Which of the following diagrams best rep resents the final state of this system

Which of the following diagrams best rep resents the

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Which of the following diagrams best rep- resents the final state of this system after equilibrium is achieved? 1. 2. 3. 4. correct 5. Explanation: Only the solvent goes through the mem- brane thus increasing the volume on the solu- tion side and decreasing volume on the solvent side. The solution side is therefore diluted slightly and the shade is lightened somewhat. The solute itself (the darker color) cannot pass through the membrane so the right side MUST also stay the same color of pure solvent (the lighest shade shown). 005 (part 1 of 1) 10 points The bond breaking process has Δ H (greater, less) than zero and Δ S (greater, less) than zero; thus it is favorable at (high, low) tem- peratures. 1. less; greater; low. 2. greater; less; high. 3. less; greater; high. 4. greater; greater; high. correct 5. less; less; low. 6. greater; less; low. 7. greater; greater; low. 8. less; less; high. Explanation: Energy input (positive Δ H , endothermic) is required to break bonds. Because you in- evitably end up with more individual particles (atoms) than what you start with (molecules), the entropy of the system increases. Processes are spontaneous when Δ G is neg- ative. Δ G = Δ H - T Δ S = (+) - T (+)
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Elbel, Brittany – Exam 1 – Due: Feb 7 2006, 3:00 pm – Inst: J A Holcombe 4 = (+) - T Δ G will be negative only at high tempera- tures. 006 (part 1 of 3) 10 points Start with a liquid of molar fraction 0 . 3. Boil it, catch the vapor, then cool the vapor until it condenses into a new liquid. 0 1 120 65 What is the composition of the new liquid? 1. 0 . 85 correct 2. 0 . 03 3. 0 . 3 4. 89 5. 68 6. 115 Explanation: 0 . 85 Each horizontal unit is 0.1. 007 (part 2 of 3) 10 points What is the boiling point of the new con- densed liquid? 1. 0 . 3 2. 0 . 85 3. 68 correct 4. 89 5. 115 6. 0 . 03 Explanation:
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Elbel, Brittany – Exam 1 – Due: Feb 7 2006, 3:00 pm – Inst: J A Holcombe 5 68 Each vertical unit is 120 - 65 10 = 5 . 5 . 008 (part 3 of 3) 0 points At what temperature would this solution boil? 1. 0 . 3 2. 0 . 85 3. 0 . 03 4. 115 5. 89 correct 6. 68 Explanation: 89 009 (part 1 of 1) 10 points Calculate the standard enthalpy change for the reaction 2 HCl(g) + F 2 (g) 2 HF( ) + Cl 2 (g) given 4 HCl(g) + O 2 (g) 2 H 2 O( ) + 2 Cl 2 (g) Δ H 0 = - 202 . 4 kJ/mol rxn 1 2 H 2 (g) + 1 2 F 2 (g) HF( ) Δ H 0 = - 600 . 0 kJ/mol rxn H 2 (g) + 1 2 O 2 (g) H 2 O( ) Δ H 0 = - 285 . 8 kJ/mol rxn 1. Δ H 0 = +1116 . 6 kJ/mol rxn 2. Δ H 0 = - 1088 . 2 kJ/mol rxn 3. Δ H 0 = - 1015 . 4 kJ/mol rxn correct 4. Δ H 0 = +1088 . 2 kJ/mol rxn 5. Δ H 0 = - 516 . 6 kJ/mol rxn 6. Δ H 0 = - 1587 . 2 kJ/mol rxn
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Elbel, Brittany – Exam 1 – Due: Feb 7 2006, 3:00 pm – Inst: J A Holcombe 6 7. Δ H 0 = +516 . 6 kJ/mol rxn 8. Δ H 0 = +1587 . 2 kJ/mol rxn 9. Δ H 0 = - 1116 . 6 kJ/mol rxn 10. Δ H 0 = +1015 . 4 kJ/mol rxn Explanation: The first equation needs to be multiplied by 1 2 in order to get the equation we’re interested in. Thus its Δ H 0 is multiplied by 1 2 as well. The second equation needs to be multiplied by two in order to get the equation we’re interested in. We also multiply its Δ H 0 by two. The third equation needs to be reversed, so the sign of its Δ H 0 should change. Then we add the equations to get the equa- tion we’re interested in. We also add the ad- justed Δ H 0 values to get the answer, - 1015 . 4 kJ/mol rxn.
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