Theory — Entropy Change at Phase Transitions (Clausius view)
At the normal transition temperature \(T_{\mathrm{tr}}\) (for example, the normal
melting point or normal boiling point) a pure substance can undergo a phase change
reversibly at essentially constant temperature and 1 bar. Using Clausius’
definition \( \mathrm{d}S=\delta q_{\mathrm{rev}}/T \) and summing over the
isothermal, reversible step gives the key relation:
\[
\Delta_{\mathrm{tr}}S
\;=\;
\frac{q_{\mathrm{rev}}}{T_{\mathrm{tr}}}
\;=\;
\frac{\Delta_{\mathrm{tr}}H}{T_{\mathrm{tr}}}
\]
The heat exchanged for a reversible phase change equals the enthalpy change for that
transition, \(q_{\mathrm{rev}}=\Delta_{\mathrm{tr}}H\). All quantities above are
per mole of substance. This expression is valid when the transition is carried
out reversibly at the usual (normal) transition temperature so that the system remains in
equilibrium between the two phases throughout the change.
Rearrangements you may need
\[
\Delta_{\mathrm{tr}}H = \Delta_{\mathrm{tr}}S\,T_{\mathrm{tr}},
\qquad
T_{\mathrm{tr}} = \frac{\Delta_{\mathrm{tr}}H}{\Delta_{\mathrm{tr}}S}
\]
Typical signs and magnitudes
- Vaporization \((\ell \to g)\): \( \Delta_{\mathrm{vap}}S > 0 \).
For many non-associating liquids at their normal boiling point,
Trouton’s rule gives
\(\Delta_{\mathrm{vap}}S^\circ \approx 87\ \mathrm{J\,mol^{-1}\,K^{-1}}\).
Strongly hydrogen-bonded or highly ordered liquids (e.g., water, ethanol) often have
larger values.
- Fusion \((s \to \ell)\): \( \Delta_{\mathrm{fus}}S > 0 \) but
typically smaller than for vaporization. Example: ice at 273.15 K has
\(\Delta_{\mathrm{fus}}S^\circ \approx 22\ \mathrm{J\,mol^{-1}\,K^{-1}}\)
from \(\Delta_{\mathrm{fus}}H^\circ=6.02\ \mathrm{kJ\,mol^{-1}}\).
- Sublimation \((s \to g)\): \( \Delta_{\mathrm{sub}}S > 0 \), generally the largest
of the three since order decreases the most.
Worked examples
1) Vaporization of water at its normal boiling point.
With \(\Delta_{\mathrm{vap}}H^\circ = 40.7\ \mathrm{kJ\,mol^{-1}}\) at
\(T_{\mathrm{vap}} = 373\ \mathrm{K}\),
\[
\begin{aligned}
\Delta_{\mathrm{vap}}S^\circ &= \frac{40.7\times 10^{3}\ \mathrm{J\,mol^{-1}}}{373\ \mathrm{K}} \\[6pt]
&= 1.09\times10^{-1}\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \\[6pt]
&= 109\ \mathrm{J\,mol^{-1}\,K^{-1}} .
\end{aligned}
\]
2) Fusion of ice at 273.15 K.
With \(\Delta_{\mathrm{fus}}H^\circ = 6.02\ \mathrm{kJ\,mol^{-1}}\),
\[
\Delta_{\mathrm{fus}}S^\circ
= \frac{6.02\times 10^{3}}{273.15}
\approx 22.0\ \mathrm{J\,mol^{-1}\,K^{-1}} .
\]
Using the calculator
- Select the transition (\(\mathrm{vap}\), \(\mathrm{fus}\), \(\mathrm{sub}\), or custom).
- Choose which quantity to solve for: \(\Delta_{\mathrm{tr}}S\), \(\Delta_{\mathrm{tr}}H\), or \(T_{\mathrm{tr}}\).
The chosen unknown is disabled.
- Enter the other two quantities. The tool accepts
\(\Delta_{\mathrm{tr}}H\) in kJ·mol\(^{-1}\) or J·mol\(^{-1}\) and temperature in K or °C
(internally converted to K).
- Click Calculate to see the result along with step-by-step LaTeX working.
If the transition is vaporization, you can optionally compare against Trouton’s rule.
Common pitfalls
- Always use the transition temperature for the phase change (normal melting/boiling point for standard values).
- Keep units consistent: J·mol\(^{-1}\) for \(\Delta_{\mathrm{tr}}H\) and K for \(T_{\mathrm{tr}}\).
The calculator handles unit conversion but your inputs must be per mole.
- Check the sign: vaporization and fusion have \( \Delta S > 0 \) (more disorder); condensation and freezing have \( \Delta S < 0 \).
- The relation assumes the latent heat \( \Delta_{\mathrm{tr}}H \) is effectively constant at \(T_{\mathrm{tr}}\)
and the process is carried out reversibly at that temperature and 1 bar.
