A two stage compressor with intercooling between the stages is used to compress

Question

A two stage compressor with intercooling between the stages is used to compress air from 1 bar and 15 degree C to 9 bar in a Gas-Turbine cycle. The high pressure turbine drives the compressor whereas low pressure turbine is coupled with an electrical generator. The compressor and each turbine efficiencies are 85%; 75%; and 75%, respectively. Heat is added to the compressed air after leaving the compressor till its temperature reaches 900 degree C. The air then expands in HPT to an intermediate pressure and reheat to the 900 degree before expanding another time in LPT to 1 bar. Find the intermediate pressure, the turbine work, the back work ratio, and the cycle thermal efficiency. Assume 15 kg/s air flow rate.

Explanation / Answer

-------------------------------------------------------------------------------- Introduction Some useful constants in thermodynamics: 1 eV = 9.6522E4 J/mol k Boltzmann's constant = 1.38E-23 J/K volume: 1 cm3 = 0.1 kJ/kbar = 0.1 J/bar mole: 1 mole of a substance contains Avogadro's number (N = 6.02E23) of molecules. Abbreviated as 'mol'. atomic weights are based around the definition that 12C is exactly 12 g/mol R gas constant = Nk = 8.314 J mol-1 K-1 Units of Temperature: Degrees Celsius and Kelvin The Celsius scale is based on defining 0 °C as the freezing point of water and 100°C as the boiling point. The Kelvin scale is based on defining 0 K, "absolute zero," as the temperature at zero pressure where the volumes of all gases is zero--this turns out to be -273.15 °C. This definition means that the freezing temperature of water is 273.15 K. All thermodynamic calculations are done in Kelvin! kilo and kelvin: write k for 1000's and K for kelvin. Never write °K. Units of Energy: Joules and Calories Joules and calories and kilocalories: A calorie is defined as the amount of energy required to raise the temperature of 1 g of water from 14.5 to 15.5°C at 1 atm. 4.184 J = 1 cal; all food 'calories' are really kcal. Many times it is easiest to solve equations or problems by conducting "dimensional analysis," which just means using the same units throughout an equation, seeing that both sides of an equation contain balanced units, and that the answer is cast in terms of units that you want. As an example, consider the difference between temperature (units of K) and heat (units of J). Two bodies may have the same temperature, but contain different amounts of heat; likewise, two bodies may contain the same heat, but be at different temperatures. The quantity that links these two variables must have units of J/K or K/J. In fact, the heat capacity C describes the amount of heat dQ involved in changing one mole of a substance by a given temperature increment dT: dQ = CdT The heat capacity C is then C = dQ/dT and must have units of J K-1 mol-1. (The specific heat is essentially the same number, but is expressed per gram rather than per mole.) Don't forget significant digits. 1*2=2; 1.1*2=2; 1.1*2.0=2.2; 1.0*2.0=2.0 Why Thermodynamics? Think about some everyday experiences you have with chemical reactions. Your ability to melt and refreeze ice shows you that H2O has two phases and that the reaction transforming one to the other is reversible--apparently the crystallization of ice requires removing some heat. Frying an egg is an example of an irreversible reaction. If you dissolve halite in water you can tell that the NaCl is still present in some form by tasting the water. Why does the NaCl dissolve? Does it give off heat? Does it require energy? How is it that diamond, a high-pressure form of C, can coexist with the low pressure form, graphite, at Earth's surface? Do diamond and graphite both have the same energy? If you burn graphite and diamond, which gives you more energy? When dynamite explodes, why does it change into a rapidly expanding gas, which provides the energy release, plus a few solids? Chemical thermodynamics provides us with a means of answering these questions and more. A Few Definitions A system is any part of the universe we choose to consider. Matter and energy can flow in or out of an open system but only energy can be added to or subtracted from a closed system. An isolated system is one in which matter and energy are conserved. A phase is a homogeneous body of matter. The components of a system are defined by a set of chemical formula used to describe the system. The phase rule: F + P = C + 2. Extensive parameters are proportional to mass (e.g., V, mass, energy). Intensive parameters are independent of mass (e.g., P, T); these are the "degrees of freedom" F contained in the phase rule. Thermodynamics: Power and Limitations Thermodynamics allows you to predict how chemical systems should behave from a supra-atomic "black-box" level--it says nothing about how chemical systems will behave. Thermodynamics also pertains to the state of a system, and says nothing about the path taken by the system in changing from one state to another. Chemical Reactions and Equations How to write chemical reactions; stoichiometry. Mass and charge balance: e.g., 2Fe3+ + 3H2O = Fe2O3 + 6H+ Reaction-Produced Change in Mass, Density, Volume The change in volume rV of a reaction is the volume V of the products minus the volume of the reactants: rV = Vproducts - Vreactants Thus, if the products are smaller than the reactants, rV < 0. In a generalized reaction such as aA + bB ... = cC + dD... rV = cVC + dVD - aVA - bVB This sort of additive relationship is true for other state variables and is usually stated as rii where i are the stoichiometric coefficients, positive for products and negative for reactants. What Actually Drives Reactions? Is it Energy? Can We Just Calculate or Measure the Energy Difference of Reactants and Products and Know Which Way the Reaction Will Go? For many years people felt that chemical reactions occurred because the reactants had some kind of energy to give up (i.e., use to do work)--and that therefore the energy of the products would be less than the energy of the reactants. However, we all know that when ice melts it consumes rather than releases heat, so there must be more to the story behind why chemical reactions occur. Le Chatelier's Principle "If a change is made to a system, the system will respond so as to absorb the force causing the change." Equilibrium A mechanical analogy for chemical change is that of a ball rolling down a slope with multiple valleys; we explain the ball's behavior by saying that mechanical systems have a tendency to reduce their potential energy. At equilibrium none of the properties of a system change with time. A system at equilibrium returns to equilibrium if disturbed. "Stable" describes a system or phase in its lowest energy state. "Metastable" describes a system or phase in any other energy state. The figure above shows the mechanical analogy for H2O at -5°C and + 5°C and 1 atm. Left: at -5°C, solid H2O has the lowest possible energy state. Right: at +5°C, liquid H2O has the lowest possible energy state. When solid H2O is actually present at +5°C, the difference between the free energy of solid H2O and liquid H2O is available to drive the reaction to form the stable solid H2O phase, and the reaction will go to completion if kinetically possible. Energy: How Do We Calculate and Measure Energy and How Can We Use this Knowledge to Predict Reaction Behavior? Thermodynamics works equally well to describe any kind of work or energy: magnetic, potential, kinetic, etc. For geological systems we typically talk about pressure-volume work, which, because mechanical work is Fx, you can imagine might be PV or VP Because we noted that rV < 0 if the products are smaller than the reactants, we choose to write the P-V work term as -PV so that a decrease in volume -V is seen as positive work or that an increase in volume +V results in a decrease in crystal energy. The absolute energy of a body can be calculated from Einstein's equation U=mc2, but the presence of the c2 term means that the energy of any system is quite large and that measuring this energy is impractical. It is more practical to measure differences in energy U, and we always discuss or measure differences relative to some arbitrary standard state. Analogous to this might be if someone in Namibia asked you to measure the elevation of the crests of waves at Campus Point--without agreement on some kind of standard, you wouldn't be able to do much more than measure the heights of individual waves. If however, you could both agree on an equivalent "sea level" at both localities, you could then compare the absolute elevations of the wave crests. A typical thermodynamic standard state is normal laboratory conditions: 25°C (298.15 K) and 1 atm (often called STP for standard temperature and pressure). The internal energy U of a mineral is the sum of the potential energy stored in the interatomic bonds and the kinetic energy of the atomic vibrations. Thus, you might expect that weakly bonded minerals have relatively low potential energy and thus low internal energy, and when a mineral is cold such that its atomic vibrations are slow it will have low kinetic energy and thus low internal energy. Internal energies are always defined relative to some non-zero standard state, so we typically talk about changes in internal energy dU. An Aside on Differences and Differentials What's the difference among , d, and ? is used to indicate any kind of difference. d is used to indicate a differential. is used to indicate a partial differential. For example, the partial differential, with respect to y, of f(x,y) = x3y4 is = 4x3y3 First Law of Thermodynamics Adding heat Q to a crystal increases its internal energy U: dU dQ ( indicates 'proportional') but if the crystal is allowed to expand, some of the added energy will be consumed by expansion dV, so the total energy of the crystal is reduced: dU = dQ - PdV This is effectively the First Law of Thermo: that total energy (heat + P-V work) is conserved. Heat Capacity Heat capacity C describes the amount of heat required to change the temperature of a substance: C = By definition, the heat capacity of water at 15°C is 1 cal K-1 g-1 or 18 cal K-1 mol-1 (i.e., the heat required to heat 1 gram of water from 14.5 to 15.5°C is 1 calorie). Heat capacities of solids approach zero as absolute zero is approached: C = 0 The heat capacity is written with a subscript P or V depending on whether it obtains for constant pressure CP or constant volume CV. As an aside, CP = CV + TV2/ where and are the expansivity and compressibility--for solids the difference between CP and CV is minimal and can be ignored as a first approximation. For gases, CP = CV + R, and is quite significant. Heat capacities are measured by calorimetry and often fit by a function of the form: CP = a + bT + cT-2 + dT-0.5 but there are other expansions for the heat capacity involving more or fewer terms. Below are some examples of heat capacities of minerals. Note how silicates have a nearly constant heat capacity of ~1 J K-1 g-1 above 400K. Enthalpy We have already talked about the familiar concept of heat as energy. Let's define another measure of energy called enthalpy H--a kind of measure of the thermal energy of a crystal. As we will see below, dH = dQ + VdP Recall that we interpreted dU = dQ - PdV to mean that the internal energy change is the heat change minus the energy lost to relaxation of the crystal. Thus, dH = dQ + VdP means that the enthalpy change is the heat change plus the energy the crystal gains by virtue of not being allowed to expand. Enthalpy includes the vibrational and bonding energy at absolute zero H0°, plus the energy required to increase temperature: H = H0° + CPdT i.e., we can find the enthalpy change H produced by changing temperature by integrating the heat capacity CP: H = CPdT Integration Reminder How to integrate the heat capacity (to determine change in enthalpy H): CP dT = (a + bT + cT-2 + dT-0.5) =aT + bT2/2 - c/T + 2dT0.5 and is evaluated as =a(T2 - T1) + b(T22 - T12)/2 - c/(T2-1 - T1-1) + 2d(T20.5 - T10.5) How to integrate the heat capacity divided by T (to determine entropy S): dT = (a/T + b + cT-3 + dT-1.5) = a ln T + b T - c T-2/2 - 2 d T -0.5 and is evaluated as a(ln T2 - ln T1) + b(T2 - T1) - c(T2-2 - T1-2)/2 - 2d(T2-0.5 - T1-0.5) As an example, let's calculate the change in enthalpy H°298-1000 that results from heating quartz from 298 K to 1000 K, given the following heat capacity expansion coefficients: a = 104.35, b = 6.07E-3, c = 3.4E+4, d = -1070 (CP dT = (a + bT + cT-2 + dT-0.5) =aT + bT2/2 - c/T + 2dT0.5 evaluated from 298 to 1000K =a*(1000-298) + b*(10002-2982)/2 - c*(1000-1-298-1) + 2d*(10000.5-2980.5) = 45.37 kJ/mol Relation Among Enthalpy, Heat, and Heat Capacity (HP=QP) An important relationship between enthalpy change H and heat change Q is revealed by differentiating H = U + PV to obtain the total differential dH = dU + PdV + VdP substituting dU = dQ - PdV we get dH = dQ + VdP dividing by dT gives = - V at constant pressure, = 0, leaving = which is equal to CP: = = CP Determining Enthalpies Thus, if we want to measure how the internal energy U of a crystal changes U with increasing temperature at constant pressure, we want to know H, and we can get that by integrating the heat capacity CP over the temperature range of interest. There's another way to measure H, though: calorimetry. By dissolving a mineral in acid and measuring the heat produced by the dissolution, we get a heat of dissolution (usually positive). The enthalpy of "formation" fH° of the mineral is then just the opposite of the heat of dissolution (usually negative). Exceptions to the "usually positive/negative" rule include CN, HCN, Cu2+, Hg2+, NO, Ag+, and S2-. Enthalpies of formation appear in tables of thermodynamic data and are usually referenced to 298 K and 1 atm. Enthalpy of Reaction To get an enthalpy of reaction rH° we can measure the enthalpies of formation of the reactants and products fH° and then take the difference between them as rH° = fH°products- fH°reactants For example, we can compute the enthalpy of the reaction anhydrite + water = gypsum: CaSO4 + 2H2O = CaSO42H2O from Ca + S + 2O2 = CaSO4 fH° = -1434.11 kJ/mol H2 + 0.5O2 = H2O fH° = -285.830 kJ/mol Ca + S + 3O2 + 2H2 = CaSO42H2O fH° = -2022.63 kJ/mol Thus, rH° = fH°gypsum - fH°anhydrite - fH°water = -16.86 kJ/mol. Exothermic vs. Endothermic If rH° < 0 the reaction produces a reduction in enthalpy and is exothermic (heat is given up by the rock and gained by the surroundings). If rH° > 0 the reaction produces an increase in enthalpy and is endothermic (heat from the surroundings is consumed by the rock). An easy way to remember this is that spontaneous reactions produce a decrease in internal energy, and because we know that UP HP a decrease in HP is also a decrease in UP. Calculating fH° at Temperatures Other Than 298 K So far we know how to calculate the change in enthalpy caused by heating and we know that we can get enthalpies of formation from tables. What if we want to know the enthalpy of formation of a mineral at a temperature other than 298 K? We do this by calculating rCP for the reaction that forms the mineral of interest: rCP = rCPproducts - rCPreactants and then integrating. Thus, for example if we want to know fH° for quartz at 1000 K, we get coefficients for the heat capacities of Si, O2 and SiO2: compound a b c d Si 31.778 5.3878E-4 -1.4654E5 -1.7864E2 O2 48.318 -6.9132E-4 4.9923E5 -4.2066E2 SiO2 104.35 6.07E-3 3.4E-4 -1070 for the reaction Si + O2 = SiO2 and we calculate a = 24.254 b = 6.2225E-3 c = -3.5E5 d = -470.7 and thus, fH°1000 - fH°298 = CP dT = a*(1000-298) + b/2*(10002-2982) - c*(1000-1-298-1) + 2d*(10000.5-2980.5) = 5.511 kJ/mol This is the change in the enthalpy of formation that results from heating. We add this to the enthalpy of formation at 298 K to get the enthalpy of formation at 1000 K: fH°1000 = (fH°1000 - fH°298) + fH°298 = 5.511 - 910.700 = -905.2 kJ/mol In other words, forming quartz from the elements at 1000 K yields slightly less heat than at 298 K. Compare this with the change in enthalpy H°298-1000 that results from heating quartz from 298 K to 1000 K, which we calculated is 45.37 kJ/mol. Entropy We have discussed the intuitive statement that reactions probably proceed because the reactants can decrease their internal energy by reacting. We also noted that internal energy scales with enthalpy, suggesting that reactions might 'go' because of a decrease in enthalpy. However, we then noted that not all reactions give off heat--some, such as the melting of ice, proceed in spite of consuming heat. Moreover, there are other processes that proceed in the apparent absence of any heat change: e.g., mixing of gases or the spreading of dye in water. What is it that causes these reactions to proceed spontaneously even if the heat change is zero or endothermic? The answer is entropy S, which is a measure of the order or disorder. Entropy has three sources: configurational, electronic, and vibrational. Configurational entropy refers to the entropy resulting from imperfect mixing of different atoms in the same site in a crystal, and is described by the Boltzmann distribution: Sconfigurational = k ln (This is engraved on Boltzmann's tomb in Vienna!) where is the probability that a given number of atoms in a given number of sites will have a particular configuration. For N atomic sites that can contain fraction XA A atoms and XB B atoms, = N is always large where moles of material are concerned, so we can simplify this (using Stirling's approximation) to S = - n R (XA ln XA + XB ln XB) where n is the number of sites per mole. For example in cordierite there are 4 Al atoms and 5 Si atoms distributed over 9 tetrahedral sites. For a random distribution the entropy is S = - 9 R (4/9 ln 4/9 + 5/9 ln 5/9) = 51.39 J mol-1 K-1 Note that the form of the configurational entropy equation (and electronic entropy as well) indicates that if XA or XB are 0 or 1, Sconfig is zero: Electronic entropy arises when an electron in an unfilled orbital can occupy more than one orbital; e.g., for Ti3+, the single 3d electron can occupy one of three possible t2g orbitals and Selectronic = 9 J mol-1 K-1. Vibrational (or calorimetric) entropy arises because the energy of lattice vibrations can only increase or decrease in discrete steps and the energy quanta (phonons) can be distributed within the possible energy steps in different ways. Vibrational entropy is very difficult to calculate from statistical mechanics but can be calculated easily from heat capacity. Here's why: The entropy of a system always increases during irreversible processes; i.e., for a reversible process, dS = 0, whereas for irreversible processes dS >0. This is the Second Law of Thermo--better known as "You can't feed s**t into the rear of a horse and get hay out the front." If a mineral becomes more ordered during a reaction, reducing its entropy, the heat liberated must increase the entropy of the surroundings by an even greater amount. Thus, we write dS > then > and recalling that CP = then > and S = dT In other words, the vibrational entropy can be found by integrating the heat capacity divided by temperature. In a perfectly ordered, pure crystalline material the entropy is zero. This is a simple statement of the Third Law of Thermo, which follows from the fact that heat capacities approach zero at zero K: C = 0 However, because the rate of atomic diffusion also goes to zero at 0 K, all compounds have some zero-point entropy S°0. Entropy is thus the only thermodynamic potential for which we can calculate an absolute value. What we typically do is determine the heat capacity from near absolute zero to ambient conditions and then integrate it to get the (absolute) entropy (in fact this gives us only the vibrational entropy and ignores configurational and electronic contributions to entropy). Entropy Change of Reaction Just like rH and rV, we can calculate entropies of reactions by using absolute entropies S and calculating a difference in entropy rS. For example, if we know that S°CaSO4 = 106.7 J mol-1 K-1 S°Ca = 41.42 J mol-1 K-1 S°S = 31.80 J mol-1 K-1 S°O2 = 205.138 J mol-1 K-1 then the entropy of the reaction Ca + S + 2O2 = CaSO4 is rS° = 106.7 - 41.42 - 31.80 - 2 * 205.138 = -376.8 J mol-1 K-1 Energy Associated With Entropy The units of entropy suggest that the energy associated with S scales with temperature: dU -TS (The minus sign is there for reasons similar to the -PV we encountered earlier.) The energy associated with configurational entropy in the Al4Si5 cordierite we talked about earlier looks like this: The energy associated with vibrational entropy in tremolite, quartz, and chalcopyrite looks like this: (Josiah Willard) Gibbs Free Energy of a Phase The Gibbs free energy G is the thermodynamic potential that tells us which way a reaction goes at a given set of physical conditions--neither the enthalpy change nor the entropy change for a reaction alone can provide us with this information. The two measures of energy (enthalpy H and entropic energy TS) are brought together in the Gibbs free energy equation: (the chemical potential is the equivalent for a component) G = U + PV - TS which says that the Gibbs free energy G is the internal energy of the crystal U plus the energy the crystal gains by virtue of not being allowed to expand minus the entropic energy TS. Recalling that H = U + PV we can write this in a more understandable way G = H - TS which says that G is the difference between the heat energy and the entropic energy. Relationship Among G, S, and V If we differentiate G = U + PV - TS to obtain dG = dU + PdV + VdP - TdS - SdT and substitute TdS = dU + PdV (this comes from dS = dQ/T and dU = dQ - PdV); we are left with dG = VdP - SdT meaning that changes in Gibbs free energy are produced by changes in pressure and temperature acting on the volume and entropy of a phase. Realize that when we write dG = VdP - SdT we are implicitly writing dG = dP - dT which means that = V and = -S These relations indicate that the change in Gibbs free energy with respect to pressure is the molar volume V and the change in Gibbs free energy with respect to temperature is minus the entropy S. Gibbs Free Energy of Formation The defining equation for Gibbs free energy G = H - TS can be written as G = H - TS such that the Gibbs free energy of formation fG° is fG° = fH° - TfS° For example, to calculate the Gibbs free energy of formation of anhydrite, we can use fH°CaSO4 = -1434.11 kJ/mole S°CaSO4 = 106.7 J mol-1 K-1 S°Ca = 41.42 J mol-1 K-1 S°S = 31.80 J mol-1 K-1 S°O2 = 205.138 J mol-1 K-1 and we calculate the entropy of formation of anhydrite fS° = S°CaSO4 - S°Ca - S°S - 2 * S°O2 = -376.796 J mol-1 K-1 and then use fG° = fH° - TfS° = -1434,110 - 298.15 * -376.796 = -1321.77 kJ/mol Gibbs Free Energy of Reaction We can write the Gibbs free energy of reaction as the enthalpy change of reaction minus the entropic energy change of reaction rG = rH - TrS If the heat energy equals the entropic energy rH = TrS then rG = 0 and there is no reaction. Finally we have come to a satisfying point--we can now determine whether a given reaction will occur if we know H and S, and both of these are measurable or can be calculated. If rG < 0, the Gibbs free energy of the products is lower than the Gibbs free energy of the reactants and the reaction moves to produce more products. If rG > 0, the Gibbs free energy of the products is greater than the Gibbs free energy of the reactants and the reaction moves to produce more reactants. For example, to calculate rG° at STP for the reaction aragonite = calcite we use rH° = 370 J rS° = 3.7 J mol-1 K-1 to calculate rG° = 370 - 298.15 * 3.7 = -733 J/mol The negative value of G tells us that calcite has lower Gibbs free energy and that the reaction runs forward (aragonite calcite). Clapeyron Relation There is a useful relation between the slope of a reaction in PT space (i.e., dP/dT) and the entropy and volume changes of the reaction that follows from rG = VrdP - SrdT At equilibrium G = 0, such that rVdP = rSdT or = So, the P-T slope of a reaction is equal to the ratio of the entropy change to the volume change. Alternatively, along the equilibrium curve, the changes in pressure times the volume change are equal to changes in temperature times the entropy change. This is the Clapeyron Equation. So, a phase diagram is a kind of free energy map. = along an equilibrium, at high T and low P. Along the equilibrium boundary the Gibbs Free energies of the reactants and products are equal and the Gibbs Free energy of reaction rG, is zero. Shortcutting H and S and Finding G Directly Like other thermodynamic potentials, we can write the change in Gibbs free energy of reaction as rG° = fG°reactants- fG°products Instead of using fH° and fS°, it is often possible to obtain fG° values for most compounds from electronic data bases. For example, if the following Gibbs free energies of formation are known: fG°CaSO42H2O = -1707.280 kJ/mol fG°CaSO4 = -1321.790 kJ/mol fG°H2O = -237.129 kJ/mol then for CaSO4 + 2H2O = CaSO42H2O rG° = -1.232 kJ/mol Gibbs Free Energy at Any Pressure and Temperature We know many ways to determine rG at STP--but how do we calculate rG for other pressures and temperatures? Recall that the changes in Gibbs free energy with pressure and temperature are given by two of Maxwell's relations = rV and = -rS If we recast these as = rVP and = -rST and integrate, we get rGdP = rGP - rGPref = rVdP or rGP = rGPref + rVdP and rGdT = rGT - rGTref = - rSdT or rGT = rGTref - rSdT thus rGPT = rGPrefTref + rVdP - rSdT Solving the Pressure Integral at Constant Temperature To a first approximation, we can ignore the expansivity and compressibility of solids and use rVsdP = rVs(P - 1) as a simplification. Don't forget that this approximation is valid for solids only! An even more common assumption for P>>1 is rVsdP = rVsP For example, calculate the change in Gibbs free energy for the reaction 2 jadeite = albite + nepheline if pressure increases from 1 bar to 10 kbar, given nepheline = 54.16 cm3 albite = 100.43 cm3 jadeite = 60.40 cm3 First we calculate rV and find r = nepheline + albite - 2jadeite = 33.79 cm3 = 3.379 J/bar and thus rGPT - rG1,T = rVsP = 33.79 kJ/mol Solving the Temperature Integral at Constant Pressure Recall that the effect of temperature on the entropy change of reaction rS depends on the heat capacity change of reaction rCP: rS = dT Thus rGT = rGTref - rSdT expands to rGT = rGTref - STref + dTdT If the form of the heat capacity expansion is CP = a + bT + cT-2 + dT-0.5 then the above double integral is a(T - T ln T) - bT2/2 - cT-1/2 + 4dT0.5 - aTref - bTref2/2 + cTref-1 - 2dTref0.5 + aTlnTref + bTTref - cTTref-2/2 - 2dTTref-0.5 - TrSTref + TrefrSTref Note that this considers only vibrational entropy and ignores configurational entropy. This means of solving for rG requires that you know rG at the reference temperature. An alternative path that requires that you know the enthalpy change rH at the reference temperature is rGT = rHTref + CPdT - TrSTref + dT Solving the Temperature and Pressure Integrals for rGP,T To calculate the Gibbs free energy change of a reaction at any pressure and temperature, we can use either of the following equations, depending on whether we know rH or rG rGP,T = rG1,Tref - rSTref + dTdT + rVsP rGP,T = rH1,Tref + CPdT - TrSTref + dT + rVsP If you don't have heat capacity data for the reaction of interest, these equations can be roughly approximated as rGP,T = rG1,Tref - rS1,Tref(T - Tref) + rVsP rGP,T = rH1,Tref - TrS1,Tref + rVsP For example, calculate rG for jadeite + quartz = albite at 800 K and 20 kbar. The data at 298 K and 1 bar are rH° = 15.86 kJ/mol rS° = 51.47 J K-1 mol-1 rVs° = 1.7342 J/bar = 17.342 cm3/mol Using rGP,T = rH1,Tref - TrS1,Tref + rVsP = 15,860 - 800 * 51.47 + 1.7342 * 20,000 = 9.37 kJ/mol If we had used the complete equation for solids, integrating the heat capacities, we would have obtained an answer of 9.86 kJ/mol--not horrifically different. Calculating the PT Position of a Reaction If we say that rG = 0 at equilibrium, then we can write our solids-only and constant-heat-capacity approximations as 0 = rG1,T - rS1,Tref(T - Tref) + rVsP 0 = rH1,Tref - TrS1,Tref + rVsP and thus we can calculate the pressure of a reaction at different temperatures by P = rG1,Tref - rS1,Tref(T - Tref) / -rVs P = rH1,Tref - TrS1,Tref / -rVs and we can calculate the temperature of a reaction at different pressures by T = Tref + rG1,Tref + rVsP / rS1,Tref T = Tref + rH1,Tref + rVsP / rS1,Tref Let's do this for the albite = jadeite + quartz reaction at T = 400 K and T = 1000 K: P = (15,860 - 5147 * 400) / -1.7342 = 2.7 kbar P = (15,860 - 5147 * 1000) / -1.7342 = 20.6 kbar Assuming that dP/dT is constant (a bad assumption, we know), the reaction looks like this Introduction to the Equilibrium Constant A bit farther down the road we will encounter a monster called the equilibrium constant K: K = exp(-rG°/RT) or ln K = -rG°/RT At equilibrium, where rG°= 0, ln K = 0 and K = 1. Let's see what K looks like for jadeite + quartz = albite at 800 K and 20 kbar: ln K = - (rH1,Tref - TrSTref + rVsP)/ RT = -(15,860 - 800 * 51.47 + 1.7342 * 20,000)/(8.314*800) = -1.4 If we do this for all of PT space, we can contour PT space in terms of lnK: Solutions Almost no phases are pure, but typically are mixtures of components. For example, olivine varies from pure forsterite Mg2SiO4 to pure fayalite Fe2SiO4, and can have any composition in between--it is a solid solution. We need a way to calculate the thermodynamic properties of such solutions. As a measure of convenience, we use mole fraction to describe the compositions of phases that are solid solutions. For example, a mix of 1 part forsterite and 3 parts fayalite yields an olivine with 25 mol% forsterite and 75 mol% fayalite, which can be written as (Mg0.25Fe0.75)2SiO4 or fo25fa75, etc. Mole fractions are denoted as Xi. We need a way of splitting up the Gibbs free energy of a phase among the various components of the phase--how for example do we decide how much of the Gibbs free energy of an olivine is related to the forsterite component and how much derives from the fayalite component? Likewise, how does the Gibbs free energy of a phase vary with composition--is the relationship linear between endmembers?? We address these issues by defining a partial Gibbs free energy for each component at constant pressure and temperature and constant composition of other components, called the partial molar Gibbs free energy or chemical potential i = where n is the amount of substance. For olivine solid solution composed of fayalite and forsterite components or endmembers, we write dG = dnfayalite + dnforsterite Volume of Mixing Imagine that mole fractions of phase A and phase B with molar volumes VA and VB, are mixed together. We can describe the volume of the mixture as V = XAVA° + XBVB° and it is a linear mixing of the two endmember volumes. We call this ideal mixing or mechanical mixing. Real solutions, however, do not behave this way, and the mixing is always non ideal, although sometimes only weakly so. The figure shows mixing that produces a smaller volume than expected, but it is not possible to predict the shapes and positions of such mixing curves. Partial Molar Volume The partial molar volume is defined as i If you mix two compounds A and B together and find a volume of mixing that is non-ideal, how can you determine the contribution that A and B each make to the volume? That is, what are the partial molar volumes of A and B, A and B?? Graphically, the partial molar volumes are the A and B axis intercepts of the tangent to the mixing curve, and can be described by the simple relationship: Vmix = XAA + XBB or Vmix = Xii The behavior of this function is such that when XA is 1, Vmix = VA and when XA is 0, Vmix = VB. Alternatively, A = (Vmix - XBB) / XA Entropy of Mixing The entropy of mixing is never zero because mixing increases entropy. As we discussed days ago, the entropy of mixing (i.e., the configurational entropy) is Smix = -R (Xi ln Xi) where i = 1..n is the number of sites over which mixing is occurring. Enthalpy of Mixing Enthalpies also do not combined ideally (linearly) in mixtures because the mixture may have stronger bonds than were present in either of the unmixed phases. The excess enthalpy is Hmix = 0.5 * N z XAXB [2AB - AA - BB] where AB is the interaction energy among A-B atoms, AA is the interaction energy among A-A atoms, and BB is the interaction energy among B-B atoms. Gibbs Free Energy of Mixing Recall that all spontaneous processes/reactions occur because of a decrease in Gibbs free energy. It should therefore not surprise you that the Gibbs free energy of mixing is always negative--otherwise mixing would not occur. The fact that A AA + BB, which makes sense because it means that the A-B bonds have a higher free energy than the sum of the free energies of separate AA and BB bonds. Activities In reality, no phases behave ideally--that is, their chemical potentials are never simple logarithmic functions of composition as A - °A = RT lnXA implies. Instead, we say that the chemical potential is a simple logarithmic function of activity and define activity as a = (X) where a is the activity of a compound, is the "site occupancy coefficient" (e.g., = 2 for Mg in Mg2SiO4), and is the activity coefficient that describes the non-ideal behavior. Thus we write A - °A = RT lnaA For pure compounds a=1 because X=1. For ideal compounds =1. As a specific example, the chemical potential of the almandine (Fe3Al2Si3O12) component of a garnet solid solution ((Fe, Mg, Ca, Mn)3Al2Si3O12) is alm = °alm + RT lnXalm To be clear, ° is the chemical potential of the component in its pure reference state and varies as a function of pressure and temperature; this we measure with calorimetry. is the chemical potential as it actually occurs and varies as a function of phase composition; this we measure with an electron microprobe. The activity forms a bridge between idealized behavior and real behavior. The Equilibrium Constant At equilibrium the sum of the Gibbs free energies of the reactants equals the sum of the Gibbs free energies of the products. Equally, the sum of the partial molar Gibbs free energies (chemical potentials) of the reactants equals the sum of the partial molar Gibbs free energies (chemical potentials) of the products. In other words, for SiO2 + 2H2O = H4SiO4 at equilibrium, H4SiO4 = SiO2 + 2H2O More generally, for aA + bB = cC + dD then cC + dD = aA + bB or, at equilibrium r = 0 = cC + dD - aA - bB which we can reformat as r = i i where i is the stochiometric coefficient of a product or reactant and is positive if for a product and negative if for a reactant. If we then remember that - ° = RT lna and rewrite it as = ° + RT lna we can reformat the earlier equation as r = 0 = c(C° + RT lnaC) + d(D° + RT lnaD) - a(A° + RT lnaA) - b(B° + RT lnaB) which looks nicer as r = r° + RT ln (aCc aDd/aAa aBb) To be completely general we write r = r° + RT ln ai ( means to multiply all i terms) This equation is invariably simplified to r = r° + RT lnQ and Q is the activity product ratio. The activities in the Q term change as the reaction progresses toward equilibrium. To be clear again, r° is the difference in the Gibbs free energies of the products and reactants when each is in its pure reference state and varies as a function of pressure and temperature. r is the difference in the Gibbs free energies of the products and reactants as they actually occur and varies as a function of phase composition. At equilibrium, the product and reactant activities have adjusted themselves such that r = 0. We write this (with K instead of Q, to signify equilibrium) as 0 = rG° = -RT ln K K is called the equilibrium constant. If K is very large (ln K positive), the combined activities of the reaction products are enormous relative to the combined activities of the reactants and the reaction will likely progress. On the other hand, if K is small (ln K negative), there is unlikely to be any reaction. The utility of K is that it tells us for any reaction and any pressure and temperature, what the activity ratios of the phases will be at equilibrium. For example, for the reaction albite = jadeite + quartz let's say that at a particular P and T, rG° = -20.12 kJ/mol Using rG° = -RT lnK we calculate log K = 3.52 This means that at equilibrium, (ajadeite aquartz / aNaAlSi3O8) = e3.52 Where ajadeite is the activity of NaAlSi2O6 in clinopyroxene and aalbite is the activity of NaAlSi3O8 in plagioclase. Alternative Route to the Equilibrium Constant When we think of mass balance in a reaction, we can explicitly write 0 = iMi where i are the stoichiometric coefficients and Mi are the masses or the phase components. Analogously, we can explicitly write a similar balance among the chemical potentials: 0 = ii For each chemical potential we can write i = °i + RT lnai Combining these two equations we find 0 = i°i + iRTln ai 0 = i°i + RTln (ai) 0 = i°i + RTln ai 0 = i°i + RTln K and eventually 0 = rG° + RT ln K or rG° = -RT ln K The equation K = ai is the law of mass action (which actually discusses the action of chemical potential rather than mass). We can also write for 298 K and 1 atm rH° - TrS° = - RT ln K and for any P and T of interest: rH1,Tref + CPdT - TrSTref + dT + rVP = - RT ln K This has been called "the most important equation in thermodynamics," so you'd better like it(!) The equilibrium constant K is a function of 1/T -ln K = (rG° / RT) = [(rH / RT) - (rS / R)] Which looks like Activity Models (Activity-Composition Relations) for Crystalline Solutions Garnets are solid solutions of component abbrev. formula pyrope prp Mg3Al2Si3O12 almandine alm Fe3Al2Si3O12 grossular grs Ca3Al2Si3O12 spessartine sps Mn3Al2Si3O12 andradite and Ca3Fe23+Si3O12 Mixing models derive from entropy considerations. In particular the relation Smix = -R Xi ln Xi although we will not go through the derivation. Mixing on a Single Site The simplest type of useful activity model is the ionic model, wherein we assume that mixing occurs on crystallographic sites. For a Mg-Fe-Ca-Mn garnet with mixing on one site, which we can idealize as (A,B,C,D)Al2Si3O12, the activities are aprp = Mg3XMg3 aalm = Fe3XFe3 agrs = Ca3XCa3 asps = Mn3XMn3 The pyrope activity is shown in the above figure. In general, for ideal mixing in a mineral with a single crystallographic site that can contain ions, ai = Xj where a, the activity of component i, is the mole fraction of element j raised to the power. For non-ideal mixing, we include an activity coefficient ai = jXj Mixing on a Several Sites For minerals with two distinct sites and the general formula (A,B)(Y,Z) there are four possible end member components AY, AZ, BY, and BZ. The ideal activities of these components are aAY = XAXY aAZ = XAXZ aBY = XBXY aBZ = XBXZ For non-ideal garnet activities we write aprp = XMg3 XAl2 or Mg3 XMg3 Al2 XAl2 aalm = XFe3 XAl2 or Fe3 XFe3 Al2 XAl2 agrs = XCa3 XAl2 or Ca3 XCa3 Al2 XAl2 asps = XMn3 XAl2 or Mn3 XMn3 Al2 XAl2 aand = XCa3 XFe3+2 or Ca3 XMn3 Fe3+2 XFe3+2 where the X3 term describes mixing on the 8-fold trivalent site and the X2 term describes mixing on the octahedral divalent site. The pyrope activity is shown in the above figure for Mg from 0 3 and Al from 0 2. It is common to modify such models that are based on completely random mixing of elements with models that consider local charge balance on certain sites or the Al-avoidance principle. Geothermometry and Geobarometry Exchange Reactions Many thermometers are based on exchange reactions, which are reactions that exchange elements but preserve reactant and product phases. For example: Fe3Al2Si3O12 + KMg3AlSi3O10(OH)2 = Mg3Al2Si3O12 + KFe3AlSi3O10(OH)2 almandine + phlogopite = pyrope + annite We can reduce this reaction to a simple exchange vector: (FeMg)gar+1 = (FeMg)bio-1 Popular thermometers include garnet-biotite (GARB), garnet-clinopyroxene, garnet-hornblende, and clinopyroxene-orthopyroxene; all of these are based on the exchange of Fe and Mg, and are excellent thermometers because rV is small, such that = is large (i.e., the reactions have steep slopes and are little influenced by pressure). Let's write the equilibrium constant for the GARB exchange reaction K = (aprpaann)/(aalmaphl) thus rG = -RT ln (aprpaann)/(aalmaphl) This equation implies that the activities of the Fe and Mg components of biotite and garnet are a function of Gibbs free energy change and thus are functions of pressure and temperature. If we assume ideal behavior ( = 1) in garnet and biotite and assume that there is mixing on only 1 site aalm = Xalm3 = [Fe/(Fe + Mg + Ca + Mn)]3 aprp = Xprp3 = [Mg/(Fe + Mg)]3 aann = Xann3 = [Fe/(Fe + Mg)]3 aphl = Xphl3 = [Mg/(Fe + Mg)]3 Thus the equilibrium constant is K = (XMggar XFebio)/(XFegar XMgbio) When discussing element partitioning it is common to define a distribution coefficient KD, which is just the equilibrium constant without the exponent (this just describes the partitioning of elements and not the partitioning of chemical potential): KD = (XMggar XFebio)/(XFegar XMgbio) = (Mg/Fe)gar /(Mg/Fe)bio = K1/3 Long before most of you were playground bullies (1978) a couple of deities named John Ferry and Frank Spear measured experimentally the distribution of Fe and Mg between biotite and garnet at 2 kbar and found the following relationship: If you compare their empirical equation ln KD = -2109 / T + 0.782 this immediately reminds you of ln K = - (rG° / RT) = -(rH / RT) - (PrV / RT) + (rS / R) and you realize that for this reaction rS = 3*0.782*R = 19.51 J/mol K (the three comes from the site occupancy coefficient; i.e., K = KD3) and -(rH / R) - (PrV / R) = -2109 or rH = 3*2109*R -2070*rV Molar volume measurements show that for this exchange reaction rV = 0.238 J/bar, thus rH = 52.11 kJ/mol The full equation is then 52,110 - 19.51*T(K) + 0.238*P(bar) + 3RT ln KD = 0 To plot the KD lines in PT space Net-Transfer Reactions Net-transfer reactions are those that cause phases to appear or disappear. Geobarometers are often based on net-transfer reactions because rV is large and relatively insensitive to temperature. A popular one is GASP: 3CaAl2Si2O8 = Ca3Al2Si3O12 + 2Al2SiO5 + SiO2 anorthite = grossular + kyanite + quartz which describes the high-pressure breakdown of anorthite. For this reaction rG = -RT ln [(aqtzaky2agrs) / aan3] = -RT ln agrs / aan3 (the activities of quartz and kyanite are one because they are pure phases). A best fit through the experimental data for this reaction by Andrea Koziol and Bob Newton yields P(bar) = 22.80 T(K) - 7317 for rV = -6.608 J/bar. Again, if we use ln K = -(rH / RT) - (PrV / RT) + (rS / R) and set ln K = 0 to calculate values at equilibrium, we can rewrite the above as (PrV / R) = -(rH / R) + (TrS / R) or P = TrS / rV - rH / rV if TrS / rV = 22.8 then rS = -150.66 J/mol K if rH / rV = 7317 then rH = -48.357 kJ/mol So, we can write the whole shmear as 0 = -48,357 + 150.66 T(K) -6.608 P (bar) + RT ln K Contours of ln K on a PT diagram for GASP look like this: Kinetics Thermodynamics places no constraints on the rate or mechanism of reaction--that is the realm of kinetics. A popular method for describing the rate at which reactions proceed is to talk of an activated state through which the reaction must pass: When a system passes from an initial to a final state it must overcome an activation energy barrier G*. The advantages of this activation energy barrier paradigm are that it qualitatively explains the i) persistence of metastable states; ii) effect of catalysts in lowering G*; iii) temperature dependence of transformation. We can draw a similar diagram for a change in enthalpy induced by the reaction rH and an activation enthalpy barrier H* (usually called an activation energy Q*). Of course, unlike rG, rH can be positive or negative: It is not easy to generalize about the activation entropy S*, however, in general, reactions with positive entropy change rS are faster. For example, evaporation is faster than condensation, melting is faster than crystallization, and disordering is faster than ordering. Because thermal energy dictates whether an atom has sufficient energy to overcome an activation energy barrier, we will write that the fraction of atoms with thermal energy greater than H* is f = exp ( -H*/RT) i.e., if f

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