Thermodynamics

Free Energy

Free energy, which is the capacity of a system to do work, can be used to determine whether a reaction is spontaneous or nonspontaneous under given conditions.

The capacity of a system to do work, or use force to move an object, is its free energy. This is often expressed as Gibbs free energy, named after the American scientist Josiah Willard Gibbs, known for his work in thermodynamics in the late 1800s. Gibbs free energy (G) is the amount of work that can be done by a system, and it is expressed as G=HTSG=H-TS, where H is enthalpy in kilojoules (which is the total internal energy of the system plus the product of the pressure and volume), T is temperature, and S is entropy.

For any chemical reaction, the change in Gibbs free energy (ΔG\Delta G) can be calculated. Assuming that temperature is standard (25°C, which is 298.15 K), the change in Gibbs free energy is ΔG=ΔHTΔS\Delta G=\Delta H-T\Delta S, with T = 298.15 K. Importantly, if ΔG\Delta G is less than 0, then the reaction is spontaneous; it does not need energy input to proceed. If ΔG\Delta G is greater than 0, then the reaction is nonspontaneous and requires energy input to proceed.

Step-By-Step Example
Calculation of Change in Gibbs Free Energy
Calculate the change in Gibbs free energy at 25°C and 1 atm for the reaction N2(g)+3H2(g)2NH3(g){\rm {N}_{2}}(g)+3{\rm {H}_{2}}( g)\rightarrow2{\rm{NH}_{3}}(g), and determine whether the reaction is spontaneous or nonspontaneous.
Step 1
First, calculate the change in enthalpy (ΔH\Delta H), which is based on the heat of formation of each molecule (ΔHf\Delta H_{f}^\circ). Elements in their gaseous form have ΔHf=0.\Delta H_{f}^\circ=0. NH3(g){\rm{NH}}_{3}(g) has ΔHf=45.9kJ\Delta H_{f}^\circ=-45.9\;{\rm{kJ}}. Subtract the enthalpy of the reactants from the enthalpy of the products.
ΔH=HprodHreac=2ΔHNH3ΔHN23ΔHH2=2(45.9kJ)0kJ3(0kJ)=91.8kJ\begin{aligned}\Delta{H}&=\sum{H}_{\rm{prod}}-\sum{H}_{\rm{reac}}\\&=2\Delta{H}_{\rm{NH}_{3}}-\Delta{H}_{\rm{N}_{2}}-3\Delta{H}_{\rm{H}_{2}}\\&=2(-45.9\;\rm{kJ)}-0\;\rm{kJ}-3(0\;\rm{ kJ})\\&=-91.8\;\rm{kJ}\end{aligned}
Step 2
Next, calculate the change in molar entropy change (ΔS\Delta{S}) using the standard molar entropy (S°) for each molecule. The standard molar entropy for N2 is 191.61 J/K·mol, for H2 is 130.68 J/K·mol, and for NH3 is 192.77 J/K·mol.
ΔS=SprodSreac=2ΔSNH3ΔSN23ΔSH2=2(192.77J/K)191.61J/K3(130.68J/K)=191.11J/K\begin{aligned}\Delta{S}&=\sum{S}_{\rm{prod}}-\sum{S}_{\rm{reac}}\\&=2\Delta{S}_{\rm{NH}_{3}}-\Delta{S}_{\rm{N}_{2}}-3\Delta{S}_{\rm{H}_{2}}\\&=2(192.77\;\rm{ J/K)}-191.61\;\rm{ J/K}-3(130.68\;\rm{ J/K})\\&=-191.11\;\rm{ J/K}\end{aligned}
Solution
When setting up the equation for ΔG\Delta{G}, it is important to convert all values to the same units. ΔH\Delta{H} is given in kilojoules, so units for ΔS\Delta{S} are converted to give –0.198 11 kJ/K. Now the equation can be solved for ΔG\Delta{G}.
ΔG=ΔHTΔS=91.8kJ(298K)(0.19811kJ/K)=32.8kJ\begin{aligned}\Delta G&=\Delta H-T\Delta S\\&=-91.8\;{\rm{kJ}}-(298\;{\rm K)}(-0.19811\;{\rm{kJ/K)}}\\&=-32.8\;{\rm{kJ}}\end{aligned}
The change in Gibbs free energy for this reaction is less than 0, so the reaction is spontaneous.

Not all reactions proceed at 298 K. Therefore, the temperature at which a reaction proceeds affects whether the reaction is spontaneous. For the reaction between nitrogen gas and hydrogen gas occurring at a temperature of 550 K, the Gibbs free energy change is:
ΔG=ΔHTΔS=91.8kJ(550K)(0.19811kJ/K)=17.2kJ\begin{aligned}\Delta G&=\Delta H - T\Delta S\\&=-91.8\;\rm{kJ}-(550\rm{ K)}(-0.19811\;\rm{kJ}/\rm K)\\&=17.2\;\rm{kJ}\end{aligned}
This reaction has a ΔG\Delta{G} greater than 0, so it is not spontaneous. Standard molar free energy change (ΔGf\Delta G_{f}^\circ ), given in kL/mol, is the energy change for a reaction at standard temperature and pressure that produces one mole of a substance from its constituent elements. Standard molar free energy change has been calculated for most compounds. It can be used to understand the relationship between free energy (ΔG\Delta{G}) for a reaction and the equilibrium constant (Keq) for that reaction as a function of temperature. It can also be used to determine whether or not a reaction is at equilibrium, in which the forward and reverse reactions are equal. Keq is the ratio of the concentration of products and reactants of a chemical reaction at equilibrium and indicates whether the reaction favors products or reactants. The relationship between ΔG\Delta{G} and Keq is given as
ΔG=ΔGf+RTlnKeq\Delta G=\Delta G_{f}^\circ+{R}{T} \;{\rm ln}\;K_{\rm{eq}}
R is the ideal gas constant, 8.314 J/mol⋅K, which relates the kinetic energy of molecules to temperature per mole, and T is temperature.
Step-By-Step Example
Using the Equilibrium Constant to Determine the Change in Standard Molar Free Energy
Use the equilibrium constant (Keq) to determine the change in standard molar free energy for the reaction N2(g)+3H2(g)2NH3(g){\rm {N}_{2}}(g)+3{\rm {H}_{2}}( g)\rightarrow2{\rm{NH}_{3}}(g) at 25°C (298.15 K) and 2.0 atm.
Step 1
ΔGf\Delta G_{f}^\circ for the reaction N2(g)+3H2(g)2NH3(g){\rm {N}_{2}}(g)+3{\rm {H}_{2}}( g)\rightarrow2{\rm{NH}_{3}}(g) at 25°C (298 K) and 2.0 atm has already been found to be –32.8 kJ. To determine Keq considering any reaction aA+bBcC+dD{{a\rm{A}}+{b\rm{B}}}\rightarrow{{c\rm{C}}+{d\rm{D}}}, the equation is as follows:
Keq=[C]c[D]d[A]a[B]bK_{\rm{eq}}=\frac{\lbrack{\rm C}\rbrack^{c}\lbrack{\rm D}\rbrack^{d}}{\lbrack{\rm A}\rbrack^{a}\lbrack{\rm B}\rbrack^{b}}
where the superscripts for each molecule represent the number of moles of that molecule in the reaction equation.
Step 2
For a reaction in which the products or reactants are gases, the partial pressure is used rather than the concentration. Therefore, it is possible to calculate Keq for the reaction of N2 and H2 using the coefficients of the balanced equation.
Keq=[NH3]2[N2][H2]3 =[2.0]2[2.0][2.0]3=0.25\begin{aligned}K_{\rm{eq}}&=\frac{\lbrack{\rm{NH}_{3}}\rbrack^{2}}{\lbrack{\rm {N}_{2}}\rbrack\lbrack{\rm {H}_{2}}\rbrack^{3}}\\\ &=\frac{\lbrack2.0\rbrack^{2}}{\lbrack2.0\rbrack\lbrack2.0\rbrack^3}\\&=0.25\end{aligned}
Solution
Now it is possible to solve for ΔG\Delta{G}.
ΔG=32.76kJ+(8.314J/K)(298K)(ln0.25)=36.20kJ\begin{aligned}\Delta G&=-32.76\;{\rm{kJ}+(8.314\;J/K})(298\;{\rm {K)}{(\rm {ln}}\;0.25)}\\&=-36.20\;\rm{kJ}\end{aligned}
Note that in this step it was necessary to convert J to kJ.