How can standard enthalpies of fusion and vaporization add up to that of sublimation when these occur at different temperatures?

Question

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

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:

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.

Answer 1

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.