Why do we say that at boiling point liquid and vapour exist in equilibrium?

Question

I understand that at boiling point, vapour pressure becomes equal to the external pressure. But in my textbook it is written that at boiling point liquid and vapour exist in equilibrium. What does it mean by 'vapour and liquid exist in equilibrium' and also why do they exist in equilibrium?

Answer 1

In a closed container at a fixed temperature, the rate at which molecules leave the liquid is equal to the rate at which they return. Boiling in a liquid occurs when the temperature at the bottom allows bubbles to form and grow. The vapor pressure in the bubble must equal that of the surrounding liquid. The bubbles are almost like closed containers.

Answer 2

The key ingredient missing from the other discussion here is the concept of chemical equilibrium. You are entirely correct that the boiling temperature is the point at which the vapor pressure of the liquid is equal to the surrounding ambient pressure. But this can also easily be framed as a chemical equilibrium problem. The equilibrium reaction of interest is the conversion of the liquid phase to the vapor phase, \begin{gather} A_{(l)} \leftrightarrow A_{(g)} \end{gather} which has the equilibrium constant $K_{eq} = P_A$ since the liquid phase is pure and does not appear in the equilibrium constant. As long as the condensed phase is present, this equilibrium must remain established, at least in the immediate vicinity of the liquid. If we use the well-known formula, \begin{align} \Delta G^\circ &= -RT \ln (K_{eq}) \\ \implies P_A &= e^{-\frac{\Delta G^\circ_{vap}}{RT}} \end{align} then the vapor pressure of substance $A$ directly above the liquid is related to the Gibbs free energy change for the vaporization reaction. Thermochemical data allows us to determine the boiling point $T_{BP}$ directly by rearranging the equation above along with $\Delta G = \Delta H - T\Delta S$ and $P_A = 1$ to find, \begin{gather} T_{BP} = \frac{\Delta H^\circ _{vap}}{\Delta S^\circ _{vap}} \end{gather} Plugging in the experimental values of the enthalpy and entropy of vaporization gives the boiling point. Note that the reason the vapor pressure cannot exceed the ambient pressure is that the the surroundings would immediately act to equilibrate a pressure gradient.

As an example of using this approach, even if we assume that $\Delta H^\circ _{vap}$ and $\Delta S^\circ _{vap}$ are temperature independent and use the following data from the CRC Handbook of Chemistry and Physics,

	$H_2O_{(l)}$	$H_2O_{(g)}$
$\Delta \bar{H}_f^\circ$ (kJ/mol)	-285.83	-241.83
$\bar{S}^\circ$ (J/mol$\cdot$K)	69.95	188.83

we obtain an estimate of the boiling point of 97 $^\circ$C, which is still rather close to the true value of around 100 $^\circ$C. Using the dependence of the thermochemical data on temperature makes the estimate spot on to the accepted value. There is sufficient data in the CRC Handbook to do just that, but doing so here would bog down the discussion.

Answer 3

Vapor and liquid can coexist in a range of temperatures, usually (under standard pressure 1atm), in range $273.15-373.15$K. When increasing temperature mentally (moving horizontally to the right on the phase diagram, while keeping pressure the same), the temperature of boiling is the last temperature at which stable coexistence is possible. One can get unstable coexistence even for higher temperatures, if bubble creation in the liquid is hampered, but this is an unstable state, and introduction of small dust will turn it into violent boiling.

Answer 4

What does it mean by 'vapour and liquid exist in equilibrium' and also why do they exist in equilibrium?

They exist in a phase equilibrium in which the boiling point is equal to the condensation point. So the rate water condenses is equal to the rate water evaporates in a closed container. At 1 atm this is 100°C

Answer 5

The boiling point of a liquid is the point till which the liquid and gaseous phases can coexist. A dynamic equilibrium is created. At higher temperatures, only one phase dominates.

The rate at which the liquid vaporizes is equal to the rate at which the gas liquefies. At equilibrium vapor pressure remains constant and is equal to the atmospheric pressure.