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Chapter 5: Q 5.83 (page 211)
Write down the equilibrium condition for each of the following reactions:
Hence, the equilibrium condition for each reaction is given.
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How can diamond ever be more stable than graphite, when it has
less entropy? Explain how at high pressures the conversion of graphite to diamond
can increase the total entropy of the carbon plus its environment.
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:
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.
Use the data at the back of this book to calculate the slope of the calcite-aragonite phase boundary (at 298 K). You located one point on this phase boundary in Problem 5.28; use this information to sketch the phase diagram of calcium carbonate.
Express in terms of the volumes of solutions of reactants and products, for a chemical reaction of dilute solutes. Plug in some reasonable numbers, to show that a pressure increase of 1 atm has only a negligible effect on the equilibrium constant.
In the previous section I derived the formula . Explain why this formula makes intuitive sense, by discussing graphs of F vs. V with different slopes.
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