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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), TΔS = 773 × (−0.198) = −153 kJ mol⁻¹, making ΔG = −92 − (−153) = +61 kJ mol⁻¹ — non-spontaneous. High temperature kills the reaction thermodynamically but is forced by kinetic necessity.

Gibbs's 1873 equation — At Yale University, Josiah Willard Gibbs derived ΔG = ΔH − TΔS. Applied to the Haber process: at 500°C, TΔS = 773 × (−0.198) = −153 kJ mol⁻¹, making ΔG = −92 − (−153) = +61 kJ mol⁻¹ — non-spontaneous. The reaction runs anyway, because kinetics demands high temperature. ΔG tells you what's possible; kinetics tells you how fast.

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The Haber Paradox

The Haber process converts N₂ and H₂ into ammonia — a reaction that is exothermic (ΔH = −92 kJ mol⁻¹) and should favour products. Yet it needs to run at 400–500°C in industrial plants. If the reaction is already thermodynamically favoured (ΔH < 0), why does it need such a high temperature?

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

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 $T_\text{cross} = \Delta H / \Delta S$

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: spontaneous; ΔG° > 0: non-spontaneous; ΔG° = 0: at equilibrium.

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 (T_cross)

T = ΔH°/ΔS° (with ΔS° in kJ K⁻¹ mol⁻¹); the temperature at which ΔG = 0 and spontaneity changes.

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?

Defining Gibbs Free Energy

core concept

Ammonia synthesis (N₂ + 3H₂ → 2NH₃) is exothermic: ΔH = −92 kJ mol⁻¹. Every calorimetry lesson says this reaction should proceed spontaneously. Yet at 500°C — the temperature required for a practical rate — it does NOT proceed spontaneously. Enthalpy cannot explain this. Entropy alone cannot explain it. 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: ΔG represents the maximum useful work available from a reaction at constant temperature and pressure.

Thermodynamic meaning

Spontaneous (ΔG < 0)

Non-spontaneous (ΔG > 0)

Equilibrium (ΔG = 0)

What it means in practice

Reaction can proceed without continuous energy input; releases free energy

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 completely independent.

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?

The Four ΔH/ΔS Combinations and Temperature Dependence

core 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 T_cross
+ (endothermic)	+ (more disorder)	+ve	+ − (+) = depends on T	Spontaneous at high T	Becomes spontaneous above T_cross

The temperature-dependent cases (rows 3 and 4) are the most commonly tested. The crossover temperature $T_\text{cross} = \Delta H° / \Delta S°$ is where ΔG = 0 — at this temperature the reaction transitions between spontaneous and non-spontaneous.

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 T_cross 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 — INTERACTIVE Interactive

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 T_cross = ΔH° ÷ ΔS° (with ΔS° in kJ K¹ mol¹) 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.

The Haber Process Paradox — Resolved

core 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 → because kinetics, not thermodynamics, dictates the operating conditions.

The Haber process is the perfect example of a thermodynamically spontaneous reaction that still needs high temperature — but the reason is kinetics, not thermodynamics.

$\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \to 2\text{NH}_3\text{(g)}$, $\Delta H° = -92 \text{ kJ mol}^{-1}$, $\Delta S° = -198.9 \text{ J K}^{-1}\text{mol}^{-1} = -0.1989 \text{ kJ K}^{-1}\text{mol}^{-1}$

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 (spontaneous) but the N≡N triple bond's E_a is so large 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 thermodynamically ΔG° is positive (non-spontaneous!). Industrial plants operate in a thermodynamically unfavourable regime purely for kinetic reasons.
The industrial compromise: Run at 400–500°C (kinetics), use an iron catalyst to lower E_a further, maintain high pressure (Le Chatelier — Module 5), and continuously remove NH₃ to drive the equilibrium forward.

HSC answer requirement: In the Haber process paradox, distinguish clearly between the thermodynamic reason (ΔG becomes less negative / positive at high T, because ΔH < 0 and ΔS < 0 → row 3) and the kinetic reason (high temperature needed for a practical rate through the high-E_a N≡N bond). Both are required for a complete answer.

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 (responsible for ~50% of the nitrogen in the proteins of every human alive today) is the most important example of this balance in history.

The Haber process (N&sub2; + 3H&sub2; → 2NH&sub3;, ΔH = −92 kJ mol¹) is spontaneous at low temperature (ΔH < 0, ΔS < 0 — row 3) but kinetically too slow. 400–500°C is used for an adequate rate; T_cross ≈ 465 K means ΔG > 0 at operating temperature — the classic case of kinetics overriding thermodynamics.

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."

Solving ΔG Problems — Step by Step

core 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.

Action

Convert T: $T(\text{K}) = T(°\text{C}) + 273$

Convert ΔS: $\Delta S(\text{kJ}) = \Delta S(\text{J}) \div 1000$

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 mol¹ K¹ 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 ____.

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 T_cross. (d) Is the reaction spontaneous at 500°C?

Convert units (mandatory first step):
$T = 25 + 273 = 298 \text{ K}$
$\Delta S° = -198.9 \div 1000 = -0.1989 \text{ kJ K}^{-1}\text{mol}^{-1}$

Both conversions done before substitution — this is non-negotiable.

$\Delta G°(298\text{ K}) = -92.4 - (298 \times -0.1989) = -92.4 - (-59.27) = -92.4 + 59.27 = \mathbf{-33.1 \text{ kJ mol}^{-1}}$

Subtracting a negative gives addition — check the sign carefully.

$\Delta G° = -33.1 < 0 \to$ 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.

$T_\text{cross} = \Delta H° / \Delta S° = -92.4 / (-0.1989) = \mathbf{464.6 \text{ K} \approx 465 \text{ K} (192°\text{C})}$
Below 465 K: ΔG < 0 (spontaneous). Above 465 K: ΔG > 0 (non-spontaneous).

The crossover temperature is where spontaneity changes — critical for understanding the Haber paradox.

$T = 500°\text{C} = 773 \text{ K} > 465 \text{ K}$
$\Delta G°(773\text{ K}) = -92.4 - (773 \times -0.1989) = -92.4 + 153.8 = \mathbf{+61.4 \text{ kJ mol}^{-1}} > 0 \to$ 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 T_cross.

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 T_cross.

Identify the combination first — this lets you sanity-check your calculated answer.

Convert: $T = 500 + 273 = 773 \text{ K}$; $\Delta S° = 250 \div 1000 = 0.250 \text{ kJ K}^{-1}\text{mol}^{-1}$

Both unit conversions completed before substitution.

$\Delta G° = 180 - (773 \times 0.250) = 180 - 193.25 = \mathbf{-13.25 \text{ kJ mol}^{-1}} < 0 \to$ spontaneous at 500°C. Consistent with prediction — high T makes ΔG negative. ✓

Answer matches qualitative prediction from step 1 — good sign check.

$T_\text{cross} = 180 / 0.250 = \mathbf{720 \text{ K} = 447°\text{C}}$
Below 720 K (447°C): non-spontaneous. Above 720 K: spontaneous. Our temperature of 773 K > 720 K → spontaneous ✓.

T_cross confirms: 773 K is above the 720 K threshold, so spontaneous — consistent throughout.

Predict then reveal +8 XP

1 · Predict

2 · Reveal

3 · Compare

A reaction has ΔH = +60 kJ mol¹ and ΔS = +200 J mol¹ K¹. At 25 °C (298 K), is this reaction spontaneous? What about at 500 K?

Confidence 50%

Formula reference · this lesson

core formulas

📐

Key Formulas — Gibbs Free Energy

$\Delta G° = \Delta H° - T\Delta S°$

ΔH° in kJ mol⁻¹ T in Kelvin (K = °C + 273) ΔS° in kJ K⁻¹ mol⁻¹ (÷ 1000 from J)

Spontaneous: $\Delta G < 0$ | Non-spontaneous: $\Delta G > 0$ | Equilibrium: $\Delta G = 0$

Crossover temperature: $T_\text{cross} = \dfrac{\Delta H°}{\Delta S°}$ (ΔS° in kJ K⁻¹ mol⁻¹)

Unit checklist: T in K not °C | ΔS in kJ not J | both conversions before substitution

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Common errors · the 3 traps that cost marks

"Spontaneous means the reaction happens quickly"

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 completely independent of ΔG.

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.

"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 T_cross. The Haber process at industrial temperatures is the textbook example — exothermic but non-spontaneous above 192°C.

Work mode · how are you completing this lesson?

Quick-fire practice · 5 reps +2 XP per reveal

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?

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 T_cross. (d) Does limestone decompose spontaneously in a kiln at 900°C?

$2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \to 2\text{SO}_3\text{(g)}$: $\Delta H° = -197 \text{ kJ mol}^{-1}$; $\Delta S° = -187 \text{ J K}^{-1}\text{mol}^{-1}$.
(i) Identify the ΔH/ΔS combination category. (ii) Predict spontaneity at low vs high T without calculating. (iii) Calculate T_cross.

$\text{N}_2\text{O}_4\text{(g)} \to 2\text{NO}_2\text{(g)}$: $\Delta H° = +57 \text{ kJ mol}^{-1}$; $\Delta S° = +175 \text{ J K}^{-1}\text{mol}^{-1}$.
Identify the combination, state spontaneity at 25°C, calculate T_cross, and explain the significance.

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 T_cross and explain its physical significance.

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.

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Interactive Tool — Hess's Law & Bond Energy Open fullscreen ↗

True or false?

According to the Hess’s Law tool, the total enthalpy change of a reaction is independent of the pathway taken.

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.

01b

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 T_cross. 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.

Short answer

ApplyBand 4

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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ApplyBand 4

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 $\Delta S° = -549 \text{ J K}^{-1}\text{mol}^{-1}$.
(a) Calculate ΔG° at 25°C. (b) Is rusting spontaneous? (c) Calculate T_cross. (d) Rusting is slow at room temperature. Distinguish between the thermodynamic and kinetic aspects of this observation. 6 MARKS

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EvaluateBand 5

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 T_cross, and distinguish between thermodynamic and kinetic factors. 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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Multiple Choice

MC 1 — B: T_cross = −50/(−0.100) = 500 K. Convert ΔS first.

MC 2 — C: Spontaneous = ΔG < 0 = thermodynamically favourable. Not fast, not exothermic necessarily.

MC 3 — D: ΔH < 0, ΔS > 0 → ΔG always negative.

MC 4 — A: ΔG > 0 means non-spontaneous thermodynamically; still runs industrially due to kinetics + Le Chatelier.

MC 5 — B: Using J instead of kJ makes TΔS 1000× too large.

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).

Crossover temperature: T_cross = ΔH/ΔS = 178/0.1607 = 1108 K = 835°C. Above 835°C: ΔG < 0 (spontaneous).

At 900°C (1173 K): TΔS = 1173 × 0.1607 = 188.5 kJ mol⁻¹; ΔG = 178 − 188.5 = −10.5 kJ mol⁻¹ < 0 → spontaneous. The lime kiln operates at 900°C specifically to exceed this crossover temperature.

Q7 — Rusting of iron (6 marks)

(a) T = 298 K; ΔS° = −549 ÷ 1000 = −0.549 kJ K⁻¹ mol⁻¹; ΔG° = −1648 − (298 × −0.549) = −1648 + 163.6 = −1484.4 kJ mol⁻¹

(b) ΔG° = −1484.4 kJ mol⁻¹ < 0 → spontaneous at 25°C. The large negative ΔH overwhelms the negative TΔS contribution.

(c) T_cross = −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 spontaneous, rusting is extremely slow at room temperature because the activation energy for the surface oxidation mechanism is high. The Fe₂O₃ layer also passivates the surface. Spontaneity tells us rusting will occur; rate tells us how fast — these are independent properties.

Q8 — Haber process evaluation (8 marks)

Statement evaluation: The statement is incorrect. The Haber process is run at high temperature despite being thermodynamically non-spontaneous at 400–500°C — the high temperature is needed for kinetic (rate) reasons, not thermodynamic ones.

ΔG° at 25°C (298 K): ΔS° = −0.1989 kJ K⁻¹ mol⁻¹; ΔG° = −92.4 − (298 × −0.1989) = −92.4 + 59.3 = −33.1 kJ mol⁻¹ → spontaneous at 25°C.

ΔG° at 450°C (723 K): ΔG° = −92.4 − (723 × −0.1989) = −92.4 + 143.8 = +51.4 kJ mol⁻¹ → non-spontaneous at 450°C.

T_cross: T = −92.4/(−0.1989) = 464.6 K (191.6°C). The reaction is thermodynamically spontaneous only below ~465 K. Above this temperature, ΔG becomes increasingly positive. The industrial operating temperature of 400–500°C lies well above T_cross.

Why then does the Haber process use 400–500°C? At 25°C, despite ΔG < 0, the rate of reaction is essentially zero — the N≡N triple bond (bond energy = 941 kJ mol⁻¹) has an enormous activation energy that prevents bond breaking even with a catalyst. Increasing temperature to 400–500°C provides sufficient energy for a practical reaction rate. An iron catalyst (with K₂O and Al₂O₃ promoters) lowers E_a further. High pressure (~200 atm) and continuous removal of NH₃ (Le Chatelier, Module 5) help drive the equilibrium forward.

Summary: The Haber process is the textbook example of a reaction run in a thermodynamically unfavourable regime for kinetic reasons. Thermodynamics says the reaction should happen below 465 K; kinetics says it needs above ~673 K for a useful rate. Industrial chemistry navigates this tension through catalysis, pressure, and product removal.

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