Heat, unlike light, is an indiscriminate agent of chemical and physical change. It feeds into all available molecular states with no preference because it is basically just a collective manifestation of a body of vibrating, rotating molecules moving with a distribution of speeds and bumping into each other with high frequency. If two reaction pathways have exactly the same energetic barrier to get to their respective products, then they will occur with equal rates. (Well, not really - there are other things that impact kinetics of a reaction. But if we pretend the rate only depends in enthalpy, and we assume both reactions are mostly irreversible...)
I mean, consider some hypothetical molecule A-B-C, which can either decompose to form AB + C or A + BC, and the "strength" of the bonds A-B and B-C are identical. Heat - transfer of kinetic or vibratory motions - is just as likely to feed energy into the A-B bond as it is the B-C bond, and because these bonds take exactly the same amount of energy to break, the decomposition of A-B-C would tend to yield equivalent amounts (all things being equal) of AB + C and A + BC products. If one bond is more stable than the other, the reaction which requires breaking the less stable bond will be favored. That doesn't mean you won't see ANY of the less favored product. All these concepts are subject to statistical distributions of energies, and all molecular processes are, at their heart, a matter of probability.
(The A-B-C reactions described above are actually very far from reality - in any molecule like A-B-C, the vibration of A-B and B-C are not independent of each other, so this kind of dichotomous reaction mechanism would be virtually impossible. In other words, any heat that goes into A-B actually goes into B-C as well, and vice-versa. Heat goes into vibrational modes, which involve all atoms/bonds in a molecule, not into individual vibrating bonds. In A-B-C, you'd have some kind of symmetric vibrational mode, an asymmetric mode and a bending mode, for example. But the point stands - in real reaction dynamics, every chemical reaction proceeds in some way through one (or more) unique vibrational modes. Feeding energy into other vibrational modes cannot yield the reaction (although they could yield other reactions), but heat is fed into all vibrational modes, so it's a pretty inefficient way to make reactions go. Light, on the other hand, can be used to excite one vibrational mode without exciting others, so you can in principle drive a chemical reaction with light a lot more efficiently than you can with heat, because it provides a way to selectively put energy into reactive vibrational modes. )
Intermolecular bonds are significantly weaker than covalent bonds, but they operate under the same kinds of rules. A molecule like water has a high specific heat capacity (takes a lot of energy to heat it up) because there are lots of intermolecular bonds that can absorb that energy. This is also the primary reason why water has a pretty high boiling point despite its low mass. (Compare to methane - almost the same mass, far lower melting and boiling point.) In physical chemistry lingo we call this factor - the amount of bonds (actually vibrational modes) and such that can absorb energy - "degrees of freedom".
Having said all that, you have to realize that the chemical decomposition of sucrose and the melting of sucrose are two nominally independent processes that happen to require around the same amount of heat to make happen. The bonds between atoms in sucrose are strong, but so is the crystal lattice energy. The reason you can't just melt sucrose and then cool it down and have sucrose again (unlike, say, water) is because while that much heat is sufficient to break down the crystal lattice of sucrose to form a liquid, it's also enough heat to excite internal sucrose vibrational modes to the extent that some of them break as well. However I'll point out that as I kind of did before that the favorability of a reaction is not only dependent on having enough energy to break bonds - it's also on the stability of the products formed. You may very well have enough energy to break bonds, but if the products aren't particularly stable, that same heat can get them to react back the other way almost instantly, which is basically just a way of saying that the equilibrium favors the reactants. In the case of sucrose, though, the products of caramelization are probably various polymers and so forth which are pretty stable, so it makes sense that the reaction is favorable. And thank the heavens for that, else we'd not have yummy caramel to put on top of our ice cream.
Anyway, that was kind of a mess of a response which threw around a lot of concepts in no particular order. I hope I answered your question without confusing you too much.
ps. Not a chef, but I like to cook. There's an incredible amount of chemistry that goes on in cooking, and virtually everyone loves food, so I always view cooking examples as a great way to engage students and teach them what would otherwise be boring chemical concepts.
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Topic: Calcium Oxide Decomposition (Read 43162 times)