Understand Entropy & Spontaneity — Understand Reactivity...

SECTION 1

Historical Context & Motivation

In the early nineteenth century, scientists assumed that all reactions needed heat to drive them forward. Exothermic reactions made sense—they released energy and seemed to "go downhill." But a puzzling set of observations challenged this simple picture: ice melts spontaneously at room temperature even though it absorbs heat, and certain salts dissolve in water while cooling the solution. Clearly, energy alone could not explain why some processes happen on their own. A deeper principle was needed—one that accounted for the natural tendency of matter to spread out and become more disordered.

1824

Carnot's Heat Engine Analysis

Sadi Carnot studied the efficiency of steam engines and showed that no engine can convert all heat into work, hinting that something is irreversibly "lost" in every real process.

1865

Clausius Coins 'Entropy'

Rudolf Clausius introduced the term entropy (from the Greek word for "transformation") and stated that the entropy of the universe tends to increase.

1877

Boltzmann's Statistical View

Ludwig Boltzmann connected entropy to the number of microscopic arrangements (microstates) available to a system, giving entropy a powerful molecular interpretation: S = k_B ln W.

1878

Gibbs Free Energy

Josiah Willard Gibbs combined enthalpy and entropy into a single quantity—Gibbs free energy (G)—that predicts spontaneity at constant temperature and pressure, the conditions most relevant to chemistry.

The central question this lesson addresses is deceptively simple: What determines whether a reaction is spontaneous? The answer lies in the interplay between enthalpy changes (ΔH), entropy changes (ΔS), and temperature (T), all unified by the Gibbs equation.

SECTION 2

Core Principles & Definitions

Before diving into calculations, you need a solid grasp of the key ideas that underpin entropy and spontaneity. Each of these principles builds on the others, so take them in order.

Entropy (S)

A measure of the number of possible microscopic arrangements (microstates) of a system. Higher entropy means greater disorder and more ways energy can be distributed among particles. Units: J K⁻¹ mol⁻¹.

Second Law of Thermodynamics

In any spontaneous process the total entropy of the universe (system + surroundings) increases: ΔS_total > 0. This is the ultimate criterion for spontaneity.

Gibbs Free Energy (G)

Defined as G = H − TS, where H is enthalpy, T is temperature in kelvin, and S is entropy. The change ΔG combines enthalpy and entropy into one number that predicts spontaneity at constant T and P.

Spontaneity Criterion

A reaction is spontaneous when ΔG < 0 (negative). When ΔG = 0 the system is at equilibrium; when ΔG > 0 the reaction is non-spontaneous in the forward direction.

Standard Entropy (S°)

The absolute entropy of a substance at 298 K and 100 kPa. Unlike enthalpy, absolute entropy values exist because the Third Law of Thermodynamics defines S = 0 for a perfect crystal at 0 K.

✦ KEY TAKEAWAY

Think of entropy like shuffling a deck of cards. A brand-new deck is perfectly ordered (low entropy). After a few shuffles, the cards are in one of billions of random arrangements (high entropy). Nature overwhelmingly favours the shuffled state because there are vastly more disordered arrangements than ordered ones. Similarly, molecules naturally spread out and their energy disperses—unless something forces them back into order.

SECTION 3

Visualising Entropy Changes

The diagram below illustrates how entropy changes with phase transitions and chemical processes. Notice how entropy increases as matter moves from the solid phase to liquid to gas, and how dissolving a solid or increasing the number of gas molecules in a reaction also raises entropy.

Entropy increases from left to right: solid → liquid → gas. Reactions that produce more gas molecules have a large positive ΔS. The particles (dots) become progressively more dispersed in each phase.

Several trends help you predict the sign of ΔS for a reaction. Entropy increases when a substance changes from solid to liquid or liquid to gas, when a solid dissolves in solution, when the number of moles of gas increases, or when temperature rises. Conversely, entropy decreases when gases condense, when fewer moles of gas form, or when a solution crystallises. Keeping these patterns in mind will let you quickly assess the entropy change for almost any reaction you encounter on the IB exam.

SECTION 4

Mathematical Framework

Three equations form the mathematical backbone of entropy and spontaneity at the AHL level. Master these, and you can tackle any Gibbs free energy problem the IB throws at you.

STANDARD ENTROPY CHANGE

ΔS°reaction = ΣS°(products) − ΣS°(reactants)

ΔS° is the standard entropy change of the reaction in J K⁻¹ mol⁻¹. Use absolute standard entropy values (S°) from the IB Data Booklet, multiplied by their stoichiometric coefficients.

GIBBS FREE ENERGY EQUATION

ΔG° = ΔH° − TΔS°

ΔG° = standard Gibbs free energy change (kJ mol⁻¹), ΔH° = standard enthalpy change (kJ mol⁻¹), T = temperature in kelvin (K), ΔS° = standard entropy change. Watch units: ΔS° is usually in J K⁻¹ mol⁻¹, so divide by 1000 (or multiply by 10⁻³) before subtracting from ΔH° in kJ.

CROSSOVER TEMPERATURE

T = ΔH° / ΔS°

When ΔG° = 0, the reaction is at equilibrium. Rearranging ΔG° = ΔH° − TΔS° gives the crossover temperature at which spontaneity switches. Both ΔH° and ΔS° must be in consistent units (both in kJ or both in J).

⚠️ Unit Trap — The #1 IB Mistake

ΔH° is almost always given in kJ mol⁻¹ while ΔS° is in J K⁻¹ mol⁻¹. You must convert ΔS° to kJ K⁻¹ mol⁻¹ (divide by 1000) before substituting into ΔG° = ΔH° − TΔS°. Forgetting this step will give you an answer that is off by a factor of 1000.

SECTION 5

The Four ΔH / ΔS Scenarios

Whether a reaction is spontaneous depends on the signs of ΔH and ΔS and on the temperature. There are exactly four combinations, and understanding all four is essential for IB exam success. The diagram below maps them out, and the table provides quick reference.

The four quadrants show every possible combination of ΔH sign and ΔS sign. Green (top right) is always spontaneous; red (bottom left) is never spontaneous. The amber and orange quadrants are temperature-dependent—these are the ones where calculating the crossover temperature T = ΔH°/ΔS° becomes important.

Summary of the four spontaneity scenarios

ΔH	ΔS	ΔG	Spontaneous?
− (exothermic)	+ (increases)	Always negative	Yes, at all T
− (exothermic)	− (decreases)	Negative at low T	Only at low T
+ (endothermic)	+ (increases)	Negative at high T	Only at high T
+ (endothermic)	− (decreases)	Always positive	Never (reverse is)

SECTION 6

Worked Example — Decomposition of CaCO₃

Calcium carbonate decomposes when heated strongly. Let's determine the standard Gibbs free energy change and the minimum temperature at which this decomposition becomes spontaneous.

Reaction: CaCO₃(s) → CaO(s) + CO₂(g)

Given data: ΔH° = +178 kJ mol⁻¹; S°(CaCO₃) = 92.9 J K⁻¹ mol⁻¹, S°(CaO) = 39.7 J K⁻¹ mol⁻¹, S°(CO₂) = 213.6 J K⁻¹ mol⁻¹.

Finding ΔG° and the Crossover Temperature

Step 1 — Calculate ΔS°Use the formula ΔS° = ΣS°(products) − ΣS°(reactants). Substituting: ΔS° = [39.7 + 213.6] − [92.9] = 253.3 − 92.9 = +160.4 J K⁻¹ mol⁻¹. The positive sign makes sense—a gas is being produced from a solid.

ΔS° = +160.4 J K⁻¹ mol⁻¹

Step 2 — Convert ΔS° to kJDivide by 1000: ΔS° = +160.4 ÷ 1000 = +0.1604 kJ K⁻¹ mol⁻¹. This ensures consistent units with ΔH° in kJ.

ΔS° = +0.1604 kJ K⁻¹ mol⁻¹

Step 3 — Calculate ΔG° at 298 KApply ΔG° = ΔH° − TΔS°. Substituting: ΔG° = +178 − (298 × 0.1604) = +178 − 47.8 = +130.2 kJ mol⁻¹. Since ΔG° is positive, the decomposition is non-spontaneous at room temperature—no surprise, limestone doesn't crumble on its own!

ΔG° (298 K) = +130.2 kJ mol⁻¹ → Non-spontaneous

Step 4 — Find the Crossover TemperatureSet ΔG° = 0: 0 = ΔH° − TΔS°, so T = ΔH° / ΔS° = 178 / 0.1604 = 1110 K (≈ 837 °C). Above this temperature, the TΔS° term exceeds ΔH° and ΔG° becomes negative.

T(crossover) = 1110 K ≈ 837 °C

Step 5 — Interpret the ResultThis is a ΔH > 0, ΔS > 0 scenario (bottom-right quadrant on our diagram). The reaction becomes spontaneous only at high temperatures. This matches industrial practice: limestone is heated in kilns above 900 °C to produce quicklime (CaO).

SECTION 7

Strengths & Limitations of the Gibbs Approach

The Gibbs free energy equation is a powerful predictive tool, but like all models it has boundaries. Understanding what it can and cannot tell you will prevent common misconceptions on the IB exam.

Strengths and limitations of using ΔG° to predict spontaneity

Strengths	Limitations
Predicts the thermodynamic feasibility (spontaneity) of a reaction in one calculation.	Says nothing about the rate (kinetics) of the reaction. A reaction can be spontaneous yet infinitely slow (e.g. diamond → graphite).
Identifies the crossover temperature where spontaneity switches, useful for industrial process design.	Assumes ΔH° and ΔS° are constant with temperature, which is only an approximation over narrow ranges.
Links directly to equilibrium (ΔG° = −RT ln K), bridging thermodynamics and equilibrium constants.	Standard values (°) refer to standard conditions; real reactions may occur under non-standard conditions requiring ΔG (not ΔG°).
Uses readily available data from the IB Data Booklet (ΔH°f, S° values).	Cannot predict reaction mechanisms or product distributions for competing reactions.

✦ KEY TAKEAWAY

Gibbs free energy tells you if a reaction can happen, not how fast it happens. Think of a ball on top of a hill: thermodynamics says it "wants" to roll down (spontaneous), but if there's a fence at the top (activation energy), it may sit there indefinitely until something gives it a push (a catalyst or a spark).

SECTION 8

Connection to Equilibrium & Advanced Theory

The relationship between Gibbs free energy and the equilibrium constant K provides one of the most elegant bridges in chemistry. At the AHL level, you should recognise how ΔG° connects to K, even though the full derivation is beyond the IB syllabus.

GIBBS–EQUILIBRIUM RELATIONSHIP

ΔG° = −RT ln K

R = 8.314 J K⁻¹ mol⁻¹, T = temperature in kelvin, K = equilibrium constant. A large negative ΔG° means a very large K (products strongly favoured). A positive ΔG° means K < 1 (reactants favoured).

SL vs. AHL expectations for entropy and spontaneity

Concept	Standard Level (SL)	Additional Higher Level (AHL)
Predicting spontaneity	Qualitative: exothermic + increased entropy → likely spontaneous	Quantitative: calculate ΔG° = ΔH° − TΔS°
Entropy	Know that entropy increases with disorder and in phase changes solid → liquid → gas	Calculate ΔS° from absolute S° values; use it in Gibbs equation
Temperature dependence	Recognise that some reactions only occur at high T	Calculate the crossover temperature T = ΔH°/ΔS°
Equilibrium link	Not required	ΔG° = −RT ln K links thermodynamics to equilibrium

At university level, you will explore how non-standard conditions modify ΔG using the equation ΔG = ΔG° + RT ln Q, where Q is the reaction quotient. This extends the framework to predict the direction a reaction will shift when it is not yet at equilibrium. For now, focus on mastering ΔG° calculations at standard conditions and understanding the four quadrant scenarios—these form the foundation for everything that follows.

SECTION 9

Practice Problems

PROBLEM 1 — CONCEPTUAL

Explain, using the concept of entropy, why a gas expands to fill its container spontaneously but never spontaneously compresses itself into one corner.

PROBLEM 2 — BASIC CALCULATION

Given ΔH° = −92.2 kJ mol⁻¹ and ΔS° = −198.7 J K⁻¹ mol⁻¹ for the synthesis of ammonia (N₂ + 3H₂ → 2NH₃), calculate ΔG° at 298 K and state whether the reaction is spontaneous.

PROBLEM 3 — INTERMEDIATE

For the reaction 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH° = −198 kJ mol⁻¹ and ΔS° = −187 J K⁻¹ mol⁻¹. (a) Calculate the crossover temperature. (b) Is this reaction spontaneous at 800 K?

PROBLEM 4 — APPLIED

The Haber process operates at approximately 450 °C despite ammonia synthesis being spontaneous at 298 K (see Problem 2). Using your knowledge of thermodynamics and kinetics, explain why the process uses such a high temperature. At what temperature does ammonia synthesis become non-spontaneous?

PROBLEM 5 — CRITICAL THINKING

Diamond spontaneously converting to graphite has ΔG° = −2.9 kJ mol⁻¹ at 298 K. Yet diamonds last for millions of years. Furthermore, synthetic diamonds are manufactured at high temperatures and pressures. Discuss how thermodynamic spontaneity and kinetics interact here. Why doesn't the negative ΔG° guarantee a rapid conversion?

SUMMARY

Lesson Summary

Entropy (S) measures the number of microstates available to a system—more disorder means higher entropy. The Second Law of Thermodynamics states that spontaneous processes always increase the total entropy of the universe. Entropy tends to increase when solids melt, liquids boil, solids dissolve, or the number of gas molecules increases. You can calculate ΔS° for a reaction from absolute standard entropy values: ΔS° = ΣS°(products) − ΣS°(reactants).

The Gibbs free energy equation, ΔG° = ΔH° − TΔS°, combines enthalpy and entropy to predict spontaneity: ΔG° < 0 means spontaneous, ΔG° = 0 means equilibrium, and ΔG° > 0 means non-spontaneous. The four ΔH/ΔS scenarios determine temperature dependence: reactions with ΔH < 0 and ΔS > 0 are always spontaneous, those with ΔH > 0 and ΔS < 0 are never spontaneous, and the other two combinations switch at the crossover temperature T = ΔH°/ΔS°. Always remember: ΔG° predicts feasibility, not speed—kinetics determines how fast a reaction occurs.