Chapter 05 Ans

# Chapter 05 Ans - Answers for Chapter 5 An Introduction to...

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Answers for Chapter 5: An Introduction to Thermodynamics. 1. The molar Gibbs free energy of formation of quartz is the energy change involved in the reaction of Si metal with O 2 gas to form a mole of quartz SiO 2 : Si (metal) + O 2 (gas) = SiO 2 (quartz). What does the value o f G = –856.3 kJ/mol tell us about the stability of quartz (as compared to Si metal and O 2 gas) at 25 o C and 1 atmosphere pressure? Explain. Because quartz has a lower Gibbs free energy than its constituent elements, quartz is a more stable compound than silicon metal in the presence of oxygen. That’s why you find quartz commonly but never find metallic silicon in nature. You can see some by paying a visit to your campus chemistry stockroom. It is metastable in this protected indoor environment. Imagine Si metal car fenders oxidizing to quartz rather than rust. 2. What would be the entropy of any phase at 0K? Explain. Can you explain this on the molecular level? According to the Third law of Thermodynamics, the entropy of any substance at zero K is zero. At this absolute zero temperature there is absolutely no atomic vibrational or rotational movement, so no entropy (randomness) whatsoever. 3. How do we determine the enthalpy of formation of a mineral? A calorimeter can measure the heat (enthalpy) change associated with the reaction by which a mineral is formed by reaction from the constituent elements (which are arbitrarily assigned an enthalpy of zero in the standard state). The enthalpy of formation is thus the enthalpy change of reaction ( H reaction - H mineral – H elements = H mineral – 0 = H mineral ). At least this is the answer I’d suggest to students, because it conforms to simple theory at this point. Many of us know that H o can be derived from the slope of K D vs. 1/T K from high temperature experiments (see equation 27.31). 4. In what order would you place crystals, gases, and liquids in terms of increasing (molar) entropy? Explain. Gases > liquids > crystals. Entropy can be envisioned as randomness, and crystals have highly ordered atomic structures (hence low entropy). Liquids have much more random arrangements, and gases exceedingly so (atoms are distributed over a much larger volume). 5. In what order would you place crystals, gases, and liquids in terms of increasing (molar) volume? Explain. Gases > liquids > crystals. The entropy increases generally correlate with increased volume. Also, looking at a phase diagram such as Figure 5.2 shows that, for any particular temperature, solids aer stable at high pressure, suggesting that they have a lower volume. Gases would be stable at even lower pressures than liquids (take a peek at Figure 6.7). 6. At constant pressure, dP = 0, so dG = -SdT. We could integrate eq. 5.3 (assuming a constant S) to produce: 2 1 T T dG S dT =- so that dG S dT = - where all properties are molar Explain in words the meaning of equation on the right (think in terms of a slope).

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