Solutions to the Extra Problems for Chapter 14

Transcription

1 Slutins t the Extra Prblems r Chapter 1 1. The H T use bnd energies, we have t igure ut what bnds are being brken and what bnds are being made, s we need t make Lewis structures r everything: + + We have t break ur C H bnds (each which is wrth 10 / and tw OO bnds (each which is wrth 98 /. We then have t make tw CO bnds (each which is wrth 732 / and ur O H bnds (each which is wrth 60 /. H ( s (10 H (Energy r breaking bnds (Energy rm making bnds + (2 s (98 H -670 (2 s (732 ( (60 2. The bnd energy is 80 /. We have bnd energies r all the bnds except the C C bnd. That means we can set up the H equatin and slve r the energy the C C bnd. The Lewis structures tell us we have t break tw C H bnd (which is wrth 10 /, ne C C bnd (which is nt knwn, and tw H H bnds (each which is wrth 36 /. We then have t make six C H bnds (each which is wrth 10 /, ne C C bnd (which is wrth 350 /. H (2 s (10 + (1 (350 (1 (C C bnd + (2 s (36 H 1,120 + (1 (C C bnd (6 s (10 Remember, we are treating these like additin/subtractin prblems, and least precise numbers (10 and 350 require us t reprt ur answer t the tens place. We knw what H is. The prblem gave it t us. S we can put that int the equatin and slve r the energy the C C bnd.

2 280 1,120 + (1 (C C bnd C C bnd , The H -1, T use Hess s Law, we need the chemical equatin and the standard enthalpies rmatin. Yu learned cmplete cmbustin in Chapter 6: 2CH O (l + 3O 2 (g 2CO 2 (g + H 2 O (g The standard enthalpies rmatin are: / r CO 2 (g, / r H 2 O (g, / r CH O (l, and 0 r O 2 (g, since it s an element in its natural phase. H (2 s ( ( s ( 21.8 (2 s ( , The H 91.3 /. The table standard enthalpies rmatin desn t have NO (g, but since we knw H, we can calculate it. The standard enthalpy rmatin r H 2 O (g is /. Fr NH 3 (g, it is -5.9 /, and r O 2 (g, it is zer, since it is an element in its natural phase. H ( s H [NO(g] + (6 s ( 21.8 ( s (-5.9 (5 s (0 H ( s H [NO(g] + 1,267.2 Since we are given H, we can plug it int that equatin and slve r H [NO (g]: ( s H [NO (g] + 1, ,267.2 H. + [NO (g] 91.3 s 5. 89,000 will be released. Since H is negative (yu calculated it t be -1,276.8 in prblem 3, the reactin releases energy, s energy is a prduct. We can put it right int the equatin: Nw we can just d stichimetry: 2CH O (l + 3O 2 (g 2CO 2 (g + H 2 O (g + 1,276.8 Mass CH O amu amu amu amu Since this tells us the number grams in ne, we need t cnvert.5 kg t,500 g.

3 ,500 g CH 1 O 1 CHO g CH O 10 s CH O The chemical equatin tells us: 2 s CH O 1,276.8 We can use that t cnvert s CH O int energy released: 10 s CH 1 O 1, s CH O 89, a. Reactin II is the endthermic reactin, since the energy the prducts is higher than that the reactants. b. The H H is the dierence between the energy the prducts and that the reactants. It is negative because the prducts are lwer in energy than the reactants. c. The H 100. H is the dierence between the energy the prducts and that the reactants. It is psitive because the prducts are higher in energy than the reactants. d. Reactin I has the highest activatin energy. The activatin energy is the dierence in energy between the tp the hill and the reactants. Fr Reactin I, that s abut 270. Fr Reactin II, it is nly abut a. S is negative ( S < 0. There are nly gases n the reactants side, but there is a slid n the prducts side. Als, there are ewer cules n the prducts side. This means the prducts side has lwer entrpy, s S is negative. b. S is psitive ( S > 0. There is nly a slid n the reactants side. Hwever, there is a gas n the prducts side as well as mre cules n the prducts side. That makes S psitive. c. S is negative ( S < 0. There are nly gases n bth sides the equatin, but there are mre cules n the reactants side (ive as ppsed t ur. That means S is negative. 8. S is /. Yu already determined the equatin: 2CH O (l + 3O 2 (g 2CO 2 (g + H 2 O (g The abslute entrpies are: / r CO 2 (g, / r H 2 O (g, / r CH O (l, and / r O 2 (g.

4 S (2 s ( ( s (188.7 (3 s (205.0 (2 s (126.8 S / 9. The reactin is spntaneus, with a G -2,883.. The cmplete cmbustin C 2 H 6 (g is: 2C 2 H 6 (g + 7O 2 (g CO 2 (g + 6H 2 O (g The G s are: -39. / r CO 2 (g, / r H 2 O (g, / r C 2 H 6 (g, and 0 r O 2 (g, since it s an element in its natural phase. G ( s ( (6 s ( (2 s ( ,883. Since G is negative, the reactin is spntaneus. 10. a. This reactin is spntaneus at T < In this ne, we can t make a statement r all temperatures, s we need t use Equatin (1.5, remembering that we have t cnvert t t make the units cnsistent: G H T S 115,600 T ( T be spntaneus, G < 0: 115,600 T ( < 0 Slving r T: T ( < 115,600 The negatives cancel, s when we divide by / we get: T < 115, T < b. In this case, G will always be psitive, because Equatin (1.5 has us adding smething t a psitive number, s this reactin is never spntaneus n matter what the temperature. c. This reactin is spntaneus at T > 1,532. In this case, we have t use Equatin (1.5, remembering t cnvert t t make the units cnsistent. T be spntaneus, G < 0:

5 Slving r T: 203,100 T (132.6 < 0 T (132.6 < 203,100 We need t divide by , but that means switching the directin the sign: T > 203, T > 1,532