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# solut10 - CHAPTER TEN SPONTANEITY ENTROPY AND FREE ENERGY...

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280 CHAPTER TEN SPONTANEITY, ENTROPY, AND FREE ENERG Y Spontaneity and Entropy 11. A spontaneous process is one that occurs without any outside intervention. 12. a. Entropy is a measure of disorder or randomness. b. The quantity T S has units of energy. 13. We draw all of the possible arrangements of the two particles in the three levels. 2 kJ x x xx 1 kJ x xx x 0 kJ xx x x Total E = 0 kJ 1 kJ 2 kJ 2 kJ 3 kJ 4 kJ The most likely total energy is 2 kJ. 14. 2 kJ AB B A B A 1 kJ AB B A A B 0 kJ AB A B A B E T = 0 kJ 2 kJ 4 kJ 1 kJ 1 kJ 2 kJ 2 kJ 3 kJ 3 kJ The most likely total energy is 2 kJ. 15. Processes a, b, d, and g are spontaneous. Processes c, e, and f require an external source of energy in order to occur since they are nonspontaneous. 16. Of the three phases, solids are most ordered and gases are most disordered. Thus, a, b, c, e, and g involve an increase in entropy. All increase disorder. 17. a. Entropy increases; there is a greater volume accessible to the randomly moving gas molecules which increases disorder.

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CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 281 b. The positional entropy doesn’t change. There is no change in volume and thus, no change in the numbers of positions of the molecules. The total entropy ( S univ ) increases because the increase in temperature increases the energy disorder ( S surr ). c. Entropy decreases because the volume decreases (P and V are inversely related). 18. a. N 2 O; It is the more complex molecule, i.e., more parts, more disorder. b. H 2 at 100 ° C and 0.5 atm; Higher temperature and lower pressure means greater volume and hence, greater positional entropy. c. N 2 ; N 2 at STP has the greater volume. d. H 2 O(l) is more disordered than H 2 O(s). 19. There are more ways to roll a seven. We can consider all of the possible throws by constructing a table. There are six ways to get a seven, more than any other number. The seven is not favored by energy; rather it is favored by probability. To change the probability we would have to expend energy (do work). 20. Arrangement I: S = k ln ; = 1; S = k ln 1 = 0 Arrangement II: = 4; S = k ln 4 = 1.38 × 10 -23 J/K ln 4, S = 1.91 × 10 -23 J/K Arrangement III: = 6; S = k ln 6 = 2.47 × 10 -23 J/K one die 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 sum of the two dice 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12
CHAPTER 10 SPONTANEITY, ENTROPY, AND FREE ENERGY 282 Energy, Enthalpy, and Entropy Changes Involving Ideal Gases and Physical Changes 21. 15.0 g He × g 4.003 mol 1 = 3.75 mol He q v = E = nC v T = 3.75 mol (12.47 J K -1 mol -1 ) (56.5 K) = 2640 J = 2.64 kJ 22. 1.00 × 10 3 g C 2 H 6 × g 30.07 mol 1 = 33.3 mol q v = E = nC v T = 33.3 mol (44.60 J mol -1 K -1 ) (48.4 K) = 7.19 × 10 4 J = 71.9 kJ At constant volume, 71.9 kJ of energy are required and E = 71.9 kJ. At constant pressure (assuming ethane acts as an ideal gas): C p = C v + R = 44.60 + 8.31 = 52.91 J mol -1 K -1 Energy required = q p = H = nC p T = 33.3 mol (52.91 J mol -1 K -1 ) (48.4 K) = 8.53 × 10 4 J = 85.3 kJ For the constant pressure process, E = 71.9 kJ as calculated previously ( E is unchanged).

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