Thermodynamics filled in class notes_Part_68

Thermodynamics - 4.11 EQUILIBRIUM IN A TWO-COMPONENT SYSTEM 143 One can also eliminate to get n N in terms of n N 2 n N = 2 n N 2 | t =0 − n N

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Unformatted text preview: 4.11. EQUILIBRIUM IN A TWO-COMPONENT SYSTEM 143 One can also eliminate ζ to get n N in terms of n N 2 : n N = 2 ( n N 2 | t =0 − n N 2 ) . (4.439) Now for the reaction, one must have, for second law satisfaction, that μ N 2 dn N 2 + μ N dn N ≤ , (4.440) μ N 2 ( − dζ ) + μ N (2 dζ ) ≤ , (4.441) ( − μ N 2 + 2 μ N ) dζ ≤ (4.442) ( − μ N 2 + 2 μ N ) dζ dt ≤ . (4.443) In order to satisfy the second law, one can usefully, but na¨ ıvely, hypothesize that the non-equilibrium reaction kinetics are given by dζ dt = − k ( − μ N 2 + 2 μ N ) , k ≥ , convenient, but na¨ ıve model (4.444) Note there are other ways to guarantee second law satisfaction. In fact, a more complicated model is well known to fit data well, and will be studied later. For the present purposes, this na¨ ıve model will suffice. With this assumption, the second law reduces to − k ( − μ N 2 + 2 μ N ) 2 ≤ , k ≥ , (4.445) which is always true. Obviously, the reaction ceases when dζ/dt = 0, which holds only when 2 μ N = μ N 2 . (4.446) Away from equilibrium, for the reaction to go forward, one must expect...
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This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Park-sou during the Fall '11 term at FSU.

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Thermodynamics - 4.11 EQUILIBRIUM IN A TWO-COMPONENT SYSTEM 143 One can also eliminate to get n N in terms of n N 2 n N = 2 n N 2 | t =0 − n N

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