1 / 3

CHEM 433 – 12/ 1 / 11

CHEM 433 – 12/ 1 / 11. VIII. Chemical Equilibrium ( 7.1-7.4 ) • Thermodynamic description EQ - More on LeChatlier’s Principle - K at another T – quick rev - Pressure dep. – quick … - Adding reactant/product IX. Kinetics – first few comments … maybe?

ziv
Télécharger la présentation

CHEM 433 – 12/ 1 / 11

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. CHEM 433 – 12/1/11 VIII. Chemical Equilibrium (7.1-7.4) • Thermodynamic description EQ - More on LeChatlier’sPrinciple - K at another T – quick rev - Pressure dep. – quick … - Adding reactant/product IX. Kinetics – first few comments … maybe? READ: Chapter 7 HW #10 due Friday – end of day is fine …

  2. • 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)

  3. LeChatlier’s Principle: A system at EQ responds to a change in a manner that counteracts the change Still working with: N2 + 3 H2 < —> 2 NH3H° = -92.2 kJ G° = -32.9 kJ/mol K (298K) = 5.9 x105 Lets examine changes in reactant product concentrations: (Calculating DrG (“not knot”) is the key here …) Suppose we have PH2 = PN2 = PNH3 = 1.0 bar Suppose:PH2 = PN2 = PNH3 = 10 bar Suppose:PH2= PN2 = 0.01 bar; PNH3= 10 bar

More Related