chemical kinetics chapter 14 chemistry: the molecular nature of matter, 6 th edition

Chemical Kinetics

CHAPTER 14

Chemistry: The Molecular Nature of Matter, 6th editionBy Jesperson, Brady, & Hyslop

2

CHAPTER 14 Chemical Kinetics

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Learning Objectives: Factors Affecting Reaction Rate:

o Concentrationo Stateo Surface Areao Temperatureo Catalyst

Collision Theory of Reactions and Effective Collisions Determining Reaction Order and Rate Law from Data Integrated Rate Laws Rate Law Concentration vs Rate Integrated Rate Law Concentration vs Time Units of Rate Constant and Overall Reaction Order Half Life vs Rate Constant (1st Order) Arrhenius Equation Mechanisms and Rate Laws Catalysts

3



Lecture Road Map:

① Factors that affect reaction rates

② Measuring rates of reactions

③ Rate Laws

④ Collision Theory

⑤ Transition State Theory & Activation Energies

⑥ Mechanisms

⑦ Catalysts

4Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E

Factors that Affect Reaction

Rates


Kinetics: Study of factors that govern oHow rapidly reactions occur and oHow reactants change into products

Rate of Reaction: o Speed with which reaction occursoHow quickly reactants disappear and

products form

Kinetics The Speed at Which Reactions Occur

Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 5

Kinetics The Speed at Which Reactions Occur


AB

Reaction rate is measured by the amount of product produced or reactants consumed per unit time.

o [B] concentration of products will increase over time

o [A] concentration of reactants will decrease over time


7

Kinetics Factors Affecting Reaction Rates

1. Chemical nature of reactantso What elements, compounds, salts

are involved?o What bonds must be formed,

broken?o What are fundamental differences in

chemical reactivity?


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2. Ability of reactants to come in contact(Reactants must meet in order to react)

The gas or solution phase facilitates thiso Reactants mix and collide with each other

easilyo Homogeneous reaction

oAll reactants in same phaseoOccurs rapidly

o Heterogeneous reactionoReactants in different phasesoReactants meet only at interface between

phasesoSurface area determines reaction rate o Increase area, increase rate; decrease area,

decrease rate


9


3. Concentrations of reactantso Rates of both homogeneous and

heterogeneous reactions affected by [X ]

o Collision rate between A and B increase if we increase [A] or increase [B ].o Often (but not always) reaction rate

increases as [X ] increases


10


4. Temperatureo Rates are often very sensitive to temperature

o Raising temperature usually makes reaction faster for two reasons:o Faster molecules collide more often and

collisions have more energyo Most reactions, even exothermic reactions,

require energy to occur

o Rule of thumb: Rate doubles if temperature increases by 10 °C (10 K)


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5. Presence of Catalystso Substances that increase rates of chemical

reactions without being used up

o Rate-accelerating agents

o Speed up rate dramaticallyoRate enhancements of 106 not uncommon

o Chemicals that participate in mechanism but are regenerated at the end


Measuring Reaction Rates



13

Rates Measuring Rate of Reaction

o Rate = ratio with time unit in denominator

o Rate of Chemical Reactiono Change in concentration per unit time.

o Always with respect to a given reactant or producto [reactants] decrease with timeo [products] increase with time


14


o Concentration in M unitso Time in s unitso Units on rate:

o [product] increases by 0.50 mol/L per second rate = 0.50 M/s

o [reactant] decreases by 0.20 mol/L per second rate = 0.20 M/s


15

Rates Rate of Reaction

Always positive whether something is increasing or decreasing in [X ]

o Reactants o Reactant consumedo So [X ] is negativeo Need minus sign to make rate positive

o Products o Produced as reaction goes alongo So [X ] is positiveo Thus rate already positive


16


Coefficients indicate the relative rates at which reactants are consumed and products are formed

o Related by coefficients in balanced chemical equationo Know rate with respect to one product or reactanto Can use equation to determine rates with respect to all

other products and reactants.

A + B C + D


17

Rates Rate of Reaction: Example

• O2 reacts 5 times as fast as C3H8

• CO2 forms 3 times faster than C3H8 consumed

• H2O forms 4/5 as fast as O2 consumed

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)


18

GroupProblem

Clorox bleach is sodium hypochlorite. It should never be mixed with acids, (like vinegar) because it forms chlorine gas:

NaClO + 2 HCl → Cl2 + H2O + NaCl

If Chlorine gas (Cl2) is formed at a rate of 5.0 x 10-4 mol/Ls what rate is HCl consumed?

HCl: Cl2 2:1

Therefore HCl will disappear twice as fast as Cl2 is formed.Rate HCl consumed = 10. x 10-4 mol/Ls


19

Rates Change of Reaction Rate with Time

Generally reaction rate changes during reaction, it isn’t constant

o Often initially fast when lots of reactant present

o Slower and slower as reactants are depletedWhy?

o Rate depends on the concentration of the reactants

o Reactants being used up, so the concentration of the reactants are decreasing and therefore the rate decreases

Measured in 3 ways: o instantaneous rate, average rate, initial rate


20

Rates Instantaneous & Initial Reaction Rate

Instantaneous rateo Slope of tangent to curve at some specific time

Initial rateo Determined at time = 0

NO2 appearance

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0 200 400 600 800

Time (s)

[NO

2]


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Rates Average Rate of Reaction

Average Rate: Slope of line connecting starting and ending coordinates for specified time frame

rateΔtimeΔ[Product]

NO2 appearance

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0 200 400 600 800

Time (s)

[NO

2]

22

Rates Example Reporting Different Types of Rates

Rate at any time t = negative slope (or tangent line) of curve at that point

Concentration vs. Time Curve for 0.005M phenolphthalein reacting with 0.61 M NaOH at room temperature

http://chemed.chem.purdue.edu


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[P] (mol/L) Time (s)

0.005 00.0045 10.50.004 22.3

0.0035 35.70.003 51.1

0.0025 69.30.002 91.6

0.0015 120.4

Initial rate = Average rate between first two data points

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Instantaneous Rate at 120.4 s

(90,0.0028)

(160,0.0018)

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Average Rate between 0 and 120.4 s


26

GroupProblem

A reaction was of NO2 decomposition was studied. The concentration of NO2 was found to be 0.0258 M at 5 minutes and at 10 minutes the concentration was 0.0097 M. What is the average rate of the reaction between 5 min and 10 min?A. 310 M/minB. 3.2 × 10–3 M/minC. 2.7 × 10–3 M/minD. 7.1 × 10–3 M/min


Rate Laws



28

Rate Laws Rates Based on All Reactants

A + B C + D

o Rate Law or Rate expressiono k is the rate constant

o Dependent on Temperature & Solvento m and n = exponents found experimentallyo No necessary connection between stoichiometric

coefficients (, ) and rate exponents (m, n)o Usually small integerso Sometimes simple fractions (½, ¾) or zero

= k[A]m[B]n


29

Rate Laws Rates Based on All Reactants

Below is the rate law for the reaction 2A +B → 3C

rate= 0.045 M–1s–1 [A][B]If the concentration of A is 0.2 M and that of B is 0.3 M, and the reaction is 1st order (m & n = 1) what will be the reaction rate?

rate=0.0027 M/s 0.003 M/s

rate=0.045 M–1 s–1 [0.2][0.3]


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Rate Laws Order of Reactions

Rate = k[A]m[B]n

Exponents specify the order of reaction with respect to each reactant

Order of Reactiono m = 1 [A]1 1st order in [A]o m = 2 [A]2 2nd order in [A]o m = 3 [A]3 3rd order in [A]o m = 0 [A]0 0th order in [A]

[A]0 = 1 means A doesn't affect rate

Overall order of reaction = sum of orders (m and n) of each reactant in rate law


31

Rate Laws Order of Reactions: Example

5Br– + BrO3– + 6H+ 3Br2 + 3H2O

x = 1 y = 1 z = 2o1st order in [BrO3

–]o1st order in [Br –] o2nd order in [H+]oOverall order = 1 + 1 + 2 = 4


32

Rate Laws Order of Reaction & Units for k


33

GroupProblem

The following rate law has been observed:Rate = k[H2SeO][I–]3[H+]2. The rate withrespect to I– and the overall reaction rate is:A. 6, 2B. 2, 3C. 1, 6D. 3, 6


34

Rate Laws Calculating k

If we know rate and concentrations, can use rate law to calculate k

From Text Example of decomposition of HI at 508 °C

• Rate= 2.5 × 10–4 M/s• [HI] = 0.0558 M


35

Rate Laws Determining Exponents in Rate Law

Experimental Determination of Exponentso Method of initial rateso If reaction is sufficiently slow

o or have very fast technique o Can measure [A] vs. time at very beginning of reaction

o Before it slows very much, then

o Set up series of experiments, where initial concentrations vary


36

Rate Laws Determining Rate Law Exponents: Example

3A + 2B productsRate = k[A]m[B]n

Expt. #

[A]0, M [B]0, M Initial Rate, M/s

1 0.10 0.10 1.2 10–4 2 0.20 0.10 4.8 10–4 3 0.20 0.20 4.8 10–4

Convenient to set up experiments so The concentration of one species is doubled or tripled And the concentration of all other species are held constant

Tells us effect of [varied species] on initial rate


37

Rate Laws Determining Rate Law Exponents

• If reaction is 1st order in [X],– Doubling [X]1 21

– Doubles the rate• If reaction is 2nd order in [X],

– Doubling [X]2 22

– Quadruples the rate• If reaction is 0th order in [X],

– Doubling [X]0 20

– Rate doesn't change• If reaction is nth order in [X]

– Doubling [X]n 2n times the initial rate


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Expt. #


1 0.10 0.10 1.2 104 2 0.20 0.10 4.8 104 3 0.20 0.20 4.8 104Comparing Expt. 1 and 2

o Doubling [A]o Quadruples rateo Reaction 2nd order in A = [A]2

4102.1108.4

1 Rate2 Rate

4

4

2m = 4 or m = 2


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Expt. # [A]0, M [B]0, M Initial Rate, M/s1 0.10 0.10 1.2 104 2 0.20 0.10 4.8 104 3 0.20 0.20 4.8 104

Comparing Expt. 2 and 3o Doubling [B]o Rate does not changeo Reaction 0th order in B = [B]0 = 1

1108.4108.4

2 Rate3 Rate

4

4

2n = 1 or n = 0



o Conclusion: rate = k[A]2

o Can use data from any experiment to determine k o Let’s choose first experiment

Expt. #


1 0.10 0.10 1.2 10–4 2 0.20 0.10 4.8 10–4 3 0.20 0.20 4.8 10–4


41

Rate Laws Determining Rate Law Exponents: Ex 2

2 SO2 + O2 2 SO3 Rate = k[SO2]m[O2]n

Expt #

[SO2] M [O2] M Initial Rate of SO3 Formation, M s–1

1 0.25 0.30 2.5 103

2 0.50 0.30 1.0 102

3 0.75 0.60 4.5 102

4 0.50 0.90 3.0 102


42


4 = 2m or m = 2


43


3 = 3n or n = 1


44


Rate = k[SO2]2[O2]1

• 1st order in [O2]• 2nd order in [SO2]• 3rd order overall• Can use any experiment to find k


45

GroupProblem

Using the following experimental data, determine the order with respect to NO and O2 .

A. 2, 0B. 3,1C. 2, 1D. 1, 1

Expt #

[NO] M

[O2] M

Initial Rate M s–1

1 0.12 0.25 1.5 10–3

2 0.24 0.25 6.0 10–3

3 0.24 0.50 1.2 10–2


46

GroupProblem

Example : Method of Initial RatesBrO3

– + 5Br– + 6H+ 3Br2 + 3H2O

Expt #

[BrO3–]

mol/L[Br–] mol/L

[H+] mol/L

Initial Rate mol/(L s)

1 0.10 0.10 0.10 8.0 10–4

2 0.20 0.10 0.10 1.6 10–3

3 0.20 0.20 0.10 3.2 10–3

4 0.10 0.10 0.20 3.2 10–3

47

Ex.: Method of Initial Rates

48

Compare 2 and 3

Compare 1 and 2

Ex.: Method of Initial Rates

• First order in [BrO3–] and [Br–]

• Second order in [H+] • Overall order = m + n + p = 1 + 1 + 2

= 4• Rate Law is: Rate = k[BrO3

–][Br–][H+]2

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Compare 1 and 4


Integrated Rate Laws



51

Integrated Rate Laws Concentration & Time

Rate law tells us how speed of reaction varies with concentrations.

Sometimes want to know o Concentrations of reactants and products at

given time during reactiono How long for the concentration of reactants to

drop below some minimum optimal value

Need dependence of rate on time


52

Integrated Rate Laws First Order Integrated Rate Law

• Corresponding to reactions – A products

• Integrating we get

• Rearranging gives

• Equation of line y = mx + b


53

Integrated Rate Laws First Order Integrated Rate Law

Yields straight lineo Indicative of first order

kineticso Slope = –ko Intercept = ln [A]0

o If we don't know already

0]ln[]ln[ AktA t Slope = –k


54

Integrated Rate Laws 2nd Order Integrated Rate Law

• Corresponding to special second order reaction – 2B products

• Integrating we get

• Rearranging gives

• Equation of line y = mx + b


55

Integrated Rate Laws 2nd Order Integrated Rate Law

Yields straight lineo Indicative of 2nd order

kineticso Slope = +ko Intercept = 1/[B]0

0][1

][1

Bkt

B t

Slope= +k


56

Integrated Rate Laws Graphically determining Order

Make two plots:1. ln [A] vs. time2. 1/[A] vs. time

o If ln [A] is linear and 1/[A] is curved, then reaction is 1st order in [A]

o If 1/[A] plot is linear and ln [A] is curved, then reaction is 2nd order in [A]

o If both plots give horizontal lines, then 0th order in [A]


57



58


Add examples of graphs


59

Time, min [SO2Cl2], M ln[SO2Cl2] 1/[SO2Cl2] (L/mol)

0 0.1000 -2.3026 10.000100 0.0876 -2.4350 11.416200 0.0768 -2.5666 13.021300 0.0673 -2.6986 14.859400 0.0590 -2.8302 16.949500 0.0517 -2.9623 19.342600 0.0453 -3.0944 22.075700 0.0397 -3.2264 25.189800 0.0348 -3.3581 28.736900 0.0305 -3.4900 32.787

1000 0.0267 -3.6231 37.4531100 0.0234 -3.7550 42.735

Example: SO2Cl2 SO2 + Cl2Integrated Rate Laws


60

First Order Plot for SO2Cl2 Decomposition

-3.8

-3.6

-3.4

-3.2

-3.0

-2.8

-2.6

-2.4

-2.2

0 200 400 600 800 1000 1200time (min)

ln[S

O2C

l 2]

Second order plot for SO2Cl2

Decomposition

10

15

20

25

30

35

40

45

0 200 400 600 800 1000 1200time (min)

1/[S

O2C

l 2]

(L/m

ol)

Reaction is 1st order in SO2Cl2

Example: SO2Cl2 SO2 + Cl2Integrated Rate Laws


61

Example: HI(g) H2(g) + I2(g)Integrated Rate Laws

Time(s)

[HI] (mol/L) ln[HI] 1/[HI]

(L/mol)0 0.1000 -2.3026 10.000

50 0.0716 -2.6367 13.9665

100 0.0558 -2.8860 17.9211

150 0.0457 -3.0857 21.8818

200 0.0387 -3.2519 25.840

250 0.0336 -3.3932 29.7619

300 0.0296 -3.5200 33.7838

350 0.0265 -3.6306 37.7358


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Example: HI(g) H2(g) + I2(g)Integrated Rate Laws

First Order Plot for HI Decomposition at 508 oC

-3.8

-3.6

-3.4

-3.2

-3.0

-2.8

-2.6

-2.4

-2.2

0 50 100 150 200 250 300 350time (s)

ln[H

I]

Second order plot for HI Decomposition at 508 oC

10

15

20

25

30

35

40

0 50 100 150 200 250 300 350time (s)

1/[H

I] (L

/mol

)

Reaction is second order in HI.


63

GroupProblem

A plot for a zeroth order reaction is shown. What is the proper label for the y-axis in the plot ?A. ConcentrationB. ln of ConcentrationC. 1/ConcentrationD. 1/ ln Concentration

0 200 400 600 800 1000 1200time (min)

Zeroth Order Plot


Half Life (t1/2) for first order reactionsIntegrated Rate Laws

Half-life = t½ We often use the half life to describe how fast a reaction takes place First Order Reactions

o Set

o Substituting into

o Gives

o Canceling gives ln 2 = kt½

o Rearranging gives


65

Half Life (t1/2) for First Order ReactionsIntegrated Rate Laws

Observe: 1. t½ is independent of [A]o

o For given reaction (and T)o Takes same time for concentration to fall from

o2 M to 1 M as from o5.0 10–3 M to 2.5 10–3 M

2. k1 has units (time)–1, so t½ has units (time)o t½ called half-life

oTime for ½ of sample to decay


66

Half Life (t1/2)Integrated Rate Laws

Does this mean that all of sample is gone in two half-lives (2 × t½)?No!

o In 1st t½, it goes to ½[A]o o In 2nd t½, it goes to ½(½[A]o) = ¼[A]o o In 3rd t½, it goes to ½(¼[A]o) = ⅛[A]o

o In nth t½, it goes to [A]o/2n


67

Half Life (t1/2)Integrated Rate Laws


68

Half Life (t1/2): First Order ExampleIntegrated Rate Laws

131I is used as a metabolic tracer in hospitals. It has a half-life, t½ = 8.07 days. How long before the activity falls to 1% of the initial value?


69

GroupProblem

The radioactive decay of a new atom occurs so that after 21 days, the original amount is reduced to 33%. What is the rate constant for the reaction in s–1?

k = 6.11 × 10–7 s–1

k = 0.0528 day–1


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GroupProblem

The half-life of I-132 is 2.295 h. What percentage remains after 24 hours?

0.302 h–1 = k

A = 0.0711% Ao

ktAAo

ln

248.7 eAeAA okt

o


71

Half Life (t1/2): Carbon-14 DatingIntegrated Rate Laws


72

Half Life (t1/2): Second Order ReactionsIntegrated Rate Laws

How long before [A] = ½[A]o?

o t½, depends on [A]o

o t½, not useful quantity for a second order reaction


73

GroupProblem

The rate constant for the second order reaction 2A → B is 5.3 × 10–5 M–1 s–1. What is the original amount present if, after 2 hours, there is 0.35 M available?

A0=0.40 M


74

GroupProblem

Add better rate law problem


Collision Theory



76

Collision Theory

Reaction Rates

Collision TheoryAs the concentration of reactants increase

oThe number of collisions increaseso Reaction rate increases

As temperature increasesoMolecular speed increasesoHigher proportion of collisions with

enough force (energy) oThere are more collisions per secondoReaction rate increases


77

Collision Theory

Reaction Rates

Rate of reaction proportional to number of effective collisions/sec among reactant moleculesEffective collision

o One that gives rise to producte.g. At room temperature and pressure

o H2 and I2 molecules undergoing 1010 collisions/seco Yet reaction takes a long timeo Not all collisions lead to reaction

Only very small percentage of all collisions lead to net change


78

Collision Theory

Molecular Orientation


79

Collision Theory

Temperature

As T increaseso More molecules have Eao So more molecules undergo reaction


80

Collision Theory

Activation Energy (Ea)

Molecules must possess certain amount of kinetic energy (KE) in order to react

Activation Energy, Ea = Minimum KE needed for reaction to occur

o Get energy from collision with other moleculeso If molecules move too slowly, too little KE, they just

bounce off each othero Without this minimum amount, reaction will not occur

even when correctly oriented


81

GroupProblem

Summerize with a pretty picture


Transition State Theory



83

Transition State

Molecular Basis of Transition State Theory

KE decreasing as PE increases

KE KE

PE

KE KE

Is the combined KE of both molecules

enough to overcome Activation Energy


84

Transition State

Molecular Basis of Transition State Theory

Reaction Coordinate (progress of reaction)

Pote

ntia

l Ene

rgy

Activation energy (Ea) = hill or barrier

between reactants and products

Heat of reaction (H) = difference in PE between

products and reactants

Hreaction = Hproducts – Hreactants Products


85

Transition State

Exothermic Reactions

Reaction Coordinate (progress of reaction)

Pote

ntia

l Ene

rgy Exothermic reaction• Products lower PE than reactants

Exothermic Reaction H = –

Products


86

Transition State

Exothermic Reactions

o Hreaction < 0 (negative)o Decrease in PE of system

o Appears as increase in KEo So the temperature of the system increases

o Reaction gives off heat o Can’t say anything about Ea from size of H

o Ea could be high and reaction slow even if Hrxn large and negative

o Ea could be low and reaction rapid


87

Transition State

Endothermic Reactions

Endothermic Reaction H = +

Hreaction = Hproducts – Hreactants


88

Transition State

Endothermic Reactions

o Hreaction > 0 (positive)o Increase in PE

o Appears as decrease in KEo So temperature of the system decreases

o Have to add E to get reaction to goo Ea Hrxn as Ea includes Hrxn o If Hrxn large and positive

o Ea must be high o Reaction very slow


89

Transition State

Activated Complex

o Arrangement of atoms at top of activation barrier o Brief moment during successful collision when

o bond to be broken is partially broken and o bond to be formed is partially formed

Example

N CH3C C NH3CH3CN

CTransition State


90

Transition State

Example

Develop example that is not NO2Cl + Cl


91

GroupProblem

Draw the transition state complex, or the activated complex for the following reaction:

CH3CH2O- + H3O+ CH3CH2OH + H2O


Activation Energies



93

Ea Arrhenius Equation

The rate constant is dependent on Temperature, which allows us to calculate Activation Energy, Ea

Arrhenius Equation: Equation expressing temperature-dependence of k

oA = Frequency factor has same units as koR = gas constant in energy units

= 8.314 J mol–1 K–1

oEa = Activation Energy—has units of J/moloT = Temperature in K


94

Ea Calculating Activation Energy

• Method 1. Graphically• Take natural logarithm of both sides

• Rearranging

• Equation for a line • y = b + mxArrhenius Plot• Plot ln k (y axis) vs. 1/T (x axis) yield a

straight line• Slope = -Ea/R• Intercept = A


95

Ea Arrhenius Equation: Graphing Example

Given the following data, predict k at 75 ˚C using the graphical approach

k (M/s) T, ˚C T, K0.00088

6 25 2980.00089

4 50 3480.00090

8 100 3980.00091

8 150 448? 75 348

ln (k) = –36.025/T – 6.908

ln (k) = –36.025/(348) – 6.908 = – 7.011


96

Ea Arrhenius Equation: Graphing Example

-7.02

-7.01

-7.00

-6.99f(x) = − 36.0249500146524 x − 6.90800232199209R² = 0.999727511024561

1/T (K–1)

ln k


97

Ea Arrhenius Equation

Sometimes a graph is not neededo Only have two k s at two Ts

Here use van't Hoff Equation derived from Arrhenius equation:


98

Ea Arrhenius Equation: Ex Vant Hoff Equation

CH4 + 2 S2 CS2 + 2 H2S

k (L/mol s) T (˚C) T (K)1.1 = k1 550 823 = T1

6.4 = k2 625 898 = T2


99

GroupProblem

Given that k at 25 ˚C is 4.61 × 10–1 M/s and that at 50 ˚C it is 4.64 × 10–1 M/s, what is the activation energy for the reaction?

Ea = 208 J/mol


100

GroupProblem

A reaction has an activation energy of 40 kJ/mol. What happens to the rate if you increase the temperature from 70˚C to 80 ˚C?A. Rate increases approximately 1.5 timesB. Rate increases approximately 5000 timesC. Rate does not increaseD. Rate increases approximately 3 times


101

GroupProblem

A reaction has an activation energy of 40 kJ/mol. What happens to the rate if you increase the temperature from 70˚C to 80 ˚C?


Mechanisms of Reactions



103

Mechanisms Overall vs Individual Steps

Sometimes rate law has simple form

– N2O5 NO2 + NO3

– NO2 + NO3 N2O5

But others are complex

– H2 + Br2 2 HBr


104


Some reactions occur in a single step, as written

Others involve a sequence of stepsoReaction Mechanism

oEntire sequence of stepso Elementary Process

oEach individual step in mechanismoSingle step that occurs as written


105


o Exponents in rate law for elementary process are equal to coefficients of reactants in balanced chemical equation for that elementary process

o Rate laws for elementary processes are directly related to stoichiometry

o Number of molecules that participate in elementary process defines molecularity of step


106

Mechanisms Unimolecular Process

o Only one molecule as reactanto H3C—NC H3C—CNo Rate = k[CH3NC]

o 1st order overallo As number of molecules increases, number that rearrange

in given time interval increases proportionally


107

Mechanisms Bimolecular Process

o Elementary step with two reactantso NO(g) + O3(g) NO2(g) + O2(g)o Rate = k[NO][O3]o 2nd order overall

o From collision theory: o If [A] doubles, number of collisions between A and B will

double o If [B] doubles, number of collisions between A and B will

double o Thus, process is 1st order in A, 1st order in B, and 2nd order

overall


108

Mechanisms Termolecular Process

o Elementary reaction with three molecules o Extremely rare

o Why? o Very low probability that three molecules

will collide simultaneouslyo 3rd order overall


109

Mechanisms Elementary Processes

Molecularity

Elementary Step Rate Law

Unimolecular

A products Rate = k[A]

Bimolecular A + A products Rate = k[A]2

Bimolecular A + B products Rate = k[A][B]Significance of elementary steps: o If we know that reaction is elementary stepo Then we know its rate law


110

Mechanisms Multi-step Mechanisms

o Contains two or more steps to yield net reactiono Elementary processes in multi-step mechanism must always

add up to give chemical equation of overall processo Any mechanism we propose must be consistent with

experimentally observed rate law

o Intermediate = species which are formed in one step and used up in subsequent stepso Species which are neither reactant nor product in overall

reactiono Mechanisms may involve one or more intermediates


111

Mechanisms Example

The net reaction is:NO2(g) + CO(g) NO(g) + CO2(g)

The proposed mechanism is:NO2(g) + NO2(g) NO3(g) + NO(g) NO3(g) + CO(g) NO2(g) + CO2(g)

2NO2(g) + NO3(g) + CO(g) NO2(g) + NO3(g) + NO(g) + CO2(g) or

NO2(g) + CO(g) NO(g) + CO2(g)

1


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Mechanisms Rate Determining Step

o If process follows sequence of steps, slow step determines rate = rate determining step.

o Think of an assembly lineo Fast earlier steps may cause intermediates to pile upo Fast later steps may have to wait for slower initial steps

o Rate-determining step governs rate law for overall reaction

o Can only measure rate up to rate determining step


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Mechanisms Example: Rate Determining Step

(CH3)3CCl(aq) + OH–(aq) (CH3)3COH(aq) + Cl–(aq) chlorotrimethylmethane trimethylmethanol

o Observed rate = k[(CH3)3CCl]o If reaction was elementary

o Rate would depend on both reactantso Frequency of collisions depends on both concentrations

o Mechanism is more complex than single stepo What is mechanism?

o Evidence that it is a two step process


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Mechanisms Rate Determining Step as Initial Step

Step 1: (CH3)3CCl(aq) (CH3)3C+(aq) + Cl–(aq) (slow)Step 2: (CH3)3C+(aq) + OH–(aq) (CH3)3COH(aq) (fast)

o Two steps each at different rateso Each step in multiple step mechanism is elementary process,

soo Has its own rate constant and its own rate law

o Hence only for each step can we write rate law directlyo Observed rate law says that step 1 is very slow compared to

step 2o In this case step 1 is rate determiningo Overall rate = k1[(CH3)3CCl]


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Mechanisms Mechanisms with Fast Initial Step

1st step involves fast, reversible reactionEx. Decomposition of ozone (No catalysts)Net reaction: 2O3(g) 3O2(g)

Proposed mechanism:

O3(g) O2(g) + O(g) (fast)

O(g) + O3(g) 2O2(g) (slow)

]O[]O[ Rate Observed

2

23k


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Mechanisms Is the Mechanism Rate Law Consistent?

o Rate of formation of O2 = Rate of reaction 2 = k2[O][O3]

o But O is intermediateo Need rate law in terms of reactants and products

o and possibly catalystso Rate (forward) = kf[O3]o Rate (reverse) = kr[O2][O]o When step 1 comes to equilibrium

o Rate (forward) = Rate (reverse) o kf[O3] = kr[O2][O]


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Mechanisms Is the Mechanism Rate Law Consistent?

o Solving this for intermediate O gives:

o Substitution into rate law for step 2 gives:

o Rate of reaction 2 = k2[O][O3] = o where

o This is observed rate lawo Yes, mechanism consistent


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GroupProblem

The reaction mechanism that has been proposed for the decomposition of H2O2 is

1. H2O2 + I– → H2O + IO– (slow)2. H2O2 + IO– → H2O + O2 + I– (fast)

What is the expected rate law?

First step is slow so the rate determining step defines the rate law

rate=k [H2O2][I–]


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GroupProblem

The reaction: A + 3B → D + F was studied and the following mechanism was finally determined:

1. A + B C (fast)2. C + B → D + E (slow)3. E + B → F (very fast)

What is the expected rate law?

Rate Step 2=k2[C][B] Rate forward = kf[A][B]Rate reverse = kr[C]kf[A][B] = kr[C][C]= kf[A][B]/kr Rate = kobs[A][B]2

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Terminal Reaction for Superoxide Radical

Miller, Anne-Frances. “Fe Superoxide Dismutase” Handbook of Metalloproteins. John Wiley & Sons, Ltd, Chinchester, 2001Rodrigues, J. V; Abreu, I. A.; Cabelli, D; Teixeira, M. Biochemistry 2006, 45, 9266-9278.

Catalyst


Catalysts



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Catalyst Definition

o Substance that changes rate of chemical reaction without itself being used up

o Speeds up reaction, but not consumed by reactiono Appears in mechanism, but not in overall reactiono Does not undergo permanent chemical changeo Regenerated at end of reaction mechanismo May appear in rate law o May be heterogeneous or homogeneous


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Catalyst Activation Energy

o By providing alternate mechanismo One with lower Ea

o Because Ea lower, more reactants and collisions have minimum KE, so reaction proceeds faster


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Catalyst Activation Energy


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Catalyst Homogeneous Catalyst

• Same phase as reactantsConsider : S(g) + O2(g) + H2O(g) H2SO4(g) S(g) + O2(g) SO2(g) NO2(g) + SO2(g) NO(g) + SO3(g) Catalytic pathwaySO3(g) + H2O(g) H2SO4(g) NO(g) + ½O2(g) NO2(g) Regeneration of catalystNet: S(g) + O2(g) + H2O(g) H2SO4(g) • What is Catalyst?

– Reactant (used up) in early step– Product (regenerated) in later step

• Which are Intermediates?


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Catalyst Heterogeneous Catalyst

o Exists in separate phase from reactantso Usually a solido Many industrial catalysts are heterogeneouso Reaction takes place on solid catalyst

Ex. 3H2(g) + N2(g) 2NH3(g)


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Catalyst Heterogeneous Catalyst

H2 and N2 approach Fe catalyst

H2 and N2 bind to Fe& bonds break

N—H bonds forming

N—H bonds forming

NH3 formation complete

NH3 dissociates

Catalyst Methane Oxidation to Methanol

CH4 CH3OH

o C-H has a high bond strength (~ 410 kJ/mol)

o Thermodynamically favorable to oxidize CH4 to CO2

o Selective catalysts can stop the oxidation at methanol

GOAL: Develop a selective and robust catalyst that produces high yields of methanol at low temperatures and pressureso Syngas synthesis of methanol (CO + H2):

o High Pressure with ZnO/Cr2O3 catalyst

o Cu-Zeolite catalysto Shilov cycle with a platinum catalyst

Catalyst Methane Monooxygenase

CH4 CH3OH

o Oxidoreductase Enzyme found in methanotrophic bacteria that help cycle carbon in anarobic sediments.

o Bioinorganic chemistry tries to replicate the highly specialized chemistry found in nature using smaller molecules that can be synthesized in a lab.

Splitting WaterCatalyst

Requirments: Very Endothermico Need a minimum of 1.23 V to split water

o Kinetically infrared light could do this, but the reaction is very slow

o The potential really needs to be at least 3.0 V to utilize the full spectrum of light

Catalyst Photosystem II

PQ + H2O --> PQH2 + O2 (g)

The overall reaction of Photosystem II is the oxidation of water and the reduction of plastoquinone.

Splitting WaterCatalyst

Increase efficiency and decrease activation energy of electrolysis of water with a catalyst that will work at room temperature: Co3+ HPO4-

chemical kinetics chapter 14 chemistry: the molecular nature of matter, 6 th edition

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