrations & Rate Laws

� Rate law - relationship between concentration of each

reactant and the reaction rate.

� aA + bB � cC + dD

� rate = k[A]x[B]y

� Rate is the rate of reaction

� k is the rate constant

� [A] and [B] are reactant concentrations

� x and y are orders of reactants (NOT necessarily the same

as coefficients).

16

Rate Laws� Rate Laws have to be determined from experimental

data (concentrations and reaction rates) using initial

rates (at time = 0)

� Br2(aq) + HCOOH(aq) � 2Br - (aq) + 2H+ (aq) + CO2(g)

� rate = k[Br2]1[HCOOH]0 = rate= k[Br2]

� Examples of balanced equations and rate laws:

17

Reaction Equation Rate Law

2 N2O5 (g) � 4 NO2 (g) + O2 (g) rate = k[N2O5]

2 H2O2 (aq) + I- (aq) � 2 H2O (l) + O2 (g) rate = k[H2O2][I-]

H2 (g) + I2 (g) � 2 HI (g) rate = k[H2][I2]

Determining a Rate Law� 5Br - (aq) + BrO3

- (aq) + 6H+ (aq) � 3Br2(aq) + 3H2O(l)

� Need two trials where only one reactant concentration changes

(all others are constant).

� Compare change in rates to change in concentration to determine

the order of the reactant (the exponent in the rate law).18

Trial [Br-](M) [BrO3-](M) [H+](M) Initial Rate (M/s)

1 0.10 0.010 0.15 1.2 x 10–3

2 0.10 0.020 0.15 4.8 x 10–3

3 0.30 0.010 0.15 3.6 x 10–3

4 0.10 0.020 0.30 4.8 x 10–3

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1 0.10 0.010 0.15 1.2 x 10–3

2 0.10 0.020 0.15 4.8 x 10–3

3 0.30 0.010 0.15 3.6 x 10–3

4 0.10 0.020 0.30 4.8 x 10–3

5Br - (aq) + BrO3- (aq) + 6H+ (aq) � 3Br2(aq) + 3H2O(l)

[Br-]

triples

[BrO3-]

constant

Rate

triples

[Br-] triples, rate triples: Br- is 1st order (x = 1)

Which two trials should be used for BrO3-?

Which two for H+?

19

Rate Laws

[H+]

constant

Rate Laws

rate = k[Br-][BrO3-]2

Now we can solve for the rate constant. How? What

information do we need to solve for k?

Rate = 120 M-2s-1 [Br-][BrO3-]2

What is the rate of reaction for trial 5? 20


1 0.10 0.010 0.15 1.2 x 10–3

2 0.10 0.020 0.15 4.8 x 10–3

3 0.30 0.010 0.15 3.6 x 10–3

4 0.10 0.020 0.30 4.8 x 10–3

5 0.20 0.015 0.20 ?

Rate LawsFour important things to remember about rate laws:

1) A rate law is an equation that is always defined in

terms of reactant concentrations. Never products.

2) The exponents (orders of reactants) in a rate law must

be determined from a table of experimental data.

3) Comparing changes in individual reactant

concentrations to changes in rate allows you to find

the order of the reactant.

4) We can use the complete rate law (with rate constant) to

calculate rate for any combination of concentrations. 21

Shortcut for determining orders:

� If the concentration doubles and the rate doesn’t

change, then the concentration has no effect on rate

and the reactant is 0 order

� If the concentration doubles and the rate doubles, then

the reactant is 1st order.

� What would happen to rate if the concentration of a

1st order reactant tripled?

� If the concentration doubles and rate quadruples, then

the reactant is 2nd order.

� What would happen to rate if the concentration of a

2nd order reactant tripled? 22

Orders of reactants

What do orders of reactants mean?

� They are exponents that indicate by what factor the rate of

reaction will change when concentration changes.

� For a zero order reactant, raising the ratio of concentrations

to the 0th power results in no change for rate.

� Raising the ratio of concentrations to the 0th power equals 1.

� For a first order reactant, the change in concentration

(double, triple, etc.) is the same as the change in rate.

� Raising the ratio of concentrations to the 1st power results in the

same ratio for rate.

� For a second order reactant, raising the ratio of

concentrations to the 2nd power is the same as squaring.

� If a reactant is 2nd order and its concentration is tripled,

how will the rate change? 23

Orders of reactants Rate Laws – Practice

� S2O82–(aq) + 3I–(aq) → 2SO4

2–(aq) + I3–(aq)

� Determine the rate law and calculate the rate constant,

including its units.

rate = 0.081 M-1s-1 [S2O82–] [I–]

24

Trial [S2O82–](M) [I–](M) Initial Rate (M/s)

1 0.080 0.034 2.2 x 10–4

2 0.080 0.017 1.1 x 10–4

3 0.16 0.017 2.2 x 10–4

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Rate Laws – Practice

Determine the rate law and calculate the rate constant (with

units): 2NO + Cl2 � 2NOCl. Calculate the rate of reaction

when [NO] and [Cl2] = 0.15 M.

rate = 0.17 M-2s-1 [NO] [Cl2]2; rate = 5.7x10-4 M/s

Trial [NO] [Cl2] Rate (M/s)

1 0.10 M 0.10 M 1.7x10-4

2 0.10 M 0.20 M 6.7x10-4

3 0.10 M 0.30 M 1.5x10-3

4 0.20 M 0.30 M 3.0x10-3

5 0.30 M 0.30 M 4.5x10-3

Rate Laws – Practice

� Determine the rate law and calculate the rate constant for: 2O3(g) � 3O2(g)

� [O3] (M) Rate (M/s)

0.00600 5.03 x 10-7

0.00300 1.28 x 10-7

0.00150 3.08 x 10-8

� What is the rate if [O3] = 0.00500 M?

� Hint: What other information is needed to solve this?

� Rate = 0.0140 M-1s-1 [O3]2; rate = 3.49 x 10-7 M/s

Examples 12.3 – 12.526

12.4 Integrated Rate Laws

� Concentration/Time relationship:

� 1st order reaction ( A � B ),

� Rate also can be written, rate = k[A]

� Setting them equal:

� Integrate to get 1st order integrated rate law:

27

� First order integrated rate law:

� ln is natural logarithm

� Linear format:

�

� y = mx + b

� Math note:

� ln(5/7) = ln 5 – ln 7

� ln [A]t = 0.23

� [A]t = 1.328

Integrated Rate Laws

IRL - First Order

� Data from slide #8: 2 H2O2(aq) � 2 H2O (l) + O2 (g)

� [H2O2] and ln [H2O2]

29

y = -0.0375x + 0.8375R² = 0.871

y = -0.1155xR² = 1

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

0 5 10 15 20 25 30

[H2O

2]

Time (hr)

[H2O] vs Time

Half-life: the time it

takes for the amount of

reactant to decrease by

one half (50%). For 1st

order reactions, each

half-life is the same

length.

ln (50/100) = -kt1/2

30

First Order Half-Life

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IRL - Second Order

� Two options for rates of second order reactions:

� Rate = k[A]2 OR Rate = k[A][B]

� 2nd option is too complex, we’ll only work with the first

example.

� Rate = k[A]2 = -∆[A]/∆t

� Integrate: 1111�A��A��A��A�tttt

= kt + 1111�A��A��A��A��

(linear eqn)

31

IRL - Second Order

32

0

0.5

1

1.5

2

2.5

3

3.5

4

0 5 10 15 20 25 30

Concentr

ation [

A]

Time (s)

Concentration Functions vs Time

[A]

1/[A]

Linear ([A])

Linear (1/[A])

33

Second Order Half-Life


takes for a reactant

to decrease by half

(50%). For 2nd order

reactions, the length

of each half life

doubles in length.

y = 0.1078x + 0.5201R² = 0.9994

0

0.5

1

1.5

2

2.5

3

3.5

4

0 5 10 15 20 25 30

Concentr

ation [

A]

Time (s)


[A]

1/[A]

Log. ([A])

Log. ([A])

Linear (1/[A])

34

Second Order Half-Life


takes for a reactant

to decrease by half

(50%). For 2nd order

reactions, the length

of each half life

doubles in length.

y = 0.1078x + 0.5201R² = 0.9994

0

0.5

1

1.5

2

2.5

3

3.5

4

0 5 10 15 20 25 30

Concentr

ation [

A]

Time (s)


[A]

1/[A]

Half-life

Linear (1/[A])

� Integrated Rate Laws can also be used to graph

functions of concentrations over time. The function

that gives the most linear graph determines the order

of the reactant (the other functions will be curved).

� [A] vs. time (zero order)

� ln [A] vs time (first order)

� 1/[A] vs. time (second order)

Consider the reaction 2NO2(g) � 2 NO(g) + O2(g).

Time and concentration are plotted to find the most

linear plot.35

IRL’s– Graphing

Consider the reaction 2NO2(g) � 2 NO(g) + O2(g)

36

IRL’s– Graphing

Time (s) [NO2], M ln [NO2] 1/[NO2], M-1

0 1.00x10-2 -4.605 100

60 6.83x10-3 -4.986 146

120 5.18x10-3 -5.263 193

180 4.18x10-3 -5.477 239

240 3.50x10-3 -5.655 286

300 3.01x10-3 -5.806 332

360 2.64x10-3 -5.937 379

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� Plot: [A] vs time (zero order):

37

IRL’s– Graphing

0.00E+00

1.00E-03

2.00E-03

3.00E-03

4.00E-03

5.00E-03

6.00E-03

7.00E-03

8.00E-03

9.00E-03

1.00E-02

0 50 100 150 200 250 300 350 400

[NO2] versus time

� Plot: ln [A] vs time (first order):

38

IRL’s– Graphing

-6.5

-6

-5.5

-5

-4.5

-4

0 50 100 150 200 250 300 350 400

ln [NO2] versus time

� Plot: 1/[A] vs time (second order):

39

IRL’s– Graphing

0

50

100

150

200

250

300

350

400

0 50 100 150 200 250 300 350 400

1/[NO2] versus time

40Image from Principles of General Chemistry http://2012books.lardbucket.org/books/principles-of-general-chemistry


IRL – Practice

� The half life for the reaction 2A → B is 8.06

minutes at 110oC. The reaction is first order in A.

How long will it take for A to decrease from 1.25

M to 0.71 M?

� k = 8.598x10-2 min-1; t = 6.6 minutes41

IRL – Practice � If the half life is 18.2 min for a first order reactant with

initial concentration of 2.4 M, what is the concentration of

A after 7.5 minutes?

� k = 0.0381 min-1, [A]t = 1.8 M

� If a first order reaction has a rate constant of 4.73 x 102 s-1,

how long would it take for 75% to react?

� t = 2.9 x 10-3 s

Examples 12.6 – 12.10 42

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IRL – Practice

� A second order reaction has a half life of 699 seconds. If the initial reaction concentration is 0.0355 M, what will the reactant concentration be after 855 seconds?

� k = 0.040299 M-1s-1, [A]t = 0.0160 M 43

12.5 Collision Theory

� Most reactions occur as a result of collisions

between reacting molecules.

� Collision theory: rate of reaction is directly

proportional to the number of molecule collisions

per time.

� As temperature and concentration increase, more

collisions occur and therefore rates of reaction

increase.

44

Collision Theory

� Effective Collisions: molecules must collide with

sufficient kinetic energy and in the correct

orientation in order to make products.

2 CO (g) + O2 (g) �� 2 CO2 (g)

45 46

Collision Theory

NO + O3 � NO2 + O2



Activation Energy� Activation Energy (Ea): energy barrier that

molecules have to overcome in order to react. The

high energy state is the transition state.

A + B �� C + D

47

� Only a small fraction of molecules have enough

kinetic energy to overcome the activation energy.

� Ea is different for each reaction.

� Ea = T.S. – R (for forward reaction)

� Ea = T.S. – P (for reverse reaction)

� Reactions with low Ea are faster because more

molecules can overcome the Ea.

� At higher temperatures, a larger fraction of molecules

have enough kinetic energy to overcome Ea. 48

Activation Energy

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� Enthalpy (∆H) is heat of reaction:

� Endothermic: Energy absorbed, ∆H > 0

� Exothermic: Energy released, ∆H < 0

49

Reaction Energy Profile



Reaction Energy Profile� Is the reaction shown on the energy profile below

endo- or exothermic?

� What is the value of Ea? What is ∆H?

Example 12.13 50

Reaction Energy Profile� Draw an energy profile diagram that has an enthalpy

value of +15 kJ and an Activation Energy = +40 kJ.

Example 12.13 51

0

10

20

30

40

50

60

Po

ten

tia

l E

ne

rgy (

kJ

)

reaction progress

Arrhenius Equation

� The Arrhenius Equation can be used to relate the

temperature of a reaction to the rate constant.

� Original form: � ��

�

� k: rate constant

� A: frequency factor (we will not use or calculate this

value)

� Ea: activation energy (kJ/mol)

� R: gas constant (8.314 J/mol·K)

� T: temperature (Kelvin)

� Rate constant changes with temperature;

Temperature increases, rate increases, k increases 52

Arrhenius Equation

� The initial equation can be modified to give a linear

variation.

� This variation works well for sets of 4 or more data

points.

y = m x + b

53

Arrhenius Equation• Worked Example 12.11: Rate constants for the

reaction 2HI(g) → 2H2(g) + I2(g) were measured at

five different temperatures. The data are shown in

the table below. Determine the activation energy for

this reaction. (This will be used in Kinetics lab!)

54

T (K) k (M-1·s-1) 1/T (1/K) ln k

555 3.52x10-7 0.001802 -14.86

575 1.22x10-6 0.001739 -13.62

645 8.59x10-5 0.001550 -9.362

700 1.16x10-3 0.001429 -6.759

781 3.95x10-2 0.001280 -3.231

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m = -Ea / R � Ea = m x -R = -22287 K x -8.314 J / K·mol

Ea = 185,294 J/mol = 185 kJ/mol (3 s.f. because of T and k) 55

Arrhenius Equation: Example 12.11

y = -22287x + 25.204

-16

-14

-12

-10

-8

-6

-4

-2

0.001 0.0011 0.0012 0.0013 0.0014 0.0015 0.0016 0.0017 0.0018 0.0019

ln k

1/T (1/K)

Arrhenius Plot

Arrhenius Equation

� The Arrhenius Equation can also be used to

determine:

� 1) the rate constant after a change in temperature

(i.e., 2 data points) or

� 2) Ea given two k’s and T’s:

56

� Using the activation energy just calculated (Ea =

185.249 kJ/mol) for the system 2HI(g) → 2H2(g) +

I2(g), determine the rate constant at 325oC, using

k1 = 3.52x10-7 M-1s-1 and T1 = 555 K.

� Convert Ea to J/mol (or R to kJ/mol)

� Convert T to Kelvin

� k2 = 6.32x10-6 M-1s-1

57

Arrhenius Equation – Practice 12.6 Reaction Mechanisms

Reaction Mechanism: the sequence of reaction steps

that describes the pathway from reactants to products.

A balanced chemical equation does not indicate how a

reaction actually takes place.

A chemical reaction takes place through a series of

simpler steps called elementary steps (typically 2 or

more). These elementary steps determine the step-

wise reaction mechanism for the overall reaction.

58

Given elementary steps in a mechanism, we can:

� Determine the molecularity and rate law for each step

� Write the overall (or net) equation

� Use relative rates (slow or fast) to identify the rate-

determining step

� Identify the intermediate(s)

� Is produced and then consumed in a later step

� Identify the catalyst (if it exists)

� Is consumed and then produced in a later step (recycled)

59

Reaction Mechanisms Reaction Mechanisms

Each step in a mechanism is defined by its molecularity: the

number of reactant molecules. Rate laws for elementary steps

are written based on the molecularity.

unimolecular (one reactant molecule)

O3 � O2 + O rate = k[O3] 1st order

bimolecular (two reactant molecules)

NO2 + CO � NO + CO2 rate = k[NO2][CO] 2nd order

2 HI � H2 + I2 rate = k[HI]2 2nd order

termolecular (three reactant molecules) – rare!

2 NO + Cl2 � 2 NOCl rate = k[NO]2[Cl] 3rd order

60

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� One step in a multistep mechanism will be slower than the others. The slow step is the rate-determining step (RDS).

� Step 1 (slow): 2 NO2 � NO3 + NO

� Step 2 (fast): NO3 + CO � NO2 + CO2

� 1) Write the overall equation.

� 2) What is the molecularity of each step?

� 3) Which step is the rate-determining step?

� 4) What is the intermediate? The catalyst?

� 5) Write the rate law for each step.61

Multistep Mechanisms

� If the first step is slow (RDS), the rate law of the first step is the rate law of the overall equation.

� If the rate law for the overall equation does NOT match the rate law for the first step, the first step is NOT the slow step (RDS).

� If the second (or third or fourth or…) step is slow, the rate law of the overall equation is more difficult to determine.

62

Rate Laws: Elementary Steps

� If the 1st step is slow, the rate law of the overall reaction is the same as the rate law of the first step. If the 2nd step is slow, rate laws are more complicated to determine.

� Step 1: Tl3+ + Fe2+ � Tl2+ + Fe3+

� Step 2: Tl2+ + Fe2+ � Tl+ + Fe3+

� Write the overall equation. Write the rate law for each step.

� If the overall rate law is: , which step is RDS?

� Second63

Rate Laws: Mechanisms

� Group Work:

� Step 1 (slow): S2O82- + I- � 2SO4

2- + I+

Step 2 (fast): I+ + I- � I2

� Which step is rate determining? � Step 1

� What is the intermediate? Catalyst?� Intermediate: I+, Catalyst: none

� What is the rate law for each step? What is the overall rate law?� rate = k[S2O8

2-][I-]; rate = k[I+][I-]

� overall: rate = k[S2O82-][I-]

64

Mechanisms – Practice

� Step 1: 2 NO � N2O2

� Step 2: N2O2 + H2 � N2O + H2O

� Step 3: N2O + H2O � N2 + H2O

� Determine: overall equation, molecularity of each step, intermediate, and catalyst.

� Overall: 2 NO + H2 �� N2 + H2O; all 3 bimolecular; Intermediates: N2O2, H2O, N2O; Catalyst: N/A

� Write the rate law for each step.

� If the overall rate law is rate = k[NO]2[H2], which step is

the rate-determining step? Not the first step65

Mechanisms – Practice

Rate = k[NO]2

Rate = k[N2O2][H2]

Rate = k[N2O][H2O]

66

Step 1:

Step 2:

H2O2 + I–

H2O2 + IO–

H2O + IO–

H2O + O2 + I–

Overall reaction: 2H2O2 2H2O + O2

Which step is slow?

Rate = k[H2O2][I–]

Rate Laws: Mechanisms

(slow)

5/14/2018

12

12.7 Catalysis

67

A catalyst is a substance that increases the rate of a

chemical reaction without itself being consumed.

A catalyst speeds up a reaction by providing a set of

elementary steps with more favorable kinetics than those

that exist in its absence; it usually does this by lowering

the activation energy.

Step 1: (slow)

Step 2:

H2O2 + I–

H2O2 + IO–

H2O + IO–

H2O + O2 + I–

Overall reaction: 2H2O2 2H2O + O2

Catalysis

� Energy profiles with catalysts.

� Figures 12.21 and 12.22.

68

Catalysts� Catalysts can be classified as heterogeneous: the

catalyst and reactants are in different phases (Figure 12.26)

69

There are four steps in the catalysis of the reaction C2H4 + H2 ⟶ C2H6 by nickel.

1) Hydrogen is adsorbed on the surface, breaking H–H bonds and forming Ni–H bonds.

2) Ethylene is adsorbed on the surface, breaking the π-bond and forming Ni–C bonds.

3) Atoms diffuse across the surface and form new C–H bonds when they collide.

4) C2H6 molecules escape from the nickel surface, since they are not strongly attracted to nickel.

� Practical Application: Catalytic Converter

70

Catalysts

A catalytic converter allows for the combustion of all carbon-

containing compounds to carbon dioxide, while at the same time

reducing the output of nitrogen oxide and other pollutants in

emissions from gasoline-burning engines.

Catalysts� Catalysts can be classified as homogeneous: the

catalyst and reactants are in the same phase.

� The advantages of homogeneous catalysts:

� Reactions can be carried out under atmospheric

conditions

� Can be designed to function selectively

� Are generally cheaper

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Catalysts� Enzymes: large protein molecules that act as

catalysts for biological reactions. Digestive

enzymes allow us to process starch (e.g., potatoes)

but not cellulose (e.g., grass).

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reaction rates 12.1 reaction rates - welcome to...

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