Final answer:
The hydrolysis of phosphoenolpyruvate (PEP) is the most thermodynamically favorable due to its very high-energy phosphate bond, which releases significant free energy when cleaved by specific enzymes such as pyruvate kinase.
Step-by-step explanation:
The hydrolysis of a phosphate from phosphoenolpyruvate (PEP) is the most thermodynamically favorable compared to ATP, ADP, glucose, or sucrose. This is due to PEP having a very high-energy phosphate bond that, when hydrolyzed by an enzyme such as pyruvate kinase, releases a significant amount of free energy. The enzyme catalyzes the transfer of a phosphate from PEP to ADP, producing ATP in a biologically irreversible, exergonic reaction. In contrast, when a phosphate group is cleaved from ATP to form ADP, the reaction releases over 7 kcal/mol of energy. However, the hydrolysis of PEP is considered to have an even greater Gibbs free energy change, making it more energetically favorable.
It is important to note that substrate-level phosphorylation is a process where the energy released from compounds such as 1,3 diphosphoglyceric acid (1,3 diPG) during their hydrolysis is used to generate ATP from ADP. Both 1,3 diPG and PEP are described as very high-energy phosphate compounds in this context, driving the synthesis of ATP in metabolic pathways such as glycolysis.