Calculating Gibbs Free Energy and Predicting Reaction Spontaneity

Welcome to the module on Energetics and Reaction Direction. Previously, you balanced redox reactions by tracking conserved atoms, charge, and electrons. Thermodynamics asks a different question: given a correctly written reaction, is there a driving force for it to proceed in the stated direction?

In this lesson, you will calculate Gibbs free-energy change from enthalpy, entropy, and temperature, then use its sign to decide whether a reaction is thermodynamically spontaneous. You will also learn why temperature sometimes changes that prediction.

---

Gibbs free energy: one criterion for reaction direction

A reaction may release heat, increase disorder, do both, or do neither. Gibbs free energy combines these two influences into a single quantity:

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

Here:

• $\Delta G$ is the Gibbs free-energy change.
• $\Delta H$ is the enthalpy change, associated with heat transfer at constant pressure.
• $\Delta S$ is the entropy change.
• $T$ is the absolute temperature in kelvin.

At constant temperature and pressure, the sign of $\Delta G$ gives the thermodynamic criterion:

Value of $\Delta G$	Meaning
$\Delta G<0$	The process is spontaneous in the stated direction
$\Delta G>0$	The process is nonspontaneous in the stated direction
$\Delta G=0$	The system is at equilibrium

“Spontaneous” does not mean fast. A reaction with negative $\Delta G$ has a thermodynamic driving force, but it may still be extremely slow because of a large activation energy. Reaction rates and catalysts will be addressed in the final module.

When thermodynamic quantities are given under standard-state conditions, we write a superscript circle:

$$\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$$

A calculation of $\Delta G^{\circ }$ predicts spontaneity under standard-state conditions at the given temperature. In the next lesson, you will see why the actual composition of a reaction mixture can change the direction predicted under standard conditions.

---

The calculation method and its most important trap

OpenStax’s 12.4 Free Energy develops the connection between Gibbs free energy and spontaneity, then illustrates a calculation using thermodynamic data.

The formula is simple, but calculations commonly fail because the units are inconsistent.

Usually:

$$\Delta H^{\circ }\text{ is reported in }\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$$

while

$$\Delta S^{\circ }\text{ is reported in }\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$$

Before substituting values, put both energy quantities in either joules or kilojoules. A convenient choice is usually to convert entropy to $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$:

$$1\mathrm{kJ}=1000\mathrm{J}$$

Also, temperature must be in kelvin:

$$T(\mathrm{K})=T{(}^{\circ }\mathrm{C})+273.15$$

A reliable calculation routine is:

1. Write $\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$.
2. Convert the temperature to kelvin.
3. Make the energy units of $\Delta H^{\circ }$ and $\Delta S^{\circ }$ compatible.
4. Calculate $T\Delta S^{\circ }$.
5. Subtract this term from $\Delta H^{\circ }$.
6. Interpret the sign of $\Delta G^{\circ }$.

Keep the negative sign before $T\Delta S$ visible. It determines how entropy affects spontaneity:

• If $\Delta S$ is positive, then $-T\Delta S$ is negative and favors spontaneity.
• If $\Delta S$ is negative, then $-T\Delta S$ is positive and opposes spontaneity.

---

Worked calculation: formation of solid ammonium chloride

Consider the reaction:

$$\mathrm{N}{\mathrm{H}}_3(\mathrm{g})+\mathrm{HCl}(\mathrm{g})⟶\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}(\mathrm{s})$$

Suppose its thermodynamic data at $298\mathrm{K}$ are:

$$\Delta H^{\circ }=-176.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$$ $$\Delta S^{\circ }=-284.8\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$$

The negative enthalpy says the reaction is exothermic. The negative entropy is reasonable: two gases form one ordered solid, reducing the system’s dispersal of matter and energy. But neither sign alone settles spontaneity. We must calculate $\Delta G^{\circ }$.

1. Convert entropy units

Because enthalpy is in kilojoules, convert entropy from joules to kilojoules:

$$\Delta S^{\circ }=-284.8\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}\left(\frac{1\mathrm{kJ}}{1000\mathrm{J}}\right)$$ $$\Delta S^{\circ }=-0.2848\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$$

2. Substitute into the Gibbs equation

$$\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$$ $$\Delta G^{\circ }=-176.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}-\left(298\mathrm{K}\right)\left(-0.2848\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}\right)$$

The temperature and kelvin units cancel in the product:

$$\left(298\mathrm{K}\right)\left(-0.2848\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}\right)=-84.9\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$$

Therefore:

$$\Delta G^{\circ }=-176.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}-\left(-84.9\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}\right)$$ $$\boxed{\Delta G^{\circ }=-91.1\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}}$$

Because $\Delta G^{\circ }$ is negative, the reaction is thermodynamically spontaneous under standard-state conditions at $298\mathrm{K}$.

Notice the logic in the arithmetic. The entropy change is unfavorable, so the term $-T\Delta S^{\circ }$ is positive. It offsets part of the favorable negative enthalpy. At room temperature, however, the exothermic contribution is larger in magnitude, leaving $\Delta G^{\circ }$ negative overall.

---

Why temperature can reverse spontaneity

The temperature dependence comes directly from the $T\Delta S$ term. The graph below treats $\Delta G$ as a function of temperature.

The equation

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

has two useful graphical features:

• At $T=0$, the free-energy change equals $\Delta H$. Thus, the vertical intercept is $\Delta H$.
• The slope is $-\Delta S$. A positive $\Delta S$ gives a downward slope; a negative $\Delta S$ gives an upward slope.

This gives four possible sign combinations.

$\Delta H$	$\Delta S$	Consequence for $\Delta G$	Spontaneity
Negative	Positive	Both terms favor negative $\Delta G$	At all temperatures
Positive	Negative	Both terms make $\Delta G$ positive	At no temperatures
Positive	Positive	Entropy term can outweigh enthalpy at sufficiently high $T$	At high temperatures
Negative	Negative	Entropy penalty becomes larger as $T$ rises	At low temperatures

The phrase “high temperature” is relative to a threshold temperature for that particular process. It does not necessarily mean a temperature that feels hot in daily life. For example, melting ice is favorable above $0{}^{\circ }\mathrm{C}$, even though $5{}^{\circ }\mathrm{C}$ would usually not be called hot.

For combinations where $\Delta H$ and $\Delta S$ have the same sign, the threshold occurs at equilibrium, where:

$$\Delta G=0$$

Substitute this into the Gibbs equation:

$$0=\Delta H-T\Delta S$$

Solving for temperature gives:

$$T_{\mathrm{eq}}=\frac{\Delta H}{\Delta S}$$

The units of $\Delta H$ and $\Delta S$ must match before this division. The resulting temperature is in kelvin.

For the ammonium chloride example:

$$T_{\mathrm{eq}}=\frac{-176.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}}{-0.2848\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}}$$ $$T_{\mathrm{eq}}\approx 618\mathrm{K}$$

Because both $\Delta H^{\circ }$ and $\Delta S^{\circ }$ are negative, the forward reaction is spontaneous below this threshold and nonspontaneous above it. At $298\mathrm{K}$, it is therefore favorable; at a sufficiently high temperature, the unfavorable entropy change dominates.

These classifications assume that $\Delta H$ and $\Delta S$ do not change much over the temperature range considered. That approximation is generally appropriate for introductory calculations unless the problem states otherwise.

---

A precise interpretation of your result

A negative result such as

$$\Delta G^{\circ }=-91.1\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$$

means that the reaction, as written, is thermodynamically favored under standard-state conditions at the stated temperature. It does not mean:

• the reaction proceeds instantly;
• every reactant molecule becomes product;
• the reaction cannot run in reverse;
• concentration and pressure never matter.

The first limitation belongs to kinetics: activation energy can make a favorable reaction slow. The remaining limitations belong to equilibrium thermodynamics. In the next lesson, the reaction quotient $Q$ and equilibrium constant $K$ will let you predict the direction a mixture shifts from its actual, nonstandard composition.

---

Key takeaways

Gibbs free energy is the central test for thermodynamic spontaneity at constant temperature and pressure:

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

Use the sign carefully:

• $\Delta G<0$: spontaneous in the forward direction.
• $\Delta G>0$: nonspontaneous in the forward direction.
• $\Delta G=0$: equilibrium.

For dependable calculations, always convert temperature to kelvin and make the units of $\Delta H$ and $\Delta S$ consistent before multiplying by $T$. Finally, the signs of $\Delta H$ and $\Delta S$ explain whether temperature has no effect on spontaneity or can reverse it at a threshold temperature.

Next, you will move beyond standard-state predictions by comparing $Q$ with $K$, which tells you the direction an actual reaction mixture will shift as it approaches equilibrium.