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The methods of this section can also be applied to reactions in which one set of solids converts to another. A geologically important example is the transformation of albite into jadeite + quartz:

NaAlSi3O8⟷NaAlSi2O6+SiO2

Use the data at the back of this book to determine the temperatures and pressures under which a combination of jadeite and quartz is more stable than albite. Sketch the phase diagram of this system. For simplicity, neglect the temperature and pressure dependence of both ∆S and ∆V.

Short Answer

Expert verified

The temperature is 100 K above room temperature, an extra 1.7 kbar of pressure is required to keep Jadeite+Quartz stable.

Step by step solution

01

Given information

The Gibbs free energy, Entropy and Molar Volumes for Albite, Jadeite, quartz is given as:

AlbiteGa=-3711.5KJSa=204.4J/KVa=100.07cm3JadeiteGj=-2852·1KJSj=133·5J/KVj=60·40cm3QuartzGq=-856.64KJSq=41·84J/KVq=22·69cm3

Albite→jadeite+quartz

For the above reaction

role="math" localid="1646935903662" ΔG=Gfinal-Ginitial=Gj+Gq-Ga=(-2852·1-856·64+3711·5)kJ=2·76kJ>0

02

Explanation

As a result, under normal settings, the Albite is more stable.

At high pressures, the jadeite + quartz combination becomes more stable.

The change in volume is calculated as

ΔV=Va-Vj-Vq=(100·07-60·40-22·69)cm3=16·98cm3=1.698×10-5m31kJkbar=10-5m3∴ΔV=1·698kJkbar

03

Explanation

At standard temperatures, the pressure at which jadeite + quartz becomes stable is:

P=ΔGΔVP=2·761·698KJKJKbarP=1.6254Kbar

The coexistence line with the P-axis in the P-T diagram is thus of relevance.

The Clausius-Clapeyron equation's slope is

dPdT=ΔSΔVΔS=Sa-Sj-Sq=(204·4-133·5-41·84)J/KΔ³§=29·06J/KSlope=Δ³§Î”³Õ=29·06J/K1·698J/bar=17·11barK

04

Conclusion

The phase diagram for Jadeite+Quartz and Albite is

Jadeite+Quartz is stable at pressures more than 1.6254kbar at room temperature. The Jadeite& Quartz-Albite phase border has a slope of 17.11 bar/K.

If the temperature is 100 K above room temperature, an extra 1.7 kbar of pressure is required to keep Jadeite+Quartz stable.

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Most popular questions from this chapter

In the previous section I derived the formula (∂F/∂V)T=-P. Explain why this formula makes intuitive sense, by discussing graphs of F vs. V with different slopes.

Is heat capacity (C) extensive or intensive? What about specific heat (c) ? Explain briefly.

Use the data at the back of this book to verify the values of ΔH and ΔGquoted above for the lead-acid reaction 5.13.

At temp 298K and pressure 1 bar.

Functions encountered in physics are generally well enough behaved that their mixed partial derivatives do not depend on which derivative is taken first. Therefore, for instance,
∂∂V∂U∂S=∂∂S∂U∂V

where each ∂/∂Vis taken with S fixed, each∂/∂S is taken with V fixed, and N is always held fixed. From the thermodynamic identity (for U ) you can evaluate the partial derivatives in parentheses to obtain

∂T∂VS=-∂P∂SV

a nontrivial identity called a Maxwell relation. Go through the derivation of this relation step by step. Then derive an analogous Maxwell relation from each of the other three thermodynamic identities discussed in the text (for H, F, and G ). Hold N fixed in all the partial derivatives; other Maxwell relations can be derived by considering partial derivatives with respect to N, but after you've done four of them the novelty begins to wear off. For applications of these Maxwell relations, see the next four problems.


Sulfuric acid, H2SO4,readily dissociates intoH+andHSO4-H+andHSO4-ions

H2SO4⟶H++HSO4-

The hydrogen sulfate ion, in turn, can dissociate again:

HSO4-⟷H++SO42-

The equilibrium constants for these reactions, in aqueous solutions at 298 K, are approximately 10 and 10*, respectively. (For dissociation of acids it is usually more convenient to look up K than ∆G°. By the way, the negative base-10 logarithm of K for such a reaction is called pK, in analogy to pH. So for the first reaction pK = -2, while for the second reaction pK = 1.9.)

(a) Argue that the first reaction tends so strongly to the right that we might as well consider it to have gone to completion, in any solution that could possibly be considered dilute. At what pH values would a significant fraction of the sulfuric acid not be dissociated?

(b) In industrialized regions where lots of coal is burned, the concentration of sulfate in rainwater is typically 5 x 10 mol/kg. The sulfate can take any of the chemical forms mentioned above. Show that, at this concentration, the second reaction will also have gone essentially to completion, so all the sulfate is in the form of SOg. What is the pH of this rainwater?

(c) Explain why you can neglect dissociation of water into H* and OH in answering the previous question. (d) At what pH would dissolved sulfate be equally distributed between HSO and SO2-?

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