# ch17 - 17 17.1 17.2 (a) spontaneous; Thermodynamics:...

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17 Thermodynamics: Entropy, Free Energy, and Equilibrium 17.1 (a) spontaneous; (b), (c), and (d) nonspontaneous 17.2 (a) H 2 O(g) H 2 O(l) A liquid is more ordered than a gas. Therefore, ΔS is negative. (b) I 2 (g) 2 I(g) ΔS is positive because the reaction increases the number of gaseous particles from 1 mol to 2 mol. (c) CaCO 3 (s) CaO(s) + CO 2 (g) ΔS is positive because the reaction increases the number of gaseous molecules. (d) Ag + (aq) + Br(aq) AgBr(s) A solid is more ordered than +1 and 1 charged ions in an aqueous solution. Therefore, ΔS is negative. 17.3 (a) A 2 + AB 3 3 AB (b) ΔS is positive because the reaction increases the number of gaseous molecules. 17.4 (a) disordered N 2 O (b) silica glass (amorphous solid, more disorder) (c) 1 mole N 2 at STP (larger volume, more disorder) (d) 1 mole N 2 at 273 K and 0.25 atm (larger volume, more disorder) 17.5 CaCO 3 (s) CaO(s) + CO 2 (g) ΔS o = [S o (CaO) + S o (CO 2 )] S o (CaCO 3 ) ΔS o = [(1 mol)(39.7 J/(K mol)) + (1 mol)(213.6 J/(K mol))] (1 mol)(92.9 J/(K mol)) = +160.4 J/K 17.6 From Problem 17.5, ΔS sys = ΔS o = 160.4 J/K CaCO 3 (s) CaO(s) + CO 2 (g) ΔH o = [ΔH o f (CaO) + ΔH o f (CO 2 )] ΔH o f (CaCO 3 ) ΔH o = [(1 mol)(635.1 kJ/mol) + (1 mol)(393.5 kJ/mol)] (1 mol)(1206.9 kJ/mol) = +178.3 kJ ΔS surr = K 298 J 178,300 _ = T H _ o = 598 J/K ΔS total = ΔS sys + ΔS surr = 160.4 J/K + (598 J/K) = 438 J/K Because ΔS total is negative, the reaction is not spontaneous under standard-state conditions at 25 o C. 17.7 (a) ΔG = ΔH TΔS = 57.1 kJ (298 K)(0.1758 kJ/K) = +4.7 kJ Because ΔG > 0, the reaction is nonspontaneous at 25 o C (298 K) (b) Set ΔG = 0 and solve for T. 233

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0 = ΔH TΔS; T = kJ/K 0.1758 kJ 57.1 = S H = 325 K = 52 o C 17.8 (a) ΔG = ΔH TΔS = 58.5 kJ/mol (598 K)[0.0929 kJ/(K mol)] = +2.9 kJ/mol Because ΔG > 0, Hg does not boil at 325 o C and 1 atm. (b) The boiling point (phase change) is associated with an equilibrium. Set ΔG = 0 and solve for T, the boiling point. 0 = ΔH vap TΔS vap ; T bp = mol) kJ/(K 0.0929 kJ/mol 58.5 = S H vap vap = 630 K = 357 o C 17.9 ΔH < 0 (reaction involves bond making - exothermic) ΔS < 0 (the reaction becomes more ordered in going from reactants (2 atoms) to products (1 molecule) ΔG < 0 (the reaction is spontaneous) 17.10 From Problems 17.5 and 17.6: ΔH o = 178.3 kJ and ΔS o = 160.4 J/K = 0.1604 kJ/K (a) ΔG o = ΔH o TΔS o = 178.3 kJ (298 K)(0.1604 kJ/K) = +130.5 kJ (b) Because ΔG > 0, the reaction is nonspontaneous at 25 o C (298 K). (c) Set ΔG = 0 and solve for T, the temperature above which the reaction becomes spontaneous. 0 = ΔH TΔS;T = kJ/K 0.1604 kJ 178.3 = S H = 1112 K = 839 o C 17.11 2 AB 2 A 2 + 2 B 2 (a) ΔS o is positive because the reaction increases the number of molecules. (b)
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## This note was uploaded on 08/27/2008 for the course CHEM 1280 taught by Professor Schmidt during the Fall '07 term at Toledo.

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ch17 - 17 17.1 17.2 (a) spontaneous; Thermodynamics:...

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