solution_pdf - Version 227 Exam 4 mccord(50970 This...

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Version 227 – Exam 4 – mccord – (50970) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Which of the following statements concerning the first law of thermodynamics is/are true? I) The internal energy of the universe is always increasing. II) Internal energy lost by a system is always gained by the surroundings. III) The universe is an isolated system. 1. II only 2. II and III only correct 3. I, II and III 4. I only 5. I and III only 6. III only 7. I and II only Explanation: Statement I is false; the first law states that the energy of the universe is conserved, in other words a constant value. Statement II and III are true; internal energy in the uni- verse is conserved, and thus energy lost by the system is always gained by the surroundings. The universe is the most obvious example of an isolated system in that energy and matter are conserved in the universe. 002 10.0 points Which one of the following thermodynamic quantities is NOT a state function? 1. heat correct 2. pressure 3. temperature 4. entropy 5. free energy Explanation: All except heat are independent of path and are dependent solely on the current state of the system. Heat (as well as work) is not a state function. It is dependent on the process or pathway. HOW you got there DOES affect the value of heat and work. 003 10.0 points Consider the following data at 25.0 C and 1 atm. Species Δ H 0 f S 0 Δ G 0 f kJ/mol J/mol · K kJ/mol N 2 (g) 0 . 0 191 . 5 0 . 0 O 2 (g) 0 . 0 205 . 0 0 . 0 N 2 O 5 (g) 11 . 0 ? 115 . 0 Calculate the absolute entropy of N 2 O 5 at 25.0 and 1 atm. 1. - 704 J · mol 1 · K 1 2. +47 . 5 J · mol 1 · K 1 3. - 349 . 0 J · mol 1 · K 1 4. +252 . 5 J · mol 1 · K 1 5. +396 . 5 J · mol 1 · K 1 6. +1053 . 0 J · mol 1 · K 1 7. +355 . 0 J · mol 1 · K 1 correct Explanation: Δ G rxn = Δ H rxn - T Δ S rxn so we must find the Δ H and Δ S of the re- action. To do this, we need to write out the balanced equation: 2 N 2 + 5 O 2 2 N 2 O 5 Δ H rxn = Δ H products - Δ H reactants = (2 mol)(11 kJ / mol) - 0 = 22 kJ
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Version 227 – Exam 4 – mccord – (50970) 2 Δ S rxn = Δ S products - Δ S reactants = 2 x - (5 mol)(205 J / mol · K) - (2 mol)(191 . 5 J / mol · K) = 2 x - 1408 J / K = 2 x - 1408 1000 kJ / K Now substitute into Δ G rxn = Δ H rxn - T Δ S rxn : (2 mol)(115 kJ / mol) = 22 - (298 K)(2 x - 1408) kJ / K 1000 Solving for x gives x = 355 . 007J/mol · K. 004 10.0 points Calculate Δ H for the formation of 1 . 13 moles of MgCl 2 at 298 K and 1 atm by the reaction Mg(OH) 2 (s) + 2 HCl(aq) MgCl 2 (s) + 2 H 2 O( ) . 1. 56.703 2. 26.738 3. 125.853 4. 52.093 5. 132.768 6. 44.256 7. 115.25 8. 135.995 9. 57.625 10. 63.618 Correct answer: 52 . 093 kJ. Explanation: Reactants: Δ H f Mg(OH) 2 (s) = - 924 . 7 kJ/mol Δ H f HCl(aq) = - 167 . 4 kJ/mol Products: Δ H f MgCl 2 (s) = - 641 . 8 kJ/mol Δ H f H 2 O(g) = - 285 . 8 kJ/mol Δ H 0 rxn = summationdisplay n Δ H 0 f prod - summationdisplay n Δ H 0 f rct = [2 ( - 285 . 8 kJ / mol) + ( - 641 . 8 kJ / mol)] - [2 ( - 167 . 4 kJ / mol) + ( - 924 . 7 kJ / mol)] = parenleftBig 46 . 1 kJ mol rxn parenrightBigparenleftBig 1 mol rxn 1 mol MgCl 2 parenrightBig × (1 . 13 moles MgCl 2 ) = 52 . 093 kJ 005 10.0 points Heat absorbed by a system at constant vol- ume is equal to 1. Δ H 2. ΔV 3. Δ E correct 4. Δ S 5. Δ G Explanation: Δ E = q
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