1. 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

2. Consider: 2 N 2 (g) + O 2 (g) 2 N 2 O (g) K = ?  f G° = 0 0 +104.2 kJ/mol Calculate K (298K). (What kind of K is this in the ideal limit?) If P O2 = 0.21 bar, and P N2 = 0.78 bar, what is P N2O ? P N2O (observed) = 3.30 x 10 -7 bar - what is up ? Good enough for government work? http://webbook.nist.gov/cgi/cbook.cgi?ID=C1002497 2&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*: N 2 + 3 H 2 2 NH 3  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  H constant)

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

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

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