6.4 Gibbs free energy and spontaneity of reactions

Gibbs Free Energy and Spontaneity

Gibbs free energy ties together enthalpy and entropy into a single value that tells you whether a reaction will happen on its own. It's one of the most useful quantities in thermodynamics because it directly predicts spontaneity at constant temperature and pressure, which is how most real-world reactions occur (open beakers, biological systems, industrial reactors).

Beyond just "will it go?", Gibbs free energy also connects to equilibrium constants, letting you quantify how far a reaction proceeds. This section covers how to calculate $\Delta G$, interpret its sign and magnitude, and relate it to $K$.

Gibbs Free Energy Fundamentals

Gibbs free energy ($G$) is defined as:

$G = H - TS$

• $H$ = enthalpy (total heat content of the system)
• $S$ = entropy (disorder or randomness of the system)
• $T$ = absolute temperature in Kelvin

For a process at constant temperature, the change in Gibbs free energy is:

$\Delta G = \Delta H - T\Delta S$

The sign of $\Delta G$ is what matters for spontaneity:

• $\Delta G < 0$: spontaneous (product-favored). Example: graphite converting to diamond at high pressure.
• $\Delta G > 0$: non-spontaneous (reactant-favored). Example: water flowing uphill requires energy input.
• $\Delta G = 0$: the system is at equilibrium. Example: a saturated sugar solution with no more dissolving or crystallizing.

Notice how $\Delta G$ depends on both enthalpy and entropy. A reaction can be spontaneous even if it's endothermic ($\Delta H > 0$), as long as the $T\Delta S$ term is large enough to make $\Delta G$ negative. Temperature acts as a "weight" on the entropy term, which is why some reactions become spontaneous only at high temperatures.

Calculation of Gibbs Free Energy

Standard Gibbs free energy of formation ($\Delta G_f^\circ$) is the change in $G$ when one mole of a compound forms from its elements in their standard states (1 atm, usually 298 K). By convention, $\Delta G_f^\circ = 0$ for any element in its standard state.

To find the standard Gibbs free energy change ($\Delta G^\circ$) for a reaction, use formation values:

$\Delta G^\circ = \sum \nu_p \Delta G_{f,p}^\circ - \sum \nu_r \Delta G_{f,r}^\circ$

Step-by-step:

1. Write a balanced equation for the reaction.
2. Look up $\Delta G_f^\circ$ for each reactant and product (from a reference table).
3. Multiply each $\Delta G_f^\circ$ by its stoichiometric coefficient ($\nu$).
4. Add up the product terms, then subtract the sum of the reactant terms.

For example, for the combustion of methane:

$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$

You'd calculate: $\Delta G^\circ = [1 \times \Delta G_f^\circ(CO_2) + 2 \times \Delta G_f^\circ(H_2O)] - [1 \times \Delta G_f^\circ(CH_4) + 2 \times \Delta G_f^\circ(O_2)]$

Since $O_2$ is an element in its standard state, its $\Delta G_f^\circ = 0$.

Spontaneity Prediction Using Gibbs Energy

The sign of $\Delta G$ tells you the direction; the magnitude tells you the driving force.

• Large negative $\Delta G$: strongly spontaneous. Combustion of methane ($\Delta G^\circ = -818 \text{ kJ/mol}$) proceeds vigorously.
• Small negative $\Delta G$: weakly spontaneous. Dissolving NaCl in water happens, but the driving force is modest.
• Small positive $\Delta G$: weakly non-spontaneous. Melting ice at just below 0°C is close to equilibrium.
• Large positive $\Delta G$: strongly non-spontaneous. Decomposing water into $H_2$ and $O_2$ requires significant energy input (electrolysis).

One common trap: spontaneous does not mean fast. Diamond formation from graphite is thermodynamically favorable at standard conditions, but it's so slow you'll never observe it on a human timescale. Spontaneity is about thermodynamic favorability, not reaction rate.

The Four $\Delta H$ / $\Delta S$ Combinations

Since $\Delta G = \Delta H - T\Delta S$, the interplay of enthalpy and entropy determines spontaneity:

The temperature-dependent cases are the ones that show up most on exams. To find the crossover temperature where $\Delta G = 0$, set $\Delta H = T\Delta S$ and solve: $T = \frac{\Delta H}{\Delta S}$.

Relationship Between $\Delta G^\circ$ and the Equilibrium Constant

Gibbs free energy connects directly to the equilibrium constant through:

$\Delta G^\circ = -RT \ln K$

• $R$ = 8.314 J/(mol·K)
• $T$ = temperature in Kelvin
• $K$ = equilibrium constant

What this equation tells you:

• $\Delta G^\circ < 0$ → $K > 1$ → products favored at equilibrium. Example: formation of water from $H_2$ and $O_2$ has a very large $K$.
• $\Delta G^\circ > 0$ → $K < 1$ → reactants favored at equilibrium. Example: decomposition of $CaCO_3$ at 298 K.
• $\Delta G^\circ = 0$ → $K = 1$ → neither side favored. Example: a liquid at exactly its boiling point (1 atm).

Be careful with units here. $R$ is in J/(mol·K), so if your $\Delta G^\circ$ is in kJ/mol, convert to J/mol before plugging in.

You can also rearrange to solve for $K$:

$K = e^{-\Delta G^\circ / RT}$

Gibbs Energy Diagrams

Gibbs free energy diagrams plot $G$ along the vertical axis against the reaction coordinate (progress of reaction) on the horizontal axis.

• Reactants and products sit at energy minima (stable states).
• The transition state sits at the energy maximum between them.
• A spontaneous reaction has products at a lower $G$ than reactants. The system naturally moves "downhill" on the diagram.
• At equilibrium, the system sits at the lowest accessible point on the curve, and there's no net change.

These diagrams are distinct from energy diagrams you may have seen in kinetics. The height of the barrier (activation energy) determines how fast the reaction goes, but the difference in $G$ between reactants and products determines whether it's spontaneous.

Applications of Gibbs Free Energy

Assessing feasibility: If $\Delta G < 0$, the process can occur without external energy. The Haber process for synthesizing ammonia ($N_2 + 3H_2 \rightarrow 2NH_3$) is feasible because $\Delta G^\circ$ is negative at moderate temperatures. If $\Delta G > 0$, you need to supply energy, as in electrolysis of water.

Comparing compound stability: Compounds with more negative $\Delta G_f^\circ$ values are more thermodynamically stable. For instance, $CO_2$ ($\Delta G_f^\circ = -394.4 \text{ kJ/mol}$) is very stable, which is why combustion reactions that produce $CO_2$ tend to be strongly spontaneous.

Shifting equilibrium: Changing conditions (temperature, pressure, concentration) alters $\Delta G$ and shifts the equilibrium position.

• Raising temperature for an endothermic reaction makes $\Delta G$ more negative, shifting equilibrium toward products.
• Decreasing pressure favors the side with more moles of gas.

These shifts follow Le Chatelier's principle, but Gibbs free energy gives you the quantitative framework behind why they happen.