EAS 4550: Geochemistry Prelim Solutions October 13, 2017

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Moses McCarthy
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1 Name Key Part I Construct a pe-ph diagram for the manganese aqueous species and solids (denoted with subscript s) at 0.1 MPa and 298 K. Use the adjacent thermodynamic data. You may assume that solid phases are pure and ideal. Assume an activity of dissolved Mn of 10-6 M for equilibrium between aqueous species and solid phases. The value of R is J/mol-K. Label stability fields. Show all work as partial credit is possible. Use any additional sheets necessary (printouts from Excel or MatLab OK when properly labelled). (One issue: the G listed for Mn(OH) 3-1 was incorrect; should have been kj; using the value given, Mn(OH) 3-1 would not be stable on the diagram). There are many possible reactions, but when we consider which species are likely to be in equilibrium, there are far fewer. Manganese is in 3 possible valence states over this range: +2, +3, and +4. Mn 2+ species will be in equilibrium with Mn 3+ species, but not Mn 4+ species. Now when we consider the reaction between MnO and Mn(OH) 2, we find that the G of that reaction is negative and independent of ph. In other words, Mn(OH) 2 is stable relative to MnO over the whole range of the diagram, so we do not consider MnO. The reactions of interest are as follows: Solid phase reactions G (kj/mol) Log K 1. MnO + H 2 O = Mn(OH) Mn(OH) 2 = Mn 3 O 4 +3H 2 O+3H+3e Mn 3 O 4 + H 2 O = 3Mn 2 O 3 + 2H+ + 2e Mn 2 O 3 + H 2 O = 2MnO 2 +2H+ + 2e Aqueous-Solid Reactions 5. Mn H 2 O = Mn(OH) 2 + 2H Mn H 2 O = Mn 3 O 4 + 8H+ + 2e Mn H 2 O = Mn 2 O 3 +6H + + 2e Mn H 2 O = MnO H + + 2e Mn(OH) 2 + H 2 O = Mn(OH) H Mn(OH) H + = Mn 3 O 4 + 5H 2 O + 2e Equilibrium Constant Expressions Solid phase reactions slope intercept 1. MnO + H 2 O = Mn(OH) 2 1

2. pe = logk/3 -ph -1 8.75 3. pe = logk -ph -1 14.04 4. Mn 2 O 3 + H 2 O = 2MnO 2 +2H+ + 2e - -1 16.5 Aqueous-Solid Reactions 5. ph=(-logk - [Mn 2+ ])/2 ph = 12.9 6.

2 2. pe = logk/3 -ph pe = logk -ph Mn 2 O 3 + H 2 O = 2MnO 2 +2H+ + 2e Aqueous-Solid Reactions 5. ph=(-logk - [Mn 2+ ])/2 ph = pe= (-logk - 3log[Mn 2+ ])/2-4pH pe=(-logk - 2log[Mn 2+ ])/2-3pH pe=(-logk -log([mn 2+ ])/2-2pH ph = -logk log([mn(oh) -1 3 ] ph = pe = (-logk log([mn(oh) -1 3 ])/2+ 1 / 2 ph The pe-ph diagram resulting is shown below: Show all work for partial credit. The value of R is J/mol-K. 2

3 Part II 1. 1.Kyanite, sillimanite, and andalusite are isomorphs of SiAl2O5. In a system where kyanite and andalusite coexist at equilibrium, how many independent variables would have to be specified to completely determine the system? The phase rule is: ƒ = C f + 2 where c is the number of components and f is the number of phases. We have 1 component, SiAl 2 O 5, and two phases, so ƒ = 1. The system is univariant and only temperature or pressure need be specified. 2. Use the following data to answer this question Phase Formula H o f S V (J/mol) (J/K-mol) (cc/mol) water H 2 O Talc Mg3Si4O10(OH) Brucite Mg(OH) Quartz SiO (note that 1 cc/mol) = 1J/MPa-mol a. Write a reaction relating quartz and brucite to talc and water. 3Mg(OH) 2 + 4SiO 2 = Mg 3 Si 4 O 10 (OH) 2 + 2H 2 O b. Which side is stable at the surface of the Earth? Why? Whichever side has the lowest G. We calculate DG as: where G for each phase is calculated as G = H TS. DG is kj, so the right side is stable. c. Which side is favored by increasing pressure? Why? Since, whichever side has the lowest V will be favored by increasing pressure; i.e., if the DV r is negative, the r.h.s. is favored. DV r calculated by Hess s Law is cc/mol, so the right side is favored by increasing pressure. 3

4 3. Shown adjacent is the phase diagram for the plagioclase liquid and solid solutions. Draw G-bar-X diagrams (X should be mole fraction An) for 1600 C, 1400 C, 1200 C, and 1000 C showing the free energy of the liquid and solid solutions. 4

5 4. The graph below shows the results a series of reaction rate experiments. The log of the rate constant, k, is plotted on the Y (vertical) axis and inverse temperature (i.e., 1/T in kelvins) is plotted on the X axis. The result of each experiment is shown as a dot; a line has been fitted to the results. Estimate the frequency factor and the activation, or barrier, energy for this reaction. In log form, the Arhenius relation is: ln K = A E RT So on this plot, the slope is -E A /R and the intercept is ln A. The slope is 9/0.004 = 2250 and E A = 2250*R = kj. The frequency factor is e 6 =

6 5. What is the ph of a solution with SCO 2 = M at the CO 2 equivalence point (the point where [H + ] = [HCO 3 ])? Use the equilibrium constants below and assume ideality. Mass balance requires [H 2 CO 3 ] + [HCO 3 ] + [CO 3 ] = ΣCO 2 With no other ions present, we can guess our solution will be neutral to acidic, which means we can ignore both the carbonate ion (and the second dissociation constant), so: [H 2 CO 3 ] + [HCO 3 ] = ΣCO 2 (We can also ignore OH, so our charge balance equation becomes [HCO 3 ] = [H + ], but this is the condition we set in any case). Solving the first dissociation constant expression for carbonic acid, we have: [H 2 CO 3 ] =[H + ][HCO 3 1- ] /K 1 =[H + ] 2 /K 1 Substituting into the mass balance equation, we have: [H + ] 2 /K 1 + [H + ] - ΣCO 2 = 0 This is a quadratic equation: [H, ] = b ± b1 4ac 2a where a = 1/K 1, b = 1, and c = -ΣCO 2 Solving, we find ph = 5.0 (In fact, at this ph, both [H+] and [HCO 3 ] << [H 2 CO 3 ], so that [H 2 CO 3 ] ΣCO 2 In that case, an even simpler solution is to use the first dissociation constant relation to solve for [H + ] 2 K 1 ΣCO 2 ). 6

Geological Sciences 4550: Geochemistry

Geological Sciences 4550: Geochemistry 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