CH302: Worksheet 1c Answer key. Examples of Advanced Thermo listed by question type. • Hess’s Law and combined reaction enthalpies 1. Given that S(s) + O2(g) → SO2(g) ∆H = -296.8 kJ/mol S(s) + 3/2 O2(g) → SO3(g) ∆H = -395.6 kJ/mol, determine the enthalpy change for the decomposition reaction 2SO3(g) → 2SO2(g) + O2(g). Answer: Multiply top equation by 2 and bottom equation by -2. The answer for ∆Hrxn = (2 x -296.8) + (-2 x -395.6) = 197.6kJ 2. Given that 2H2 + O2 → 2H2O ∆H = -571.6 kJ/mol C3H4 + 4O2 → 3CO2 + 2H2O ∆H = -1937 kJ/mol C3H8 + 5O2 → 3CO2 + 4H2O ∆H = -2220 kJ/mol, determine the heat of the reaction for C3H4 + 2H2 → C3H8. Answer: Multiply the bottom equation by -1. Everything then cancels. The answer for ∆Hrxn = (-571.6) + (-1937) + (-1 x -2220 ) = -288.6 kJ • Internal Energy calculations 3. The standard molar heat of freezing for water is -6020 J/mol. Calculate q, w, and ∆E for freezing 1.0 mol of water at 0oC and 1 atm pressure. (Hint: there is no need to use a calculator.) Answer: q = -6020 J/mol w = 0 (no gas involved) ∆E = q + w = -6020 J/mol 4. For a reaction in which more moles of gas are produced than are consumed (at constant pressure), ∆H is a) Equal to ∆E b) Less than ∆E c) Greater than ∆E correct from equation ∆H = ∆E + P∆V since P∆V will be positive • Statistical thermodynamics: internal energy theory 5. What is the total motional contribution to the molar internal energy of CO2? (Express your answer in amounts of RT.) Answer: 3/2RT + 2/2RT = 5/2RT for linear molecule 6. What is the total motional contribution to the molar internal energy of NH3? (Express your answer in Amounts of RT.) Answer: 3/2RT + 3/2RT = 3RT for linear molecule• Calculation of the entropy change at a phase transition 8. What is the entropy change for the freezing of 3.33 grams of an alcohol, C2H3OH, at 373.2K given that ∆H = -40,700 J/mol? Answer: ∆G = ∆H – T∆S but ∆G = 0 for a phase transition so ∆H = T∆S. Substitute T = 373.2K and -40700 J/mol to find ∆S = -109 J/K for one mole. The question is ambiguous about the quantity of lcohol so I am finding ∆S per mole so I don’t have to do any more work. • Statistical thermodynamics: Boltzmann formula 9. Use the Boltzmann formula to calculate the entropy at T = 0 for a) a mole of BCl3 that can be oriented one way b) a mole of BCl2Br that can be oriented three ways Answer: a) S = 0 because there is only one orientation and ln 1 will = 0 in the equation S = k ln W no matter how many molecules. b) S = 1.38 x 10-23 ln 36.02 x 1023 = 9.1 J/Kmole • Statistical thermodynamics: Third Law 10. Based on the structures of each of the following molecules, which are most likely to have a residual energy in their crystal forms at T = 0? a) CO2 b) O3 c) HCl d) Cl2 Answer: O3 and HCl have residual entropy because they have more than one orientation. Symmetrical CO2and Cl2 do not and have S = 0. • Entropy Change and the surrounding 11. When a sugar cube dissolves in a cup of coffee (an endothermic process), what the the signs of the entropy change for the system, surroundings and universe, respectively. a) -,-,- b) -,+,+ c) +,-,+ correct. The system is getting more disordered and the universe must get more disordered because of second law. d) +,+,+ e) none are correct • Calculating the change in free energy 12. Calculate ∆Go for the reaction 2N2(g) + 3O3 (g) Æ 2 N2O3 (g) at 25oC ∆Hfo So N2 0 191.5 O3 0 205 N2O3 83.72 312.2 a) 540 kJ/mol rxn b) 278.7 kJ/mol rxn c) -561 kJ/mol rxn d) -540 kJ/mol rxn e) +56 kJ/mol rxn Answer: Calculate ∆H = 167 kJ and T∆S = 111kJ. So ∆H – T∆S = ∆G = 56
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