What is Gibbs free energy?

S: Δ G/T = Δ H/T - Δ S. Whaddya mean, look like entropy change? Δ S is q/T. "Entropy change (1) = Entropy change (2) – Entropy change (3)" BUT we’d better be a lot more specific and talk about what those three entropy changes really mean. S: Darn right. Divide by T and I admit everything in that Gibbs looks like entropy change. But that just confuses me. What happens to the fight between enthalpy and entropy if enthalpy turns into entropy? Do I have to learn another mysterious phys chem equation? P: No way, no mystery. Let’s give it the full court press – you’ll be amazed at how neat everything comes out (because now that "fight" between enthalpy and entropy will make sense). It'll give you a much better feel for entropy itself. To start, let’s think about a system in which a chemical reaction is occurring. There can be thermal energy transferred ("heat") from the system to its surroundings or vice versa. Well, let’s really think big by saying that nothing else is happening in the entire universe but the reaction in our system. Look at the entropy changes involved: ΔS(surroundings) + ΔS(system) (1) Now, if chemicals are mixed in the system (at constant T and, normally, constant P) and a reaction occurs, some thermal energy transfer ("heat"), q, takes place in the reaction. How much q? That's easy: q is the change in enthalpy; q = ΔH(system). (The sign of ΔH can be + or - , but we're just talking broadly and generally, so let's start simply with + ΔH.) However, from the viewpoint of the surroundings, the sign of ΔH changes when thermal energy is transferred from the system and becomes absorbed by the surroundings, i.e., a + ΔH(system) when transferred to the surroundings becomes - ΔH(surroundings) -- (and, of course, a -ΔH(system) when transferred out of the system becomes ΔH(surroundings). As a general equation, simply to express that change of sign, here's (2): -ΔH(surroundings) =ΔH(system) (2) What does this have to do with entropy? To answer that, let’s divide equation (2) by T: -ΔH/T(surroundings) = ΔH/T(system) (3) Then, since ΔS = q/T, and the only q in the surroundings right now is - ΔH, that means that - ΔH/T(surroundings) = - ΔS(surroundings). Therefore, inserting this result in equat. (3): (Assuming that the surroundings are far larger than the system, i.e., reversible conditions.) -ΔS(surroundings) = ΔH/T(system) (4) or, changing signs merely so we have a + ΔS to work with in a moment, ΔS(surroundings) = - ΔH/T(system) (5) Now, replacing DS(surroundings) in equat. (1) with -ΔH/T(system) as just justified by (5): ΔS(universe) = - ΔH/T(system) + ΔS(system) (6) ΔS(universe)?? Just to say that aloud seems like a really big mouthful -- and head-full! But remember that we started out originally by saying that the only reaction happening in the whole universe was the one in our constant T, constant P system. So any DS(universe) would be perfectly measured by what happens only in our system. Let's see. To put it in the most general terms, the energy change that occurred in the reaction in the system -- and which entropy measures by ΔS = q/T -- has been spread out, some remaining in the system as ΔS and some being transferred to the surroundings and identifiable as the ΔH/T(surroundings). (It can be used for work, but right now we're looking at it either as though it was all dissipated as unavailable thermal energy or as though the work itself had resulted in the same amount of waste energy, i.e., as ΔS(surroundings). in the universe, and because energy flow (thermal energy transfer, "heat") divided by T is entropy, ΔG/T(system) is equal to the total entropy change in the entire universe due to this reaction! ΔS(universe) = ΔG/T(system) ΔS(universe) = - ΔG/T(system) and equation (6) -- derived from that "way-back-there" equation (1) -- becomes - ΔG/T(system) = - ΔH/T(system) + ΔS(system

It's nothing but a quantity that defines spontaneity of a reaction. Say you have to push a stone up a hill...it is non spontaneous. But if you have to push it down it is spontaneous as it happens itself after initiation. Same goes a chemical reaction :) Answer mine? http://answers.yahoo.com/question/index?qid=20120314074732AAEnmgc

It's nothing but a quantity that defines spontaneity of a reaction. Say you have to push a stone up a hill...it is non spontaneous. But if you have to push it down it is spontaneous as it happens itself after initiation. Same goes a chemical reaction :) Answer mine? http://answers.yahoo.com/question/index?qid=20120314074732AAEnmgc