Module 4 · L13 of 13~45 min⚡ +65 XP in Learn · +25 to complete
Gibbs Free Energy & Spontaneity
In 1873, Josiah Willard Gibbs at Yale University published "A Method of Geometrical Representation of the Thermodynamic Properties of Substances", deriving the equation ΔG = ΔH − TΔS and showing that a reaction is spontaneous when ΔG < 0. Gibbs's equation finally resolved the paradox of the Haber process: ammonia synthesis has ΔH = −92 kJ mol⁻¹ (favourable) and ΔS = −198 J K⁻¹ mol⁻¹ (unfavourable). At 500°C (773 K), the standard free-energy change is ΔG° = ΔH° − TΔS° = −92 − 773 × (−0.198) = +61 kJ mol⁻¹, so the standard-state equilibrium constant K is small (it favours reactants). Industry still makes ammonia by running at high pressure and continuously removing NH₃, which keeps the actual ΔG negative; the high temperature is chosen for an acceptable rate, not because it helps the thermodynamics.
Gibbs's 1873 equation, At Yale University, Josiah Willard Gibbs derived ΔG = ΔH − TΔS. Applied to the Haber process: at 500°C the standard ΔG° = −92 − 773 × (−0.198) = +61 kJ mol⁻¹, so the standard equilibrium favours reactants (K is small). Plants still make ammonia using high pressure and continuous NH₃ removal (keeping the actual ΔG < 0); high temperature is chosen for rate. ΔG° tells you about K; the actual ΔG tells you which way the real mixture goes; kinetics tells you how fast.
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You're here
Warm up, context and priorities
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
Lesson question: what actually decides whether a reaction "goes"? You already know some reactions release heat and some increase disorder, but which one wins? This lesson combines both into a single number that answers it.
1Combine the two drivers. Fold enthalpy and entropy into one quantity: ΔG = ΔH − TΔS (with T in kelvin).
2Check the sign. ΔG < 0 = spontaneous; ΔG > 0 = non-spontaneous; ΔG = 0 = at equilibrium.
3Judge temperature. Because of the TΔS term, temperature can flip spontaneity, decide when it does.
Quick prerequisite: you need ΔH (enthalpy, L01–L10) and ΔS (entropy, L11–L12), and you must convert between J and kJ (÷ 1000). If either driver is shaky, revisit those lessons or start with the Supported route below. This is also the module capstone, it unifies every driver studied in Module 4. Syllabus reference: solve problems using the Gibbs free energy equation to classify reactions as spontaneous or non-spontaneous (ACSCH127).
Convert ΔS from J to kJ (÷ 1000) before substituting
Should know
The four ΔH/ΔS sign combinations and their outcomes
How temperature controls the two mixed-sign cases
The crossover temperature Tcross = ΔH°/ΔS°
Going deeper
ΔG predicts feasibility, not rate, feasible ≠ fast (kinetics)
How ΔG links to the equilibrium constant K in later modules
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Recall, mastery and key terms
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Recall, your gut answer first
+5 XP warm-up
The Haber Paradox
The Haber process converts N₂ and H₂ into ammonia, a reaction that is exothermic (ΔH = −92 kJ mol⁻¹). Exothermicity alone does not guarantee that products are favoured, entropy and the actual composition matter too. Industrial plants run it at 400–500°C. Why, if the reaction is already exothermic, is such a high temperature needed, and how do plants still get a good yield?
Your instinct might be "to make it go faster", and that's partly right, but the full story involves how temperature affects something deeper about whether the reaction can proceed at all.
What thermodynamic concept are we missing? Write your prediction before reading on.
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What you'll master, and the words for it
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What you'll master
Know
Key Facts
The definition of Gibbs free energy and the equation $\Delta G = \Delta H - T\Delta S$
The spontaneity criteria: ΔG < 0, ΔG > 0, ΔG = 0
All four ΔH/ΔS sign combinations and their spontaneity outcomes
Understand
Concepts
Why spontaneous does NOT mean fast, thermodynamics vs kinetics
How temperature determines spontaneity in mixed-sign combinations
The Haber process paradox resolved using ΔG analysis
Can Do
Skills
Calculate ΔG° at a given temperature with full unit conversions
Classify reactions as spontaneous, non-spontaneous, or at equilibrium
Calculate the crossover temperature $T_\text{cross} = \Delta H / \Delta S$
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Key terms
Gibbs free energy (ΔG°)
ΔG° = ΔH° − TΔS°; the thermodynamic potential that determines if a reaction is spontaneous at constant temperature and pressure.
Spontaneity criterion
ΔG < 0: proceeds forward at the current composition; ΔG > 0: proceeds in reverse; ΔG = 0: at equilibrium. ΔG° is the standard-state value and sets K via ΔG° = −RT ln K; use ΔG = ΔG° + RT ln Q for the actual mixture.
ΔH–ΔS combinations
(−ΔH, +ΔS): always spontaneous; (+ΔH, −ΔS): never spontaneous; mixed signs: spontaneity depends on temperature.
Temperature dependence
At low T, ΔH dominates; at high T, TΔS dominates; reactions can switch from non-spontaneous to spontaneous as T changes.
Crossover temperature (Tcross)
T = ΔH°/ΔS° (with ΔS° in kJ K⁻¹ mol⁻¹); an estimate of the temperature at which ΔG° = 0, assuming ΔH° and ΔS° stay roughly constant with temperature.
Standard vs non-standard ΔG
ΔG° refers to standard conditions (25°C, 100 kPa); ΔG = ΔG° + RT ln Q accounts for actual concentrations.
Cross-lesson links: Gibbs free energy is the culmination of Module 4. ΔG = ΔH − TΔS draws on enthalpy (L01–L10) and entropy (L11–L12) simultaneously. When ΔH dominates (low T), the lessons on calorimetry and bond energies predict spontaneity. When TΔS dominates (high T), the entropy lessons predict it. This lesson unifies both drivers into a single spontaneity criterion, completing the module's central question: what makes a reaction go?
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Gibbs free energy and temperature dependence
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Defining Gibbs Free Energy
Must knowcore concept
Ammonia synthesis (N₂ + 3H₂ → 2NH₃) is exothermic: ΔH = −92 kJ mol⁻¹. Exothermicity alone is not enough to guarantee a reaction is favoured, entropy and composition also matter. At 500°C the standard free-energy change ΔG° is unfavourable, yet industrial plants still produce ammonia. Enthalpy alone cannot explain this, and entropy alone cannot either. Something else is needed: a single number that combines both, and changes with temperature.
That number is Gibbs free energy. Named after Josiah Willard Gibbs (Yale, 1873), Gibbs free energy is defined as $G = H - TS$, and for a reaction at constant temperature and pressure:
$\Delta G = \Delta H - T\Delta S$
Interpretation: at constant temperature and pressure, −ΔG gives the maximum non-expansion (useful) work obtainable from a reaction when it is carried out reversibly.
Thermodynamic meaning
Spontaneous (ΔG < 0)
Non-spontaneous (ΔG > 0)
Equilibrium (ΔG = 0)
What it means in practice
Thermodynamically favourable: ΔG < 0 for the change at the current composition (works even for endothermic processes)
Reaction requires energy input; reverse reaction is spontaneous
Forward and reverse reactions proceed at equal rates; no net change
Most common Module 4 misconception: "Spontaneous means the reaction happens quickly." This is wrong. Spontaneous means thermodynamically favourable, not fast. Diamond converting to graphite has ΔG < 0 (spontaneous) but takes millions of years (kinetically frozen by a very high activation energy). Spontaneity is a thermodynamic property; rate is a kinetic property. They are distinct criteria: ΔG does not determine the rate, and the activation energy does not determine the equilibrium position.
Unit checklist, before every calculation:
1. Convert T from °C to K: $T(\text{K}) = T(°\text{C}) + 273$
2. Convert ΔS from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹: $\Delta S(\text{kJ}) = \Delta S(\text{J}) \div 1000$
Then substitute into ΔG = ΔH − TΔS. Both conversions must happen before substitution.
Gibbs free energy: ΔG = ΔH − TΔS (T in Kelvin). ΔG < 0 → spontaneous; ΔG > 0 → non-spontaneous; ΔG = 0 → equilibrium. ΔG combines enthalpy and entropy into a single criterion for spontaneity at constant temperature and pressure.
Pause, copy the highlighted definition into your book before moving on.
Quick check: Which of the following statements about a reaction with ΔG° = −45 kJ mol⁻¹ is correct?
Correct. ΔG° = −45 kJ mol⁻¹ < 0 means spontaneous (thermodynamically favourable). It says nothing about rate, some spontaneous reactions are extremely slow due to high activation energy.
The Four ΔH/ΔS Combinations and Temperature Dependence
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The Four ΔH/ΔS Combinations and Temperature Dependence
Should knowcore concept
We just saw that ΔG = ΔH − TΔS combines both enthalpy and entropy into a single spontaneity criterion. That raises a question: how do the four possible sign combinations of ΔH and ΔS determine whether temperature controls spontaneity? This card answers it → with the four cases that define when temperature controls the outcome.
Whether a reaction is spontaneous depends on the signs of both ΔH and ΔS, and in two of the four combinations, the answer changes with temperature.
ΔG = ΔH − TΔS across all four sign combinations
ΔH
ΔS
TΔS term
ΔG = ΔH − TΔS
Spontaneous?
Temperature effect
− (exothermic)
+ (more disorder)
+ve
− − (+) = always −ve
Always spontaneous
More spontaneous at higher T
+ (endothermic)
− (less disorder)
−ve
+ − (−) = always +ve
Never spontaneous
Never changes sign
− (exothermic)
− (less disorder)
−ve
− − (−) = depends on T
Spontaneous at low T
Becomes non-spontaneous above Tcross
+ (endothermic)
+ (more disorder)
+ve
+ − (+) = depends on T
Spontaneous at high T
Becomes spontaneous above Tcross
The temperature-dependent cases (rows 3 and 4) are the most commonly tested. The crossover temperature $T_\text{cross} = \Delta H° / \Delta S°$ is an estimate of where ΔG° = 0; it assumes T > 0 and that ΔH° and ΔS° stay roughly constant over the temperature range with no phase change, so treat the always/never entries in the table as bounded by these assumptions.
For temperature-dependent cases: Always calculate $T_\text{cross} = \Delta H° / \Delta S°$ (with ΔS° in kJ K⁻¹ mol⁻¹) to find where the reaction transitions. Then determine which side of Tcross the given temperature falls on.
Common error: Concluding "exothermic reactions are always spontaneous", only true when ΔS is also positive (row 1). If ΔS < 0 (row 3), the reaction becomes non-spontaneous at high temperature. The Haber process is the perfect example of this.
GIBBS FREE ENERGY, INTERACTIVEInteractive
Calculator: adjust ΔH, ΔS, T and watch ΔG change, Quadrant Map shows all 4 combinations, Haber shows the crossover temperature
Four ΔH/ΔS combinations: (−, +) always spontaneous; (+, −) never spontaneous; (−, −) spontaneous at low T; (+, +) spontaneous at high T. Crossover temperature Tcross = ΔH° ÷ ΔS° (with ΔS° in kJ K⁻¹ mol⁻¹) is where ΔG = 0.
Add the highlighted point to your notes before the check below.
Explain it: A reaction has ΔH = +120 kJ mol⁻¹ and ΔS = +180 J K⁻¹ mol⁻¹. Without calculating, predict whether it is spontaneous at room temperature and at high temperature. Explain your reasoning using the ΔH/ΔS combination table.
ΔH > 0 and ΔS > 0 → Row 4 (spontaneous at high T). At room temperature: ΔH dominates, making ΔG positive (non-spontaneous). At high temperature: the TΔS term becomes large enough to outweigh ΔH, flipping ΔG negative (spontaneous). Tcross = 120 / 0.180 = 667 K (394°C).
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The Haber Process Paradox, Resolved
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The Haber Process Paradox, Resolved
Should knowcore concept
We just saw the four ΔH/ΔS combinations and how temperature controls spontaneity. That raises a question: the Haber process should be thermodynamically favoured at low temperature, so why is it run at 400–500°C? This card answers it → high temperature is chosen for rate; high pressure and continuous NH₃ removal then keep the actual ΔG negative even where ΔG° is positive.
The Haber process is the perfect example of keeping three ideas separate: the standard-state thermodynamics (ΔG° and K), the actual ΔG of the real operating mixture, and the rate.
ΔG° for the Haber process at different temperatures
Temperature
Calculation
ΔG° (kJ mol⁻¹)
Spontaneous?
25°C (298 K)
−92 − (298 × −0.1989) = −92 + 59.3
−32.7
Yes ✓ (but very slow)
190°C (463 K)
−92 − (463 × −0.1989) ≈ −92 + 92.0
≈ 0
At crossover point
500°C (773 K)
−92 − (773 × −0.1989) = −92 + 153.8
+61.8
No ✗ (but faster!)
The paradox resolved:
At 25°C: ΔG° = −32.7 kJ mol⁻¹ (standard-state favourable) but breaking the strong N≡N triple bond gives such a high activation barrier that the rate is essentially zero, even with a catalyst, nothing happens at room temperature.
At 400–500°C: The rate is now practical, but ΔG° is positive, so the standard-state equilibrium constant K is small. The high temperature is chosen for kinetics; the plant then compensates with high pressure and continuous NH₃ removal, which keep the reaction quotient Q small so the actual ΔG = ΔG° + RT ln Q stays negative and ammonia keeps forming.
The industrial compromise: Run at 400–500°C (kinetics), use an iron catalyst to lower Ea further, maintain high pressure (Le Chatelier, Module 5), and continuously remove NH₃ to drive the equilibrium forward.
HSC answer requirement: In the Haber process, distinguish the standard-state thermodynamics (raising T makes ΔG° less favourable and lowers K, because ΔH° < 0 and ΔS° < 0 → row 3) from the kinetic reason high temperature is used (a practical rate through the high-Ea N≡N pathway). Plants offset the unfavourable ΔG° with high pressure and continuous NH₃ removal, keeping the actual ΔG negative. A complete answer separates ΔG° (and K) from the actual ΔG and from rate.
Why this matters: This is why thermodynamics and kinetics are studied together in Module 4. Thermodynamics tells you if a reaction can happen; kinetics tells you how fast. Industrial chemists must balance both simultaneously, and the Haber process (synthetic nitrogen fertiliser is estimated to support roughly half of current global food production) is the most important example of this balance in history.
The Haber process (N₂ + 3H₂ → 2NH₃, ΔH = −92 kJ mol⁻¹) is spontaneous at low temperature (ΔH < 0, ΔS < 0, row 3) but kinetically too slow. 400–500°C is used for an adequate rate; Tcross ≈ 465 K means ΔG° > 0 at operating temperature, so plants keep the actual ΔG negative with high pressure and continuous NH₃ removal.
Pause, write the highlighted definition into your book.
True or false: "The Haber process is run at 400–500°C because the reaction is thermodynamically more spontaneous at higher temperatures."
False. The Haber process has ΔH° < 0 and ΔS° < 0 (row 3), so its standard ΔG° rises (and K falls) as temperature increases, the standard yield drops above Tcross ≈ 465 K (192°C). High temperature is used for kinetic reasons (a practical rate through the high-activation-energy N≡N pathway); plants recover yield with high pressure and continuous NH₃ removal, keeping the actual ΔG < 0.
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Solving ΔG Problems, Step by Step
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Solving ΔG Problems, Step by Step
Must knowcore concept
We just saw the Haber process as a case study of kinetics overriding thermodynamics. That raises a question: what specific calculation errors appear most often in ΔG problems, and how can they be avoided? This card answers it → with the three unit and sign errors that HSC marking schemes consistently penalise.
Three common errors in ΔG calculations, Celsius instead of Kelvin, J instead of kJ for ΔS, wrong sign, are all avoidable with a careful, step-by-step approach.
Substitute into $\Delta G = \Delta H - T\Delta S$, tracking signs carefully
State sign of ΔG and classify: spontaneous / non-spontaneous / equilibrium
If asked about temperature effects, calculate $T_\text{cross} = \Delta H / \Delta S$
Why it matters
T in Celsius gives completely wrong TΔS term
J vs kJ mismatch inflates TΔS by 1000×
Subtracting a negative = adding; easy to slip
The answer requires interpretation, not just a number
Identifies where spontaneity changes
Write out each conversion step explicitly don't try to combine steps in your head. The two unit conversions (°C → K and J → kJ) must both happen before substitution.
The 1000× error: Using ΔS = −198.9 J K⁻¹ mol⁻¹ instead of −0.1989 kJ K⁻¹ mol⁻¹ in the Haber example gives TΔS = 298 × (−198.9) = −59,272 kJ mol⁻¹ instead of −59.3 kJ mol⁻¹, wrong by a factor of 1000. Always convert J → kJ before substitution.
Three critical errors in ΔG = ΔH − TΔS: (1) T in Celsius instead of Kelvin (add 273); (2) ΔS in J mol⁻¹ K⁻¹ not converted to kJ (divide by 1000, or the TΔS term is 1000× too large); (3) sign error when subtracting a negative ΔS term. Always write out each conversion step explicitly before substituting.
Add the highlighted point to your notes before the check below.
Fill in the blanks: Before substituting into ΔG = ΔH − TΔS, temperature must be converted from °C to ____ by adding ____. Entropy must be converted from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ by dividing by ____. Failure to convert ΔS inflates the TΔS term by a factor of ____.
Kelvin (K) | 273 | 1000 | 1000. Both unit conversions are mandatory, they must both be completed before any numbers are substituted into ΔG = ΔH − TΔS.
Same science, choose your support
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Same science, choose your support
pick one
All three routes practise the same Must-Know skill: compute ΔG = ΔH − TΔS and decide spontaneity, including a temperature-dependence case. Do one; try another if you want more practice.
Supported word bank + sentence frame start here if unsure
Word bank: 0.050 · −35 · negative · spontaneous
A reaction has ΔH = −20 kJ mol⁻¹ and ΔS = +50 J K⁻¹ mol⁻¹ at 300 K. Complete the frame:
"Convert ΔS = 50 ÷ 1000 = ______ kJ K⁻¹ mol⁻¹. Then ΔG = −20 − (300 × 0.050) = ______ kJ mol⁻¹. Because ΔG is ______, the reaction is ______."
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Show a hint
ΔG < 0 means spontaneous. Here ΔH is already negative and you subtract a positive TΔS term, so ΔG stays negative: −20 − 15 = −35 kJ mol⁻¹.
Core independent HSC-standard the exam standard
A reaction has ΔH° = +120 kJ mol⁻¹ and ΔS° = +150 J K⁻¹ mol⁻¹. (a) Calculate ΔG° at 25 °C and classify it. (b) Calculate the crossover temperature Tcross. (c) State whether the reaction is spontaneous at 1000 K. Show both unit conversions.
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Stretch reasoning in a new context push yourself
The thermite reaction (2Al + Fe₂O₃ → Al₂O₃ + 2Fe) has ΔH° = −852 kJ mol⁻¹ and ΔS° = −38 J K⁻¹ mol⁻¹, yet it must be lit with a burning magnesium ribbon before it starts. Using ΔG, Tcross and activation energy, evaluate the claim that "a spontaneous reaction always begins on its own".
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Worked examples, key idea and formula reference
Worked examples · reveal as you go
Worked example 1+5 XP on full reveal
Full Haber Process ΔG Analysis. For $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \to 2\text{NH}_3\text{(g)}$: $\Delta H° = -92.4 \text{ kJ mol}^{-1}$; $\Delta S° = -198.9 \text{ J K}^{-1}\text{ mol}^{-1}$.
(a) Calculate ΔG° at 25°C. (b) Is the reaction spontaneous at 25°C? (c) Find Tcross. (d) Is the reaction spontaneous at 500°C?
Subtracting a negative gives addition, check the sign carefully.
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$\Delta G° = -33.1 < 0 \to$ spontaneous at 25°C. Thermodynamically favourable. However, essentially no reaction occurs at room temperature, kinetically limited by the N≡N bond.
Spontaneous ≠ fast. Kinetics and thermodynamics are independent.
The crossover temperature is where spontaneity changes, critical for understanding the Haber paradox.
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$T = 500°\text{C} = 773 \text{ K} > 465 \text{ K}$ $\Delta G°(773\text{ K}) = -92.4 - (773 \times -0.1989) = -92.4 + 153.8 = \mathbf{+61.4 \text{ kJ mol}^{-1}} > 0 \to$ non-spontaneous at 500°C The Haber process runs in a thermodynamically unfavourable regime at industrial temperatures, for kinetic reasons only.
This is the resolution of the Haber paradox: thermodynamically unfavourable at operating temperature, kinetically necessary.
Worked example 2+5 XP on full reveal
Temperature-Dependent Spontaneity. A reaction has $\Delta H° = +180 \text{ kJ mol}^{-1}$ and $\Delta S° = +250 \text{ J K}^{-1}\text{mol}^{-1}$.
(a) Predict qualitatively whether this is spontaneous at low or high temperature. (b) Calculate ΔG° at 500°C. (c) Find Tcross.
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ΔH > 0, ΔS > 0 → row 4 in the combination table. Spontaneous only at high T the +TΔS term must grow large enough to outweigh the positive ΔH. Qualitative prediction: not spontaneous at room temperature; becomes spontaneous above Tcross.
Identify the combination first, this lets you sanity-check your calculated answer.
Students write ΔG < 0 and conclude the reaction will occur rapidly at that temperature.
Fix: Spontaneous means thermodynamically favourable (ΔG < 0), it says nothing about rate. Diamond converting to graphite is spontaneous (ΔG < 0) but takes geological time. Rate is determined by activation energy, a kinetic property distinct from ΔG (ΔG does not determine rate).
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Using T in °C or ΔS in J, the unit errors
Students substitute T = 25 (°C) instead of T = 298 (K), or use ΔS = −198.9 J without converting to kJ.
Fix: Always write two conversion lines before substitution: (1) T(K) = T(°C) + 273, (2) ΔS(kJ) = ΔS(J) ÷ 1000. The J → kJ error inflates TΔS by 1000× and produces a completely nonsensical answer. Show these conversions explicitly in HSC.
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"Exothermic reactions are always spontaneous"
Students assume ΔH < 0 guarantees ΔG < 0 at all temperatures.
Fix: Only true when ΔS > 0 as well (row 1). If ΔH < 0 and ΔS < 0 (row 3), the reaction becomes non-spontaneous above Tcross. The Haber process at industrial temperatures is the textbook example, exothermic but non-spontaneous above 192°C.
Drill, then revisit
Quick-fire practice · 5 reps +2 XP per reveal
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The combustion of methane: $\Delta H° = -890 \text{ kJ mol}^{-1}$; $\Delta S° = -243.1 \text{ J K}^{-1}\text{mol}^{-1}$.
(a) Calculate ΔG° at 25°C. (b) Is the reaction spontaneous? (c) What category does this fall into from the ΔH/ΔS combination table?
(a) T = 298 K; ΔS = −0.2431 kJ K⁻¹ mol⁻¹; ΔG° = −890 − (298 × −0.2431) = −890 + 72.4 = −817.6 kJ mol⁻¹ (b) ΔG° = −817.6 < 0 → spontaneous at 25°C. Strongly so, the large negative ΔH dominates. (c) ΔH < 0, ΔS < 0 → Row 3 (spontaneous at low T). Tcross = −890/(−0.2431) = 3661 K, remains spontaneous over all practical temperatures.
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Decomposition of limestone: $\text{CaCO}_3\text{(s)} \to \text{CaO(s)} + \text{CO}_2\text{(g)}$; $\Delta H° = +178 \text{ kJ mol}^{-1}$; $\Delta S° = +160.7 \text{ J K}^{-1}\text{mol}^{-1}$.
(a) Calculate ΔG° at 25°C. (b) Calculate ΔG° at 900°C. (c) Find Tcross. (d) Does limestone decompose spontaneously in a kiln at 900°C? (Take P(CO₂) = 1 bar; the actual decomposition temperature depends on the CO₂ partial pressure, which is why kilns are ventilated.)
$2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \to 2\text{SO}_3\text{(g)}$: $\Delta H° = -197 \text{ kJ mol}^{-1}$; $\Delta S° = -187 \text{ J K}^{-1}\text{mol}^{-1}$.
(i) Identify the ΔH/ΔS combination category. (ii) Predict spontaneity at low vs high T without calculating. (iii) Calculate Tcross.
(i) ΔH < 0, ΔS < 0 → Row 3. Spontaneous at low T; becomes non-spontaneous above Tcross.
(ii) Spontaneous at low T; above Tcross the TΔS term outweighs ΔH making ΔG positive.
(iii) Tcross = −197/(−0.187) = 1053 K (780°C). Spontaneous at all typical lab/industrial temperatures below 780°C.
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$\text{N}_2\text{O}_4\text{(g)} \to 2\text{NO}_2\text{(g)}$: $\Delta H° = +57 \text{ kJ mol}^{-1}$; $\Delta S° = +175 \text{ J K}^{-1}\text{mol}^{-1}$.
Identify the combination, state spontaneity at 25°C, calculate Tcross, and explain the significance.
ΔH > 0, ΔS > 0 → Row 4. Non-spontaneous at 25°C (ΔH dominates, ΔG > 0). Tcross = 57/0.175 = 326 K (53°C). Spontaneous above 53°C, this is why NO₂ (brown gas) forms preferentially at higher temperatures in the N₂O₄/NO₂ equilibrium.
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Freezing of water at −10°C: $\Delta H° = -6.0 \text{ kJ mol}^{-1}$; $\Delta S° = -22 \text{ J K}^{-1}\text{mol}^{-1}$.
Calculate ΔG° at −10°C (263 K) and verify that freezing is spontaneous below 0°C. Then find Tcross and explain its physical significance.
ΔH < 0, ΔS < 0 → Row 3. At T = 263 K: ΔS = −0.022 kJ K⁻¹ mol⁻¹; ΔG° = −6.0 − (263 × −0.022) = −6.0 + 5.79 = −0.21 kJ mol⁻¹ < 0 → spontaneous at −10°C ✓.
Tcross = −6.0/(−0.022) = 273 K (0°C). Below 273 K: freezing is spontaneous. Above 273 K: melting is spontaneous. This elegantly confirms why 0°C is the freezing point of water, it is the crossover temperature where ΔG = 0 for the liquid–solid transition.
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Revisit your thinking
Go back to your Think First response about the Haber paradox. Now that you've studied Gibbs's 1873 equation from Yale University, ΔG = ΔH − TΔS:
The missing concept was temperature dependence of ΔG. For the Haber process: ΔH = −92 kJ mol⁻¹ (favourable), ΔS = −198 J K⁻¹ mol⁻¹ (unfavourable). At 25°C: ΔG = −92 − 298(−0.198) = −92 + 59 = −33 kJ mol⁻¹ (spontaneous). At 500°C: ΔG = −92 − 773(−0.198) = −92 + 153 = +61 kJ mol⁻¹ (non-spontaneous). Gibbs's equation explains both.
The Haber process is run at high T purely because of kinetics, the rate at 25°C is immeasurably slow even with catalyst. Engineers accept thermodynamic non-spontaneity at 500°C because the iron catalyst provides enough rate. ΔG tells you what's possible; kinetics tells you how fast.
Gibbs free energy, derived at Yale University in 1873, unifies enthalpy (L01–L10) and entropy (L11–L12) into the single spontaneity criterion that completes Module 4. Every reaction in this module can now be evaluated with ΔG = ΔH − TΔS.
According to the Hess’s Law tool, the total enthalpy change of a reaction is independent of the pathway taken.
✓
Multiple choice
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence. That tells the system what to drill next.
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Misconceptions to fix before short answer
Wrong: A negative ΔG means a reaction will happen quickly.
Right: A negative ΔG means a reaction is thermodynamically spontaneous, it can occur without external energy input. It says nothing about reaction rate. Thermodynamic spontaneity and kinetic rate are independent. Some spontaneous reactions are extremely slow.
Wrong: Exothermic reactions (ΔH < 0) are always spontaneous.
Right: Only true when ΔS > 0 as well. If ΔH < 0 and ΔS < 0, the reaction becomes non-spontaneous above Tcross. Always check both signs.
Wrong: Using T = 25°C or ΔS in J directly in the ΔG formula.
Right: Convert T to Kelvin (+273) and ΔS to kJ (÷1000) before every calculation. Show both conversion steps explicitly in HSC responses.
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Short answer
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Q6. (6 marks) Explain, using the equation $\Delta G = \Delta H - T\Delta S$, why the decomposition of calcium carbonate ($\text{CaCO}_3 \to \text{CaO} + \text{CO}_2$) is non-spontaneous at room temperature but spontaneous in a lime kiln at 900°C. In your answer, identify the ΔH/ΔS combination category and calculate the crossover temperature. ($\Delta H° = +178 \text{ kJ mol}^{-1}$; $\Delta S° = +160.7 \text{ J K}^{-1}\text{mol}^{-1}$)
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Q7. (6 marks) The rusting of iron: $4\text{Fe(s)} + 3\text{O}_2\text{(g)} \to 2\text{Fe}_2\text{O}_3\text{(s)}$ has $\Delta H° = -1648 \text{ kJ mol}^{-1}$ and $\Delta S° = -549 \text{ J K}^{-1}\text{mol}^{-1}$. (Real rust is a porous mixture of hydrated iron oxides; Fe₂O₃ is used here as a simplified thermodynamic model.)
(a) Calculate ΔG° at 25°C. (b) Is rusting spontaneous? (c) Calculate Tcross. (d) Rusting is slow at room temperature. Distinguish between the thermodynamic and kinetic aspects of this observation. 6 MARKS
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Q8. (8 marks) Evaluate the statement: "The Haber process is run at 400–500°C because the reaction is thermodynamically spontaneous at high temperatures." Is this statement correct? In your answer, calculate ΔG° at both 25°C and 450°C, explain the concept of Tcross, distinguish between thermodynamic and kinetic factors, and explain how high pressure and continuous NH₃ removal allow the plant to keep producing ammonia. Use: $\Delta H° = -92.4 \text{ kJ mol}^{-1}$; $\Delta S° = -198.9 \text{ J K}^{-1}\text{mol}^{-1}$. 8 MARKS
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Comprehensive Answers
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Q6, CaCO₃ decomposition (6 marks)
ΔH/ΔS combination: ΔH° = +178 kJ mol⁻¹ (>0, endothermic); ΔS° = +160.7 J K⁻¹ mol⁻¹ (>0, CO₂ gas produced, Δn(gas) = +1). Row 4: ΔH > 0, ΔS > 0. Spontaneous only at high T.
At room temperature (298 K): ΔS° = 0.1607 kJ K⁻¹ mol⁻¹; TΔS = 298 × 0.1607 = 47.9 kJ mol⁻¹; ΔG = 178 − 47.9 = +130.1 kJ mol⁻¹ > 0 → non-spontaneous. The T·ΔS term (+47.9) is insufficient to overcome the large positive ΔH (+178).
(b) ΔG° = −1484.4 kJ mol⁻¹ < 0 → spontaneous at 25°C. The large negative ΔH overwhelms the negative TΔS contribution.
(c) Tcross = −1648/(−0.549) = 3002 K (2729°C). Above 3002 K: non-spontaneous. This temperature exceeds the melting point of iron, rusting is spontaneous at all practical temperatures.
(d) Thermodynamic aspect: ΔG° = −1484.4 kJ mol⁻¹ ≪ 0 at room temperature, thermodynamically, rusting is strongly favoured (a large driving force exists). Kinetic aspect: Despite being thermodynamically favourable, rusting is slow at room temperature because the surface oxidation mechanism has a high activation energy and depends on slow transport of water, oxygen and ions to the metal. Unlike a true passivating film (such as Al₂O₃ on aluminium), ordinary rust is porous and flaky, so it does not protect the iron underneath. Spontaneity tells us rusting is favoured; rate tells us how fast, these are distinct properties.
Q8, Haber process evaluation (8 marks)
Statement evaluation: The statement is incorrect. High temperature is needed for kinetic (rate) reasons, not because the reaction becomes more thermodynamically favourable. A positive standard ΔG° at 400–500°C means the standard equilibrium constant K is small; it does NOT mean the forward reaction cannot proceed. Industry keeps the actual ΔG negative by other means (see below).
ΔG° at 450°C (723 K): ΔG° = −92.4 − (723 × −0.1989) = −92.4 + 143.8 = +51.4 kJ mol⁻¹ → ΔG° > 0, so the standard-state equilibrium (K) now favours reactants. This describes K, not a ban on the forward reaction.
Tcross: T = −92.4/(−0.1989) = 464.6 K (191.6°C, an estimate assuming roughly constant ΔH° and ΔS°). The standard ΔG° is negative only below ~465 K; above it K falls. The industrial temperature of 400–500°C lies well above Tcross, so the standard yield is poor, which is why other conditions are used.
Why then does the Haber process use 400–500°C? At 25°C, despite ΔG° < 0, the rate is essentially zero: breaking the strong N≡N triple bond (bond energy ≈ 941 kJ mol⁻¹) gives the reaction a very high activation barrier, so little happens even with a catalyst. Raising the temperature to 400–500°C provides a practical rate, and an iron catalyst (with K₂O and Al₂O₃ promoters) lowers Ea further. High pressure (~200 atm) and continuous removal of NH₃ keep the reaction quotient Q small, so the actual ΔG = ΔG° + RT ln Q stays negative and ammonia keeps forming even though ΔG° is positive at this temperature (Le Chatelier, Module 5).
Summary: The Haber process is the textbook example of separating standard-state thermodynamics (ΔG°, K) from the actual operating conditions and from kinetics. The standard ΔG° is unfavourable above ~465 K, but high pressure and continuous NH₃ removal keep the actual ΔG negative, while high temperature provides a usable rate. Industrial chemistry navigates this through catalysis, pressure and product removal.
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