<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Skip Navigation
Our Terms of Use (click here to view) have changed. By continuing to use this site, you are agreeing to our new Terms of Use.

20.7: Changes of State and Free Energy

Difficulty Level: At Grade Created by: CK-12
Atoms Practice
Estimated10 minsto complete
Practice Changes of State and Free Energy
This indicates how strong in your memory this concept is
Estimated10 minsto complete
Estimated10 minsto complete
Practice Now
This indicates how strong in your memory this concept is
Turn In

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

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 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 ΔSfus represents the entropy change during the melting process, while Tf is the freezing point of water.

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 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 ΔSvap would then be as follows:

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 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 ΔH?
  2. Why is temperature selection important?
  3. Why would the entropy of vaporization be so much larger than the entropy of fusion?

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

Image Attributions

Show Hide Details
Difficulty Level:
At Grade
Date Created:
Jun 25, 2013
Last Modified:
Sep 11, 2016
Save or share your relevant files like activites, homework and worksheet.
To add resources, you must be the owner of the Modality. Click Customize to make your own copy.
Please wait...
Please wait...
Image Detail
Sizes: Medium | Original