Ch 6 Mon Nov 7 Prof Reid

Ch 6 Mon Nov 7 Prof Reid - 11/2/2011 Chem 142: Lecture 17...

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11/2/2011 1 Chem 142: Lecture 17 Pressures and K (6.3) Heterogeneous Equilibrium (6.5) Elementary Applications (6.6) vs KQ In the previous lecture we expressed K for this reaction with using concentrations: At equilibrium [NH 3 ] = 3.1 x 10 2 M, [N 2 ] = 8.5 x 10 1 M, and [H 2 ] = 3.1 x 10 3 M. What is the equilibrium constant? Pressure and K   3 22 2 3 2 33 NH 1 M NH NH C a K aa     3 N( ) 3H( ) 2NH( ) gg g 
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11/2/2011 2 Continuing on Notice the effect of dividing by a standard state is to render each component of K unitless. Pressure and K (cont.)   3 22 2 3 2 33 2 2 4 3 13 NH 1 M NH 3.1 10 M 3.8 10 8.5 10 M NH C a K aa       Pressure and K (cont.) Recall the ideal gas law: Focusing on concentrations for each species: Substituting in pressure for concentration: PV nRT nP VR T nm o l VL 2 3 2 3 NH NH C K  3 3 2 2 2 3 3 NH NH NH C P P P RT K RT K RT PP RT RT
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11/2/2011 3 For the generic reaction (all gaseous species): Notice, this is the connection between the expression for K C (the equilibrium constant using concentrations alone) and K P (the equilibrium constant using pressures alone). nis the difference in moles (products minus reactants) of gaseous species only! Pressure and K (cont.)  () de cb n CP P KK R T K R T    bB cC dD eE  Pressure and K (cont.) Example: Determine K P for the following reaction at 1000. Notice all (RT) n is doing is providing a conversion factor between concentration and pressure units. -1 22 3 2SO ( ) O ( ) 2SO ( ) K 280. M c gg g   nn PC K KR T K T   1 -1 -1 -1 -1 280. L mol 3.4 atm 0.0821 L atm mol K 1000. K R T
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11/2/2011 4 In a chemical reaction we can have reactants and products in a variety of phases.
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This note was uploaded on 01/18/2012 for the course CHEM 142B taught by Professor John during the Fall '11 term at University of Washington.

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Ch 6 Mon Nov 7 Prof Reid - 11/2/2011 Chem 142: Lecture 17...

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