What is Gibbs Free Energy & When is a Reaction Spontaneous?

What Is Gibbs Free Energy?

When chemists ask "will this reaction happen?", they aren't asking about speed — they want to know if the reaction is thermodynamically feasible (spontaneous). The answer lies in Gibbs Free Energy ( $\Delta G$ ).

$\Delta G$ combines two competing drives in nature:

Enthalpy ( $\Delta H$ ): The tendency toward lower energy (exothermic reactions release heat).
Entropy ( $\Delta S$ ): The tendency toward greater disorder (systems become more disordered over time).

Learning Goals: By the end of this guide, you should be able to:

Use $\Delta G = \Delta H - T\Delta S$ to predict spontaneity.

Identify when temperature determines whether a reaction is spontaneous.

Calculate the temperature at which spontaneity changes.

Interpret all four $\Delta H / \Delta S$ combinations.

Avoid common unit-conversion and sign errors.

The Gibbs Equation

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

Symbol	Meaning	Unit
$\Delta G$	Change in Gibbs Free Energy	$kJ\ mol^{-1}$
$\Delta H$	Change in Enthalpy	$kJ\ mol^{-1}$
$T$	Temperature	$K$ (Kelvin)
$\Delta S$	Change in Entropy	$J\ mol^{-1}K^{-1}$

The Three Outcomes

$\Delta G < 0$ → Spontaneous (thermodynamically feasible). The reaction favours products.
$\Delta G > 0$ → Non-spontaneous. The reaction favours reactants under these conditions.
$\Delta G = 0$ → Equilibrium. The system is at balance.

Critical warning: Spontaneous does NOT mean fast. Diamond converting to graphite is thermodynamically spontaneous ( $\Delta G < 0$ ) but takes millions of years. Speed is determined by kinetics, not thermodynamics.

The Four Scenarios

The signs of $\Delta H$ and $\Delta S$ together determine how temperature affects spontaneity:

$\Delta H$	$\Delta S$	$\Delta G$	Spontaneity
Negative (exothermic)	Positive (more disorder)	Always negative	Spontaneous at all temperatures
Positive (endothermic)	Negative (less disorder)	Always positive	Never spontaneous
Negative (exothermic)	Negative (less disorder)	Depends on $T$	Spontaneous at low $T$ (enthalpy-driven)
Positive (endothermic)	Positive (more disorder)	Depends on $T$	Spontaneous at high $T$ (entropy-driven)

The last two cases are the most interesting — temperature acts as the "switch" that can flip spontaneity.

Finding the Crossover Temperature

When $\Delta G = 0$ :

0 = \Delta H - T\Delta S \quad \Rightarrow \quad T = \frac{\Delta H}{\Delta S}

Above or below this temperature, the reaction switches between spontaneous and non-spontaneous.

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Entropy: A Closer Look

Entropy ( $S$ ) measures the number of ways energy can be distributed in a system — more possible arrangements = higher entropy.

Rules for Predicting $\Delta S$

Change	$\Delta S$	Why
Solid → Liquid → Gas	Positive	Particles become more disordered
Fewer moles → More moles of gas	Positive	More gas particles = more arrangements
Dissolving a solid in water	Usually positive	Ions/molecules spread out in solution
More moles → Fewer moles of gas	Negative	Fewer ways to arrange fewer particles

Example:

CaCO_3(s) \rightarrow CaO(s) + CO_2(g)

$\Delta S > 0$ because a gas is produced from a solid (1 solid → 1 solid + 1 gas).

Worked Examples

Example 1: Basic Calculation

Given: $\Delta H = -92\ kJ\ mol^{-1}$ , $\Delta S = -198\ J\ mol^{-1}K^{-1}$ , $T = 298\ K$

Step 1: Convert $\Delta S$ to kJ: $-198 \div 1000 = -0.198\ kJ\ mol^{-1}K^{-1}$

Step 2: Apply the equation:

\Delta G = -92 - (298)(-0.198) = -92 + 59.0 = -33.0\ kJ\ mol^{-1}

Result: $\Delta G < 0$ , so the reaction is spontaneous at 298 K.

Example 2: Finding the Crossover Temperature

Given: $\Delta H = +178\ kJ\ mol^{-1}$ , $\Delta S = +161\ J\ mol^{-1}K^{-1}$

This is the thermal decomposition of calcium carbonate ( $CaCO_3$ ). Both $\Delta H$ and $\Delta S$ are positive (endothermic, more disorder), so it becomes spontaneous at high temperature.

T = \frac{\Delta H}{\Delta S} = \frac{178}{0.161} = 1106\ K\ (833°C)

Below 833°C: non-spontaneous. Above 833°C: spontaneous. This is why limestone must be heated to high temperatures in a lime kiln.

Example 3: Which Scenario?

Question: The synthesis of ammonia: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ , $\Delta H = -92\ kJ\ mol^{-1}$

Analysis:

$\Delta H < 0$ (exothermic ✓)
4 moles of gas → 2 moles of gas → $\Delta S < 0$ (fewer gas molecules, less disorder)
This is the "Exothermic + Less Disorder" scenario: spontaneous at low temperature, non-spontaneous at high temperature.

This explains the Haber process compromise: low temperature favours yield but slows the reaction, so an intermediate temperature (~450°C) with a catalyst is used.

Common Mistakes

Forgetting to convert $\Delta S$ from J to kJ — This is the #1 error. $\Delta H$ is in kJ, $\Delta S$ is in J. You must divide $\Delta S$ by 1000 before plugging into the equation.
Using Celsius instead of Kelvin — Temperature must be in Kelvin ( $K = °C + 273$ ). Using °C gives completely wrong answers.
Confusing spontaneous with fast — $\Delta G < 0$ means the reaction is thermodynamically feasible, NOT that it happens quickly. Rusting is spontaneous but slow; explosions are spontaneous and fast. Rate depends on activation energy and kinetics.
Getting the sign of $T\Delta S$ wrong — Remember: subtracting a negative makes a positive. If $\Delta S$ is negative, then $-T\Delta S$ is positive, which makes $\Delta G$ more positive (less spontaneous).
Thinking "non-spontaneous" means impossible — Non-spontaneous reactions can still be driven by coupling them with spontaneous ones, or by applying external energy (e.g., electrolysis).

Exam Tips (A-Level / AP / IB)

Show your unit conversion explicitly: " $\Delta S = +161\ J\ mol^{-1}K^{-1} = +0.161\ kJ\ mol^{-1}K^{-1}$ ". Examiners award marks for this step.
For the crossover temperature, always state: "At $T = \Delta H / \Delta S$ , $\Delta G = 0$ and the system is at equilibrium."
When given a scenario question (e.g., "explain why reaction X is spontaneous at high T"), identify $\Delta H$ and $\Delta S$ signs first, then justify using the four-box table.
Remember: $\Delta G$ predicts the direction of reaction, not the position of equilibrium. $\Delta G° < 0$ means $K > 1$ (products favoured), not that the reaction goes to completion.

Frequently Asked Questions

What does Gibbs Free Energy actually measure?

$\Delta G$ measures the maximum amount of non-expansion work a system can do at constant temperature and pressure. In practical terms, it tells you whether a reaction will proceed spontaneously (without external energy input).

Can a reaction with positive ΔG ever happen?

Yes, but not on its own. A non-spontaneous reaction can be driven by coupling it to a more spontaneous reaction, or by supplying external energy. Electrolysis, for example, drives non-spontaneous decomposition of water using electrical energy.

Why does temperature affect spontaneity?

The $T\Delta S$ term in the Gibbs equation means that entropy's contribution grows with temperature. At high $T$ , the entropy term ( $-T\Delta S$ ) dominates. At low $T$ , the enthalpy term ( $\Delta H$ ) dominates.

What is the relationship between ΔG and equilibrium?

At standard conditions: $\Delta G° = -RT\ln K$ . If $\Delta G° < 0$ , then $K > 1$ (products favoured at equilibrium). If $\Delta G° > 0$ , then $K < 1$ (reactants favoured). If $\Delta G° = 0$ , then $K = 1$ .

Is ΔG the same as ΔG°?

No. $\Delta G°$ is the standard Gibbs energy change (at standard conditions: 298 K, 1 bar, 1 mol/L). $\Delta G$ is the actual free energy change under non-standard conditions, calculated using $\Delta G = \Delta G° + RT\ln Q$ .

Hess's Law — Calculate $\Delta H$ values using energy cycles, which feed into Gibbs calculations.
Born-Haber Cycles — Apply energy cycle analysis to ionic compound formation.
Le Chatelier's Principle — Understand how temperature shifts equilibrium — directly related to the $\Delta G$ sign.

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