Chemical reaction Masatsugu Sei Suzuki Department of Physics,...

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Chemical reaction Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: October, 14, 2017) Equilibrium in reactions We may write the equation of a chemical reaction as 0 j j A where j A denotes the chemical species, and the j ’s are the coefficients of the species in the reaction equation. ((Example)) H2 +Cl2 = 2HCl 1.H2+1 Cl2 = 2 HCl 1 1 , A1 = H2, 1 2 , A2 = Cl2 2 3 , A3 = HCl The discussion of the chemical equilibria is usually presented for reactions under conditions of constant pressure and temperature. In equilibrium G is an extremum and dG must be zero. VdP SdT dN dG j j j Since 0 dT dP , j j j dN dG . The change in the Gibbs free energy in reaction depends on the chemical potentials of the reactants. In equilibrium G is an extremum and dG must be zero

Transcript of Chemical reaction Masatsugu Sei Suzuki Department of Physics,...

Page 1: Chemical reaction Masatsugu Sei Suzuki Department of Physics, …bingweb.binghamton.edu/~suzuki/ThermoStatFIles/9.7 CP Chemical... · QA QB where EA ... C. Kittel and H. Kroemer,

Chemical reaction

Masatsugu Sei Suzuki

Department of Physics, SUNY at Binghamton

(Date: October, 14, 2017)

Equilibrium in reactions

We may write the equation of a chemical reaction as

0 jjA

where jA denotes the chemical species, and the j ’s are the coefficients of the species in the

reaction equation.

((Example))

H2 +Cl2 = 2HCl

1.H2+1 Cl2 = 2 HCl

11 , A1 = H2,

12 , A2 = Cl2

23 , A3 = HCl

The discussion of the chemical equilibria is usually presented for reactions under conditions of

constant pressure and temperature. In equilibrium G is an extremum and dG must be zero.

VdPSdTdNdGj

jj

Since 0 dTdP ,

j

jjdNdG .

The change in the Gibbs free energy in reaction depends on the chemical potentials of the

reactants. In equilibrium G is an extremum and dG must be zero

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0j

jjdNdG

We may write jdN in the form

NddN jjˆ

where Nd ˆ indicates how many times the reaction takes place

0ˆ)( j

jj NddG

In equilibrium, 0dG , so that

0j

jjdG

2. Equilibrium for ideal gases

We obtain a simple and useful form of the general equilibrium condition

0j

jj

When we assume that each of the constituents acts as an ideal gas. The single molecule energy of

the molecule consisting of gas splits into a term representing the translational motion of the

molecular center of mass (CM) and another portion that reflects the internal state; rotation,

vibration, and spin multiplicity;

(int))( jjj ZCMZZ

use the chemical potential of species j as

j

j

BjC

nTk ln

Note that

)]exp((int)[ jjQjj EZnC

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where )(CMZ j contributes to the factor, )exp( jQj En , jE is the energy of the ground state of

the translational motion and (int)jZ is the internal partition function. For the spin multiplicity,

we have

12(int) SZ j .

where S is the spin. Then the equation 0j

jj can be rearranged as

j

jj

j

jj Cn lnln

or

j

j

j

jjj Cn

lnln

or

)(.......... 43214321

43214321 TKCCCCnnnn

where )(TK is called the equilibrium constant, and is a function of T. Then

....)](exp[

.....}(int)](int)][(int)](int)].....]{[

.....

.....)(

2211

43214321

4321

4321

43214321

4321

4321

EE

ZZZZnnnn

CCCC

nnnnTK

QQQQ

where

(int)(int)]ln[(int)]ln[ jjjjj FZZ j

or

(int)]exp[(int)][ jjj FZ j

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where

(int)]ln[(int) jBj ZTkF

Then we have

....](int)(int)[

.....)](exp[.....)(

.....]}(int)ln(int)lnexp{[

.....)](exp[.....)(

.....]}(int)(int)[exp{

.....)](exp[.....)(

.....]}(int)(int)[exp{.....)(

.....)(

21

4321

4321

4321

4321

4321

21

22114321

2211

22114321

2211

22114321

22114321

4321

ZZ

EEnnnn

ZZ

EEnnnn

FF

EEnnnn

FFnnnn

nnnnTK

QQQQ

QQQQ

QQQQ

QQQQ

or

.....]}(int)(int)[exp{

.....)](exp[.....)(

.....)(

2211

22114321

4321

4321

4321

FF

EEnnnn

nnnnTK

QQQQ

((Example))

Chemical reaction:

0 CBA (reaction)

int)]}(int)(int)[exp{)](exp[ CBACBA

QC

QBQAFFFEEE

n

nn

C

BAK

Suppose that

12(int) SZ j with S = 0. In other words, we have

0)1ln((int)ln(int) TkZTkF BjBj .

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Then we have

)exp( H

n

nn

C

BAK

QC

QBQA

where

HEEE CBA

[A], [B], and [C] denote the concentration of A, B, and C. The energy measures the energy

involved in the reaction and determine the equilibrium concentration ratio,

CBA

.

The activation energy is the height of the potential barrier to be negotiated before the reaction

can proceed, and it determines the rate at which the reaction takes place.

3. Problem and Solution (1)

C. Kittel and H. Kroemer, Thermal Physics, second edition (W.H. Freeman and Company, 1980).

Problem 9-2

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((Solution))

(a)

Thermal ionization of hydrogen

HHe

The law of mass action:

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)](exp[][

]][[HHe

QH

QHQEEE

n

nn

H

He

Here we define the ionization energy

HHe EEEI .

Note that

QHQH nn

)exp(][

]][[In

H

HeQ

with

2/3

22

ℏTmk

n BQ

If all the electrons and protons arise from the ionization from hydrogen atoms; ][][ He , then

we get

)exp(][][ 2 InHe Q

or

)2

exp(][][2/12/1 I

nHe Q

where

2310][ H , 6.13I eV. T = 5000 K.

2/3

22

ℏTmk

n BQ

= 8.53721 x 1020 cm-3

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5643.31Tk

I

B

29291.1][ e x 1015 cm-3

(b) )]([ excH denotes the equilibrium concentration of H atoms in the first excited electronic

state

]4

3exp[][4)]([

IHexcH =4x5.23436 x 1012 cm-3=2.0937 x 1013 cm-3

The factor 4 is needed. Note that the first excited state is n = 2. Since l = 1 (3 states; d-orbitals)

and l = 0 (1 state; s-orbital). It is four-fold degenerate.

((Mathematica))

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______________________________________________________________________________

4. Problem and Solution (2)

C. Kittel and H. Kroemer, Thermal Physics, second edition (W.H. Freeman and Company, 1980).

Problem 9-3

Clear "Global` " ;

rule1 kB 1.3806504 10 16, NA 6.02214179 1023,

c 2.99792 1010, 1.054571628 10

27, me 9.10938215 10

28,

qe 4.8032068 10 10, eV 1.602176487 10 12,

I1 13.6 eV, T1 5000, NH 1023 ;

nQme kB T1

2 2

3 2

. rule1

8.53721 1020

I1

kB T1. rule1

31.5643

Ne NH1 2 nQ1 2 ExpI1

2 kB T1. rule1

1.29291 1015

4 NH Exp3 I1

4 kB T1. rule1

2.09375 1013

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((Solution))

Energy separation between the donor level and the conduction band is

56.131 *

m

m

eV

For Si, 3.0*

m

m, 7.11 .

0297.0 eV.

The law of mass action:

)exp()(

)(

)](exp[)(

)(

][

]][[

Dn

nDn

EEEDn

nDn

D

eD

Q

QQ

DeDQ

QQ

Here we use the relation )()( DnDn QQ

2/3

2

2/3

2 2

2

ℏ Tmk

h

Tmkn BBQ

We use

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enD ][ ene ][ , ed nnD ][ .

T = 300 K, 1710dn .

2/3

22

ℏTmk

n BQ

= 2.41469 x 1018 cm-3

We have the equation to determine the value of en .

)exp(

2

Q

ed

e nnn

n=7.69216 x 1016

and get the value

161073066.5 en cm-3.

((Mathematica))

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____________________________________________________________________________

5. Problem and Solution (3)

C. Kittel and H. Kroemer, Thermal Physics, second edition (W.H. Freeman and Company, 1980).

Problem 9-4

Clear "Global` " ;

rule1 kB 1.3806504 1016, NA 6.02214179 10

23,

c 2.99792 1010, 1.054571628 10

27, me 9.10938215 10

28,

qe 4.8032068 1010, eV 1.602176487 10

12,

I1 13.6 eV, T1 100, 0.0297 eV, nd 1017

;

nQme kB T1

2 2

3 2

. rule1

2.41469 1018

s1 nQ ExpkB T1

. rule1

7.69216 1016

eq1ne2

nd nes1 . rule1 Simplify

7.69216 1033 7.69216 1016 ne 1. ne2

1. 1017

1. ne0

Solve eq1, ne Simplify , ne 0 &

Solve: Solve was unable to solve the system with inexact coefficients. The answer was

obtained by solving a corresponding exact system and numericizing the result.

ne 1.34228 1017

, ne 5.73066 1016

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((Solution))

Monomer:

N = 1

N = 2

N = 3

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………………………………………………………………………………………………

Monomer + Nmer =(N+1)mer

]1[

]][1[

N

NKN

Similarly we have

][

]1][1[1

N

NKN

]1[

]2][1[2

N

NKN

……………………………………………..

]4[

]3][1[3 K

]3[

]2][1[2 K

]2[

]1][1[1 K

From these equations, we get

]1[

]1[

]1[

]][1[

][

]1][1[

]1[

]2][1[.......

]4[

]3][1[

]3[

]2][1[

]2[

]1][1[.....

1

12321

N

N

N

N

N

N

NKKKKKK

N

NNN

Thus we obtain

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NNN

N

KKKKKKN

12321

1

.....

]1[]1[

We consider

]1[

]][1[

N

NKN

From the definition, we have

)exp()1(

)()1(

)](exp[)1(

)()1(11

FNn

Nnn

FFFNn

NnnK

Q

QQ

NN

Q

QQ

N

with

1`1 FFFF NN

2/3

22)(

ℏTkM

Nn BNQ

.

NM is the mass of merN molecules and NF . is the free energy of one N merN molecule.

(c) Assume 1N

)()1( NnNn QQ

Then

)](exp[)1( 11 NNQN FFFnK

The concentration ratio

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)exp()1(

]1[

)](exp[)1(

]1[

][1[

][

]1[

11

Fn

FFFn

KN

N

Q

NN

Q

N

If 011 FFFF NN

2/3

22]1[

)1(

]1[

][

]1[

ℏTMk

nN

N B

Q

Here we have

2010]1[ cm-3. T = 300 K.

Molecular weight of the monomer 200 g per mol

ANM

200

271076203.2)1( Qn cm-3

81062058.3][

]1[ N

N

(d)

)exp()1(

]1[

][

]1[F

nN

N

Q

For the reaction to go in the direction of long molecules

1][

]1[

N

N

or

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FnQ

)1(

]1[ln

or

Fn

TkQ

B )1(

]1[ln

44295.0F eV

____________________________________________________________________________

6. Problem and Solution (4)

C. Kittel and H. Kroemer, Thermal Physics, second edition (W.H. Freeman and Company, 1980).

Problem 9-5

((Solution))

(a)

0 AA

)0(exp[ EEnnnn QQ

When

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EE ,

2/3

22

ℏTMk

nn BQQ

Then we get

)2

exp(2

2/3

2

ℏTMk

nnn B

(b) In the semiconductor

mM (mass of electron and hole)

2/3

22

ℏTmk

n BQ

When 20 , T = 300 K

Qn =1.25471 x 1019 cm-3

n 5.69637 x 1014 cm-3.

When )12(int SZ ,

)exp()12(

(int)(int))exp(

2

21

QQ

QQ

nnS

ZZnnnn

For nnn , QQQ nnn , 2

1S

)exp()12(222 QnSn

or

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)2

exp(2

)2

exp()12(

Q

Q

n

nSn

((Mathematica))

7. Electron positron pair (R. Baierlein, Thermal Physics (Cambridge, 2001).

In the extremely hot interior of stars, gamma rays can form an electron-positron pair,

ee

and the pair can mutually annihilate to form two gamma rays. Moreover, this process occurred in

the early evolution of our universe. In thermal equilibrium, we have

0 ee

since the chemical potential of photon is zero. Then we have

)2exp(2

4)2exp(4]][[ 2

3

3

2 mch

Tmkmcnnee B

ee

Clear "Global` " ;

rule1 kB 1.3806504 1016, NA 6.02214179 10

23,

c 2.99792 1010, 1.054571628 10

27, me 9.10938215 10

28,

qe 4.8032068 1010, eV 1.602176487 10

12,

I1 13.6 eV, T1 300, 20 ;

nQme kB T1

2 2

3 2

. rule1

1.25471 1019

n1 nQ Exp2

. rule1

5.69637 1014

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where 2mc is the rest energy of the electron and positron and the factor 4 comes from the spin

factor 2)12( S for electron and positron (both spin 1/2).

The core of a hot, massive star (of mass equal to 12 solar masses, say) may have a

temperature of 910T K. The right side has the value 1.64877 x 1053 cm-6, a reasonable value

for such a star. If 27103][ e cm-3, we have 25104959.5][ e cm-3. That is, there is one

positron for every 54.586≈ 55 electrons.

If ][][ ee , we have

06.4][][ ee x 1026 cm-3.

((Mathematica))

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((Note))

At extremely high temperatures, comparable with the rest energy 2mc of the electron,

collisions of particles in matter may be accompanied by the formation of electron-positron pairs.

The number of particles itself then causes to be a given quantity, and depends on the conditions

of thermal equilibrium. Pair production (and the reverse process, annihilation) can be regarded

thermodynamically as a chemical reaction

ee

Clear "Global` " ;

rule1 kB 1.3806504 1016, NA 6.02214179 10

23,

c 2.99792 1010, h 2 , 1.054571628 10 27,

me 9.10938215 10 28, mp 1.672621637 10 24,

mn 1.674927211 10 24, qe 4.8032068 10 10,

eV 1.602176487 10 12 ;

nQ2 me kB T

h2

3 2

. rule1

2.41469 1015

T3 2

f1 T1 : 4 nQ2 Exp2 me c2

kB T. T T1 . rule1

k1 f1 109

1.64877 1053

g1 3 1027; g2 k1 g1

5.49591 1025

g1 g2

54.586

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where the symbols e and e denote a positron and an electron, and denotes one or more

photons. The chemical potential of the photon gas is zero. The condition of equilibrium for pair

production is therefore

0

where and are the chemical potentials of the electron and positron gases. It should be

noted that m here denotes the relativistic expression for the chemical potential, including the rest

energy of the particles, which plays an important part in pair production.

REFERENCES

C. Kittel and H. Kroemer, Thermal Physics, second edition (W.H. Freeman and Company, 1980).

R. Baierlein, Thermal Physics (Cambridge, 2001).

L.D. Landau and E.M. Lifshitz, Statistical Physics (Pergamon Press 1976).

_____________________________________________________________________________

APPENDIX

Saha equation (Fermi)

In Na vapor at very high temperatures, Na atoms become ionized. They lose one of their

electrons, and thus changed into ions.

)electron()ion(Na(atom) Na e .

It is found that, at any given temperature, this ionization reaction reaches a state of thermal

equilibrium which is quite analogous to the chemical equilibrium for ordinary chemical reactions.

)exp()2(

)]()Na()Na([exp{)()Na(

)Na(

]][Na[

]Na[

2/3

2/3

3

Em

h

eEEEenn

n

e QQ

Q

where we use

)Na()Na( QQ nn

Let x be the degree of ionization, that is, the fraction of atoms that are ionized;

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]Na[]Na[

]Na[

x

and let ]Na[]Na[ n be the total concentration of the sodium (atoms and ions). Then we

have

nx ]Na[ , )1(]Na[ xn

Since there is obviously one electron present for each sodium ion, we have

nx ]Na[]e[

and we finally obtain

)exp()2(

11 2/3

2/3

3

2E

m

h

x

x

n

or

)exp(1041469.2

)exp()2(

1

2/315

2/3

3

2/32

Tk

ET

Tk

ET

h

mk

x

xn

B

B

B

(Saha equation)

where n is the total concentration of the sodium (atoms and ions).

((Note)) Saha equation

The Saha ionization equation, also known as the Saha–Langmuir equation, is an expression

that relates the ionization state of a gas in thermal equilibrium to the temperature and pressure.

The equation is a result of combining ideas of quantum mechanics and statistical mechanics and

is used to explain the spectral classification of stars. The expression was developed by the Indian

astrophysicist Meghnad Saha in 1920, and later (1923) by Irving Langmuir.

https://en.wikipedia.org/wiki/Saha_ionization_equation