chem.414 - physical chemistry ii spring 2012/2013/2014

Chem.414 - Physical Chemistry IIChem.414 - Physical Chemistry II

T ranslational

M echanism s

Collisions

Diffusion Ideal/Dilute

E-Chem (Lab)

Henry/Raoult T herm odynam ics Phase diagram s

Solutions

Boltzm ann Distributions T ransiton State

T heories of Reaction Rates

Rate Law s T -Dependence

Kinetics

Operator Algebra Applications

Quantum Chem istry

Spring 2012/2013/2014

Chemical KineticsChemical Kinetics

0th Order

M echanism s

T -Dependence of Rxn Rates

Steady State Approxim ation Experim ental Techniques

Order & M olecularity

2nd Order nth Order

Concepts of Rxn RatesFirst Order

Study of Chemical KineticsStudy of Chemical Kinetics

• Rate of reaction

• Dependence of concentration of species

• Dependence of temp., pressure, catalyst

• Control of reactions

• Mechanisms [Dominating step (fast vs. slow)]

• Guide to chemical intuition

Reaction Rate and Stoichiometry• For the reaction

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

we know

• In general for

aA + bB cC + dD

Reaction RatesReaction Rates

dt

d

dt

d OHHCClHCRate 9494

dt

d

ddt

d

cdt

d

bdt

d

a

DCBARate

1111

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

Time / s [N2O5] / M ln [N2O5] d[N2O5]/dt(tangential slope)

1 0 1.00

2 200 0.88

3 400 0.78

4 600 0.69

5 800 0.61

6 1000 0.54

7 1200 0.48

8 1400 0.43

9 1600 0.38

10 1800 0.34

11 2000 0.30

4. Consider the following N2O5 reaction: 2 N2O5(soln) ----> 4 NO2(soln) + O2(g)

Let: C = [N2O5]

(a) Using a graph of C vs. t, obtain tangential slopes and plot dC/dt vs. C. Calculate k after fitting with linear regression. (b) Plot ln C vs. t. Calculate k after fitting with linear regression. (c) Plot C vs. t. Fit the data with an appropriate function. Display the equation in standard IRL form with the appropriate variable names for this reaction. (d) Calculate half-live (t2) and life-time (). Compare them to the interpolated values from the plot of C vs. t.

EXCEL

First Order Reactions (to one component)

The Change of Concentration with TimeThe Change of Concentration with Time

0lnln CktC CNCHNCCH Co

39.198

3

Isomeric Transformation of Methyl Isonitrile to Acetonitrile

Differential and Integrated Rate LawsDifferential and Integrated Rate Laws

n-th Order to One Component (Generalized Rate Laws)

Let: C = concentration of reactant A remaining at time tCo = initial concentration of reactant A (i.e. t=0)k = rate constant (units depends on n)

DRL:

IRL:

Differential and Integrated Rate LawsDifferential and Integrated Rate Laws

Rate Law: First Order to One ComponentRate Law: First Order to One Component

Second Order Reactions

The Change of Concentration with TimeThe Change of Concentration with Time

0

11

Ckt

C

)(2

1)()( 2

3002 gOgNOgNO Co

Rate Law: Second Order to One ComponentRate Law: Second Order to One Component

Gas-Phase Decomposition of Nitrogen DioxideGas-Phase Decomposition of Nitrogen Dioxide

)(2

1)()( 2

3002 gOgNOgNO Co

Time / s [NO2] / M

0.0 0.01000

50.0 0.00787

100.0 0.00649

200.0 0.00481

300.0 0.00380

k = 0.543 unit?k = 0.543 unit?Is this reaction first or second order?

Half-Lives, Rate Constants and CHalf-Lives, Rate Constants and Coo

Half-Lives, Rate Constants and CHalf-Lives, Rate Constants and Coo - II - II

Zeroth Order to One Component - CatalysisZeroth Order to One Component - Catalysis

1. Provide the DRL.

2. Determine the IRL.

3. Sketch the IRL: Co=1.00 mol L-1 , k = 5.00x10-3 mol L-1 s-1 .

4. Use Mathcad (or EXCEL) to generate the IRL graph.

5. Obtain the half-life expression.

6. How many half-lives would it take for the reaction to reach equilibrium (i.e. completion)? [ Hint: Solve the IRL for time when C=0. Confirm by graph. ]

Summary of Rate Laws to One-ComponentSummary of Rate Laws to One-Component

First-Order Second-Order Zeroth-Order

DRL

(-dC/dt)kC kC2 k

IRLC = Co·e-kt

ln C = -kt + ln Co

1/C = kt + 1/Co C = -kt + Co

Linear Equation

ln C vs. t 1/C vs. t C vs. t

Linear Plot

Half-Life ln(2)/k 1/kCo Co/2k

Units on k time-1 M-1 time-1 M time-1

m = -k

b = ln Co

m = k

b = 1/Co

m = -k

b = Co

Exponents in the Rate Law• For a general reaction with rate law

we say the reaction is mth order in reactant 1 and nth order in reactant 2.

• The overall order of reaction is m + n + ….• A reaction can be zeroth order if m, n, … are zero.• Note the values of the exponents (orders) have to be determined

experimentally. They are not simply related to stoichiometry.

Concentration and RateConcentration and Rate

nmk ]2reactant []1reactant [Rate

Method of Initial/Comparative RatesMethod of Initial/Comparative Rates

Expt # [NH4+]o / M [NO2

-]o / M (Rate)o / M s-1

1 0.100 0.0050 1.35x10-7

2 0.100 0.0100 2.70x10-7

3 0.200 0.0100 5.40x10-7

)(2)()()( 2224 OHgNaqNOaqNH

Three Component Rate LawThree Component Rate Law

Expt # [BrO3-]o / M [Br-]o / M [H+]o / M (Rate)o / M s-1

1 0.10 0.10 0.10 8.0x10-4

2 0.20 0.10 0.10 1.6x10-3

3 0.20 0.20 0.10 3.2x10-3

4 0.10 0.10 0.20 3.2x10-3

)(3)(3)(6)(5)( 223 OHBraqHaqBraqBrO

Techniques for Multiple Component Rate LawsTechniques for Multiple Component Rate Laws

1. Integration Approach:

Second Order – First Order to each of two components

2. Flooding Technique:

Rate = k [A]x [B]y [C]z

Second Order: First order to each of two components

Consider: A + B ---> Products (D)

ratedCA

dt

dCB

dt

dCD

dt

At t=0; Let: CA = a ; C B = b

At some time t; Let: x (mol/L) of A and B be reacted

CA a x CB b x CD x

dx

dtk a x( ) b x( )

dx

a x( ) b x( )k dt

expands in partial fractions to1

a x( ) b x( )1

b a( ) a x( )1

b a( ) b x( )

by integration, yields1

a x( ) b x( )1

b aln a x( )

1

b aln b x( )

by integration, yields1b a( ) a x( )

1

b a( ) b x( )

1

b aln a x( )

1

b aln b x( )

Re-writing to give the following: 1

a bdx

a xdx

b x

k dt

1

a b0

x

x1

a x1

b x

d k0

t

t1

d

Integrating yields: 1

a bln

b a x( )a b x( )

k t

Z1

a bln

b a x( )a b x( )

Z versus t yields straight line.

Applications of First-Order ProcessesApplications of First-Order Processes

1. Radioactive Decay

2. Bacterial Growth

3. Interest and Exponential Growth

[Credit Card]

4. Loan Balance

Interest and Exponential Growth

P = future valueC = initial depositr = interest rate (expressed as a fraction: eg. 0.06)n = # of times per year interest is compoundedt = number of years invested

C 10000 r 0.06 n 1P t( ) C 1

r

n

n t

Continuous Compound Interestn

1rate

n

n tlim

erate t

n ---> infinity

P 30( ) 57 103

PE t( ) C er t

0 3 6 9 12 15 18 21 24 27 300

8000

1.6 104

2.4 104

3.2 104

4 104

4.8 104

5.6 104

6.4 104

7.2 104

8 104

P t( )

PE t( )

t

PE 30( ) 60 103

PE 30( ) P 30( ) 3061.56

Loan Balance A person initially borrows an amount A and in return agreesto make n repayments per year, each of an amount P.While the person is repaying the loan, interest isaccumulating at an annual percentage rate of r, and thisinterest is compounded n times a year (along with eachpayment). Therefore, the person must continue payingthese installments of amount P unitl the original amountand any accumulated interest is repayed. The equation B(t)gives the amount B that the person still needs to repay aftert years.

B = balance after t yearsA = amount borrowedn = number of payments per yearP = amount paid per paymentr = annual percentage rate (APR)

A 115000 n 12 P 800 r 0.05B t( ) A 1

r

n

n t P

1r

n

n t1

1r

n

1

0 2 4 6 8 10 12 14 16 18 205 10

4

1 104

3 104

7 104

1.1 105

1.5 105

B t( )

t

The Arrhenius Equation• Arrhenius discovered most reaction-rate data obeyed the

Arrhenius equation:

– k is the rate constant, Ea is the activation energy, R is the gas constant (8.3145 J K-1 mol-1) and T is the temperature in K.

– A is called the frequency factor.

– A is a measure of the probability of a favorable collision.

– Both A and Ea are specific to a given reaction.

Temperature and RateTemperature and Rate

RTEa

eAk

Temperature and RateTemperature and Rate

• The balanced chemical equation provides information about the beginning and end of reaction.

• The reaction mechanism gives the path of the reaction.• Mechanisms provide a very detailed picture of which bonds

are broken and formed during the course of a reaction.

Elementary Steps• Elementary step: any process that occurs in a single step.

Reaction MechanismsReaction Mechanisms

Elementary Steps• Molecularity: the number of molecules present in an

elementary step.– Unimolecular: one molecule in the elementary step,

– Bimolecular: two molecules in the elementary step, and

– Termolecular: three molecules in the elementary step.

• It is not common to see termolecular processes (statistically improbable).


Rate Laws for Elementary Steps• The rate law of an elementary step is determined by its

molecularity:– Unimolecular processes are first order,

– Bimolecular processes are second order, and

– Termolecular processes are third order.

Rate Laws for Multistep Mechanisms• Rate-determining step is the slowest of the elementary

steps. [example]


Rate Laws for Elementary Steps

Reaction MechanismsReaction MechanismsReaction MechanismsReaction Mechanisms

Rate ExpressionsRate Expressions

yDxCwBvAk

k

1

1

dt

d

ydt

d

xdt

d

wdt

d

v

DCBARate

1111

If elementary steps:

-d[A]/dt = vk1[A]v[B]w – vk-1[C]x[D]y

-d[B]/dt = wk1[A]v[B]w – wk-1[C]x[D]y

d[C]/dt = xk1[A]v[B]w – xk-1[C]x[D]y

d[D]/dt = yk1[A]v[B]w – yk-1[C]x[D]y

Mechanisms with an Initial Fast Step

2NO(g) + Br2(g) 2NOBr(g)

• The experimentally determined rate law can be:

d[NOBr]/dt = kobs[NO]2[Br2] (or) = kobs’[NO][Br2]

• Consider the following mechanism


NO(g) + Br2(g) NOBr2(g)k1

k-1

NOBr2(g) + NO(g) 2NOBr(g)k2

Step 1:

Step 2:

(fast)

(slow)

NO(g) + Br2(g) NOBr2(g)k1

k-1

NOBr2(g) + NO(g) 2NOBr(g)k2

Step 1:

Step 2:

(fast)

(slow)

Spring 2014

General MechanismGeneral Mechanism

DCBA Overall Reaction:

Proposed Mechanism:

DBM

CMA

k

k

k

2

1

1Where: D = observable product

M = intermediate

DBM

CMA

k

k

k

2

1

1

DBM

CMA

k

k

k

2

1

1

Spring 2014

)(

)(

2

1

1

DBM

CMA

k

k

k

Spring 2014

Hydrogen-Iodine ReactionHydrogen-Iodine Reaction

HIIH 222 Overall Reaction:

Proposed Mechanism:

HIIH

II

k

k

k

22

2

3

1

2

2

2

Where: I• = free radical

?][

dt

HId

HIIH

II

k

k

k

22

2

3

1

2

2

2

HIIH

II

k

k

k

22

2

3

1

2

2

2

Spring 2012

Rice-Hertzfeld Free Radical Chain Reaction Mechanism

Rice-Hertzfeld Free Radical Chain Reaction Mechanism

)()()( 43 gCOgCHgCHOCH Overall Reaction:

Proposed Mechanism:

ationterchainHCCH

npropagatiochainCHCOCHCHCHOCH

initiationchainCHOCHCHOCH

k

k

k

min2 623

3433

33

3

2

1

?][ 4

dt

CHd

ationterchainHCCH

npropagatiochainCHCOCHCHCHOCH

initiationchainCHOCHCHOCH

k

k

k

min2 623

3433

33

3

2

1

KineticsKinetics

optial rotationabsorption/em issiondielectric constantrefractive indexdilatom etric (Vol.)pressure jum ptem perature jum pelectric fie ldconductivity

Hardw are"Instrum entation"

reaction rates vs. tim econcentrations vs. tim einitia l rates vs. tim e"lives" vs. tim eguess type of ordercom puter fitsflooding/isolationcatalystsm echanism s

Softw are"Brainm entation"

Experim ental T echniquesData to Conclusions

CatalysisCatalysis

Heterogeneous Catalysis• Consider the hydrogenation of ethylene:

C2H4(g) + H2(g) C2H6(g), H = -136 kJ/mol.– The reaction is slow in the absence of a catalyst.– In the presence of a metal catalyst (Ni, Pt or Pd) the reaction occurs

quickly at room temperature.– First the ethylene and hydrogen molecules are adsorbed onto active

sites on the metal surface.– The H-H bond breaks and the H atoms migrate about the metal

surface.

CatalysisCatalysis

CatalysisCatalysis

Enzymes• Enzymes are biological catalysts.• Most enzymes are protein molecules with large molecular

masses (10,000 to 106 amu).

• Enzymes have very specific shapes.• Most enzymes catalyze very specific reactions.• Substrates undergo reaction at the active site of an enzyme.• A substrate locks into an enzyme and a fast reaction occurs.• The products then move away from the enzyme.

CatalysisCatalysis

Enzymes• Only substrates that fit into the enzyme lock can be

involved in the reaction.• If a molecule binds tightly to an enzyme so that another

substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors).

• The number of events (turnover number) catalyzed is large for enzymes (103 - 107 per second).

CatalysisCatalysis

Enzymes

CatalysisCatalysis

][

][][)( 2

SK

SEkMentenMichaelisRate

M

t

Mechanism: Two IntermediatesMechanism: Two Intermediates

ClIOOClIOverall Reaction:

Proposed Mechanism:

rdsslowOIOHHOIOH

fastClHOIHOClI

mequilibriufastveryOHHOClOHOCl

k

k

k

k

2

2

3

2

1

1

][

][][][

OH

OClIk

dt

OId obsExperimentally found:

Show that the proposed mechanism is consistent with the observed RL.

MechanismMechanism

)(3)(2 23 gOgO

Overall Reaction:

Proposed Mechanism:

slowOOO

fastOOO

k

k

k

23

23

22

1

1

][

][][

2

233

O

Ok

dt

Odobs

Observed Rate Law:

slowOOO

fastOOO

k

k

k

23

23

22

1

1

Chemical KineticsChemical Kinetics

0th Order

M echanism s

T -Dependence of Rxn Rates

Steady State Approxim ation Experim ental Techniques

Order & M olecularity

2nd Order nth Order

Concepts of Rxn RatesFirst Order

nCkdt

dCRateDRL tversusCsIRL '

0lnln CktC 0

11

Ckt

C

RTEa

eAk

DBM

CMA

k

k

k

2

1

1

chem.414 - physical chemistry ii spring 2012/2013/2014

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