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Changes of State and Free Energy

Demonstrates calculations used to demonstrate entropy change at phase transition temperatures.

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Changes of State and Free Energy

An increase in the level of energy can cause a change in state

Credit: Jon Sullivan
Source: http://commons.wikimedia.org/wiki/File:Geysers_steam_boiling_Yellowstone.jpg
License: CC BY-NC 3.0

How are energy and changes of state related?

Energy in a body of water can be gained or lost depending on conditions.  When water is heated above a certain temperature steam is generated.  The increase in heat energy creates a higher level of disorder in the water molecules as they boil off and leave the liquid.

Changes of State and Free Energy

At the temperature at which a change of state occurs, the two states are in equilibrium with one another. For an ice-water system, equilibrium takes place at 0°C, so \begin{align*}\Delta G°^\circ\end{align*}ΔG° is equal to 0 at that temperature. The heat of fusion of water is known to be equal to 6.01 kJ/mol, and so the Gibbs free energy equation can be solved for the entropy change that occurs during the melting of ice. The symbol \begin{align*}\Delta S_{\text{fus}}\end{align*}ΔSfus represents the entropy change during the melting process, while \begin{align*}T_{\text{f}}\end{align*}Tf is the freezing point of water.

\begin{align*}\Delta G & = 0 = \Delta H - T \Delta S\\ \Delta S_{\text{fus}} & = \frac{\Delta H_{\text{fus}}}{T_{\text{f}}} = \frac{6.01 \text{ kJ/mol}}{273 \text{ K}} = 0.0220 \text{ kJ/K} \cdot \text{mol} = 22.0 \text{ J/K} \cdot \text{mol}\end{align*}ΔGΔSfus=0=ΔHTΔS=ΔHfusTf=6.01 kJ/mol273 K=0.0220 kJ/Kmol=22.0 J/Kmol

The entropy change is positive as the solid state changes into the liquid state. If the transition went from the liquid to the solid state, the numerical value for \begin{align*}\Delta S\end{align*}ΔS would be the same, but the sign would be reversed since we are going from a less ordered to a more ordered situation.

A similar calculation can be performed for the vaporization of liquid to gas. In this case we would use the molar heat of vaporization. This value would be 40.79 kJ/mol. The \begin{align*}\Delta S_{\text{vap}}\end{align*}ΔSvap would then be as follows:

\begin{align*}\Delta S = \frac{40.79 \text{ kJ/mol}}{373 \text{ K}} = 0.1094 \text{ kJ/K} \cdot \text{mol} = 109.4 \text{ J/K} \cdot \text{mol}\end{align*}ΔS=40.79 kJ/mol373 K=0.1094 kJ/Kmol=109.4 J/Kmol

The value is positive, again reflecting the increase in disorder going from liquid to vapor. Condensation from vapor to liquid would give a negative value for \begin{align*}\Delta S\end{align*}ΔS.


  • Calculations are shown for determining entropy changes at transition temperatures (ice → water or water → vapor and reverse).


  1. What precautions need to be taken in selecting a value for \begin{align*}\Delta H\end{align*}ΔH?
  2. Why is temperature selection important?
  3. Why would the entropy of vaporization be so much larger than the entropy of fusion?

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Image Attributions

  1. [1]^ Credit: Jon Sullivan; Source: http://commons.wikimedia.org/wiki/File:Geysers_steam_boiling_Yellowstone.jpg; License: CC BY-NC 3.0

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