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, ΔG°, according to the following relation.
$$ΔG^°=ΔH^°-TΔS^°$$Using Standard Enthalpy and Entropy Changes to Calculate ΔG°Use standard enthalpy and entropy data from the Appendices 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:
The standard change in free energy may be calculated using the following equation:
$$ΔG^°=ΔH^°-TΔS^°$$From the Appendix:
| Substance | $ΔH_f^°$ (kJ/mol) | $S°$ (J/K·mol) |
|---|---|---|
| H2O(l) | −286.83 | 70.0 |
| H2O(g) | −241.82 | 188.8 |
Using the appendix data to calculate the standard enthalpy and entropy changes yields:
$$ΔH^°=ΔH_f^°\;(H_2O(g))-ΔH_f^°\;(H_2O(l))$$ $$ΔH^°=[-241.82\;kJ/mol-(-286.83\;kJ/mol)]=45.01 kJ$$ $$ΔS^°=1\;mol\times S^°\;(H_2O(g))-1\;mol\times S^°\;(H_2O(l))$$ $$ΔS^°=(1\;mol)188.8\;J/mol·K-(1\;mol)70.0\;J/mol·K=118.8\;J/mol·K$$Substitution into the standard free energy equation yields:
$$ΔG^°=ΔH^°-TΔS^°$$ $$ΔG^°=45.01\;kJ-(298\;K\times 118.8\;J/K)\times \frac{1\;kJ}{1000\;J}$$ $$ΔG^°=45.01\;kJ-35.4\;kJ=9.6\;kJ$$At 298 K (25 °C) $ΔG^°>0$, so boiling is nonspontaneous (not spontaneous).
Check Your Learning
Use standard enthalpy and entropy data from Appendix to calculate the standard free energy change for the reaction shown here (298 K). What does the computed value for ΔG° say about the spontaneity of this process?
Answer:
$ΔG^°=102.0\;kJ/mol$; the reaction is nonspontaneous (not spontaneous) at 25 °C.
The standard free energy change for a reaction may also be calculated from standard free energy of formation ΔGf° 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,$ΔG_f^°$ is by definition zero for elemental substances in their standard states. The approach used to calculate $ΔG^°$ for a reaction from $ΔG_f^°$ values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction
$$m\text{A}+n\text{B}⟶x\text{C}+y\text{D}$$the standard free energy change at room temperature may be calculated as
$$ΔG^°=∑νΔG^°\text{(products)}-∑νΔG^°\text{(reactants)}$$ $$[xΔG_f^°\text{(C)}+yΔG_f^°\text{(D)}]-[mΔG_f^°\text{(A)}+nΔG_f^°\text{(B)}]$$Using Standard Free Energies of Formation to Calculate ΔG°
Consider the decomposition of yellow mercury(II) oxide.
Calculate the standard free energy change at room temperature, $ΔG^°$, 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 Appendix and are shown here.
| Compound | $ΔG_f^°$ (kJ/mol) | $ΔH_f^°$ (kJ/mol) | $S^°$ (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:
$$ΔG^°=∑νΔG_f^°\text{(products)}-∑νΔG_f^°\text{(reactants)}$$ $$=[1ΔG_f^°\text{Hg}(l)+\frac{1}{2}ΔG_f^°O_2(g)]-1ΔG_f^°HgO(s\text{, yellow})$$ $$=[1\;mol(0\;kJ/mol)+\frac{1}{2}\;mol(0\;kJ/mol)]-1\;mol(-58.43\;kJ/mol)=58.43\;kJ/mol$$(b) Using enthalpies and entropies of formation:
$$ΔH^°=∑νΔH_f^°\text{(products)}-∑νΔH_f^°\text{(reactants)}$$ $$=[1ΔH_f^°\text{Hg}(l)+\frac{1}{2}ΔH_f^°O_2(g)]-1ΔH_f^°HgO(s\text{, yellow})$$ $$=[1\;mol(0\;kJ/mol)+\frac{1}{2}\;mol(0\;kJ/mol)]-1\;mol(-90.46\;kJ/mol)=90.46\;kJ/mol$$ $$ΔS^°=∑νΔS^°\text{(products)}-∑νΔS^°\text{(reactants)}$$ $$=[1ΔS^°\text{Hg}(l)+\frac{1}{2}ΔS^°O_2(g)]-1ΔS^°HgO(s\text{, yellow})$$ $$=[1\;mol(75.9\;J/mol·K)+\frac{1}{2}\;mol(205.2\;J/mol·K)]-1\;mol(71.13\;J/mol·K)=107.4\;J/mol·K$$ $$ΔG^°=ΔH^°-TΔS^°=90.46\;kJ-(298.15\;K\times 107.4\;J/mol·K)\times \frac{1\;kJ}{1000\;J}$$ $$ΔG^°=(90.46-32.01)\;kJ/mol=58.45\;kJ/mol$$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.
Check Your Learning
Calculate ΔG° using (a) free energies of formation and (b) enthalpies of formation and entropies (Appendix). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?
Answer:
(a) 140.8 kJ/mol, nonspontaneous
(b) 141.5 kJ/mol, nonspontaneous