Chapter 3 · Equilibrium Isotope Fractionation

Answers · Sharp, Principles of Stable Isotope Geochemistry, 1st ed. (2007) · ← Problems

Answer 1

See discussion in Sections 3.2.7 and 3.3.1.

Answer 2

See discussion in Sections 3.2–3.4.

Answer 3

  1. ΔG°r = −1.464 kJ, equal to ΔG°f of HD since the reactants are both zero. reconstructed [The printed part (a) line is lost in the extraction; the value −1464 J is used in part (b) as printed.]
  2. We know that ΔG°r = −RT·ln(K), so with ΔG°r = −1464 J, K = 1.8055. For initial fractions of H2 and D2 of 0.5, the final quantity of each is 0.5 − x and the amount of HD is 2x, where K = [HD]/([H2]½[D2]½). Finally, we solve the equation 1.8055 = 2x/(0.5 − x) to get x = 0.237. The percentage of HD in equilibrium with equal quantities of H2 and D2 is therefore 23.7%. [As printed. With x = 0.237 the mole fraction of HD is 2x = 47.4%, so the printed 23.7% corresponds to x rather than 2x; check.]
  3. HD, while not stable as a pure phase, is metastable. HDO will quickly react to equilibrium with H2O and D2O.

[Worked grid (constructed). At full precision K = 1.805 vs the printed 1.8055, and the HD mole fraction is 2x = 47.4%; the printed 23.7% corresponds to x. Check vs print.]

Answer 4

The exchange reaction is given by H2(g) + HDO(g) = HD(g) + H2O(g). In this case, the free-energy change for the reaction is given by

ΔG°r,25°C = ΣΔG°f(products) − ΣΔG°f(reactants) = (−0.35 − 54.634) − (0.0 − 55.719) = 0.735 kJ, or 735 J
K = [H2O][HD] / ([HDO][H2]),   ΔG°r = −RT·ln(K)

where α relates to K through equation (3.4). Solving, we get α(H2–H2O(g)) = 0.289. From equation (2.14), δD of the H2O(g) is +384‰. This is in fairly good agreement with a published value of 0.2824 (Suess, 1949, Zeitschr. Naturforschung 4, 328–332).

[Worked grid (constructed). The α row carries the printed 0.289; the printed equation-(3.4) step from K = 0.743 to α is lost in the extraction, check vs print.]

Answer 5

Equation (3.12) supplies the answer to this problem. Rearranging to solve for ks gives

ks = (2πν)²·μ, with μ = mamb/(ma + mb) and ν = c·ṽEq 3.12 rearranged

Given that c (speed of light) = 2.998·1010 cm/sec and ṽ = 2167.4 cm−1, we obtain ν = 6.4979·1013 sec−1. For ma = 12 and mb = 16, ks = 1.143·1030. Question (as printed): what are the units of ks?

[Worked grid (constructed). With μ left in amu the units of ks are amu·s−2, which is the point of the printed closing question; the last row converts to 1898 N/m.]

Answer 6

In the first case, the two phases fayalite and quartz would be intergrown at a fine scale and probably physically impossible to separate. In both cases, although the reactions may go to completion, there is no guarantee that the phases will be in isotopic equilibrium, because the reactions are driven by more energetic chemical changes rather than isotopic ones.

Answer 7

Albite and microcline should be the same, because substitution between Na+ and K+ should have a negligible effect. Albite has Si:Al in the proportion 3:1 and anorthite 1:1. Because there is a higher Si/Al ratio in albite vs. anorthite, the former should concentrate more 18O relative to the latter. (b) Sillimanite has the highest Si/(Si+Al) ratio, followed by mullite and corundum. They should concentrate 18O in that order.

Answer 8

Equation (3.13) relates the vibrational frequency of the isotopically substituted molecule to the reduced mass, defined as μ = mamb/(ma + mb). For 16O16O the reduced mass is 8. For 18O18O the reduced mass is 9. The vibrational frequency of 18O18O is calculated to be 1490.0 cm−1.

Answer 9

The molecular weights of calcite and water are 100.087 and 18.015, respectively. The fraction of oxygen in calcite is 0.4796 and 0.888 for water. 12 mg of calcite has 5.76 mg of O. For 50 mg of water, there are 44.41 mg of O. In general, the overall δ value of a mixture is given by δmix = (maδa + mbδb)/(ma + mb), where ma is the mass (or number of moles) of oxygen in phase a and δa is the δ value of phase a (similar notation for phase b). For our experimental conditions δmix = (5.76 × 24.32 + 44.41 × (−3.65))/50.17 = −0.44‰. The δ value of our mixture does not change during the course of the reaction. At equilibrium, δcal = 1.00490·δw + 4.90 (from the definition of α). Plugging into the mixture formula gives δ18Owater ≈ −1.0‰ and δ18Ocalcite ≈ +3.9‰. reconstructed [The printed final values are lost in the extraction; computed from the printed setup.]

[Worked grid (constructed). G4 solves the mass balance combined with the equilibrium relation; click it to see the rearranged formula.]

Transcribed from Z. Sharp, Principles of Stable Isotope Geochemistry, 1st ed. (Pearson Prentice Hall, 2007), end-of-chapter problem set (published source). Reference-scale names are written VSMOW, VPDB and VCDT where the original used SMOW, PDB and CDT, with the original numerical constants unchanged; historical usages (e.g. Craig's definition of SMOW) are kept as printed.

The source PDF's text layer scrambles mathematics, so equations and some values were reconstructed during transcription. Bracketed notes and anything tagged reconstructed or check vs print are editorial (constructed), not part of the original, and the warn-tagged items should be verified against the print copy.