Chapter 11 · Igneous Petrology

Answer 1

Figure 11.12 is based on equation (11.3),

This equation can be solved for either hydrogen or oxygen. The only additional variable to consider is that the W/R ratio for a given quantity of water exchanging with the rock is not the same for oxygen and hydrogen. While there are approximately equal molar fractions of oxygen in water and rock by volume, there may be at least 20 times the molar fraction of H in water compared with a relatively anhydrous rock. Therefore, the W/R ratio of hydrogen will be 20 times that of oxygen for the same passage of water. Given the initial variables δ¹⁸O_i,r = 8.5, δ¹⁸O_i,w = 0, δD_i,r = −65, δD_i,w = 0, Δ¹⁸O_{feldspar–water} = 2, and ΔD_{biotite–water} = −25, equation (11.P1) reduces to

[The printed solution figure (Fig. 11.P1) is not reproduced. The printed text also contains a clause reading “if the volumetric abundance of H is 1/30 that of O” whose placement is garbled in the extraction; check vs print.]

[Worked grid (constructed). The plot's model curves draw both exchange trajectories and stand in for Fig. 11.P1; the hydrogen system is buffered by W/R ≈ 0.2 (oxygen basis) while oxygen takes several W/R units.]

Answer 2

The molecular weight of Mg_1.5Fe_0.5SiO₄ is 156.48. For every mole of olivine, we get 2 moles of O₂. Therefore, 156.48 mg of olivine gives 2 mmol of O₂. This translates to 12.8 µmol of O₂ per mg olivine. reconstructed [The printed final value is lost in the extraction; 2/156.48 mmol/mg = 12.78 µmol/mg.]

Answer 3

Not printed in the 1st ed. [Constructed: 0·x + (−25)(1 − x) = −5.7 gives 1 − x = 0.228, i.e. roughly 77% carbonate to 23% reduced carbon.]

Answer 4

Not printed in the 1st ed.

Answer 5

The total amount of carbon (hosted in diamonds) in a mantle rock is at trace levels. Therefore, any addition, mixing, devolatilization, or incorporation of exotic material can change the δ¹³C value of the reservoir from which the diamond ultimately forms. Carbon is a major phase in carbonatites, which can be quite massive. Therefore, slight degassing, mixing with an exotic reservoir, or the addition of small amounts of methane, etc., will have very little effect on the δ¹³C value of carbonatites.

Answer 6

1. The printed temperature is lost in the extraction. [Constructed: Δ¹³C(CO₂–CH₄) = 3.65‰, i.e. 10³lnα = 3.68; solving the given equation yields T ≈ 1500 K, about 1230 °C.]
2. The high temperatures are reasonable, and the reason the values could be preserved during cooling is that (i) cooling was rapid and (ii) exchange between methane and CO₂ is very slow. The latter is true because, in order for exchange to occur, a molecule of CH₄ and CO₂ would need to collide, and during the collision all bonds would have to be broken and re-form around the other carbon. The activation energy for such a reaction is too high without a catalyst. Methane–CO₂ carbon isotope exchange will not occur in the gas phase.

[Worked grid (constructed). A trial T of 1504 K zeroes the residual, so T ≈ 1230 °C; the printed value is lost in the extraction.]

Answer 7

The fractionation between quartz and water at 450 °C is 3.05‰. For a rock of 7.5‰, the initial water should have a δ¹⁸O value of 4.45‰. With time, however, the rock will see ever higher amounts of water, so the W/R ratio will increase. Eventually (if the system remains hot for a long enough period of time), the rock will be buffered by the water, so that the geothermal waters will retain their initial values. Overall, the waters should decrease from an initial composition of 4.45‰ to −8‰.