Standard Free energy of hydrolysis of phosphoenolpyruvate (PEP) In ATP we have two anhydride bonds, each with a Standard Gibbs free energy of hydrolysis ~ -30 kJ / mol and one phosphoesterbond of ~ -20 kJ/ mol, if my notes are correct. In phosphoenolpyruvate (PEP) we also have a phosphoester bond, but the Standard Gibbs free energy of hydrolysis is three times higher, at ~ -62 kJ / mol. How does the double bond / enol form contribute to this high free energy of hydrolysis? ![enter image description here](

An answer to this is given in the online version of Berg _et al._

> Why does phosphoenolpyruvate have such a high phosphoryl-transfer potential? The phosphoryl group traps the molecule in its unstable enol form. When the phosphoryl group has been donated to ATP, the enol undergoes a conversion into the more stable ketone—namely, pyruvate.

![Hydrolysis of PEP](

> Thus, the high phosphoryl-transfer potential of phosphoenolpyruvate arises primarily from the large driving force of the subsequent enol-ketone conversion

Note that Berg _et al._ use the term “high phosphoryl-transfer potential” or “high phosphoryl group-transfer potential”, as explained in sections 14.1.4 and 14.1.5. These sections are worth reading, and this terminology is worth following.