04CHM114_0426

04CHM114_0426 - MP.“- CHEM 114 Spring 2004. April 26....

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Unformatted text preview: MP.“- CHEM 114 Spring 2004. April 26. Lecture No. 34 CHEMICAL EQUILIBRIUM (cont’d) (CHAPTER 15) ll Classic case of equilibrium in the gas state is the "ammonia synthesis reaction" N2(g) + 3 H2(g) 69" 2NH3(g) 4 moles gas 2 moles gas AH = - 91kJ/mole 9X (at STP) = —44.8 Li Why is this the value of AV? (is it exactly correct? If not why not? (which is the component responsible for it not being exactly correct) Which way will increase of pressure affect the equilibrium? Why important? Haber process. . .. N2 + 3 H2 69" ZNH3 4 moles 2 moles AH = — 91kJ/mole (i) what is the equilibrium constant? M, 8'5 0 ' c lit, = (""33 - g nodu- [NQ Lat] 3 n“ ‘ (ii) what will be the effect of temperature on the reaction #9.” £th «me-cum- Ow'ug M- inn '7' (iii) "what will be the effect of pressure? . Ski" \- s at..un I ""‘ 'h V... r ssh- A? Ln“; 1' Lehuhnw 1t; rcfiolfih. (I #a. 3500* (iv) optimizing the process: the need for catalysts. (What is a catalyst?) Equilibrium constants in terms of partial pressures (partial pressure???) read p 566 Kc (concentrations) E A 1 K1, (partial pressures) NO2 example ’ 1:40.. (1) —~ N's-94(5) R % PNVOU P 1. Pan, 3 HETEROGENEOUS EQUILIBRIA. These are equilibria between substances in different states of matter, or different phases of the same state. examples: 1. Liquid-solid 2HC03'(aq) + Ca2t(aq) 6-) H200) + CaCO3(s) ' w...» said 2. Gas-solid CaCO3(g) 6-) CaO(S) + C02(g) limestone lime 3. Gas-liquid-solid C be». NJv-ahm B V ‘1 0 fig CaCO3(s) + 2HCl(aq) (--) C02(g) + H20 + CaC12(aq) #’ “do 3a.: M“ We consider examgle 2 in more detail. Practical case (“liming”of limestone) CaCO3(g) 6-) Ca0_(s) + COEgg) *For equilibria involving Eure solids or liquids, there is a rule. We write [pure solid or liquid] = 1.0 so it drops out of the expression befits K condition for reaction "going" is PC02 > 1 because then KP '> 1 $500.- hv‘ ‘T‘ lOQC.I- Why does reaction “go” when it makes the system increase in energy?????? What is the other driving force involved??? We have seen it before, of course. It is entropy! (Subject of detailed study in our next chapter) Preliminary: Thinking about the CaCO3 decomposition, isn't it very like boiling??? "Why do liquids boil?”, then, is the same question but we already know the reason for that. It is G(gas) < Gfliq) for T > Tb , and Th is the temperature above which the Pmp) is greater th 11 1 atm. condition for reaction "going" is Pm, > 1 because then Kp > 1. It is exactly the same condition we had for our earlier definition of the "normal" boiling point. Why do liquids boil?" then, is the same question, as why does CaC03 decompose, when both involve an increase of energy. but we already know the reason for that. It is that the free energy of the liquid state crosses over that of the solid, and so the liquid becomes stable. It crosses over because it has the steeper slope because it has the larger disorder (entropy) and G (stability index) = H — TS Q0 4- CO v I Cato, I K“.=|'° So the reason that t1}; equilibrium in the decomposition of CaCO3 shifts to the right (i.e Keq increases) with increasing temperature is going to haver something to do with the high entropy of gaseous carbon dioxide.... See the connection?? Now let’s do it in detail Ch. 19. ...
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This note was uploaded on 11/01/2009 for the course CHM 114 taught by Professor Aksyonov during the Fall '08 term at ASU.

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04CHM114_0426 - MP.“- CHEM 114 Spring 2004. April 26....

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