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Solutions Chapter 18 2009

Solutions Chapter 18 2009 - CHAPTER 18 ENTROPY FREE ENERGY...

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CHAPTER 18 ENTROPY, FREE ENERGY, AND EQUILIBRIUM 18.6 The probability ( P ) of finding all the molecules in the same flask becomes progressively smaller as the number of molecules increases. An equation that relates the probability to the number of molecules is given in the text. 1 2 = N P where, N is the total number of molecules present. Using the above equation, we find: (a) P = 0.02 (b) P = 9 × 10 19 (c) P = 2 × 10 181 18.9 (a) This is easy. The liquid form of any substance always has greater entropy (more microstates). (b) This is hard. At first glance there may seem to be no apparent difference between the two substances that might affect the entropy (molecular formulas identical). However, the first has the O H structural feature which allows it to participate in hydrogen bonding with other molecules. This allows a more ordered arrangement of molecules in the liquid state. The standard entropy of CH 3 OCH 3 is larger. (c) This is also difficult. Both are monatomic species. However, the Xe atom has a greater molar mass than Ar. Xenon has the higher standard entropy. (d) Same argument as part (c). Carbon dioxide gas has the higher standard entropy (see Appendix 3). (e) O 3 has a greater molar mass than O 2 and thus has the higher standard entropy. (f) Using the same argument as part (c), one mole of N 2 O 4 has a larger standard entropy than one mole of NO 2 . Compare values in Appendix 3. Use the data in Appendix 3 to compare the standard entropy of one mole of N 2 O 4 with that of two moles of NO 2 . In this situation the number of atoms is the same for both. Which is higher and why? 18.10 In order of increasing entropy per mole at 25 ° C: (c) < (d) < (e) < (a) < (b) (c) Na( s ): ordered, crystalline material. (d) NaCl( s ): ordered crystalline material, but with more particles per mole than Na(s). (e) H 2 : a diatomic gas, hence of higher entropy than a solid. (a) Ne( g ): a monatomic gas of higher molar mass than H 2 . (b) SO 2 ( g ): a polyatomic gas of higher molar mass than Ne. 18.11 Using Equation (18.7) of the text to calculate rxn Δ S o (a) rxn 2 2 (SO ) [ (O ) (S)] Δ = ° ° + ° S S S S o rxn (1)(248.5 J/K mol) (1)(205.0 J/K mol) (1)(31.88 J/K mol) Δ = = 11.6 J/K mol S o
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CHAPTER 18: ENTROPY, FREE ENERGY, AND EQUILIBRIUM 523 (b) rxn 2 3 (MgO) (CO ) (MgCO ) Δ = ° + ° ° S S S S o rxn (1)(26.78 J/K mol) (1)(213.6 J/K mol) (1)(65.69 J/K mol) Δ = + = 174.7 J/K mol S o 18.12 Strategy: To calculate the standard entropy change of a reaction, we look up the standard entropies of reactants and products in Appendix 3 of the text and apply Equation (18.7). As in the calculation of enthalpy of reaction, the stoichiometric coefficients have no units, so rxn Δ S o is expressed in units of J/K mol. Solution: The standard entropy change for a reaction can be calculated using the following equation. rxn (products) (reactants) Δ = Σ ° − Σ ° S nS mS o (a) 2 2 (Cu) (H O) [ (H ) (CuO)] = ° + ° ° + ° rxn S S S S Δ S o = (1)(33.3 J/K mol) + (1)(188.7 J/K mol) [(1)(131.0 J/K mol) + (1)(43.5 J/K mol)] = 47.5 J/K mol (b) 2 3 (Al O ) 3 (Zn) [2 (Al) 3 (ZnO)] = ° + ° ° + ° rxn S S S S Δ S o = (1)(50.99 J/K mol) + (3)(41.6 J/K mol) [(2)(28.3 J/K mol) + (3)(43.9 J/K
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