1 / 14

Chapter 8 (CIC) and Chapter 4, 20 (CTCS)

Chapter 8 (CIC) and Chapter 4, 20 (CTCS). Read in CTCS Chapter 4.4 (pgs 120-123), and 20.1-2 Problems in CTCS: 4.37, 39 and 20.3, 5, 7, 9. New Energy Sources – Why?. Run out of fossil fuels Hydrocarbons not available for plastics or medicines Fossil fuel wastes

madison
Download Presentation

Chapter 8 (CIC) and Chapter 4, 20 (CTCS)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 8 (CIC) and Chapter 4, 20 (CTCS) • Read in CTCS Chapter 4.4 (pgs 120-123), and 20.1-2 • Problems in CTCS: 4.37, 39 and 20.3, 5, 7, 9

  2. New Energy Sources – Why? • Run out of fossil fuels • Hydrocarbons not available for plastics or medicines • Fossil fuel wastes • CO2 generation gives greenhouse • NOx and SOx give acid rain problems • Smog gives tropospheric ozone problems • CO and other particulates • Nuclear power wastes

  3. H2 as a Fuel • H2 + 1/2O2 H2O ΔH = -286 kJ • That’s 143 kJ per gram of H compared to • 30 kJ/g for coal • 46 kJ/g of octane (11 kcal/g) • 54 kJ/g of methane • How do we get H2? • M + 2 H+ M2+ + H2 • Very expensive for industrial scale

  4. Most Abundant Element in the Universe • H2O  H2 + 1/2O2ΔH = +286 kJ • Electrolysis – but where do we get electricity? • Fossil fuel combustion • Pollutants • 60% efficiency (at best) – 2nd law of thermodynamics • CH4(g) + 2H2O(g) 4H2(g) + CO2(g) ΔH = +165 kJ • Find better catalysts?

  5. Storage and Combustion • 1 g of H2 occupies 12L of volume at atmospheric pressure • Condense it at -253ºC • Li(s) + 1/2 H2(g) LiH(s) • Occupies 1 teaspoon of volume • LiH(s) + H2O(l) H2(g) + LiOH(s) • Hindenburg and space shuttle Challenger

  6. Electrochemistry (Redox) • Electricity and chemical reactions • Zn(s) + 2 H+(aq) H2(g) + Zn2+(aq) • Zinc loses 2e- while each H gains an e- • Zinc is oxidized (oxidation is a loss of e-) • Hydrogen is reduced (reduction is a gain of e-) • OIL RIG or LEO goes GER • How did oxidation/reduction terms come about? • Zn(s) + 1/2 O2(g) ZnO(s) • ZnO(s) + H2(g) H2O(l) + Zn(s)

  7. Oxidation Numbers (States) • Based on nonsharing of electrons • Oxidation Number (O.N.) Rules • 1. O.N. of any element is zero (Li, H2, P4, S8, ...) •   2. O.N. is equal to the charge on an individual atomic ion • (Li+ = +1, Cl- = -1, Ca2+ = +2, P3- = -3, ...) •   3. O.N. of oxygen equals -2 except OF2, H2O2, and other peroxides (O22-; -1) and superoxides (O2- ; -1/2)

  8. 4. O.N. of fluorine in a compound is -1   5. O.N. of hydrogen is +1 unless it is combined with an element which has a lower electronegativity than itself (NaH, CaH2, B2H6, ...)   6. Sum of O.N. must equal charge on species. H3PO4 H = +1 P = +5 O = -2 3(+1) + 1(+5) + 4(-2) = 0 7. Charges must balance on both sides of the equation

  9. Zn(s) + 2 H+(aq) H2(g) + Zn2+(aq) Zn = 0, H+ = +1, Zn2+ = +2, H2 = 0 Q: Determine the O.N. of each element in the following compounds: NaCl BaI2 FeBr3 MnO2 H2O2 HNO3 AlPO4 SO42-

  10. Redox/Travel Agents • If one species is oxidized, another must be reduced (electrons have to go somewhere!) • The species that is oxidized is a reducing agent (oftentimes contains O) • The species that is reduced is an oxidizing agent Zn(s) + 2 H+(aq) H2(g) + Zn2+(aq) Q: Which is the reducing agent and which is the oxidizing agent?

  11. Balancing Redox Equations • Not only must atoms balance, but e-’s must too • Two methods: half-reaction and O.N. method •  Write skeletal eqn and separate into 2 half-rxns CH3CH2OH + Cr2O72- CH3CO2H + Cr3+ Oxdn: Redn:   Balance atoms by adding H+, OH-, or H2O 3. Add electrons and equilize e- for the two reactions 4. Add half-reactions and cancel equalities 5. Verify elemental and charge balance

  12. Balance this reaction in an acid CH3CH2OH + Cr2O72- CH3CO2H + Cr3+

  13. In a basic solution, Br2, disproportionates to bromate (BrO3-) and bromide ions. Balance this equation.

  14. In a basic solution, iron(III) hydroxide reacts with hypochlorite (OCl-) ion to produce FeO42- and chloride. Balance this reaction.

More Related