General Chemistry 2 • Thermochemistry & Thermodynamics

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

Learn how Gibbs free energy unites enthalpy and entropy to predict whether a process will occur without external intervention.

An atom with laboratory flask and electron orbit for General Chemistry 2
Inside this lesson 10 sections 5 questions 9 figures

SECTION 1

Historical Context & Motivation

The question of why certain chemical reactions proceed on their own while others require continuous energy input has occupied scientists for centuries. Early investigators like Antoine Lavoisier focused almost exclusively on heat transfer, assuming that exothermic reactions were inherently favored by nature. This view seemed intuitive—burning wood releases heat, iron rusts with a warm glow—but it failed spectacularly for endothermic processes that nonetheless occur spontaneously, such as the dissolution of ammonium nitrate in water, which absorbs heat yet proceeds without any external driving force. The resolution of this puzzle required an entirely new thermodynamic quantity that could account for both energy and disorder simultaneously.

1824 • Carnot's Heat Engine Analysis

Sadi Carnot published Réflexions sur la puissance motrice du feu, establishing that not all heat can be converted to work. This insight laid the conceptual groundwork for the Second Law of Thermodynamics and the idea that nature imposes directionality on energy flow.

1850–1865 • Clausius Defines Entropy

Rudolf Clausius formalized the concept of entropy (S), declaring that the entropy of the universe tends toward a maximum. He introduced the inequality δq/T ≤ dS, providing a quantitative criterion for irreversibility and spontaneous change in isolated systems.

1873–1876 • Gibbs Free Energy Formulated

Josiah Willard Gibbs published a series of papers introducing the function G = H − TS, now called Gibbs free energy. This single state function allowed chemists to predict spontaneity at constant temperature and pressure—the conditions under which most laboratory and biological reactions occur.

1923 • Lewis and Randall Systematize Chemical Thermodynamics

Gilbert N. Lewis and Merle Randall published Thermodynamics and the Free Energy of Chemical Substances, making standard free energy tables accessible to practicing chemists and establishing the notation conventions still used today.

The central question that Gibbs resolved was deceptively simple: given a chemical reaction at constant temperature and pressure, will it proceed in the forward direction without outside intervention? Answering this requires balancing two competing thermodynamic tendencies—the drive toward lower enthalpy and the drive toward greater entropy—within a single criterion, ΔG. The remainder of this lesson develops that criterion rigorously and shows how to apply it.

SECTION 2

Core Principles & Definitions

Before diving into calculations, it is essential to establish precise definitions for the three state functions that govern spontaneity. Each captures a different facet of a system's thermodynamic behavior: energy exchange with the surroundings, molecular randomness, and the net thermodynamic driving force. Together they form the foundation of the Gibbs equation, which serves as the universal criterion for spontaneity under laboratory conditions.

1

Enthalpy (ΔH)

Enthalpy is the heat absorbed or released at constant pressure. An exothermic reaction (ΔH < 0) releases heat, lowering the system's energy and generally favoring spontaneity. An endothermic reaction (ΔH > 0) absorbs heat, opposing spontaneity from the energy standpoint alone.
2

Entropy (ΔS)

Entropy quantifies the dispersal of energy and matter among accessible microstates. Systems tend toward greater disorder: a positive ΔS (more microstates in products than reactants) favors spontaneity. Processes that decrease entropy (ΔS < 0) require an enthalpic or external driving force to proceed.
3

Gibbs Free Energy (ΔG)

Gibbs free energy combines enthalpy and entropy into a single criterion: ΔG = ΔH − TΔS. When ΔG < 0, the process is spontaneous (thermodynamically favorable). When ΔG > 0, the reverse process is spontaneous. When ΔG = 0, the system is at equilibrium.
4

Temperature as the Arbiter

Temperature (T in kelvins) scales the entropy contribution via the TΔS term. At low temperatures, ΔH dominates the sign of ΔG; at high temperatures, ΔS dominates. This temperature dependence explains why some reactions become spontaneous only upon heating or cooling.
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KEY TAKEAWAY

Think of ΔG as the balance in a thermodynamic bank account. Enthalpy (ΔH) represents incoming or outgoing cash—energy deposits or withdrawals. Entropy (TΔS) represents a bonus or penalty that depends on how widely the energy is spread. A reaction is spontaneous when the account finishes in the black (ΔG < 0). Just as a large bonus can offset a withdrawal, a sufficiently large positive TΔS can overcome an unfavorable endothermic ΔH, especially at high temperatures.

SECTION 3

Visualizing the Four Spontaneity Scenarios

The sign of ΔG depends on the interplay between ΔH, ΔS, and T. Because each of ΔH and ΔS can be positive or negative, there are four distinct thermodynamic scenarios. The diagram below maps these scenarios onto a two-by-two grid, showing how the sign of ΔG changes with temperature in each case. Pay particular attention to the two temperature-dependent cases, where a crossover temperature exists at which ΔG switches sign.

Four Scenarios for Spontaneity: ΔG = ΔH − TΔSΔS > 0ΔS < 0ΔH < 0ΔH > 0CASE 1ΔH < 0, ΔS > 0ΔG < 0 at ALL temperaturesAlways Spontaneous ✓Example: 2 H₂O₂(l) → 2 H₂O(l) + O₂(g)(−TΔS negative, ΔH negative → both favor ΔG < 0)CASE 2ΔH < 0, ΔS < 0ΔG < 0 only at LOW TSpontaneous at Low TExample: H₂O(l) → H₂O(s) (freezing)(ΔH wins over −TΔS when T is small)CASE 3ΔH > 0, ΔS > 0ΔG < 0 only at HIGH TSpontaneous at High TExample: CaCO₃(s) → CaO(s) + CO₂(g)(−TΔS overcomes +ΔH at high T)CASE 4ΔH > 0, ΔS < 0ΔG > 0 at ALL temperaturesNever Spontaneous ✗Example: 3 O₂(g) → 2 O₃(g)(Both terms make ΔG positive)
The four spontaneity scenarios arranged by the signs of ΔH (rows) and ΔS (columns). Case 1 (always spontaneous) and Case 4 (never spontaneous) are temperature-independent, while Case 2 and Case 3 depend on temperature. The crossover temperature where ΔG = 0 is T = ΔH/ΔS.

The diagram makes clear that enthalpy and entropy are not independent criteria—they compete through the temperature term. In Cases 2 and 3, there exists a crossover temperature (Tcrossover = ΔH / ΔS) at which ΔG transitions from negative to positive or vice versa. Below this temperature, the enthalpy term dominates; above it, the entropy term takes control. Recognizing which case applies to a given reaction is the first step in any spontaneity assessment.

SECTION 4

Mathematical Framework

The mathematical framework for assessing spontaneity begins with the fundamental Gibbs equation and extends to standard-state calculations and the relationship between ΔG° and the equilibrium constant. Each equation below is presented with its assumptions and the physical meaning of every variable.

GIBBS EQUATION

ΔG=ΔH−TΔSΔG = ΔH − TΔS

ΔG = change in Gibbs free energy (J or kJ); ΔH = change in enthalpy (J or kJ); T = absolute temperature (K); ΔS = change in entropy (J/K or kJ/K). Critical: T must be in kelvins, and ΔH and TΔS must share the same energy units before combining.

STANDARD FREE ENERGY OF REACTION

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

ΔG°f = standard free energy of formation of each substance; n = stoichiometric coefficient. This parallels the Hess's law approach used for ΔH°rxn calculations. Standard conditions: 1 atm, 298 K, 1 M concentration.

CROSSOVER TEMPERATURE

Tcrossover=ΔH/ΔST_crossover = ΔH / ΔS

At this temperature, ΔG = 0 and the system is at equilibrium. This expression is valid when ΔH and ΔS have the same sign (Cases 2 and 3). Ensure that ΔH and ΔS are in compatible units (e.g., both in kJ and kJ/K, or both in J and J/K).

FREE ENERGY AND EQUILIBRIUM

ΔG°=−RTlnKΔG° = −RT ln K

R = 8.314 J/(mol·K); K = thermodynamic equilibrium constant. When ΔG° < 0, K > 1 (products favored at equilibrium). When ΔG° > 0, K < 1 (reactants favored). This equation bridges thermodynamics and chemical equilibrium.

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Common Unit Pitfall

The most frequent error in ΔG calculations is a unit mismatch. Enthalpy values from tables are typically given in kJ/mol, while entropy values are given in J/(mol·K). Before computing ΔG = ΔH − TΔS, convert ΔS to kJ/(mol·K) by dividing by 1000, or convert ΔH to J/mol by multiplying by 1000. Failure to do so will produce an answer that is off by a factor of 1000.

SECTION 5

Temperature Dependence of ΔG

The Gibbs equation can be viewed as a linear function of temperature: ΔG = ΔH − TΔS is analogous to y = b + mx, where ΔG plays the role of y, T is the independent variable x, −ΔS is the slope, and ΔH is the y-intercept. This perspective makes it straightforward to visualize how ΔG varies with temperature for each of the four cases, and it reveals the crossover temperature geometrically as the x-intercept of the line.

ΔG vs. Temperature for the Four CasesT (K)ΔG0ΔG > 0 (non-spontaneous)ΔG < 0 (spontaneous)Case 1ΔH<0, ΔS>0 → always below axisCase 4ΔH>0, ΔS<0 → always above axisCase 2T_crossCase 3T_crossCase 1 (always spont.)Case 2 (low T spont.)Case 3 (high T spont.)Case 4 (never spont.)
Plot of ΔG versus temperature for the four spontaneity cases. Case 1 (solid green) remains entirely below the ΔG = 0 axis. Case 4 (solid red) remains entirely above it. Cases 2 (dashed violet) and 3 (dashed amber) cross the axis at the crossover temperature Tcross = ΔH / ΔS, where the system reaches equilibrium.
CaseΔHΔSΔG Spontaneous?Example
1< 0 (exo)> 0 (↑ disorder)Always (all T)Combustion of CH₄
2< 0 (exo)< 0 (↓ disorder)Low T onlyFreezing of H₂O
3> 0 (endo)> 0 (↑ disorder)High T onlyCaCO₃ decomposition
4> 0 (endo)< 0 (↓ disorder)Never (no T)3 O₂(g) → 2 O₃(g)

Summary of the four ΔG spontaneity cases

When confronted with a new reaction, the strategy is straightforward: determine the signs of ΔH and ΔS, identify the applicable case, and—if the case is temperature-dependent—calculate the crossover temperature. If the question specifies a particular temperature, substitute into ΔG = ΔH − TΔS and check the sign directly. The linear relationship ensures that these calculations are algebraically simple, but unit vigilance is essential: always convert ΔS from J/(mol·K) to kJ/(mol·K) or ΔH from kJ to J before combining terms.

SECTION 6

Worked Example

Consider the thermal decomposition of calcium carbonate, a classic example of a Case 3 reaction:

CaCO3(s) → CaO(s) + CO2(g)

Given: ΔH° = +178.3 kJ/mol, ΔS° = +160.5 J/(mol·K). Determine (a) whether the reaction is spontaneous at 25 °C, (b) the crossover temperature, and (c) whether it is spontaneous at 1500 K.

Decomposition of CaCO₃

  1. 1

    Step 1 — Identify the Case

    ΔH° is positive (endothermic) and ΔS° is positive (entropy increases because a gas is produced from a solid). This matches Case 3: the reaction should be spontaneous only at sufficiently high temperatures, where the TΔS term overwhelms the positive ΔH.
  2. 2

    Step 2 — Convert Units

    ΔS° = +160.5 J/(mol·K) must be converted to kJ/(mol·K) to match the enthalpy units: ΔS° = +160.5 / 1000 = +0.1605 kJ/(mol·K). This is a critical step that prevents order-of-magnitude errors.
    ΔS° = +0.1605 kJ/(mol·K)
  3. 3

    Step 3 — Calculate ΔG° at 25 °C (298 K)

    ΔG° = ΔH° − TΔS° = +178.3 kJ/mol − (298 K)(0.1605 kJ/(mol·K)) = +178.3 − 47.8 = +130.5 kJ/mol. Since ΔG° > 0, the reaction is non-spontaneous at 298 K.
    ΔG°(298 K) = +130.5 kJ/mol → non-spontaneous
  4. 4

    Step 4 — Calculate the Crossover Temperature

    At the crossover temperature, ΔG = 0: Tcross = ΔH° / ΔS° = 178.3 kJ/mol ÷ 0.1605 kJ/(mol·K) = 1111 K (≈ 838 °C). Above this temperature, ΔG becomes negative and the reaction is spontaneous.
    Tcross = 1111 K
  5. 5

    Step 5 — Calculate ΔG° at 1500 K

    ΔG° = +178.3 − (1500)(0.1605) = +178.3 − 240.8 = −62.5 kJ/mol. The negative sign confirms that the decomposition is spontaneous at 1500 K, consistent with the industrial lime-burning process that operates above 900 °C.
    ΔG°(1500 K) = −62.5 kJ/mol → spontaneous
ℹ

Assumption Check

This calculation assumes that ΔH° and ΔS° are approximately temperature-independent over the range considered. In reality, heat capacities cause both to drift with temperature, but for most general chemistry applications this approximation is acceptable. In physical chemistry and engineering, Kirchhoff's equation and heat capacity integrals refine the estimate.

SECTION 7

Spontaneity vs. Rate: Strengths & Limitations

A common misconception is that a spontaneous reaction must occur quickly. In reality, thermodynamics tells us whether a reaction can occur, not how fast it will proceed. The distinction between thermodynamic favorability and kinetic feasibility is one of the most important in all of chemistry. The table below contrasts the capabilities and limitations of the Gibbs free energy criterion with kinetic considerations.

CriterionΔG (Thermodynamics)Eₐ / Rate (Kinetics)
What it answersWill the reaction occur? (direction & feasibility)How fast will it occur? (rate & mechanism)
State function?Yes — depends only on initial and final statesNo — depends on the pathway (mechanism)
Temperature effectChanges sign of ΔG via the TΔS termIncreases rate exponentially (Arrhenius equation)
Catalyst effectNo effect on ΔGLowers Eₐ, increases rate
Classic exampleDiamond → graphite: ΔG < 0, but immeasurably slowH₂ + O₂ mixture: thermodynamically favorable, kinetically stable until sparked

Thermodynamic spontaneity vs. kinetic rate

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KEY TAKEAWAY

Imagine you are standing on a hill. Thermodynamics (ΔG) tells you whether the valley below is lower than your current position—if so, rolling downhill is spontaneous. But kinetics (Eₐ) determines whether there is a barrier or a fence between you and the valley. A large activation energy is like a tall fence: even though downhill is thermodynamically favored, you cannot get there without enough energy to clear the obstacle. This is why diamonds persist indefinitely at room temperature despite being thermodynamically less stable than graphite.

SECTION 8

Connections to Equilibrium & Non-Standard Conditions

The standard Gibbs free energy (ΔG°) applies strictly at standard-state concentrations and pressures (1 M, 1 atm, 298 K unless otherwise specified). Real systems rarely operate under these conditions, so chemistry extends the framework to non-standard states using the reaction quotient Q. The relationship ΔG = ΔG° + RT ln Q adjusts the free energy for actual concentrations, connecting the thermodynamic criterion to real-world chemical conditions. Furthermore, the equation ΔG° = −RT ln K provides a direct bridge between free energy and the equilibrium constant, enabling prediction of equilibrium position from thermodynamic data alone.

ConceptStandard (ΔG°)Non-Standard (ΔG)
EquationΔG° = ΔH° − TΔS°ΔG = ΔG° + RT ln Q
ConditionsAll species at 1 M or 1 atmActual concentrations / pressures
EquilibriumΔG° = −RT ln K (relates to K)ΔG = 0 when Q = K
Sign interpretationTells whether K > 1 or K < 1Tells direction of net reaction at given Q
Use in courseworkGeneral chemistry, tabulated dataPhysical chemistry, biochemistry, electrochemistry

Standard vs. non-standard free energy

In more advanced courses—physical chemistry, biochemistry, and electrochemistry—the non-standard form ΔG = ΔG° + RT ln Q becomes indispensable. In electrochemistry, it gives rise to the Nernst equation (E = E° − (RT/nF) ln Q), which adjusts cell potentials for non-standard concentrations. In biochemistry, a modified standard state (ΔG°', pH 7) is used because biological reactions occur under physiological conditions rather than pure standard states. Understanding the standard-state Gibbs equation thoroughly is therefore not merely an end in itself but a gateway to these more specialized applications.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A reaction has ΔH < 0 and ΔS < 0. Without performing any calculation, explain at what temperatures this reaction will be spontaneous and provide the physical reasoning behind your answer.
Show answer
This is a Case 2 reaction: spontaneous only at low temperatures. At low T, the negative ΔH dominates ΔG = ΔH − TΔS, making ΔG negative. As temperature increases, the magnitude of the TΔS term (which is positive because ΔS is negative, making −TΔS positive) grows and eventually overwhelms the favorable ΔH. Physically, the reaction releases energy but decreases disorder; nature 'allows' the entropy decrease only when the temperature is low enough that the energy benefit outweighs the disorder penalty. The crossover temperature T = ΔH/ΔS marks the boundary. A common example is freezing water: exothermic but entropy-decreasing, and spontaneous only below 273 K at 1 atm.

Question 2: BASIC CALCULATION

For the reaction N₂(g) + 3 H₂(g) → 2 NH₃(g), ΔH° = −92.2 kJ/mol and ΔS° = −198.7 J/(mol·K). Calculate ΔG° at 298 K and determine whether the reaction is spontaneous under standard conditions.
Show answer
First, convert ΔS° to kJ/(mol·K): −198.7 J/(mol·K) ÷ 1000 = −0.1987 kJ/(mol·K). Then apply the Gibbs equation: ΔG° = ΔH° − TΔS° = −92.2 kJ/mol − (298 K)(−0.1987 kJ/(mol·K)) = −92.2 + 59.2 = −33.0 kJ/mol. Since ΔG° < 0, the synthesis of ammonia is spontaneous at 298 K under standard conditions. Note that this is a Case 2 reaction (ΔH < 0, ΔS < 0), so at sufficiently high temperatures it will become non-spontaneous. The crossover temperature is T = 92.2 / 0.1987 ≈ 464 K.

Question 3: INTERMEDIATE

The vaporization of ethanol (C₂H₅OH) has ΔH° = +38.6 kJ/mol and ΔS° = +110.0 J/(mol·K). Calculate the normal boiling point of ethanol (the temperature at which liquid and vapor are in equilibrium at 1 atm).
Show answer
At the boiling point, liquid and vapor coexist in equilibrium, meaning ΔG = 0. Setting ΔG = ΔH − TΔS = 0 and solving: T = ΔH / ΔS. Converting units: ΔS = 110.0 J/(mol·K) = 0.1100 kJ/(mol·K). Therefore T = 38.6 kJ/mol ÷ 0.1100 kJ/(mol·K) = 351 K, which corresponds to 78 °C. This agrees well with the experimentally measured boiling point of ethanol (78.37 °C). The small discrepancy arises because ΔH and ΔS are slightly temperature-dependent, but the approximation is excellent for this purpose.

Question 4: APPLIED

In a steel blast furnace, iron(III) oxide is reduced by carbon monoxide: Fe₂O₃(s) + 3 CO(g) → 2 Fe(s) + 3 CO₂(g). Given ΔH° = −24.8 kJ/mol and ΔS° = +15.2 J/(mol·K), calculate ΔG° at the furnace operating temperature of 1200 K. Is the reaction spontaneous? Would you expect this reaction to also be spontaneous at room temperature?
Show answer
Convert ΔS°: +15.2 J/(mol·K) = +0.0152 kJ/(mol·K). At 1200 K: ΔG° = −24.8 − (1200)(0.0152) = −24.8 − 18.2 = −43.0 kJ/mol. Since ΔG° < 0, the reaction is spontaneous at 1200 K. This is a Case 1 reaction (ΔH < 0, ΔS > 0), so ΔG is negative at all temperatures. At 298 K: ΔG° = −24.8 − (298)(0.0152) = −24.8 − 4.5 = −29.3 kJ/mol, confirming spontaneity at room temperature as well. The high furnace temperature is used not because it is thermodynamically required, but because it increases the reaction rate and allows efficient kinetic throughput—a perfect illustration of the distinction between thermodynamic spontaneity and kinetic feasibility.

Question 5: CRITICAL THINKING

A biochemist measures ΔG° = +13.8 kJ/mol for the isomerization of dihydroxyacetone phosphate (DHAP) to glyceraldehyde-3-phosphate (G3P) at 298 K. Despite the positive ΔG°, this reaction proceeds readily in glycolysis. Explain how this is thermodynamically possible, and calculate the ratio [G3P]/[DHAP] at equilibrium. (R = 8.314 J/(mol·K))
Show answer
A positive ΔG° means the reaction is non-spontaneous under standard conditions (1 M concentrations), but it does not mean the reaction cannot proceed under cellular conditions. Using ΔG° = −RT ln K: 13800 J/mol = −(8.314)(298) ln K → ln K = −13800 / 2477.6 = −5.57 → K = e−5.57e^{−5.57} ≈ 0.0038. At equilibrium, [G3P]/[DHAP] ≈ 0.004, meaning only about 0.4% exists as G3P. However, in glycolysis, the next enzyme (glyceraldehyde-3-phosphate dehydrogenase) rapidly consumes G3P, keeping its concentration far below the equilibrium value. This makes the actual reaction quotient Q much smaller than K, which makes ΔG = ΔG° + RT ln Q negative, driving the reaction forward. This is a powerful example of how metabolic flux—through Le Chatelier-like coupling—can drive thermodynamically unfavorable reactions in the cell.

SUMMARY

Lesson Summary

Summary

The Gibbs free energy equation ΔG = ΔH − TΔS provides a single, decisive criterion for spontaneity at constant temperature and pressure. A reaction is spontaneous when ΔG < 0, non-spontaneous when ΔG > 0, and at equilibrium when ΔG = 0. The interplay between enthalpy (ΔH) and entropy (ΔS) produces four cases: always spontaneous (ΔH < 0, ΔS > 0), spontaneous at low T only (ΔH < 0, ΔS < 0), spontaneous at high T only (ΔH > 0, ΔS > 0), and never spontaneous (ΔH > 0, ΔS < 0).

For temperature-dependent cases, the crossover temperature T = ΔH/ΔS identifies the boundary between spontaneous and non-spontaneous regimes. The relationship ΔG° = −RT ln K connects free energy to the equilibrium constant, while ΔG = ΔG° + RT ln Q extends the analysis to non-standard conditions. Always remember that spontaneity is not the same as speed—a thermodynamically favorable reaction may still require a catalyst or high temperature to proceed at an observable rate.

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