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Thermodynamics filled in class notes_Part_68

# Thermodynamics filled in class notes_Part_68 - 4.11...

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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 0 , (4.440) μ N 2 ( ) + μ N (2 ) 0 , (4.441) ( μ N 2 + 2 μ N ) 0 (4.442) ( μ N 2 + 2 μ N ) dt 0 . (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 dt = k ( μ N 2 + 2 μ N ) , k 0 , 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 0 , k 0 , (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 dζ/dt> 0, and then one must have μ N 2 + 2 μ N 0 , (4.447) 2 μ N

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