Free Energy and Equilibrium

Learning Objective 9.5.A Explain whether a process is thermodynamically favored using the relationships between the equilibrium constant (K), standard free energy change (ΔG°), and temperature (T).

Quick Notes

A reaction is thermodynamically favored when ΔG° < 0.
At equilibrium: ΔG = 0

The relationship between ΔG°, K, and T is:
ΔG° = −RT ln K or K = e^−ΔG°/RT).

If K > 1, products are favored (ΔG° < 0).
If K < 1, reactants are favored (ΔG° > 0).
If K ≈ 1, the system has significant amounts of both reactants and products (ΔG° ≈ 0).
Both ΔG° and K depend on temperature.

Full Notes

Recap - Free Energy and Thermodynamic Favorability

Gibbs free energy change under standard conditions, represented by ΔG°, tells us whether a reaction is thermodynamically favored. For a reversible reaction, this means whether it has a natural tendency to proceed in the forward direction without needing continuous external energy.

When ΔG° < 0, the reaction is thermodynamically favored, meaning it tends to proceed forward and form products.

When ΔG° > 0, the reaction is not favored in the forward direction; instead, the reverse reaction is more likely.

When ΔG° = 0, the system is at equilibrium under standard conditions – neither reactants nor products are favored.

In simple terms: negative ΔG° means the reaction "wants" to happen; positive ΔG° means it doesn’t.

Connecting ΔG° to Equilibrium (K)

There is a direct relationship between ΔG°, the equilibrium constant (K), and temperature (T):

AP Chemistry graphic showing ΔG° = −RT ln K and K = e to the power of minus ΔG° over RT, with definitions of R, T and K.

ΔG° = −RT ln K and K = e^−ΔG°/RT

R is the gas constant = 8.314 J/mol·K
T is the temperature in kelvins
ln is the natural logarithm

These equations connect thermodynamics (ΔG°) with equilibrium (K).

They allow us to predict how product or reactant favored a reaction is.

How ΔG° Relates to K: Qualitative Understanding

If ΔG° < 0, then ln K > 0, so K > 1: products are favored at equilibrium.
If ΔG° > 0, then ln K < 0, so K < 1: reactants are favored at equilibrium.
If ΔG° ≈ 0, then K ≈ 1: significant amounts of both reactants and products are present.

Examples:

If ΔG° = −30 kJ/mol, K is large (≫ 1): the reaction strongly favors products.
If ΔG° = +40 kJ/mol, K is very small (≪ 1): the reaction barely proceeds; reactants dominate.
If ΔG° ≈ 0, K is close to 1: the reaction reaches a balanced equilibrium between reactants and products.

These relationships allow you to estimate K without a calculator, just by reasoning from the sign and magnitude of ΔG°.

Temperature Dependence

Both ΔG° and K depend on temperature, especially for reactions where ΔH° (enthalpy) and ΔS° (entropy) are significant.

As temperature changes, the value of ΔG° changes (based on ΔG° = ΔH° − TΔS°), which in turn shifts the value of K.

For example:

In reactions with ΔH° > 0 and ΔS° > 0, increasing temperature makes ΔG° more negative, K increases and more products are favored.
In reactions with ΔH° < 0 and ΔS° < 0, increasing temperature makes ΔG° more positive, K decreases and more reactants are favored.

Summary

ΔG° < 0 → K > 1 → products are favored
ΔG° > 0 → K < 1 → reactants are favored
ΔG° ≈ 0 → K ≈ 1 → mixture of products and reactants
These relationships help us predict not just whether a reaction is favorable, but how far it will go toward completion at a given temperature.
Remember: thermodynamic favorability does not imply rate – only possibility under given conditions.