This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 4.11. EQUILIBRIUM IN A TWO-COMPONENT SYSTEM 143 One can also eliminate Î¶ to get n N in terms of n N 2 : n N = 2 ( n N 2 | t =0 âˆ’ n N 2 ) . (4.439) Now for the reaction, one must have, for second law satisfaction, that Î¼ N 2 dn N 2 + Î¼ N dn N â‰¤ , (4.440) Î¼ N 2 ( âˆ’ dÎ¶ ) + Î¼ N (2 dÎ¶ ) â‰¤ , (4.441) ( âˆ’ Î¼ N 2 + 2 Î¼ N ) dÎ¶ â‰¤ (4.442) ( âˆ’ Î¼ N 2 + 2 Î¼ N ) dÎ¶ dt â‰¤ . (4.443) In order to satisfy the second law, one can usefully, but naÂ¨ Ä±vely, hypothesize that the non-equilibrium reaction kinetics are given by dÎ¶ dt = âˆ’ k ( âˆ’ Î¼ N 2 + 2 Î¼ N ) , k â‰¥ , convenient, but naÂ¨ Ä±ve 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 naÂ¨ Ä±ve model will suffice. With this assumption, the second law reduces to âˆ’ k ( âˆ’ Î¼ N 2 + 2 Î¼ N ) 2 â‰¤ , k â‰¥ , (4.445) which is always true. Obviously, the reaction ceases when dÎ¶/dt = 0, which holds only when 2 Î¼ N = Î¼ N 2 . (4.446) Away from equilibrium, for the reaction to go forward, one must expect...
View Full Document
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.
- Fall '11