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Matt Hanson
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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{gather} \Delta G^\circ = -RT \ln (K_{eq}) \\ \implies P_A = e^{-\frac{\Delta G^\circ_{vap}}{RT}} \end{gather}\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.

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{gather} \Delta G^\circ = -RT \ln (K_{eq}) \\ \implies P_A = e^{-\frac{\Delta G^\circ_{vap}}{RT}} \end{gather} 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.

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.

Source Link
Matt Hanson
  • 3.4k
  • 5
  • 14
  • 33

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{gather} \Delta G^\circ = -RT \ln (K_{eq}) \\ \implies P_A = e^{-\frac{\Delta G^\circ_{vap}}{RT}} \end{gather} 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.