Back to Ron's home page

Geochemical Thermodynamics Lecture Notes for Ron Stoessell

The course is concerned with predicting the direction of chemical reactions and predicting the pressure and temperature conditions for chemical equilibrium. In geology, these reactions take place on the earth's surface and within the interior of the earth. To do this, we must be able to define the lowest possible energy of a combination of reactants and products at the pressure and temperature of interest. Please note that although we may be able to predict the direction a chemical reaction will take, we will not be able to predict the chemical kinetics governing the reaction. The actual attainment of equilibrium depends on the kinetics and is more likely to occur with increasing time and at higher temperatures.

Energy units

Energy has units of (force)(distance) or (mass)(acceleration)(distance). An erg is one g cm2/sec2. A joule is one kg m2/sec2 or 107 ergs. The watt is one joule sec. The heat energy unit, a calorie is equivalent to 4.184 joules, 4.184 x 107 ergs, and 1.162 x 10-6 kilowatt hr.

Definition of different types of systems.

A system is a definable part of the universe and can be either open, closed, or isolated.

An open system exchanges both matter and energy with its surroundings, e.g., the space within an open beaker.

A closed system can only exchange energy with its surroundings, e.g., the space within a closed metal container. It has diathermal walls through which heat can pass.

An isolated system cannot exchange energy or matter with its surroundings, e.g., the space within a thermo. It has adiabatic walls through which heat cannot pass.

Definition of intensive and extensive variables.

Intensive variables do not depend on size of the system, e.g., temperature (T, in degrees Kelvin in this study), pressure (P, in bars in this study) and density (mass/unit volume); whereas, extensive variables do depend upon the size of the system; e.g., volume (V), total mass.

Definition of a phase.

A phase is a part of the system having uniform properties, e.g., a gas phase, an aqueous solution, a magma, a mineral.

Definition of reversible and irreversible processes.

A reversible process is one that occurs under equilibrium conditions. The reversible process is never attained in the real world but is approximated by a series of successive near-equilibrium small changes such as system expansion in which the opposing pressure is only slightly less than the system pressure or heat flow down a near zero temperature gradient.

An irreversible process is any natural process such as expansion into a vacuum, heat flow down a large thermal gradient, the flow of electricity, diffusion of matter.

Internal energy of a system.

For a closed system of constant composition whose potential energy and kinetic energy remains constant, the change in internal energy dU will be the result of the exchange of heat and work done on or by the system.

dU = dq + dw (1)

where dq and dw are the heat exchanged across the boundary of the system and the work done on or by the system. dw includes all work, e.g., pressure volume work (-PdV), electrical work (dwe), etc. dq and dw are positive terms if energy is added to the system and negative if energy is removed from the system.

Separating dw into PV work and other work and assuming no other work produces:

dU = dq - PdV (2)

which is the usual statement for the change in U with heat exchange and pressure volume work on or by a closed system.

As the closed system moves from one state to another state (defined by any two of the variables P, V, and T), the internal energy will change. A system may move from one state to another along different paths; however, the overall energy change will be the same regardless of the path. U is a function of state so the integral of dU will provide the change in U, U, between two states. Because heat and mechanical energy can be converted back and forth, dq and -PdV are not functions of state (nor is dwe), i.e., there are many different paths leading between any two states.

The main application of thermodynamics to geology is to determine if a chemical reaction will occur and which direction it will go. If a chemical reaction occurs within a closed system, the U is considered to be a function of not only the heat exchange and the work done on or by a system but also the composition of the system. A summation term is added to dU in equation (2) to account for changes due to a chemical reaction. (Note this is a point that your instructor does not understand - since energy cannot be created or destroyed in a chemical reaction (1st law below), it follows that only the exchange of energy with the system's environment should account for the change in internal energy of a closed system, not internal chemical reactions?)

The change in internal energy due to a chemical reaction is just the internal energy of the products formed minus the internal energy of the reactants destroyed. The computation of dU for a closed system with a chemical reaction is

dU = dq + dw + (U/ni)dni (3)

(U/ni)dni is a summation over all reactants and products in a chemical reaction where ni refers to the moles of the ith reaction component and dni is the change in moles of this component in the reaction. Reactants have negative dni and products have positive dni.

If dq and dw are both zero, equilibrium in the chemical reaction occurs when dU = (U/ni)dni = 0

The zeroth law of thermodynamics states that if system A is in thermal equilibrium with system C and if system B is also in thermal equilibrium with system C than system A and system B are in thermal equilibrium, i.e., have the same temperature.

The first law of thermodynamics states that energy cannot be created or destroyed by ordinary chemical changes but can be converted from one form of energy to another. The first law does not hold if nuclear reactions occur in which matter and energy are converted.

The interesting point is that although energy can be converted from one form to another, it is impossible to convert all heat energy into work energy. In the process of doing work, some of the heat taken from one reservoir must be transferred to another reservoir. This is because heat must flow in order for work to occur. If q1 heat is gained by the system from reservoir 1 at temperature T1 and used to do work during which heat is lost to reservoir 2 at temperature T2, the minimum amount of heat -q2 lost to reservoir 2 is (q1/T1)T2 where T is in absolute degrees Kelvin. In other words, -q2/T2 = q1/T1. The negative sign for -q2 makes the heat loss positive, because heat leaving the system is always counted as negative.

The above relationship can be calculated from a Carnot cycle which traces a complete path such that a system comes back to its initial state. Typically (in four steps), the system expands isothermally at T1 in which q1 heat is absorbed to maintain a constant temperature, followed by an adiabatic expansion (without heat exchange so the temperature decreases from T1 to T2), followed by an isothermal contraction at T2 in which -q2 heat is evolved to maintain a constant temperature, followed by an adiabatic contraction (without heat exchange so the temperature rises back to T1) back to its initial state. This minimum exchange of heat occurs only if the process is reversible (near equilibrium).

For irreversible processes, -q2/T2 > q1/T1. The heat fraction available for work for reversible processes is

[q1-(-q2]/q1 = 1 - T2/T1 = (T1 - T2)/T1.

The higher T1 is compared to T2, the greater the heat fraction available as work. This is why modern engines extract work at high temperatures (T1) compared to the atmospheric temperature (T2).

For a reversible process in a closed system, the ratio of dq/T is called dS, the change in entropy in the system. System A loses entropy as heat flows into the system B at temperature T and system B gains entropy at temperature T. For a reversible process, the overall change in entropy is zero. If the process is irreversible, dS > dq/T, and the overall change in entropy is positive. S like U is a function of state. The change in S from one state to another can be evaluated by integrating dS between state I and state II. However, the change in entropy can only be evaluated for processes that approach reversibility, because otherwise it generally cannot be calculated.

The above definition of entropy is a statement of the second law of thermodynamics which defines the direction of heat flow, from a hotter object to a colder object. Consider the flow of heat q from system A to system B which are at temperatures TA and TB, respectively. The flow of heat is an irreversible process so the overall change in entropy must be positive. If TA and TB, are close together, so that the heat flow approaches reversibility, we can compute the overall change in entropy as

q/TB - q/TA. For this change to be positive, TB < TA, i.e., heat flowed from the hotter to the colder object

Using the definition of entropy for closed system undergoing reversible processes, with only PdV work, heat exchange and an internal chemical reaction in equation (3), we have

dU = TdS - PdV + (U/ni)S,V,nj dni (4)

U/ni)S,V,nj was defined by Gibbs (the father of thermodynamics) as the chemical potential (i).

The usefulness of equation (4) is that at constant S and V:

dU = i dni (5)

So if we can evaluate the change in U at constant S and V, we can test for chemical equilibrium which occurs when dU = 0. Unfortunately, we need an equation to evaluate chemical equilibrium at constant T and P which are the two variables easiest to hold constant in the natural earth environment. So the above equation is not very useful and needs to be modified.

Because U is a function of state, a total differential equation can be written, using partial derivatives, describing dU as a function of its three variables: S, V, and ni.

dU = (U/Si)V,nj dS + (U/Vi)S,nj dV + (U/ni)S,V,nj dni (6)

Comparison with dU = TdS - PdV + i dni shows that

T = (U/Si)V,nj, (7a)

-P = (U/Vi)S,nj, (7b)

and we have already given by definition that

i = (U/ni)S,V,nj (7c)

The definition of U can be modified to define new functions that contain i which can determine chemical equilibrium at constant T and P (G, Gibbs free energy), constant T and V (A, Helmholtz free energy), and constant S and P (H, enthalpy). These functions are defined below and evaluated by substituting in the definition of dU in equation (4).

H = U + PV, dH = dU + PdV + VdP = TdS + VdP + i dni

(8a, 8b, 8c)

A = U - TS, dA = dU - TdS - SdT = -SdT - PdV + i dni

(9a, 9b, 9c)

G = U+PV-TS, dG = dU + PdV+VdP - TdS-SdT = -SdT + VdP + idni

(10a, 10b, 10c)

What are the natural variables for U, H, A, and G?

U(S,V,ni), H(S,P,ni), A(T,V,ni), and G(T,P,ni)

What are the coefficients in each energy function equal to in terms of partial derivatives?

T = (U/S)V,nj, -P = (U/V)S,nj, i = (U/ni)S,V,nj

(11a, 11b, 11c)

T = (H/S)P,nj, V = (H/P)S,nj, i = (H/ni)S,P,nj

(12a, 12b, 12c)

-S = (A/T)V,nj, -P = (A/V)T,nj, i = (A/ni)T,V,nj

(13a, 13b, 13c)

-S = (G/T)P,nj, V = (G/P)T,nj, i = (G/ni)T,P,nj

(14a, 14b, 14c)

Because the order of differentiation is not important in taking second partial derivatives, new relationships are present, e.g.,

if df = M dx + N dy + O dz, then (M/y)x,z = (N/x)y,z

The equalities that result from using the first two terms of the dU, dH, dA, dG equations (i.e., closed systems without chemical reactions) are called the Maxwell relations?

(T/V)S,nj = (-P/S)V,nj from dU expression (15a)

(T/P)S,nj = (V/S)P,nj from dH expression (15b)

(-S/V)T,nj = (-P/T)V,nj from dA expression (15c)

(-S/P)T,nj = (V/T)P,nj from dG expression (15d)

We need to define what is meant by enthalpy H, entropy S, and Gibbs free energy G. Let's begin with enthalpy.

H = U + PV, so dH = TdS + VdP + i dni

For a closed system with reversible processes and without a chemical reaction, dH = TdS + VdP. And at constant pressure, dH = TdS = dq, the heat exchanged with the system's surroundings.

If a reversible chemical reaction occurs within this closed system within a constant pressure calorimeter, the change in enthalpy resulting from the reaction Hr will be the heat released or absorbed due to the chemical reaction. A negative Hr means heat is released and the reaction is exothermic. A positive Hr means heat is absorbed and the reaction is endothermic. The Hr can be calculated from summing the molar enthalpies of formation, Hf,i, of the reaction components using the reaction coefficients, ni (positive for a product and negative for a reactant):

Hr = ni Hf,i (16)

where Hf,i is change in enthalpy in the reaction forming the ith component from the elements. Use of the convention "formation from the elements" is explained later but follows from the inability to measure an absolute energy. As shown later, for the ith component, energy values such as enthalpy, internal energy, Helmholtz free energy and Gibbs free energy are referenced from reactions forming the component from the elements. Because all chemical reactions are balanced with regard to elements, the contributions from the elements cancel out in the summation term in equation (16). The same can be done for entropy; however, absolute values are available for entropy due to the third law of thermodynamics which is also explained later.

In general, we will never do the calculation in equation (16) unless all the reaction components are in their standard states (also explained below). This is because the ultimate goal is to calculate changes in Gibbs free energy of a reaction and calculating the change in the standard state enthalpy of a reaction is often used to calculate the standard state Gibbs free energy of a reaction. The calculation that we want to do is

Hro = ni Hof,i (17)

where the superscript o indicates a standard state property.

For example, consider the reaction aA + bB = cC + dD

Hro = cHof,C + dHof,D - aHof,A - bHof,B

Suppose we knew Hof,i to calculate Hro at a reference temperature Tref and want to calculate Hro at a different temperature. We have to move Hro from Tref to T.

It is conceptionally simpler to consider several closed systems, each containing one mole of one of the reactants or products. The change in Hof,i in each of these one component systems as T changes is computed using the definition of molar heat capacity of a component in its standard state at constant pressure, Cpio.

Cpio = (dqi/dT)P (18)

The reactant and product heat capacities are combined into a summation for the overall change in heat capacity in the reaction, Cpro.

Cpro = ni Cpio (19)

The change in Hro as T changes at constant pressure, is from

dHr = TdSr = dqr. By substitution,

dHro = CprodT (20)

And by integration, Hor,T,P = Hor,Tref,P + TrefTCprodT (21)

The expression used in equation (21) for Hor,T contains the reaction summation in equation (19) of heat capacities of the individual reaction components in Cpro. These are the heat capacities of the components, not the reaction heat capacities of the components being formed from the elements. We can use the individual heat capacities because the contributions from the elements will cancel out in a balanced reaction.

The standard state molar enthalpies of formation of all substances, Hof,i, have been tabulated in reference books at 25oC (298.15oK) and one bar. They represent the change in enthalpy in forming a substance in its standard state from the elements in their standard states, all at 25oC and 1 bar. They are called the molar or molal (if a solute in a an aqueous fluid) standard state enthalpies of formation. For solids and water, the standard state is the pure stable form for the solid and the pure liquid for water. For gases, the standard state is the pure gas behaving as an ideal gas. For dissolved components in water, the standard state is a one molal concentration (one mole per kilogram of water) behaving as though the dissolved component were at infinite dilution. The standard states of solids, water, and dissolved salts will move with changes in P and T; and the standard state of gases move with T; however, the standard state energies of formation are only tabulated at 25oC and one bar.

Since only changes in enthalpies can be measured, not absolute values, the tabulated values are actually relative values, based upon the arbitrary convention of reporting them relative to the formation from the elements in their standard state. This convention results in a zero standard state enthalpy of formation of each element in its standard state. Hence, the standard state enthalpy of formation of a substance is the measured enthalpy in a reaction forming it in its standard state from the elements in their standard states. This same convention is followed for U, G, A, and S; however, absolute values of S can be calculated, as shown later.

One important note is that in this course, standard state energies of formation for U, G, A, S, and H are only used at 25oC and 1 bar. At other PT values, we will use apparent standard state energies of formation, indicated by an * rather than a o. An apparent standard state energy of formation refers to the change in energy in forming the substance at a particular P and T in its standard state from the elements in their standard states at 25OC and 1 bar. Hence at pressures and temperatures different from 25oC and 1 bar, the apparent standard state values of the elements in their standard states will not be zero.

Using apparent standard state energies of formation of the reaction components to calculate the standard state energy change in a reaction gives the same energy change as using standard state energies of formation of the reaction components. This follows because the contributions from the elements cancel out in a balanced reaction. Using apparent values is simpler because we do not have to change the energies of the elements from 25oC and 1 bar. However, you must be careful not to mix standard state energies of formation with apparent standard state energies of formation in computing the change in energy in a reaction not at 25oC and 1 bar.

Do a sample reaction of Hor,25 C for calcite and quartz reacting to produce wollastonite and carbon dioxide.

CaCO3 (calcite) + SiO2 (quartz) = CaSiO3 (wollastonite) + CO2 (gas)

The standard state enthalpies of formation at 25oC and 1 bar in calories per mole are listed below and the heat capacity of each component can be described by a Maier Kelly 3 term expansion.

Cpio = ai + biT +ci/T2 (22)

The coefficients for the Cpio equation for each component are also listed below. Calculate Hro at 100oC (273.15oK) and one bar, using equations (19), (20), (21), and (22).

Hof,i Cpio

25oC, 1 Bar ai bi ci

cal/mole cal/mole/oK cal/mole/oK2 caloK/mole

CaCO3 -288,772. 24.98 0.00524 -620,000.

SiO2 -217,650. 11.22 0.00820 -270,000.

CaSiO3 -389,810. 26.64 0.0036 -652,000.

CO2 - 94,054. 10.57 0.00210 -206,000.

We can repeat the calculation for the entropy change. Remembering the definition of entropy for a reversible process

in a closed system,

TdSr = dqr (23)

Similar to Hro, Sro at constant temperature and pressure can be calculated from the molar entropies, Sio, of the reaction components using the reaction coefficients, ni:

Sro = niSio (24)

Again consider the reaction aA + bB = cC + dD

Sro = cSCo + dSDo - aSAo - bSBo

Suppose we knew Sio to calculate Sro at a reference temperature Tref and want to calculate Sro at a different temperature. We have to move Sio from Tref to T at constant P.

Again, it is conceptionally simpler to consider several closed systems, each containing one mole of one of the reactants or products. The change in Sio in each of these one component systems as T changes is computed using the definition of molar heat capacity of a component in its standard state at constant pressure, Cpio, given in equation (18). Again, the reactant and product heat capacities are combined into a summation for the overall change in heat capacity in the reaction, Cpro, as in equation (19). The change in Sro (in equation 24) as T changes at constant pressure, is computed using equation (23). By substitution,

dSro = [Cpro/T]dT. (25)

By integration Sor,T = Sor,Tref + TrefT[Cpro/T]dT (26)

In the above exercises, we have computed Hor,T and Sor,T using Cpro. Instead we could have used Cpio and integrated the individual heat capacity expressions to move Hof,i and Sof,i from Tref to T and then combined them in a summation for the chemical reactions. However, mathematically its simpler to combine them in one summation expression and then integrate that one expression.

The reference molar or molal entropies of all substances in their standard states, called third law entropies, are tabulated in reference books at the reference temperature and pressure of 25oC (298.15oK) and one bar. These are not molar or molal standard state entropies of formation but are for the actual component. The o superscript indicates that they are in their standard states. The standard state conditions are as previously described for enthalpy.

What are third law entropies? These are molar or molal entropies calculated on the basis of the third law of thermodynamics which states that at absolute zero, the entropy of any perfectly crystalline substance is zero. This is a postulate, like the first and second laws, based on experiments. If the substance is not perfectly crystalline at absolute zero, then the third law entropy is equal to the configurational entropy (due to disorder) which is calculated from Boltzmann's law.

S = k ln (27)

k is Boltzmann's constant which is equal to R (the gas constant in the ideal gas law) divided by No, Avogadro's number of particles in a mole of particles. is the total number of distinguishable configurations of one mole of atoms or molecules in the substance. We will use the binomial theorem later on to calculate in solid solutions to obtain configurational entropy.

However, for now, the important point is that the third law allows either a zero or calculated value of any substance at absolute zero. The tabulated values for Soi at 25oC (298.15oK) and 1 bar for the ith substance have been obtained by integrating at constant pressure from absolute zero (equation 26 for one component).

Soi,298.15oK = Soi,OoK + [Cpoi/T]dT

We will actually use Sio like Hof,i as part of an overall calculation of Gor,T, the change in Gibbs free energy of a reaction at a particular temperature and pressure. As discussed later, the Gibbs free energy of a substance in its standard state, io, like enthalpy, is reported in terms of formation from the elements, i.e., as the change in Gibbs free energy in forming the substances from the elements in their standard states. (This is because there is no way to measure an absolute Gibbs free energy of a substance, only the change in Gibbs free energy in a reaction.) Hence, third law entropies cannot be used to directly calculate io because they are not the entropies of formation from the elements. However, entropies of formation from the elements can be calculated from a reaction forming the substance from the elements, using the tabulated third law entropies.

In actuality, entropies of formation from the elements are rarely computed because the third law entropies can be used to directly calculate Sor,T which is then used calculate Gor,T. Third law entropies will produce the same value for Sor,T, as calculated with entropies of formation, because the contributions from the elements are the same for both the products and reactants and cancel out in a balanced chemical reaction.

Do a sample calculation of Sor,25 C for the reactants and products in their standard states: calcite and quartz reacting to produce wollastonite and carbon dioxide. using the molar third law entropies in the table below.

CaCO3 (solid) + SiO2 (solid) = CaSiO3 (solid) + CO2 (gas)

The heat capacity of each component can be described by the Maier Kelly three term expansion where T is in degrees Kelvin (oC + 273.15) of the form of equation (22). The coefficients for the Cpio equation for each component were listed above in the discussion on enthalpy.

Calculate Sor,T at 100oC and one bar using the third law entropies listed below, the heat capacity coefficients in the previous table, and equations (19), (22), (24), and (26).

Sio Sio

25oC, 1 Bar 25oC, 1 Bar

cal/mole/oK cal/mole/oK

CaCO3 22.15 C 1.362

SiO2 9.88 O2 49.003

CaSiO3 19.60 Si 4.493

CO2 51.072 Ca 9.90

Values of third law entropies Sio of the elements that make up the above substances are also listed above for 25oC and one bar. Calculate entropies of formation from the elements at 25oC and one bar for the above substances. Use these entropies of formation to calculate Sor,25oC for the reaction on the previous page and compare with Sor,25oC computed using the third law entropies.

Those procedures for calculating reaction changes in standard state enthalpies and entropies with temperature can be used to compute standard state Gibbs free energies of reactions. In addition, the change with pressure can also be calculated.

As previously presented in equations (10a, 10b, and 10c), the Gibbs free energy is defined by

G = U+PV-TS, and dG = -SdT + VdP + idni leading to

-S = (G/T)P,nj, V = (G/P)T,nj, i = (G/ni)T,P,nj

i represents the molar or molal Gibbs free energy associated with the ith component. This term shows the overall increase in Gibbs free energy at constant temperature and pressure and all other components, when the component is added to or removed from the system in a chemical reaction (and also in a open system). By comparing the combined Gibbs free energy associated with the products and that associated with the reactants, we can determine if they are identical, i.e., at reaction equilibrium.

dG = idni (28)

To calculate the change in Gibbs free energy in a reaction at a temperature and pressure of interest, add the i (for that PT point) for all the products and subtrack the i of all the reactants, each multiplied by its reaction coefficient.

Gr = nii (29)

At equilibrium the change in Gibbs free energy of reaction Gr is zero. If the change is negative the reaction goes to the right and if positive, to the left. The reaction direction is towards lowering the Gibbs free energy of the system.

The trick in testing reaction equilibrium is to determine the correct value of i to use in the test. Values of io,(Gof,i) the standard state molar or molal free energy of formation of each substance, are listed in reference books at 25oC and 1 bar. The standard states are the same as previously described for enthalpy and for entropy. And the concept of a standard state Gibbs free energy of formation carries the same meaning as previously discussed for the standard state enthalpy of formation. io is the Gibbs free energy change in forming the substance in its standard state from the elements in their standard states at a particular pressure and temperature.

Gro = niio (30)

If we need to calculate Gro at a PT point different from 25oC and 1 bar, we use

Gro = nii* (31)

where i*, the apparent standard state Gibbs free energy of formation, is the Gibbs free energy change from forming the substance in its standard state at the PT point of interest from the elements in their standard states at 25oC and 1 bar. actually only move the substance, not the elements. Because the contribution of the elements to Gro cancels out in a balanced reaction, it doesn't matter if apparent standard state values are used so long as they are not mixed with true standard state values in calculating Gro. Note that by definition, i* and io are identical at 25oC and 1 bar.

One point of possible confusion occurs when a reaction contains constituents formed from elements of which some but not all are gases in their standard states. Gas standard states are limited to one bar; however this restriction is not shared by the standard states of other elements that are solids or liquids. In a balanced reaction, the contributions from the elements will still cancel out because the standard state for each element (used in the reaction of formation from the elements of each constituent) is identical on both sides of the reaction.

Again consider the reaction aA + bB = cC + dD

Gr = cC + dD - aA - bB

The first step in calculating Gr, is to calculate Gro

Gro = cCo + dDo - aAo - bBo

Suppose we knew io to calculate Gro at Tref and Pref and want to calculate Gro at a different temperature and a different pressure. We want to move i* from Tref to T (at Pref) and then from Pref to P (at T).

There are two ways to move apparent standard state Gibbs free energies through PT space.

First Method

Remember that G = U + PV - TS, and that H = U + PV.

Hence, by substitution, G = H - TS and dG = dH - TdS - SdT.

At constant T,

dG = dH - TdS (32)

and at any PT point,

GP,T = HP,T - TSP,T (33)

If the reactants in a reaction form one state at a particular P and T, and the products form another state at the same P and T, then the difference in free energy between the two states will be the free energy of reaction.

Gr,P,T = Hr,P,T - TSr,P,T (34)

If the reactants and products are present in their standard states.

Gor,P,T = Hor,P,T - TSor,P,T (35)

We can write a similar expression for each reactant and product component in the reaction. Visualize a system composed of a single component i in its standard state at a PT point and use equation (36) in equation (31) to calculate Gor,P,T.

i* = H*f,i - TS*f,i (36)

For the above expression, S*f,i is an apparent standard state entropy of formation from the elements, not a third law entropy, to be consistent with i* and H*f,i. However, since we are going to use i* in a balanced chemical reaction, the contributions from the elements will cancel out, so we can use third law entropies to calculate Sro for calculating Gro.

Moving i* with temperature from Tref to T at constant Pref

H*f,i and S*f,i in equation (36) can be rewritten as their values at 25oC (Tref) and the integration of the change in those values up to temperature T, i.e.,

H*f,i,T,Pref = Hof,i,Tref,Pref + (Hof,i,Pref/T)dT (37)

S*f,i,T,Pref = Sof,i,Tref,Pref + (Sof,i,Pref/T)dT (38)

Substituting equations (37) and (38) into equation (36) gives

*i,T,Pref = Hof,i,Tref,Pref + (Hio/T)Pref dT

- T[Sof,i,Tref,Pref + (Sio/T)PrefdT] (39)

From the previous discussions on enthalpy and entropy, we know that

(Hio/T)Pref = Cpio (40)

and (Sio/T)Pref = Cpio/T. (41)

Substitution of equations (40) and (41) into equation (39),

*i,T,Pref = Hof,i,Tref,Pref + (CpiodT - T[Sof,i,Tref,Pref + (Cpio/T)dT] (42)

Again, note that Soi,Tref,Pref can be used in place of S*f,i,Tref,Pref if we are going to use *i,T,Pref in a balanced reaction (equation 31) to calculate Gor,T,Pref.

Since the intent is to compute Gro, it is simpler to do the summation of the reaction constituents prior to the integration. This is identical to previous problems computing Hro and Sro. Substituting equation (42) into equation (31), yields

Gor,T,Pref = Hor,Tref,Pref + (CprodT

- T[Sor,Tref,Pref + (Cpro/T)dT] (43)

The integrations were previously done for heat capacities expressed in Maier-Kelly algorithms in calculating Hro and Sro .

Moving *i,T with pressure

The above equation moves Gro from Tref to T at Pref. To move Gro from Pref to P at T, we make use of the following relationship previously derived.

(G/P)T,nj = V or dG = VdP

For a system constituent whose standard state moves with pressure (i.e., non-gas component), the equation becomes

d*i,T = Voi,TdP (44)

where Vio is the molar or molal volume of the ith component in its standard state. For a gas component whose standard state stays at 1 bar,

d*i,T = 0

As previously discussed with entropy and enthalpy, a reaction summation is done using the molar or molal volumes of the non-gas reaction constituents at Pref,T to come up with Vro, the volume change of reaction for all non-gas constituents. Again, gases aren't included because their standard state does not vary with pressure. Vro is then integrated from Pref,T to P,T. An alternative would be to first compute i* at P by integration for each component and then do a reaction summation; however, this would require more steps. For the reaction,

dGor,T = Vor,TdP (45)

or Gor,T,P - Gor,T,Pref = Vor,TdP (46)

Adding equations (43) and (46) together gives the overall expression for moving the Gro through PT space.

Gor,T,P = Hor,Tref,Pref + (Cpro)PrefdT - TSor,Tref,Pref

- T(Cpro/T)PrefdT + Vor,TdP (47)

where the volumes of the solids can usually be considered constant. If the Vor,T is not constant then an appropriate expression is used to describe it that can be integrated through P space.

Note that if Cpro is constant with temperature then

Gor,T,P = Hor,Tref,Pref - TSor,Tref,Pref

+Cpor,Pref[T-Tref + T ln(Tref/T)] + Vor,TdP (47a)

Second Method

The second way to move standard state Gibbs free energies through PT space uses the partial derivative of G with T, resulting in a double integral to move through T space. However, the change with pressure is handled as described above.

For a closed system, dG = -SdT + VdP

For a system composed of several components in their standard states, we can write the above equation individually for the apparent standard state chemical potential of each of the individual components making up the system. This follows because the Gibbs free energy is additive and conserved.

d*i = -S*f,idT + V*f,idP

Where S*f,i and V*f,i are the standard state partial molar or molal entropy and volume of formation from the elements for the ith component. We want to move *i through PT space from Pref and Tref.

*i = - S*f,idT + V*f,idP where *i = *i,T,P - *i,Tref,Pref

and then combine these components equation (31) in a reaction to compute Gor,T,P. Note that in the integrations for changes in entropy with temperature and changes in volume with pressure that we can use third law entropies and volumes of the constituents in their standard states because these are apparent values (as indicated by the *).

The computational effort will be less if we combine all the reaction components and integrate them together.

Gor,T,P - Gor,Tref,Pref = - Sor,PrefdT + Vor,TdP

where Sor is integrated with T at constant 1 bar pressure, followed by the integration of Vor with P at constant T.

Following equation (38), Sor can be written as

Sor,Pref = Sor,Tref,Pref + (Sor/T)PrefdT

and remembering that (Sor/T)PrefdT = (Cpor/T)PrefdT. Hence,

Gor,T,P = Gor,Tref,Pref - Sor,Tref,PrefdT

+ [Cpor/T]PrefdT)dT + Vor,TdP

= Gor,Tref,Pref - Sor,Tref,Pref(T-Tref)

- [Cpor/T]PrefdTdT + Vor,TdP (48)

Note that if Cpor,Pref is constant with temperature then

Gor,T,P = Gor,Tref,Pref + Sor,Tref,Pref(Tref-T)

+Cpor,Pref[T-Tref + T ln(Tref/T)] + Vor,TdP (48a)

Sor,Tref,Pref can be computed directly from 3rd law entropies in reference tables using

Sor,Tref,Pref = niSoi,Tref,Pref (49)

and Gor,Tref,Pref = nioi,Tref,Pref (50)

Gor,Tref,Pref can also be computed from standard state enthalpies of formation and third law entropies in reference tables at Tref and Pref, i.e.,

Gor,Tref,Pref = -niTSoi,Tref,Pref + niHof,i,Tref,Pref (51)

[Cpor,Pref/T]dTdT can be integrated by substituting an

appropriate algorithm for Cpor,1 bar, such as the Maier Kelly expression. Note that the T term containing the upper temperature from the first integral is considered a variable and the Tref term is considered a constant in the second integration. For solids, Vor is often considered constant. If Vor is not constant than it also has to be integrated in a double integral similar to that done above for entropy. However, remember that the volumes of gases are not included in Vor.

Calculate Gor,T,P for the example reaction at 100oC and 300 bars pressure. Note that 1 cm3bar = 0.02390 calories. Assume constant volume of the solids. Use the following molar volumes: CaCO3, 36.934 cm3/mole; CaSiO3, 39.93 cm3/mole; and SiO2, 22.688 cm3/mole.

Computation of Gr,T,P and the Equilibrium Constant

Now that we've moved Gor,T,P through PT space, we still have to combine it with the non-standard state properties of the reaction components in order to compute Gr,T,P. To do this we have to substitute the definition of the chemical potential of each reaction component (equation 52) into equation for Gr,T,P in equation (29).

i = io + RT ln ai (52)

where ai is the activity of the ith component.

Gr,Tref,Pref = niio + RT aini (53)

and at P and T other than Pref and Tref

Gr,T,P = nii* + RT ln ( aini) (54)

By convention, the symbol Q is used to represent aini. Q is called the reaction quotient. Using the definition and equation (31) in equation (54),

Gr,T,P = Gor,T,P + RT ln Q. (55)

And at reaction equilibrium,

Gr,T,P = 0 and Q is called K, the equilibrium constant.

so, Gor,T,P = -RT ln K, and using ln x = 2.303 log x,

log K = -Gor,T,P/(2.303RT) (56)

The important point is that the equilibrium constant for any reaction is a standard state property and dependent on how the standard state is defined. The K values used in geochemistry are consistent with the following standard states in which standard state properties of any substance moved from 25oC and 1 bar are moved as apparent standard state properties, i.e., without moving the elements:

solid - pure solid at the temperature and pressure of interest;

liquid - pure liquid at the temperature and pressure of interest;

dissolved component in water - one molal concentration in which the component behaves as though it is at infinite dilution, at the temperature and pressure of interest; and the

gas component - pure gas phase, behaving the ideal gas law, at one bar and at the temperature of interest.

The activity expression of each component is used to correct from the standard state to the actual state of the component, i.e., bring it to reality. If a component is in its standard state, the activity is one, resulting in the activity portion dropping out of the chemical potential expression. The activity expressions that are used with the above standard states are as follows:

solid and liquid components, ai = Xii (57)

however, in a solid solution the Xi may be modified as shown later.

dissolved aqueous component, ai = mii (58)

gas component, ai = PXii (59)

where Xi is the mole fraction of i in the phase and i is its activity coefficient if it is a solid or a liquid component and i is its activity coefficient if it is a gas component, mi is the molality of i in the aqueous phase and i is its activity coefficient. ai for a gas is also called the fugacity or fi.

The fugacity is the vapor pressure of a substance and is defined by the expression

di = RT dln fi (60)

Note that at constant pressure and temperature, we can differentiate equation (52) and eliminate i* because standard states can vary only with T and P for non-gas components and T for a gas component.

di = RT dln ai

If we integrate from a standard state to some other state at constant pressure and temperature,

dln ai = dln fi or ai/aio = fi/fio

where aio is the activity in the standard state which is one. Hence,

ai = fi/fio

For a gas in its standard state, the vapor pressure is one bar, hence fio is one and ai = fi for a gas.

The above list of standard states and activity expressions does not include liquid magma. Igneous petrologists have developed their own conventions, and your instructor doesn't know what is commonly used and accepted in the geochemical literature.

Each activity expression contains a concentration unit, either mole fraction, molality, and an activity coefficient (sometimes calculated and sometimes used as a fudge factor) to modify the concentration unit to produce the correct chemical potential. In addition, the gas activity expression includes the pressure because the gas standard state does not vary with pressure. If an activity coefficient is one, the component is said to behave ideally. For a component in its standard state, its concentration unit would be one, so its activity coefficient would also have to be one, for the activity to be one.

For our hypothetical reaction,

CaCO3 (solid) + SiO2 (solid) = CaSiO3 (solid) + CO2 (gas)

Gr,T,P = Gor,T,P + RT ln[(aCaSiO3)1(aCO2)1(aCaCO3)-1(aCaSiO3)-1]

= Gor,T,P + RT ln[(XCaSiO3)1(XCO2)1(XCaCO3)-1(XCaSiO3)-1

+ RT ln[(CaSiO3)1(CO2)1(CaCO3)-1(CaSiO3)-1 + RT ln P

We can generally assume the solid phases are in their standard states as pure solids with activities of unity, i.e., mole fractions and activity coefficients are also unity. However, a CO2 gas phase at higher pressure and temperature is not going to be pure because it will contain some water vapor and will not behave ideally. We can probably neglect a small fraction of water vapor in the gas phase, letting XCO2 becomes one.

Lets calculate the change in Gibbs free energy of the reaction, Gr,T,P, at 100oC and 300 bars. We have already used equations (47) or (48) to calculate Gor,100oC, 300 bars. Assume the solids are in their standard states, i.e., unit activities and the CO2 gas phase is a pure CO2.

Gr,100oC,300 bars = Gor,100oC,300 bars + RT ln[(CO2)P]

where R, the gas constant has a value of 1.987 cal/oK mole. In order to evaluate the above expression, we have to be able to calculate CO2.

Calculation of Gas Fugacity Coefficients for Pure Gas Phases

The aCO2 is a measure of how CO2 deviates from an ideal gas at the temperature and pressure of interest. For a gas phase containing CO2 at a PT point,

aCO2 = P(CO2)XCO2

and we can write the chemical potential of pure CO2 as

CO2 = *CO2 + RT ln[P(CO2)] = *CO2 + RT[ln P + ln CO2 + ln XCO2]

Holding T constant and taking the partial derivative of the above expression with respect to P at constant composition gives

(CO2/P)T = RT/P + RT(ln CO2/P)T

where the partial derivative of *CO2 was zero because the standard state of a gas doesn't vary with P. Remembering that the first partial derivative of the chemical potential of the ith substance is its partial molal volume VCO2, and substituting into the above expression gives

V-CO2 = RT/P + RT(ln CO2/P)T

Integrate both sides of the equation with respect to P at constant T from Po to P where Po approaches zero so that the gas behaves ideally.

V-CO2dP = (RT/P)dP + RT(dlnCO2)

By combining terms, and writing the integrated form of (dlnCO2)

ln [CO2,P,T/CO2,Po,T] = -(1/RT)[(RT/P) - V-CO2]dP

CO2,Po,T is one since the pressure is low enough for the gas to behave ideally. Therefore,

ln CO2,P,T = -(1/RT)[(RT/P) - V-CO2]dP

where the bracketed term is the difference between the ideal gas volume and the real volume. For a pure gas phase of one component, it is sometimes tabulated in the literature as i. For a pure gas phase, the partial molal volume is the molar volume. The general expression for a gas phase composed of a single i component is

ln i,P,T = -(1/RT)[(RT/P) - V-i]dP (61)

There are several ways to evaluate the integral. An expression for V-i can be used to integrate it as a function of P and T. Experimental data on the molar volume can also be integrated by graphical means, i.e., plot (1/RT)[(RT/P) - V-i] from Po to P at constant T and determine the area under the curve. Note that negative areas have to be subtracted from positive areas. Note also that the value of (1/RT)[(RT/P) - V-i]

will be zero when the gas behaves ideally. If the gas behaves ideally at 1 bar and we are integrating from Po = 1 bar, the above expression reduces to

ln i,P,T = -ln P + (1/RT)V-idP (62)

Use the compressibility z data (PV-/RT) from the attached chart for pure gases to compute the V-CO2 as a function of P from 1 to 300 bars at 100oC (373.15oK). This chart from Lewis and Randall (1961) shows the best fit molar z data for a group of nonpolar gases as a function of reduced variables (T/Tc,i where Tc,CO2 = 304.2oK) and P/Pc,i in atmospheres where Pc,CO2 = 73.7 atm

and 100 bars = 296.1 atm, i.e., 1 bar = 0.987 atm). Plot V-CO2/RT (y axis) versus P in bars (x axis) on graph paper and graphically integrate by determining the area under the curve. The area units will be in term of (delta P) in bars times (delta Vi/RT) in 1/bars. V-i is in cm3/mol and R is 83.144 cm3 bar mol-1 (84.239 cm3 atm mol-1). Use the above equation to determine CO2,P,T at 300 bars and 100oC (373.15oK). Because the data doesn't extend down to 1 bar, you need to extrapolate molar volumes to one bar. Compare the value of CO2,P,T at 300 bars and 100oC with that from the second chart from Garrels and Christ (1965) which directly gives the value in terms of the reduced variables Tc and Pc (also in atm not bars).

Calculation of Activities of Solids in Ideal Solid Solutions

A solid solution is a mixture of components. The total free energy of the solid solution is the sum of the chemical potential of each component multiplied by the number of moles of the component in the solid solution.

Gsolid solution = (nii) (63)

This expression can be rewritten using equation (52)

Gsolid solution = (niio) + RT (ni ln aio) + RT (ni ln i) (64)

where aio is used to represent the departure from the standard state when the component is mixing ideally in the solid solution. Previously, for solids we have used the expression

ai = Xii; however, now we are using ai = aioi

where it will be shown that aio = Xi for most but not for all cases.

The three terms in the expression for Gsolid solution can be considered as follows: (niio), the mechanical mixture of the components, Gmechanical mixture; RT (ni ln aio), the contribution from ideal mixing, Gideal mixing; and RT (ni ln i), the contribution from non-ideal mixing or excess mixing, Gexcess. Remembering that G = H - TS, we can write for Gideal mixing at constant T and P,

Gideal mixing = Hideal mixing - TSideal mixing (65)

where Hideal solid solution = 0 because no heat is released or absorbed in ideal mixing. Hence

Gideal mixing = -TSideal mixing = RT (ni ln aio) (66)

Boltzmann postulated that Sideal mixing = k ln (equation (27)

where is the increase in distinguishable configurations on the sites resulting from mixing, i.e., where atoms of the same element are not distinguishable but lattice sites are distinguishable. k is Boltzmann's constant which is R/A, the gas constant divided by Avogadro's number.

Consider the mixing of one mole of CY and BY molecules where the distinguishable configurations result from the placement of C and B on the cation sites of the lattice. Note that placement of Y on the anion sites generated no distinguishable permutations because it is the same anion in both molecules. Also note that there is only one distinguishable configuration for pure CY or for pure BY. Let NC and NB be the numbers of C and B cations, respectively, and their sum will then total one mole of cations, i.e., Avogadro's number A. The binomial formula gives the number of distinguishable configurations.

= (NC + NB)!/(NC!NB!)

Taking the natural log of and applying Stirling's approximation for the factorial of large numbers of objects, e.g., ln Z! = Z ln Z - Z, yields

ln = -NC ln[NC/(NC + NB)] - NB ln[NB/(NC + NB)]

Since XC = NC/(NC + NB), XB = NB/(NC + NB), NC + NB = A, and k = R/A

Tk ln = - RT XC ln XC - RT XB ln XB

Note that the mole fractions of C and B on the cation sites are equal to the mole fractions of CY and BY molecules on the lattice. Substituting the above into equation (27) gives

-TSideal mixing = RT (XCY ln XCY + XBY ln XBY) (67)

Remembering also that -TSideal mixing = RT (ni ln aio) where ni = Xi if the number of moles of all components sum to one mole, i.e., the number of molecules sum to Avogadro's number A. Hence,

-TSideal mixing = RT XCY ln aoCY + XBY ln aoBY (68)

By comparison between equations (67) and (68), the relations aoCY = XCY and aoBY = XBY are recognized and these correspond to the conventional case of representing the ideal activity of a liquid or solid component by its mole fraction.

However, what if mixing occurs over two cation sites per molecule, i.e., the molecule has the formula C2Y and B2Y. In this case, it can be shown by comparison in the above procedure that

aoC2Y = (XC2Y)2 and aoB2Y = (XB2Y)2

Indeed for mixing molecules of the type CzY and BzY to form a solid solution, the ideal activities are

aoCzY = (XCzY)z and aoBzY = (XBzY)z (69)

for use in the activity expressions for the chemical potentials

aCzY = (XCzY)zCzY and aBzY = (XBzY)zBzY (70)

rather than aCzY = (XCzY)CzY and aBzY = (XBzY)BzY which are incorrect

for molecules of this type in a solid solution.

Show that aoC2Y = (XC2Y)2 and aoB2Y = (XB2Y)2 for a solid solution of C2Y and B2Y.

Regular Solutions

A regular solution on a crystalline lattice is one in which there is zero Sexcess, excess entropy of mixing, but nonzero excess enthalpy of mixing. The free energy of mixing becomes that due to ideal mixing (-TSideal mixing) plus that due to the excess enthalpy of mixing (Hexcess). Note that Guggenheim called this a "strictly" regular solution; whereas, Hildebrand, who devised the model, called it a regular solution. The model was devised for nonpolar liquid mixtures. Applying it to crystalline solids makes it less likely that Sexcess could be zero.

Gmixing = Hexcess - TSideal mixing (71)

This implies that mixing is random even though interactions between adjacent atoms on sites depend on the site occupancies. For example, suppose NCY and NBY molecules are mixed on the lattice. Each C atom and each B atom have z adjacent neighbors. Let the interaction energy between two adjacent C atoms be eCC, the interaction energy between two adjacent B atoms be eBB, and the interaction energy between a B and C atom be eBC. Within the lattice there are z(NCY + NBY)/2 adjacent site interactions on the lattice. The change in the number of CB interactions on the lattice resulting from random mixing is NCB. From simple probability,

NCB = (z/2)(NCY + NBY)2[NCY/(NCY + NBY)][NBY/(NCY + NBY)]


Note that there are no CB adjacent interactions in the pure end members. The change in the number of CC and BB adjacent interactions due to mixing can be deduced from mass constraints. Two CB interactions result from the breaking of one CC and one BB interaction in the end members. Hence,

NCC = -NCB/2 and NBB = -NCB/2

The expression for Hex is just the sum of all the changes in energy due to interactions between atoms on adjacent sites as the result of mixing. The mixing is done at constant pressure which actually makes the sum of these energy changes equal to Uex but this is approximately equivalent to Hex as long as the mixing is done with little or no change in volume of the lattice.


= z[NCYNBY/(NCY + NBY)][eBC - 0.5eCC - 0.5eBB]


where EBC equals [eBC - 0.5eCC - 0.5eBB]. Dividing the top and bottom by Av2, Avogadro's number squared, changes N, the number of molecules, to n the number of moles. Let E-BC be EBCAv.

Hex = z[nCYnBY/(nCY + nBY)]E-BC (72)

Gex = Hex because Sex is zero. (73)

Remembering from equation (14c), i = (G/ni)T,P,nj

The free energy can be split up into that due to the mechanical mixture, that due to ideal mixing, and that due to excess mixing, We can use the same expression but with the partial of Gex and set it equal to i,ex.

i,ex = (Gex/ni)T,P,nj and since i,ex = RT ln i,

ln i = (1/RT)(Gex/ni)T,P,nj (74)

Equation (74) can be used with equations (73) and (72) to define the activity coefficient expressions in a regular solution, i.e.,

ln BY = (XCY)2zW/RT and ln CY = (XBY)2zW/RT.

Phase Changes

Two components having the same chemical composition undergo a phase change at equilibrium, e.g., ice and water. Using H2O as an example, we note that at equilibrium, the change in free energy for the reaction

H2Oice = H2Owater

is zero. Hence,

ice = water

or *ice + RT ln Xiceice = *water + RT ln Xwaterwater

If the phases are pure phases, their activities are unity and

*ice = *water

If P and T are changed and equilibrium is maintained.

d*ice = d*water

(*ice/T)dT + (*ice/P)dP = (*water/T)dT + (*water/P)dP

or - Soice dT + VoicedP = - Sowater dT + VowaterdP

or Sor dT = VordP and recognizing that since equilibrium is being maintained.

Gor = 0 = Hor - TSor, substituting for Sor gives

Hor/(TVor) = dP/dT (75)

This is one example of the Clapeyron equation which describes the univariant equilibrium curve between two phases, i.e., if the pressure is changed, we can change temperature until equilibrium is again reached and vice versa.

If the above reaction had been written between water and water vapor, both pure phases, and equilibrium had been maintained with changing pressure and temperature

water = vapor and dwater = dvapor

(water/T)dT + (water/P)dP = (vapor/T)dT + (vapor/P)dP

or - SowaterdT + VowaterdP = - SvapordT + VvapordP

and Sr dT = VrdP, or (Hr/T)dT = VrdP or Hr/(TVr) = dP/dT

where the superscript was removed from Vvapor to indicate the vapor is not in its standard state and furthermore, the standard state of a gas doesn't move with pressure. However, Vvapor corresponds to the molar volume of pure water vapor.

If the components of the phase change are not pure (so they can't be in their standard states) but the composition remains fixed, the above equations will become

Sr dT = VrdP, or (Hr/T)dT = VrdP or Hr/(TVr) = dP/dT

where these are now the changes in partial molal entropies, partial molal enthalpies, and partial molal volumes. These equations hold for any reaction in which the phases have a fixed composition, e.g., our example reaction.

CaCO3 (solid) + SiO2 (solid) = CaSiO3 (solid) + CO2 (gas)

They are derived by the above procedure, by setting dGr = 0 as the pressure and temperature are changed.

Sr/Vr = Hr/(TVr) = dP/dT (76)

is the general form of the Clapeyron equation. For our example reaction, the volume change in the reaction is approximately equal to the volume of the gas component because the volumes of the solids will be insignificant.

Example Problem

Find two PT points for equilibrium for our "example" reaction in which the CO2 gas phase is composed of pure CO2. For the first point, hold the temperature constant at 500oC and vary the pressure until Gr is zero. For the second point, hold the pressure constant at one bar and vary the temperature until Gr is zero. It is easiest to do this in Excel by setting the equation for Gr = 0 as a function of P and T and then varying these variables.

Gr = Gor,T,1 bar + Vor(P-1) + RT ln CO2P = 0

where the first term is a function only of temperature and the second two terms are functions only of pressure. Remember that Vor does not include the gas component and has to multiplied by 0.0239 to convert to calories. To get the first PT point at 500oC, assume that CO2 is one until you find a pressure setting Gr to zero. The compute CO2 for that pressure (as you did in your previous homework) and substitute back into the equation to solve again for the pressure. To get the second PT point at one bar, you will have to write Gor,T,1 bar in terms of its temperature dependence going from Gor,25 C,1 bar to Gor,T,1 bar, using one of the two methods previously done in your homework. Because the pressure is one bar, CO2 can be set to one.

Apply the Clapeyron equation to your two PT points to calculate an average Sr/Vr for the reaction between the two PT points. Also, compute this ratio at 1 bar using the computed values of Sr and Vr. (Note that Vr includes all the components in the reaction, not just the solid volumes previously given to you that were used in Vor. Use an estimated volume for the gas component based on the ideal gas law, RT/P). Why is the computed ratio at 1 bar so different from the average ratio calculated from the two PT points?

The Phase Rule

If a system contains c components and p phases then each phase will contain some amount of each component, even if it is only a trace amount. If the composition of each phase is described in mole fractions, then the total number of unknowns in each phase is c-1 because the sum of the mole fractions must be one. In p phases, we have p(c-1) variables needed to describe the system's composition. An additional two variables such as temperature and pressure or temperature and volume or volume and pressure are needed to define the system, leading to

p(c-1) + 2 unknowns

However, if the phases are in equilibrium with each other, the chemical potential of each component must be the same in all phases. Otherwise, one component would move from one phase to another to reach chemical equilibrium. Hence in p phases, each component will have p-1 relationships to be maintained at equilibrium or

c(p-1) equilibrium relationships for all the components.

Subtracting the number of relationships from the number of unknowns gives the number of variables that can be independently changed and equilibrium still be maintained. The number of variables is called the degrees of freedom f.

f = c - p + 2 (77)

which is the well known phase rule.

Consider our example reaction of

CaCO3 (solid) + SiO2 (solid) = CaSiO3 (solid) + CO2 (gas)

Suppose our system consisted of these four phases. The number of components are the minimum number of components needed to describe the system. These are CaO, SiO2, and CO2. We have three components and four phases. Hence, by the above formula for f, the number of degrees of freedom are one. We can arbitrarily vary either the pressure or temperature but not both because we only have one degree of freedom. For example, in your homework, you arbitrarily held T constant at 500oC and then found the P that equilibrium occurred. You also arbitrarily held P constant at 1 bar and then found the T that equilibrium occurred. You were only able to arbitrarily fix one variable.

If we add water to our system, we now have four components: H2O, CaO, SiO2, and CO2, and five phases. The number of degrees of freedom are still one. It's true that the gas phase will have a certain amount of water vapor. However, the mole fraction of water vapor in the gas phase cannot be arbitrarily varied because it will be fixed at any PT point by equilibrium with the aqueous solution. We still have only the option of setting either temperature or pressure and then varying the other variable to reach equilibrium.

Suppose the system consists only of water. We have one component H2O and one phase, so we can arbitrarily vary both P and T and still maintain water in a state of equilibrium (divariant system). If we add water vapor to the system, then we can only vary P or T (univariant system). If we add ice to the system, then there is only one PT point at which equilibrium exists (invariant system).

The number of components used in the phase rule is the number of thermodynamic components. This is the minimum number of components needed to describe the system. For example, within water, the H2O molecule can split into ions such as H+ and OH-. These are not additional thermodynamic components because they can be calculated from the equilibrium constant and reaction quotient of the reaction forming them from H2O:

H2O = H+ + OH- where K = aH+aOH-/aH2O

plus the charge-balance constraint that requires the concentration of negative charges to equal the concentration of positive charges.

If we have CO2 dissolved in water, we have the additional species of H2CO3, HCO3-, and CO32-. However, the concentrations of all these species can be calculated with the equilibrium constants and reaction quotients for the reactions controlling their formation plus the charge balance constraint. Hence, only two components are needed to describe CO2 dissolved in water, i.e., CO2 and H2O.

If we have a system consisting of CaCO3 and MgCO3 minerals, we only have two components: CaCO3 and MgCO3; not 3 components: CaO, MgO, and CO2. This is because there is always a one to one ratio of CaO to CO2 and MgO to CO2 in the phases.

Picking the correct number of thermodynamic components will become clearer as the course proceeds.

Redox Reactions, pe and Eh

Redox reactions involve a flow of electrons from the oxidized component (reductant) to the reduced component (oxidant). The most familiar reductant and oxidant are carbon and oxygen, respectively, e.g., in respiration,

CH2O + O2 = H2O + CO2

The reaction really consists of two half cell reactions in which the electrons cancel out when the reactions are added together.

CH2O = H2O + C4+ + 4e-

O2 + 4e- = 2O2-

CH2O + O2 = H2O + 2O2- + C4+ = H2O + CO2

We can't actually work with an individual half cell reaction in a laboratory. We always have to use a balanced reaction in which the electrons cancel out, e.g., an electrochemical cell, in which the flow of electrons is balanced by the movement of ions through a salt bridge.

However, we can use a half-cell reaction to define an activity of an electron in an aqueous solution, ae-, even though aqueous electrons probably do not exist. Consider the following reaction in an aqueous solution

Mn4+ + 2e- = Mn2+

Beginning with equation (55),

Gr,T,P = Gor,T,P + RT ln Q. And at reaction equilibrium,

Gr,T,P = 0 and Q is called K, the equilibrium constant.

This leads to equation (56), log K = -Gor,T,P/(2.303 RT), where

log K = log {aMn2+/[aMn4+(ae-)2]}

Defining -log ae- = pe, makes log K = log (aMn2+/aMn4+) + 2 pe

or pe = [log K - log (aMn2+/aMn4+)]/2

Hence, if a redox reaction in aqueous solution is at equilibrium, we can calculated pe from the measured activity ratio of the reduced and oxidized forms and the computed log K. The computed pe in a particular aqueous solution should be the same, regardless of which equilibrium redox reaction is used to compute it. However, if more than one measured activity ratio of redox couples, e.g., aFe2+/aFe3+ and aHS-/aSO4(2-), from the same solution is used to compute the pe, the values are generally different. The general conclusion is that redox reactions are often not at equilibrium in an aqueous solution, particularly near earth-surface temperatures.

Another way of computing an aqueous electron activity is to use an electrochemical cell. In this case, the aqueous solution makes up one half cell in which contact with an inert platinum electrode is used to make contact with a reference half cell. The procedure consists of measuring the potential (voltage) necessary to prevent the flow of electrons between the solution and the reference half cell. As shown below, the pe can be calculated from this potential.

Consider a closed system, in which a reaction occurs resulting in the flow of electrons, i.e., electricity.

Remembering equation (1), dU = dq + dw, where if electrical work is occurring together with PV work, dw = -PdV + dwelec and dq = TdS, if processes are reversible.

By definition (equation 10a), G = U + PV - TS

and dG = dU + PdV + VdP - TdS - SdT,

by substitution for dU,

dG = -SdT + VdP + dwelec + idni (78)

and at constant T and P,

dG = dwelec + idni (79)

For a redox reaction in which n moles of electrons are used to reduce one mole of oxidant, welec is computed using the electric potential E in volts.

welec = -nFE (80)

where F is the Faraday constant, 23,060 calories/volt/mole of electrons. The minus sign refers to the convention of assigning a positive potential for a spontaneous half cell reaction in which the product (right side) contains the reduced component and the reactant (left side) contains the oxidized component.

In an electrochemical cell, the cell potential E is the opposite of the potential applied to prevent a chemical reaction from occurring. The free energy change is computed from E,

G = -nFE (81)

The potential is measured relative to a reference half cell within the electrochemical cell. For example, for

Mn4+ + 2e- = Mn2+ where n = 2. The conventional reference half

cell is H2 gas = 2e- + 2H+ where the aH+ and the aH2 gas are unity,

i.e., the hydrogen ion concentration in the solution is that of an HCl solution with unit activity of H+ ions and the hydrogen gas bubbling through the water is being maintained at a partial pressure of 1 bar for H2, such that its activity is unity. This reference half cell is called the Standard Hydrogen Electrode or SHE and potential measured relative to it are called Eh.

Note that the chemical potentials of H+ and H2 are equal to their standard state chemical potentials, because their activities are unity. By convention, the standard state chemical potential of H+ is set to zero at all PT points because it cannot be measured independently of an anion. The apparent standard state chemical potential of H+ is also zero at all PT points because by convention H+ has zero heat capacity and zero volume. Again this is because the aqueous heat capacity and volume of H+ can't be measured independently of an anion. The standard state chemical potential of H2 is zero because this is the change in Gibbs free energy in forming H2 from elemental hydrogen which happens to be H2. Because, a voltage cannot be measured for a half cell, i.e., only for two half cells, the E and Eo of SHE are arbitrarily set to zero. Since GoSHE = -nFEoSHE = zero, then GoSHE is zero, and since GSHE = -nFESHE = zero, then GSHE is zero, therefore the standard state chemical potential of e- is zero and the ae- is unity in SHE. This is shown below.

GoSHE = 2oH+ + 2oe- - oH2 = 0 and oH+ = 0 and oH2 = 0, so oe- = 0.

Since GSHE = 0, log Q = log K = -GoSHE /(2.303 RT) = 0, or K = 1.

K = (aH+)2(ae-)2/aH2 = 1, and aH+ = 1, aH2 = 1, so ae- = 1.

Hence the activities of all components are unity in SHE.

In our example Mn is being reduced or oxidized in one half cell and SHE is the other half cell. The overall reaction is

Mn4+ + H2 = 2H+ + Mn2+ where

G = Go + 2.303 RT log {(aMn2+/aMn4+)solution([aH+]2/aH2)SHE}

The activities of H+ and H2 are unity in SHE so they drop out of the activity expression, leaving

G = Go + 2.303 RT log (aMn2+/aMn4+)solution.

Go equals the sum of the Go for each half cell reaction;

however, GoSHE is zero, so Go is for Mn4+ = Mn2+ + 2e- half cell.

G = -2FEh where Eh is the voltage measured relative to SHE, and

Go = -2FEoh where Eoh is the voltage measured relative to SHE if

Mn4+, Mn2+, and e- were in their standard states in the solution. By substitution,

Eh = Eoh + (2.303 RT/2F) log (aMn4+/aMn2+)solution

Now how does Eh relate to pe? Consider the half cell reaction in the solution making up half the electrochemical cell.

Mn4+ + 2e- = Mn2+

G = Go + 2.303 RT log {aMn2+/(aMn4+ae-2)}solution = 0 because the

half cell reaction is assumed to be in redox equilibrium. Hence,

Go = -2.303 RT log {aMn2+/(aMn4+ae-2)}solution

And, as previously shown, Go for the electrochemical cell is the same as for this half cell. Using Go = -2FEoh and substituting

Eoh = 2.303 (RT/2F) log {aMn2+/(aMn4+ae-2)}solution

from the above half cell at equilibrium into the previous expression for Eh for the electrochemical cell gives

Eh = -(2.303 RT/F) log (ae-) = (2.303 RT/F)pe

or pe = (F/2.303 RT) Eh (82)

In practice, the potential is measured versus a convenient reference electrode other than SHE. The calomel electrode is commonly used as a reference electrode. The redox half cell is

Hg2Cl2 solid + 2e- = 2Hgliq + 2Cl-aq

which has an Eh of 0.2444 volts at 25oC if the aqueous reference solution is saturated with respect to KCl, i.e., contains solid KCl. By adding the Eh of the reference electrode to the E of the solution, the Eh of the solution is obtained, i.e.,

Eh solution = Esolution + Eh ref. electrode

pH definition

pH = -log aH+ in an aqueous solution. Like pe, it can be measured with an electrochemical cell by using an electrode combined with a reference cell. Unless you are doing a mass balance involving H+, there is no need to calculate H+ since the activity can be measured directly.

Also like pe, the pH can be calculated from other solution activities, e.g., by measuring the molalities of HCO3- and CO32- in solution and calculating their activity coefficients for use in the following reaction:

HCO3- = H+ + CO32- leads to K = aH+aCO3(2-)/aHCO3- or

pH = -log aH+ = -log K - log aHCO3- + log aCO3(2-)

Unlike pe, the pH calculated from different reactions in the same solution is always the same, unless these are redox reactions in which the uncertainity in pe will produce differences in pH.

pe and pH Relations in Biological Systems

Animal-like organisms can use different molecules to accept electrons to oxidize carbon in their metabolism. A general reaction sequence (listed below) is followed going from oxidizing (high pe) to reducing (low pe) in which the amount of energy obtained by the organism decreases in the sequence. For this reason, an organism living in a more oxidizing metabolism is always favored over an organism living in a more reducing environment. The overall change in Gibbs free energy can be calculated from equation (55).

Gr,T,P = Gor,T,P + RT ln Q where Gor,T,P = -RT ln K

and by substituting reasonable activity values for the reaction components in Q.

aerobic oxidation, log K25 C & 1 bar = 20.75

0.25 O2 gas + H+ + e- = 0.5 H2O

denitrification, log K25 C & 1 bar = 21.05

0.20 NO3- + 1.2 H+ + e- = 0.1 N2 + 0.6 H2O

manganese reduction, log K25 C & 1 bar = 17.90

0.5 MnO2 solid + 0.5 HCO3- + 1.5 H+ + e- = 0.5 MnCO3 solid + H2O

nitrate reduction, log K25 C & 1 bar = 14.90

0.125 NO3- + 1.25 H+ + e- = 0.125 NH4+ + 0.375H2O

iron reduction, log K25 C & 1 bar = 10.20

FeOOHsolid + HCO3- + 2 H+ + e- = FeCO3 solid + 2 H2O

fermentation, log K25 C & 1 bar = 3.99

0.5 CH2O + H+ + e- = 0.5 CH3OH

sulfate reduction, log K25 C & 1 bar = 4.12

0.125 SO42- + 1.125 H+ + e- = 0.125 HS- + 0.5 H2O

methane fermentation, log K25 C & 1 bar = 2.87

0.125 CO2 gas + H+ + e- = 0.125 CH4 gas + 0.25 H2O

nitrogen fixation, log K25 C & 1 bar = 4.65

0.167 N2 + 1.333 H+ + e- = 0.333 NH4+

Osmotic Pressure

Consider two aqueous salt solutions, A and B, in a U-shaped tube, separated at the bottom by a semi-permeable membrane that only water can pass through. Each salt solution extends up one arm of the U tube and initially to the same height as the other solution. Because of different amounts of salt in the two solutions, the chemical potential of water will not be the same in both solutions. Water will move across the membrane from the more dilute solution (A) to the more concentrated solution (B), diluting B and also increasing the pressure on water molecules in (B) at the membrane because the height of solution (B) will increase in the U tube. The difference in pressure, PB - PA, across the membrane at equilibrium is called the osmotic pressure.

A,H2O = B,H2O at equilibrium

oH2O,PA + 2.303 RT log XA,H2OA,H2O = oH2O,PB + 2.303 RT log XB,H2OB,H2O

where oH2O,PB = oH2O,PA + VoH2OdP and as an approximation

= oH2O,PA + VoH2O(PB - PA). The osmotic pressure

P = PB - PA = 2.303 (RT/VoH2O) log [XA,H2OA,H2O/XB,H2OB,H2O] (83)

Plot the osmotic pressure, assuming solution A is pure water, and XB varies from 0.9 to 0.99 and assuming B,H2O is unity.

In sediments, there is a general increase in water salinity with depth. Assume a shale unit acts as a semipermeable membrane between two sandstone units. How thick must the shale unit be

to produce equilibrium between the waters in the two sandstone units, if the lower sandstone unit has a mole fraction of water of 0.90 and the upper sandstone unit has a mole fraction of water of 0.95. Assume H2O is unity, the density of water is 1.05 g/cm3, and that the difference in pressure across the shale is due to the column of water in the shale. What would happen if the shale unit were thicker than the equilibrium value? Is this consistent with the observed increase in salinity with depth.

Aqueous Activity Coefficients

Measurement of aqueous activity coefficients is done by using the Gibbs-Duhem equation. From equation (10c), for any system at constant T and P.

dG = idni (84)

We also know that the Gibbs free energy of any system at constant T and P is represented by the sum of the Gibbs free energies of its components, i.e.,

G = ini (85)

hence, dG = idni + nidi. (86)

A comparison between equations (84) and (86) gives the Gibbs-Duhem equation at constant T and P.

nidi = 0 (87)

Remember i = io + RT ln ai. Since T and P are constant, dio = 0, and

niRT dln ai = 0 (88)

which allows determining the activity of one component in a solution if the activities can be measured for the other components. Activity coefficients of solutes can be determined in aqueous binary solutions easily because the activity of water can be determined from its measured vapor pressure. The procedure uses the definition of fugacity of water fH2O (equation 60) and its subsequent relation to the activity, fH2O/foH2O = aH2O where fH2O and foH2O are the vapor pressures, respectively, of water in the solution and in its standard state. The procedure becomes much more complicated for ternary solutions.

Within aqueous solutions, the solvent H2O follows Raoult's law for ideal mixing at infinite dilution:

ai = Xi as Xi => 1 (89)

because i => 1 as the standard state is approached.

From the Gibbs-Duhem equation, Henry's law for a solute

ai = khXi as Xi => 0 (90)

can be shown to follow from Raoult's Law. kh, the Henry's law constant, equals i on the mole fraction scale for a solute and is the non-zero limiting value of the activity coefficient at infinite dilution.

However, we use a molality scale, rather than a mole fraction scale for solutes in aqueous solution. The mole fraction, Xi, of the ith aqueous solute component is related to the molality mi by

Xi = mi/(mk + 55.56) (91)

where 55.56 is the number of moles of water in one kilogram of water. We also expect solutes on a molality scale to follow Henry's law,

ai = khmmi as mi => 0, (92)

where khm is the Henry's law constant on a molality scale, equals i and is the non-zero limiting value of the activity coefficient at infinite dilution. Note that the constant is the slope of a plot of activity versus molality. Combining equations (90), (91), and (92) shows the two Henry's law constants are related by the factor mk + 55.56 which becomes 55.56 at infinite dilution.

khm = kh/(mk + 55.56)

Experiments show that Henry's law is followed for aqueous solutes that do not disassociate into ions, e.g., CH4 aq or CO2 aq. If we plot their activity versus their molality, the Henry's law constant that is obtained has a non-zero limiting value that is constant over a large concentration range and is only a function of the total ionic strength of the solution at constant T and P. Hence, for neutral components in aqueous solution, their activity coefficients can generally be represented by the following type of expression:

log i = ksI (93)

where ks, the salting-out coefficient is determined from experimental measurement and

I = 0.5mk(zk)2 (94)

is the ionic strength of the aqueous solution. Note that only charged species contribute to the ionic strength. The ionic strength can be computed on a stoichiometric basis, using ions such as Na+ and CO32-, and neglecting ion pairing such as NaCO3-, or as a true ionic strength which includes all ion pairs, depending on how I was defined during the measurement of ks.

For dissolved components that disassociate into ions, e.g., a salt, plotting the activity of the neutral salt versus the molality of the salt yields a Henry's law constant of zero at infinite dilution. This is a violation of Henry's law which predicts a non-zero limiting constant and occurs because of the disassociation process. Plotting the activity of the neutral salt versus the molality of the salt raised to the number of ions into which it disassociates, yields a non-zero Henry's law constant at infinite dissolution. We have to take into account disassociation in dealing with salts.

Let i represent the salt Cv+Av- which dissolves into water to produce (v+)C cations and (v-)A anions. Note that electrical neutrality requires that the charge on C is (v-) and the charge on A is -(v+). The chemical potential of i is represented by

i = io + RT ln ai where ai = mii

and at infinite dilution, the salt is totally disassociated. We can also represent i as the sum of the contributions from its parts.

i = (v+)C + (v-)A = (v+)Co + (v-)Ao + RT ln(aC)v+(aA)v-

where ai = (aC)v+(aA)v- = (mC)v+(mA)v-(C)v+(A)v-

For a single-salt solution, mC = (v+)mi and mA = (v-)mi

or ai = mi(v+ + v-) (v+)v+(v-)v-(C)v+(A)v- (95)

which shows that if Henry's Law is followed (equation 92) then a plot of ai versus mi(v+ + v-) yields a Henry's Law constant at infinite dilution of

khm = (v+)v+(v-)v-(C)v+(A)v- = (v+)v+(v-)v-i (96)

Experimental data show this to have a non-zero limiting value at infinite dilution.

We can define a mean ionic activity (ai), mean ionic molality (mi), and mean ionic activity coefficient (i) of i such that when raised to (v+ + v-), the product of the mean ionic molality and the mean ionic activity coefficient equals the activity of the neutral salt in a single salt aqueous solution (equation 95). Likewise, the mean ionic activity when raised to (v+ + v-) is equal to the activity of the neutral single salt in a single salt aqueous solution, i.e.,

ai = ai[1/(v+ + v-)] = [(aC)v+(aA)v-][1/(v+ + v-)] (97)

mi = mi[1/(v+ + v-)] = mi[(v+)v+(v-)v-][1/(v+ + v-)] (98)

i = i[1/(v+ + v-)] = [(C)v+(A)v-][1/(v+ + v-)] (99)

In calculating the activity of a neutral dissolved salt, such as CaSO4, we can use total (stoichiometric) molality or the free (uncomplexed) molality in the expression; however, the computed activity must be the same. (This is because the free energy change in forming the complexes is zero at equilibrium in the solution.) Hence, the activity coefficient will change depending on the use of total or free molality in the expression. Measurements of solution activities are fitted to expressions that use either total molalities or free molalities. We have no theoretical expression for aqueous activity coefficients that holds at high concentrations to calculate activities in equation (99), only approximations.

Debye and Huckel formulated a limiting law for activity coefficients of ions in aqueous solution based on the electrical work of interactions between the ions at infinite dilution at constant T and P. Expressed in the form for the mean ionic activity coefficient of the salt,

log i = -Az+z-I (100)

where A is an H2O parameter, z+ and z- are the charges of the cation and anion respectively, and I is the ionic strength previously defined. Because the equation holds at infinite dilution only, there are no complexes and the stoichiometric ionic strength is the same as the true ionic strength.

At higher concentrations, The modified Debye-Huckel expression becomes

log i = -Az+z-I/(1 + BI) (101)

where B is another H2O parameter. And at still higher concentrations, the equation becomes

log i = -Az+z-I/(1 + aoiBI) + biI (102)

where aoi and bi are a salt-specific parameters accounting for the size of the ions and complexing between the ions, respectively.

Equations (100) and (101) for i can be split up into single ion activity coefficients which can be combined back into the mean ionic activity coefficient definition, e.g.,

log i = -A(zi)2I/(1 + BI) (103)

where zi is the ionic charge of the ith ion. Substitution of this expression for C and A into equation (99) defining i for the salt will yield equation (101) for i.

In ground water and sea water type solutions, equation (102) is commonly split up into single ion activity coefficients of the form

log i = -A(zi)2I/(1 + aoiBI) + biI (104)

where aoi and bi are fit parameters for the ith ion. However, the expression cannot be combined back into equation (99) to give equation (102) unless bi and aoi are the same for both cations and anions. In practice, different values are used so there is a lack of thermodynamic consistency in using equation (104). The more complicated viral equations used in Pitzer's method of calculating aqueous activity coefficients in brines are thermodynamically consistent in that you can combine the cation and anion activity coefficient expressions to come out with the correct expression for the mean ionic activity coefficient of the neutral salt. (More on this later).

Experimentally, the activity of a dissolved salt can be measured and the activity coefficient back calculated using either the stoichiometric molality or the free molality. The activity coefficient of the individual ions making up the salt can also be calculated by splitting up the measured activity of dissolved salt into ionic activities (using one of several conventions) and then calculating individual stoichiometric or free ion activity coefficients by using either the stoichiometric ion molalities or free ion molalities. These computed activity coefficients of the individual ions are then fit to a general expression such as the modified Debye-Huckel expression given above for subsequent use in calculating activities in other aqueous solutions. This was the procedure used by Truesdell and Jones in the text. In geochemistry, we often use free ion activity coefficients so the modified Debye-Huckel expression above would use a true ionic strength, not a stoichiometric ionic strength.

Phase Diagrams

Solution Activity Phase Diagrams

Using the system SiO2, Na2O, Al2O3, H2O, and HCl, calculate the phase diagram for minerals in equilibrium with water in which aluminum is conserved in reactions written between the minerals. Conservation of aluminum in the solids is suggested by low concentrations of aluminum in aqueous solution. Assume the aH2O is unity. Use the log aSiO2 aq on the x axis and the log aNa+/aH+ on the y axis. Combining two solution variables, H+ and Na+, on one axis is possible because charge balance in the reaction links them together.

What is the maximum number of phases that can coexist at equilibrium on the diagram? We have 5 thermodynamic components. The phase rule is f = c - p + 2 in which c is 5. The minimum number of degrees of freedom is three because we have fixed T and P for constructing the diagram and arbitrarily held the aH2O at unity. If f is three then p is 4. These must include water so the total number of mineral phases is three.

What mineral phases could these be? They must have Al in them in order to write reactions between them that balance on Al. One potential list is

NaO0.5/Al2O3 SiO2/Al2O3

gibbsite Al(OH)3 0 0

kaolinite Al2Si2O5(OH)4 0 2

paragonite NaAl3Si3O10(OH)2 2/3 2

pyrophyllite Al2Si4O10(OH)2 0 4

albite NaAlSi3O8 2 6

One graphical way to determine which possible phases could be in equilibrium together is to take the ratio of the number of moles of NaO0.5/Al2O3 and SiO2/Al2O3 in each of the above minerals and plot them on a graph of NaO0.5/Al2O3 (y axis) versus SiO2/Al2O3 (x axis). On this plot, NaO0.5/Al2O3 is used rather than Na2O/Al2O3 because the y axis on the phase diagram contains the log aNa+, not 2 log aNa+. Connect the points on the plot. The connecting lines cannot cross other connecting lines. You will find there will be a limited number of possibilities to connect points without crossing other connecting lines, eliminating many possibilities of equilibrium pairs. The actual slope of the boundaries on our phase diagram between any two phases will be the negative reciprocal of the slope of the connecting line. This type of approach will allow you to eliminate metastable phase boundaries.

Another way to predict which possible phases could be in equilibrium together is to realize that the most Na-rich phase, (paragonite) will form at the highest activities of Na+, and the most Si-rich phase (albite) will form at the highest activities of SiO2 aq. Those with no Na or SiO2 in their formulas will form at low solution activities of both components, near the origin, and so on. Note that quartz and amorphous silica can only be plotted on the diagram as saturation lines because they don't contain aluminum.

Using the following standard state free energy data for the minerals and aqueous components at 25oC and 1 bar, calculate the phase diagram described above, assuming the aH2O is one. Write reactions between phases, balancing on aluminum, and calculating the log K of the reaction. Write the log K in terms of the variables in the activity quotient and plot the boundaries on the phase diagram, disregarding all metastable boundaries. Plot the saturation lines for quartz and amorphous silica on the diagram.

Go in cal/mole

gibbsite Al(OH)3 - 276,339

kaolinite Al2Si2O5(OH)4 - 908,167

paragonite NaAl3Si3O10(OH)2 -1,331,352

pyrophyllite Al2Si4O10(OH)2 -1,258,811

albite NaAlSi3O8 - 886,734

quartz SiO2 - 204,646


silica SiO2 - 202,892

Na+ aqueous - 62,591

H+ aqueous 0

SiO2 aq - 199,190

H2O liquid - 56,686

pH-pe Phase Diagrams

pH-pe diagrams are diagrams showing the stability fields of minerals in the presence of an aqueous solution and aqueous species within the solution. The diagrams use pe, the -log ae-, as the y variable and with pH, the -log aH+, as the x variable. There are three types of stability boundaries on this phase diagram.

1) Reactions are written between aqueous species such as Fe2+ and Fe3+, in which the boundary between dominant species is taken at equal activities, e.g., aFe2+/aFe3+ = 1. (generally low pH region)

2) Reactions are written between aqueous components and a solid phase, in which the saturation boundary is taken at some stated activity of the aqueous components, e.g., aFe2+ of 10-5 in equilibrium with Fe(OH)2. (generally low to intermediate pH regions)

3) Reactions are written between solid phases to determine a stability boundary between solids, e.g., between Fe(OH)2 and Fe(OH)3 and conserving Fe in the reaction. (generally intermediate to high pH region)

Because an aqueous solution is present on the diagram, the stability boundaries of water are used to define the limits of the diagram. The upper boundary of water is marked by the oxidation of oxygen in H2O to O2 gas. The lower boundary is marked by the reduction of H+ ions in water to H2 gas. The two reactions are given below together with their 25oC and 1 bar log K values.

upper boundary O2 gas + 4H+ + 4e- = 2H2O log K = 83.1

lower boundary 2H+ + 2e- = H2 (gas) log K = 0

Construct a 25oC and 1 bar pe-pH diagram for the FeO-Fe2O3- HCl-H2O system. Consider the following aqueous components: Fe3+, Fe(OH)2+, Fe(OH)2+, Fe(OH)4-, Fe2+, Fe(OH)+, and the following two solids: Fe(OH)3(s) and Fe(OH)2(s). Assume aH2O is unity. Balance the reaction between the two solids with Fe being conserved in the reaction. Set the stability boundary between aqueous species at equal activities. Set the saturation boundary between a solid phase and an aqueous phase by assuming the sum of all the aqueous iron activities in solution is 10-5.

Use the following reactions with the log K values given for 25oC and 1 bar. You may have to add and subtract these reactions together to obtain a particular reaction. The equilibrium constant for the new reaction can be obtained from the equilibrium constants of the added or subtracted reactions. For example, if reaction (1) is added to reaction (2), K = K1K2 and log K = log K1 + log K2. If reaction (2) is subtracted from reaction (1), K = K1/K2 and log K = log K1 - log K2. If a reaction is simply reversed, then the new K is the reciprocal of the old K and the new log K is the negative of the old log K.

Fe3+ + 3H2O = Fe(OH)3(s) + 3H+ log K = - 3.00

Fe3+ + H2O = Fe(OH)2+ + H+ log K = - 2.17

Fe3+ + 2H2O = Fe(OH)2+ + 2H+ log K = - 7.18

Fe3+ + 4H2O = Fe(OH)4- + 4H+ log K = -22.2

Fe3+ + e- = Fe2+ log K = 13.01

Fe2+ + 2H2O = Fe(OH)2(s) + 2H+ log K = -11.67

Fe2+ + H2O = Fe(OH)+ + H+ log K = - 6.70

This is actually a simple diagram. The most complicated aspect is determining the saturation boundary between the aqueous solution and a solid. Write your saturation reaction to the dominant iron aqueous species in that region of the diagram and set its activity at 10-5. If the saturation boundary crosses into another region dominated by a different iron aqueous species, then use a new reaction written to the new aqueous species with its activity set at 10-5. Where the saturation boundary crosses between the two regions, use either reaction and set the activity of the aqueous species at 0.5 x 10-5 to allow for equal activities of the two iron species at the crossover point. The saturation line will be a straight line in each region dominated by a particular species and will curve as it crosses between regions.

Keep the diagram between the stability limits of water. Assume an activity of O2 gas of 1 at the upper boundary and an activity of H2 gas of 1 at the lower boundary.

Gas Activity (Fugacity) Diagrams

The pe-pH diagram can be recast in terms of O2 activity and pH by using the reduction of O2 as a half cell reaction to balance out electrons. The y axis variable becomes the log aO2. For example, in the homework diagram:

O2 gas + 4H+ + 4e- = 2H2O log K = 83.1

+ 4Fe2+ = 4e- + 4Fe3+ log K = -52.04

O2 gas + 4H+ + 4Fe3+ = 4Fe2+ + 2H2O log K = 31.06

The use of O2 introduces a pH dependence in the oxidation of iron because O2 reduction in aqueous solution involves H+.

The use of log aH2 on the y axis can follow from using the H2 half-cell reaction to balance out electrons. For example in the homework diagram:

2H+ + 2e- = H2 (gas) log K = 0

+ 2Fe2+ = 2e- + 2Fe3+ log K = -26.02

2H+ + 2Fe2+ = H2 (gas) + 2Fe3+ log K = -26.02

Similarly, reactions involving inorganic carbon and sulfur can be written to CO2 and H2S, allowing variables to be aCO2 and the aH2S, respectively.

Metal Solubilities

The mobility of any metal is constrained by its solubility which limits its concentration in aqueous solutions. For this reason, metal solubilities are a primary concern in environmental pollution studies. Similarly, aqueous aluminum and calcium solubilities control the amount of porosity that can be produced by dissolution reactions in aluminum silicates and in limestones, respectively.

When a solution is in contact with two or more minerals containing the same metal, equilibrium can only be maintained between the solution and the mineral producing the lowest metal concentration in solution. Only if the reaction kinetics are too slow for that mineral to equilibrate, will another mineral control the solubility of the metal. For example, amorphous silica can maintain equilibrium with an aqueous solution at low temperatures (and control aqueous silica) because quartz doesn't react fast enough to reach equilibrium.

Computing metal solubility is fairly simple. Assume equilibrium between a mineral and a solution. Using the log K of the dissolution reaction and the log K values of the complexes forming in the solution which contain the metal, calculate the total concentration of the metal and all its complexes. Plot this sum versus the pH or some other variable. Do this for each mineral. Plot the curves on top of each other. The true solubility curve consists of the portions of the curves giving the lowest total metal concentration.

The major difficulty in computing metal solubilities is the lack of knowledge (thermodynamic data) on aqueous complexes containing the metal, particularly organic complexes and the general lack of solution equilibrium with minerals under earth-surface conditions. The lack of equilibrium is due to slow reaction kinetics at low temperatures. Hence, metal solubility calculations provide only minimum solubility limits under earth-surface conditions. However, at temperatures of 100oC and higher, equilibrium usually exists between minerals and an aqueous solution.

In the exercise below, plot the total solubility of aluminum versus pH at 200oC and 650 bars. Assume activity coefficients are unity and that the solution is in equilibrium with quartz. Assume that the solution is always in equilibrium with one additional phase which can be either gibbsite, kaolinite, pyrophyllite, or albite. Assume the activity of Na+ is 0.01 and of H2O is one.

Use the following information at 200oC and 650 bars.

aqueous complexes log K

Al(OH)2+ + H+ = Al3+ + H2O 0.355

Al(OH)2+ + 2H+ = Al3+ + 2H2O 2.870

Al(OH)4- + 4H+ = Al3+ + 4H2O 13.738

minerals log K

gibbsite Al(OH)3 + 3H+ = Al3+ + 3H2O 1.01

kaolinite Al2Si2O5(OH)4 + 6H+ = 2Al3+ + 2SiO2 qtz + 5H2O -0.48

pyrophyllite Al2Si4O10(OH)2 + 6H+ = 2Al3+ + 4SiO2 qtz + 5H2O -0.16

albite NaAlSi3O8 + 4H+ = Na+ + Al3+ + 3SiO2 qtz + 2H2O 4.67

Movement of log K with T and P

With Changes in T

Beginning with equation (56), log K = -Gor/(2.303RT)

Taking the partial derivative with respect to T at constant P

(log K/T)P = Gor/(2.303RT2) + Sor/(2.303RT)

Using equation (35), Gro = Hro - TSro, and substituting into the above eqt.,

(log K/T)P = Hor/(2.303RT2)

and by integrating at constant P, from Ta to Tb

d log K = [Hor,T,P/(2.303RT2)]dT

and substituting in Hor,Tb,P = Hor,Ta,P + (Hro/T)P dT

= Hor,Ta,P + Cpro dT

or log KTb = log KTa + [Hor,Ta,P/(2.303RT2)]dT + Cpro dT

= log KTa - [Hor,Ta,P/(2.303R)][1/Tb - 1/Ta] + Cpro dT (105)

The last term can be integrated using an appropriate algorithm for Cpro. This is the equation that is used to predict the change in the equilibrium constant with temperature.

With Changes in P

Again, starting with log K = -Gor/(2.303RT)

Taking the partial derivative with respect to P at constant T

(log K/P)T = -Vor/(2.303RT)

and integrating from Pa to Pb at constant T

log KPb = log KPa - [Vor,Ta,P/(2.303RT)]dP (106)

Dealing with First Order Phase Transitions in Moving o through PT Space

A phase transition occurs as a component changes from one phase to another as it moves through PT space. A first-order phase transition occurs when when there is no change in Gibbs free energy in the transformation but changes in everything else, e.g., first partial derivatives of Gibbs free energy (entropy, enthalpy, volume) and second partial derivatives of Gibbs free energy (heat capacity, compressibility), etc. This is the normal phase transition, e.g., ice to water, water to steam, alpha quartz to beta quartz, calcite to aragonite. A second-order phase change would have no changes in entropy, enthalpy, and volume, as well as Gibbs free energy, but changes in everything else, e.g., second order partial derivatives (heat capacity and compressibility), etc. Second-order phase transitions result from changes in bond orientation, ordering, magnetic properties and are commonly called lambda transitions.

A first order transition has changes in thermodynamic properties computed from 1st partial derivatives and higher of Gibbs free energy. A second order transition has changes in thermodynamic properties computed from 2nd partial derivatives and higher for Gibbs free energy. And so on.

A first-order phase transition is equivalent to a chemical reaction at equilibrium. Note that changes in enthalpy and entropy cancel out in the following equation.

Gr = Hr - TSr = 0

If the phases are in their standard states

Gro = Hro - TSro = 0

If your intent is to compute the Gibbs Free Energy at a particular PT point using the most stable phase for each component, then you must take into account that one or more of the components have undergone a phase transition prior to reaching the PT point. You must first complete your calculation of * for the initial phase to the point of the phase change. Then begin your temperature or pressure integration from the point of the phase transition to the PT point of interest using the standard state third law entropy, enthalpy, heat capacity, and volume respectively of the new phase produced in the phase transition.

To obtain the standard state third law entropy of the new phase, correct the value computed for the old phase at the phase transition point by adding Sro to it. Similarly, to obtain the standard state enthalpy of formation, correct the value computed for the old phase at the phase transition point by adding Hro to it. Do the same for the standard state volume of the component. In integrating with temperature and using Maier-Kelly heat capacity coefficients, simply substitute the coefficients for the new phase after the phase transition for the coefficients of the old phase.

Perhaps you are moving the standard state free energy of a reaction and the component undergoing a phase transition is just one of the reaction components. Then you must use the procedure described above in which the properties of the component have been modified. You must first complete your integration to the point of the phase transition and then begin your temperature or pressure integration again after using the above procedure to modify the standard state third law entropy of reaction, the standard state enthalpy of formation of reaction, the standard state volume of reaction, and the standard state heat capacity of reaction.

Tardy-Garrels approach to estimating the standard state Gibbs free energy of formation of a mineral.

The approach is best explained using clay minerals. The formula of a clay mineral can be written as a summation of metal oxides in the interlayer (I), octahedral (O), and tetrahedral (T) sites and H2O in between the layers. In general, these contributions will be from Na2O, K2O, CaO in the interlayer sites; MgO, FeO, Fe2O3, Al2O3 in the octahedral sites; Al2O3, SiO2, TiO2, in the tetrahedral sites and H2O. We assume the standard state Gibbs free energy of formation of the clay can also be written as a summation of the contributions of the metal oxides in the interlayer, octahedral, and tetrahedral sites and water within the clay. Each contribution is multiplied by its molar amount in the unit formula of the clay.

For example, we can write for a beidellite having a formula


obeid. = 0.165oNa2O,I + oAl2O3,O + 0.167oAl2O3,T + 3.67oSiO2,T + oH2O,

We use a set of clay minerals, each of which the standard state Gibbs free energy has already been measured, and set each measured value equal to a summation of contributions from the metal oxides. We need as many clay minerals as they are different metal oxides, 11 in the above paragraph. The set of standard state Gibbs free energies associated with these metal oxides is solved for. These values are then used to estimate Gibbs free energies of other clays having different compositions.

Margules Equations

We have previously shown that the Gibbs free energy of a solid solution (ss) is the sum of the mechanical mixture (mm), ideal mixing (im), and excess mixing (ex).

Gss = Gmm + Gim + Gex where the three terms can be

rewritten as

Gss = (niio) + RT (ni ln aio) + RT (ni ln i) (107)

The Gibbs free energy of mixing (mix) is the sum of the last two terms and does not include the mechanical mixture.

Previously, the relationship has been derived between aio and the number of sites over which mixing is occurring. For simple cases where mixing between two end-members involves mixing of ions on one type of site, containing y sites per unit formula, we have shown (equation 69) that

aio = (Xi)y (108)

A regular solution assumes ideal mixing with a non-zero excess mixing. The excess mixing is assumed to be due to the change in bond energies resulting from an increase in the number of ions on adjacent sites that are different. Because ideal mixing is assumed, the number of adjacent sites occupied by different ions can be calculated from random mixing. The excess free energy of mixing for a binary regular solid solution was given by equations (72) and (73). The Gibbs free energy of a regular solid solution is symmetrical about the midpoint in composition. Real solid solutions are not symmetrical about a midpoint. In addition, the limitation that zE-ex (the exchange energy between the two atoms on the sites) cannot exceed RT in order for random mixing to occur is an unrealistic assumption.

A more practical way to represent Gex is to use a Margules expansion and to calculate Gmix by adding the ideal mixing term to it. This is an empirical approach in which the Gibbs free energy of a solid solution is first measured as a function of composition, temperature, and pressure and then fit to the Margules expansion. The fitting involves finding best-fit values of the parameters in the equation.

For a binary solid solution involving the mixing of two different ions on a lattice.

Gex = a + bX2 + cX22 + dX23 + ... (109)

where a, b, c, d,... are Margules parameters which can be functions of temperature and pressure. Note that if a three parameter equation is used in which a = 0 and b = -c, then

Gex = -cX2 + cX22 = X2X1b (110)

which is identical to the previously derived results for a regular solution, i.e., substitution of equation (73) into equation (72) for one mole of components in a binary mixture. The b term above is equivalent to zE-ex in equation (73).

To obtain an asymmetrical binary solid solution we use a four parameter Margules expansion in which a = 0. The resulting expression for Gex contains 3 parameters (b, c, d) which can be combined into 2 new parameters WG1 and WG2 by letting

b = WG2

c = WG1 - 2WG2

d = WG2 - WG1

Gex = X1X2(WG1X2 + WG2X1) (111)

In a solid solution, a phase is unstable if its composition falls between two minimum in a plot of Gmix versus composition (at constant temperature and pressure). These two minimum points mark the solvus and can be calculated by setting the first partial derivative of Gmix with respect to composition to zero. The actual unmixing only occurs if the phase's composition falls between the two inflection points between the two minimum and the maximum in a plot of Gmix versus composition. These two points mark the spinoidal and can be calculated by setting the second partial derivative of Gmix with respect to composition to zero. The fact that phases having a composition between the solvus and the spinoidal do not undergo unmixing is because unmixing will produce even more unstable phases in this region. For a constant pressure, the critical temperature and critical composition above which all compositions are stable is found by setting the first, second, and third partial derivatives equal to zero and solving for temperature and composition.

Surface Free Energy and the Arrhenius Equation

The rate k of a reaction can often be represented by the empirical Arrhenius equation which contains two fit constants, Ea, the activation energy and A, the pre-exponential factor:

k = Ae-Ea/RT (112)

A has the same units as the rate constant or sec-1 for a first order rate constant (e.g., radiaoactive decay) and is sometimes called the frequency factor. Note that taking the natural log of equation (112) and converting to base 10 log gives

log k = [Ea/(2.303R)](1/T) + log A (113)

A plot of experimental k values versus (1/T) is often linear. In this case, A and Ea are determined, respectively, from the y intercept which is log A and the line slope which is [Ea/(2.303R)].

The surface free energy of nucleation can sometimes be linked to the rate of nucleation. The Gibbs surface free energy is given by the surface tension which is the energy (per cm2) required to increase the surface area at constant T and P in the absence of a chemical reaction. We can modify equation (10c) by adding , as the partial of G with respect to a change in surface area, i.e.,

dG = -SdT + VdP + idni + idAreai (114)

i and Gr, the free energy change in the reaction forming a mole of the component making up the particle, can be used to compute Gn, the change in Gibbs free energy resulting from nucleating a particle of component i.

For example, for a particle of cubic shape with edges of length d in cm,

Gn = 6d2i + d3(/Mw)Gr (115)

where and Mw are the density in g/cm3 and the molecular weight in g/mole of the component being nucleated. The first term is positive and the second term is negative in equation (115). At very small values of d, the first term dominates over the second term. The situation is reversed at large values of d. A plot of Gn versus d has a maximum which is often assumed to equal the activation energy in the Arrhenius equation. dc, the d corresponding to this maximum value, is found by taking the first partial derivative of Gn with respect to d and setting this equal to zero to solve for dc. For the situation of cubic particles,

dc = - 4Mw/(Gr) for cubic particles. (116)

Substituting dc from equation (116) into equation (115) gives

(Gn)max = 323[Mw/(Gr)]2 for cubic particles. (117)

For a spherical particle, what are the expressions for the critical radius and (Gn)max? Water has a surface tension of 71.97 ergs/cm2 (1.739 x 10-6 calories) in contact with the atmosphere at 25oC and 1 bar. What Gr is necessary for raindrops of water to form from saturated atmosphere with (Gn)max at a particle diameter of 1 micron (10-4 cm) 25oC and 1 bar. What activity ratio of aH2O liquid/aH2O vapor corresponds to this Gr. Assume the water in the raindrops is pure water and the water vapor in the atmosphere behaves ideally, i.e., with unit activity coefficients. The standard state chemical potentials of water and water vapor are, respectively, -56.687 and -54.630 kcal/mole. Is the computed mole fraction of H2O in the vapor phase reasonable for the atmosphere?

When rain falls into a body of water, the surface area of the rain drops is destroyed which releases heat as the result in the change in enthalpy associated with the surface area. We can write equation (33) for the contribution of surface area to G or Garea to calculate Harea, the contribution of surface area to H.

Garea = Harea - TSarea = Harea + T(Garea/T) (119)

and since Garea = Area, we have Area = Harea + T(/T)Area from which Harea can be solved. For water, the partial of with respect to T is -0.148 ergs/oKcm2. Plot the release of heat in calories per mole of rainwater as a function of the size of raindrops being absorbed into a lake.

Equations of State for Molar Volume in a Pure Gas Phase

The ideal gas law is obeyed when gas atoms or molecules do not interact. This condition is usually satisfied at low P and/or high T when the particle interaction energy is much less that the thermal energy of the particle.

PV- = RT (120)

The compressibility factor z is defined from the ideal gas law and is a convenient scale to determine how ideal a gas is behaving.

z = PV-/RT (121)

Van der Waals equation takes into account the volume b of the gas atoms or molecules at absolute zero and attractive forces between particles. The former correction to the ideal gas law is made by changing the molar volume V- to V- - b, which corrects the volume of a gas to be nonzero, i.e., b, at absolute zero. The latter correction is made by addition to the ideal gas law of the term -a/V-2. The 1/V-2 term is proportional to the square of the particle concentration, i.e., n/V-2, which in turn is proportional to the probability of two particles interacting. The proportionality constant a is a function of temperature and to a lesser extent of pressure. The negative sign occurs because attraction reduces volume.

P = RT/(V- - b) - a/V-2 (122)

The Redlich-Kwong equation is a modification of the Van der Waals equation with a more complicated attraction term

P = RT/(V- - b) - a/[T1/2V-(V- + b)] (123)

Sometimes a virial equation is used to represent the compressibility factor as a function of V-.

z = PV-/RT = 1 + B/V- + C/V-2 + D/V-3 + ... (124)

The form of the virial equation can be derived from statistical thermodynamics in which B represents deviations from ideality due to 2 particle interactions, C represents deviations due to 3 particle interactions, etc.

Solving by algebra for V- from van der Waals equation, the Redlich-Kwong equation, or the truncated three term virial equation involves the solution of a cubic equation - not easy algebra. An easy way is solve for V- in any of these three equations is to iterate using Newton's method.

At the critical point the properties of a gas pass without a phase transition to those of the liquid. In a plot of P (y axis variable) versus V- (x axis variable), the first and second partial derivatives of P with respect to V- are zero at constant T, i.e., the inflection point has moved to a maximum point on the curve. These two conditions allow for the evaluation of the a and b parameters in the Van der Waals and Redlich Kwong equations (incredible algebra). These parameters, calculated at the critical point, are assumed to be constant and are used with fair success at other PT points for non-polar gases.

aVdW = 0.4219 R2Tc2/Pc (125)

bVdW = 0.1250 RTc/Pc (126)

aR-K = 0.4275 R2Tc2.5/Pc (127)

bR-K = 0.0866 RTc/Pc (128)

The critical point compressibility factors (zc) for non-polar gases (e.g., N2, CO2, He, O2, SO2, H2S) are between 0.27 and 0.29, i.e., the gases behave similarly at their critical points. This fact led to the idea that an unknown equation of state, using reduced variables Tr, Pr, and V-r equal to their respective values divided by their critical point values, i.e., T/Tc, P/Pc, and V-/V-c, could describe the properties of a pure gas phase. This is called the theory of corresponding states. To this end, charts have been drawn relating the compressibility factors z (and also activity coefficients) as a function of Pr for isotherms of Tr. The charts have been averaged to find one that gives the best fit to all the non-polar gas data. This type of best-fit chart is often used to predict the activity coefficient for a pure gas phase in calculations. Unfortunately, H2O is a polar molecule and doesn't fit the data well, i.e., zc is 0.229.

Computing Gas Fugacities in a Mixture of Non-Ideal Gases

For a single component gas phase, we previously derived equation (61) to compute the fugacity of gas component.

ln i,P,T = -(1/RT)[(RT/P) - V-]dP (61)

For a gas phase of more than one component, the above equation is the same except the molar volume becomes the partial molar volume V-i in a mixture. The equations of state are written to isolate P not V of the gas phase, making it difficult to take the partial derivative V with respect to ni to derive V-i. The equation used for this purpose is derived below as equation (129), using one of Maxwell Relations from the differential of Helmholtz equation, i.e., makes use of the fact that the order of differentiation can be reversed to set two second partial derivatives equal.

We begin with the definition of the chemical potential.

i = *i + RT ln[XiP(i)]

Holding constant both T and moles of all components while taking the partial derivative of the above expression with respect to V gives

(i/V)T,n = RT(ln XiPi/V)T,n

where the partial derivative of *i was zero because the standard state of a gas doesn't vary with P, i.e., V cannot be varied at constant T and n without varying P. The term on the left is given by one of Maxwell relations, using the second partial derivatives of the the first partial derivatives in equations (13b) and (13c) of the expression for dA in equation (9c).

(i/V)T,n = -(P/ni)T,nj

Combining the two equations and integrating with respect to V from V = (gas behaves ideally because of low pressure) to V gives

V(P/ni)T,nj dV = RT ln XiPi/V)T,n - RT ln XiPV=>

The negative sign is gone from the integral on the left hand side because the order of integration was reversed. Note that the last term on the right is infinity, i.e., - ln 0 = . Also, the activity coefficient from the last term on the right is gone because it is one when the gas is behaving ideally.

The following terms are added to the right hand side.

RT ln RT - RT ln RT + RT V(l/V)dV - RT V(l/V)dV

This gives

V[(P/ni)T,nj dV - RT/V] dV = RT ln XiPi - RT ln XiPV=>

- RT ln (/RT) + RT ln (V/RT)

Remembering that the ideal gas law holds as V => , Xi = ni/ntot, and V = V-ntot

-RT ln XiPV=> = -RT ln niRT/VV=>) = RT ln /RT + RT ln (1/ni)

By subsitution into the previous equation,

RT ln XiPi = V[(P/ni)T,nj dV - RT/V] dV - RT ln (V/niRT)

Add - RT ln XiPi to both sides of the above equation and note

that XiV/ni = V- gives equation (129)

ln i,P,T = (1/RT)V[(P/ni)V,T,ni - RT/V]dV - ln[PV-/(RT)] (129)

van der Waals Equation for Gas Mixtures

The van der Waals equation for a mixture (equation 122) can

be rewritten with the molar volume V- of the mixture replaced with V/ntot. This is done to integrate equation (129).

P = RT/(V- - b) - a/V-2 (122)

P = ntotRT/(V - bntot) - a(ntot)2/V2 (130)

Equation (130) can be used to take the partial derivative of P with respect to ni and then integrated in equation (129) to determine ln i,P,T. We have to decide on mixing rules for computing a and b of the mixture. The usual mixing rules are linear for b over all gas components and geometric for a over all possible pairs of gas components.

b = i=1Xibi (131)

and a = i=1 j=1 XiXjaij (132)

where aii and bi are for the a and b values for the ith component, previously derived from the pure gas properties, using the van der Waals's equation, at the critical point (equations 125 and 126). If i =/ j,

aij = (aiaj)0.5 (133)

Equation (129) is integrated, using the above mixing rules and van der Waals equation, yielding

ln i,P,T = ln[V-/(V--b)] + bi/(V--b) - ln[PV-/(RT)]

- [1/(RTV-)][2ai0.5j=1 Xjaj0.5] (134)

V- for the mixture for use in equation (134) is determined from the equation of state, equation (123), using a procedure such as Newton's method.

Redlich-Kwong Equation for Gas Mixtures

The above procedure can be repeated for the Redlich-Kwong equation to get a more accurate representation of the gas activity coefficient than given by van der Waals equation. Equation (123) was the Redlich-Kwong equation.

P = RT/(V- - b) - a/[T1/2V-(V- + b)] (123)

As with van der Waals equation, for integration purposes in equation (129), the molar mixture volume was initially replaced with the mixture volume using the total moles present in the mixture (ntot). We use the two mixing relations given by equations (131) and (132) for b and a. However, ai and bi are now given by modifications of equations (127) and (128) which were previously derived from critical point properties with the Redlich-Kwong equation. The universal constants in those equations are replaced below with component-specific constants.

ai = ai R2Tc2.5/Pc (135)

bi = bi RTc/Pc (136)

For the mixing relation in equation (132) used when i =/ j, equation (133) is replaced with

aij = (0.5/Pcij)(ai + aj)R2T2.5cij (137)

where Pcij, ai, and Tcij are defined later. Note the subscript c refers to a critical point parameter.

The integrated form of the Redlich-Kwong equation in equation (129) is

ln i,P,T = ln[V-/(V--b)] + bi/(V--b) - ln[PV-/(RT)]

+ [abi/(RT1.5b2)]{ln[(V-+b)/V-] - b/(V-+b)}

-[2/(RT1.5b)]{ln[(V-+b)/V-]}j=1 Xjaji (138)

As was the case for the van der Waals equation, V- for the mixture for use in equation (138) is determined from the equation of state, equation (123), using a procedure such as Newton's method.

The constants in equation (137) are defined below for use in equation (138).

Pcij = 8[0.291 - 0.04(i + j)]RTcij/(V-ci1/3 + V-cj1/3)3 (139)

Tcij = (TciTcj)0.5(1-kij) (140)

where V-ci and Tci are the molar volume and temperature at the critical point for the ith component. i, ai, and bi are constants for the ith component and kij represents the deviation from the geometric mean for Tcij. These latter four constants are tabulated in the book by Prausnitz (1969, p. 157-159) entitled "Molecular Thermodynamics of Fluid-Phase Equilibria"

Using equations (123) and (138) is straightforward but complicated because of the component constants and binary constants for binary interactions.

A complication arises for quantum gases (H2, He, Ne) in which ai and bi are 0.4278 and 0.0867, respectively, for all quantum components in equations (135), (136), and (137). If either i or j is a quantum gas component, equation (140) is replaced with

Tcij = (TcioTcjo)0.5(1-kij)/[1 + 21.8/(MwijT)] (141)


Mwij = 2MwiMwj/(Mwi + Mwj) (142)

in which Mwi and Mwj are the molecular weights of components i and j, respectively. If i is for a quantum gas, Tcio is the classical critical temperature; otherwise, Tcio is the critical temperature of the ith non-quantum component. Tcjo is defined in a similar fashion.

Also, if either i or j is a quantum gas, equation (139) is replaced with

Pcij = 8[0.291-0.04(i+j)]R(TcioTcjo)0.5(1-kij)/(V-cio1/3 + V-cjo1/3)3 (143)

in which Tcio and Tcjo are defined as in equation (141). If i is for a quantum gas, i is zero and

V-cio = 0.291RTcio/Pcio (144)

Otherwise, V-cio is the critical volume of the ith component at the critical point. Pcio is the classical critical pressure of the ith quantum component or the critical pressure of the ith nonquantum component. j, V-cj, and Pcjo are defined in a similar fashion for use in equation (143).

Virial Equation for Gas Mixtures

The virial equation of state commonly used for gases was given in equation (124). The form of the equation can be derived from statistical mechanics, making it theoretically more significant than van der Waals or the Redlich-Kwong equations. The equation is commonly truncated after the third term, eliminating higher order terms.

z = PV-/RT = 1 + B/V- + C/V-2 (145)

From statistical mechanics, B and C are the temperature-dependent second and third virial coefficients, for two particle and three particle interactions, respectively. At high densities (approaching the critical point) higher order virial coefficients, corresponding to 4 particle and 5 particle interactions may be necessary.

The virial coefficients for the mixture are related to the virial coefficients for the binary and ternary interactions by the following mixing rules which follow from statistical mechanics.

B = i=1 j=1 XiXjBij (146)

C = i=1 j=1 k=1XiXjXkCijj (147)

The integrated form of equation (129) for the virial equation becomes

ln i,P,T = (2/V-)j=1 XjBij + (2/V-2) j=1 k=1 XjXkCijk - ln[PV-/(RT)] (148)

As with van der Waals and the Redlich-Kwong equations, V- in equation (148) is calculated from the equation of state (equation 145). The second and third virial coefficients for binary and ternary interactions, respectively, can be calculated from intermolecular potentials of two and three particles. Equation (148) will be used more as more virial coefficients are available from theoricaal calculations and experimental measurements.

Kinetics and Diffusion

Fick's first Law

flux = - D(C/x)T,P

where D is the diffusion coefficient in cm2/sec, x is the distance in cm, and C is the concentration in g/cm3. In a liquid, D is of the order of 10-5cm2/sec and in a solid D is < 10-10cm2/sec. In porous fluid-saturated sediment,

Dsed Dfluid/2

Where is the porosity fraction and is the tortuosity or the ratio of the true path length L to the distance x along the x axis. The porosity factor reduces the flux to that moving through the liquid phase. The reciprocal of one of the tortuosity factors reduces the flux because of lower concentration gradient along the flow path C/L, rather than C/x. The other tortuosity reciprocal corrects the flux which is parallel to the L direction to the component parallel to the x axis (uses the cosine of the angle between the L direction and the x direction). Often the estimation for Dsed in the literature is

Dsed Dfluid/

eliminating the cosine correction which may not be necessary (?).

In sediments, the ratio of /, is estimated from the formation factor F, the ratio of the specific electrical resistance of the saturated sediment (rsed) to the that of the fluid within the sediment (rfluid). The specific electrical resistance is measured using a cube of specified geometry for both the fluid and saturated sediment. The assumption is made that the electrical current is carried almost entirely within the fluid in the sediment. The increase in specific electrical resistance of the cube of saturated sediment over a cube of fluid is proportional to L/x, the increased path length and inversely proportional to , the porosity fraction (because current only flows through the cross-sectional area of the pores).

F = rsed/rfluid = (L/x)/ = /

Diffusion away from a dissolving grain as a rate-limiting step, i.e., the rate of dissolution is faster than diffusion so diffusion becomes rate limiting.

Let x be the width of the diffusion layer in cm in solution. Let Co and C be the aqueous concentrations (g/cm3) near the surface of the grain and in the bulk solution, respectively. Let Dfluid be the diffusion coefficient (cm2/sec) of the components in the fluid, t be the time in sec, and V be the stirred volume in cm3 the solution. We can set the flux diffusing away from the grain (across x, a constant boundary thickness), equal to the change in bulk solution composition around the grain and integrate from time t = 0 to time t, assuming Co remains constant and C in solution was initially zero at time t = 0.

-Dfluid(C-Co)/x = V(C/t)T,P

t=0t (Dfluid/Vx) dt = C=0C (1/(Co-C) dC

(Dfluid/Vx)t = - ln[(Co-C) - [-ln(Co-0)]

exp (tDfluid/Vx) = Co/(Co-C) or

C = Co[1 - 1/exp (tDfluid/Vx)]

At t = 0, C = 0, at t => , C => Co

Diffusion through a surface layer of thickness x which forms as the mineral surface dissolved.

x increases in thickness with increasing time. How does the concentration in the surrounding solution change with time assuming the surrounding solution is stirred. We set the flux diffusing out to the flux of released component resulting from the increasing thickness of the layer x. The terms are defined as before except x is no longer constant and is the g/cm3 in the dissolving mineral of the component being released as the surface layer forms and Asurf is the surface area of the dissolving component.

Dfluid(C-Co)/x = Asurf(x/t)T,P

-t=0t [Dfluid(C-Co)/(Asurf)]dt = x=0x x dx

Co is constant (set by a steady state balance between dissolution and diffusion) and very large and C is very small. Hence, even though C is slowly increasing with time, we can to a first approximation assume C-Co is constant with time.

tDfluid(Co-C)/(Asurf) = x2/2

or x = [2tDfluid(Co-C)/(Asurf)]0.5

Noting that the change in aqueous concentration C is equal to

the change in concentration divided by V, the solution volume.

dC = -t=0t [Dfluid(C-Co)/x]dt

= t=0t [Dfluid(Co-C)/[2tDfluid(Co-C)/(Asurf)]0.5]dt

= t=0t {Dfluid(Co-C)(Asurf)]/(2t)]}0.5]dt

= {2Dfluid(Co-C)(Asurf)t}0.5

Hence, the change in concentration is proportional to the square root of time!

Listed below are several problems. Do not work together. To receive any credit, show all your work in a logical sequence. You can use a spread sheet for your computations but you must show the logical sequence and all equations.

First Problem

Construct a pe-pH diagrams for the system: FeO, Fe2O3, CO2, SiO2, H2S, and H2O at 100oC and 300 bars, 200oC and 650 bars (Mark), and 300oC and 1000 bars. Consider these solid phases: Fe2O3, Fe3O4, FeSiO3, FeCO3, FeS2, in which reactions written between solids are balanced on Fe. Consider the following aqueous species: Fe2+, Fe3+, Fe(OH)2+, and Fe(OH)+. Assume aH2O = 1, aCO2 = 0.1P, aH2S = 0.05P, and constant molar volume of solids. Assume quartz is present and in equilibrium with the aqueous solution, so any reactions releasing SiO2 can be written directly to quartz. Set the aqueous activity of the dominant iron species in any reaction written to a solid to be 10-4.5 at 100oC and 300 bars; 10-4 at 200oC and 650 bars; and 10-3.5 at 300oC and 1000 bars. Use the following data to calculate your diagram where the a, b, and c parameters are Maier Kelley heat capacity coefficients.

If assuming a constant activity of H2S results in replacing hematite with pyrite, you have the option of removing pyrite from your considered phases.

o in calories/mole

25oC 100oC 200oC 300oC

1 bar 300 bars 650 bars 1,000 bars

Fe2O3 -178,155

Fe3O4 -242,574

FeSiO3 -267,160

FeCO3 -162,414

FeS2 - 38,293

CO2 gas - 94,262

H2S gas - 8,016

SiO2 quartz -204,646

H2O liquid - 57,964 - 60,085 - 62,563

Fe2+ aq. - 20,080 - 17,530 - 14,840

Fe3+ aq. + 91 + 6,834 + 14,260

Fe(OH)2+ aq. - 56,946 - 54,745 - 52,058

Fe(OH)+ aq. - 66,310 - 65,261 - 64,208

e- and H+ aq. 0 0 0

So25 C, 1 bar a b c V

cal/oK/mole cm3/mole

CO2 gas 51.072 10.57 0.00210 -206,000

H2S gas 49.16 7.81 0.00296 -46,000

Fe2O3 20.94 23.49 0.01860 -355,000 30.274 Fe3O4 34.83 21.88 0.04820 0 44.524

FeSiO3 22.6 26.49 0.00507 -555,000 32.952

FeCO3 25.1 11.63 0.02680 0 29.378

FeS2 12.65 17.88 0.00132 -305,000 23.940

SiO2 qtz 9.88 11.22 0.00820 -270,000 22.688

Second Problem

SO2 is a frequent waste product gas in industry processes. The gas can be removed by reacting with portlandite Ca(OH)2 or with limestone CaCO3 at higher than surface temperatures to produce anhydrite CaSO4 in the presence of O2 and producing either CO2 or H2O.

reaction CaCO3 + SO2 + 1/2 O2 = CaSO4 + CO2

reaction Ca(OH)2 + SO2 + 1/2 O2 = CaSO4 + H2O

Write out the expression for Gr,T,P for your reaction. Assume a constant T (one of the 2 T values given below) and a constant reaction Q for the components in the reaction and calculate Gr,T,1 for the T and then calculate the P at which Gr,T,P is zero. Repeat this procedure for a range of values of the reaction quotient Q which give a positive pressure at 0 to a few thousand bars. Plot this equilibrium isotherm as a function of the Q versus P. Do this for each of the following temperatures: 500oC and 1,000oC.

Assume a gas phase has XSO2 = 0.5 and the solids are pure solids in their standard states. Plot your 500oC isotherm data (Gene) and your 1000oC isotherm data as the ratio of the other two gas components, XCO2/XO2 and XCO2/XH2O, assuming the activity of each gas component can be represented by the Lewis rule where i is equal to the activity coefficient of the ith gas component in a pure gas of i at the pressure P. Determine i from the corresponding state activity coefficient chart in your textbook.

Use the following thermodynamic data at 25oC and 1 bar to move through PT space.

o in calories/mole

25oC and 1 bar

CaCO3 -270,100

Ca(OH)2 -214,724

CaSO4 -315,925

SO2 gas - 71,750

O2 gas 0

CO2 gas - 94,262

H2O steam - 54,634

So25 C, 1 bar a b c V

cal/oK/mole cm3/mole

CaCO3 22.15 24.98 0.00524 -620,000 36.93

Ca(OH)2 19.93 19.07 0.01080 0 33.06

CaSO4 25.5 16.78 0.02360 0 45.94

SO2 gas 59.33 11.04 0.00188 -184,000

O2 gas 49.03 7.16 0.001 - 40,000

CO2 gas 51.07 10.57 0.00210 -206,000

H2O steam 45.10 7.30 0.00246 0

3rd Problem

Listed below are three different reactions used in geothermometers, each involving two solid solutions in which reactions are written between end-member components in the solid solutions. Write out the expression for Gr,T,P for your reaction. Assume a constant T (one of the 4 T values given below) and a constant reaction Q for the components in the reaction and calculate Gr,T,1 for the T and then calculate the P at which Gr,T,P is zero. Repeat this procedure for a range of values of the reaction quotient Q which result in P values between 0 and a few thousand bars. Try Q between 0.01 and 100. Plot this equilibrium isotherm as a function of the Q reaction quotient. Do this for the following temperatures: 100oC, 400oC, 700oC and 1000oC. Use the four curves to find 4 potential PT points representing equilibrium for the actual assemblage composition given for each reaction. Assume ideal mixing for the assemblage composition.

Reaction 1: pyrope + annite = almandine + phlogopite

Mg3Al2Si3O12 + KFe3AlSi3O10(OH)2 = Fe3Al2Si3O12 + KMg3AlSi3O10(OH)2

The assemblage composition to be used has [Fe2Mg]Al2Si3O24 for ideal site mixing between almandine and pyrope in garnet and K[Mg2Fe]AlSi3O10(OH)2 for ideal site-mixing between annite and phlogopite in biotite. Use Karpov's o25 C, 1 bar data which hopefully are internally consistent.

Reaction 2: muscovite + albite = paragonite +


KAl2(AlSi3)O10(OH)2 + NaAlSi3O8 = NaAl2(AlSi3)O10(OH)2 + KAlSi3O8

The assemblage to be used has [K0.95Na0.05]Al2(AlSi3)O10(OH)2 for ideal site mixing between muscovite and paragonite in a mica and [K0.9Na0.1]AlSi3O8 for ideal site mixing between albite and microcline.

Reaction 3: hydroxy-apatite + 0.5fluoro-phlogopite = fluoro-apatite + 0.5 hydroxy-phlogopite

Ca5(PO4)3(OH) + 0.5KMg3(AlSi3)O10F2 =

Ca5(PO4)3F + 0.5KMg3(AlSi3)O10(OH)2

The assemblage to be used has Ca5(PO4)3[(OH)0.8F0.2] for ideal site mixing within apatite and KMg3(AlSi3)O10[(OH)0.9F0.1] for ideal site mixing within phlogopite. Use Robie"s and Helgeson's o25 C, 1 bar data which hopefully are internally consistent.

The thermodynamic data needed for the problems are listed below.

o in calories/mole

25oC and 1 bar

Ca5(PO4)3OH -1,502,412 Robie

Ca5(PO4)3F -1,542,968 Robie

KMg3(AlSi3)O10F2 -1,446,725 Robie

KMg3(AlSi3)O10(OH)2 -1,396,187 Helgeson, -1,403,585 Karpov

Fe3Al2Si3O12 -1,187,811 Karpov

Mg3Al2Si3O12 -1,428,776 Karpov

KFe3AlSi3O10(OH)2 -1,147,156 Helgeson, -1,159,082 Karpov

KAl2(AlSi3)O10(OH)2 -1,339,249 Berman

NaAl2(AlSi3)O10(OH)2 -1,331,352 Berman

NaAlSi3O8 - 886,734 Berman

KAlSi3O8 - 896,776 Berman

So25 C, 1 bar a b c V

cal/oK/mole cm3/mole

Ca5(PO4)3OH 93.30 114.26 0.01981 -2,500,000 157.56

Ca5(PO4)3F 92.70 113.39 0.01362 -2,453,000 159.60

KMg3(AlSi3)O10F2 75.90 96.13 0.02674 -2,041,000 146.37

KMg3(AlSi3)O10(OH)2 76.10 100.61 0.02878 -2,150,000 149.66

Fe3Al2Si3O12 75.60 97.50 0.03364 -1,873,000 115.28

Mg3Al2Si3O12 62.32 91.69 0.03264 -2,092,000 113.15

KFe3AlSi3O10(OH)2 95.20 106.43 0.02977 -1,931,000 154.32

KAl2(AlSi3)O10(OH)2 70.07 97.56 0.02638 -2,544,000 140.87

NaAl2(AlSi3)O10(OH)2 66.37 97.40 0.02450 -2,644,000 132.16

NaAlSi3O8 53.64 61.70 0.01390 -1,501,000 100.83

KAlSi3O8 51.18 63.83 0.01290 -1,705,000 108.69

4th Problem

Construct an activity phase diagram at 300oC and 1 kilobar for the Na2O, K2O, SiO2, Al2O3, HCl, H2O system. Assume quartz saturation so the aSiO2 is not a variable. Also assume aH2O is one. Compute the maximum number of phases (including water and quartz) that can be at equilibrium (from the phase rule) on this diagram. Consider the phases listed below and elevate the following 25oC and 1 bar standard state data from Helgeson to 300oC and 1 kilobar. If nephline and kalsilite do not show up on the phase diagram than pick a lower activity of SiO2 aq than at quartz saturation.

o in calories/mole

25oC and 1 bar 300oC and 1 kbar

nepheline NaAlSiO4 - 472,872

kalsilite KAlSiO4 - 481,750

microcline KAlSi3O8 - 895,374

albite NaAlSi3O8 - 884,509

muscovite KAl3Si3O10(OH)2 -1,336,301

paragonite NaAlSi3O10(OH)2 -1,326,012

quartz SiO2 - 204,646

kaolinite Al2Si2O5(OH)4 - 905,614

pyrophyllite Al2Si4O10(OH)2 -1,255,997

water H2O - 62,563

SiO2 aq -203,100

Na+aq - 67,590

K+aq - 74,260

H+aq 0

So25 C, 1 bar a b c V

cal/oK/mole cm3/mole

NaAlSiO4 29.72 35.91 0.00646 - 732,797 54.16

KAlSiO4 31.85 29.43 0.01736 - 532,000 59.89

KAlSi3O8 51.13 63.83 0.01290 -1,705,000 108.74

NaAlSi3O8 52.30 61.70 0.01390 -1,501,000 100.07

KAl3Si3O10(OH)2 68.8 97.56 0.02638 -2,544,000 140.71

NaAlSi3O10(OH)2 66.40 97.43 0.02450 -2,644.000 132.53

SiO2 9.88 11.22 0.0082 - 270,000 22.69

Al2Si2O5(OH)4 48.53 72.77 0.02920 -2,152,000 99.52

Al2Si4O10(OH)2 57.2 79.43 0.03921 -1,728,200 126.6

molar CO2 gas phase at 100oC from F. Din (1962) Thermodynamic Functions of Gases, Volume 1


0.5 atm 0.5066 bars 61,163.5 cm3 1.971

1 1.013 30,545 0.9845

2 2.026 15,235.75 0.4911

3 3.040 10,132.4 0.3266

4 4.053 7,580.7 0.2443

5 5.066 6,050 0.1950

6 6.079 5,029 0.1621

8 8.105 3,573.6 0.11518

10 10.13 2,988.3 0.09632

12 12.16 2,477.3 0.07985

15 15.20 1,966.8 0.06339

20 20.26 1,456.3 0.04694

25 25.33 1,150 0.03707

30 30.40 945.8 0.03048

40 40.53 690.5 0.02226

50 49.35 536.9 0.01731

60 60.79 434.3 0.01400

80 81.05 305.8 0.00986

100 101.3 229.6 0.00740

150 152.0 129.74 0.00418

200 202.6 89.82 0.00290

300 304.0 66.01 0.00213

The integrated first method is

Gor,T,P = Hor,Tref,Pref + (Cpro)PrefdT - TSor,Tref,Pref

- T(Cpro/T)PrefdT] + Vor,TdP

= Hor,Tref,Pref - TSor,Tref,Pref + Vor,TdP

+ (ar + brT + cr/T2)PrefdT

- T(ar/T + br + cr/T3)PrefdT]

= Hor,Tref,Pref - TSor,Tref,Pref + Vor,TdP

+ ar(T - Tref) + 0.5br(T2 - Tref2) - cr(1/T - 1/Tref)

- T[ar ln(T/Tref) + br(T - Tref) - 0.5cr(1/T2 - 1/Tref2)]

= Hor,Tref,Pref - TSor,Tref,Pref + Vor,TdP

+ ar[T-Tref-T ln(T/Tref)] + br[0.5(T2 - Tref2) - T(T-Tref)]

+ cr[-(1/T - 1/Tref) + 0.5(1/T2 - 1/Tref2)]

If Vor,T is constant with pressure then

= Hor,Tref,Pref - TSor,Tref,Pref + Vor,T(P-Pref)

+ ar[T-Tref-T ln(T/Tref)] + br[0.5(T2 - Tref2) - T(T-Tref)]

+ cr[-(1/T - 1/Tref) + 0.5(1/T2 - 1/Tref2)]

Note that if Cpor,Pref had been constant, then the first method gives

Gor,T,P = Hor,Tref,Pref - TSor,Tref,Pref + Vor,T(P-Pref)

+ Cpro dT - TCpro(1/T)dT]

= Hor,Tref,Pref - TSor,Tref,Pref +Vor,T(P-Pref)

+ Cpor,Pref[T-Tref + T ln(Tref/T)]

= Gor,Tref,Pref + Sor,Tref,Pref(Tref-T) + Vor,T(P-Pref)

+ Cpor,Pref[T-Tref + T ln(Tref/T)]

For the second method, if Cpor,Pref had been constant

Gor,T,P = Gor,Tref,Pref - (Sor,Tref,Pref + Vor,T(P-Pref)

+ Cpor(1/T)dT)dT

= Gor,Tref,Pref - Sor,Tref,Pref(T-Tref) + Vor,T(P-Pref)

- Cpor(ln T - ln Tref)dT

= Gor,Tref,Pref - Sor,Tref,Pref(T-Tref) + Vor,T(P-Pref)

- Cpor[T ln(T) - T - Tref ln(Tref) + Tref - ln(Tref)(T-Tref)]

= Gor,Tref,Pref - Sor,Tref,Pref(T-Tref) + Vor,T(P-Pref)

+ Cpor[T-Tref + T ln(Tref/T)]

which is identical to the first method (above).

Example Midterm

Home sewage systems allow for the breakdown of organic matter by bubbling air through the sludge to stimulate bacteria growth. In the process, the organic matter is converted to CO2, organic nitrogen is transformed from ammonium (NH4+) to nitrate, (NO3-) and organic polyphosphates (Hn+2PnO2n+2) are transformed to orthophosphate species (PO43-). In St. Tammany Parish, the nitrogen conversion is usually incomplete because of the high water table which retards oxidation of the nitrogen. The nutrients (nitrogen and phosphorous species) in the effluent eventually cause problems in lakes and streams by stimulating the growth of algae and other plants. Your midterm examines some ways of removing the nutrients in the effluent of a home sewage system. Complete A-D below, using the thermodynamic data provided and show your work. From your results, write a one page typewritten report, describing the aqueous speciation and the best way to remove the nutrients.

A) In the output from a home sewage system, nutrients are released as forms of oxidized or reduced nitrogen (NO3-, HNO3-, NO2-, HNO2-, NH3 and NH4+) and forms of orthophosphate (H3PO4, H2PO4-, HPO42-, and PO43-). Construct a 25oC and 1 bar pH-pe diagram with aH2O = 1 for the aqueous nitrogen system and overlay it with the pH diagram for the aqueous phosphorous system. Use the stability boundaries of water to define the upper and lower stability lines on the diagram. Use different colored lines for the phosphorous boundaries and the nitrogen boundaries.

B) We want to get rid of the nitrogen and the phosphorous. If the nitrogen is in the form of NO3- (this is desirable but unusual in St. Tammany Parish), we can set up an oxygen depleted environment to let the bacteria reduce the nitrate to N2 gas which will escape to the atmosphere (denitrification). We do not have to generally worry about reducing the nitrate all the way down to ammonium (nitrogen fixation) because only a few (actually aerobic) bacteria can do this.

2H+ + 2NO3- + CH2O = N2 + CO2 + H2O

or in terms of pe,

12H+ + 2NO3- + 10e- = N2 + 6H2O

Draw in the stability boundary on your pe-pH diagram in (A) for the boundary between a TOTAL aqueous nitrogen activity of 10-6. Assume the activity of N2 gas in the atmosphere is 0.8 bars and that the solution is in equilibrium with the atmosphere. Use a different colored line than the two colors used in Part A above. Note that the reaction above between N2 and the aqueous nitrogen species should be written to the dominant nitrogen species in each of the regions that it crosses on the diagram. This boundary marks the pe as a function of pH that is needed to remove most of the nitrate by inorganic denitrification. However, because the denitrification is controlled by the metabolism of bacteria, the true boundary in nature is somewhat different.

C) To remove the aqueous orthophosphate, we need to precipitate it. Some possibilities are Mg oxyapatite [Mg4O(PO4)2], magnesium hydroxyapatite [Mg5OH(PO4)3], and farringtonite [Mg3(PO4)2]. All of these minerals precipitate more readily as the pH rises but the reaction kinetics are not well known. If we add granular brucite [Mg(OH)2]as a source of Mg2+, we also raise the pH. Assume equilibrium of each of these three Mg phosphate minerals with brucite and compute and plot the sum of the aqueous activities of H3PO4 + H2PO4- + HPO42- + PO43- as a function of pH between 7 and 14 to determine which phosphate mineral has the lowest phosphate solubility. Balance your reactions by keeping Mg in the solid phases so you don't have to assume anything about the aqueous activity of Mg.

D) What if the nitrogen is not in the form of nitrate, but occurs as ammonium, which unfortunately is the usual case in St. Tammany Parish where the oxidation system hasn't oxidized the ammonium to nitrate. There is the possibility of precipitating a magnesium ammonium phosphate mineral which might serve to remove both ammonium and phosphate, e.g., struvite MgNH4PO4.6H2O. Repeat the procedure in (C) with the additional constraint of equilibrium with struvite. Balance both the Mg and PO4 between the three solids so that only the nitrogen species is an aqueous variable. Calculate the activity sum of NH4+ and NH3 as a function of pH for the three Mg phosphate minerals to see which scenario gives the lowest activity sum of nitrogen species over this pH range. Refer to the diagram in (C) to see what the actual activity sum of the phosphate species was as a function of pH for each of the three Mg phosphate minerals. Why isn't the best case scenario in (C) consistent with the best case scenario in (D).

Other minerals could be used as a sink for phosphate such as hydroxyapatite (Ca5OH(PO4)3) in which the dissolution of calcite (CaCO3) would provide a source for calcium and the dissolution of brucite (Mg(OH)2 would raise the pH to provide for precipitation of hydroxyapatite. What might prevent the precipitation of hyroxyapatite.

Write a short 1 page typewritten report on your results?

Thermodynamic data Gof,25oC,1 bar

in kj/mol

aqueous species

PO43- -1,018.8

HPO42- -1,089.3

H2PO4- -1,130.4

H3PO4 -1,142.6

NO3- - 111.3

HNO3- - 111.3

NO2- - 37.2

HNO2- - 42.97

NH3 - 26.57

NH4+ - 79.37

Mg2+ - 454.8

e- 0

H+ 0

H2O - 237.18

(liquid water)

N2 0


H2 0


O2 0



Mg(OH)2 - 833.5


Mg5OH(PO4)3 -5,758.11

(Mg hydroxyapatite)

Mg4O(PO4)2 -4,172.7

(Mg oxyapatite)

Mg3(PO4)2 -3,538.7


MgNH4PO4.6H2O -3,051.1


Note that R = .0083136 kj/mol/oK and 25oC = 298.15oK

Data Source of aqueous species, brucite, gases, and liquid water was Stumm and Morgan "Aquatic Chemistry" (1981) and phosphate minerals are from Woods and Garrels "Thermodynamic Values ..." (1987) and Dixon and Weed "Minerals in Soil Environments" (1977)

Comment on Example Midterm test on part D.

Mg(OH)2 + Mg3(PO4)2 = Mg4O(PO4)2 + H2O

brucite farringtonite Mg oxyapatite water

-833.5 kj/mol -3,538.7 kj/mol -4,172.7 kj/mol -237.18 kj/mol

All the components are in their standard states so the Greaction is -37.68 kj, i.e., farringtonite cannot control the ammonium solubility because it will convert to Mg oxyapatite in the presence of brucite unless reaction kinetics are too slow for the reaction to occur.