Gibbs Free Energy & the Equilibrium Criterion

At constant temperature and pressure, reactions proceed in the direction that lowers Gibbs free energy until ΔG = 0.

Material & Energy BalancesChemical Engineering Year 1Free preview

⏱️ About 16 min

When a reaction “wants” to go forward, what is the exact thermodynamic quantity it is trying to lower?

💡

The big idea: At constant T and P, the reaction direction is the one that decreases Gibbs free energy: ΔG < 0 spontaneous, ΔG = 0 equilibrium.

🎯 By the end, you'll be able to

Define Gibbs free energy G and interpret its terms (H and TS)
State the equilibrium/spontaneity criterion at constant T and P using ΔG
Compute ΔG from ΔH and ΔS with consistent units
Avoid confusing ΔG (actual conditions) with ΔG° (standard-state quantity)

📎 Helpful to know first

Second-Law Applications to Process Equipment

Gibbs Free Energy: A “Potential” for Constant T, P Processes

The Gibbs free energy is defined as:

G = H − TS

It combines enthalpy (H) and entropy (S) into a single quantity that is especially useful for systems at constant temperature and pressure (a very common situation for reacting mixtures in contact with the environment).

\[ G = H - TS \]

Definition of Gibbs free energy.

The Equilibrium Criterion at Constant T and P

For a reaction occurring at constant T and P:

ΔG < 0 → spontaneous in the forward direction
ΔG = 0 → equilibrium
ΔG > 0 → not spontaneous forward (the reverse direction would be spontaneous)

Intuition: the system “rolls downhill” in Gibbs free energy until it cannot decrease any further.

\[ \Delta G = \Delta H - T\,\Delta S \]

At constant temperature, ΔG can be computed from ΔH and ΔS (use consistent units).

⚠️ Pitfall: ΔG is for the actual conditions, not standard state

The spontaneity criterion uses ΔG at the actual reaction conditions (actual composition/activities).

ΔG° is a standard-state quantity. ΔG° alone does not tell you whether a reaction mixture at some arbitrary composition is currently spontaneous.

📝 Worked example: At T = 300 K, a reaction has ΔH = −20.0 kJ/mol and ΔS = −50.0 J/(mol·K). Compute ΔG and classify whether the forward reaction is spontaneous, at equilibrium, or non-spontaneous.

Convert ΔS to kJ/(mol·K): ΔS = −50.0 J/(mol·K) = −0.0500 kJ/(mol·K)
Compute TΔS: TΔS = (300 K)(−0.0500 kJ/(mol·K)) = −15.0 kJ/mol
Compute ΔG = ΔH − TΔS: ΔG = (−20.0) − (−15.0) = −20.0 + 15.0 = −5.0 kJ/mol
Since ΔG < 0, the forward direction is spontaneous at these conditions.

✓ ΔG = −5.0 kJ/mol, so the forward reaction is spontaneous (ΔG < 0).

✏️ Practice: At T = 400 K, a reaction has ΔH = +10.0 kJ/mol and ΔS = +30.0 J/(mol·K). Compute ΔG (kJ/mol).

kJ/mol

Solution

Convert ΔS: +30.0 J/(mol·K) = +0.0300 kJ/(mol·K)
Compute TΔS: (400)(0.0300) = 12.0 kJ/mol
ΔG = ΔH − TΔS = 10.0 − 12.0 = −2.0 kJ/mol
ΔG < 0, so forward is spontaneous.

Check your understanding

1. For a reaction at constant temperature and pressure, the correct equilibrium criterion is:

At constant T and P, equilibrium corresponds to ΔG = 0 for the reaction progress variable.

2. A negative ΔG for a reaction mixture at its current composition means:

ΔG < 0 indicates a driving force for the forward direction; it says nothing about rate.

✅ Key takeaways

Gibbs free energy is defined by G = H − TS
At constant T and P: ΔG < 0 forward spontaneous, ΔG = 0 equilibrium, ΔG > 0 forward non-spontaneous
A reaction proceeds in the direction that lowers G until it reaches the minimum (equilibrium)
ΔG is evaluated at actual conditions; ΔG° is a standard-state quantity used to relate to K

➡️ Now that ΔG tells you the direction and equilibrium condition, we connect standard Gibbs energy change ΔG° to the equilibrium constant K via a simple logarithm/exponential relationship.

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