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Triple point defines a situation of simultaneous equilibrium between the solid, the liquid and the gas phases.
For such an equilibrium, you simply write:
$$ \Delta S= \frac{\Delta H}{T} $$
that rises from the fact that $\Delta G=0$.
Considering water molar ($m$) enthalpies for each phase transition at $273\,K$:
$$ \Delta H_{melting,m}=6.01\,kJ\,mol^{-1} $$ $$ \Delta H_{vaporisation,m}=45.05\,kJ\,mol^{-1} $$ $$ \Delta H_{sublimation,m}=51.06\,kJ\,mol^{-1} $$
one realises that
In this respect the triple point has no peculiar behaviour compared to other points where a two-phase equilibrium is established.
$$\Delta S= \frac{\Delta H}{T}$$
$$\Delta H_{melting,m}=6.01\,kJ\,mol^{-1}$$$$\Delta H_{vaporisation,m}=45.05\,kJ\,mol^{-1}$$$$\Delta H_{sublimation,m}=51.06\,kJ\,mol^{-1}$$
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2026-06-19
Triple point defines a situation of simultaneous equilibrium between the solid, the liquid and the gas phases.
For such an equilibrium, you simply write:
$$ \Delta S= \frac{\Delta H}{T} $$
that rises from the fact that $\Delta G=0$.
Considering water molar ($m$) enthalpies for each phase transition at $273\,K$:
$$ \Delta H_{melting,m}=6.01\,kJ\,mol^{-1} $$ $$ \Delta H_{vaporisation,m}=45.05\,kJ\,mol^{-1} $$ $$ \Delta H_{sublimation,m}=51.06\,kJ\,mol^{-1} $$
one realises that
In this respect the triple point has no peculiar behaviour compared to other points where a two-phase equilibrium is established.
Triple point defines a situation of simultaneous equilibrium between the solid, the liquid and the gas phases.
For such an equilibrium, you simply write:
$$\Delta S= \frac{\Delta H}{T}$$
that rises from the fact that $\Delta G=0$.
Considering water molar ($m$) enthalpies for each phase transition at $273\,K$:
$$\Delta H_{melting,m}=6.01\,kJ\,mol^{-1}$$$$\Delta H_{vaporisation,m}=45.05\,kJ\,mol^{-1}$$$$\Delta H_{sublimation,m}=51.06\,kJ\,mol^{-1}$$
one realises that
In this respect the triple point has no peculiar behaviour compared to other points where a two-phase equilibrium is established.
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VOTE