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

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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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• What is that "equilibrium" at 100ºC, really? I'll tell you what: it is when both states coexist. Say, you have 1kg of vapor and 10kg of liquid at 100ºC, and these are in equilibrium. You add some heat, it goes straight into breaking those bonds, so after a while you have 10kg of vapor and 1kg of liquid - and that's still an equilibrium. – Ivan Neretin Apr 7 '16 at 13:45
• While the free energies of water and vapor are the same at 100C, to convert liquid to vapor requires adding the enthalpy of the phase transition (latent heat). So, you have on phase with one (H,S) combination, and a second phase with a different (H,S) combination. – Jon Custer Apr 7 '16 at 13:46
• @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

## 1 Answer

At equilibrium, not only are molecular bonds of the liquid breaking to form vapor. There are also molecular bonds of the vapor reforming to produce liquid. At equilibrium, the rates of the two processes are equal. But, if you add heat to the liquid, the rate of breaking bonds exceeds to rate of reforming bonds (i.e., the temperature and vapor pressure of the liquid increases slightly), and some of the liquid changes to vapor. But once you stop adding heat, equilibrium is re-established.