Chemical reaction Masatsugu Sei Suzuki Department of Physics,...
Transcript of Chemical reaction Masatsugu Sei Suzuki Department of Physics,...
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
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
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
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 .
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
((Solution))
(a)
Thermal ionization of hydrogen
HHe
The law of mass action:
)](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
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))
______________________________________________________________________________
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
((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
DeDQ
Here we use the relation )()( DnDn QQ
2/3
2
2/3
2 2
2
ℏ Tmk
h
Tmkn BBQ
We use
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))
____________________________________________________________________________
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
((Solution))
Monomer:
N = 1
N = 2
N = 3
………………………………………………………………………………………………
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
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
NN
Q
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
)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
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
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
nnS
ZZnnnn
For nnn , QQQ nnn , 2
1S
)exp()12(222 QnSn
or
)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
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))
((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
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;
]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