Chemistry 1A, Fall 2009 Midterm Exam #3 November 10, 2009 (90 min, closed book) Name: SID: GSI Name: The test consists of 7 short answer questions and 18 multiple choice questions. Put your written answers in the boxes provided. Answers outside the boxes may not be considered in grading. Show all your work and reasoning. Useful Equations and constants: Write your name on every page of the exam. G = - T S = n f (products) - n f (reactants) Question Page Points Score S = ns (products) - ns (reactants) Multiple Choice 2-5 85 G = n G f (products) - n G f (reactants) S = k B lnw S = q rev /T Question 19 5 6 E = q + w w = - P ext V Question 20-21 6 12 G = - RTln K G = G + RTln Q G = - nf Eº E = Eº - (RT/nF) lnq Question 22-24 7 17 q = n C p T Total 120 N 0 = 6.02214 10 23 mol -1 T (K) = T (C) + 273.15 F = 96,485 C / mol 1 V = 1 J / C R = 8.31451 J K -1 mol -1 R = 8.20578 10-2 L atm K -1 mol
A turkey and potatoes are roasting in the oven in an aluminum pan with salt and oil. Use the data in the table to answer the questions below. Specific heat capacity (J/g deg) eat of combustion (kj/g) Protein (meat) ( RC 2 2 NO) n 2.8 17 Carbohydrates (potato) ( C 2 O) n 3.4 16 Fat (olive oil) C 18 34 O 2 1.3 37 Salt NaCl 1.2 no reaction Aluminum Al 0.9 16 1. The dinner is placed in a preheated oven. After 3 minutes, none of the materials have equilibrated to the oven temperature. Which substance will be at the lowest temperature? Assume the masses are similar and there is no heat transfer between substances. A) potato B) salt C) Al 2. Which food has the most Calories per gram? A) meat B) potato C) oil 3. Which substance has the largest standard entropy per mole? A) oil B) salt C) Al 4. ow do the calories of a potato with a lot of salt compare with a potato with no salt? A) The potato with salt has more calories B) The potato with salt has fewer calories. C) Both have the same calories. 5. Which transfers more heat? A) Combustion of 1 g fat. B) Cooling 1 g salt by 100 degrees. C) Cooling 1 g aluminum by 100 degrees. 6. Instead of eating of potatoes, you exercise and burn fat instead. ow much fat would you burn to use the same energy that is generated from eating 150 g of potatoes? A) 392 g B) 65 g C) 260 g Page 2 of 7
Solid carbon can be found as diamond, graphite, or fullerene. Graphite and diamond are network covalent solids, while fullerene is a molecular covalent solid. Use the data in the table to answer the questions below. All values are reported per mole of carbon atoms for easier comparison. T = 298K º f (kj/mol) Sº (J/mol K) C (s) graphite 0 +5.7 C (s) diamond +1.9 +2.4 C 60 (s) fullerene +38.0 +7.1 7. The most stable form of elemental carbon at 25 C is A) graphite B) diamond C) fullerene D) There is not enough information to determine 8. Which of the conversions shown below is the most exothermic? A) diamond to graphite B) diamond to fullerene C) graphite to fullerene D) graphite to diamond 9. The conversion of diamond to graphite is favored at A) low temperature B) high temperature C) all temperatures D) no temperature 10. The best reason to explain why it is favorable for diamonds to convert to graphite at 25 C is A) The change in entropy is negative. B) The change in enthalpy is negative. C) The change in free energy is negative. D) The energy of activation is very large. 11. The best reason to explain the claim that diamonds last forever is A) The change in entropy is negative. B) The change in enthalpy is negative. C) The change is free energy is negative. D) The energy of activation is very large. Page 3 of 7
The combustion of methane is used to warm water in home water heaters, heat food on gas stoves, and spin turbines to generate electricity in power plants. The combustion reaction and the thermodynamic data at 298K are given below. C 4 (g) + 2 O 2 (g) CO 2 (g) + 2 2 O (l) ΔG o rxn = 818 kj/mol C 4 Δ o rxn = 890 kj/mol C 4 ΔS o rxn = 242 J/mol-K C 4 12. ow do you know that ΔS universe is increasing? A) Δ o rxn < 0 B) ΔS o rxn < 0 C) ΔS surroundings > 0 D) ΔS o rxn + > 0 13. Suppose that you start with 1 atm C 4, 0.2 atm O 2, and 0.00038 atm CO 2. ow does the maximum work compare with standard conditions? A) The maximum work is greater because Δ = ΔE + PΔV B) The maximum work is slightly greater because ΔG rxn < ΔG o rxn. C) The maximum work is slightly smaller because ΔG rxn < ΔG o rxn. D) The maximum work is slightly smaller because ΔG rxn > ΔG o rxn. 14. The enthaply of combustion of C 4 is used to heat 1mole N 2 gas. The temperature of the gas increases from 300 K to 450 K. The N 2 gas is allowed to expand. ow much work is done by the system (N 2 gas) for expansion against an external pressure of 1 atm? A) +24.6 L atm B) +6.15 J C) 12.3 L atm D) 24.6 L atm 15. The value for Δ o rxn for the methane combustion calculated with average bond enthalpies is -802 kj/mol. The main reason for the difference with the experimentally measured value given above is: A) The zero of energy is the atoms, so Δ o rxn is less negative. B) Calculations using average bond enthalpies are approximate. C) The condensation of water releases heat, which is not taken into account when using average bond enthalpies. D) The change in pressure is not taken into account when using average bond enthalpies. Page 4 of 7
Use the standard potentials to answers questions 16-21 below. All materials are present in their standard states unless specified. Standard half-cell reaction E ΔG f Zn 2+ (aq) + 2e Zn(s) 0.76 V +147 kj/mol Ni 2+ (aq) + 2e Ni(s) 0.28 V +54 kj/mol Cu 2+ (aq) + 2e Cu(s) +0.34 V 65 kj/mol Fe 3+ (aq) + e Fe 2+ (aq) +0.77 V 74 kj/mol Ag + (aq) + e Ag(s) +0.80 V 77 kj/mol Pt 2+ (aq) + 2e Pt(s) +1.18 V 227 kj/mol Au + (aq) + e Au(s) +1.69 V 163 kj/mol Q16 E =+0.62 V - + Cu(s) Q17 Q18 anode cathode A galvanic cell with two metals is shown above. Choose the correct labels below for the diagram. 16. Which substance is Q16? A) Zn(s) B) Cu(s) C) Ag(s) D) Ni(s) 17. Which substance is Q17? A) Zn 2+ (aq) B) Cu 2+ (aq) C) Ag + (aq) D) Ni 2+ (aq) 18. Which substance is Q18? A) Zn 2+ (aq) B) Cu 2+ (aq) C) Ag + (aq) D) Ni 2+ (aq) 19. Short Answer Examine the galvanic cell shown above. Both solutions are diluted so the concentration is 0.50M. Explain what happens to the cell voltage and why. Use words and equations. Page 5 of 7
You can create photographs from negatives because the photographic paper turns dark when exposed to light. Chemicals embedded in a gel in the photographic paper undergo a two step oxidation-reduction process. Step 1) Fe 3+ (aq) + C 2 O 2-4 (aq) Fe 2+ (aq) + 2CO 2 (g) E rxn = +1.20 V Step 2) Fe 2+ (aq) reacts with Pt 2+ (aq) to make a dark solid E rxn =? 20. Step 1 does not occur very quickly unless light hits the paper to initiate the reaction. What role does the light play in the process? Complete the reaction diagram below and explain using words. Potential Energy Reactants Reaction Progress 21. In Step 2 of the reaction Fe 2+ (aq) reacts with Pt 2+ (aq) to produce a dark solid. Write a balanced equation for the reaction below and compute E for the reaction using data from the previous page. Indicate which species is the dark solid. Page 6 of 7
Polyethylene is the polymer used to make plastic shopping bags. The reactions below show the formation of polyethylene by the two step process of making ethylene from octane and then linking ethylene molecules together. Step 1 ¼ C 8 18 (l) ¼ 2 (g) + C C (g) Δ rxn = +115 kj/mol ethylene Step 2 2 (g) + 500 C C (g) C C (s) Δ rxn =? 500 ethylene polyethylene Use the information in the table to answer the questions below. Physical properties Average bond enthalpies Melting point 120 C C C 347 kj/mol C p (polyethylene) 28 kj/mol K C=C 611 kj/mol Δ melting (polyethylene) 3,000 kj/mol C 412 kj/mol 435 kj/mol 22. What is the heat of reaction (kj) for Step 2, the formation of one mole of polyethylene from 500 moles of ethylene molecules? 23. Recycling polyethylene requires melting the polymer. ow much heat is required per mole of polyethylene to raise the temperature of polyethylene from 20 C to the melting point and also melt the polyethylene? 24. Based on your calculations, is it better to recycle polyethylene or make new polyethylene starting from octane (Step 1 and 2)? Explain your reasoning. Page 7 of 7