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# 20.7: Changes of State and Free Energy

Difficulty Level: At Grade Created by: CK-12
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Practice Changes of State and Free Energy

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#### 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°\begin{align*}\Delta G°^\circ\end{align*} 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\begin{align*}\Delta S_{\text{fus}}\end{align*} represents the entropy change during the melting process, while Tf\begin{align*}T_{\text{f}}\end{align*} is the freezing point of water.

ΔGΔSfus=0=ΔHTΔS=ΔHfusTf=6.01 kJ/mol273 K=0.0220 kJ/Kmol=22.0 J/Kmol\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*}

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\begin{align*}\Delta S\end{align*} 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\begin{align*}\Delta S_{\text{vap}}\end{align*} would then be as follows:

ΔS=40.79 kJ/mol373 K=0.1094 kJ/Kmol=109.4 J/Kmol\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*}

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\begin{align*}\Delta S\end{align*}.

### Summary

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

### Review

1. What precautions need to be taken in selecting a value for ΔH\begin{align*}\Delta H\end{align*}?
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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