I understand that this question is similar to a few that were asked before, but I read over them and had difficulty understanding a lot of the terminology. My question also incorporates dehydration: breaking bonds requires energy and forming bonds releases energy, so why does dehydration synthesis absorb energy and hydrolysis release energy? Does it have something to do with he strength of the bonds, entropy, high and low energy states, repulsion between atoms, etc.? Please try to make the answer sort of simple because I just started taking biochemistry this year. Thanks!
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There are many hydrolysis reactions, and so the answer depends on the specific reaction. However, you can see some common themes using ester hydrolysis.
The number of reactants and products is the same. The bonds broken and made are the same (an O-H bond is broken and one is made, and a C-O bond is broken and one is made). So there is no expectation of a large enthalpy or entropy of reaction. In fact, you can either make esters (in non-aqueous solvent, with removal of water) or hydrolyse them (in water, with acid or base as catalyst). In biochemical settings, the concentration of water is much higher than that of the other species, and that drives hydrolysis.
Not all hydrolysis reactions have a near-zero reaction enthalpy and entropy, and in those cases, there are additional factors favoring hydrolysis.
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$\begingroup$ Your example might be a bit confusing, since in aqueous non-acidic solution, the subsequent partial deprotonation of the propanoic acid changes the result and your statement that the bonds broken and made are the same. $\endgroup$– AndrewCommented Nov 8 at 22:26
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$\begingroup$ @Andrew Excellent point, and that would be another factor driving hydrolysis. And you provide a fix as well - look at it in a solution with pH of, say, 3. $\endgroup$– Karsten ♦Commented Nov 8 at 22:28