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In Atkins' Physical Chemistry, the following diagram appears in the chapter on thermochemistry:

Standard enthalpies of fusion, vaporization, and sublimation

The definition of standard enthalpy change is also given:

The standard state of a substance at a specified temperature is its pure form at 1 bar.

My doubt is regarding the temperature at which these changes occur. Say the fusion occurs at the melting point, vaporization at the boiling point. According to the definition of standard enthalpy change, these need to occur at a fixed temperature. How can we add the standard enthalpies associated with the two to obtain the standard enthalpy of sublimation, which occurs at a different temperature and may not be possible at the standard $1\pu{bar}$ pressure? Shouldn't the heats associated with changing the temperature of the system be included in the equation?

The phase diagram of a general solid-liquid-gas may be represented by the following:

general solid-liquid-gas phase diagram

According to the diagram, solid, liquid, and gas phases can only exist together at the triple point, leading me to the conclusion that the equation $\Delta_\text{fus}H^\ominus+\Delta_\text{vap}H^\ominus=\Delta_\text{sub}H^\ominus$ is only valid at the triple point.

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    $\begingroup$ (...)''can only exist together'' in equilibrium ''at the triple point'', but who says there's got to be equilibrium? $\endgroup$
    – Mithoron
    Nov 11, 2023 at 1:59
  • $\begingroup$ @Mithoron thank you for the comment. Say the fusion occurs at the melting point, vaporization at the boiling point, and sublimation, which may not be possible at standard pressure, occurs at a different temperature. According to the definition of standard enthalpy change, these need to occur at a fixed temperature. Shouldn't the heats associated with changing the temperature of the system be included in the equation? $\endgroup$
    – ananta
    Nov 11, 2023 at 2:07
  • $\begingroup$ Say the standard temperature were closen as the triple point. Would you be comfortable with the explanation then? $\endgroup$ Nov 11, 2023 at 11:50
  • $\begingroup$ @ChetMiller I think changing the definitions set by IUPAC is just bypassing the problem and not really solving it. $\endgroup$
    – ananta
    Nov 12, 2023 at 4:31
  • $\begingroup$ That is not what I asked. $\endgroup$ Nov 12, 2023 at 11:52

1 Answer 1

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You can certainly have liquid and vapor at equilibrium at temperatures other than the atmospheric boiling point. And you can certainly have solid and vapor at equilibrium at temperatures less than the triple point. In establishing the standard enthalpy change for vaporization at a temperature less than the atmospheric boiling point (say 25 C), the vapor state is hypothetical, and an extrapolation of ideal gas behavior from the equilibrium vapor pressure to 1 bar. Something similar (extrapolation) is done with the heat of fusion to estimate the heat of fusion at 1 bar.

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  • $\begingroup$ According to your answer, the formula adding standard enthalpies of fusion and vaporization to that of sublimation is an approximation derived from extrapolation. However, as far as I know, this formula is exact. I think this requires further explanation. Do you know where I can read about all the extrapolation? $\endgroup$
    – ananta
    Nov 12, 2023 at 0:15
  • $\begingroup$ Do you think that water vapor at 25 C and 1 atm. is a thermodynamic equilibrium state of water? See Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics. $\endgroup$ Nov 12, 2023 at 11:51
  • $\begingroup$ Yes, it is definitely in equilibrium with liquid water, and thanks for the reference. $\endgroup$
    – ananta
    Nov 12, 2023 at 16:44

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