PS8Solutions

PS8Solutions - Chemistry 120B SP11 Problem Set 8 Solutions...

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Unformatted text preview: Chemistry 120B SP11 Problem Set 8 Solutions Total: 50 points 1. See MQS Solutions Manual. 2. (i) ρ P (g) ρ R (g) = K eq (g) = exp h- β μ (0) P (g)- μ (0) R (g) i ρ P ( ‘ ) ρ R ( ‘ ) = K eq ( ‘ ) = exp h- β μ (0) P ( ‘ )- μ (0) R ( ‘ ) i (ii) ρ R (g) ρ R ( ‘ ) = α R = exp h- β μ (0) R (g)- μ (0) R ( ‘ ) i ρ P (g) ρ P ( ‘ ) = α P = exp h- β μ (0) P (g)- μ (0) P ( ‘ ) i (iii) K eq (g) K eq ( ‘ ) = ρ P (g) ρ R (g) ρ R ( ‘ ) ρ P ( ‘ ) = ρ P (g) ρ P ( ‘ ) ρ R ( ‘ ) ρ R (g) = α P α R = exp h- β μ (0) P (g)- μ (0) P ( ‘ ) i exp h- β μ (0) R (g)- μ (0) R ( ‘ ) i (iv) The quantity μ (0) R (g)- μ (0) R ( ‘ ) represents the free energy change when one reactant molecule is moved from the liquid phase to the gas phase at the standard state. Similarly for the product term. (v) K eq (g) K eq ( ‘ ) = exp( β ˜ w P ) exp( β ˜ w R ) = exp h β ( ˜ w P- ˜ w R ) i ˜ w P- ˜ w R = k B T ln K eq (g) K eq ( ‘ ) = 8 . 31 J mol K...
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This note was uploaded on 09/12/2011 for the course CHEM 107B taught by Professor Jamesames during the Spring '09 term at UC Davis.

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PS8Solutions - Chemistry 120B SP11 Problem Set 8 Solutions...

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