Termodinâmica e Entropia

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Chapter 18 Free Energy and Thermodynamics 419 18.113 (b) has the largest decrease in the number of microstates from the initial to the final state. In (a), there are initially = 90 microstates and 9! = 1680 microstates at the end; so > 0. In (b), there are initially 9! 9! = 1260 microstates and 9! = 84 microstates at the end; so AS 1, the natural log of K is positive and When Q = 336, the second term in = + RT In Q is positive; so at high temperature, this term can dominate and make Questions for Group Work 18.119 The average of all of the dice will be 3.5 (one-sixth will be 1, one-sixth will be 2, one-sixth will be 3, ...) or + + + + + = It is very unlikely that the sum would be one million after the earthquake because they would all have to be It is just as unlikely that the sum would be six million after the earthquake because they would all have to be 6s. The total would be close to 3.5 million (1 million times the average); this illustrates the second law because the numbers on the dice are maximizing their dispersal. 18.121 3 + 6 H₂(g) + 6 C(s, graphite) glucose) S°(J/K mol) 0 205.2 H₂(g) 0 130.7 C(s, graphite) 0 5.7 C₆H₁₂O₆(s, glucose) -1273.3 212.1 = (products) (reactants) = glucose))] + (H₂(g)) + (C(s, graphite))] = - + 6(0.0 kJ) + = [-1273.3 kJ] = 1273.3 kJ = (s, glucose) = - = glucose))] - + 6(S°(H₂(g)) + graphite))] = [1(212.1 J/K)] - [3(205.2 J/K) + 6(130.7 J/K) + 6(5.7 J/K)] = [-212.1 - [1434.0 J/K] = -1221.9 kJ is zero for an element in its standard state at 25 °C.S° is zero only for a perfect crystal at 0 K. Copyright © 2017 Pearson Education, Inc.