Class Notes Feb 19

Class Notes Feb 19 - 2/19/2010 Announcements: Friday,...

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Unformatted text preview: 2/19/2010 Announcements: Friday, February 19, 2010 Homework WebAssign #7- Due Wednesday, February 24 at 8:00 AM Recitation Next Week Week 7 – Redox Reactions Lab Next Week- Hypochlorite in Bleach Complete your prelab before recitation. Suggested Textbook Questions Finish working Chapter 20 problems. Begin reading Chapter 4.5 and Chapter 21. (ClO-) You are cordially invited to attend an Informational Session to learn more about becoming a Learning Assistant with the STEM-Colorado Program. When: Wednesday, February 24, 2010, at 6 p.m. Where: UMC 235 (hall right of Reception Desk) RSVP: By Feb. 19 to Olivia Holzman (olivia.holzman@colorado.edu) Refreshments will be served, while they last. Applications for Fall 2010 available Feb 24 - Mar 10 at http://stem.colorado.edu/Apps Get more information from faculty and LAs in these departments: CHEM will be hiring for: Applied Math Math • General Chemistry MCDB Astronomy • Physical Chemistry with Physics Chemistry Biochemistry Applications Geological Sciences February 19, 2010 - CT#1 If a reaction has a DH° of +3.0 kJ/mol and a DS° of +30 J/mol·K, above what temperature is the reaction spontaneous at standard conditions? Temperature Dependence of DG and K DG = DH - TDS and K = e-DG/RT Both DG and K depend on T. We only have tables for T = 298 K. So assume DH and DS are temperature independent. Calculate DH and DS from Appendix B at 298 K then solve for DG and K at the other temperature. A. 0.1 K B. 1 K C. 10 K D. 100 K K E. 1000 K At DG° = 0, the reaction becomes spontaneous. 0 = DH° -TDS° Rearrange: DH° = TDS° T = DH°/DS° T = (3.0 kJ/mol)/(0.030 kJ/mol·K) T = 100 K Temperature Dependence of DG and K Estimate DG and K for the following reaction at 500 K given that DH = -483.64 kJ/mol (exothermic) and DS = -88.77 J/molK ( entropy). 2H2(g) + O2(g) 2H2O(g) DG500K = DH298K - TDS298K DG500K = (-483.64 kJ) – (500 K)(-0.08877kJ/K) DG500K = -439.26 kJ/mol Rxn is spontaneous! K500K = e K500K = -(-439.26 kJ/mol)/(0.008314 kJ/ molK)(500 K) e-DG500K/RT RECAP To correct for non-equilibrium conditions: DGT = DGT + RTln(Q) To correct for temperatures other than 298K: DGT = DH298 - TDS298 To convert between DGo K DGo = -RTln(K) or K = e -DG/RT K500K = 7.78 1045 K > 1 for spontaneous rxns! 1 2/19/2010 Electrochemistry- reactions that involve the transfer of electrons. Electron acceptor Electron donor Ag+ + e– Ag(s) Reduction Oxidation Is L oss R eduction Is G ain Electrochemistry- Redox Reactions Oxidizing agent Reducing agent Ag+ + e– Ag(s) Cu(s) Cu2+ + 2e– Reduction Oxidation Cu(s) Cu2+ + 2e– Oxidation Silver ion gained an electron- “reduced” Copper lost electrons- “oxidized” “REDOX” reactions Oxidizing agent- oxidizes something else by being reduced. SILVER ION Reducing agent- reduces something else by being oxidized. COPPER Redox couple- the oxidized and reduced forms of a species (ox/red). Examples: Ag+/Ag, Cu2+/Cu, Fe3+/Fe For every reduction, there must be an oxidation and vice versa. The electrons have to come from somewhere and go somewhere. February 19, 2010 - CT#2 What is the oxidation number of chlorine in HClO 2? February 19, 2010 - CT#3 In the following redox reaction, which element(s) is being reduced? PbO(s) + CO(g) → Pb(s) + CO2(g) (+2)(-2) (+2)(-2) (0) (+4)(2x-2) A. -3 B. -1 C. 0 D. +1 HClO2 is neutral. So (+1) + (x) + 2(-2) = 0 x = +3 Also recall that the oxidation number of any atom in its elemental form is zero! Ex. Cl2 Review oxidation number rules in Table 4.3, page 131. lead A. lead B. oxygen C. carbon Pb goes from +2 to 0. The oxidation number went down = reduction. C goes from +2 to +4. The oxidation number went up = oxidation. E. +3 D. lead and carbon E. None of the elements is being reduced 2 ...
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This note was uploaded on 03/23/2010 for the course CHEM 1131 taught by Professor Redin,kend during the Spring '08 term at Colorado.

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