Physical Chemistry Problem (Atkin's Physical Chemistry P7.36)?

Suppose that an iron catalyst at a particular manufacturing plant produces ammonia in the most cost-effective manner at 450 C when the pressure is such that deltaG for the reaction 1/2 N2 (g) + 3/2 H2 (g) --%26gt; NH3 (g) is equal to -500 Jmol^-1. (A) What is the pressure needed? (B) Now suppose that a new catalyst is developed that is most cost-effective at 400 C when the pressure gives the same value of deltaG. What pressure is needed when the new catalyst is used? What are the advantages of the new catalyst? Assume that (i) all gases are perfect gases or that (ii) all gases are van der Walls gases. Isotherms of deltaG(T,p) in the pressure range 100 atm %26lt; p %26lt; 400 atm are needed to derive the answer. (c) Do the isotherms you plotted confirm Le Chatelier's principle concerning the response of equilibrium changes in the temperature and pressure?

Physical Chemistry Problem (Atkin's Physical Chemistry P7.36)?
Which version of Physical Chemistry are you using? I don't seem to have Problem 7.36. I have Physical Chemistry, Atkins %26amp; DePaula, 7th Edition. Anyway....





Ugh, I thought PChem would be the death of me. I'll warn you it has been some time since I've taken the class, but I don't see any other answers on Yahoo and I'm not sure how many people on here know much about Pchem, so I'll give you my best shot.


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The reactions given are of course equilibria, and not linear reactions. You will probably need to use


[dG = -RTlnK]





(just take dG to mean Delta G). Since you're given dG and T and know R, you can find lnK and hence K. Now, they want you to solve for a pressure. In general, K is an equilibrium, where you would use K = [c/][a][c]. This is a GAS PHASE reaction, so instead of a conventional "concentration", you'll use partial pressures.





The equilibrium you want is [P_NH3] / [P_H][P_N2]. Since all of these are presumably at the same pressure, your equilibrium constant will presumably reduce to just one pressure, the one you want for your reaction.





The "new catalyst" will not give you any fundamentally new reaction, and it won'g change your delta G's or your delta H's or delta S's. What it WILL do, however, is at any given conditions, give you a MORE FAVORABLE EQUIBRIUM, or value for K.





At this point I'll have to remind you I've been out of college for about a year and haven't dealt with the likes of PChem in about 4 years, but I hope that should be enough info to get you started. Best of luck.