Chem 260 Final Exam 4/23/01 1INSTRUCTIONSThis exam consists of two parts, a 100 point exam on phases, equilibrium, electrochemistry andchemical kinetics and a 100 point cumulative final exam. BUDGET YOUR TIME WISELY. This exam has an instruction sheet, 16 numbered pages including the cover page, 13 pages ofquestions, and two information sheets, and a blank scratch sheetPART I: 100 point exam on free energy, chemical equilibrium, electrochemistry andchemical kinetics. There are four pairs of problems. 1. Phase Phenomena A and B2. Chemical Equilibrium A and B3. Chemical Kinetics A and B4. Electrochemistry A and BDo one problem from each pair, i.e. A or B. There are also three short answer questions, 5. A, B and C. You should answer two of the threeshort answer questions. Put the letters of the questions that you choose on the answer sheet.PART II: 100 point cumulative final exam.Multiple Choice (60 pts): 20 questions. Choose the letter corresponding to the best single answerfor each question. Put your answers in the numbered boxes on the cover sheet of this exam.Short Answer and Simple Problems (40 pts.): In the space available answer the questions orsolve the problems. For the short answer use complete sentences! Use equations and diagramswhere helpful.Chem 260 Final Exam 4/23/01 1Name: Score: Multiple Choice Answers:1. 6. 11. 16.2. 7. 12. 17.3. 8. 13. 18.4. 9. 14. 19.5. 10. 15. 20.SCORE: For Part I Indicate the Problems Solved or Attempted.Part I Scores: Part II Scores:1 ( ) /20 1-20 /602 ( ) /20 21 /6 25 /83 ( ) /20 22 /6 26 /64 ( ) /20 23 /65 ( ) /10 24 /85 ( ) /10Total: /100 Total: /100Chem 260 Final Exam 4/23/01 15Name Chem 260 Final Exam 4/23/01 15∆S  ‹BAdQrevT  k ln(WBWA) w  ‹VBVAPexdVhν  Φ mv22 λ  hmv HΨ  EΨ ∆p∆x Ap2Evib/rothc  ν n 12 BJ(J 1) xν n 122 D J(J 1)2 α n 12J(J 1) ν 12πkµ B h8π2cI E  hν C  νλ E  hν(n 12)A  … cl ψn  2LsinnπLx En  n2h28mL2∆rG  ∆rGô RT lnQ ∆rG  νFE E  Eô  RTνFlnQln K  νFEôRT  ∆rGôRT k  AQeEa/RT k  kBThe∆G‡/RTPotentially Useful Information:(P + n2a/V2)(V - nb) = nRT ∆H = ∆U + ∆(PV) ∆U = q + w = q - P∆V ∆G = ∆H - T∆SdH = CpdT[A]o - [A] = kt [A] = [A]oexp(-kt) [A]-1 - [A]o-1 = ktChem 260 Final Exam 4/23/01 16Name Chem 260 Final Exam 4/23/01 16Possibly Useful Constants and Quantities and Equations:Constant or Quantity SI unitsAvogadro's Number (No) 6.02214×1023 mol-1Boltzmann's Constant (k) 1.38066×10-23 J K-1Faraday Constant (F) 9.6485×104 C mol-1Electron Charge (e) 1.602177×10-19 CElectron Mass (me) 9.10939×10-31 kgGas Constant (R) 8.31451 J K-1 mol-10.0820578 L atm K-1 mol-1Planck Constant (h) 6.62608×10-34 J sProton Mass (mp) 1.67262×10-27 kgRydberg Constant (R) 1.09677×105 cm-1Speed of Light in a Vacuum (c) 2.99792458×108 m s-1Atomic Mass Unit 1.66054×10-27 kgAcceleration of Gravity (g) 9.80665 m s-2Bohr Radius (ao) 5.29177×10-11 mSome Thermodynamic Data:M g mol-1∆fHô kJ mol-1∆fGô kJ mol-1∆Sô J K-1 mol-1C(s, graphite) 12.011 0 0 5.740C(s, diamond) 12.011 1.895 2.900 2.377CH4 (g) 16.04 -74.81 -50.72 197.67C3H8 (g) 42.10 -103.85 -23.49 269.91CO2(g) 44.010 -393.51 -394.36 213.74H2(g) 2.016 0 0 130.684H2O (l) 18.015 -285.83 -237.13 69.91H2O (g) 18.015 -241.82 -228.57 188.83N2(g) 28.013 0 0 191.61O2 (g) 31.999 0 0