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CHEM 433 - 11 /30/11

CHEM 433 - 11 /30/11. VIII. Chemical Equilibrium ( 7.1-7.4 ) • Thermodynamic description EQ - K from data - a case study (N 2 O) - LeChatlier’s Principle - K at another T - Molecular-Level considerations READ: Chapter 7 HW #10 due Friday . Consider:

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CHEM 433 - 11 /30/11

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  1. CHEM 433 - 11/30/11 VIII. Chemical Equilibrium (7.1-7.4) • Thermodynamic description EQ - K from data - a case study (N2O) - LeChatlier’s Principle - K at another T - Molecular-Level considerations READ: Chapter 7 HW #10 due Friday

  2. Consider: 2 N2 (g) + O2(g) <—> 2 N2O (g) K = ? fG° = 0 0 +104.2 kJ/mol • Calculate K (298K). (What kind of K is this in the ideal limit?) If PO2 = 0.21 bar, and PN2 = 0.78 bar, what is PN2O? PN2O (observed) = 3.30 x 10-7 bar - what is up ? Good enough for government work? http://webbook.nist.gov/cgi/cbook.cgi?ID=C10024972&Units=SI

  3. LeChatlier’s Principle: A system at EQ responds to a change in a manner that counteracts the change (a.k.a. “negative feedback…”) We will consider this rxn*: N2 + 3 H2 < —> 2 NH3H = -92.2 kJ K (298K) = 5.9 x 10-5 Know your history: http://www.chemheritage.org/discover/chemistry-in-history/themes/early-chemistry-and-gases/haber.aspx Let’s calculate K at 750 K (assuming DH constant)

  4. For an exothermic reaction, w/ S ~0, rG° ( + or - ) K > or < 1

  5. Now - how can an exothermic reaction be product favored? (DS could be +)

  6. • How/why does an increase in temperature make an endothermic reaction more favorable ? (left)(DG probably still + in this pic.) • Opposite effect for exothermic (right)

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