20.7: Changes of State and Free Energy
- Page ID
- 53927
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Energy in a body of water can be gained or lost, dependent 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 \(0^\text{o} \text{C}\), so \(\Delta G^\text{o}\) is equal to 0 at that temperature. The heat of fusion of water is known to be equal to \(6.01 \: \text{kJ/mol}\), and so the Gibbs free energy equation can be solved for the entropy change that occurs during the melting of ice. The symbol \(\Delta S_\text{fus}\) represents the entropy change during the melting process, while \(T_\text{f}\) is the freezing point of water.
\[\begin{align*} \Delta G &= 0 = \Delta H - T \Delta S \\ \Delta S_\text{fus} &= \frac{\Delta H_\text{fus}}{T_\text{f}} = \frac{6.01 \: \text{kJ/mol}}{273 \: \text{K}} = 0.0220 \: \text{kJ/K} \cdot \text{mol} = 22.0 \: \text{J/K} \cdot \text{mol} \end{align*}\nonumber \]
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 \(\Delta S\) would be the same, but the sign would be reversed, since the phase changes indicates 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 use the molar heat of vaporization. This value is \(40.79 \: \text{kJ/mol}\). The \(\Delta S_\text{vap}\) is as follows:
\[\Delta S = \frac{40.79 \: \text{kJ/mol}}{373 \: \text{K}} = 0.1094 \: \text{kJ/K} \cdot \text{mol} = 109.4 \: \text{J/K} \cdot \text{mol}\nonumber \]
The value is positive, reflecting the increase in disorder going from liquid to vapor. Condensation from vapor to liquid would give a negative value for \(\Delta S\).