[tjalfe89]

i often hear that the product of a reaction with the strongest bonds is the termodynamic and most stable product. fx. 1,4 addition to alpha-beta-unsatured carbonyl compunds, i heard that addition to the dobbelt bond is the termodynamic product because of the high bonding energy of the carbonyl group, u thereby keep. and fx. bonding between an enzyme and substrate lowers the energy of transition state and thereby the activation anergy. how can this match the equation for gibbs free energy: G = H -DeltaS ?? according to this, the product must have lower bonding energy and the reactants higher anergy bonds to get a negative DeltaG.. can someone explain for me please.

[jusy1]

Hi tjalfe89,

I think you're making confusion with the relation between strongest bonds and higher energy. Forgetting about the entropy effect, when the most stable product forms in a chemical reaction, the newly formed bonds will be stronger than the reactants's, but that doesn't mean they will be higher in energy. What you can say is that you will need more energy to break those bonds (I think maybe this is what was making confusion), because they are more stable and therefore lower in energy (the change in enthalpy will be negative).

[tjalfe89]

so the more stable bonds have more negative entalpi and thereby the change in gibbs free energi will be negative (lower energi bonds) ?

[jusy1]

The products of the reaction, with more stable bonds than the reactants, will be lower in enthalpy (not necessarly negative) and yes the gibbs free energy change, not accouting here for the entropy effect, will be negative.

Because they are more stable you require more energy to break those bonds (high bond dissociation energy).