[Burningkrome]

A tutorial says:
"1. ΔH < 0, ΔS > 0:

This is an exothermic reaction with an increase in entropy. Thus ΔS_univ > 0 and ΔG < 0. The Second Law of Thermodynamics says that such a reaction is product-favored, so a ΔG less than zero also means a product-favored reaction.

Example:     2CH₄(g) + 3O₂(g)    2CO₂(g) + 2H₂O(g)"

Why would this reaction have a POSITIVE ΔS? It intuitively seems like it is getting more ordered (I.e. 5 molecules recombine to form 4 molecules.)

Can someone help me gain a better understanding of what constitutes "more disorder"?

Thanks! :-)

[Dan]

There are a number of sources of confusion here. In my opinion, the material you quoted is poorly constructed:

1. The reaction equation is incorrect (not balanced).
2. ΔS for the combustion of methane is actually negative - you can calculate it from the molar entropies of the reactants and products (but balance the chemical equation first)
3. ΔS_univ (= -ΔG/T) is not the same as ΔS (entropy change of the reaction/system, a.k.a ΔS_sys).

ΔS_univ = ΔS - ΔH/T

I don't really understand what point the tutorial is trying to make... perhaps that if a reaction has ΔH < 0; ΔS > 0, then ΔS_univ is always > 0 and "product-favoured"? That is fine, but combustion of methane is not an example of such a reaction.

Some links:

http://digipac.ca/chemical/mtom/contents/chapter5/chap5_4.htm
http://en.wikipedia.org/wiki/Gibbs_free_energy#Free_energy_of_reactions
http://2ndlaw.oxy.edu/gibbs.html

[Burningkrome]

Thanks for the better links. As for the tutorial...you know, it were on the internets, so it mustá been true ;-)