Geological Sciences 4550:
Geochemistry
Problem Set Three Solutions
Due Sept. 18, 2015
1. Consider the following hypothetical gaseous solution: gases 1 and 2 form an ideal binary solution; at 1000 K, the free energies of formation from the elements are -­‐‑50kJ/mol for component 1 and -­‐‑40kJ/mol for component 2. a.) Calculate ∆Gmixing for the solution at 0.1 increments of X2. Plot your results. –
b.) Calculate G for ideal solution at 0.1 increments of X2. Plot your results. c.) Using the method of intercepts, find µμ1 and µμ2 in the solution at X2 = 0.2 This problem also involves repeated calculations of the same formulae, so it is
ideal for a spreadsheet (below). For part c, we take the derivative dG/dX to get
the slope, which is µ2+µ1 + RT ln(X2/X1). (We need to make sure the units are
right as the Gf’s are in kJ, but R is in J). The chemical potentials are calculated
as µ1 = Gideal(X2) – slope*X2 and µ2 = Gideal(X2) + slope (1-X2).
X1 R T mu1 mu2 X2 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 ∆Gid mix
∆Gid. Mix. ∆Gideal ln(X1) ln(X2) KJ KJ µ1 0 0.000 0.00 -­‐50.00 0.1 -­‐0.105 -­‐2.303 -­‐2.70 -­‐51.70 0.2 -­‐0.223 -­‐1.609 -­‐4.16 -­‐52.16 0.3 -­‐0.357 -­‐1.204 -­‐5.08 -­‐52.08 0.4 -­‐0.511 -­‐0.916 -­‐5.60 -­‐51.60 0.5 -­‐0.693 -­‐0.693 -­‐5.76 -­‐50.76 0.6 -­‐0.916 -­‐0.511 -­‐5.60 -­‐49.60 0.7 -­‐1.204 -­‐0.357 -­‐5.08 -­‐48.08 0.8 -­‐1.609 -­‐0.223 -­‐4.16 -­‐46.16 0.9 -­‐2.303 -­‐0.105 -­‐2.70 -­‐43.70 1 0.000 0.00 -­‐40.00 = RT S (xlnx) ∆Gideal = S(xµ)+∆Gid mix slope = dG/dX2 = µ2-­‐µ1+RTln(X2/X1) 8.314 J/K-­‐mol 1000 -­‐50 kJ/mol -­‐40 kJ/mol -­‐52.97 -­‐54.25 -­‐55.76 -­‐57.62 -­‐60.01 -­‐63.38 -­‐69.14 -­‐40.00 1
slope µ1 µ2 -­‐50.00 -­‐50.88 -­‐51.86 -­‐1.53 -­‐51.86 -­‐53.38 Geological Sciences 4550:
Geochemistry
Problem Set Three Solutions
Due Sept. 18, 2015
0
-10
∆Gbar
∆Gbar mixing
G ideal (kJ)
-20
-30
-40
-50
-60
0
0.2
0.4
0.6
0.8
1
X2
2. The fol lowing is an analysis of Acqua di Nepi, bottled water from the Italian province of Viterbo: -­‐‑
2-­‐‑
HCO3 451 ppm SO4 38 ppm F– 1.3 ppm -­‐‑
Cl-­‐‑ 20 ppm NO3 9 ppm Ca2+ 82 ppm Mg2+ 27 ppm Na+ 28 ppm K+ 50 ppm a. Calculate the ionic strength of this water. (Recall that concentrations in ppm are equal to con-­‐‑
centrations in mmol kg-­‐‑1 multiplied by formula weight.) b. Using the Debye-­‐‑Hückel equation and the data in Table 3.2, calculate the practical activity co-­‐‑
efficients for each of these species at 25°C. See the spreadsheet below.
2
Geological Sciences 4550:
Geochemistry
Problem Set Three Solutions
Due Sept. 18, 2015
HCO3 Cl Ca Na SO4 NO3 Mg K F z A B -­‐1 -­‐1 2 1 -­‐2 -­‐1 2 1 -­‐1 conc ppm mol. wt. conc mmol 451 61 7.393 20 35.45 0.564 82 40.08 2.046 28 23 1.217 38 96.07 0.396 9 62 0.145 27 24.3 1.111 50 39.1 1.279 1.3 19 0.068 mmol*z^2 7.393 0.564 8.184 1.217 1.582 0.145 4.444 1.279 0.068 0.5092 0.3283 I, mol root I, mol 0.01244 0.11152941 4 å log gamma gamma 4.25 -­‐0.049 0.893 3 -­‐0.051 0.889 6 -­‐0.186 0.651 4.25 -­‐0.049 0.893 4.25 -­‐0.197 0.636 3 -­‐0.051 0.889 8 -­‐0.176 0.667 3 -­‐0.051 0.889 3.5 -­‐0.050 0.891 3. Consider the following analysis of a pyroxene: Site Ion Ions per site Tetrahedral Si 1.96 Al 0.04 Octahedral M1 Al 0.12 Mg 0.88 Octahedral M2 Fe 0.06 Ca 0.82 Na 0.12 Use the mixing on site model of ideal activities to calculate the activity of jadeite (NaAlSi2O6) and diopside (CaMgSi2O6) in this mineral. The relevant equation is 3.80. To calculate mole fraction, we need to divide the
ions per site, by the stoichiometric coefficient, which is 1 for the octahedral
sites, but 2 for the tetrahedral one (the one with Si). So the mole of Si is 1.96/2
and its contribution to activity of both phase components is (1.96/2)2.
Activities are as shown below.
3
Geological Sciences 4550:
Geochemistry
Problem Set Three Solutions
M1
M2
Si
Al
Al
Mg
Fe
Ca
Na
atoms/site Coeff
1.96
0.04
0.12
0.88
0.06
0.82
0.12
jadeite act
diopside act.
2
2
1
1
1
1
1
Due Sept. 18, 2015
mole frac.
0.98
0.9604
0.02
0.12
0.12
0.88
0.06
0.82
0.12
0.12
0.0138
0.9604
0.88
0.82
0.6930
Notice that the Al in the tetrahedral site as well as iron are effectively dilutants
and don’t contribute to the activity of either jadeite or diopside. Also note that
to do this properly, we need some knowledge of the lattice structure of the
mineral.
4. Write the equilibrium constant expression for the reaction: +
CaCO3(s ) + 2H (aq)
+ SO42− + H 2O(liq) ! CaSO4 ⋅ 2H 2O + CO2(g)
assuming the solids are pure crystalline phases and that the gas is ideal. A general form would be:
K=
agypsum aCO2 ( g )
acalciteaH2 + aSO2+ aH 2O
4
Assuming the solids are pure phases means their activities are 1. If CO2 is ideal,
then its activity may be replaced by its partial pressure, hence:
K=
pCO2 ( g )
aH2 + aSO2+ aH 2O
4
Provided the solution is reasonably dilute, we would commonly also assume the
activity of water was one, so that:
pCO
K = 2 2(g)
aH + aSO2+
4
5. The equilibrium constant for the dissolution of galena: PbSsolid + 2H+ ® Pb 2aq+ + H2Saq
is 9.12 × 10-­‐‑7 at 80° C. Using the γPb2+ = 0.11 and γH2S = 1.77, calculate the equilibrium concentration of Pb2+ in aqueous solution at this temperature and at pH’s of 6, 5 and 4. Assume the dissolution of galena is 4
Geological Sciences 4550:
Geochemistry
Problem Set Three Solutions
Due Sept. 18, 2015
the only source of Pb and H2S in the solution and that there is no significant dissociation of H2S. Hint: mass balance requires that [H2S] = [Pb2+]. Since galena is a solid, we assume an activity of 1, so the equilibrium constant
expression is 1:
a 2+ a
a 2+ a
K = 2Pb H 2S = Pb 2 H 2S
aH + aPbSsolid
aH +
Expressed in log form, the equilibrium constant expression for this reaction is:
log K = 2 pH + log aPb2+ + log aH S = 2 pH + log[Pb2+ ] + log γ Pb2+ + log[H 2 S] + log γ H S
2
2
(assuming PbS is a pure phase). Since [H2S] =
[Pb2+],
we may write:
2+
log K = 2 pH + 2log[Pb ] + log γ Pb2+ + log γ H S
2
Rearranging:
log[Pb2+ ] =
log K − log γ Pb2+ + log γ H S
2
2
We obtain the following answers:
pH
[Pb2+] M
6
2.16 × 10-9
5
2.16 × 10-8
4
2.16 × 10-7
5
− pH