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 cm^{2}/sec^{2}. A joule is one kg m^{2}/sec^{2} or 10^{7} ergs. The watt is one joule sec. The heat energy unit, a calorie is equivalent to 4.184 joules, 4.184 x 10^{7} 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 (dw_{e}), 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 dw_{e}), 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/n_{i})dn_{i} (3)
(U/n_{i})dn_{i} is a summation over all reactants and products in a chemical reaction where n_{i} refers to the moles of the ith reaction component and dn_{i} is the change in moles of this component in the reaction. Reactants have negative dn_{i} and products have positive dn_{i}.
If dq and dw are both zero, equilibrium in the chemical reaction occurs when dU = (U/n_{i})dn_{i} = 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 q_{1} heat is gained by the system from reservoir 1 at temperature T_{1} and used to do work during which heat is lost to reservoir 2 at temperature T_{2}, the minimum amount of heat -q_{2} lost to reservoir 2 is (q_{1}/T_{1})T_{2} where T is in absolute degrees Kelvin. In other words, -q_{2}/T_{2} = q_{1}/T_{1}. The negative sign for -q_{2} 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 T_{1} in which q_{1} heat is absorbed to maintain a constant temperature, followed by an adiabatic expansion (without heat exchange so the temperature decreases from T_{1} to T_{2}), followed by an isothermal contraction at T_{2} in which -q_{2} heat is evolved to maintain a constant temperature, followed by an adiabatic contraction (without heat exchange so the temperature rises back to T_{1}) back to its initial state. This minimum exchange of heat occurs only if the process is reversible (near equilibrium).
For irreversible processes, -q_{2}/T_{2} > q_{1}/T_{1}. The heat fraction available for work for reversible processes is
[q_{1}-(-q_{2}]/q_{1} = 1 - T_{2}/T_{1} = (T_{1} - T_{2})/T_{1}.
The higher T_{1} is compared to T_{2}, the greater the heat fraction available as work. This is why modern engines extract work at high temperatures (T_{1}) compared to the atmospheric temperature (T_{2}).
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 T_{A} and T_{B}, respectively. The flow of heat is an irreversible process so the overall change in entropy must be positive. If T_{A} and T_{B}, are close together, so that the heat flow approaches reversibility, we can compute the overall change in entropy as
q/T_{B} - q/T_{A}. For this change to be positive, T_{B} < T_{A}, 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/n_{i})_{S,V,nj} dn_{i} (4)
U/n_{i})_{S,V,nj} was defined by Gibbs (the father of thermodynamics) as the chemical potential (µ_{i}).
The usefulness of equation (4) is that at constant S and V:
dU = µ_{i} dn_{i} (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 n_{i}.
dU = (U/S_{i})_{V,nj} dS + (U/V_{i})_{S,nj} dV + (U/n_{i})_{S,V,nj} dn_{i} (6)
Comparison with dU = TdS - PdV + µ_{i} dn_{i} shows that
T = (U/S_{i})_{V,nj}, (7a)
-P = (U/V_{i})_{S,nj}, (7b)
and we have already given by definition that
µ_{i} = (U/n_{i})_{S,V,nj} (7c)
The definition of U can be modified to define new functions that contain µ_{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 + µ_{i} dn_{i}
(8a, 8b, 8c)
A = U - TS, dA = dU - TdS - SdT = -SdT - PdV + µ_{i} dn_{i}
(9a, 9b, 9c)
G = U+PV-TS, dG = dU + PdV+VdP - TdS-SdT = -SdT + VdP + µ_{i}dn_{i}
(10a, 10b, 10c)
What are the natural variables for U, H, A, and G?
U(S,V,n_{i}), H(S,P,n_{i}), A(T,V,n_{i}), and G(T,P,n_{i})
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}, µ_{i} = (U/n_{i})_{S,V,nj}
(11a, 11b, 11c)
T = (H/S)_{P,nj}, V = (H/P)_{S,nj}, µ_{i} = (H/n_{i})_{S,P,nj}
(12a, 12b, 12c)
-S = (A/T)_{V,nj}, -P = (A/V)_{T,nj}, µ_{i} = (A/n_{i})_{T,V,nj}
(13a, 13b, 13c)
-S = (G/T)_{P,nj}, V = (G/P)_{T,nj}, µ_{i} = (G/n_{i})_{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 + µ_{i} dn_{i}
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 H_{r} will be the heat released or absorbed due to the chemical reaction. A negative H_{r} means heat is released and the reaction is exothermic. A positive H_{r} means heat is absorbed and the reaction is endothermic. The H_{r} can be calculated from summing the molar enthalpies of formation, H_{f,i}, of the reaction components using the reaction coefficients, n_{i} (positive for a product and negative for a reactant):
H_{r} = n_{i} H_{f,i} (16)
where H_{f,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
H_{r}^{o} = n_{i} H^{o}_{f,i} (17)
where the superscript o indicates a standard state property.
For example, consider the reaction aA + bB = cC + dD
H_{r}^{o} = cH^{o}_{f,C} + dH^{o}_{f,D} - aH^{o}_{f,A} - bH^{o}_{f,B}
Suppose we knew H^{o}_{f,i} to calculate H_{r}^{o} at a reference temperature T_{ref} and want to calculate H_{r}^{o} at a different temperature. We have to move H_{r}^{o} from T_{ref} 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 H^{o}_{f,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, Cp_{i}^{o}.
Cp_{i}^{o} = (dq_{i}/dT)_{P} (18)
The reactant and product heat capacities are combined into a summation for the overall change in heat capacity in the reaction, Cp_{r}^{o}.
Cp_{r}^{o} = n_{i} Cp_{i}^{o} (19)
The change in H_{r}^{o} as T changes at constant pressure, is from
dH_{r} = TdS_{r} = dq_{r}. By substitution,
dH_{r}^{o} = Cp_{r}^{o}dT (20)
And by integration, H^{o}_{r,T,P} = H^{o}_{r,Tref,P} + TrefTCp_{r}^{o}dT (21)
The expression used in equation (21) for H^{o}_{r,T} contains the reaction summation in equation (19) of heat capacities of the individual reaction components in Cp_{r}^{o}. 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, H^{o}_{f,i}, have been tabulated in reference books at 25^{o}C (298.15^{o}K) 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 25^{o}C 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 25^{o}C 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 25^{o}C 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 25^{O}C and 1 bar. Hence at pressures and temperatures different from 25^{o}C 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 25^{o}C 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 25^{o}C and 1 bar.
Do a sample reaction of H^{o}_{r,25 C} for calcite and quartz reacting to produce wollastonite and carbon dioxide.
CaCO_{3} _{(calcite)} + SiO_{2} _{(quartz)} = CaSiO_{3} _{(wollastonite)} + CO_{2} _{(gas)}
The standard state enthalpies of formation at 25^{o}C 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.
Cp_{i}^{o} = a_{i} + b_{i}T +c_{i}/T^{2} (22)
The coefficients for the Cp_{i}^{o} equation for each component are also listed below. Calculate H_{r}^{o} at 100^{o}C (273.15^{o}K) and one bar, using equations (19), (20), (21), and (22).
H^{o}_{f,i} Cp_{i}^{o}
25^{o}C, 1 Bar a_{i} b_{i} c_{i}
cal/mole cal/mole/^{o}K cal/mole/^{o}K^{2} cal^{o}K/mole
CaCO_{3} -288,772. 24.98 0.00524 -620,000.
SiO_{2} -217,650. 11.22 0.00820 -270,000.
CaSiO_{3} -389,810. 26.64 0.0036 -652,000.
CO_{2} - 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,
TdS_{r} = dq_{r} (23)
Similar to H_{r}^{o}, S_{r}^{o} at constant temperature and pressure can be calculated from the molar entropies, S_{i}^{o}, of the reaction components using the reaction coefficients, n_{i}:
S_{r}^{o} = n_{i}S_{i}^{o} (24)
Again consider the reaction aA + bB = cC + dD
S_{r}^{o} = cS_{C}^{o} + dS_{D}^{o} - aS_{A}^{o} - bS_{B}^{o}
Suppose we knew S_{i}^{o} to calculate S_{r}^{o} at a reference temperature T_{ref} and want to calculate S_{r}^{o} at a different temperature. We have to move S_{i}^{o} from T_{ref} 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 S_{i}^{o} 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, Cp_{i}^{o}, 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, Cp_{r}^{o}, as in equation (19). The change in S_{r}^{o} (in equation 24) as T changes at constant pressure, is computed using equation (23). By substitution,
dS_{r}^{o} = [Cp_{r}^{o}/T]dT. (25)
By integration S^{o}_{r,T} = S^{o}_{r,Tref} + TrefT[Cp_{r}^{o}/T]dT (26)
In the above exercises, we have computed H^{o}_{r,T} and S^{o}_{r,T} using Cp_{r}^{o}. Instead we could have used Cp_{i}^{o} and integrated the individual heat capacity expressions to move H^{o}_{f,i} and S^{o}_{f,i} from T_{ref} 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 25^{o}C (298.15^{o}K) 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 N_{o}, 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 S^{o}_{i} at 25^{o}C (298.15^{o}K) and 1 bar for the ith substance have been obtained by integrating at constant pressure from absolute zero (equation 26 for one component).
S^{o}_{i,298.15oK} = S^{o}_{i,OoK} + [Cp^{o}_{i}/T]dT
We will actually use S_{i}^{o} like H^{o}_{f,i} as part of an overall calculation of G^{o}_{r,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, µ_{i}^{o}, 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 µ_{i}^{o} 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 S^{o}_{r,T} which is then used calculate G^{o}_{r,T}. Third law entropies will produce the same value for S^{o}_{r,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 S^{o}_{r,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.
CaCO_{3} _{(solid)} + SiO_{2} _{(solid)} = CaSiO_{3} _{(solid)} + CO_{2} _{(gas)}
The heat capacity of each component can be described by the Maier Kelly three term expansion where T is in degrees Kelvin (^{o}C + 273.15) of the form of equation (22). The coefficients for the Cp_{i}^{o} equation for each component were listed above in the discussion on enthalpy.
Calculate S^{o}_{r,T} at 100^{o}C 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).
S_{i}^{o} S_{i}^{o}
25^{o}C, 1 Bar 25^{o}C, 1 Bar
cal/mole/^{o}K cal/mole/^{o}K
CaCO_{3} 22.15 C 1.362
SiO_{2} 9.88 O_{2} 49.003
CaSiO_{3} 19.60 Si 4.493
CO_{2} 51.072 Ca 9.90
Values of third law entropies S_{i}^{o} of the elements that make up the above substances are also listed above for 25^{o}C and one bar. Calculate entropies of formation from the elements at 25^{o}C and one bar for the above substances. Use these entropies of formation to calculate S^{o}_{r,25oC} for the reaction on the previous page and compare with S^{o}_{r,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 + µ_{i}dn_{i} leading to
-S = (G/T)_{P,nj}, V = (G/P)_{T,nj}, µ_{i} = (G/n_{i})_{T,P,nj}
µ_{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 = µ_{i}dn_{i} (28)
To calculate the change in Gibbs free energy in a reaction at a temperature and pressure of interest, add the µ_{i} (for that PT point) for all the products and subtrack the µ_{i} of all the reactants, each multiplied by its reaction coefficient.
G_{r} = n_{i}µ_{i} (29)
At equilibrium the change in Gibbs free energy of reaction G_{r} 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 µ_{i} to use in the test. Values of µ_{i}^{o},(G^{o}_{f,i}) the standard state molar or molal free energy of formation of each substance, are listed in reference books at 25^{o}C 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. µ_{i}^{o} 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.
G_{r}^{o} = n_{i}µ_{i}^{o} (30)
If we need to calculate G_{r}^{o} at a PT point different from 25^{o}C and 1 bar, we use
G_{r}^{o} = n_{i}µ_{i}^{*} (31)
where µ_{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 25^{o}C and 1 bar. actually only move the substance, not the elements. Because the contribution of the elements to G_{r}^{o} 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 G_{r}^{o}. Note that by definition, µ_{i}^{*} and µ_{i}^{o} are identical at 25^{o}C 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
G_{r} = cµ_{C} + dµ_{D} - aµ_{A} - bµ_{B}
The first step in calculating G_{r}, is to calculate G_{r}^{o}
G_{r}^{o} = cµ_{C}^{o} + dµ_{D}^{o} - aµ_{A}^{o} - bµ_{B}^{o}
Suppose we knew µ_{i}^{o} to calculate G_{r}^{o} at T_{ref} and P_{ref} and want to calculate G_{r}^{o} at a different temperature and a different pressure. We want to move µ_{i}^{*} from T_{ref} to T (at P_{ref}) and then from P_{ref} 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,
G_{P,T} = H_{P,T} - TS_{P,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.
G_{r,P,T} = H_{r,P,T} - TS_{r,P,T} (34)
If the reactants and products are present in their standard states.
G^{o}_{r,P,T} = H^{o}_{r,P,T} - TS^{o}_{r,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 G^{o}_{r,P,T}.
µ_{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 µ_{i}^{*} and H^{*}_{f,i}. However, since we are going to use µ_{i}^{*} in a balanced chemical reaction, the contributions from the elements will cancel out, so we can use third law entropies to calculate S_{r}^{o} for calculating G_{r}^{o}.
Moving µ_{i}^{*} with temperature from T_{ref} to T at constant P_{ref}
H^{*}_{f,i} and S^{*}_{f,i} in equation (36) can be rewritten as their values at 25^{o}C (T_{ref}) and the integration of the change in those values up to temperature T, i.e.,
H^{*}_{f,i,T,Pref} = H^{o}_{f,i,Tref,Pref} + (H^{o}_{f,i,Pref}/T)dT (37)
S^{*}_{f,i,T,Pref} = S^{o}_{f,i,Tref,Pref} + (S^{o}_{f,i,Pref}/T)dT (38)
Substituting equations (37) and (38) into equation (36) gives
µ^{*}_{i,T,Pref} = H^{o}_{f,i,Tref,Pref} + (H_{i}^{o}/T)_{Pref} dT
- T[S^{o}_{f,i,Tref,Pref} + (S_{i}^{o}/T)_{Pref}dT] (39)
From the previous discussions on enthalpy and entropy, we know that
(H_{i}^{o}/T)_{Pref} = Cp_{i}^{o} (40)
and (S_{i}^{o}/T)_{Pref} = Cp_{i}^{o}/T. (41)
Substitution of equations (40) and (41) into equation (39),
µ^{*}_{i,T,Pref} = H^{o}_{f,i,Tref,Pref} + (Cp_{i}^{o}dT - T[S^{o}_{f,i,Tref,Pref} + (Cp_{i}^{o}/T)dT] (42)
Again, note that S^{o}_{i,Tref,Pref} can be used in place of S^{*}_{f,i,Tref,Pref} if we are going to use µ^{*}_{i,T,Pref} in a balanced reaction (equation 31) to calculate G^{o}_{r,T,Pref}.
Since the intent is to compute G_{r}^{o}, it is simpler to do the summation of the reaction constituents prior to the integration. This is identical to previous problems computing H_{r}^{o} and S_{r}^{o}. Substituting equation (42) into equation (31), yields
G^{o}_{r,T,Pref} = H^{o}_{r,Tref,Pref} + (Cp_{r}^{o}dT
- T[S^{o}_{r,Tref,Pref} + (Cp_{r}^{o}/T)dT] (43)
The integrations were previously done for heat capacities expressed in Maier-Kelly algorithms in calculating H_{r}^{o} and S_{r}^{o} .
Moving µ^{*}_{i,T} with pressure
The above equation moves G_{r}^{o} from T_{ref} to T at P_{ref}. To move G_{r}^{o} from P_{ref} 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
dµ^{*}_{i,T} = V^{o}_{i,T}dP (44)
where V_{i}^{o} 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,
dµ^{*}_{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 P_{ref},T to come up with V_{r}^{o}, the volume change of reaction for all non-gas constituents. Again, gases aren't included because their standard state does not vary with pressure. V_{r}^{o} is then integrated from P_{ref},T to P,T. An alternative would be to first compute µ_{i}^{*} at P by integration for each component and then do a reaction summation; however, this would require more steps. For the reaction,
dG^{o}_{r,T} = V^{o}_{r,T}dP (45)
or G^{o}_{r,T,P} - G^{o}_{r,T,Pref} = V^{o}_{r,T}dP (46)
Adding equations (43) and (46) together gives the overall expression for moving the G_{r}^{o} through PT space.
G^{o}_{r,T,P} = H^{o}_{r,Tref,Pref} + (Cp_{r}^{o})_{Pref}dT - TS^{o}_{r,Tref,Pref}
- T(Cp_{r}^{o}/T)_{Pref}dT + V^{o}_{r,T}dP (47)
where the volumes of the solids can usually be considered constant. If the V^{o}_{r,T} is not constant then an appropriate expression is used to describe it that can be integrated through P space.
Note that if Cp_{r}^{o} is constant with temperature then
G^{o}_{r,T,P} = H^{o}_{r,Tref,Pref} - TS^{o}_{r,Tref,Pref}
+Cp^{o}_{r,Pref}[T-T_{ref} + T ln(T_{ref}/T)] + V^{o}_{r,T}dP (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.
dµ^{*}_{i} = -S^{*}_{f,i}dT + V^{*}_{f,i}dP
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 µ^{*}_{i} through PT space from P_{ref}_{} and T_{ref}_{}.
µ^{*}_{i} = - S^{*}_{f,i}dT + V^{*}_{f,i}dP where µ^{*}_{i} = µ^{*}_{i,T,P} - µ^{*}_{i,Tref,Pref}
and then combine these components equation (31) in a reaction to compute G^{o}_{r,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.
G^{o}_{r,T,P} - G^{o}_{r,Tref,Pref} = - S^{o}_{r,Pref}dT + V^{o}_{r,T}dP
where S^{o}_{r} is integrated with T at constant 1 bar pressure, followed by the integration of V^{o}_{r} with P at constant T.
Following equation (38), S^{o}_{r} can be written as
S^{o}_{r,Pref} = S^{o}_{r,Tref,Pref} + (S^{o}_{r}/T)_{Pref}dT
and remembering that (S^{o}_{r}/T)_{Pref}dT = (Cp^{o}_{r}/T)_{Pref}dT. Hence,
G^{o}_{r,T,P} = G^{o}_{r,Tref,Pref} - S^{o}_{r,Tref,Pref}dT
+ [Cp^{o}_{r}/T]_{Pref}dT)dT + V^{o}_{r,T}dP
= G^{o}_{r,Tref,Pref} - S^{o}_{r,Tref,Pref}(T-T_{ref}_{})
- [Cp^{o}_{r}/T]_{Pref}dTdT + V^{o}_{r,T}dP (48)
Note that if Cp^{o}_{r,Pref} is constant with temperature then
G^{o}_{r,T,P} = G^{o}_{r,Tref,Pref} + S^{o}_{r,Tref,Pref}(T_{ref}_{}-T)
+Cp^{o}_{r,Pref}[T-T_{ref} + T ln(T_{ref}/T)] + V^{o}_{r,T}dP (48a)
S^{o}_{r,Tref,Pref} can be computed directly from 3rd law entropies in reference tables using
S^{o}_{r,Tref,Pref} = n_{i}S^{o}_{i,Tref,Pref} (49)
and G^{o}_{r,Tref,Pref} = n_{i}µ^{o}_{i,Tref,Pref} (50)
G^{o}_{r,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.,
G^{o}_{r,Tref,Pref} = -n_{i}TS^{o}_{i,Tref,Pref} + n_{i}H^{o}_{f,i,Tref,Pref} (51)
[Cp^{o}_{r,Pref}/T]dTdT can be integrated by substituting an
appropriate algorithm for Cp^{o}_{r,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 T_{ref} term is considered a constant in the second integration. For solids, V^{o}_{r} is often considered constant. If V^{o}_{r} 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 V^{o}_{r}.
Calculate G^{o}_{r,T,P} for the example reaction at 100^{o}C and 300 bars pressure. Note that 1 cm^{3}bar = 0.02390 calories. Assume constant volume of the solids. Use the following molar volumes: CaCO_{3}, 36.934 cm^{3}/mole; CaSiO_{3}, 39.93 cm^{3}/mole; and SiO_{2}, 22.688 cm^{3}/mole.
Computation of G_{r,T,P} and the Equilibrium Constant
Now that we've moved G^{o}_{r,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 G_{r,T,P}. To do this we have to substitute the definition of the chemical potential of each reaction component (equation 52) into equation for G_{r,T,P} in equation (29).
µ_{i} = µ_{i}^{o} + RT ln a_{i} (52)
where a_{i} is the activity of the ith component.
G_{r,Tref,Pref} = n_{i}µ_{i}^{o} + RT a_{i}^{ni} (53)
and at P and T other than P_{ref}_{} and T_{ref}_{}
G_{r,T,P} = n_{i}µ_{i}^{*} + RT ln ( a_{i}^{ni}) (54)
By convention, the symbol Q is used to represent a_{i}^{ni}. Q is called the reaction quotient. Using the definition and equation (31) in equation (54),
G_{r,T,P} = G^{o}_{r,T,P} + RT ln Q. (55)
And at reaction equilibrium,
G_{r,T,P} = 0 and Q is called K, the equilibrium constant.
so, G^{o}_{r,T,P} = -RT ln K, and using ln x = 2.303 log x,
log K = -G^{o}_{r,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 25^{o}C 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, a_{i} = X_{i}_{i} (57)
however, in a solid solution the X_{i} may be modified as shown later.
dissolved aqueous component, a_{i} = m_{i}_{i} (58)
gas component, a_{i} = PX_{i}_{i} (59)
where X_{i} 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, m_{i} is the molality of i in the aqueous phase and _{i} is its activity coefficient. a_{i} for a gas is also called the fugacity or f_{i}.
The fugacity is the vapor pressure of a substance and is defined by the expression
dµ_{i} = RT dln f_{i} (60)
Note that at constant pressure and temperature, we can differentiate equation (52) and eliminate µ_{i}^{*} because standard states can vary only with T and P for non-gas components and T for a gas component.
dµ_{i} = RT dln a_{i}
If we integrate from a standard state to some other state at constant pressure and temperature,
dln a_{i} = dln f_{i} or a_{i}/a_{i}^{o} = f_{i}/f_{i}^{o}
where a_{i}^{o} is the activity in the standard state which is one. Hence,
a_{i} = f_{i}/f_{i}^{o}
For a gas in its standard state, the vapor pressure is one bar, hence f_{i}^{o} is one and a_{i} = f_{i} 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,
CaCO_{3} _{(solid)} + SiO_{2} _{(solid)} = CaSiO_{3} _{(solid)} + CO_{2} _{(gas)}
G_{r,T,P} = G^{o}_{r,T,P} + RT ln[(a_{CaSiO3})^{1}(a_{CO2})^{1}(a_{CaCO3})^{-1}(a_{CaSiO3})^{-1}]
= G^{o}_{r,T,P} + RT ln[(X_{CaSiO3})^{1}(X_{CO2})^{1}(X_{CaCO3})^{-1}(X_{CaSiO3})^{-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 CO_{2} 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 X_{CO2} becomes one.
Lets calculate the change in Gibbs free energy of the reaction, G_{r,T,P}, at 100^{o}C and 300 bars. We have already used equations (47) or (48) to calculate G^{o}_{r,100oC, 300 bars}. Assume the solids are in their standard states, i.e., unit activities and the CO_{2} gas phase is a pure CO_{2}.
G_{r,100oC,300 bars} = G^{o}_{r,100oC,300 bars} + RT ln[(_{CO2})P]
where R, the gas constant has a value of 1.987 cal/^{o}K 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 a_{CO2} is a measure of how CO_{2} deviates from an ideal gas at the temperature and pressure of interest. For a gas phase containing CO_{2} at a PT point,
a_{CO2} = P(_{CO2})X_{CO2}
and we can write the chemical potential of pure CO_{2} as
µ_{CO2} = µ^{*}_{CO2} + RT ln[P(_{CO2})] = µ^{*}_{CO2} + RT[ln P + ln _{CO2} + ln X_{CO2}]
Holding T constant and taking the partial derivative of the above expression with respect to P at constant composition gives
(µ_{CO2}/P)_{T} = RT/P + RT(ln _{CO2}/P)_{T}
where the partial derivative of µ^{*}_{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 V_{CO2}, 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 P_{o} to P where P_{o} approaches zero so that the gas behaves ideally.
V-_{CO2}dP = (RT/P)dP + RT(dln_{CO2})
By combining terms, and writing the integrated form of (dln_{CO2})
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 P_{o} 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 P_{o} = 1 bar, the above expression reduces to
ln _{i,P,T} = -ln P + (1/RT)V-_{i}dP (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 100^{o}C (373.15^{o}K). 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/T_{c,i} where T_{c,CO2} = 304.2^{o}K) and P/P_{c,i} in atmospheres where P_{c,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 V_{i}/RT) in 1/bars. V-_{i} is in cm^{3}/mol and R is 83.144 cm^{3 }bar mol^{-1} (84.239 cm^{3} atm mol^{-1}). Use the above equation to determine _{CO2,P,T} at 300 bars and 100^{o}C (373.15^{o}K). 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 100^{o}C with that from the second chart from Garrels and Christ (1965) which directly gives the value in terms of the reduced variables T_{c} and P_{c} (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.
G_{solid solution} = (n_{i}µ_{i}) (63)
This expression can be rewritten using equation (52)
G_{solid solution} = (n_{i}µ_{i}^{o}) + RT (n_{i} ln a_{i}^{o}) + RT (n_{i} ln _{i}) (64)
where a_{i}^{o} 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
a_{i} = X_{i}_{i}; however, now we are using a_{i} = a_{i}^{o}_{i}
where it will be shown that a_{i}^{o} = X_{i} for most but not for all cases.
The three terms in the expression for G_{solid solution} can be considered as follows: (n_{i}µ_{i}^{o}), the mechanical mixture of the components, G_{mechanical mixture}; RT (n_{i }ln a_{i}^{o}), the contribution from ideal mixing, G_{ideal mixing}; and RT (n_{i} ln _{i}), the contribution from non-ideal mixing or excess mixing, G_{excess}. Remembering that G = H - TS, we can write for G_{ideal mixing} at constant T and P,
G_{ideal mixing} = H_{ideal mixing} - TS_{ideal mixing} (65)
where H_{ideal solid solution} = 0 because no heat is released or absorbed in ideal mixing. Hence
G_{ideal mixing} = -TS_{ideal mixing} = RT (n_{i} ln a_{i}^{o}) (66)
Boltzmann postulated that S_{ideal 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 N_{C} and N_{B} 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.
= (N_{C} + N_{B})!/(N_{C}!N_{B}!)
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 = -N_{C} ln[N_{C}/(N_{C} + N_{B})] - N_{B} ln[N_{B}/(N_{C} + N_{B})]
Since X_{C} = N_{C}/(N_{C} + N_{B}), X_{B} = N_{B}/(N_{C} + N_{B}), N_{C} + N_{B} = A, and k = R/A
Tk ln = - RT X_{C} ln X_{C} - RT X_{B} ln X_{B}
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
-TS_{ideal mixing} = RT (X_{CY} ln X_{CY} + X_{BY} ln X_{BY}) (67)
Remembering also that -TS_{ideal mixing} = RT (n_{i} ln a_{i}^{o}) where n_{i} = X_{i} 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,
-TS_{ideal mixing} = RT X_{CY} ln a^{o}_{CY} + X_{BY} ln a^{o}_{BY} (68)
By comparison between equations (67) and (68), the relations a^{o}_{CY} = X_{CY} and a^{o}_{BY} = X_{BY} 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 C_{2}Y and B_{2}Y. In this case, it can be shown by comparison in the above procedure that
a^{o}_{C2Y} = (X_{C2Y})^{2} and a^{o}_{B2Y} = (X_{B2Y})^{2}
Indeed for mixing molecules of the type C_{z}Y and B_{z}Y to form a solid solution, the ideal activities are
a^{o}_{CzY} = (X_{CzY})^{z} and a^{o}_{BzY} = (X_{BzY})^{z} (69)
for use in the activity expressions for the chemical potentials
a_{CzY} = (X_{CzY})^{z}_{CzY} and a_{BzY} = (X_{BzY})^{z}_{BzY} (70)
rather than a_{CzY} = (X_{CzY})_{CzY} and a_{BzY} = (X_{BzY})_{BzY} which are incorrect
for molecules of this type in a solid solution.
Show that a^{o}_{C2Y} = (X_{C2Y})^{2} and a^{o}_{B2Y} = (X_{B2Y})^{2} for a solid solution of C_{2}Y and B_{2}Y.
Regular Solutions
A regular solution on a crystalline lattice is one in which there is zero S_{excess}, excess entropy of mixing, but nonzero excess enthalpy of mixing. The free energy of mixing becomes that due to ideal mixing (-TS_{ideal mixing}) plus that due to the excess enthalpy of mixing (H_{excess}). 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 S_{excess} could be zero.
G_{mixing} = H_{excess} - TS_{ideal mixing} (71)
This implies that mixing is random even though interactions between adjacent atoms on sites depend on the site occupancies. For example, suppose N_{CY} and N_{BY} 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 e_{CC}, the interaction energy between two adjacent B atoms be e_{BB}, and the interaction energy between a B and C atom be e_{BC}. Within the lattice there are z(N_{CY} + N_{BY})/2 adjacent site interactions on the lattice. The change in the number of CB interactions on the lattice resulting from random mixing is N_{CB}. From simple probability,
N_{CB} = (z/2)(N_{CY} + N_{BY})2[N_{CY}/(N_{CY} + N_{BY})][N_{BY}/(N_{CY} + N_{BY})]
= z[N_{CY}N_{BY}/(N_{CY} + N_{BY})]
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,
N_{CC} = -N_{CB}/2 and N_{BB} = -N_{CB}/2
The expression for H_{ex} 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 U_{ex} but this is approximately equivalent to H_{ex} as long as the mixing is done with little or no change in volume of the lattice.
H_{ex} = N_{CB}e_{BC} + N_{CC}e_{CC} + N_{BB}e_{BB}
= z[N_{CY}N_{BY}/(N_{CY} + N_{BY})][e_{BC} - 0.5e_{CC} - 0.5e_{BB}]
= z[N_{CY}N_{BY}/(N_{CY} + N_{BY})]E_{BC}
where E_{BC} equals [e_{BC} - 0.5e_{CC} - 0.5e_{BB}]. Dividing the top and bottom by A_{v}^{2}, Avogadro's number squared, changes N, the number of molecules, to n the number of moles. Let E-_{BC} be E_{BC}A_{v}.
H_{ex} = z[n_{CY}n_{BY}/(n_{CY} + n_{BY})]E-_{BC} (72)
G_{ex} = H_{ex} because S_{ex} is zero. (73)
Remembering from equation (14c), µ_{i} = (G/n_{i})_{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 G_{ex} and set it equal to µ_{i,ex}.
µ_{i,ex} = (G_{ex}/n_{i})_{T,P,nj} and since µ_{i,ex} = RT ln _{i},
ln _{i} = (1/RT)(G_{ex}/n_{i})_{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} = (X_{CY})^{2}zW/RT and ln _{CY} = (X_{BY})^{2}zW/RT.
Phase Changes
Two components having the same chemical composition undergo a phase change at equilibrium, e.g., ice and water. Using H_{2}O as an example, we note that at equilibrium, the change in free energy for the reaction
H_{2}O_{ice} = H_{2}O_{water}
is zero. Hence,
µ_{ice} = µ_{water}
or µ^{*}_{ice} + RT ln X_{ice}_{ice} = µ^{*}_{water} + RT ln X_{water}_{water}
If the phases are pure phases, their activities are unity and
µ^{*}_{ice} = µ^{*}_{water}
If P and T are changed and equilibrium is maintained.
dµ^{*}_{ice} = dµ^{*}_{water}
(µ^{*}_{ice}/T)dT + (µ^{*}_{ice}/P)dP = (µ^{*}_{water}/T)dT + (µ^{*}_{water}/P)dP
or - S^{o}_{ice} dT + V^{o}_{ice}dP = - S^{o}_{water} dT + V^{o}_{water}dP
or S^{o}_{r} dT = V^{o}_{r}dP and recognizing that since equilibrium is being maintained.
G^{o}_{r} = 0 = H^{o}_{r} - TS^{o}_{r}, substituting for S^{o}_{r} gives
H^{o}_{r}/(TV^{o}_{r}) = 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
µ_{water} = µ_{vapor} and dµ_{water} = dµ_{vapor}
(µ_{water}/T)dT + (µ_{water}/P)dP = (µ_{vapor}/T)dT + (µ_{vapor}/P)dP
or - S^{o}_{water}dT + V^{o}_{water}dP = - S_{vapor}dT + V_{vapor}dP
and S_{r} dT = V_{r}dP, or (H_{r}/T)dT = V_{r}dP or H_{r}/(TV_{r}) = dP/dT
where the superscript was removed from V_{vapor} to indicate the vapor is not in its standard state and furthermore, the standard state of a gas doesn't move with pressure. However, V_{vapor} 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
S_{r} dT = V_{r}dP, or (H_{r}/T)dT = V_{r}dP or H_{r}/(TV_{r}) = 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.
CaCO_{3} _{(solid)} + SiO_{2} _{(solid)} = CaSiO_{3} _{(solid)} + CO_{2} _{(gas)}
They are derived by the above procedure, by setting dG_{r} = 0 as the pressure and temperature are changed.
S_{r}/V_{r} = H_{r}/(TV_{r}) = 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 CO_{2} gas phase is composed of pure CO_{2}. For the first point, hold the temperature constant at 500^{o}C and vary the pressure until G_{r} is zero. For the second point, hold the pressure constant at one bar and vary the temperature until G_{r} is zero. It is easiest to do this in Excel by setting the equation for G_{r} = 0 as a function of P and T and then varying these variables.
G_{r} = G^{o}_{r,T,1 bar} + V^{o}_{r}(P-1) + RT ln _{CO2}P = 0
where the first term is a function only of temperature and the second two terms are functions only of pressure. Remember that V^{o}_{r} does not include the gas component and has to multiplied by 0.0239 to convert to calories. To get the first PT point at 500^{o}C, assume that _{CO2} is one until you find a pressure setting G_{r} 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 G^{o}_{r,T,1 bar} in terms of its temperature dependence going from G^{o}_{r,25 C,1 bar} to G^{o}_{r,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 S_{r}/V_{r} for the reaction between the two PT points. Also, compute this ratio at 1 bar using the computed values of S_{r} and V_{r}. (Note that V_{r} includes all the components in the reaction, not just the solid volumes previously given to you that were used in V^{o}_{r}. 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
CaCO_{3} _{(solid)} + SiO_{2} _{(solid)} = CaSiO_{3} _{(solid)} + CO_{2} _{(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, SiO_{2}, and CO_{2}. 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 500^{o}C 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: H_{2}O, CaO, SiO_{2}, and CO_{2}, 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 H_{2}O 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 H_{2}O 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 H_{2}O:
H_{2}O = H^{+} + OH^{-} where K = a_{H+}a_{OH-}/a_{H2O}
plus the charge-balance constraint that requires the concentration of negative charges to equal the concentration of positive charges.
If we have CO_{2} dissolved in water, we have the additional species of H_{2}CO_{3}, HCO_{3}^{-}, and CO_{3}^{2-}. 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 CO_{2} dissolved in water, i.e., CO_{2} and H_{2}O.
If we have a system consisting of CaCO_{3} and MgCO_{3} minerals, we only have two components: CaCO_{3} and MgCO_{3}; not 3 components: CaO, MgO, and CO_{2}. This is because there is always a one to one ratio of CaO to CO_{2} and MgO to CO_{2} 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,
CH_{2}O + O_{2} = H_{2}O + CO_{2}
The reaction really consists of two half cell reactions in which the electrons cancel out when the reactions are added together.
CH_{2}O = H_{2}O + C^{4+} + 4e^{-}
O_{2} + 4e^{-} = 2O^{2-}^{}
CH_{2}O + O_{2} = H_{2}O + 2O^{2-} + C^{4+} = H_{2}O + CO_{2}
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, a_{e-}, even though aqueous electrons probably do not exist. Consider the following reaction in an aqueous solution
Mn^{4+} + 2e^{-} = Mn^{2+}
Beginning with equation (55),
G_{r,T,P} = G^{o}_{r,T,P} + RT ln Q. And at reaction equilibrium,
G_{r,T,P} = 0 and Q is called K, the equilibrium constant.
This leads to equation (56), log K = -G^{o}_{r,T,P}/(2.303 RT), where
log K = log {a_{Mn2+}/[a_{Mn4+}(a_{e-})^{2}]}
Defining -log a_{e-} = pe, makes log K = log (a_{Mn2+}/a_{Mn4+}) + 2 pe
or pe = [log K - log (a_{Mn2+}/a_{Mn4+})]/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., a_{Fe2+}/a_{Fe3+} and a_{HS-}/a_{SO4(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 + dw_{elec} 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 + dw_{elec} + µ_{i}dn_{i} (78)
and at constant T and P,
dG = dw_{elec} + µ_{i}dn_{i} (79)
For a redox reaction in which n moles of electrons are used to reduce one mole of oxidant, w_{elec} is computed using the electric potential E in volts.
w_{elec} = -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
Mn^{4+} + 2e^{-} = Mn^{2+} where n = 2. The conventional reference half
cell is H_{2 gas} = 2e^{-} + 2H^{+} where the a_{H+} and the a_{H2 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 H_{2}, 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 E_{h}.
Note that the chemical potentials of H^{+} and H_{2} 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 H_{2} is zero because this is the change in Gibbs free energy in forming H_{2} from elemental hydrogen which happens to be H_{2}. Because, a voltage cannot be measured for a half cell, i.e., only for two half cells, the E and E^{o} of SHE are arbitrarily set to zero. Since G^{o}_{SHE} = -nFE^{o}_{SHE} = zero, then G^{o}_{SHE} is zero, and since G_{SHE} = -nFE_{SHE} = zero, then G_{SHE} is zero, therefore the standard state chemical potential of e^{-} is zero and the a_{e-} is unity in SHE. This is shown below.
G^{o}_{SHE} = 2µ^{o}_{H+} + 2µ^{o}_{e-} - µ^{o}_{H2} = 0 and µ^{o}_{H+} = 0 and µ^{o}_{H2} = 0, so µ^{o}_{e-} = 0.
Since G_{SHE} = 0, log Q = log K = -G^{o}_{SHE} /(2.303 RT) = 0, or K = 1.
K = (a_{H+})^{2}(a_{e-})^{2}/a_{H2} = 1, and a_{H+} = 1, a_{H2} = 1, so a_{e-} = 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
Mn^{4+} + H_{2} = 2H^{+} + Mn^{2+} where
G = G^{o} + 2.303 RT log {(a_{Mn2+}/a_{Mn4+})_{solution}([a_{H+}]^{2}/a_{H2})_{SHE}}
The activities of H^{+} and H_{2} are unity in SHE so they drop out of the activity expression, leaving
G = G^{o} + 2.303 RT log (a_{Mn2+}/a_{Mn4+})_{solution}.
G^{o} equals the sum of the G^{o} for each half cell reaction;
however, G^{o}_{SHE} is zero, so G^{o} is for Mn^{4+} = Mn^{2+} + 2e^{-} half cell.
G = -2FE_{h} where E_{h} is the voltage measured relative to SHE, and
G^{o} = -2FE^{o}_{h} where E^{o}_{h} is the voltage measured relative to SHE if
Mn^{4+}, Mn^{2+}, and e^{-} were in their standard states in the solution. By substitution,
E_{h} = E^{o}_{h} + (2.303 RT/2F) log (a_{Mn4+}/a_{Mn2+})_{solution}
Now how does E_{h} relate to pe? Consider the half cell reaction in the solution making up half the electrochemical cell.
Mn^{4+} + 2e^{-} = Mn^{2+}
G = G^{o} + 2.303 RT log {a_{Mn2+}/(a_{Mn4+}a_{e-}^{2})}_{solution} = 0 because the
half cell reaction is assumed to be in redox equilibrium. Hence,
G^{o} = -2.303 RT log {a_{Mn2+}/(a_{Mn4+}a_{e-}^{2})}_{solution}
And, as previously shown, G^{o} for the electrochemical cell is the same as for this half cell. Using G^{o} = -2FE^{o}_{h} and substituting
E^{o}_{h} = 2.303 (RT/2F) log {a_{Mn2+}/(a_{Mn4+}a_{e-}^{2})}_{solution}
from the above half cell at equilibrium into the previous expression for E_{h} for the electrochemical cell gives
E_{h} = -(2.303 RT/F) log (a_{e-}) = (2.303 RT/F)pe
or pe = (F/2.303 RT) E_{h} (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
Hg_{2}Cl_{2 solid} + 2e^{-} = 2Hg_{liq} + 2Cl^{-}_{aq}
which has an E_{h} of 0.2444 volts at 25^{o}C if the aqueous reference solution is saturated with respect to KCl, i.e., contains solid KCl. By adding the E_{h} of the reference electrode to the E of the solution, the E_{h} of the solution is obtained, i.e.,
E_{h solution} = E_{solution} + E_{h ref. electrode}
pH definition
pH = -log a_{H+} 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 HCO_{3}^{-} and CO_{3}^{2-} in solution and calculating their activity coefficients for use in the following reaction:
HCO_{3}^{-} = H^{+} + CO_{3}^{2-} leads to K = a_{H+}a_{CO3(2-)}/a_{HCO3-} or
pH = -log a_{H+} = -log K - log a_{HCO3-} + log a_{CO3(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).
G_{r,T,P} = G^{o}_{r,T,P} + RT ln Q where G^{o}_{r,T,P} = -RT ln K
and by substituting reasonable activity values for the reaction components in Q.
aerobic oxidation, log K_{25 C & 1 bar} = 20.75
0.25 O_{2 gas} + H^{+} + e^{-} = 0.5 H_{2}O
denitrification, log K_{25 C & 1 bar} = 21.05
0.20 NO_{3}^{-} + 1.2 H^{+} + e^{-} = 0.1 N_{2} + 0.6 H_{2}O
manganese reduction, log K_{25 C & 1 bar} = 17.90
0.5 MnO_{2 solid} + 0.5 HCO_{3}^{-} + 1.5 H^{+} + e^{-} = 0.5 MnCO_{3 solid} + H_{2}O
nitrate reduction, log K_{25 C & 1 bar} = 14.90
0.125 NO_{3}^{-} + 1.25 H^{+} + e^{-} = 0.125 NH_{4}^{+} + 0.375H_{2}O
iron reduction, log K_{25 C & 1 bar} = 10.20
FeOOH_{solid} + HCO_{3}^{-} + 2 H^{+} + e^{-} = FeCO_{3 solid} + 2 H_{2}O
fermentation, log K_{25 C & 1 bar} = 3.99
0.5 CH_{2}O + H^{+} + e^{-} = 0.5 CH_{3}OH
sulfate reduction, log K_{25 C & 1 bar} = 4.12
0.125 SO_{4}^{2-} + 1.125 H^{+} + e^{-} = 0.125 HS^{-} + 0.5 H_{2}O
methane fermentation, log K_{25 C & 1 bar} = 2.87
0.125 CO_{2 gas} + H^{+} + e^{-} = 0.125 CH_{4 gas} + 0.25 H_{2}O
nitrogen fixation, log K_{25 C & 1 bar} = 4.65
0.167 N_{2} + 1.333 H^{+} + e^{-} = 0.333 NH_{4}^{+}
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, P_{B} - P_{A}, across the membrane at equilibrium is called the osmotic pressure.
µ_{A,H2O} = µ_{B,H2O} at equilibrium
µ^{o}_{H2O,PA} + 2.303 RT log X_{A,H2O}_{A,H2O} = µ^{o}_{H2O,PB} + 2.303 RT log X_{B,H2O}_{B,H2O}
where µ^{o}_{H2O,PB} = µ^{o}_{H2O,PA} + V^{o}_{H2O}dP and as an approximation
= µ^{o}_{H2O,PA} + V^{o}_{H2O}(P_{B} - P_{A}). The osmotic pressure
P = P_{B} - P_{A} = 2.303 (RT/V^{o}_{H2O}) log [X_{A,H2O}_{A,H2O}/X_{B,H2O}_{B,H2O}] (83)
Plot the osmotic pressure, assuming solution A is pure water, and X_{B} 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/cm^{3}, 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 = µ_{i}dn_{i} (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 = µ_{i}n_{i} (85)
hence, dG = µ_{i}dn_{i} + n_{i}dµ_{i}. (86)
A comparison between equations (84) and (86) gives the Gibbs-Duhem equation at constant T and P.
n_{i}dµ_{i} = 0 (87)
Remember µ_{i} = µ_{i}^{o} + RT ln a_{i}. Since T and P are constant, dµ_{i}^{o} = 0, and
n_{i}RT dln a_{i} = 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 f_{H2O} (equation 60) and its subsequent relation to the activity, f_{H2O}/f^{o}_{H2O} = a_{H2O} where f_{H2O} and f^{o}_{H2O} 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 H_{2}O follows Raoult's law for ideal mixing at infinite dilution:
a_{i} = X_{i} as X_{i} => 1 (89)
because _{i} => 1 as the standard state is approached.
From the Gibbs-Duhem equation, Henry's law for a solute
a_{i} = k_{h}X_{i} as X_{i} => 0 (90)
can be shown to follow from Raoult's Law. k_{h}, 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, X_{i}, of the ith aqueous solute component is related to the molality m_{i} by
X_{i} = m_{i}/(m_{k} + 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,
a_{i} = k_{hm}m_{i} as m_{i} => 0, (92)
where k_{hm} 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 m_{k} + 55.56 which becomes 55.56 at infinite dilution.
k_{hm} = k_{h}/(m_{k} + 55.56)
Experiments show that Henry's law is followed for aqueous solutes that do not disassociate into ions, e.g., CH_{4 aq} or CO_{2 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} = k_{s}I (93)
where k_{s}, the salting-out coefficient is determined from experimental measurement and
I = 0.5m_{k}(z_{k})^{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 CO_{3}^{2-}, and neglecting ion pairing such as NaCO_{3}^{-}, or as a true ionic strength which includes all ion pairs, depending on how I was defined during the measurement of k_{s}.
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 C_{v+}A_{v-} 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
µ_{i} = µ_{i}^{o} + RT ln a_{i} where a_{i} = m_{i}_{i}
and at infinite dilution, the salt is totally disassociated. We can also represent µ_{i} as the sum of the contributions from its parts.
µ_{i} = (v+)µ_{C} + (v-)µ_{A }= (v+)µ_{C}^{o} + (v-)µ_{A}^{o} + RT ln(a_{C})^{v+}(a_{A})^{v-}
where a_{i} = (a_{C})^{v+}(a_{A}^{)v-} = (m_{C})^{v+}(m_{A}^{)v-}(_{C})^{v+}(_{A})^{v-}
For a single-salt solution, m_{C} = (v+)m_{i} and m_{A} = (v-)m_{i}
or a_{i} = m_{i}^{(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 a_{i} versus m_{i}^{(v+ + v-)} yields a Henry's Law constant at infinite dilution of
k_{hm} = (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 (a±_{i}), mean ionic molality (m±_{i}), 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.,
a±_{i} = a_{i}^{[1/(v+ + v-)]} = [(a_{C})^{v+}(a_{A}^{)v-}]^{[1/(v+ + v-)]} (97)
m±_{i} = m_{i}^{[1/(v+ + v-)]} = m_{i}[(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 CaSO_{4}, 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 H_{2}O 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 H_{2}O parameter. And at still higher concentrations, the equation becomes
log ±_{i} = -Az_{+}z_{-}I/(1 + a^{o}_{i}BI) + b_{i}I (102)
where a^{o}_{i} and b_{i} 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(z_{i})^{2}I/(1 + BI) (103)
where z_{i} 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(z_{i})^{2}I/(1 + a^{o}_{i}BI) + b_{i}I (104)
where a^{o}_{i} and b_{i} are fit parameters for the ith ion. However, the expression cannot be combined back into equation (99) to give equation (102) unless b_{i} and a^{o}_{i} 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 SiO_{2}, Na_{2}O, Al_{2}O_{3}, H_{2}O, 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 a_{H2O} is unity. Use the log a_{SiO2 aq} on the x axis and the log a_{Na+}/a_{H+} 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 a_{H2O} 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
NaO_{0.5}/Al_{2}O_{3} SiO_{2}/Al_{2}O_{3}
gibbsite Al(OH)_{3} 0 0
kaolinite Al_{2}Si_{2}O_{5}(OH)_{4} 0 2
paragonite NaAl_{3}Si_{3}O_{10}(OH)_{2} 2/3 2
pyrophyllite Al_{2}Si_{4}O_{10}(OH)_{2} 0 4
albite NaAlSi_{3}O_{8} 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 NaO_{0.5}/Al_{2}O_{3} and SiO_{2}/Al_{2}O_{3} in each of the above minerals and plot them on a graph of NaO_{0.5}/Al_{2}O_{3} (y axis) versus SiO_{2}/Al_{2}O_{3} (x axis). On this plot, NaO_{0.5}/Al_{2}O_{3} is used rather than Na_{2}O/Al_{2}O_{3} because the y axis on the phase diagram contains the log a_{Na+}, not 2 log a_{Na+}. 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 SiO_{2 aq}. Those with no Na or SiO_{2} 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 25^{o}C and 1 bar, calculate the phase diagram described above, assuming the a_{H2O} 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.
G^{o} in cal/mole
gibbsite Al(OH)_{3} - 276,339
kaolinite Al_{2}Si_{2}O_{5}(OH)_{4} - 908,167
paragonite NaAl_{3}Si_{3}O_{10}(OH)_{2} -1,331,352
pyrophyllite Al_{2}Si_{4}O_{10}(OH)_{2} -1,258,811
albite NaAlSi_{3}O_{8} - 886,734
quartz SiO_{2} - 204,646
amorphous
silica SiO_{2} - 202,892
Na^{+} aqueous - 62,591
H^{+} aqueous 0
SiO_{2 aq} - 199,190
H_{2}O 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 a_{e-}, as the y variable and with pH, the -log a_{H+}, as the x variable. There are three types of stability boundaries on this phase diagram.
1) Reactions are written between aqueous species such as Fe^{2+} and Fe^{3+}, in which the boundary between dominant species is taken at equal activities, e.g., a_{Fe2+}/a_{Fe3+} = 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., a_{Fe2+} 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 H_{2}O to O_{2} gas. The lower boundary is marked by the reduction of H^{+} ions in water to H_{2} gas. The two reactions are given below together with their 25^{o}C and 1 bar log K values.
upper boundary O_{2 gas} + 4H^{+} + 4e^{-} = 2H_{2}O log K = 83.1
lower boundary 2H^{+} + 2e^{-} = H_{2 (gas)} log K = 0
Construct a 25^{o}C and 1 bar pe-pH diagram for the FeO-Fe_{2}O_{3}- HCl-H_{2}O system. Consider the following aqueous components: Fe^{3+}, Fe(OH)^{2+}, Fe(OH)_{2}^{+}, Fe(OH)_{4}^{-}, Fe^{2+}, Fe(OH)^{+}, and the following two solids: Fe(OH)_{3}(s) and Fe(OH)_{2}(s). Assume a_{H2O} 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 25^{o}C 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 = K_{1}K_{2} and log K = log K_{1} + log K_{2}. If reaction (2) is subtracted from reaction (1), K = K_{1}/K_{2} and log K = log K_{1} - log K_{2}. 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.
Fe^{3+} + 3H_{2}O = Fe(OH)_{3}(s) + 3H^{+} log K = - 3.00
Fe^{3+} + H_{2}O = Fe(OH)^{2+} + H^{+} log K = - 2.17
Fe^{3+} + 2H_{2}O = Fe(OH)_{2}^{+} + 2H^{+} log K = - 7.18
Fe^{3+} + 4H_{2}O = Fe(OH)_{4}^{-} + 4H^{+} log K = -22.2
Fe^{3+} + e^{-} = Fe^{2+} log K = 13.01
Fe^{2+} + 2H_{2}O = Fe(OH)_{2}(s) + 2H^{+} log K = -11.67
Fe^{2+} + H_{2}O = 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 O_{2} gas of 1 at the upper boundary and an activity of H_{2} gas of 1 at the lower boundary.
Gas Activity (Fugacity) Diagrams
The pe-pH diagram can be recast in terms of O_{2} activity and pH by using the reduction of O_{2} as a half cell reaction to balance out electrons. The y axis variable becomes the log a_{O2}. For example, in the homework diagram:
O_{2 gas} + 4H^{+} + 4e^{-} = 2H_{2}O log K = 83.1
+ 4Fe^{2+} = 4e^{-} + 4Fe^{3+}^{} log K = -52.04
O_{2 gas} + 4H^{+} + 4Fe^{3+} = 4Fe^{2+}^{} + 2H_{2}O log K = 31.06
The use of O_{2} introduces a pH dependence in the oxidation of iron because O_{2} reduction in aqueous solution involves H^{+}.
The use of log a_{H2} on the y axis can follow from using the H_{2} half-cell reaction to balance out electrons. For example in the homework diagram:
2H^{+} + 2e^{-} = H_{2 (gas)} log K = 0
+ 2Fe^{2+} = 2e^{-} + 2Fe^{3+}^{} log K = -26.02
2H^{+} + 2Fe^{2+} = H_{2 (gas)} + 2Fe^{3+} log K = -26.02
Similarly, reactions involving inorganic carbon and sulfur can be written to CO_{2} and H_{2}S, allowing variables to be a_{CO2} and the a_{H2S}, 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 100^{o}C and higher, equilibrium usually exists between minerals and an aqueous solution.
In the exercise below, plot the total solubility of aluminum versus pH at 200^{o}C 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 H_{2}O is one.
Use the following information at 200^{o}C and 650 bars.
aqueous complexes log K
Al(OH)^{2+} + H^{+} = Al^{3+} + H_{2}O 0.355
Al(OH)_{2}^{+} + 2H^{+} = Al^{3+} + 2H_{2}O 2.870
Al(OH)_{4}^{-} + 4H^{+} = Al^{3+} + 4H_{2}O 13.738
minerals log K
gibbsite Al(OH)_{3} + 3H^{+} = Al^{3+} + 3H_{2}O 1.01
kaolinite Al_{2}Si_{2}O_{5}(OH)_{4} + 6H^{+} = 2Al^{3+} + 2SiO_{2 qtz} + 5H_{2}O -0.48
pyrophyllite Al_{2}Si_{4}O_{10}(OH)_{2} + 6H^{+} = 2Al^{3+} + 4SiO_{2 qtz} + 5H_{2}O -0.16
albite NaAlSi_{3}O_{8} + 4H^{+} = Na^{+} + Al^{3+} + 3SiO_{2 qtz} + 2H_{2}O 4.67
Movement of log K with T and P
With Changes in T
Beginning with equation (56), log K = -G^{o}_{r}/(2.303RT)
Taking the partial derivative with respect to T at constant P
(log K/T)_{P} = G^{o}_{r}/(2.303RT^{2}) + S^{o}_{r}/(2.303RT)
Using equation (35), G_{r}^{o} = H_{r}^{o} - TS_{r}^{o}, and substituting into the above eqt.,
(log K/T)_{P} = H^{o}_{r}/(2.303RT^{2})
and by integrating at constant P, from T_{a} to T_{b}
d log K = [H^{o}_{r,T,P}/(2.303RT^{2})]dT
and substituting in H^{o}_{r,Tb,P} = H^{o}_{r,Ta,P} + (H_{r}^{o}/T)_{P} dT
= H^{o}_{r,Ta,P} + Cp_{r}^{o} dT
or log K_{Tb} = log K_{Ta} + [H^{o}_{r,Ta,P}/(2.303RT^{2})]dT + Cp_{r}^{o} dT
= log K_{Ta} - [H^{o}_{r,Ta,P}/(2.303R)][1/T_{b} - 1/T_{a}] + Cp_{r}^{o} dT (105)
The last term can be integrated using an appropriate algorithm for Cp_{r}^{o}. 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 = -G^{o}_{r}/(2.303RT)
Taking the partial derivative with respect to P at constant T
(log K/P)_{T} = -V^{o}_{r}/(2.303RT)
and integrating from P_{a} to P_{b} at constant T
log K_{Pb} = log K_{Pa} - [V^{o}_{r,Ta,P}/(2.303RT)]dP (106)
Dealing with First Order Phase Transitions in Moving µ^{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.
G_{r} = H_{r} - TS_{r} = 0
If the phases are in their standard states
G_{r}^{o} = H_{r}^{o} - TS_{r}^{o} = 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 µ^{*} 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 S_{r}^{o} 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 H_{r}^{o} 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 H_{2}O in between the layers. In general, these contributions will be from Na_{2}O, K_{2}O, CaO in the interlayer sites; MgO, FeO, Fe_{2}O_{3}, Al_{2}O_{3} in the octahedral sites; Al_{2}O3, SiO_{2}, TiO_{2}, in the tetrahedral sites and H_{2}O. 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
Na_{0.33}Al_{2}(Al_{0.33}Si_{3.67})O_{10}(OH)_{2}:
µ^{o}_{beid.} = 0.165µ^{o}_{Na2O,I} + µ^{o}_{Al2O3,O} + 0.167µ^{o}_{Al2O3,T} + 3.67µ^{o}_{SiO2,T} + µ^{o}_{H2O,}
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).
G_{ss} = G_{mm} + G_{im} + G_{ex} where the three terms can be
rewritten as
G_{ss} = (n_{i}µ_{i}^{o}) + RT (n_{i} ln a_{i}^{o}) + RT (n_{i} 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 a_{i}^{o} 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
a_{i}^{o} = (X_{i})^{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 G_{ex} is to use a Margules expansion and to calculate G_{mix} 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.
G_{ex} = a + bX_{2} + cX_{2}^{2} + dX_{2}^{3} + ... (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
G_{ex} = -cX_{2} + cX_{2}^{2} = X_{2}X_{1}b (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 G_{ex} contains 3 parameters (b, c, d) which can be combined into 2 new parameters WG_{1} and WG_{2} by letting
b = WG_{2}
c = WG_{1} - 2WG_{2}
d = WG_{2} - WG_{1}
G_{ex} = X_{1}X_{2}(WG_{1}X_{2} + WG_{2}X_{1}) (111)
In a solid solution, a phase is unstable if its composition falls between two minimum in a plot of G_{mix} 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 G_{mix} 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 G_{mix} versus composition. These two points mark the spinoidal and can be calculated by setting the second partial derivative of G_{mix} 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, E_{a}, 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 = [E_{a}/(2.303R)](1/T) + log A (113)
A plot of experimental k values versus (1/T) is often linear. In this case, A and E_{a} are determined, respectively, from the y intercept which is log A and the line slope which is [E_{a}/(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 cm^{2}) 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 + µ_{i}dn_{i} + _{i}dArea_{i} (114)
_{i} and G_{r}, the free energy change in the reaction forming a mole of the component making up the particle, can be used to compute G_{n}, 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,
G_{n} = 6d^{2}_{i} + d^{3}(/M_{w})G_{r} (115)
where and M_{w} are the density in g/cm^{3} 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 G_{n} versus d has a maximum which is often assumed to equal the activation energy in the Arrhenius equation. d_{c}, the d corresponding to this maximum value, is found by taking the first partial derivative of G_{n} with respect to d and setting this equal to zero to solve for d_{c}. For the situation of cubic particles,
d_{c} = - 4M_{w}/(G_{r}) for cubic particles. (116)
Substituting d_{c} from equation (116) into equation (115) gives
(G_{n})_{max} = 32^{3}[M_{w}/(G_{r})]^{2} for cubic particles. (117)
For a spherical particle, what are the expressions for the critical radius and (G_{n})_{max}? Water has a surface tension of 71.97 ergs/cm^{2} (1.739 x 10^{-6} calories) in contact with the atmosphere at 25^{o}C and 1 bar. What G_{r} is necessary for raindrops of water to form from saturated atmosphere with (G_{n})_{max} at a particle diameter of 1 micron (10^{-4} cm) 25^{o}C and 1 bar. What activity ratio of a_{H2O liquid}/a_{H2O vapor} corresponds to this G_{r}. 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 H_{2}O 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 G_{area} to calculate H_{area}, the contribution of surface area to H.
G_{area} = H_{area} - TS_{area} = H_{area} + T(G_{area}/T) (119)
and since G_{area} = Area, we have Area = H_{area} + T(/T)Area from which H_{area} can be solved. For water, the partial of with respect to T is -0.148 ergs/^{o}Kcm^{2}. 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/[T^{1/2}V-(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.
a_{VdW} = 0.4219 R^{2}T_{c}^{2}/P_{c} (125)
b_{VdW} = 0.1250 RT_{c}/P_{c} (126)
a_{R-K} = 0.4275 R^{2}T_{c}^{2.5}/P_{c} (127)
b_{R-K} = 0.0866 RT_{c}/P_{c} (128)
The critical point compressibility factors (z_{c}) for non-polar gases (e.g., N_{2}, CO_{2}, He, O_{2}, SO_{2}, H_{2}S) 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 T_{r}, P_{r}, and V-_{r} equal to their respective values divided by their critical point values, i.e., T/T_{c}, P/P_{c}, 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 P_{r} for isotherms of T_{r}. 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, H_{2}O is a polar molecule and doesn't fit the data well, i.e., z_{c} 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 n_{i} 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.
µ_{i} = µ^{*}_{i} + RT ln[X_{i}P(_{i})]
Holding constant both T and moles of all components while taking the partial derivative of the above expression with respect to V gives
(µ_{i}/V)_{T,n} = RT(ln X_{i}P_{i}/V)_{T,n}
where the partial derivative of µ^{*}_{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).
(µ_{i}/V)_{T,n} = -(P/n_{i})_{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/n_{i})_{T,nj} dV = RT ln X_{i}P_{i}/V)_{T,n} - RT ln X_{i}P_{V}_{=>}
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/n_{i})_{T,nj} dV - RT/V] dV = RT ln X_{i}P_{i} - RT ln X_{i}P_{V}_{=>}
- RT ln (/RT) + RT ln (V/RT)
Remembering that the ideal gas law holds as V => , X_{i} = n_{i}/n_{tot}, and V = V-n_{tot}
-RT ln X_{i}P_{V}_{=>} = -RT ln n_{i}RT/V_{V}_{=>}) = RT ln /RT + RT ln (1/n_{i})
By subsitution into the previous equation,
RT ln X_{i}P_{i} = _{V}[(P/n_{i})_{T,nj} dV - RT/V] dV - RT ln (V/n_{i}RT)
Add - RT ln X_{i}P_{i} to both sides of the above equation and note
that X_{i}V/n_{i} = V- gives equation (129)
ln _{i,P,T} = (1/RT)_{V}[(P/n_{i})_{V,T,n}_{i} - 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/n_{tot}. This is done to integrate equation (129).
P = RT/(V- - b) - a/V-^{2} (122)
P = n_{tot}RT/(V - bn_{tot}) - a(n_{tot})^{2}/V^{2} (130)
Equation (130) can be used to take the partial derivative of P with respect to n_{i} 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=1}X_{i}b_{i} (131)
and a = _{i=1 j=1 }X_{i}X_{j}a_{ij} (132)
where a_{ii} and b_{i} 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,
a_{ij} = (a_{i}a_{j})^{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)] + b_{i}/(V--b) - ln[PV-/(RT)]
- [1/(RTV-)][2a_{i}^{0.5}_{j=1} X_{j}a_{j}^{0.5}] (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/[T^{1/2}V-(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 (n_{tot}). We use the two mixing relations given by equations (131) and (132) for b and a. However, a_{i} and b_{i} 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.
a_{i} = _{a}_{i} R^{2}T_{c}^{2.5}/P_{c} (135)
b_{i} = _{b}_{i} RT_{c}/P_{c} (136)
For the mixing relation in equation (132) used when i =/ j, equation (133) is replaced with
a_{ij} = (0.5/P_{cij})(_{a}_{i} + _{a}_{j})R^{2}T^{2.5}_{cij} (137)
where P_{cij}, _{a}_{i}, and T_{cij} 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)] + b_{i}/(V--b) - ln[PV-/(RT)]
+ [ab_{i}/(RT^{1.5}b^{2})]{ln[(V-+b)/V-] - b/(V-+b)}
-[2/(RT^{1.5}b)]{ln[(V-+b)/V-]}_{j=1} X_{j}a_{ji} (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).
P_{cij} = 8[0.291 - 0.04(_{i} + _{j})]RT_{cij}/(V-_{ci}^{1/3} + V-_{cj}^{1/3})^{3} (139)
T_{cij} = (T_{ci}T_{cj})^{0.5}(1-k_{ij}) (140)
where V-_{ci} and T_{ci} are the molar volume and temperature at the critical point for the ith component. _{i}, _{a}_{i}, and _{b}_{i} are constants for the ith component and k_{ij} represents the deviation from the geometric mean for T_{cij}. 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 (H_{2}, He, Ne) in which _{a}_{i} and _{b}_{i} 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
T_{cij} = (T_{ci}^{o}T_{cj}^{o})^{0.5}(1-k_{ij})/[1 + 21.8/(M_{wij}T)] (141)
where
M_{wij} = 2M_{wi}M_{wj}/(M_{wi} + M_{wj}) (142)
in which M_{wi} and M_{wj} are the molecular weights of components i and j, respectively. If i is for a quantum gas, T_{ci}^{o} is the classical critical temperature; otherwise, T_{ci}^{o} is the critical temperature of the ith non-quantum component. T_{cj}^{o} is defined in a similar fashion.
Also, if either i or j is a quantum gas, equation (139) is replaced with
P_{cij} = 8[0.291-0.04(_{i}+_{j})]R(T_{ci}^{o}T_{cj}^{o})^{0.5}(1-k_{ij})/(V-_{ci}^{o1/3} + V-_{cj}^{o1/3})^{3} (143)
in which T_{ci}^{o} and T_{cj}^{o} are defined as in equation (141). If i is for a quantum gas, _{i} is zero and
V-_{ci}^{o} = 0.291RT_{ci}^{o}/P_{ci}^{o} (144)
Otherwise, V-_{ci}^{o} is the critical volume of the ith component at the critical point. P_{ci}^{o} is the classical critical pressure of the ith quantum component or the critical pressure of the ith nonquantum component. _{j}, V-_{cj}, and P_{cj}^{o} 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 }X_{i}X_{j}B_{ij} (146)
C = _{i=1 j=1 k=1}X_{i}X_{j}X_{k}C_{ijj} (147)
The integrated form of equation (129) for the virial equation becomes
ln _{i,P,T} = (2/V-)_{j=1} X_{j}B_{ij} + (2/V-^{2}) _{j=1 k=1 }X_{j}X_{k}C_{ijk} - 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 cm^{2}/sec, x is the distance in cm, and C is the concentration in g/cm^{3}. In a liquid, D is of the order of 10^{-5}cm^{2}/sec and in a solid D is < 10^{-10}cm^{2}/sec. In porous fluid-saturated sediment,
D_{sed} D_{fluid}/^{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 D_{sed} in the literature is
D_{sed} D_{fluid}/
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 (r_{sed}) to the that of the fluid within the sediment (r_{fluid}). 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 = r_{sed}/r_{fluid} = (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 C^{o} and C be the aqueous concentrations (g/cm^{3}) near the surface of the grain and in the bulk solution, respectively. Let D_{fluid} be the diffusion coefficient (cm^{2}/sec) of the components in the fluid, t be the time in sec, and V be the stirred volume in cm^{3} 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 C^{o} remains constant and C in solution was initially zero at time t = 0.
-D_{fluid}(C-C^{o})/x = V(C/t)_{T,P}
_{t}_{=0}t (D_{fluid}/Vx) dt = _{C}_{=0C} (1/(C^{o}-C) dC
(D_{fluid}/Vx)t = - ln[(C^{o}-C) - [-ln(C^{o}-0)]
exp (tD_{fluid}/Vx) = C^{o}/(C^{o}-C) or
C = C^{o}[1 - 1/exp (tD_{fluid}/Vx)]
At t = 0, C = 0, at t => , C => C^{o}
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/cm^{3} in the dissolving mineral of the component being released as the surface layer forms and A_{surf} is the surface area of the dissolving component.
D_{fluid}(C-C^{o})/x = A_{surf}(x/t)_{T,P}
-_{t}_{=0t} [D_{fluid}(C-C^{o})/(A_{surf})]dt = _{x}_{=0x} x dx
C^{o} 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-C^{o} is constant with time.
tD_{fluid}(C^{o}-C)/(A_{surf}) = x^{2}/2
or x = [2tD_{fluid}(C^{o}-C)/(A_{surf})]^{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} [D_{fluid}(C-C^{o})/x]dt
= _{t}_{=0t} [D_{fluid}(C^{o}-C)/[2tD_{fluid}(C^{o}-C)/(A_{surf})]^{0.5}]dt
= _{t}_{=0t} {D_{fluid}(C^{o}-C)(A_{surf})]/(2t)]}^{0.5}]dt
= {2D_{fluid}(C^{o}-C)(A_{surf})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, Fe_{2}O_{3}, CO_{2}, SiO_{2}, H_{2}S, and H_{2}O at 100^{o}C and 300 bars, 200^{o}C and 650 bars (Mark), and 300^{o}C and 1000 bars. Consider these solid phases: Fe_{2}O_{3}, Fe_{3}O_{4}, FeSiO_{3}, FeCO_{3}, FeS_{2}, in which reactions written between solids are balanced on Fe. Consider the following aqueous species: Fe^{2+}, Fe^{3+}, Fe(OH)^{2+}, and Fe(OH)^{+}. Assume a_{H2O} = 1, a_{CO2} = 0.1P, a_{H2S} = 0.05P, and constant molar volume of solids. Assume quartz is present and in equilibrium with the aqueous solution, so any reactions releasing SiO_{2} 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 100^{o}C and 300 bars; 10^{-4} at 200^{o}C and 650 bars; and 10^{-3.5} at 300^{o}C 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 H_{2}S results in replacing hematite with pyrite, you have the option of removing pyrite from your considered phases.
µ^{o} in calories/mole
25^{o}C 100^{o}C 200^{o}C 300^{o}C
1 bar 300 bars 650 bars 1,000 bars
Fe_{2}O_{3} -178,155
Fe_{3}O_{4} -242,574
FeSiO_{3} -267,160
FeCO_{3} -162,414
FeS_{2} - 38,293
CO_{2} gas - 94,262
H_{2}S gas - 8,016
SiO_{2} quartz -204,646
H_{2}O liquid - 57,964 - 60,085 - 62,563
Fe^{2+} aq. - 20,080 - 17,530 - 14,840
Fe^{3+} 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
S^{o}_{25 C, 1 bar} a b c V
cal/^{o}K/mole cm^{3}/mole
CO_{2} gas 51.072 10.57 0.00210 -206,000
H_{2}S gas 49.16 7.81 0.00296 -46,000
Fe_{2}O_{3} 20.94 23.49 0.01860 -355,000 30.274 Fe_{3}O_{4} 34.83 21.88 0.04820 0 44.524
FeSiO_{3} 22.6 26.49 0.00507 -555,000 32.952
FeCO_{3} 25.1 11.63 0.02680 0 29.378
FeS_{2} 12.65 17.88 0.00132 -305,000 23.940
SiO_{2} qtz 9.88 11.22 0.00820 -270,000 22.688
Second Problem
SO_{2} is a frequent waste product gas in industry processes. The gas can be removed by reacting with portlandite Ca(OH)_{2} or with limestone CaCO_{3} at higher than surface temperatures to produce anhydrite CaSO_{4} in the presence of O_{2} and producing either CO_{2} or H_{2}O.
reaction CaCO_{3} + SO_{2} + 1/2 O_{2} = CaSO_{4} + CO_{2}
reaction Ca(OH)_{2} + SO_{2} + 1/2 O_{2} = CaSO_{4} + H_{2}O
Write out the expression for G_{r,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 G_{r,T,1} for the T and then calculate the P at which G_{r,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: 500^{o}C and 1,000^{o}C.
Assume a gas phase has X_{SO2} = 0.5 and the solids are pure solids in their standard states. Plot your 500^{o}C isotherm data (Gene) and your 1000^{o}C isotherm data as the ratio of the other two gas components, X_{CO2}/X_{O2} and X_{CO2}/X_{H2O}, 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 25^{o}C and 1 bar to move through PT space.
µ^{o} in calories/mole
25^{o}C and 1 bar
CaCO_{3} -270,100
Ca(OH)_{2} -214,724
CaSO_{4} -315,925
SO_{2} gas - 71,750
O_{2} gas 0
CO_{2} gas - 94,262
H_{2}O steam - 54,634
S^{o}_{25 C, 1 bar} a b c V
cal/^{o}K/mole cm^{3}/mole
CaCO_{3} 22.15 24.98 0.00524 -620,000 36.93
Ca(OH)_{2} 19.93 19.07 0.01080 0 33.06
CaSO_{4} 25.5 16.78 0.02360 0 45.94
SO_{2} gas 59.33 11.04 0.00188 -184,000
O_{2} gas 49.03 7.16 0.001 - 40,000
CO_{2} gas 51.07 10.57 0.00210 -206,000
H_{2}O 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 G_{r,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 G_{r,T,1} for the T and then calculate the P at which G_{r,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: 100^{o}C, 400^{o}C, 700^{o}C and 1000^{o}C. 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
Mg_{3}Al_{2}Si_{3}O_{12} + KFe_{3}AlSi_{3}O_{10}(OH)_{2} = Fe_{3}Al_{2}Si_{3}O_{12} + KMg_{3}AlSi_{3}O_{10}(OH)_{2}
The assemblage composition to be used has [Fe_{2}Mg]Al_{2}Si_{3}O_{24} for ideal site mixing between almandine and pyrope in garnet and K[Mg_{2}Fe]AlSi_{3}O_{10}(OH)_{2} for ideal site-mixing between annite and phlogopite in biotite. Use Karpov's µ^{o}_{25 C, 1 bar} data which hopefully are internally consistent.
Reaction 2: muscovite + albite = paragonite +
microcline
KAl_{2}(AlSi_{3})O_{10}(OH)_{2} + NaAlSi_{3}O_{8} = NaAl_{2}(AlSi_{3})O_{10}(OH)_{2} + KAlSi_{3}O_{8}
The assemblage to be used has [K_{0.95}Na_{0.05}]Al_{2}(AlSi_{3})O_{10}(OH)_{2} for ideal site mixing between muscovite and paragonite in a mica and [K_{0.9}Na_{0.1}]AlSi_{3}O_{8} for ideal site mixing between albite and microcline.
Reaction 3: hydroxy-apatite + 0.5fluoro-phlogopite = fluoro-apatite + 0.5 hydroxy-phlogopite
Ca_{5}(PO_{4})_{3}(OH) + 0.5KMg_{3}(AlSi_{3})O_{10}F_{2} =
Ca_{5}(PO_{4})_{3}F + 0.5KMg_{3}(AlSi_{3})O_{10}(OH)_{2}
The assemblage to be used has Ca_{5}(PO_{4})_{3}[(OH)_{0.8}F_{0.2}] for ideal site mixing within apatite and KMg_{3}(AlSi_{3})O_{10}[(OH)_{0.9}F_{0.1}] for ideal site mixing within phlogopite. Use Robie"s and Helgeson's µ^{o}_{25 C, 1 bar} data which hopefully are internally consistent.
The thermodynamic data needed for the problems are listed below.
µ^{o} in calories/mole
25^{o}C and 1 bar
Ca_{5}(PO_{4})_{3}OH -1,502,412 Robie
Ca_{5}(PO_{4})_{3}F -1,542,968 Robie
KMg_{3}(AlSi_{3})O_{10}F_{2} -1,446,725 Robie
KMg_{3}(AlSi_{3})O_{10}(OH)_{2} -1,396,187 Helgeson, -1,403,585 Karpov
Fe_{3}Al_{2}Si_{3}O_{12} -1,187,811 Karpov
Mg_{3}Al_{2}Si_{3}O_{12} -1,428,776 Karpov
KFe_{3}AlSi_{3}O_{10}(OH)_{2} -1,147,156 Helgeson, -1,159,082 Karpov
KAl_{2}(AlSi_{3})O_{10}(OH)_{2} -1,339,249 Berman
NaAl_{2}(AlSi_{3})O_{10}(OH)_{2} -1,331,352 Berman
NaAlSi_{3}O_{8} - 886,734 Berman
KAlSi_{3}O_{8} - 896,776 Berman
S^{o}_{25 C, 1 bar} a b c V
cal/^{o}K/mole cm^{3}/mole
Ca_{5}(PO_{4})_{3}OH 93.30 114.26 0.01981 -2,500,000 157.56
Ca_{5}(PO_{4})_{3}F 92.70 113.39 0.01362 -2,453,000 159.60
KMg_{3}(AlSi_{3})O_{10}F_{2} 75.90 96.13 0.02674 -2,041,000 146.37
KMg_{3}(AlSi_{3})O_{10}(OH)_{2} 76.10 100.61 0.02878 -2,150,000 149.66
Fe_{3}Al_{2}Si_{3}O_{12} 75.60 97.50 0.03364 -1,873,000 115.28
Mg_{3}Al_{2}Si_{3}O_{12} 62.32 91.69 0.03264 -2,092,000 113.15
KFe_{3}AlSi_{3}O_{10}(OH)_{2} 95.20 106.43 0.02977 -1,931,000 154.32
KAl_{2}(AlSi_{3})O_{10}(OH)_{2} 70.07 97.56 0.02638 -2,544,000 140.87
NaAl_{2}(AlSi_{3})O_{10}(OH)_{2} 66.37 97.40 0.02450 -2,644,000 132.16
NaAlSi_{3}O_{8} 53.64 61.70 0.01390 -1,501,000 100.83
KAlSi_{3}O_{8} 51.18 63.83 0.01290 -1,705,000 108.69
4th Problem
Construct an activity phase diagram at 300^{o}C and 1 kilobar for the Na_{2}O, K_{2}O, SiO_{2}, Al_{2}O_{3}, HCl, H_{2}O system. Assume quartz saturation so the a_{SiO2} is not a variable. Also assume a_{H2O} 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 25^{o}C and 1 bar standard state data from Helgeson to 300^{o}C and 1 kilobar. If nephline and kalsilite do not show up on the phase diagram than pick a lower activity of SiO_{2 aq} than at quartz saturation.
µ^{o} in calories/mole
25^{o}C and 1 bar 300^{o}C and 1 kbar
nepheline NaAlSiO_{4} - 472,872
kalsilite KAlSiO_{4} - 481,750
microcline KAlSi_{3}O_{8} - 895,374
albite NaAlSi_{3}O_{8} - 884,509
muscovite KAl_{3}Si_{3}O_{10}(OH)_{2 }-1,336,301
paragonite NaAlSi_{3}O_{10}(OH)_{2} -1,326,012
quartz SiO_{2} - 204,646
kaolinite Al_{2}Si_{2}O_{5}(OH)_{4} - 905,614
pyrophyllite Al_{2}Si_{4}O_{10}(OH)_{2} -1,255,997
water H_{2}O - 62,563
SiO_{2 aq} -203,100
Na^{+}_{aq} - 67,590
K^{+}_{aq} - 74,260
H^{+}_{aq} 0
S^{o}_{25 C, 1 bar} a b c V
cal/^{o}K/mole cm^{3}/mole
NaAlSiO_{4} 29.72 35.91 0.00646 - 732,797 54.16
KAlSiO_{4} 31.85 29.43 0.01736 - 532,000 59.89
KAlSi_{3}O_{8} 51.13 63.83 0.01290 -1,705,000 108.74
NaAlSi_{3}O_{8} 52.30 61.70 0.01390 -1,501,000 100.07
KAl_{3}Si_{3}O_{10}(OH)_{2 }68.8 97.56 0.02638 -2,544,000 140.71
NaAlSi_{3}O_{10}(OH)_{2} 66.40 97.43 0.02450 -2,644.000 132.53
SiO_{2} 9.88 11.22 0.0082 - 270,000 22.69
Al_{2}Si_{2}O_{5}(OH)_{4} 48.53 72.77 0.02920 -2,152,000 99.52
Al_{2}Si_{4}O_{10}(OH)_{2 }57.2 79.43 0.03921 -1,728,200 126.6
molar CO_{2} gas phase at 100^{o}C from F. Din (1962) Thermodynamic Functions of Gases, Volume 1
V_{CO2} V_{CO2}/RT
0.5 atm 0.5066 bars 61,163.5 cm^{3} 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
G^{o}_{r,T,P} = H^{o}_{r,Tref,Pref} + (Cp_{r}^{o})_{Pref}dT - TS^{o}_{r,Tref,Pref}
- T(Cp_{r}^{o}/T)_{Pref}dT] + V^{o}_{r,T}dP
= H^{o}_{r,Tref,Pref} - TS^{o}_{r,Tref,Pref} + V^{o}_{r,T}dP
+ (a_{r} + b_{r}T + c_{r}/T^{2})_{Pref}dT
- T(a_{r}/T + b_{r} + c_{r}/T^{3})_{Pref}dT]
= H^{o}_{r,Tref,Pref} - TS^{o}_{r,Tref,Pref} + V^{o}_{r,T}dP
+ a_{r}(T - T_{ref}) + 0.5b_{r}(T^{2} - T_{ref}^{2}) - c_{r}(1/T - 1/T_{ref})
- T[a_{r} ln(T/T_{ref}) + b_{r}(T - T_{ref}) - 0.5c_{r}(1/T^{2} - 1/T_{ref}^{2})]
= H^{o}_{r,Tref,Pref} - TS^{o}_{r,Tref,Pref} + V^{o}_{r,T}dP
+ a_{r}[T-T_{ref}-T ln(T/T_{ref})] + b_{r}[0.5(T^{2} - T_{ref}^{2}) - T(T-T_{ref})]
+ c_{r}[-(1/T - 1/T_{ref}) + 0.5(1/T^{2} - 1/T_{ref}^{2})]
If V^{o}_{r,T} is constant with pressure then
= H^{o}_{r,Tref,Pref} - TS^{o}_{r,Tref,Pref} + V^{o}_{r,T}(P-P_{ref})
+ a_{r}[T-T_{ref}-T ln(T/T_{ref})] + b_{r}[0.5(T^{2} - T_{ref}^{2}) - T(T-T_{ref})]
+ c_{r}[-(1/T - 1/T_{ref}) + 0.5(1/T^{2} - 1/T_{ref}^{2})]
Note that if Cp^{o}_{r,Pref} had been constant, then the first method gives
G^{o}_{r,T,P} = H^{o}_{r,Tref,Pref} - TS^{o}_{r,Tref,Pref} + V^{o}_{r,T}(P-P_{ref})
+ Cp_{r}^{o} dT - TCp_{r}^{o}(1/T)dT]
= H^{o}_{r,Tref,Pref} - TS^{o}_{r,Tref,Pref} +V^{o}_{r,T}(P-P_{ref})
+ Cp^{o}_{r,Pref}[T-T_{ref} + T ln(T_{ref}/T)]
= G^{o}_{r,Tref,Pref} + S^{o}_{r,Tref,Pref}(T_{ref}-T) + V^{o}_{r,T}(P-P_{ref})
+ Cp^{o}_{r,Pref}[T-T_{ref} + T ln(T_{ref}/T)]
For the second method, if Cp^{o}_{r,Pref} had been constant
G^{o}_{r,T,P} = G^{o}_{r,Tref,Pref} - (S^{o}_{r,Tref,Pref} + V^{o}_{r,T}(P-P_{ref})
+ Cp^{o}_{r}(1/T)dT)dT
= G^{o}_{r,Tref,Pref} - S^{o}_{r,Tref,Pref}(T-T_{ref}_{}) + V^{o}_{r,T}(P-P_{ref})
- Cp^{o}_{r}(ln T - ln T_{ref})dT
= G^{o}_{r,Tref,Pref} - S^{o}_{r,Tref,Pref}(T-T_{ref}_{}) + V^{o}_{r,T}(P-P_{ref})
- Cp^{o}_{r}[T ln(T) - T - T_{ref} ln(T_{ref}) + T_{ref} - ln(T_{ref})(T-T_{ref})]
= G^{o}_{r,Tref,Pref} - S^{o}_{r,Tref,Pref}(T-T_{ref}_{}) + V^{o}_{r,T}(P-P_{ref})
+ Cp^{o}_{r}[T-T_{ref} + T ln(T_{ref}/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 CO_{2}, organic nitrogen is transformed from ammonium (NH_{4}^{+}) to nitrate, (NO_{3}^{-}) and organic polyphosphates (H_{n+2}P_{n}O_{2n+2}) are transformed to orthophosphate species (PO_{4}^{3-}). 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 (NO_{3}^{-}, HNO_{3}^{-}, NO_{2}^{-}, HNO_{2}^{-}, NH_{3} and NH_{4}^{+}) and forms of orthophosphate (H_{3}PO_{4}, H_{2}PO_{4}^{-}, HPO_{4}^{2-}, and PO_{4}^{3-}). Construct a 25^{o}C and 1 bar pH-pe diagram with a_{H2O} = 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 NO_{3}^{-} (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 N_{2} 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^{+} + 2NO_{3}^{-} + CH_{2}O = N_{2} + CO_{2} + H_{2}O
or in terms of pe,
12H^{+} + 2NO_{3}^{-} + 10e^{-} = N_{2} + 6H_{2}O
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 N_{2} 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 N_{2} 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 [Mg_{4}O(PO_{4})_{2}], magnesium hydroxyapatite [Mg_{5}OH(PO_{4})_{3}], and farringtonite [Mg_{3}(PO_{4})_{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 Mg^{2+}, 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 H_{3}PO_{4} + H_{2}PO_{4}^{-} + HPO_{4}^{2-} + PO_{4}^{3-} 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 MgNH_{4}PO_{4}.6H_{2}O. Repeat the procedure in (C) with the additional constraint of equilibrium with struvite. Balance both the Mg and PO_{4} between the three solids so that only the nitrogen species is an aqueous variable. Calculate the activity sum of NH_{4}^{+} and NH_{3} 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 (Ca_{5}OH(PO_{4})_{3}) in which the dissolution of calcite (CaCO_{3}) 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 G^{o}_{f,25oC,1 bar}
in kj/mol
aqueous species
PO_{4}^{3-} -1,018.8
HPO_{4}^{2-} -1,089.3
H_{2}PO_{4}^{-} -1,130.4
H_{3}PO_{4} -1,142.6
NO_{3}^{-} - 111.3
HNO_{3}^{-} - 111.3
NO_{2}^{-} - 37.2
HNO_{2}^{-} - 42.97
NH_{3} - 26.57
NH_{4}^{+} - 79.37
Mg^{2+} - 454.8
e^{-} 0
H^{+} 0
H_{2}O - 237.18
(liquid water)
N_{2} 0
(gas)
H_{2} 0
(gas)
O_{2} 0
(gas)
solids
Mg(OH)_{2} - 833.5
(brucite)
Mg_{5}OH(PO_{4})_{3} -5,758.11
(Mg hydroxyapatite)
Mg_{4}O(PO_{4})_{2} -4,172.7
(Mg oxyapatite)
Mg_{3}(PO_{4})_{2} -3,538.7
(farringtonite)
MgNH_{4}PO_{4}.6H_{2}O -3,051.1
(struvite)
Note that R = .0083136 kj/mol/^{o}K and 25^{o}C = 298.15^{o}K
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} + Mg_{3}(PO_{4})_{2} = Mg_{4}O(PO_{4})_{2} + H_{2}O
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 G_{reaction} 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.