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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?

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  • $\begingroup$ I have edited your question so that it is clear that it refers to free energy of hydrolysis. Chemical compounds do not have energy in isolation, so the term "high energy" should be avoided. $\endgroup$ – David Mar 27 '17 at 11:04
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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.

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  • $\begingroup$ Thank you very much and also tomd for the adding comment! $\endgroup$ – idkfa Mar 28 '17 at 7:58

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