EqCnstEx

# EqCnstEx - / kT g AB g C g A g BC e – ∆ E o / RT 1 / B...

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Colby College Equilibrium Constants and Statistical Mechanics Example Example: diatom-atom exchange A + BC AB + C E o = E o (AB) – E o (BC) A R BC R AB C ν ~ o (BC) ν ~ o (AB) D o (BC) D o (BC) K p = q o AB / N A q o C / N A q o A / N A q o BC / N A e E o / RT q o t = (2 π mkT) 3 / 2 h 3 V o m q r = kT σ B ~ hc B ~ = h 4 π μ R 2 AB c q v = 1 1 – e –h ν o /kT = 1 1 – e –h ν ~ o c / kT K p = m AB m C m A m BC 3 / 2 1 σ B ~ AB 1 σ B ~ BC 1 1– e –h ν ~ o (AB) c / kT 1 1– e –h ν ~ o (BC) c
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Unformatted text preview: / kT g AB g C g A g BC e – ∆ E o / RT 1 / B ~ AB 1 / B ~ BC = μ AB R 2 AB μ BC R 2 BC calculate R AB , ν ~ o (AB), and E o (AB) from MO theory from spectroscopic data: since D o (spectroscopic) = – E o with D o &gt; 0 ∆ E o = [–D o (AB)] – [–D o (BC)]...
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## This document was uploaded on 02/22/2012.

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