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04CHM114_0503

04CHM114_0503 - CHEM 114 Spring 2004 Mon May 3rd Lecture N0...

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Unformatted text preview: CHEM 114 Spring 2004 Mon May 3rd. Lecture N0. 39 Do Problems: LAST LECTURE for $2004: Connections: Free energies, electrical decomposition, and portable energy (Batteries). (see Chapter 20, section 20.9 including sample exercise 17) Last lecture we showed how, with increasing temperature, reactions that we usually think of as being favorable, such as the reaction of magnesium metal with oxygen, can be reversed. This happens because these reactions occur with a decrease in entropy due to gaseous oxygen being converted to an entropy-poor solid, MgO. i.e. AS" is negative. All one has to do is to raise the temperature high _ enough for the TAS" term to become high enough to cancel the AH" term. ”3‘6” = AH0 — [email protected] 0 when T = AHOIAS0 _h=—I_‘I’_ -——-——_— "I- _ _ F. .. _a"' as“: :- AM? @03— GO (EH/firgi- AppxC: 602,000(meol) 7—‘1 _ 1 ° ' ”‘ * 107(Jlmol.K) l a 7 ‘37:; Ly. . l , K = 5,5530 K i.e. for decomposition of MgO (based on data tabulated for 298K) we need to raise temperature to about 5630K For A1203 the same decomposition process requires a slightly lower temperature, about 5040K. Let's see what it takes: How to do it? Heating? temperature needed to get oxygen pressure "pumped up” to standard value of 1 atm (when AG" = 0) is ~5,000K, no way! iY (remember TmG" = 0) = AH°IAS° = 5,000K Fortunately there is another way to do it: ‘We can pump it up electrically: fJGood news: “'itjonly needs a few volts!!!!! i.___; Bad news: . hut it needs a lot of current! .‘. h' Lat-”'9' ‘" &Al(s) + 3/202(g) —) A1203(s) AG" =-157SkJ/mol: AH" = -1669.8kJ/mol 1.findAS° ,-_— Mfg M" .—__.=fl Lore 01‘ [00K d-fi Sc: gain; 5". (mafia—5.7V c _:- Egofipmlm-fl— 2. find T(decomp) '3— 5; —[2s-5 ,2 +935“ We'll calculate the number of volts in a minute, but first let's talk about the principles behind it. "Free energy”of a chemical reaction is the energy of a chemical reaction that is free to do work. If we want to get a reaction with a negative free energy to go in reverse we have to do at least that amount of work on the system (Usually it's bit more than that that is needed because of kinetic factors). Now the only sort of work that we have talked about so far is PV work.... work done when a volume change occurs against an external pressure. Let's talk about a different sort of work....work done when a ”volume of electrons” (meaning a number of moles of electrons) flows against an electron pressure, measured in volts. P )4 V 1‘- ,Q/w “‘1 3: PV work: liters moved against P = energy PV work: liters moved against P = energy QE work: moles e'(g_c_ougomhs) moved vs. E(volts) = " . -—-—‘ —" So this is the way we get aluminum. Using electrical work to "push" the electrons back onto the aluminum from the oxygen. The minimum work needed is QE = - AG or for standard states, QE0 = -AG° For cell voltages we use a special symbol for the voltage to avoid confusion with energy: it is E". _‘E_._° WHAT IS Q? It is the number of electrons needed to restore neutrality to the cations. In the case of A1203, which has two Al3+ cations, it is 2 x 3 moles. 3. TbTfiL. [mi LNQLE b 7‘ NM n5 H-A .— T01 ueflflai‘fixg 9* L-L nit M3‘* in; ting Whats. C .3) ~ ~ Remember CaCO3 decomposition? T(decomp) = T where Keq = 1, meaning T where pressure of CO2 reaches one atm. So, what is pressure of oxygen in equilm with A] at room temperature?? Sb 3 P033” ”L i {in} V3 -1576 kJ/mol SO Keq = e —[-1§7,600/(3.!314X 298)] =___:36 = 10276 -——— means p02 in equilm with pure Al is about 10'184 atm (1!!) at 298K . Some vacuumII To get pure metallic Al back, then, we have to do something to pump that oxygen pressure back up to 1 atm so it will go bubbling off like gas from a freshly Opened soda can. How can we do it? Where does all that Al for the soda cans come from??? to do it, electrochemically. Big cells built where electricity is cheap.. ...Norway. This latter type of extraction, getting chemicals by using electri sag called electrolysis, or elecgolytic a? decomposition. m “g The opposite Laof electrolysis IS getting electrical energy from chemicals. i.e. batteries .' ..... unfortunately no time to tell about it. e.g. the “aluminum-air battery” releases the free energy of formation of A1203 as electrical energy. In its practical version it releases energy by forming Al(OH)3 and it is a bit less energy than if sapphire itself formed. - ...
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