LemonyCricket wrote:∆G = ∆H – T∆S
This equation tells you one thing: Among two possible states, which one is thermodynamically favored at the current temperature?
To answer this question you look first at the 'internal' energy of the two states (∆H). If more energy gets released when the bonds in state A are formed than in state B, A is energetically favorable, for example.
Second, you look at the entropy: If the system has more degrees of freedom in state A, for example because state A are more molecules than state B, or because the molecule in state B has lots of interaction surface and thus adsorbs lots of water molecules whose movement is consequently limited, A has higher entropy than B. To get an energy from the entropy, you multiply with the current temperature. This temperature is unaffected by whatever energy is liberated/used from forming the bonds, because what we're doing here is equilibrium thermodynamics, where only the end result counts, and not you get there (or whether you can get there, for example, if the molecules of state A are in different bottles, they can never form B, even though B might be thermodynamically favorable).
The interesting case is if ∆H and ∆S have the same sign, i.e. one is energetically favorable, and the other state is entropically favorable. For example, polymerization of tubulin into microtubules is good for entropy, because the tubulin 'binds' lots of water that gets liberated when the polymer is formed. However the polymer is not very favorable enrgetically. At low temperatures (4C), the T∆S term is comparatively small, and thus, tubulin is in monomer form. The more you raise the temperature, the more microtubules are formed (until you go too high and the proteins fall apart).