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CEE 680 Lecture #5 1/29/2020 1 Lecture #5 Kinetics and Thermodynamics: Fundamentals of Kinetics and Analysis of Kinetic Data (Stumm & Morgan, Chapt.2 ) (pp.1620; 6981) David Reckhow CEE 680 #5 1 (Benjamin, 1.6) Updated: 29 January 2020 Print version Elementary Reactions When reactant molecules collide with the right orientation and energy level to form new bonds Many “observable” reactions are really just combinations of elementary reactions F E B A F C D A E C D C B A 2 2 David Reckhow CEE 680 #5 2 fast slow fast Starting out with some A and B, we observe that E and F are the end products

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CEE 680 Lecture #5 1/29/2020

1

Lecture #5

Kinetics and Thermodynamics:  Fundamentals of Kinetics and  Analysis of Kinetic Data

(Stumm & Morgan, Chapt.2 )

(pp.16‐20; 69‐81)

David Reckhow CEE 680 #5 1

(Benjamin, 1.6)

Updated: 29 January 2020 Print version

Elementary Reactions When reactant molecules collide with the right orientation and energy level to form new bonds

Many “observable” reactions are really just combinations of elementary reactions

FEBA

FCDA

EC

DCBA

2

2

David Reckhow CEE 680 #5 2

fast

slow

fast

Starting out with some A and B, we observe that E and F are the end products

CEE 680 Lecture #5 1/29/2020

2

Cont.

Elementaryreactions

A single step in areaction sequence

Involves 1 or 2 reactants and 1 or 2 products

Can be described by classical chemical kinetics

David Reckhow CEE 680 #5 3

S&M: Fig. 2.11Pg. 72

Kinetics Examples

Fe+2 oxidation by O2

almost instantaneous at high pH

quite slow at low pH

high D.O. may help

Oxidation of organic material

Formation of solid phases Aluminum hydroxide

Quartz sand

David Reckhow CEE 680 #5 4

CEE 680 Lecture #5 1/29/2020

3

Kinetics

Base Hydrolysis of dichloromethane (DCM)

Forms chloromethanol (CM) and chloride

Classic second order reaction (molecularity of 2)

First order in each reactant, second order overall

David Reckhow CEE 680 #5 5

dt

Cld

dt

CMd

dt

OHd

dt

DCMdOHDCMkRate

][][][][]][[

Reaction Kinetics Irreversible reaction

is one in which the reactant(s) proceed to product(s), but there is no significant backward reaction,

In generalized for, irreversible reactions can be represented as:

aA + bB Products

David Reckhow CEE 680 #5 6

i.e., the products do not recombine or change to form reactants in any appreciable amount. An example of an irreversible reaction is hydrogen and oxygen combining to form water in a combustion reaction. We do not observe water spontaneously separating into hydrogen and oxygen.

CEE 680 Lecture #5 1/29/2020

4

Reaction Kinetics: Reversibility A reversible reaction

is one in which the reactant(s) proceed to product(s), but the product(s) react at an appreciable rate to reform reactant(s).

aA + bB pP + qQ

Most reactions must be considered reversible

David Reckhow CEE 680 #5 7

An example of a reversible biological reaction is the formation of adenosine triphosphate (ATP) and adenosine diphosphate (ADP). All living organisms use ATP (or a similar compound) to store energy. As the ATP is used it is converted to ADP, the organism then uses food to reconvert the ADP to ATP.

Kinetic principles Law of Mass Action

For elementary reactions

David Reckhow CEE 680 #5 8

where,CA = concentration of reactant species A, [moles/liter]CB = concentration of reactant species B, [moles/liter]a = stoichiometric coefficient of species A

b = stoichiometric coefficient of species Bk = rate constant, [units are dependent on a and b]

productsbBaA kbB

aACkCrate

CEE 680 Lecture #5 1/29/2020

5

Reaction Kinetics (cont.) Reactions of order “n” in reactant “c”

When n=0, we have a simple zero‐order reaction

David Reckhow CEE 680 #5 9

dc

dtkcn

dc

dtk

ktcc o

0

10

20

30

40

50

60

70

80

90

0 20 40 60 80

Co

nce

ntr

atio

n

Time (min)

k mg l 10 / / min

Slope

Reaction Kinetics (cont.) When n=1, we have a simple first‐order reaction

This results in an “exponential decay”

David Reckhow CEE 680 #5 10

dc

dtkc 1

c c eokt

0102030405060708090

0 20 40 60 80

Time (min)

Co

nc

en

tra

tio

n

k 0 032 1. min

CEE 680 Lecture #5 1/29/2020

6

Reaction Kinetics (cont.)

This equation can be linearized

good for assessment of “k” from data

David Reckhow CEE 680 #5 11

dc

dtkc 1

10

100

0 20 40 60 80

Time (min)

Co

nc

en

tra

tio

n (

log

sc

ale

)

ktcc o lnln

k 0 032 1. min

Slope

0102030405060708090

0 20 40 60 80

Time (min)

Co

nc

en

tra

tio

n

Reaction Kinetics (cont.)

This results in an especially wide range in rates

More typical to have 2nd order in each of two different reactants

David Reckhow CEE 680 #5 12

dc

dtkc 2

c ckc to

o

1

1k L mg 0 0015. / / min

When n=2, we have a simple second-order reaction

CEE 680 Lecture #5 1/29/2020

7

Reaction Kinetics (cont.) Again, the equation can be linearized to estimate “k” from data

David Reckhow CEE 680 #5 13

dc

dtkc 2

0

0.02

0.04

0.06

0.08

0.1

0.12

0 20 40 60 80

Time (min)

1/C

on

ce

ntr

ati

onkt

cc o

11

k L mg 0 0015. / / min

Slope

Comparison of Reaction Orders

Curvature as order changes: 2nd>1st>zero

David Reckhow CEE 680 #5 14

0102030405060708090

0 20 40 60 80

Time (min)

Co

nc

en

tra

tio

n

Zero Order

First Order

Second Order

CEE 680 Lecture #5 1/29/2020

8

Reaction Kinetics (cont.) For most reactions, n=1 for each of two different reactants, thus a second‐order overall reaction

David Reckhow CEE 680 #5 15

12

11ckc

dt

dc

115 min109.3 Lmgxk

Many of these will have one reactant in great excessThese become “pseudo‐1st order in the limiting reactant, as the reactant in excess really doesn’t change in concentration

2c

1c

0102030405060708090

0 20 40 60 80

Time (min)

Co

nc

en

tra

tio

n

Reaction Kinetics (cont.)

Since C2 changes little from its initial 820 mg/L, it is more interesting to focus on C1

C1 exhibits simple 1st

order decay, called pseudo‐1st order The pseudo‐1st order rate constant is just the “observed rate” or kobs

David Reckhow CEE 680 #5 16

12

11ckc

dt

dc

tko

obsecc 11

1

52

min032.0

)820(109.3

xkckobs

CEE 680 Lecture #5 1/29/2020

9

Variable Kinetic Order Any reaction order, except n=1

David Reckhow CEE 680 #5 17

nkcdt

dc

11

111

1

nno

o

tkcncc

ktncc n

on

111

11

Half‐lives Time required for initial concentration to drop to half, i.e.., c=0.5co For a zero order reaction:

For a first order reaction:

David Reckhow CEE 680 #5 18

c c kto 2

15.0 ktcc oo k

ct o5.0

21

c c eokt 2

1

5.0kt

oo ecc

k

kt

693.0

)2ln(2

1

CEE 680 Lecture #5 1/29/2020

10

Kinetic problem If the half‐life of bromide in the presence of excess chlorine is 13 seconds (pseudo‐1st order reaction, 

What is the pseudo‐1st order rate constant

how long does it take for 99% of the bromide to be oxidized?

David Reckhow CEE 680 #5 19

HOCl Br HOBr Clk

Reactions in Series

David Reckhow CEE 680 #5 20

DCBA kkk 321

Fig. 2.9Pg. 68

k1=k2=k3=0.1 day-1

CEE 680 Lecture #5 1/29/2020

11

Reversible reaction kinetics

David Reckhow CEE 680 #5 21

For a general reversible reaction: f

b

k

aA + bB pP + qQ

k

And the rate law must consider both forward and reverse reactions:

A f Aa

Bb

b Pp

Qqr = k C C - k C C

where,kf = forward rate constant, [units depend on a and b]kb = backward rate constant, [units depend on a and b]CP = concentration of product species P, [moles/liter]CQ = concentration of product species Q, [moles/liter]p = stoichiometric coefficient of species Pq = stoichiometric coefficient of species Q

Reversible 1st order reactions Kinetic law

Eventually the reaction slows and,

Reactant concentrations approach the equilibrium values

David Reckhow CEE 680 #5 22

Fig. 2.10Pg. 69

][][ 21 BkAkdt

dB

eqKk

k

A

B

BkAkdt

dB

2

1

21

][

][

][][0

CEE 680 Lecture #5 1/29/2020

12

Analysis of Rate Data

Integral Method Least squares regression of linearized form

Differential Method estimate instantaneous rate at known time and reactant concentration

Initial rate Method more rigorous, but slow

Method of Excess only when 2 or more reactants are involved

David Reckhow CEE 680 #5 23

Kinetic model for equilibrium

DCBA

DCBA

b

f

k

k

}}{{ BAkr ff

David Reckhow CEE 680 #5 24

Consider a reaction as follows:

Since all reactions are reversible, we have two possibilities

The rates are:

And at equilibrium the two are equal, rf=rb

We then define an equilibrium constant (Keq)

}}{{ DCkr bb

A + B = C + D

}}{{

}}{{

BA

DC

k

kK

b

feq

}}{{}}{{ DCkBAk bf

CEE 680 Lecture #5 1/29/2020

13

Kinetic model with moles

BAff BAkr

David Reckhow CEE 680 #5 25

In terms of molar concentrations, the rates are:

And at equilibrium the two are equal, rf=rb

And solving for the equilibrium constant (Keq)

DCbb DCkr

BA

DC

BA

DC

b

feq BA

DC

BA

DC

k

kK

DCbBAf DCkBAk

To next lecture

David Reckhow CEE 680 #5 26