Premed · Premed · General Chemistry 2
Lecture 12: Gibbs Free Energy and Spontaneity
General Chemistry II
Learning Objectives
By the end of this lecture, students will be able to:
- Define Gibbs free energy and use it to predict spontaneity
- Calculate Delta G^0 using standard free energies of formation, or from Delta H^0 and Delta S^0
- Determine the temperature at which a reaction becomes spontaneous
- Relate Delta G^0 to the equilibrium constant K
- Distinguish between Delta G^0 (standard) and Delta G (non-standard conditions)
- Interpret the four thermodynamic cases based on signs of Delta H and Delta S
Lecture Content
I. Gibbs Free Energy
Gibbs free energy (G) unifies enthalpy and entropy into a single criterion for spontaneity at constant temperature and pressure: G = H - TS. The change in Gibbs free energy for a process is Delta G = Delta H - T * Delta S. The sign of Delta G directly determines spontaneity: Delta G < 0 means the process is spontaneous (exergonic), Delta G > 0 means it is nonspontaneous (endergonic), and Delta G = 0 means the system is at equilibrium. Beyond predicting direction, Delta G also represents the maximum useful work obtainable from a process at constant T and P, with w_max = Delta G for a reversible process.
II. The Four Thermodynamic Cases
The signs of Delta H and Delta S together determine how spontaneity depends on temperature. In Case 1, when Delta H < 0 and Delta S > 0, Delta G is negative at all temperatures and the reaction is always spontaneous. Combustion reactions, such as 2H2 + O2 -> 2H2O, exemplify this case. In Case 2, when Delta H > 0 and Delta S < 0, Delta G is positive at all temperatures and the reaction is never spontaneous in the forward direction. The formation of ozone from oxygen (3O2 -> 2O3) falls into this category.
Case 3 arises when Delta H < 0 and Delta S < 0. Here the reaction is spontaneous at low temperatures, where the favorable enthalpy term dominates, but becomes nonspontaneous at high temperatures, where the unfavorable entropy term takes over. The freezing of water is a classic example. Case 4 involves Delta H > 0 and Delta S > 0. The reaction is nonspontaneous at low temperatures but becomes spontaneous at high temperatures, where the favorable entropy term outweighs the unfavorable enthalpy. Melting of ice and the dissolving of ammonium nitrate illustrate this behavior.
<image>A four-quadrant diagram showing the relationship between Delta H and Delta S. The x-axis represents Delta S (negative on the left, positive on the right) and the y-axis represents Delta H (positive on top, negative on bottom). Quadrant 1 (top right, Delta H > 0, Delta S > 0): "Spontaneous at high T, T_crossover = Delta H / Delta S." Quadrant 2 (top left, Delta H > 0, Delta S < 0): "Never spontaneous (Delta G always > 0)." Quadrant 3 (bottom left, Delta H < 0, Delta S < 0): "Spontaneous at low T, T_crossover = Delta H / Delta S." Quadrant 4 (bottom right, Delta H < 0, Delta S > 0): "Always spontaneous (Delta G always < 0)." Each quadrant includes a representative example reaction.</image>
III. Crossover Temperature
For Cases 3 and 4, where Delta H and Delta S have the same sign, there exists a crossover temperature at which Delta G = 0 and the system is at equilibrium. This temperature is given by T_crossover = Delta H / Delta S, provided both quantities are expressed in consistent units.
Consider the decomposition of calcium carbonate: CaCO3(s) <=> CaO(s) + CO2(g), with Delta H^0 = +178.3 kJ/mol and Delta S^0 = +160.5 J/(molK) = 0.1605 kJ/(molK). The crossover temperature is 178.3 / 0.1605 = 1111 K, or approximately 838 degrees C. Below this temperature, the decomposition is nonspontaneous; above it, the reaction proceeds spontaneously. This calculation assumes that Delta H and Delta S are approximately constant with temperature, which is a reasonable approximation over moderate temperature ranges.
IV. Calculating Standard Free Energy Change
There are three methods for calculating Delta G^0. The first method combines enthalpy and entropy data directly: Delta G^0 = Delta H^0 - T * Delta S^0. Care must be taken with units, as Delta H is typically given in kJ while Delta S is in J/K.
The second method uses tabulated standard free energies of formation: Delta G^0_rxn = Sum(n Delta G_f^0_products) - Sum(n Delta G_f^0_reactants). As with enthalpies of formation, Delta G_f^0 for elements in their standard states is zero.
The third method applies Hess's law: Delta G values are additive when reactions are combined, just as Delta H values are.
V. Delta G^0 and the Equilibrium Constant
The relationship between the standard free energy change and the equilibrium constant is one of the most important equations in chemistry: Delta G^0 = -RT ln(K), or equivalently K = e^(-Delta G^0 / RT). When Delta G^0 is negative, K is greater than 1 and products are favored at equilibrium. When Delta G^0 is positive, K is less than 1 and reactants are favored. When Delta G^0 is exactly zero, K equals 1.
In applying this equation, use R = 8.314 J/(mol*K), T in Kelvin, and Delta G^0 in joules (not kJ) to maintain consistent units. This equation provides the quantitative bridge between thermodynamics and equilibrium.
<image>A graph showing the relationship between Delta G^0 and K. The x-axis shows ln(K) (or equivalently K on a logarithmic scale below), and the y-axis shows Delta G^0. A straight line passes through the origin with slope -RT. The left side (K < 1, ln K < 0) corresponds to Delta G^0 > 0 (labeled "Reactants favored"). The right side (K > 1, ln K > 0) corresponds to Delta G^0 < 0 (labeled "Products favored"). The point where the line crosses zero on both axes (K = 1, Delta G^0 = 0) is labeled "Equilibrium with equal amounts." Specific K values (10^-5, 10^-1, 1, 10, 10^5) and their corresponding Delta G^0 values at 298 K are marked.</image>
VI. Delta G Under Nonstandard Conditions
The standard free energy change Delta G^0 describes behavior under standard conditions (1 M, 1 atm). Under nonstandard conditions, the actual free energy change is given by Delta G = Delta G^0 + RT ln(Q), where Q is the reaction quotient calculated from current concentrations or pressures.
At equilibrium, Q = K and Delta G = 0, which gives 0 = Delta G^0 + RT ln(K), confirming the relationship Delta G^0 = -RT ln(K). When Q < K, Delta G is negative and the reaction proceeds spontaneously in the forward direction. When Q > K, Delta G is positive and the reverse direction is favored. In essence, Delta G tells you the direction a reaction will proceed under current conditions, while Delta G^0 tells you about the equilibrium position.
VII. Coupled Reactions and Free Energy
A nonspontaneous reaction can be driven forward by coupling it with a highly spontaneous reaction, provided the total Delta G for the coupled process is negative. In biological systems, this principle is exemplified by ATP hydrolysis: ATP + H2O -> ADP + Pi, with Delta G^0 = -30.5 kJ/mol. This strongly favorable reaction is coupled to energetically unfavorable biosynthetic reactions, and as long as the combined Delta G is negative, the overall process proceeds spontaneously.
In industrial settings, the same principle applies. The extraction of metals from their ores, for instance, couples the thermodynamically unfavorable reduction of metal oxides with the strongly favorable oxidation of carbon.
<image>A diagram illustrating coupled reactions. Two reaction coordinate diagrams are shown side by side and then combined. Left: Reaction A (nonspontaneous), showing products at higher energy than reactants (Delta G_A > 0, uphill). Middle: Reaction B (highly spontaneous, such as ATP hydrolysis), showing products much lower than reactants (Delta G_B << 0, downhill). Right: The combined coupled reaction A + B, showing the net Delta G = Delta G_A + Delta G_B < 0 (net downhill). Arrows and energy levels clearly indicate that B provides enough free energy to drive A forward.</image>
