Problem Set Four Solution Due October 3, 2007

Size: px

Start display at page:

Download "Problem Set Four Solution Due October 3, 2007"

Hilary Clarke
6 years ago
Views:

1 Problem Set Four Solution Due October, Construct G-bar X diagrams for a regular solution with W= 1 kj (W is the interaction parameter in a non-ideal solution) at 100 temperature intervals from 00 to 700 C. Sketch the corresponding T-X phase diagram. The relevant equations are: G ex = X 1 X W G and G mixing = G ideal mixing + G ex, where G ideal mixing = + RT! i X i ln X i Phase diagram shown below: calculations are shown on the spreadsheet following: solution solutions X

2 Problem Set Four Solution Due October, 007 R 8.14 Gex=X1*X*W W 1000 Gideal=RT(X1*ln(X1)+X*ln(X)) Xi Xj T Gex Gid 47 Gmix Gmix Gmix Gmix Gmix Gmix C G-mix C 400 C 500 C 600 C 700 C X1

3 Problem Set Four Solution Due October, 007. Show that: G excess = W G1 X + W G X 1 X 1 X may be written as a 4 term power expansion, i.e.: G ex = A+ BX + CX +DX Substituting X 1 = (1 X ) G excess = W G1 X + W G [1 X ] (1 X )X Rearranging: G excess = W G1 X + WG X W G X ] (1 X ) G ex = W G X + (W G1 W G )X + (W G W G 1 ) X Letting A = 0, B = W G, C = W G1 W G and D = W G W G1, then G ex = A+ BX + CX +DX. Kyanite, andalusite, and sillimanite (all polymorphs of Al SiO 5 ) are all in equilibrium at 500 C and 76 MPa. Use this information and the following to construct an approximate temperature-pressure phase diagram for the system kyanite-sillimanite-andalusite. Assume V and S are independent of temperature and pressure. Label each field with the phase present. V S φ (cm ) (J/K-mol) kyanite andalusite sillimanite The easiest way to do an approximate P-T phase diagram is to use the Clayperon slope The reactions of interest and their slopes (dp/dt) are: Andelusite-Kyanite: ( )/( ) = 1.19 MPa/K Sillmanite-Kyanite: ( )/( ) = 1.85 MPa/K Andelusite-Sillmanite: ( ) /( )= MPa/K We can form equations of the form: P = 76 + slope (T - 500) to calculate the phase boundaries.

4 Problem Set Four Solution Due October, Kyanite P, MPa Sillmanite 00 0 Andelusite T, K 4. Sketch G-bar X diagrams for 1600 C, 1500 C, 100 C, and 150 C for the system Diopside-Anorthite (Figure 4.8). Draw tangents connecting the equilibrium liquids and solids C 1500 C 150 C 100 C Points show the free energy of the pure solids. Curve shows the free energy of the liquid.

5 Problem Set Four Solution Due October, Given the following analyses of basaltic glass and coexisting olivine phenocrysts, determine the K D for the MgO FeO exchange reaction, and calculate the temperatures at which the olivine crystallized using both MgO and FeO. Assume Fe O to be 10 mole% of total iron (the analysis below includes only the total iron, calculated as FeO; you need to calculate from this the amount of FeO by subtracting an appropriate amount to be assigned as Fe O ). Note that the mole % Fo in olivine is equivalent to the mole % Mg or MgO. (HINT: you will need to calculate the mole fraction of MgO and FeO in the liquid). Glass (liquid) composition: Sample TRD-1 DS-D8A (wt % oxide) (wt % oxide) SiO Al O ΣFe as FeO MgO CaO Na O..8 K O MnO TiO olivine Mole % Fo (=mole % Mg) TRD1 DS-D8A wt% w/10%fe+ Mol. wt moles mol frac. wt% w/10%fe+ moles mol frac. SiO AlO total FeO FeO FeO MgO CaO NaO KO MnO TiO Total XMgO-Ol XFeO-Ol KD 0.07 KD TMgO 180 kelvin 1107 C TMgO C TFeO 168 kelvin 1095 C TFeO C

6 Problem Set Four Solution Due October, Starting from equations 4.54, 4.56 and 4.18, use the fundamental relationships between free energy, entropy, enthalpy, and the equilibrium constant to derive the temperature dependence of the titanomagnetite ilmenite exchange (equation 4.57) We want to show that AW Usp T(K) = H! BW Mt H! CW Il H + DW Hem H + "H o AW Usp S! BW Mt S! CW Il S + DW Hem S + "S o! RlnK exch given W G = W H -TW S (4.56) and Gex = (W G1 X + W G X 1 )X 1 X (4.16) for the reaction Fe O 4 + FeTiO Fe TiO 4 + Fe O 4.5 We are dealing with two solutions: FeTiO Fe O and Fe TiO 4 Fe O 4, so the species in the above reactions are actually components in the two solutions. There will be one version of 4.16 to express the excess free energy of each solution and one version of 4.56 for each component. The free energy change of reaction ΔG r is:!g r = G Usp + G Hem " G Mag " G Ilm Let s begin by reorganizing 4.57: TAW Usp S! TBW Mt S! TCW Il S + TDW Hem S + T"S o! RT lnk exch = AW Usp H! BW Mt H! CW Il H + DW Hem H + "H o and!rt lnk exch = AW Usp H! BW Mt H! CW Il H + DW Hem H + "H o! TAW Usp S + TBW Mt S + TCW Il S! TDW Hem S! T"S o Collecting terms, we have:!rt lnk exch = "H o! T"S o + A(W Usp H! TW Usp S )! B(W Mt H! TW Mt S )! C(W Il H! TW Il S ) + D(W Hem H! TW Hem S ) Substituting equation 4.56, this becomes:!rt lnk exch = "H o! T"S o + AW Usp G! BW Mt Il Hem G! CW G + DW G Since!G =!H " T!S We have!rt lnk exch = "G o + AW Usp G! BW Mt G! CW Il Hem G + DW G Or!G o = "RT lnk exch " AW G Usp + B + CW G Il " DW G Hem (1) According to equation 4.54:! "G RT = ln $ X # Usp(1! X Ilm ) # & (1! X Usp ) # # % X Ilm ( ) + ln $ & * Usp % * Mt # # * Hem # # * Ilm ) (4.54) (

7 Problem Set Four Solution Due October, 007 Τhe first term on the right is simply lnk exch as Spencer and Lindsley defined it (since (1-X Ilm ) = X Hem ), so this can be rewritten as:!g = "RT lnk exch "#RT ln $ Usp [ ] + #RT ln [ $ Mt ] "#RT ln [ $ Hem ] + #RT ln [ $ Ilm ] () Comparing equation () with equation (1) above, we see that the problem is now simply to show that:!"rt ln # Usp [ ] + "RT ln [# Mt ]!"RT ln [# Hem ] + "RT ln [# Ilm ] =!AW Usp G + BW Mt G + CW Il Hem G! DW G According to equation 4.18:!RT ln " i = (W G j #W Gi )X j + (W Gi #W G j )X j (4.18) Let s concentrate on the first two terms on the left. Substituting equation 4.18, these two terms become:!"rt ln # Usp [ ] + "RT ln [# Mt ] =! (! )X Mt [ + (! )X Mt ] [ + (! ] + (!i We now make the substitution X MT = (1-X USP ): Expanding terms: [ ] [ + ( ] =! ( )(1! X Usp ) + ( )(1! X Usp ) + ( i =![( )(1! X Usp + X Usp ) + ( )(1! X Usp + X Usp! X Usp )] [ + ( ] + ( i and =!( ) + X Usp ( )! X Usp ( )!( ) + X Usp ( )! X Usp ( ) +X Usp ( ) + ( i + ( and =! X Usp! X Usp! X Usp + X Usp! + + 6X Usp! 6X Usp! 6X Usp W + 6X GUsp Usp +X Usp W! X GUsp Usp + X Usp W! X GUsp Uspi + X Usp! X Usp Now simplifying, this becomes =! + 4 X Usp! X Usp! X Usp + X Usp

8 Problem Set Four Solution Due October, 007 Collecting terms we have: =! (X Usp! 4X Usp +1) + (X Usp! X Usp ) =!A + B Now focusing on the last two terms:!"rt ln # Hem [ ] + "RT ln [# Ilm ] =![(W G Ilm + (W G Hem ] + [(W G Hem )X Hem + (W G Ilm )X Hem ] We now make the substitution X Hem = (1-X Ilm ): =! (W G Ilm + (W G Hem [ ] [ ] + (W G Hem )(1! X Ilm ) + (W G Ilm )(1! X Ilm ) Expanding: =![(W G Ilm + (W G Hem ] and: and: [! X Ilm ] + (W G Hem )(1! X Ilm + X Ilm ) + (W!W )(1! X + X G Ilm G Hem Ilm Ilm [ W G Ilm ] =! X Ilm W! X W + X W! X G Ilm Ilm G Ilm G Ilm Hem Hem "(W G Hem )! X Ilm (W G Hem ) + X Ilm (W!W ) % G Hem G $ Ilm + $ +(W G Ilm )! 6(W G Ilm $ # $ +6(W G Ilm! (W G Ilm & [ W G Ilm ] =! X Ilm W! X W + X W! X G Ilm Ilm G Ilm G Ilm Hem Hem " W G Hem! 4X Ilm W G Hem + X Ilm W G Ilm + X Ilm W! X G Hem Ilm $ + $ +W G Ilm! W G Hem! 6X Ilm W G Ilm + 6X Ilm W G Hem $ # $ +6X Ilm W! 6X G Ilm IlmW! X G Hem IlmW + X G Ilm IlmW ) G Hem ow collecting terms: = W G Ilm! 4X Ilm W G Ilm + X Ilm W G Hem + X Ilm W! X G Ilm Ilm [ W G Hem ] Rearranging = W G Ilm (1! 4X Ilm + X Ilm )!W (X G Hem Ilm! X Ilm ) = CW! DW G Ilm G Hem Thus!"RT ln # Usp Q.E.D. W G Ilm [ ] + "RT ln [# Mt ]!"RT ln [# Hem ] + "RT ln [# Ilm ] =!AW Usp G + BW Mt G + CW Il Hem G! DW G % &

EAS 4550: Geochemistry Problem Set 4 Solutions Due Sept. 27, 2017

EAS 4550: Geochemistry Problem Set 4 Solutions Due Sept. 27, 2017 1. Interaction parameters for the enstatite diopside solid solution have been determined as follows: WH-En = 34.0 kj/mol, WH-Di = 24.74 kj/mol (assume WV and WS are 0). (a) Use the asymmetric solution