My lecture notes state that

Condensation reactions, like all biosynthetic reactions, are endergonic.

I am unsure as to why this would be the case. I thought bond formation was an exothermic reaction, so you would get a negative enthalpy change from the bond being formed. I appreciate that you wold have a decrease in entropy when going from two small molecules to one larger molecule, but would this really counter-balance the negative enthalpy change every time? Perhaps the small molecule produced in the biosynthetic reaction, e.g. $\ce{H2O}$ for condensation reactions, is important here?

In a nutshell, I am just confused as to what ensures that all biosynthetic reactions (and condensation reactions) are always endergonic?


2 Answers 2


One of the most simple biosynthetic reactions that generates ‘oligomers’ from ‘monomers’ (between inverted commas because that description is not fully correct here) is the ester formation used to synthesise lipids. It can be broken down to:

$$\ce{R-O-H + H'-O'-C(=O)-R' -> R-O-C(=O)-R' + H'-O'-H}$$

With each important bond highlighted and the oxygen and hydrogen atoms made distinguishable by primes. A simplistic way of deriving standard reaction enthalpies is to use Hess’ law and add up the bond dissociation enthalpies of bonds that are broken and subtract that from the bond dissociation enthalpies of bonds that are formed. Doing that in this case gives:

  • Broken:

    • $\ce{C-O'}$
    • $\ce{O-H}$
  • Formed:

    • $\ce{C-O}$
    • $\ce{O'-H}$

Since the bonds being formed and those being broken are identical, at first approximation $\Delta_\mathrm{r} H^0 \approx 0$, meaning the reaction depends entirely on the entropic term.

On the entropic side of things, you are creating one huge, dissolved entity out of two not-as-huge dissolved entities. Making an exact call here is difficult, as it depends on a number of things, most notably how large the hydration sphere is for all three of these compounds. If each of them has a hydration sphere of approximately the same diameter, the product side will have more entropy since less water molecules are needed for an identical hydration sphere (akin to the hydrophobic effect). The opposite effect is also possible, though.

I would like to challenge the ‘all condensations are …’ It seems to simplistic to me. Many condensation reactions are probably endergonic but likewise I expect a few to be overall exergonic. The cell just takes no chances and uses an $\ce{ATP}$-dependent pathway in all cases to ensure the outcome it desires.


That all condensation reactions are endergonic is certainly not true. And as can be seen by the Helmholtz equation $\Delta G = \Delta H - T\cdot \Delta S$, this also depends on the employed temperature. One example is cupper(II)sulfate penta hydrate that spontaneously releases water upon heating, so there you have an exergonic condensation reaction. Forcing T higher will always favor a condensation reaction as typically $\Delta S_\text{condensation} > 0$ due to release of water.

That all (or most, anyway) biological condensation reactions are endergonic is hopefully the case, as otherwise biology would be unable to control the conversion by providing the energy to enable it. If there is an enzyme that converts A to B in an exergonic fashion, all of A will be converted to B eventually (one could argue that the activation barrier is high so that A to B conversion is prevented, but even when the activation barrier is too high for the process being spontaneous the cell would not be able to withdraw the enzyme so the enzyme still catalyzes the reaction in an uncontrolled fashion). But typically the cell will employ pyrophosphates (e.g. ATP) hydrolysis to drive and control those reactions. These will generally be endergonic, so that the cell can convert the endergonic process into an exergonic one by ATP hydrolysis and thereby control the process.

So, indeed my guess would be that most enzymes catalyze endergonic processes. However, there are also exergonic processes like the hydrogen peroxide hydrolysis catalyzed by the catalase enzyme. In that case, there is no need of the cell for A, so it gets converted to B as quickly as possible (and with equilibrium far on the side of B).

So my bet would be that most condensation reactions in the cell are ATP driven, and are thus endergonic.

  • $\begingroup$ $\text{exergonic}\ne\text{no activation energy}$. And the dehydration of copper(II) sulphate pentahydrate would not be a condensation in the strict sense. Your argument for $\Delta S$ is also not correct. $\endgroup$
    – Jan
    Dec 11, 2016 at 13:42
  • $\begingroup$ I know that exergonic $\neq$ no act en what has that to do with my answer? $\endgroup$ Dec 11, 2016 at 13:44
  • $\begingroup$ Because most processes in the cell are not spontaneous because they require activation; not because they are endergonic. $\endgroup$
    – Jan
    Dec 11, 2016 at 13:45
  • $\begingroup$ Any elementary step requires activation, its only a question how much. But yeah sure its not the question if it happens, but to what degree. E.g. catalase will convert all $\ce{H_2O_2}$ to $\ce{H_2O}$ and thats fine with the cell. However if e.g. peptide formation happens spontaneously in the cell, and all amino acids would convert completely to 'the product side of the reaction' it would be a big problem for the cell. So I guess that it is a design principle of the cell to employ endergonic reactions to prevent all monomers to completely react to the condensation product. $\endgroup$ Dec 11, 2016 at 13:49
  • $\begingroup$ So its not a kinetic but a thermodynamic argument... $\endgroup$ Dec 11, 2016 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.