volume the reaction (as written) has proceeded,

the irreversible entropy production per unit volume and per unit time due

to the advancement of the reaction can be written as

1 dξ

σ= A. (16.107)

T dt

The rate of advancement dξ/dt can be viewed as the reaction ¬‚ux and A/T

then is the driving force for the reaction. Note that reaction ¬‚uxes and forces

are scalar quantities, contrary to the vector quantities we encountered thus

far. The rate of advancement is equal to the usual net velocity of the reaction

1 d[Ci ]

vreact = . (16.108)

νi dt

In equilibrium, the a¬nity is zero. For reactions that deviate only slightly

from equilibrium, the velocity is linear in the a¬nity; far from equilibrium

no such relation with a¬nity exists.

13 We use the same notation as in (16.64).

16.10 Thermodynamics of irreversible processes 449

16.10.3 Phenomenological and Onsager relations

For small deviations from equilibrium, i.e., small values of the driving forces

X, one may assume that the ¬‚uxes are proportional to the driving forces.

Such linear relations are the ¬rst term in a Taylor expansion in the driving

forces,14 and they are justi¬ed on a phenomenological basis. The main

driving force for a ¬‚ux is its conjugated force, but in general the linear

relations may involve any other forces as well:

Jk = Lkl Xl or J = LX. (16.109)

l

Here Lkl are the phenomenological coe¬cients. This implies that for the

entropy production

σ = XT LT X. (16.110)

From the second law we know that the irreversible entropy production must

always be positive for any combination of driving forces. Mathematically

this means that the matrix L must be positive de¬nite with only positive

eigenvalues.

The diagonal phenomenological coe¬cients relate to ¬‚uxes resulting from

their conjugate forces, such as heat conduction (heat ¬‚ow due to temper-

ature gradient), viscous ¬‚ow, e.g., through a membrane or through porous

material (¬‚uid ¬‚ow due to hydrostatic pressure di¬erence), electrical conduc-

tion (current due to electric ¬eld), and di¬usion (particle ¬‚ow with respect

to solvent due to concentration gradient). O¬-diagonal coe¬cients relate to

¬‚uxes that result from other than their conjugate forces, such as:

• thermoelectric e¬ect (current due to temperature gradient) and Peltier

e¬ect (heat ¬‚ow due to electric ¬eld);

• thermal di¬usion or Soret e¬ect (particle separation due to temperature

gradient) and Dufour e¬ect (heat ¬‚ow due to concentration gradient);

• osmosis (volume ¬‚ow due to concentration gradient) and reverse osmosis

(particle separation due to pressure gradient);

• electro-osmosis (volume ¬‚ow due to electric ¬eld) and streaming potential

(current due to pressure gradient);

• di¬usion potential (current due to concentration gradient) and electro-

phoresis (particle separation due to electric ¬eld).

On the basis of microscopic reversibility and by relating the phenomeno-

logical coe¬cients to ¬‚uctuations, Onsager (1931b) came to the conclusion

14 Note that this does not imply that the coe¬cients themselves are constants; they may depend

on temperature, pressure and concentration.

450 Review of thermodynamics

that the matrix L must be symmetric:

Lkl = Llk . (16.111)

These are Onager™s reciprocal relations that relate cross e¬ects in pairs. For

example, the thermoelectric e¬ect is related to the Peltier e¬ect as follows:

1 E

Jq = Lqq grad + Lqe , (16.112)

T T

1 E

j = Leq grad + Lee . (16.113)

T T

The thermoelectric coe¬cient is de¬ned as the ratio of the potential dif-

ference, arising under current-free conditions, to the externally maintained

temperature di¬erence (in V/K). This is equal to

Leq

grad ¦

=’ . (16.114)

grad T T Lee

j=0

The Peltier e¬ect is de¬ned as the heat ¬‚ux carried per unit current density

under conditions of constant temperature (in J/C). This is equal to

Jq Lqe

=’ . (16.115)

j Lee

grad T =0

Onsager™s relation Leq = Lqe implies that the thermoelectric coe¬cient

equals the Peltier coe¬cient, divided by the absolute temperature.

16.10.4 Stationary states

When there are no constraints on a system, it will evolve spontaneously into

an equilibrium state, in which the irreversible entropy production becomes

zero. The direction of this process is dictated by the second law; a ¬‚ow Ji

diminishes the conjugate force, and in course of time the entropy production

decreases. With external constraints, either on forces or on ¬‚uxes, or on

combinations thereof, the system will evolve into a stationary state (or steady

state), in which the entropy production becomes minimal. This is easily seen

as follows.

Assume that forces X± are constrained by the environment. X± are a

subset of all forces Xi . The system develops in the direction of decreasing

entropy production, until the entropy production is minimal. Minimizing

XT LX under constraints Xk = constant, requires that

‚

( Lij Xi Xj + »± X± ) = 0, (16.116)

‚Xk ±

ij

Exercises 451

for all k, and where »± are Lagrange undetermined multipliers. This implies

that:

• J± = constant;

• Jk = 0 for k = ±.

Thus, the system evolves into a steady state with constant ¬‚uxes; ¬‚uxes

conjugate to unconstrained forces vanish.

Exercises

16.1 Show that the following concentration dependencies, valid for ideal

solutions, satisfy the Gibbs“Duhem relation:

μw = μ0 + RT ln(1 ’ x),

w

μs = μ0 + RT ln x.

s

16.2 Prove the Gibbs“Helmholtz equation, (16.32).

16.3 Show that another Gibbs“Helmholtz equation exists, replacing G by

A and H by U .

16.4 At ¬rst sight you may wonder why the term i μi dni occurs in

(16.25) to (16.28). Prove, for example, that (‚H/‚ni )p,T = μi .

16.5 Prove (16.40) and (16.41).

16.6 Prove (16.42) by starting from the pressure de¬nition from (16.27),

and then using the Maxwell relation derived from (16.27).

16.7 Estimate the boiling point of water from the Clausius“Clapeyron

equation at an elevation of 5500 m, where the pressure is 0.5 bar.

The heat of vaporization is 40 kJ/mol.

16.8 Estimate the melting point of ice under a pressure of 1 kbar. The

densities of ice and water are 917 and 1000 kg/m3 ; the heat of fusion

is 6 kJ/mol.

16.9 Rationalize that the cryoscopic constant (the ratio between freezing

point depression and molality of the solution) equals 1.86 K kg mol’1

for water.

17

Review of statistical mechanics

17.1 Introduction