Geological Sciences 4550: Geochemistry

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1 1. a.) Using the enthalpies of formation given in able 2.02, find H in Joules for the reaction: albite jadite + quartz NaAl2Si2O8 NaAl2SiO6+ SiO2 b.) Which assemblage (side of the equation) is stable at 600 C and 1 MPa? Use the thermodynamic data in able 2.2 and assume the volume change of the reaction is independent of pressure and temperature. a. H = J/mol. bhis question is similar to example 2.8. Whichever side of the reaction has the lowest Gibbs Free Energy will be the stable assemblage. If G for the reaction is positive, the reactants (left side) are stable, if it is negative, the products (right side) are stable. So we want to calculate G at 873 K and 1 MPa (= 10 bars). Since G is a state function, we can do the pressure and temperature calculations G,Pref = G o S ref a + 2 c b 2 ref aln ref separately, and the order doesn t matter. o determine how G r changes with temperature, we can use equation (since the heat capacities are given in Meier-Kelley form): We need to calculate G o and a, b, and c. Let s set up a spreadsheet to do this calculation 1 1

2 Here we have set up the parameters H, S, a, b, c, and V in a matrix. o calculate the s, we just sum the values for Ja and Qz and subtract from that the values for Al. G is then easy to calculate from the H and S. We have assigned names to each of the s, e.g., H is DH, etc. his allows us to set up an equation to calculate G() that makes some sense. Since the answer is positive, the assemblage on the left (albite) is stable. 2. Calcite and aragonite are two forms of CaCO3 that differ only their crystal lattice structure. he reaction between them is thus simply: Calcite Aragonite Using the data in able 2.02, a.) Determine which of these forms is stable at the surface of the earth (25 C and 0.1 MPa). he side of the reaction having the lowest Gibbs Free Energy is stable. According to able 2.2, G Cal = and G Arag = , G = 940 J/mol so calcite is stable. b.) Which form is favored by increasing temperature? If ΔG decreases with increasing temperature, then the product (left) side of the reaction will be favored and visa versa. he trick here is to note that G r = S, so if ΔS is positive, aragonite will be favored and visa versa. Using r P Hess s Law, we find ΔS r = ΔS will be negative, so calcite is favored by increasing. c.) Which form is favored by increasing pressure? Similarly, we note that G r P = V r. If ΔG decreases with increasing pressure, then the product (left) side of the reaction will be favored and visa versa. ΔV r = ΔV is negative, so aragonite is favored by increasing pressure. d.) At what pressure are calcite and aragonite in equilibrium at 298 K? Calcite and aragonite will be in equilibrium when pressure increases enough that ΔG decreases from the standard state value of 0.94 kj/mole to 0. o find this point, we integrate: 2 2

3 ΔG r o = 0.94kJ = P ' 0.1 ΔV r dp and solve for P. Since we have no other information, we assume ΔV is independent of pressure and hence the integral reduces to: -940J=ΔV(P -0.1) Rearranging: -940/ΔV = P Substituting cc/mol (which converts to J/MPa), we find P = 338 MPa. 4. Using your result from part d. of question 2 and the Clapeyron slope, construct an approximate phase diagram showing the stability fields of calcite and aragonite as a function of and P. So the phase boundary is located at 298K and 338 MPa. he slope of the phase boundary is given by the Clapeyron equation (equ. 3.3): d dp = ΔV r ΔS r Substituting values obtained in Problem Set 1, we find the slope is K/MPa. 3. Consider a system consisting of olivine of variable composition ((Mg,Fe)2SiO4) and orthopyroxene of variable composition ((Mg,Fe)SiO3). What is the minimum number of components needed to describe this system? his is a particularly simple case: three. A graphical approach might be easiest. We could chose them in a variety of ways. MgO, FeO, SiO 2 are the obvious and perhaps conventional choices. Other possibilities would be Mg 2 SiO 4, Fe 2 SiO 4, SiO 2 or Fe 1 Mg -1, Mg 2 SiO 4, SiO 2. One the graph below, the olivine and pyroxene compositions lie along the two joins. 3 3

4 SiO 2 MgSiO 3 FeSiO 3 Mg 2 SiO4 Fe 2 SiO 5. Consider a box partitioned into equal volumes, with the left half containing 1 mole of Ne and the right half containing 1 mole of He. When the partition is removed, the gases mix. Show, using a classical thermodynamic approach (i.e., macroscopic) that the entropy change of this process is S = 2R ln 2. Hint: Assume that He and Ne are ideal gases and that temperature is constant. he simplest approach to this problem is to consider that each gas expands into the volume of the other; in other words, the volume of each doubles. he dependence of entropy on volume change is given by equation According to this equation, ( S/ V) = α/β. he gases are ideal, α=1/ and β=1/p, so " S % # $ V & ' = α β = 1/ 1/ P = P (1) We know neither nor P, so it would be nice to have this expressed in terms of other variables. Since the gases are ideal, PV=NR. And P = NR (2) V 4 4

5 Substituting into (1) above, # S & % ( $ V ' = NR V (3) o obtain the entropy change, ΔS, we simply rearrange and integrate. Doing so, we have: V 2 NR ΔS = V dv = NR ln V 2 V 1 V 1 he ratio V 2 /V 1 = 2 and since there is one mole, we get ΔS = R ln 2. his is the entropy change for each of the gases. For both the gases, the entropy change is twice this, or 2R ln

CHAPTER 9: INTRODUCTION TO THERMODYNAMICS. Sarah Lambart

CHAPTER 9: INTRODUCTION TO THERMODYNAMICS Sarah Lambart RECAP CHAP. 8: SILICATE MINERALOGY Orthosilicate: islands olivine: solid solution, ie physical properties vary between 2 endmembers: Forsterite (Mg