Assessing Spontaneity — Use ΔG, ΔH, and ΔS to assess spontaneity - Chemistry 2

SECTION 1

Historical Context & Motivation

For centuries, scientists wondered why some chemical reactions happen effortlessly — like iron rusting in moist air — while others refuse to proceed without a constant supply of energy. Early chemists believed that all spontaneous reactions were exothermic, meaning they released heat. However, this idea fell apart when researchers observed endothermic processes that still happened on their own, such as ice melting at room temperature. Clearly, heat flow alone could not explain the full picture.

The search for a better answer spanned over a century and involved some of the greatest minds in physics and chemistry. Their work eventually produced three powerful thermodynamic quantities — enthalpy (ΔH), entropy (ΔS), and Gibbs free energy (ΔG) — that together give a definitive answer to the question: will this reaction happen spontaneously?

1824 • Sadi Carnot & Heat Engines

French physicist Sadi Carnot studied the efficiency of steam engines and laid the groundwork for the Second Law of Thermodynamics, showing that heat naturally flows from hot to cold.

1850 • Clausius Defines Entropy

Rudolf Clausius introduced the concept of entropy (S) to quantify the dispersal of energy in a system, formalizing the idea that the universe tends toward greater disorder.

1876 • Gibbs Free Energy

Josiah Willard Gibbs combined enthalpy and entropy into a single function, G = H − TS, that predicts whether a process is spontaneous at constant temperature and pressure.

1923 • Lewis & Randall's Textbook

Gilbert N. Lewis and Merle Randall published a widely used textbook that standardized free-energy calculations, making Gibbs's ideas accessible to working chemists around the world.

The central question this lesson addresses is straightforward: How do we combine ΔH, ΔS, and temperature to predict whether a reaction is spontaneous? By the end, you will be able to calculate ΔG and interpret its sign to determine spontaneity under any conditions.

SECTION 2

Core Principles & Definitions

Before diving into calculations, you need to understand three thermodynamic quantities and one key definition. A spontaneous process is one that occurs without continuous outside intervention once started. Spontaneity does not mean a reaction happens quickly — it only means the reaction is thermodynamically favorable. A diamond slowly converting to graphite is spontaneous but incredibly slow.

Enthalpy (ΔH)

Enthalpy measures the heat exchanged at constant pressure. A negative ΔH (exothermic) means heat is released; a positive ΔH (endothermic) means heat is absorbed. Exothermic reactions tend to favor spontaneity.

Entropy (ΔS)

Entropy measures the dispersal of energy and the number of possible microstates. A positive ΔS means the system becomes more disordered (more possible arrangements). Nature generally favors increasing entropy.

Gibbs Free Energy (ΔG)

ΔG combines ΔH and ΔS into one value. When ΔG < 0, the process is spontaneous. When ΔG > 0, it is nonspontaneous. When ΔG = 0, the system is at equilibrium.

Temperature (T)

Temperature acts as a weighting factor for entropy. In the Gibbs equation, T is always in kelvins (K). Higher temperatures amplify the entropy term, potentially changing whether a reaction is spontaneous.

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KEY TAKEAWAY

Think of spontaneity like a ball on a hill. A ball naturally rolls downhill (spontaneous) — it doesn't need you to push it continuously. However, two forces compete: the tendency for energy to decrease (ΔH) and the tendency for disorder to increase (ΔS). ΔG is the net effect. When ΔG is negative, the 'thermodynamic hill' slopes downward, and the reaction proceeds on its own.

SECTION 3

Visual Explanation — The Gibbs Equation at a Glance

This diagram shows how enthalpy (ΔH) and entropy (ΔS) feed into the Gibbs equation. Temperature scales the entropy term. The sign of the resulting ΔG determines whether the process is spontaneous, at equilibrium, or nonspontaneous.

The diagram above shows the two competing thermodynamic drivers. On the left, enthalpy (ΔH) captures whether the reaction releases or absorbs heat. On the right, entropy (ΔS) captures whether the system becomes more or less disordered. Both quantities feed into the Gibbs equation at the center, where temperature acts as a multiplier on the entropy term. Notice how the sign of ΔG — not ΔH or ΔS alone — is the ultimate judge of spontaneity.

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Unit Watch

ΔH is typically given in kilojoules (kJ), while ΔS is usually in joules per kelvin (J/K). Before plugging into the Gibbs equation, you must convert one so that both use the same energy unit. The most common approach is to divide ΔS by 1000 to convert it to kJ/K.

SECTION 4

Mathematical Framework

The central equation for assessing spontaneity is the Gibbs–Helmholtz relationship. It connects the three quantities you've already met — ΔG, ΔH, and ΔS — with temperature. Below are the key equations and their variable definitions.

GIBBS FREE ENERGY EQUATION

ΔG = ΔH − TΔS

ΔG = change in Gibbs free energy (kJ); ΔH = change in enthalpy (kJ); T = absolute temperature (K); ΔS = change in entropy (kJ/K after unit conversion).

STANDARD GIBBS FREE ENERGY OF REACTION

ΔG°rxn = Σ ΔG°f(products) − Σ ΔG°f(reactants)

ΔG°f = standard Gibbs free energy of formation (kJ/mol). This lets you calculate ΔG° from tabulated values, similar to Hess's law for enthalpy.

TEMPERATURE AT EQUILIBRIUM

T = ΔH / ΔS (when ΔG = 0)

Setting ΔG = 0 and solving for T gives the crossover temperature — the temperature at which the reaction shifts between spontaneous and nonspontaneous. Both ΔH and ΔS must have the same sign for a meaningful crossover temperature.

When you use the Gibbs equation, always check your units. If ΔH is given in kJ and ΔS in J/K, divide ΔS by 1000 before multiplying by T. Also, remember that T must be in kelvins: T(K) = T(°C) + 273.15. A sign error or unit mismatch is the most common mistake students make on exams.

SECTION 5

The Four Sign Combinations of ΔH and ΔS

Because ΔG depends on both ΔH and ΔS, there are four possible combinations of their signs. Each combination has a different implication for spontaneity and its dependence on temperature. Understanding these four cases is one of the most practical skills in thermodynamics.

The four quadrants show every possible combination of ΔH and ΔS signs. Case 1 (always spontaneous) and Case 2 (never spontaneous) are temperature-independent. Cases 3 and 4 depend on temperature, switching spontaneity at a crossover temperature T = ΔH/ΔS.

Case	ΔH	ΔS	ΔG Spontaneous?	Example
1	− (exo)	+ (more disorder)	Always (all T)	Combustion of methane
2	+ (endo)	− (less disorder)	Never (all T)	Ozone from oxygen (3O₂ → 2O₃)
3	− (exo)	− (less disorder)	Low T only	Water freezing
4	+ (endo)	+ (more disorder)	High T only	Ice sublimating

Summary of the four sign combinations and their spontaneity conditions

The temperature-dependent cases (3 and 4) are the most interesting. For these reactions, there is a specific crossover temperature where ΔG equals zero and the system sits at equilibrium. Below or above that temperature (depending on the case), the reaction switches from spontaneous to nonspontaneous. You can find this crossover temperature by setting ΔG = 0 and solving: T = ΔH / ΔS.

SECTION 6

Worked Example — Assessing Spontaneity

Let's work through a complete problem. Consider the decomposition of calcium carbonate:

CaCO₃(s) → CaO(s) + CO₂(g)

Given: ΔH° = +178.3 kJ and ΔS° = +160.5 J/K. Determine whether this reaction is spontaneous at 25 °C (298 K) and find the crossover temperature.

Calculating ΔG and Crossover Temperature

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Step 1 — Identify Given Values

ΔH° = +178.3 kJ (endothermic, positive). ΔS° = +160.5 J/K (entropy increases — a gas is produced from a solid). T = 298 K. This is a Case 4 reaction (ΔH positive, ΔS positive), so we expect it to be spontaneous only at high temperatures.
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Step 2 — Convert Units

ΔS is in J/K, but ΔH is in kJ. Convert ΔS to kJ/K by dividing by 1000.

ΔS° = 160.5 J/K ÷ 1000 = 0.1605 kJ/K
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Step 3 — Substitute into ΔG = ΔH − TΔS

ΔG° = 178.3 kJ − (298 K)(0.1605 kJ/K). First, compute TΔS: 298 × 0.1605 = 47.83 kJ. Then: ΔG° = 178.3 − 47.83.

ΔG° = +130.5 kJ
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Step 4 — Interpret the Sign of ΔG

Because ΔG° is positive (+130.5 kJ), the reaction is nonspontaneous at 298 K. This makes physical sense — limestone doesn't decompose at room temperature.
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Step 5 — Find the Crossover Temperature

Set ΔG = 0 and solve for T: T = ΔH / ΔS = 178.3 kJ / 0.1605 kJ/K.

T ≈ 1111 K (≈ 838 °C). Above this temperature, the reaction becomes spontaneous. This aligns with industrial lime kilns operating above 900 °C.

SECTION 7

Strengths & Limitations of ΔG Analysis

Gibbs free energy is an incredibly powerful tool, but it has boundaries. Understanding what ΔG can and cannot tell you will help you avoid common misconceptions and use the concept correctly on assignments and in real-world reasoning.

Strengths	Limitations
Provides a definitive yes/no answer for spontaneity at a given T and P.	Says nothing about reaction speed — a spontaneous reaction can still be extremely slow (e.g., diamond → graphite).
Combines enthalpy and entropy into one value, simplifying analysis.	Standard values (ΔG°) apply at 25 °C and 1 atm. Nonstandard conditions require additional corrections (ΔG = ΔG° + RT ln Q).
Crossover temperature calculation shows when a reaction switches spontaneity.	Assumes ΔH and ΔS are roughly constant over the temperature range, which is an approximation for large T ranges.
ΔG° values can be calculated from tabulated formation data.	Does not account for the effect of catalysts, which affect rate but not spontaneity.

Strengths and limitations of using ΔG to assess spontaneity

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KEY TAKEAWAY

Think of ΔG as a traffic light for reactions. Green (ΔG < 0) means go — the reaction is thermodynamically allowed. Red (ΔG > 0) means stop — the reaction won't happen on its own. But even a green light doesn't tell you how fast traffic will move — that depends on kinetics (activation energy), a separate topic entirely.

SECTION 8

Connection to Advanced Theory

The standard Gibbs equation you've learned is a powerful starting point, but chemistry courses at the AP and college level extend it in important ways. The table below previews how the concepts you've mastered connect to more advanced treatments.

This Lesson	Advanced Extension
ΔG = ΔH − TΔS at standard conditions	ΔG = ΔG° + RT ln Q relates free energy to the reaction quotient Q under nonstandard conditions.
ΔG < 0 → spontaneous	At equilibrium, ΔG = 0, leading to ΔG° = −RT ln K, which links free energy to the equilibrium constant K.
Crossover T = ΔH / ΔS	The van 't Hoff equation shows how K changes with temperature: ln(K₂/K₁) = −ΔH/R × (1/T₂ − 1/T₁).
ΔH and ΔS assumed constant	Kirchhoff's law and temperature-dependent heat capacities (Cₚ) let you adjust ΔH and ΔS for large temperature changes.

How this lesson connects to AP and college-level thermodynamics

Don't worry about mastering these extensions now. The key idea is that ΔG = ΔH − TΔS is the foundation upon which all of equilibrium thermodynamics is built. Once you're comfortable predicting spontaneity with ΔG, you'll be well prepared to tackle nonstandard conditions and the relationship between ΔG° and K in later courses.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A reaction has a negative ΔH and a positive ΔS. Without doing any calculation, is this reaction spontaneous at all temperatures, never spontaneous, or temperature-dependent? Explain your reasoning.

Show answer

This reaction is spontaneous at all temperatures. Using ΔG = ΔH − TΔS, the ΔH term is negative (which makes ΔG more negative) and the −TΔS term is also negative (since ΔS is positive and T is always positive in kelvins). Both terms push ΔG in the negative direction, so ΔG is always negative regardless of T. This is Case 1 in the four-case framework.

Question 2: BASIC CALCULATION

A reaction has ΔH = −92.0 kJ and ΔS = −198.0 J/K. Calculate ΔG at 25 °C (298 K). Is the reaction spontaneous at this temperature?

Show answer

First, convert ΔS: −198.0 J/K ÷ 1000 = −0.1980 kJ/K. Then apply the Gibbs equation: ΔG = −92.0 − (298)(−0.1980) = −92.0 − (−59.0) = −92.0 + 59.0 = −33.0 kJ. Since ΔG is negative, the reaction is spontaneous at 298 K. This is a Case 3 reaction (both ΔH and ΔS negative), and 298 K is below the crossover temperature of 465 K (92.0/0.1980), so it is spontaneous at this relatively low temperature.

Question 3: INTERMEDIATE

The formation of nitrogen dioxide from its elements has ΔH° = +33.2 kJ/mol and ΔS° = +249.0 J/(mol·K). Calculate the crossover temperature and determine whether the reaction is spontaneous at 200 °C.

Show answer

Convert ΔS°: 249.0 J/(mol·K) ÷ 1000 = 0.2490 kJ/(mol·K). The crossover temperature is T = ΔH/ΔS = 33.2/0.2490 ≈ 133.3 K (≈ −139.8 °C). Since both ΔH and ΔS are positive (Case 4), the reaction is spontaneous above 133.3 K. At 200 °C (473 K), the temperature is well above the crossover, so the reaction is spontaneous. You can verify: ΔG = 33.2 − (473)(0.2490) = 33.2 − 117.8 = −84.6 kJ, which is indeed negative.

Question 4: APPLIED

In steelmaking, iron(III) oxide is reduced by carbon monoxide: Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g). Given ΔH° = −24.8 kJ and ΔS° = +15.2 J/K, calculate ΔG° at 25 °C and explain why blast furnaces operate at approximately 1500 °C even though the reaction is spontaneous at room temperature.

Show answer

Convert ΔS°: 15.2 J/K ÷ 1000 = 0.0152 kJ/K. ΔG° = −24.8 − (298)(0.0152) = −24.8 − 4.53 = −29.3 kJ. The reaction is spontaneous at 25 °C (Case 1: ΔH negative, ΔS positive — spontaneous at all temperatures). However, spontaneity does not mean fast. At room temperature, the reaction proceeds extremely slowly because the activation energy barrier is very high. Blast furnaces operate at 1500 °C to overcome this kinetic barrier and dramatically increase the reaction rate, not to change the thermodynamic favorability.

Question 5: CRITICAL THINKING

Consider two reactions at 298 K. Reaction A has ΔG° = −50 kJ and Reaction B has ΔG° = −200 kJ. A student claims that Reaction B must proceed faster than Reaction A because it has a more negative ΔG°. Evaluate this claim. Then, describe a scenario where ΔG can equal exactly zero for a process that is clearly still occurring.

Show answer

The student's claim is incorrect. ΔG indicates the thermodynamic driving force (how favorable a reaction is), not its rate. Reaction rate depends on kinetics — specifically the activation energy (Eₐ) and the presence or absence of catalysts. A reaction with a very negative ΔG can be extremely slow if it has a high activation energy barrier (e.g., diamond → graphite has ΔG° ≈ −2.9 kJ/mol but essentially never happens at room temperature). For the second part, ΔG = 0 at equilibrium. At equilibrium, the forward and reverse reactions are still occurring — they just proceed at equal rates so there is no net change. A classic example is a saturated solution: dissolving and crystallization happen continuously, but the concentration of the dissolved solute remains constant.

SUMMARY

Lesson Summary

Summary

The Gibbs free energy equation, ΔG = ΔH − TΔS, is the definitive tool for assessing spontaneity at constant temperature and pressure. Enthalpy (ΔH) represents the heat flow, while entropy (ΔS) captures the change in disorder. A reaction is spontaneous when ΔG < 0, nonspontaneous when ΔG > 0, and at equilibrium when ΔG = 0.

There are four sign combinations. When ΔH is negative and ΔS is positive (Case 1), the reaction is always spontaneous. When ΔH is positive and ΔS is negative (Case 2), it is never spontaneous. Cases 3 and 4 are temperature-dependent, and you can find the crossover temperature with T = ΔH / ΔS. Always convert ΔS to kJ/K before calculating, and remember that spontaneity ≠ speed — ΔG tells you whether a reaction is favorable, not how fast it occurs.