MOOREHW10 - Chapter 18: Iherrnodynanncs: Directionahtv of...

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Unformatted text preview: Chapter 18: Iherrnodynanncs: Directionahtv of Chemical Reactions 41. Answer: (a) —120.64 JJK; negative AS prediction confirmed (b) 156.9 JJK; positive AS prediction confirmed (c) —19.T6 JJK; negative 115 prediction confirmed (d) 160.6 JJK; positive as prediction confirmed Strategy and Eipianafl'on: Adapt the method described in the solution to Question 39(a). (a) 018" = {1 mol) x S°{C2H5) — (1 tool) x 3°{C2H4) — {1 mol) 1-: 3°{H2} = {1 6161) x (229.60 I K-1m61-1}— (1 11101} x (219.56 J K-lmol-l} — (1 6161) 3-: (130.634 J K-1m61-1}= —120.64 J K-1 This confirms the negative 118 prediction from Question 31. (b) 1118“ = (l niol} x 8013303} + {2 mol) 1-: 80(1-120{g)) — {1 11101) x S°{CH30H{Q} — moi) x 93032) = {1 6161) x (213.24 I K-1m61-1}+(2 mol} x {133.325 J K-lmol-I) —(1m61)x(126.8 J K-lmol-lj — {i 6161} 2-: (205.133 JK-1m61-1)= 156.9 J K-1 This confirms the positive .118 prediction from Question 31. (c) 018" = {2 mol) x 3°{NH3J — {1 11101] x 8°(N3) — {3 mol) x 300-12) = {2 6161) x (192.45 J K-1m61-1}— (1 11101)): (191.61 J K-lmol-l} — (3 6161) 3-: (130.634 J K-1m61-1}= 498.26 J K-1 This confirms the negative AS prediction from Question 31. (d) AS“ = {1 mol) x S°{Ca0} + {l Inol) x 5°[C202j — {1 mol) x S°£CaCO3) =(1m61}x(39.15 J K-1m61-1)+(1 6161) x (213.74 J K-lmol-l) —(1m61)x{92.9 J K-1m61-1)= 160.6 J K-1 This confirms the positive AS prediction Erorn Question 31. 52. Await-er: (a) AS“ is negative; AHc is negative. (b) AHc = 4203.40 kJ, which is negative as predicted; A33 = —116.62 JJ'K which is negative as predicted. Sfi'atng-' and Eipfanmfon.‘ (a) The reactants of this reaction have a larger number of gas—phase products {1 mol) than gas—pha se products {0 mol), so we predict the entropy change. 318°, is negative. Magnesium reacts violently. suggesting that a great amount of heat is produced in the reaction. We will predict the enthale change. EH“. is negative. (b) Adapt the method described in the solution to Question 41. 3H” = {2 tool) 1-: AH °f{Mg'O) — {1 mol}xfi1—I}{Mg} — {1 mol}x nH :f(03{g)) = {2 Inol} x {— 601.10 klirnol} — {2 Inol) x {0 Isl-"11101) — {1 mol) x {O kIfrnol} = 4203.40 kl 315° = {2 11101} x 8°{Mg0} — {2 rnol) x AS°{I\.'1g}—{1 rnol)1< fine-3(9) = (2 mol} x 26.94 1 14416614} — {2 11101) x (32.63JK‘1n1o1‘1)—{1 1.1101) x (205.1313 1 K-1m61-1}= —216.62 J K-1 The enthalpy change and entropy change are both negative. as predicted in (a). 53. Amu-‘or.’ A50 is positive; AH“ is positive. The reaction is reactant—favored. See explanation below. Strategy andEipianarion: The equation for the chemical reaction is: H100”) —:- Egg] - % Ojg) The reaction written here is the reverse of the highly exothennic reaction described in the first sentence of the Question. so this reaction is highly endothermic. with a positive enthalpy change {AI-13]. With more gas products (1 % inol] than gas reactants {D iiiol). the entropy change [$33) is positive. Because we do not see water spontaneously decomposing. we Will conclude that the reaction is reactant-favored. The entropy increase is insufficient to drive this highly endothermic reaction to form products without assistance from the surroundings at the temperature of 25 "C. 'J'I UI . .{H511-'Q!'.' (a) —851.5 M; —3'?.52 JEK; product—favored at low temperatures (b) 66.36 M; — 2LT? JIK; never product-favored Strategy anfi’Eapfanarion: Adapt the method described in the solution to Question 45 and use Table 18.2. (a) AH3 = {2 inol) x AHHFe) - (1 mol) x AH}(A1303)— (1 mol) >c AH}(Fe103)— (1 niol) >< AI] ‘i-(Al) = {2 tool) 3-: (CI lei-"11101) + {1 11101) x (— 16??? lei-"11101) — (1 mol] x (— $24.2 kJ.-"inol) — (2 mol) x: {D lei-"mob = —851.5 l-tJ ASC = {2 tool) 3-: 3"{Fej-(1rnoljx 3°(A1103) — (1 tool) 3-: S°(Fe203) — (2 mol) >c 3°(Al) = {2 inol) x (BITS JK‘l-inol‘AJ - {1 inol) x (50.92 J K_11110l._‘-j — (1 mol) :4 [8140 J K-imoi-ij—{zmonx(23.3JK-1mol-1}= —3J.s: J K-1 The reaction is exothermic, but its entropy change is negative. so it would stop being product-favored above a specific temperature. (Calculated from Equation 13.5, this temperature is 21.700 K.) (b) AH” = (2 inol.) x AH‘ffNOfi— (1 mol] >4 AH H33) —(2 1110].) x QHHOE) = (I inol) x [33.13 Isl-"11101) — (1 mol] 3-: (CI kJ.-"mo1} — [3 mol] 3-: (O l-tJ.-"mo1]= 66.36 H ASC = (2 inol) x $96102) — (l niol} )4 830(3) — [3 11101)): SGEOEJ = {2 inol) x (340.06 J K-1m01-1}— (1 mol} x [191.61 J K-lmol-l) 4211101) 3-: (305.133 J K-lmol-IJ = — :13? J 31-1 The reaction is endothermic; the entropy change is negative. so it Will never be product—favored. 65. Amati-tar: ACE = 462.18 M, so reaction is not a good way to make pure Si. Strategy and Ernianarion: Adapt the method described in the solution to Question STUD). A c =(ln1ol]x AGHSi) - [1 mol} >< d6 EfiCOJ] — [1 mol] >< iiG HSiOfl— (1 inol) x AGE-[C] = (1 mol] >4 {0 l-tJ.-"mol) - (1 mol] >4 {—394.359 kl-"inolj — (1 niol} >c [—856.64 kJ.-"inol) — (1 mol) x {D ltfinol) = 462.28 kJ The reaction is not product—favored. so this would not be a good way to make pure silicon. 6T. Anni-w: (a) AG“ = —141.05 R] (13) AG" = 141.73 R] (c) AG“ = —959.43 R]; reactions (:1) and (c) are product-favored. Strategy andEapiauan‘on: Adapt the method described in the solution to Question 5?{b). (a) .563 = {1 moi) x AG}{C1H4) — {1 mol) at AG}(CJHE)— {1 mol) at AGE-(H3) = {1 mol) x {68.15 kl-‘inolj — {1 mol) x {209.20 kJ.-"ntol) — {1 mol) at [0 lolfntol) = —l41.05 kJ (b) AG” = {2 mol) x AG}[SO:)+(lmo1)x so H03) —{2 mol) 3-: AGHSOSJ = {2 moi) x [—3 00.194 kJ.-"mo1] + {1 mol) x [0kl:’inoi) — (2 mol] 2-: (—321.06 kJ'.-"inoi) = 141.?3kl (c) .563 = {4 moi) x AG‘ILGWID) + {I5 mol] >4 sotmgmgj) — {4 moi) a: AGEU‘CH3] — (5 mol] x AG H03) = {4 moi) x {36.55 kl-‘inolj - (5 mol) >< {—228.52}! kJ.-"mo1) — {4 mol) x: [—16.45 kJ'.-"mol] — {5 moi) x [0 kJ'.-"inol) = —959.43 kJ 95. Answer: (a}AH° E 4873 Id; Broken: S 0—H, T (1—0, 1' C—H, 5 C—C, 6 0:0; formed: 12 (1:0, 12 0—H (b) It is close; intermolecular forces in solid glucose and liquid water are being neglected. Strategy ann’EAplanan'on: Adapt the method described in Section 3.6 as described in Question 8.42. (a) Looking at the given ball—and—stick model structure we see that we must break five mol ofO—H bonds, seven moi of (3—D bonds. seyen mol of C—H bonds. and five moi of C—C bonds in glucose. We must also break six mol of 0:0 bonds. Two C=D bonds in each of six moi of CD: and two 0—H bonds in each of six 1110:10in0 are formed. The enthalpy of a reaction is approximated by adding the energy required to break one mole of bonds of each type ofbond broken, described by the bond energies {20), to the energy required to form one mole of bonds of each type of bonds produced in the reactants: described by the negative of the bond energies [—D). AH) a [bonds broken in glucose) - {bonds broken in 02) - {bonds formed in C02) - {bonds formed in H30] = (5 mol x DO_E— Fmol x D00 + }' mol 3-: DC_H— Smoi a: DC_C) - 6 mol x D0=O — 6 x {1 mol 2: Dig-=0) — 6 x {2 mol x D0_H] Notice that the number of moles multiplied by the energy per mol {in kJFmol) gives the result for each term in kl. sH=s5no_H+ Him)— a DC_E- sum—6 D0=O—12llc=0 — 13 oCHE = ?DC_O—TDC_H+S DH: -6Do=O—11Dc=O—TDO_E Now use the data in Table 8.3. = 2x {swim—7‘ x {416kJJ- 5 x {356kJ}+6 x [49313) — 1: a: (803ka — t x {46? kl] = 4373 kl s AH" (b) The actual AH” [—2316 kl) is close to the estimated value in [a). Intermolecular forces in condensed phases {solid glucose and liquid water) are being neglected in this calculation. which could explain the discrepancy. Also, the bond enthalpies given in Table 8.2 are average yalues: the actual enthalpy of a particular bond in a particular molecule may be slightly greater or less depending upon other factors. 115. Aimren-‘Eiyfanafion: The tabulated yalues for enthalpy changes are the enthalpy changes for formation reactions at 25.00 3C. Formation is defined as producing one mol of a substance from standard state elements. By this definition, the formation of an element at standard state has a zero enthalpy change. In contrast. the tabulated values for the entropy changes are absolute eiitropies at 25.00 3C. Because everything has some amount of entropy at 25.00 3C, we cannot ignore the entropies of the standard state elements. 116. Answer: exothermic; product—favored Strategy and £1plnimtton.‘ Many of the oxides have negatiye enthalpies of reaction, which means their oxidations are exothermic. These are probably product-favored reactions. 123. Aimpen-Explanation; AG =1 0 means products are favored; however: the equilibrium state will always have some reactants present, too. To get all the reactants to go away requires the removal of the products from the reactants, so that the reaction continues forward. ...
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MOOREHW10 - Chapter 18: Iherrnodynanncs: Directionahtv of...

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