Calculating Free Energy Change
Calculating Free Energy Change
Free energy is a state function, so its value depends only on the conditions of the initial and final states of the system. A convenient and common approach to the calculation of free energy changes for physical and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes according to the following relation as demonstrated in the example above.
\(\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}\)
Example
Evaluation of ΔG° from ΔH° and ΔS°
Use standard enthalpy and entropy data from this appendix to calculate the standard free energy change for the vaporization of water at room temperature (298 K). What does the computed value for ΔG° say about the spontaneity of this process?Solution
The process of interest is the following:\({\text{H}}_{2}\text{O}(l)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}(g)\)
The standard change in free energy may be calculated using the following equation:
\(\text{Δ}{G}_{298}^{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}\)
From this appendix, here is the data:
| Substance | \(\text{Δ}{H}_{\text{f}}^{°}\text{(kJ/mol)}\) | \({S}_{298}^{°}\text{(J/K·mol)}\) |
|---|---|---|
| H2O(l) | −286.83 | 70.0 |
| H2O(g) | −241.82 | 188.8 |
Combining at 298 K:
\(\begin{array}{l}\text{Δ}H\text{°}=\text{Δ}{H}_{298}^{°}=\text{Δ}{H}_{\text{f}}^{°}({\text{H}}_{2}\text{O}(g))\phantom{\rule{0.2em}{0ex}}-\text{Δ}{H}_{\text{f}}^{°}({\text{H}}_{2}\text{O}(l))\\ =[\text{−241.82 kJ}-(\text{−286.83})]\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}=\text{44.01 kJ/mol}\end{array}\)
\(\begin{array}{c}\text{Δ}S\text{°}=\text{Δ}{S}_{298}^{°}={S}_{298}^{°}({\text{H}}_{2}\text{O}(g))\phantom{\rule{0.2em}{0ex}}-{S}_{298}^{°}({\text{H}}_{2}\text{O}(l))\\ =188.8\phantom{\rule{0.2em}{0ex}}\text{J/mol·K}-70.0\phantom{\rule{0.2em}{0ex}}\text{J/K}=118.8\phantom{\rule{0.2em}{0ex}}\text{J/mol·K}\end{array}\)
\(\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}\)
Converting everything into kJ and combining at 298 K:
\(\begin{array}{c}\text{Δ}{G}_{298}^{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}\\ \\ =\text{44.01 kJ/mol}-(\text{298 K}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}118.8\phantom{\rule{0.2em}{0ex}}\text{J/mol·K})\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1 kJ}}{\text{1000 J}}\end{array}\)
\(\text{44.01 kJ/mol}-\text{35.4 kJ/mol}=\text{9.6 kJ/mol}\)
At 298 K (25 °C) \(\text{Δ}{G}_{298}^{°}>0,\) and so boiling is nonspontaneous (not spontaneous).
The standard free energy change for a reaction may also be calculated from standard free energy of formation\((\text{Δ}{G}_{\text{f}}^{°}),\) values of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states.
Similar to the standard enthalpy of formation, \(\text{Δ}{G}_{\text{f}}^{°}\) is by definition zero for elemental substances under standard state conditions. The approach used to calculate \(\text{Δ}{G}^{°}\) for a reaction from \(\text{Δ}{G}_{\text{f}}^{°}\) values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction
\(m\text{A}+n\text{B}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}x\text{C}+y\text{D},\)
the standard free energy change at room temperature may be calculated as
\(\begin{array}{}\\ \\ \text{Δ}{G}_{298}^{°}=\text{Δ}G\text{°}=\sum \nu \text{Δ}{G}_{298}^{°}(\text{products})\phantom{\rule{0.2em}{0ex}}-\sum \nu \text{Δ}{G}_{298}^{°}(\text{reactants})\\ \\ =[x\text{Δ}{G}_{\text{f}}^{°}(\text{C})+y\text{Δ}{G}_{\text{f}}^{°}(\text{D})]\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}[m\text{Δ}{G}_{\text{f}}^{°}(\text{A})+n\text{Δ}{G}_{\text{f}}^{°}(\text{B})].\end{array}\)
Example
Calculation of \(\text{Δ}{G}_{298}^{°}\)
Consider the decomposition of yellow mercury(II) oxide.
\(\text{HgO}(s,\phantom{\rule{0.2em}{0ex}}\text{yellow})\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{Hg}(l)\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\cfrac{1}{2}{\text{O}}_{2}(g)\)
Calculate the standard free energy change at room temperature, \(\text{Δ}{G}_{298}^{°},\) using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions?
Solution
The required data are available in this appendix and are shown here.
| Compound | \(\text{Δ}{G}_{\text{f}}^{°}\phantom{\rule{0.2em}{0ex}}\text{(kJ/mol)}\) | \(\text{Δ}{H}_{\text{f}}^{°}\phantom{\rule{0.2em}{0ex}}\text{(kJ/mol)}\) | \({S}_{298}^{°}\phantom{\rule{0.2em}{0ex}}\text{(J/K·mol)}\) |
|---|---|---|---|
| HgO (s, yellow) | −58.43 | −90.46 | 71.13 |
| Hg(l) | 0 | 0 | 75.9 |
| O2(g) | 0 | 0 | 205.2 |
(a) Using free energies of formation:
\(\text{Δ}{G}_{298}^{°}=\sum \nu G{S}_{f}^{°}\text{(products)}\phantom{\rule{0.2em}{0ex}}-\sum \nu \text{Δ}{G}_{f}^{°}\text{(reactants)}\)
\(=[1\text{Δ}{G}_{f}^{°}\text{Hg}(l)+\phantom{\rule{0.2em}{0ex}}\cfrac{1}{2}\text{Δ}{G}_{f}^{°}{\text{O}}_{\text{2}}(g)]\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}1\text{Δ}{G}_{f}^{°}\text{HgO}(s,\phantom{\rule{0.2em}{0ex}}\text{yellow})\)
\(=[1\phantom{\rule{0.2em}{0ex}}\text{mol}\text{(0 kJ/mol)}+\phantom{\rule{0.2em}{0ex}}\cfrac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{mol(0 kJ/mol)}]\phantom{\rule{0.2em}{0ex}}-\text{1 mol(−58.43 kJ/mol)}=\text{58.43 kJ/mol}\)
(b) Using enthalpies and entropies of formation:
\(\text{Δ}{H}_{\text{298}}^{°}=\sum \nu \text{Δ}{H}_{\text{f}}^{°}\text{(products)}\phantom{\rule{0.2em}{0ex}}-\sum \nu \text{Δ}{H}_{\text{f}}^{°}\text{(reactants)}\)
\(=[1\text{Δ}{H}_{f}^{°}\text{Hg}(l)+\cfrac{1}{2}\text{Δ}{H}_{f}^{°}{\text{O}}_{2}(g)]\phantom{\rule{0.2em}{0ex}}-1\text{Δ}{H}_{f}^{°}\text{HgO}(s,\phantom{\rule{0.2em}{0ex}}\text{yellow})\)
\(=[\text{1 mol}(\text{0 kJ/mol})\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\cfrac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{mol}(\text{0 kJ/mol})]\phantom{\rule{0.2em}{0ex}}-\text{1 mol}(\text{−90.46 kJ/mol})=\text{90.46 kJ/mol}\)
\(\Delta {S}_{298}^{°}=\sum \text{ν}\text{Δ}{S}_{298}^{°}\text{(products)}-\sum \text{ν}\text{Δ}{S}_{298}^{°}\text{(reactants)}\)
\(=[1\text{Δ}{S}_{298}^{°}\text{Hg}(l)\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\cfrac{1}{2}\text{Δ}{S}_{298}^{°}{\text{O}}_{2}(g)]\phantom{\rule{0.2em}{0ex}}-1\text{Δ}{S}_{298}^{°}\text{HgO}(s,\phantom{\rule{0.2em}{0ex}}\text{yellow})\)
\(=[\text{1 mol}\phantom{\rule{0.2em}{0ex}}(\text{75.9 J/mol K})\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\cfrac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{mol}(\text{205.2 J/mol K})]\phantom{\rule{0.2em}{0ex}}-\text{1 mol}(\text{71.13 J/mol K})=\text{107.4 J/mol K}\)
\(\text{Δ}G\text{°}=\text{Δ}H\text{°}-T\text{Δ}S\text{°}=\text{90.46 kJ}-\text{298.15 K}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{107.4 J/K·mol}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1 kJ}}{\text{1000 J}}\)
\(\text{Δ}G\text{°}=(90.46-32.01)\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}=\text{58.45 kJ/mol}\)
Both ways to calculate the standard free energy change at 25 °C give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous (not spontaneous) at room temperature.
This lesson is part of:
Thermodynamics