CHAPTER 17 1. Read Chapter 17, sectins 1,2,3. End f Chapter prblems: 25 2. Suppse yu are playing a game that uses tw dice. If yu cunt the dts n the dice, yu culd have anywhere frm 2 t 12. The ways f prducing the varius ttal number f dts is shwn in the table belw. ttal number f dts ways f achieving this result 2 1,1 3 1,2 r 2,1 4 1,3 r 3,1 r 2,2 5 1,4 r 4,1 r 2,3 r 3,2 6 1,5 r 5,1 r 2,4 r 4,2 r 3,3 7 1,6 r 6,1 r 2,5 r 5,2 r 3,4 r 4,3 8 2,6 r 6,2 r 3,5 r 5,3 r 4,4 9 3,6 r 6,3 r 4,5 r 5,4 10 4,6 r 6,4 r 5,5 11 5,6 r 6,5 12 6,6 Rank the ttal number f dts frm highest t lwest entrpy. Answer: 7 (highest); 8=6; 9=5; 10=4; 11=3; 12=2 (lwest)
3. Cnsider tw systems whse mlecules have the quantized energy states shwn in the diagram belw. In System 1 the energy states are spaced in 5 kj/ml intervals. A given mlecule culd have an energy f 5 kj/ml r 10 kj/ml, r 15 kj/ml, etc., but nt sme intermediate value like 8 kj/ml. In System 2, the energy states are spaced in 2 kj/ml intervals. Mlecules in this system culd have energies f 2, 4, 6, 8 kj/ml, etc. Nte, that in each system, it is nt pssible fr a mlecule t have 0 energy. A mlecule must have at least 5 kj/ml in System 1 and at least 2 kj/ml in System 2. Suppse each system cntains 4 mlecules and the sum f the energy f the 4 mlecules in each system ttals 20 kj/ml. The fur mlecules may have identical energies r the energies may be different as lng as the ttal is exactly 20 kj/ml. a. Which system has the highest entrpy? b. What is the entrpy f System 1? (I need a number.) Answer: a. System 2 has the higher entrpy. There is nly 1 way fr system 1 t ccur: all 4 mlecules must have 5 kj/ml energy. In system 2 there are many ways fr the system t ccur: 3 mlecules with 2 kj/ml and 1 with 14 kj/ml, 3 mlecules with 6 kj/ml and 1 with 2 kj/ml, 2 mlecules with 4 kj/ml and 2 mlecules with 8 kj/ml, etc, etc, etc. b. 0, since there is nly ne way fr system 1 t ccur.
4. End f Chapter prblems: 29, 37, 38, 39, 40 5. Match the entrpy value t the cmpund: 203 J/K ml, 161 J/Kml, 214 J/K ml CH Cl, CCl, CHCl 2 2 4 3 6. Which liquid wuld have the lwer entrpy? 7. One f these values is the entrpy f liquid ethanl and the ther is gaseus ethanl. Which ne is gaseus ethanl? 68 cal/k ml and 38 cal cal/k ml. Answers: 5. CCl 4 = 214 J/K ml, CHCl 3 = 203 kj/ml, CH2Cl 2 = 161 J/K ml (same # atms s heavier mlecule has higher entrpy) 6. C6H 6 since it has fewer number f atms and less massive. 7. gas is higher value End Exam 3 -------------------------------------------------------------------------------------------------------------------- 8. Calculate the value f ÄS fr the disslving f ne mle f magnesium chlride: If yu had predicted the sign f the entrpy change befre calculatin yu prbably are surprised t see what it really is. Ins have the ability t tie up a lt f lse water mlecules due t the attractin f the water fr the ins (Chem 105) and thus ften reduce entrpy. 9. Calculate the value f ÄS,ÄS surrundings, and ÄS universe at 298 K fr the fllwing reactin. Is the reactin pssible under these cnditins?
10. Given the fllwing values f ÄS fr these reactins: Calculate the ÄS fr this reactin: 8. -94.5 J/K 9. 216 J/K, -286 J/K, -70 J/K, N 10. -126 J/K 11. Read sectin 4,5,6. End f Chapter prblems: 41, 43, 49
298K 12. Calculate the value f ÄG fr the abve reactin frm the values f ÄG 13. Is the reactin abve spntaneus at 25 C? 14. Calculate the value f ÄG at 200 C. Is the abve reactin spntaneus at this temperature? 15. Calculate the temperature in Celsius at which the abve reactin first becmes spntaneus. 298K f 16. Is the abve reactin entrpy r enthalpy driven? In ther wrds, is it the heat exchanged r the entrpy change f the system that allws this reactin t ccur? 17. In Chem 105 yu may have investigated ther decmpsitin reactins f sdium bicarbnate besides the reactin abve. Anther pssibility is: 298K Calculate the value f ÄG and the temperature at which this reactin wuld becme spntaneus. Na2 O(s) 18. When baking sda is heated with a bunsen burner (T <1000 C) it experimentally prduces sdium carbnate nt sdium xide. Explain why sdium carbnate desn t change int sdium xide when heated by calculating the temperature at which the reactin belw wuld becme spntaneus. ÄH f ÄG f S 298K 298K 298K -414.2 kj/ml -375.5 kj/ml 75.1 J/K ml 19. Given the value f the free energy change fr this reactin: Using this and the ther reactins in the previus few prblems, predict whether NaOH wuld react at rm temperature with CO in the air t cnvert t sdium carbnate. 2
Answers: 12. +33 kj 13. n 14. ÄH = 129 kj, ÄS = 0.335 kj/k, ÄG = - 29 kj, yes 15. abut 113 C 16. entrpy driven 17. 298K Ä G = 311 kj, abut 654 C (ÄH = 452 kj, ÄS = 0.488 kj/k) 18. Abut 1840 C which is htter than the burner. (ÄH = 323 kj, ÄS = 0.153 kj/k) 19. 298K Yes. Ä G = -123 kj, 20. Read sectins 7, 8. End f Chapter prblems: 51, 53, 57 19. Cnsider the reactin: 298 K a. The value f Ä G fr the reactin is 79.9 kj. Suppse yu thught that pure water prduced n ins. Calculate the value f ÄG at 298 K and shw that it must be negative if the cncentratins were 0. Since yu can t take the ln(0), chse a really small value fr the cncentratins f hydrnium and -99 hydrxide...maybe 10 M. b. Prve that if the hydrnium and hydrxide in cncentratins were 1 x 10-9 M each that the ÄG wuld be negative, the frward reactin is still spntaneus, and the in cncentratins wuld cntinue t increase. -6 c. Prve that if the hydrnium and hydrxide in cncentratins were 1 x 10 M that the sign f ÄG wuld be psitive and the frward reactin wuld nt be spntaneus (the reverse reactin wuld be spntaneus s the in cncentratins wuld begin t decrease). -14 d. The value f K fr the reactin is 1 x 10 at 298 K. Calculate the cncentratins f the tw ins at equilibrium and prve that if the hydrnium and hydrxide had thse cncentratins, that the ÄG wuld be very clse t 0 and the system is indeed at equilibrium. 20. End f Chapter Prblems: 45, 87, 59, 61, 63, 65 (the answers are actually K p values nt K values), 67
21. Estimate the value f K (nt K p) at 100 C fr the reactin f page 8 f Chapter 13 ntes and verify it is clse t the value given in the prblem n that page. Ans: 0.40 vs 0.48 22. Graphically determine the value f ÄS and ÄH fr this reactin using the values f the equilibrium cnstant at different temperatures shwn in the table. + - 2 H2 O (l) H3 O (aq) + OH (aq) T ( C) K 0 1.14 x 10-15 20 6.81 x 10-15 40 2.92 x 10-14 60 9.61 x 10-14 Answer: ÄS = -80.8 J/mlK, ÄH = 55.9 kj/ml End f Quiz 6 --------------------------------------------------------------------------------------------------------------------