Entropy (ΔS), Gibbs Free Energy (ΔG = ΔH - TΔS) & Temperature Feasibility
Why do some endothermic reactions, such as dissolving ammonium nitrate in water or the thermal decomposition of calcium carbonate, occur spontaneously by absorbing heat from their surroundings? The answer lies in entropy (S), a thermodynamic measure of the disorder or number of ways particles and their quanta of energy can be arranged within a system. By combining enthalpy change (ΔH, in kJ mol⁻¹) and entropy change (ΔS, converted from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹) at absolute temperature T (in Kelvin), the Gibbs Free Energy equation (ΔG = ΔH - TΔS) determines whether any chemical reaction or phase change is thermodynamically feasible (ΔG ≤ 0). Examiners constantly test unit conversions (dividing ΔS by 1000), calculating the critical threshold temperature (T = ΔH / ΔS), and interpreting linear graphs of ΔG against T (y = mx + c).
- Entropy (S) & Physical State Disorder
- Entropy (S, units: J K⁻¹ mol⁻¹) is a measure of the degree of disorder or randomness of particles and energy within a system. Standard entropy follow the order: S°(Solid) < S°(Liquid) ≪ S°(Gas), because particles in a gas move freely and randomly in 3D space with many more microstates. Unlike ΔfH° (which is zero for elements), absolute standard entropy S° of any substance at 298 K is always positive; S = 0 only for a perfect crystalline solid at absolute zero (0 K).
- Predicting the Sign of ΔS & Calculating ΔS°
- A reaction has a positive entropy change (ΔS > 0, increase in disorder) when there is: 1. A change of state from solid → liquid → aqueous → gas, or 2. An increase in the total number of moles of gas (or total particles) from reactants to products (e.g. CaCO₃(s) → CaO(s) + CO₂(g) goes from 0 mol gas to 1 mol gas, so ΔS > 0). Calculated from standard entropies using: ΔS° = ΣS°(Products) - ΣS°(Reactants).
- Gibbs free energy equation: ΔG = ΔH - TΔS
- Determines whether a reaction is thermodynamically feasible (spontaneous): ΔG = ΔH - TΔS, where ΔG is in kJ mol⁻¹, ΔH is in kJ mol⁻¹, T is absolute temperature in Kelvin (K = °C + 273), and ΔS must be converted to kJ K⁻¹ mol⁻¹ by dividing J K⁻¹ mol⁻¹ by 1000! A reaction is feasible if and only if ΔG ≤ 0 (negative or zero).
Entropy (ΔS), Gibbs Free Energy (ΔG = ΔH - TΔS) & Temperature Feasibility: Reaction Profile (ΔH & E_a) + Le Chatelier Yield Simulator
Toggle a catalyst on the enthalpy profile diagram or adjust temperature and pressure to see how dynamic equilibrium shifts.
Modelled on N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (ΔH = −92 kJ/mol) (4 moles gas → 2 moles gas):
Higher pressure shifts equilibrium to the side with fewer gas moles (right, 4 → 2). Lower temperature favours the exothermic forward reaction.
Key Concepts & Exam Definitions (6)
- Entropy (S) & Physical State Disorder
- Entropy (S, units: J K⁻¹ mol⁻¹) is a measure of the degree of disorder or randomness of particles and energy within a system. Standard entropy follow the order: S°(Solid) < S°(Liquid) ≪ S°(Gas), because particles in a gas move freely and randomly in 3D space with many more microstates. Unlike ΔfH° (which is zero for elements), absolute standard entropy S° of any substance at 298 K is always positive; S = 0 only for a perfect crystalline solid at absolute zero (0 K).
- Predicting the Sign of ΔS & Calculating ΔS°
- A reaction has a positive entropy change (ΔS > 0, increase in disorder) when there is: 1. A change of state from solid → liquid → aqueous → gas, or 2. An increase in the total number of moles of gas (or total particles) from reactants to products (e.g. CaCO₃(s) → CaO(s) + CO₂(g) goes from 0 mol gas to 1 mol gas, so ΔS > 0). Calculated from standard entropies using: ΔS° = ΣS°(Products) - ΣS°(Reactants).
- Gibbs free energy equation: ΔG = ΔH - TΔS
- Determines whether a reaction is thermodynamically feasible (spontaneous): ΔG = ΔH - TΔS, where ΔG is in kJ mol⁻¹, ΔH is in kJ mol⁻¹, T is absolute temperature in Kelvin (K = °C + 273), and ΔS must be converted to kJ K⁻¹ mol⁻¹ by dividing J K⁻¹ mol⁻¹ by 1000! A reaction is feasible if and only if ΔG ≤ 0 (negative or zero).
- Four Combinations of ΔH and ΔS Signs for Feasibility
- 1. ΔH < 0 (exothermic) and ΔS > 0 (more disorder): -TΔS is always negative, so ΔG < 0 at all temperatures (always feasible). 2. ΔH > 0 (endothermic) and ΔS < 0 (less disorder): ΔG > 0 at all temperatures (never feasible). 3. ΔH > 0 (endothermic) and ΔS > 0 (more disorder): feasible only at high temperatures when TΔS > ΔH. 4. ΔH < 0 (exothermic) and ΔS < 0 (less disorder): feasible only at low temperatures when |ΔH| > |TΔS|.
- Threshold Temperature (T = ΔH / ΔS) & Phase Changes (Melting/Boiling)
- At the exact temperature where a reaction becomes feasible (or at equilibrium during a state change like melting at Tm or boiling at Tb), ΔG = 0. Setting 0 = ΔH - TΔS gives the critical threshold temperature: T (in K) = ΔH (in kJ mol⁻¹) ÷ ΔS (in kJ K⁻¹ mol⁻¹) [or T = (ΔH × 1000) ÷ ΔS in J K⁻¹ mol⁻¹].
- Graphical Analysis of ΔG against T (y = mx + c) & Kinetic Stability
- Writing ΔG = (-ΔS)T + ΔH in the linear form y = mx + c where y = ΔG and x = T (in K): 1. The y-intercept (c, at T = 0 K) equals ΔH. 2. The gradient (m) equals -ΔS (so a negative gradient means ΔS is positive; a positive gradient means ΔS is negative). 3. The x-intercept (where ΔG = 0) is the threshold temperature T at which the reaction becomes feasible. Note: Even if ΔG < 0 (feasible), a reaction may not occur noticeably at room temperature if the activation energy (Eₐ) is very high (kinetically inert).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
State the units of ΔH, ΔS, and ΔG, and write down the conversion needed before substituting into ΔG = ΔH - TΔS.
Why does the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) have a negative entropy change (ΔS < 0)?
Under what conditions of ΔH and ΔS is a chemical reaction feasible at all temperatures?
How can an endothermic reaction (ΔH > 0) occur spontaneously at room temperature?
On a straight-line graph of ΔG (y-axis) plotted against temperature T in Kelvin (x-axis), what do the y-intercept and the gradient represent?
Why does a mixture of hydrogen and oxygen gas not react at 298 K even though ΔG for forming water is strongly negative (-237 kJ mol⁻¹)?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Thermal decomposition of calcium carbonate occurs according to the equation: CaCO₃(s) → CaO(s) + CO₂(g), where ΔH° = +178 kJ mol⁻¹. Standard entropies S° (in J K⁻¹ mol⁻¹) are: CaCO₃(s) = 93, CaO(s) = 40, CO₂(g) = 214. Calculate ΔS°, ΔG° at 298 K, and the minimum temperature (in °C) at which this decomposition becomes feasible. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Calculate ΔS° = ΣS°(Products) - ΣS°(Reactants) = (40 + 214) - 93 = +161 J K⁻¹ mol⁻¹ [1 mark]. 2. Convert ΔS° to kJ K⁻¹ mol⁻¹: +161 ÷ 1000 = +0.161 kJ K⁻¹ mol⁻¹ [1 mark]. 3. Calculate ΔG° at 298 K: ΔG° = ΔH° - TΔS° = +178 - (298 × 0.161) = 178 - 47.98 = +130 kJ mol⁻¹ (since ΔG° > 0, the reaction is not feasible at 298 K) [1 mark]. 4. At the minimum temperature for feasibility, ΔG = 0, so T = ΔH° ÷ ΔS° = 178 ÷ 0.161 = 1106 K (1105.6 K) [1 mark]. 5. Convert Kelvin to °C: 1106 - 273 = 833°C [1 mark].
- Q2
Iron(III) oxide can be reduced by carbon monoxide in a blast furnace: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g) (Reaction 1, ΔH = -25 kJ mol⁻¹, ΔS = +15 J K⁻¹ mol⁻¹) or by solid carbon: Fe₂O₃(s) + 3C(s) → 2Fe(s) + 3CO(g) (Reaction 2, ΔH = +493 kJ mol⁻¹, ΔS = +543 J K⁻¹ mol⁻¹). Explain why Reaction 1 is feasible at all temperatures, whereas Reaction 2 has a much larger positive ΔS and only becomes feasible above a certain temperature. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. In Reaction 1, ΔH is negative (-25 kJ mol⁻¹) and ΔS is positive (+15 J K⁻¹ mol⁻¹), so in ΔG = ΔH - TΔS the term -TΔS is always negative, meaning ΔG < 0 at all temperatures [1 mark]. 2. Reaction 1 has only a very small positive ΔS because 3 moles of gas (3CO) produce 3 moles of gas (3CO₂) (zero change in gas moles) [1 mark]. 3. Reaction 2 has a very large positive ΔS (+543 J K⁻¹ mol⁻¹) because 4 moles of solid reactants react to produce 3 moles of gas (3CO(g)) and 2 moles of solid, greatly increasing disorder [1 mark]. 4. Because Reaction 2 is strongly endothermic (ΔH = +493 kJ mol⁻¹), ΔG is only negative (feasible) at high temperatures above T = 493 ÷ 0.543 = 908 K, where TΔS exceeds ΔH [1 mark].
- Q3
Ammonium chloride dissolves endothemically in water: NH₄Cl(s) + aq → NH₄⁺(aq) + Cl⁻(aq). Given that ΔLEH°(dissociation) = +705 kJ mol⁻¹, ΔhydH°[NH₄⁺(g)] = -307 kJ mol⁻¹, ΔhydH°[Cl⁻(g)] = -381 kJ mol⁻¹, and ΔS° = +75.8 J K⁻¹ mol⁻¹, calculate ΔsolH° and ΔG° at 298 K, and explain why ammonium chloride dissolves readily in water at room temperature despite cooling the solution. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Calculate ΔsolH° = ΔLEH°(dissociation) + ΔhydH°[NH₄⁺] + ΔhydH°[Cl⁻] = +705 + (-307) + (-381) = +17 kJ mol⁻¹ [1 mark]. 2. Convert ΔS° to kJ K⁻¹ mol⁻¹ (+75.8 ÷ 1000 = +0.0758 kJ K⁻¹ mol⁻¹) and calculate ΔG° at 298 K: ΔG° = +17 - (298 × 0.0758) = +17 - 22.59 = -5.59 kJ mol⁻¹ [1 mark]. 3. Dissolving breaks up the ordered solid ionic lattice into freely moving aqueous ions (1 mol solid → 2 mol aqueous ions), giving a large positive entropy change (ΔS° > 0) [1 mark]. 4. At 298 K, TΔS° (+22.59 kJ mol⁻¹) is greater than the small endothermic ΔsolH° (+17 kJ mol⁻¹), so ΔG° is negative (-5.59 kJ mol⁻¹) and dissolving is thermodynamically feasible [1 mark].
- Q4
For the vaporisation of liquid bromine, Br₂(l) ⇌ Br₂(g), ΔH° = +30.9 kJ mol⁻¹ and ΔS° = +93.1 J K⁻¹ mol⁻¹. Estimate the boiling point of bromine in °C and explain why the value of ΔG° for this process is zero at the boiling point. (3 marks)
Click “Check Mark Scheme” to compare your answer, 1. At the boiling point, liquid bromine and bromine vapour are in dynamic equilibrium, so the forward and reverse processes are equally feasible and ΔG = 0 [1 mark]. 2. Set ΔG = 0 in ΔG = ΔH - TΔS: Tb = ΔH° ÷ ΔS° = 30.9 kJ mol⁻¹ ÷ (93.1 ÷ 1000 kJ K⁻¹ mol⁻¹) = 30.9 ÷ 0.0931 = 331.9 K (332 K) [1 mark]. 3. Convert to °C: 331.9 - 273 = 58.9°C (59°C) [1 mark].
Common Exam Questions (4)
Why is the standard entropy S° of an element not zero, whereas ΔfH° of an element IS zero?
Standard enthalpy of formation (ΔfH°) measures a change in enthalpy relative to elements at 298 K (so forming an element from itself involves zero change). Standard entropy (S°), however, is an absolute quantity measured relative to a perfect crystal at absolute zero (0 K, the Third Law of Thermodynamics); at 298 K, every element and compound has thermal energy and disorder, so S° > 0 for all substances.
How is Gibbs Free Energy (ΔG°) related to the equilibrium constant K?
ΔG° is linked to the equilibrium constant K by the equation ΔG° = -RT ln(K). When ΔG° < 0, ln(K) > 0 so K > 1 (products predominate at equilibrium); when ΔG° > 0, K < 1 (reactants predominate); and when ΔG° = 0, K = 1.
Why does the gradient of a ΔG vs T graph abruptly change steepness at the melting point or boiling point of a reactant or product?
Because the gradient of a ΔG vs T graph equals -ΔS, and when one of the reactants or products melts or boils, its entropy S° jumps dramatically, changing the overall ΔS of the reaction and therefore changing the slope of the line.
If a reaction has ΔG > 0 at 298 K, does that mean zero product is formed?
Not necessarily zero: ΔG° > 0 means that under standard conditions (all concentrations at 1 M / pressures at 100 kPa), the equilibrium constant K is less than 1, so the equilibrium position lies to the left (favouring reactants), though a small equilibrium concentration of products may still exist.
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