S1
Supporting Information for
The Connection between the Second Law of Thermodynamics and the Principle of
Microscopic Reversibility
Gábor Lente
Department of Inorganic and Analytical Chemistry, University of Debrecen, P.O.B. 21,
Debrecen, H-4010 Hungary. Fax: + 36 52 489667; Tel: + 36 52 512900/22373; E-mail:
lenteg@delfin.unideb.hu
S2
Derivation of eq. 6 based on eq. 5
The starting point of the derivation is eq. 5.
M
 Ai ri  0
(5)
i 1
Substituting the definition of chemical affinity (A) gives:
M  G 
 ri  0
  
i 1   i  T , p
(S1)
From the standard thermodynamics of chemical equilibrium [18], it is known that
 G 


  RT ln K i  RT ln Qi
  i T , p
(S2)
Combining eqs S1 and S2 gives:
M
 ( RT ln K i  RT ln Qi )ri  0
i 1
Eq. S3 can be rearranged as follows:
(S3)
S3
M
RT  ri ln
i 1
Qi
0
Ki
As RT is always positive, eq. 6 follows directly form eq. S4.
(S4)
S4
Alternative deduction of eq. 6
Thermodynamics shows that G is a function of the amounts of different substances, pressure and
temperature. According to the derivation rule for composite functions:
dG G dT G dp N G dn j



dt
T dt p dt j 1 n j dt
(S5)
The temperature and pressure are constant so the first two terms in eq. S5 are zero. The partial
derivatives in the sum are, by definition, the chemical potentials (j). Also by definition, [Xj] =
nj/V. Substituting these into eq. S5
N
d [X j ]
dG
 V  j
dt
dt
j 1
(S6)
Writing the actual chemical potential in terms of the activity of the components gives:
N
d [X j ]
dG
 V  ( j *  RT ln a j )
dt
dt
j 1
(S7)
The definition of the rate of reaction ensures that
d [X j ]
dt
M
   ij ri
i 1
(S8)
S5
Changing the order in which the sum is calculated and using the definition of reaction quotients
(eq. 4 in the main text of the manuscript) yields:
M 
N

dG
 V  ri  RT ln Q ji    ij  j *
dt
i 1 
j 1

(S9)
In the standard thermodynamics of chemical equilibrium [18], the following equation is shown to
be true (rG* is standard Gibbs energy change for the reaction):
N
  ij  j *  r G*   RT ln K i
(S10)
j 1
Combining eqs S9 and S10 gives:
M
Q
dG
 VRT  ri ln i
dt
Ki
i 1
(S11)
The term VRT is always positive, therefore eq. 6 follows from eq. S11 and the second law of
thermodynamics, which says dG/dt  0 for any closed system at constant pressure and
temperature.
S6
Comment on the definition of the rate of reaction
On page 181 of reference 17, the rate of reaction is defined for a general reactions as:
aA + bB + ....  .... + yY + zZ
r
1 dnA
1 dnB
1 dnY
1 dn Z



aV dt
bV dt
yV dt
zV dt
(S12)
(S13)
The original reference contains an error in the sign of the last two terms, which is also obvious
from the comparison with the definitions of ‘rate of formation’ and the ‘rate of consumption’.
The original glossary only emphasizes the fact the rate of reaction can only be defined for a for a
reaction of known and time-independent stoichiometry. Strictly speaking, only elementary
processes satisfy this condition. Another way of thinking is the recognition that the appearence of
intermediates in a non-elementary process results in

1 dnB
1 dnY

bV dt
yV dt
Consequently, the rate of reaction cannot be defined in this case.
(S14)
S7
Proof of eq. 11
It will be shown that
(1  α) ln α  (α  β ) ln β  0
(11)
for any real values of  and . There are six possible cases:
1. 0 <   β  1
In this case, 0  lnβ  ln and 0  β    1  . Therefore lnβ(β  )  ln(1  ), which
can be rearranged into eq. 11.
2. 0 <   1  β
In this case, ln  0  lnβ and   β  0  1  , therefore (1  )ln  0 and (  β)lnβ  0, the
sum of which is eq. 11.
3. 1    β
In this case, 0 < ln < lnβ,   β < 0 and 1   < 0, therefore (1  )ln < 0 and (  β)lnβ < 0,
the sum of which is eq. S1.
4. 0 < β    1
In this case, lnβ  ln  0, 0    β and 0  1  , therefore (1  )ln  0 and (  β)lnβ  0,
the sum of which is eq. 11.
5. 0 < β  1  
S8
In this case, lnβ  0  ln and 1    0    β, therefore (1  )ln  0 and (  β)lnβ  0, the
sum of which is eq. 11.
6. 1  β  
In this case, 0  lnβ  ln and 0    β    1. Therefore lnβ(  β)  ln(  1), which can
be rearranged into eq. 11.
All cases have been exhausted, therefore eq. 11 is proven.
REFERENCES
Numbers are the same as used in the main text.
17 K. J. Laidler, Pure & Appl. Chem. 68 (1996) 149.
18 P. W. Atkins, Physical Chemistry, 6th ed. (Oxford University Press, Oxford, 1998).