⚖️ Gibbs Free Energy & Spontaneity
Gibbs free energy combines enthalpy and entropy into a single number that tells you whether a reaction will happen — and it solves the Haber process paradox once and for all.
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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.
Key Formulas — Gibbs Free Energy
By the end of this lesson you will be able to:
1. Defining Gibbs Free Energy
Gibbs free energy (G) combines enthalpy and entropy into a single quantity that predicts whether a process is spontaneous at a given temperature.
Named after Josiah Willard Gibbs (1870s), Gibbs free energy is defined as: $G = H - TS$, and for a reaction at constant temperature and pressure:
Interpretation: ΔG represents the maximum useful work available from a reaction at constant temperature and pressure.
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.
2. The Four ΔH/ΔS Combinations and Temperature Dependence
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.
| 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 where ΔG = 0 — at this temperature the reaction transitions between spontaneous and non-spontaneous.
3. The Haber Process Paradox — Resolved
The Haber process is the perfect example of a thermodynamically spontaneous reaction that still needs high temperature — but the reason is kinetics, not thermodynamics.
$\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \to 2\text{NH}_3\text{(g)}$, $\Delta H° = -92 \text{ kJ mol}^{-1}$, $\Delta S° = -198.9 \text{ J K}^{-1}\text{mol}^{-1} = -0.1989 \text{ kJ K}^{-1}\text{mol}^{-1}$
| 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 Ea 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 Ea further, maintain high pressure (Le Chatelier — Module 5), and continuously remove NH₃ to drive the equilibrium forward.
4. Solving ΔG Problems — Step by Step
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.
Worked Example 1 — 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?
Try It Now
For the reaction $\text{C}_2\text{H}_4\text{(g)} + \text{H}_2\text{(g)} \to \text{C}_2\text{H}_6\text{(g)}$ (hydrogenation of ethene):
$\Delta H° = -137 \text{ kJ mol}^{-1}$ | $\Delta S° = -120.7 \text{ J K}^{-1}\text{mol}^{-1}$
(a) Calculate ΔG° at 25°C (298 K). (b) Is the reaction spontaneous? (c) At what temperature does ΔG = 0?
(a) Convert: T = 298 K; ΔS° = −120.7 ÷ 1000 = −0.1207 kJ K⁻¹ mol⁻¹
ΔG° = −137 − (298 × −0.1207) = −137 − (−35.97) = −137 + 35.97 = −101.0 kJ mol⁻¹
(b) ΔG° = −101.0 kJ mol⁻¹ < 0 → spontaneous at 25°C. ΔH < 0 and ΔS < 0 (row 3) — spontaneous at low T.
(c) Tcross = ΔH°/ΔS° = −137 / (−0.1207) = 1135 K (862°C). Above 1135 K, ΔG becomes positive (non-spontaneous). Well above typical lab conditions — this reaction is spontaneous over a very wide temperature range.
Worked Example 2 — 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.
Activity 1 — ΔG Calculations
For each reaction, perform the full ΔG calculation. Show all unit conversions explicitly.
-
The combustion of methane: $\Delta H° = -890 \text{ kJ mol}^{-1}$; $\Delta S° = -243.1 \text{ J K}^{-1}\text{mol}^{-1}$ (from L12 Q8).
(a) Calculate ΔG° at 25°C. (b) Is the reaction spontaneous? (c) What category does this fall into from the ΔH/ΔS 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 Tcross. (d) Does limestone decompose spontaneously in a kiln at 900°C?
1(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
1(b): ΔG° = −817.6 kJ mol⁻¹ < 0 → spontaneous at 25°C. Strongly so — the large negative ΔH dominates.
1(c): ΔH < 0, ΔS < 0 → row 3 (spontaneous at low T, becomes non-spontaneous at high T). Tcross = −890 / (−0.2431) = 3661 K — extremely high, so combustion of methane remains spontaneous over all practical temperatures.
2(a): T = 298 K; ΔS = 0.1607 kJ K⁻¹ mol⁻¹; ΔG° = 178 − (298 × 0.1607) = 178 − 47.89 = +130.1 kJ mol⁻¹ → non-spontaneous at 25°C. ΔH > 0, ΔS > 0 → row 4 (spontaneous at high T).
2(b): T = 900 + 273 = 1173 K; ΔG° = 178 − (1173 × 0.1607) = 178 − 188.5 = −10.5 kJ mol⁻¹ → spontaneous at 900°C.
2(c): Tcross = 178 / 0.1607 = 1108 K = 835°C. Above 835°C: spontaneous.
2(d): Yes — 900°C > 835°C, so ΔG° < 0 at kiln temperature. This is exactly why lime kilns operate above 900°C.
Activity 2 — The Four Combinations Drill
For each reaction, (i) identify the ΔH/ΔS combination category, (ii) predict spontaneity without calculating, and (iii) if temperature-dependent, state whether spontaneous at low T or high T.
- $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \to 2\text{SO}_3\text{(g)}$: $\Delta H° = -197 \text{ kJ mol}^{-1}$; $\Delta S° = -187 \text{ J K}^{-1}\text{mol}^{-1}$
- $\text{N}_2\text{O}_4\text{(g)} \to 2\text{NO}_2\text{(g)}$: $\Delta H° = +57 \text{ kJ mol}^{-1}$; $\Delta S° = +175 \text{ J K}^{-1}\text{mol}^{-1}$
-
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.
1: ΔH < 0, ΔS < 0 → Row 3. Spontaneous at low T; becomes non-spontaneous above Tcross. Tcross = −197 / (−0.187) = 1053 K (780°C). Spontaneous at all typical lab/industrial temperatures below 780°C.
2: ΔH > 0, ΔS > 0 → Row 4. Spontaneous only at high T. Tcross = 57 / 0.175 = 326 K (53°C). Spontaneous above 53°C — this reaction (N₂O₄ → 2NO₂) is why NO₂ (brown gas) forms preferentially at higher temperatures.
3: 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.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Misconceptions to Fix
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 concepts. Some spontaneous reactions are extremely slow at room temperature.
Pay close attention to the conditions and reagents in each reaction. These determine the product formed and are commonly tested in HSC exams.
Multiple Choice
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer Questions
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.
ΔH/ΔS combination: ΔH° = +178 kJ mol⁻¹ (>0, endothermic — energy needed to break CaCO₃); 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. At 298 K, the T·ΔS term (+47.9) is insufficient to overcome the large positive ΔH (+178).
Crossover temperature: Tcross = 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. At this temperature, the entropy term (−T·ΔS) outweighs the endothermic ΔH, making ΔG negative. The lime kiln operates at 900°C specifically to exceed this crossover temperature.
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 Tcross. (d) Rusting is slow at room temperature. Distinguish between the thermodynamic and kinetic aspects of this observation.
(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) 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 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, further slowing the process. Spontaneity (thermodynamics) tells us rusting will occur if given sufficient time; rate (kinetics) tells us how fast it occurs. These are independent properties: ΔG < 0 says nothing about rate.
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, and distinguish between thermodynamic and kinetic factors. Use: $\Delta H° = -92.4 \text{ kJ mol}^{-1}$; $\Delta S° = -198.9 \text{ J K}^{-1}\text{mol}^{-1}$.
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.
Tcross: 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 (673–773 K) lies well above Tcross — the reaction is thermodynamically non-spontaneous at these temperatures.
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 to break N≡N bonds and achieve a practical reaction rate. An iron catalyst (with K₂O and Al₂O₃ promoters) lowers Ea further. Additionally, high pressure (~200 atm) and continuous removal of NH₃ (Le Chatelier, Module 5) help drive the equilibrium toward products.
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.
Revisit Your Prediction
Look back at your Think First response. Were you able to predict the role of entropy in the Haber paradox?
- The missing concept was the temperature dependence of ΔG — specifically that ΔH < 0 and ΔS < 0 means the reaction loses spontaneity as temperature rises.
- The Haber process is run at high T purely because of kinetics — it is thermodynamically unfavourable at industrial temperatures.
- Gibbs free energy unifies enthalpy and entropy into a single spontaneity criterion for the first time in Module 4.
Common Misconceptions — Gibbs Free Energy
Answers
MC 1 — B: Tcross = −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: Row 4 (ΔH >0, ΔS >0). At 298 K: ΔG = +130.1 kJ mol⁻¹ (non-spontaneous). Tcross = 178/0.1607 = 1108 K = 835°C. At 1173 K: ΔG = −10.5 kJ mol⁻¹ (spontaneous). Lime kiln must exceed 835°C.
Q7: (a) ΔG° = −1484.4 kJ mol⁻¹. (b) Spontaneous. (c) Tcross = 3002 K — above melting point of iron. (d) Thermodynamically strongly spontaneous; kinetically slow (high Ea, passivating Fe₂O₃ layer).
Q8: Statement incorrect. ΔG° at 25°C = −33.1 kJ mol⁻¹ (spontaneous); at 450°C = +51.4 kJ mol⁻¹ (non-spontaneous). Tcross = 465 K. High T used for kinetic reasons (N≡N Ea); thermodynamics actually becomes unfavourable. Paradox resolved by distinguishing thermodynamics (spontaneity) from kinetics (rate).
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