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Thermo%20_Challenge_KEY - Thermodynamics Challenge Problem...

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Thermodynamics Challenge Problem, Nov. 7-13 Part I. Free energy and the direction of change Consider the now familiar reaction of the vaporization of water. H 2 O (l) ' H 2 O (g) Table 1 : f and S° values for water. H 2 O (g) H 2 O (l) H ° f (kJ mol -1 ) -241.8 -285.8 S ° (J mol -1 K -1 ) 188.8 69.9 1) Calculate rxn and S° from the values in the table above. Using the values in the table, we can calculate H ° rxn = (-241.8)-(-285.5) = 44 kJ Similarly, S ° = 188.8 – 69.9 = 119 J K -1 2) Using your answers from question 1, a) Fill in the Table 2 below. b) What trends do you notice for H, S, -T S and G? Explain your values. H° remains constant over all temperatures. Changing from liquid water to gaseous water is always going to require the same amount of energy regardless of the temperature. S° also remains constant over all temperatures, which means that the entropy change associated with vaporizing water is always the same. In contrast, Τ∆ S° has a clear dependence on temperature. This makes sense because the disordered energy associated with entropy increases with higher temperatures. Think of gaseous water molecules at 400 K and 500 K; at a higher temperature, the molecules have more kinetic energy. As a result of the temperature dependence for the energy associated with entropy, G° also depends on temperature. We see that for this reaction changes from positive values at low temperatures to a negative values at higher temperatures. Table 2 : Temperature dependence for the vaporization of water T T (K) (kJ) (J/K) -T (J) (kJ) 0°C 273 44 119 -32487 11.5 25°C 298 44 119 -35462 8.54 50°C 323 44 119 -38437 5.56 100°C 373 44 119 -44387 -0.39 250°C 523 44 119 -62237 -18.24 500°C 773 44 119 -91987 -47.99 3) What conditions are optimal to make this reaction product favored?
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