1 (600)

Prévia do material em texto

19 Chemical Thermodynamics Solutions to Exercises 19.72 (a) = 1/2 O₂(g) = = 237.6 - 186.3 1/2(205.0) = = -0.0512 kJ/K (b) = - -TAS° is positive, so becomes more positive as temperature (c) = - TAS° = -126.4 K(-0.0512 kJ/K) = -111.1 kJ The reaction is spontaneous at 298 K because is negative at this temperature. In this case, could have been calculated from values in Appendix C, since these values are tabulated at 298 K. (d) The reaction is at equilibrium when = 0. = - = 0. = T = T = -126.4 kJ/K = = 2470 K. This temperature is so high that the reactants and products are likely to decompose. At standard conditions, equilibrium is functionally unattainable for this reaction. 19.73 Analyze/Plan. Follow the logic in Sample Exercise 19.10. Solve. (a) = = Tb = 33.9 10³ J/96.4J/K = 351.66 = 352 K = 79°C (b) From the Handbook of Chemistry and Physics, 74th Edition, Tb = 80.1°C. The values are remarkably close; the small difference is due to deviation from ideal behavior by and experimental uncertainty in the boiling point measurement and the thermodynamic data. 19.74 (a) As in Sample Exercise 19.10, = Use Data from Appendix C to calculate and for I₂(s). I₂(s) I₂(1) melting I₂(g) boiling I₂(s) sublimation = = = = In making this estimate, we assume that at equilibrium, both I₂ (s) and I₂(g) are present in their standard state of pure solid and vapor at 1 atm and consequently, 594