Gibbs free energy and spontaneity

Why does an ice cube melt at room temperature but water does not spontaneously freeze there? Both processes obey the first law of thermodynamics equally well, so energy conservation alone cannot decide which one actually happens. The answer needs a single number that folds together how much heat a process releases or absorbs and how much disorder it creates — without forcing you to separately track the surroundings every time. That number is the Gibbs free energy, $G$ , and its change $\Delta G$ at constant temperature and pressure is the definitive spontaneity test used throughout chemistry: negative means "go," positive means "no," zero means "balanced." This article builds $\Delta G = \Delta H - T\Delta S$ from the second law, shows how the signs of $\Delta H$ and $\Delta S$ decide spontaneity at different temperatures, and links $\Delta G^\circ$ to the equilibrium constant $K$ .

The core insight

Gibbs free energy $G = H - TS$ combines the system's enthalpy and entropy into one quantity, so that $\Delta G < 0$ at constant temperature and pressure is the complete, self-contained test for whether a process happens spontaneously.

The core idea

A single system-only criterion. The second law says a process is spontaneous only if it increases the total entropy of the universe, $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} \geq 0$ . That requires knowing the surroundings' entropy change too, which is inconvenient. Gibbs free energy repackages the same condition using only system properties.

Definition. Gibbs free energy is a state function,

G = H - TS

where: $H$ is enthalpy, $T$ is the absolute (Kelvin) temperature, and $S$ is entropy, all of the system. At constant temperature, the change is

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

where: $\Delta H$ is the enthalpy change (heat exchanged at constant pressure), $\Delta S$ is the entropy change of the system, and $T$ is the constant absolute temperature at which the process occurs.

Spontaneity criteria at constant $T$ and $P$ :

$\Delta G < 0$ — the process is spontaneous (exergonic); it proceeds on its own once started.
$\Delta G = 0$ — the system is at equilibrium; forward and reverse rates balance, and no net change occurs.
$\Delta G > 0$ — the process is non-spontaneous in the forward direction; the reverse process is the spontaneous one.

How $\Delta H$ and $\Delta S$ signs decide spontaneity. Because $\Delta G = \Delta H - T\Delta S$ , the two terms can either agree (fixing the sign of $\Delta G$ for every temperature) or fight it out (making spontaneity temperature-dependent):

Figure

A two by two grid showing how the signs of delta H and delta S combine to determine when a reaction is spontaneous

The crossover temperature where $\Delta G$ switches sign — found by setting $\Delta H = T\Delta S$ , i.e. $\Delta G = 0$ — is

T = \frac{\Delta H}{\Delta S}

where: $T$ is the threshold temperature in kelvin at which the reaction changes from spontaneous to non-spontaneous (or vice versa).

Link to the equilibrium constant. At standard conditions, the standard Gibbs energy change of a reaction relates directly to its equilibrium constant $K$ :

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

where: $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ is the gas constant, $T$ is the absolute temperature, and $K$ is the equilibrium constant. A large negative $\Delta G^\circ$ gives a large $K$ (products favoured); $\Delta G^\circ = 0$ gives $K = 1$ ; a large positive $\Delta G^\circ$ gives a small $K$ (reactants favoured). This is an introductory conceptual link only — no derivation of it is required at this level, but it is exactly the bridge that connects thermodynamics to chemical equilibrium in the next unit.

Derivation

Derivation AFrom the second law to ΔG less than zero

The goal of this derivation is to show, starting only from the second law's statement that total entropy of the universe cannot decrease during a spontaneous process, that the single system-only quantity $\Delta G = \Delta H - T\Delta S$ correctly reproduces $\Delta G \leq 0$ as the spontaneity criterion at constant temperature and pressure, so that a student never again needs to separately compute the entropy change of the surroundings.

Step 1. Write the second law for a process occurring in a system in thermal contact with its surroundings: the total entropy change must satisfy $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} \geq 0$ , with equality only at equilibrium and strict inequality for any spontaneous (irreversible) process.

Step 2. Express the surroundings' entropy change in terms of the system. At constant pressure, the heat released by the system into the surroundings equals $-\Delta H_{sys}$ (heat leaving the system enters the surroundings), and because the surroundings are effectively an infinite reservoir at constant $T$ , their entropy change is $\Delta S_{surr} = -\Delta H_{sys}/T$ .

Step 3. Substitute Step 2 into Step 1: $\Delta S_{sys} - \dfrac{\Delta H_{sys}}{T} \geq 0$ . This is now written entirely in terms of the system, but entropy is on one side and enthalpy on the other, in awkward units.

Step 4. Multiply both sides by $-T$ (a positive quantity, since $T$ is an absolute temperature), which flips the inequality: $\Delta H_{sys} - T\Delta S_{sys} \leq 0$ . The left side is exactly the definition $\Delta G_{sys} = \Delta H_{sys} - T\Delta S_{sys}$ , so the condition becomes $\Delta G_{sys} \leq 0$ .

Step 5. Read off the result. A spontaneous process at constant $T$ and $P$ must have $\Delta G \leq 0$ , with $\Delta G < 0$ for genuinely spontaneous change and $\Delta G = 0$ exactly at equilibrium — recovering the three-way criterion from a single system-only quantity, with the surroundings folded in through the $-T\Delta S$ term.

Worked example

For a reaction at 298 K, $\Delta H = -92.4\ \text{kJ mol}^{-1}$ and $\Delta S = -198.7\ \text{J K}^{-1}\text{mol}^{-1}$ . Calculate $\Delta G$ at 298 K, comment on spontaneity, and find the temperature above which the reaction becomes non-spontaneous.

Step 1. Match units before combining the two terms: $\Delta S = -198.7\ \text{J K}^{-1}\text{mol}^{-1} = -0.1987\ \text{kJ K}^{-1}\text{mol}^{-1}$ , so both $\Delta H$ and $T\Delta S$ will come out in kJ mol $^{-1}$ .

Step 2. Apply $\Delta G = \Delta H - T\Delta S$ at $T = 298\ \text{K}$ : $\Delta G = -92.4 - (298)(-0.1987) = -92.4 + 59.21 = -33.19\ \text{kJ mol}^{-1}$ .

Step 3. Interpret the sign. Since $\Delta G < 0$ at 298 K, the reaction is spontaneous at this temperature. Because $\Delta H < 0$ and $\Delta S < 0$ here, this is the mixed-sign case that is spontaneous only below a threshold temperature.

Step 4. Find that threshold using $T = \Delta H/\Delta S$ : $T = \dfrac{-92.4\ \text{kJ mol}^{-1}}{-0.1987\ \text{kJ K}^{-1}\text{mol}^{-1}} = 465.1\ \text{K}$ .

Step 5. State the final answer. Below 465.1 K the reaction is spontaneous ( $\Delta G < 0$ , confirmed by the −33.19 kJ mol⁻¹ value at 298 K); above 465.1 K it becomes non-spontaneous. This data set closely matches the industrial synthesis of ammonia, $\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ , a classic reaction favoured by lower temperature — which is also why the Haber process is run at only moderately high temperature, balancing this thermodynamic preference against a practical reaction rate.

Competitive stretch

A reaction has $\Delta G^\circ = -11.4\ \text{kJ mol}^{-1}$ at 298 K. Estimate the equilibrium constant $K$ ( $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ ).

Step 1. Convert to consistent SI units: $\Delta G^\circ = -11{,}400\ \text{J mol}^{-1}$ .

Step 2. Rearrange $\Delta G^\circ = -RT\ln K$ for $\ln K$ : $\ln K = \dfrac{-\Delta G^\circ}{RT} = \dfrac{11{,}400}{(8.314)(298)} = \dfrac{11{,}400}{2477.6} = 4.601$ .

Step 3. Exponentiate: $K = e^{4.601} \approx 99.6$ . Since $\Delta G^\circ$ is fairly negative, $K \gg 1$ , confirming the reaction strongly favours products at equilibrium.

Common traps

Unit mismatch — $\Delta H$ is usually quoted in kJ but $\Delta S$ in J K $^{-1}$ ; forgetting to convert both to the same energy unit before combining them in $\Delta G = \Delta H - T\Delta S$ is the single most common numerical error on this topic.
Confusing spontaneity with speed — a negative $\Delta G$ only says a reaction can happen on its own, not that it happens fast; spontaneity is a thermodynamic statement and says nothing about kinetics or the rate of reaction.
Assuming one sign combination applies to every reaction — spontaneity is temperature-independent only when $\Delta H$ and $\Delta S$ have opposite signs; when they share a sign, spontaneity flips at the threshold temperature $T = \Delta H/\Delta S$ .
Flipping the equilibrium-constant relationship — a large $K$ corresponds to a negative $\Delta G^\circ$ , not a positive one; keep the minus sign in $\Delta G^\circ = -RT\ln K$ firmly in mind.

Ask the tutor: Want to practise reading spontaneity off the signs of $\Delta H$ and $\Delta S$ , or work through a few more $\Delta G^\circ$ – $K$ conversions together?

Why do we need Gibbs free energy at all if the second law already tells us what is spontaneous?

The second law's condition, $\Delta S_{total} \geq 0$ , is correct but inconvenient — it forces you to compute the entropy change of the surroundings every single time, which usually means tracking heat flow into a reservoir. Gibbs free energy absorbs that surroundings term into $-T\Delta S_{sys}$ , so $\Delta G = \Delta H - T\Delta S$ uses only properties of the system itself. It is the same physics, repackaged into a form you can actually calculate from tabulated enthalpies and entropies of reactants and products.

Can a reaction with delta G less than zero still not seem to happen when I run it?

Yes, and this trips up many students. Spontaneity (delta G less than zero) only tells you the reaction is thermodynamically favourable — it says nothing about how fast it proceeds. Many spontaneous reactions have a high activation energy barrier and are kinetically extremely slow at room temperature; diamond converting to graphite is a classic example, spontaneous by thermodynamics yet imperceptibly slow in practice. Rate is decided by kinetics, not by the sign of delta G.

How is delta G standard different from just delta G?

Delta G (no superscript) is the free energy change under whatever actual conditions a process is occurring in, and it is what directly decides spontaneity via delta G less than zero. Delta G standard, written delta G with a degree symbol, is the free energy change specifically at standard conditions, typically 298 K and 1 bar pressure with all species at unit activity or concentration. Delta G standard is a fixed reference value for a reaction, used mainly to calculate the equilibrium constant K through delta G standard equals minus RT ln K, whereas the everyday spontaneity test always uses the actual delta G of the moment.