# Is water committed to vapor state at boiling point or after? [duplicate]

Quite early into my studies, I learnt about phase changes with this graph.

I’m just going to focus on vaporization/condensation. I learnt that the plateau was caused because all the heat was going into breaking the bonds between the molecules, freeing them from a liquid state to a gaseous state. Thus, no energy was going into raising the kinetic energy of the substance or the temperature.

Later, I learnt about thermodynamics and Gibbs free energy. Spontaneity is dependent on temperature—some processes are spontaneous at some temperatures, and some are not. For a temperature above 100 °C, the boiling of water is spontaneous. For a temperature below 100 °C, the condensation of water is spontaneous. At 100 °C, both states coexist in equilibrium (i.e. $\Delta G = 0$). It is only when you raise/decrease the temperature ever so slightly that water is committed to either state.

I fail to reconcile this with what I learnt. At 100 °C, energy is going into breaking intermolecular bonds. That suggests to me that at 100 °C, water is being committed to vapor. According to the thermodynamic interpretation, at 100 °C, both liquid and gas coexist in equilibrium, and water is not committed to vapor until the temperature exceeds 100 °C.

Can anyone clear up my confusion?

## marked as duplicate by Klaus-Dieter Warzecha, ringo, Martin - マーチン♦Apr 8 '16 at 6:02

• @JonCuster Perhaps it would aid me to see these concepts put in terms of the equation $\Delta G = \Delta H - T\Delta S$. I’m not quite seeing it at the moment. – lightweaver Apr 7 '16 at 13:56
• Start with 100% liquid at 100C. End with 100% vapor at 100C. The free energy of the whole system remains constant. So, $\Delta G$ for the whole system has to equal zero. Yet you put enthalpy in. – Jon Custer Apr 7 '16 at 14:07