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Name: Ahmed
Status: student
Age: 18
Location: N/A
Country: N/A
Date: 2000

When two molecules of water ionizes to give an hydroxide ion and a oxonium cation, it is one O-H bond that is broken as H2O becomes a H^+ and an OH^- and the same O-H bond is formed as the H^+ combines with the other H2O to form H3O^+ as free H^+ protons can't exist. According to this the enthalpy for the ionization of water should be zero as enthalpy of bond breaking is same as the enthalpy of bond formation. But it a well understood fact that the ionization of water is endothermic. Why?

The problem lies in the assumption that the "enthalpy of bond breaking is same as the enthalpy of bond formation." This is only true if it is the SAME BOND being broken or formed. (Actually, it is strictly NEVER true-I think you meant to say that the enthalpy of bond formation is the negative of the enthalpy of bond breaking.) Enthalpy is a state function - enthalpy differences between the same states are always identical. In the case you present, the enthalpy differences are between the states OH- and H2O in one case, and between H2O and H3O+ in the other case. In each case, you are simply adding a proton to the first species to produce the second one. However, the initial and final states are different, so there is no reason to suspect that the enthalpy changes are the same. In fact, based on the principle that "things are either the same or they are different," we are forced to predict that the two protonation processes will have different reaction enthalpies.

Richard E. Barrans Jr., Ph.D.
Assistant Director
PG Research Foundation, Darien, Illinois

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