1. Definitions: (From Smith, 1982, and Anderson and Crerar, 1994)
  1. Thermodynamics: Energy differences and transfers between systems.
  2. Systems:
  1. Equilibrium:
  1. State Variables:
  1. Work (w): "The transfer of energy from one mechanical system to another. It is always completely convertible to the lifting of a weight". "The energy that flows across a system boundary in response to a force moving through a distance (such as happens when a system changes volume".
  2. Heat (q): "The transfer of energy that results from temperature differences". "The energy that flow across a system boundary in response to a temperature gradient." "that part of any energy transfer that is not accounted for by mechanical work (FxD)." q=DU-w
  3. Heat Capacity: The relation between heat transferred to a body and the change in T.
  4. Enthalpy: The increase in enthalpy of a system is equal to the heat absorbed at constant pressure, assuming the system does only PV work.
  5. Entropy: A measure of the loss of capacity of the system to do work.
  6. Zeroth Law: "Two bodies in thermal equilibrium with a third are in thermal equilibrium with each other" (Basis of the concept of temperature)
  7. First Law: "The alegraic sum of all energy changes in and isolated system is zero" (Conservation of energy). Energy can be converted from one form to another but cannot be created or destroyed. DU=q+w. This is based only on observation
  8. Second Law: "Spontaneous changes are those which, if carried out under the proper conditions, can be made to do work. If carried out reversibly they yield a maximum amount of work. In natural processes, maximum work is never obtained.


      Basic Principles

      Thermodynamics is the study of energy in systems, and the distribution of energy among components. In chemical systems, it is the study of chemical potential, reaction potential, reaction direction, and reaction extent

      1. First Law of Thermodynamics:

dU=dq + dw

where U is the internal energy, q is the heat transferred to a system from the surroundings, and w is the work done on a system by the surroundings.

DU=q + w

      1. Work and Reversible reactions.:

dU = dq + dw

dU = dq - PdV

where PdV is the reversible expansion work done on the system by the surroundings.

      1. Second Law of Thermodynamics

The second law is concerned with entropy (S), which is that part of the total energy of the system that is not available to do useful work.  Entropy is a fundamental variable in thermodynamics, and is responsible for such phenomena as molecular diffusion.  For an interesting view of entropy, check out the movies at The U.C. San Diego.

Statement One: "The total entropy change of the closed system is dSsys and is the sum of changes inside the system, dSint, and entropy transferred to the system from its surroundings, dSsur.

dSsys= dSint + dSsur

dSsur = dq/T

For a reversible process or equilibrium state of the system, the second law states that the internal entropy change is zero:

dSint = 0

For a spontaneous or natural process in the system

dSint > 0

So for any spontaneous natural process

dSsys > dq/T

The internal entropy change for a closed system is zero at equilibrium and positive for a spontaneous process. The heat transferred to a closed system, divided by T, is equal to or less than the entropy increase for any possible process. The entropy of an isolated system increases in any spontaneous process and attains a maximum for any reversible process. (Stumm and Morgan, p. 19).

    1. Free Energy

The Gibbs Free energy is that portion of the total system energy that is available for useful work.



At equilibrium, and with fixed composition (no addition of material):

dq = TdS(sys)


dU = TdS - PdV

So the internal energy is a function of entropy and volume. Another expression for total system energy is Enthalpy (H):

H = U + PV


dH = dU + PdV

When T and P are constant, we can use Gibbs expression for free energy

dG = dU + PdV - TdS

dG = dH - TodS


G = Gibbs free energy, kcal/mol

T = absolute temperature, oK

S = entropy, cal/oK

H = enthalpy, kcal/mol

This equation tells us if a reaction will happen, and in what direction.

a) Enthalpy: the total energy content of acompound (dH = dU+PdV) in kJ/mol; .

b) Free Energy: That part of the total energy that is available to perform "useful work", that is, other than PV work, such as electrical work, in kJ/mol.

  1. Entropy: the degree of disorder of a system. The product TS is that part of the free energy which is NOT available for useful work, in joules/oK.
      1. Heat Capacity:
      2. Heat capacity is defined as the change in temperature of a system with a change of heat transferred to the system:

        Heat capacity can be measured directly by calorimetry (at constant pressure), and can provide information on enthalpy for a closed-system process.

      4. Law of Mass Action:

The driving force of a reaction is related to the concentrations of reactants and products.

aA + bB ---> cC + dD

At equilibrium this ratio is equal to the equilibrium constant, K.

      1. Equilibrium

The TOTAL FREE ENERGY is the sum of the free energies of each component.

GT = nAGA + nBGB + nCGC + nDGD

Where Gi = Free energy / mole of i

ni = number of moles of i

DG = (SuiGi)products - (SuiGi)reactants

Where: ui = the stoichiometric coefficient

Gi= the free energy per mole

a) DG < 0, and GT decreases as the reaction proceeds, the reaction is spontaneous.

b) DG = 0, the reaction is at equilibrium, and GT is at a minimum

c) dG > 0, the reaction is not spontaneous as written, but proceeds in the opposite direction.

DG = DG + RT lnQ

Where: DG = (SuiGi)products - (SuiGi)reactants


Gi = free energy per mole at standard state.

      1. Free energy of formation:
      1. Free Energy of Reaction:

DG = (SuiGf,i)products - (SuiGf,i)reactants

Gi = DGf,i + RT ln {i}

Where {i} is the effective concentration or activity of species I

        1. Example Calcium carbonate:

Cgraphite + 3/2 O2 + Ca ---> Calcite DGf = -269.78 kcal/mol


Cgraphite + 3/2 O2 + Ca ----> Calcite DGf = -269.78 kcal/mol
H+ ------> H+ D Gf = 0
Ca-----> Ca++ DGf = -132.18
H+ + C + 3/2 O2 ------> HCO3- DGf = -140.31

DG = (SviGf,i)products - (SviGf,i)reactants
3 + H+ --------> Ca++ + HCO3- Gr = -2.71


Reacting 1 mole of acid with one mole of calcite releases free energy, so the reaction is spontaneous.

    1. Activity vs. Concentration


The value of the activity of a substance is dependent on the choice of standard state conditions, or the conditions that result in unit activity.

  1. For ions and molecules in solution, (i) is related to the molar concentration [i], by (i) = gi[i], where g is the activity coefficient. As the solution becomes dilute gi approaches 1.
  2. For the solvent in a solution (i) = giXi, where Xi is the mole fraction. As the solution becomes more dilute, gi approaches 1. The activity generally is assumed to be 1 for the dilute solutions of concern here.
  3. For pure solids or liquids in equilibrium with a solution, (i) = 1.
  4. For gases in equilibrium with a solution, (i) = gi Piwhere Pi = is the partial pressure of the gas in atmospheres. As the total pressure decreases, gi approaches 1. When reactions take place at atmospheric pressure, the activity of a gas can be approximated closely by its partial pressure.

6 For mixtures of liquids, (i) = Xi

Returning to the example:

DG = [DGf,HCO3 + RTln{HCO3-} + DGf,Ca++ + RTln{Ca++}] –

Collecting Terms:

- [DGf,H+ + RTln{H+} + DGf,CaCO3 + RTln(CaCO3)]

W can state

and since the activity of a solid is unity:

DG = DG + RT ln Q

When the value of Q is identical to that of the equilibrium constant K, and DG = 0,

0= DG + RT ln K; DG= -RT ln K

When not at equilibrium, substituting we can get

DG = -RT ln K + RTlnQ

= +RT ln Q/K

when Q/K>1, not spontaneous

Q/K =1, equilibrium

Q/K < 1, spontaneous

      2. Example 3-1, S&J
      3. Determine the equilibrium constant for the dissociation of water at 25C:

        H2O <----> H+ + OH-

        1. Find the DGf of the reactants and products:

        Species DGf,

        H2O -56.69

        H+ 0

        OH- -37.60

        2. Sum the energies of formation to get a standard free energy of reaction

        DG = (1)(0) + (1)(-37.6) - (1)(-56.69) = 19.09 kcal

        19.09 = -RTlnK

        K = [H+]{OH-] = 10-14

        3. Use reaction rates to find equilibrium:


        H+ + OH- <----> H2O


        with k1 = 1.4x1011 liter/mole sec

        k2 = 2.5x10-5 /sec

        These are the fastest rates known in aqueous solutions, where the reaction is essentially instantaneous, with the overall rate limited by the rate of diffusion of reactants.

        Since the reaction is elementary, we can write:

        Forward rate = k1[H+][OH-]

        Reverse rate = k2[H2O]

        Since at equilibrium, the forward rate = reverse rate:

        k1[H+][OH-] = k2[H2O]

        k2[H2O]/k1 = [H+][OH-] (mole2/liter2)

        Substituting for k's, and using 5.5 moles/l for water concentration:

      5. Chemical Potential

a) The chemical potential is the molar free energy G/ni, and describes the system for components not in the standard state, especially for components at concentrations other than 1 molar.

mi = (dGi/dni)T,P

b) Example: Multi Phase system:

1) When two or more phases coexist at equilibrium, the chemical potential of all components in the system must be identical in each phase

at equilibrium, mA = mB


if mA is greater than mB, than some reaction will occur to bring the system to equilibrium, ie mA = mB, resulting in the release of energy (-DG). Usually, this means the transfer of some component i from A to B.

Any combination of phases will do.

    1. Temperature And Pressure Dependence OF K, G, H
    2. The driving force of a reaction is DG, as an expression of useful work (non-PV work). This is comprised of 2 components, enthalpy and entropy. At equilibrium:

      DG = DH - TDS

      So the magnitude and direction of the driving force depends on the magnitude and direction of H and S. Neither by themselves controls the reaction.

    3. Enthalpy And The Temperature Dependence Of K

a) Endothermic = positive DH

b) Exothermic = negative DH

DH can be expressed in terms of G, S, and T as: DH = DG + T DS and represents the TOTAL energy of the system, both usable and unusable. At equilibrium, DH=+TDS

DHo = S (DHproducts) - S (DHreactants)


dG = - SdT + VdP

Which at constant pressure can be written as

so the change in free energy with temperature is related to entropy. For a finite change in state:

divide through by T2:


giving the Gibbs-Helmholtz equations


By differentiating the equation: DG = -RT lnK, we obtain

an equation called the Van't Hoff Isochore

So the equilibrium constant varies with temperature as a function of the Enthalpy of the system (Assuming dH does not vary with temperature.).


Plotting log K against 1/T results in a line whose slope is --DH/2.3R

a) For an exothermic reaction, K must decrease as the temperature increases....


    1. Ideal And Non Ideal Solutions
      1. Chemical Potentials In Phases:

Chemical Potential: G = m = m(T,P)

m = m(T)

      1. 2.Chemical potential of an ideal gas.
      2. For n moles of pure gas at constant temperature

        dG = VdP

        From the ideal gas law:

        V = nRT1/P


        dG = nRT dP/P


        dG = nRTdlnP

        Integrating from a standard pressure P to a pressure P yields

        G-G=nRTln P/P

        Dividing by n

        G/n - G/n = RTln P/P

        Since the chemical potential for a one-component phase is G/n:

        m(g) - m(g) = RTln P/P

        When P = P, m(g) = m(g), or

        mi(g) = mi(g) + RT lnP/P

        If P = 1

        mi(g) = mi(g) + RT lnP

        and in an ideal gas mixture:

        mi(g) = mi(g) + RT lnPi

        where the value of m at 1 atm is m standard state, and Pi is the partial pressure of component i. If xi is the more fraction of component i: Pi = XiP, leading to:

        mi(g) = mi(g) + RT lnP + RTln Xi

      3. Chemical potentials in ideal solutions:

a) This closely follows ideal gas mixtures: (159)

mi(sol) = mi(sol) + RT lnXi

Where: Xi = the mole fraction of i in solution

mi(sol) = mi(g)

mi(sol) + RT ln Xi = mi(g) + RT lnPi


1) This equation states that the vapor pressure of component i is proportional to its mole fraction in a solution behaving ideally. This is the base equation for Raoults and Henrys law.:

      1. Raoult's Law:
      2. The ideal gas vapor pressure of an ideal solution component is equal to the vapor pressure of the pure liquid component multiplied by the mole fraction of the component in the solution, or:

        Pi = XiPi

        where Pi is the partial pressure,

        Xi is the mole fraction, and

        Pi is the partial pressure of pure component i.

        and when Xi = 1:

        mi(sol) = mi(g) + RT lnPi

        Thus the standard state chemical potential for the ideal solution component depends on the ideal gas state and the convention that Xi =1 is the standard state for the component in solution.

        --For real solutions, Raoult's law behavior is approached as Xi ---> 1.


      3. Henry's Law:


For a dilute solution the ideal gas vapor pressure of a volatile solute is proportional to its mole fraction in the solution; that is, escaping tendency of the solute molecule is proportional to their mole fraction, ie:

Pi = KiXi

This means that Ki is the slope of the plot of Pi vs Xi for sufficiently dilute solutions, or where solute-solute interactions are negligible.. So:

For real solutions, Henry's law behavior is approached as Xi approaches 0.

Important: mi(sol) for a Henrys law solute reflects the properties of a dilute solution, and for a Raoults law component it reflects the properties of a pure component, in other words, the solvent.

    2. Nonideal Behavior Of Ions And Molecules In Soluitons:
      1. For Real Solutions:
      2. 1) The term activity is used for idealized concentration, vs actual concentration:

        mi = mi + RT lnai


        ai = giXi

        giving the chemical potential in a real solution:

        mi = mi + RTlngiXi


        mi = mi + RTlngi+ RTlnXi

        Where the expression RTlngi is the partial molar free energy of the interactions that occur in nonideal mixtures.

      3. Volatile Solution Components

If mi(sol) = mi(g) then:

mi(sol) + lngiXi = mi(g) + RT ln Pigfi

If the pressure of the vapor is low enough so that gfi = 1, the solution-phase activity coefficient is:

g1 = P1/P1

a1 = g1X1 = P1/P1

    2. Ionic Strength Effects On Equilibrium:


ai = gi Xi or {i} = g[i]

--The activity coefficient becomes less than one and the chemical equilibrium is affected, forcing the rewriting of the mass action equation to:

This describes the magnitude of the interionic effects, or the intensity of the electric field in a solution.

      1. Example 3-6:

Compute the ionic strength of a solution containing the following concentration of ions:

[Ca2-] =10-4M, [CO32-] =10-5M, [HCO3-] =10-3M, [SO42-] =10-4M, [Na+] =10-3M

m = ([10-4 x 22]+[10-5x22]+10-3x12]+10-4x22]+[10-3x12])

m = 1.43x10-3

I ~ 2.5x10-5xTDS; I = 1.6x10-5 x specific Conductance


      1. DeBye-Huckel limiting law:
      2. This is the simplest equation, and is good up only up to I = 0.005

        Based on the effect ionic interactions should have on free energy. In an ionic solution, positive ions will tend to have a cloud of negative ions around them, and negative ions a cloud of positive ions.

        If it is assumed that the ions are point charges, the interactions are entirely electrostatic, and the ions around any particular ion follow a Boltzmann distribution.


      3. Extended DeBye-Huckel approximation of the DeBye-Huckel limiting law,


A = a constant that relates to the solvent, and varies with temperature. (A = 1.82 x 106(eT)-3/2, e = dielectric constant.)

A = 0.5 at 15C; 0.509 at 25C

B = a constant that relates to the solvent and varies with temperature. B=50.3(eT)-1/2

B = 0.326x108 at 15C; and 0.328x108 at 25C.

ai =a constant that relates to the diameter of the hydrated ion. For monovalent ions except H+ this is about 3-4 x 10-8

Zi = the charge of the ion

      1. The Guntelberg approximation

This approximation is useful in solutions of mixed ions with varying charges, since the limiting law is not satisfactory with mixed charge types, and is useful up to I = 0.1.


      2. Davies equation
      3. This is best at higher ionic strengths; to about I = 0.5m

      4. Electroneutrality:
      5. All solutions are charge balanced, with the meq of anions exactly balanced by the meq of cations. The standard equation is:

        The usual QA cutoff is +/- 5% on the CB, or with the alternative equation, 10%. This is often exceeded with very dilute waters and brines.


      6. Relation of ionic strength to TDS
      7. m = 2.5 x 10-5 x TDS

        m = 1.6 x 10-5 x specific conductance (mmho/cm)

      8. Activity coefficients of nonelectrolytes in aqueous solutions:

This is not as well developed as for electrolytes, and is based on empirical equations such as

log g = ks m


        1. Example: dissolved oxygen in NaCl:

ks = 0.132
m = 0.05
log g = 0.132 x 0.05 = 0.0066
g = 1.02
so as the ionic strength increases, the activity coefficient increases.




1. Mass Action aA + bB ---> cC + dD

2. Free Energy G = H - ToS

DG = (SuiGi)products - (SuiGi)reactants

DG = DH - TDS (at constant T,P)

DG = DG + RT lnQ

0 = DG + RT ln K


DG= -RT ln K

DG = -RT ln K + RTlnQ

= +RT ln Q/K

3. Temperature effects


4. Gas Laws Raoult Pi = XiPi

Hanks Pi = KiXi

5. Activity mi = mi + RT lnai

and ai = giXi

6. Activity Coef. ai = gi Xi or {i} = g[i]