Thermodynamics Part II: Free Energy, Spontaneity, and Equilibrium

Review of Thermodynamics and Spontaneity

A spontaneous reaction is defined as a change that occurs by itself under specified conditions without requiring an additional continuous input of energy from outside the system. While enthalpy change ($\Delta H_{system}$) or entropy change ($\Delta S_{system}$) are factors in a chemical process, neither value alone is sufficient to characterize whether a reaction will be spontaneous. According to the Second Law of Thermodynamics, spontaneity is governed by the change in standard free energy ($\Delta G^\circ$), which is expressed by the equation $\Delta G^\circ = \Delta H^\circ_{system} - T\Delta S^\circ_{system} = -T\Delta S_{total}$. A reaction is considered spontaneous if \Delta G^\circ < 0. Understanding thermodynamics allows us to address why change occurs in the first place, why there is a drive for a system to attain equilibrium, and what $\Delta G$ reveals about the equilibrium position or the extent to which a reaction goes toward completion.

The Function and Properties of Free Energy (G)

Free energy ($G$) is a thermodynamic function directly related to spontaneity that assists in managing the temperature dependence of a process. It is defined by the fundamental equation $G = H - TS$. For any process occurring at a constant temperature, the change in free energy is calculated as $\Delta G = \Delta H - T\Delta S$. In a system held at constant temperature and pressure, a process is spontaneous only in the direction in which the free energy decreases. Therefore, a negative value for $\Delta G$ corresponds to a positive value for the entropy of the universe ($\Delta S_{univ}$).

Concept Check: Vaporization at the Boiling Point

When a liquid is vaporized at its normal boiling point, several thermodynamic signs can be predicted. In this scenario, the work ($w$) is negative because the system is expanding against the atmosphere as it turns from liquid to gas. The heat ($q$) is positive because vaporization is an endothermic process requiring energy. Consequently, the change in enthalpy ($\Delta H$) is positive. The entropy of the system ($\Delta S$) is positive because a gas has more microstates and disorder than a liquid. The entropy of the surroundings ($\Delta S_{surr}$) is negative because the surroundings lose heat to the system ($\Delta S_{surr} = -\frac{q}{T_{surr}}$). Finally, the change in free energy ( $\Delta G$) is zero because the system is at equilibrium at the boiling point.

Example: Spontaneity and Temperature for Bromine Vaporization

Consider the process of liquid bromine turning into gaseous bromine: $Br_2(l) \rightarrow Br_2(g)$, with $\Delta H^\circ = 31.0\,kJ/mol$ and $\Delta S^\circ = 93.0\,J/K \cdot mol$. To determine the temperatures at which this process is spontaneous at $1\,atm$, we analyze the balance between enthalpy and entropy. If T > 333\,K, the term $T\Delta S^\circ$ dominates, meaning the increase in entropy during vaporization is the controlling factor, and the process is spontaneous. If T < 333\,K, the process is spontaneous in the opposite direction (condensation) where it is exothermic, and the $\Delta H^\circ$ term controls the spontaneity. At exactly $T = 333\,K$, the opposing driving forces are balanced, resulting in $\Delta G^\circ = 0$. At this temperature, the liquid and gaseous phases of bromine coexist in equilibrium, identifying $333\,K$ as the normal boiling point of liquid bromine.

Concept Check: Synthesis of Gas AB

In a reaction where gas $A_2$ reacts with gas $B_2$ to form gas $AB$ ($A_2(g) + B_2(g) \rightarrow 2AB(g)$) at constant temperature and pressure, thermodynamic signs can be predicted based on bond energies. If the bond energy of $AB$ is much greater than that of either reactant, the reaction is releasing a significant amount of energy, making $\Delta H$ negative (exothermic). Since the number of moles of gas is equal on both sides (2 moles of reactants for 2 moles of product), $\Delta S_{sys}$ is likely near zero or very small. Given the strongly exothermic nature, $\Delta G$ will be negative, making the reaction spontaneous. Consequently, $\Delta S_{univ}$ will be positive.

The Third Law of Thermodynamics and Standard Entropy

The Third Law of Thermodynamics states that the entropy of a perfect crystal at $0\,K$ is exactly zero. As the temperature of a substance increases, its entropy also increases. Standard entropy values ($S^\circ$) quantify the entropy increase that occurs when one mole of a substance is heated from $0\,K$ to $298\,K$ at $1\,atm$ of pressure. The standard entropy change for a reaction ($\Delta S^\circ_{reaction}$) is calculated using the formula: $\Delta S^\circ_{reaction} = \sum n_p S^\circ_{products} - \sum n_r S^\circ_{reactants}$, where $n_p$ and $n_r$ represent the number of moles of products and reactants, respectively.

Example: Calculating Standard Entropy Change

To calculate the $\Delta S^\circ$ for the reduction of aluminum oxide by hydrogen gas ($Al_2O_3(s) + 3H_2(g) \rightarrow 2Al(s) + 3H_2O(g)$), we use the following standard entropy values ($J/K \cdot mol$): $Al_2O_3(s) = 51$, $H_2(g) = 131$, $Al(s) = 28$, and $H_2O(g) = 189$. The calculation is performed as follows: $\Delta S^\circ = [2(28) + 3(189)] - [51 + 3(131)] = [56 + 567] - [51 + 393] = 623 - 444 = 179\,J/K \cdot mol$.

Calculating Standard Free Energy Change (\Delta G^∘)

The standard free energy change ($\Delta G^\circ$) is the change in free energy that occurs when reactants in their standard states are converted to products in their standard states. There are two primary methods for calculating this value. The first involves the equation $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ at a constant temperature. The second method involves calculating the $\Delta G^\circ$ for each individual step in a complex reaction and summing those values. This is possible because free energy is a state function, allowing for an approach similar to Hess's Law used for enthalpy.

Standard Free Energy of Formation and Reaction Coupling

The standard free energy of formation ($\Delta G^\circ_f$) is the change in free energy that accompanies the formation of one mole of a substance from its constituent elements, with all substances in their standard states. The reaction free energy is determined by: $\Delta G^\circ = \sum n_p \Delta G^\circ_{f(products)} - \sum n_r \Delta G^\circ_{f(reactants)}$. For Example: The combustion of methanol is given by $2CH_3OH(g) + 3O_2(g) \rightarrow 2CO_2(g) + 4H_2O(g)$. Using $\Delta G^\circ_f$ values ($kJ/mol$) of $CH_3OH(g) = -163$, $O_2(g) = 0$, $CO_2(g) = -394$, and $H_2O(g) = -229$, the total change is $\Delta G^\circ = [2(-394) + 4(-229)] - [2(-163) + 3(0)] = [-788 - 916] - [-326] = -1704 + 326 = -1378\,kJ/mol$.

Example: Determining the Free Energy of Formation for Phosphoric Acid

To find the standard free energy of formation for phosphoric acid ($P(s) + \frac{3}{2}H_2(g) + 2O_2(g) \rightarrow H_3PO_4(l)$), we couple provided reactions:

1. $P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s)$ where $\Delta G^\circ_{298} = -2697.0\,kJ/mol$

2. $2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$ where $\Delta G^\circ_{298} = -457.18\,kJ/mol$

3. $6H_2O(g) + P_4O_{10}(s) \rightarrow 4H_3PO_4(l)$ where $\Delta G^\circ_{298} = -428.66\,kJ/mol$ By appropriately scaling and summing these reactions, the cumulative free energy change yields the standard free energy of formation for the target acid.

Free Energy and Pressure Dependence

The free energy of a system is dependent on the pressure of gases or the concentration of species in solution. This means $\Delta G$ changes as the reaction proceeds. For an ideal gas, entropy depends on volume; specifically, S_{large\,volume} > S_{small\,volume}. Because pressure and volume are inversely related, entropy is higher at lower pressures: S_{low\,p} > S_{high\,p}. The free energy of an ideal gas at any pressure $P$ is related to its standard state free energy by the equation $G = G^\circ + RT \ln(P)$. For a general reaction $A(g) \rightarrow B(g)$, the change in free energy is expressed as $\Delta G = G_B - G_A = (G_B^\circ + RT \ln P_B) - (G_A^\circ + RT \ln P_A)$. This simplifies to $\Delta G = \Delta G^\circ + RT \ln\left(\frac{P_B}{P_A}\right)$, or more generally, $\Delta G = \Delta G^\circ + RT \ln Q$, where $Q$ is the reaction quotient.

Relationship Between Free Energy and Equilibrium

Equilibrium represents the lowest possible free energy value available to a chemical system. At equilibrium, the change in free energy ($\Delta G$) is zero. Substituting this into the pressure equation gives $0 = \Delta G^\circ + RT \ln K$, which leads to the vital relationship $\Delta G^\circ = -RT \ln K$, where $K$ is the equilibrium constant. The magnitude and sign of $\Delta G^\circ$ determine the value of $K$:

• If $\Delta G^\circ = 0$, then $K = 1$.

• If \Delta G^\circ > 0, then K < 1 (the reaction is reactant-favored).

• If \Delta G^\circ < 0, then K > 1 (the reaction is product-favored).

Example: Calculating Non-Standard Free Energy for Methanol Synthesis

Consider the reaction $CO(g) + 2H_2(g) \rightarrow CH_3OH(l)$ at $25\,^\circ C$. We calculate $\Delta G$ when the partial pressures are non-standard: $P_{CO} = 5.0\,atm$ and $P_{H_2} = 3.0\,atm$. Using the relationship $\Delta G = \Delta G^\circ + RT \ln Q$, we first determine $\Delta G^\circ$ from standard tables and then calculate $Q = \frac{1}{P_{CO} \times P_{H_2}^2} = \frac{1}{5.0 \times 3.0^2} = \frac{1}{45}$. Plugging these values into the equation (with $R = 8.314\,J/K \cdot mol$ and $T = 298.15\,K$) allows for the calculation of the specific free energy change under these distinct pressure conditions.

Thermodynamics and the Reaction Profile

The total free energy of a system ($G_{system}$) can be visualized as a curve relative to the reaction progress. For any reaction $A(g) \rightarrow B(g)$, if $\Delta G^\circ = 0$, the standard free energies of pure $A$ and pure $B$ are equal ($G_A^\circ = G_B^\circ$), and the equilibrium position ($Q = K = 1$) occurs when partial pressures are equal ($P_A = P_B$). In cases where \Delta G^\circ > 0, the system is reactant-favored (K < 1), meaning the minimum of the free energy curve lies closer to the reactants (P_A > P_B). Conversely, when \Delta G^\circ < 0, the system is product-favored (K > 1), and the minimum free energy is found closer to the products (P_B > P_A). Regardless of the starting point, if Q < K, then \Delta G < 0 and the reaction proceeds forward. If Q > K, then \Delta G > 0 and the reaction proceeds in reverse toward the equilibrium position, which is always the point of lowest system free energy.