Total entropy change and feasibilityEdexcel International A Level Chemistry: Revision notes
Section 1
Entropy change of the system
Entropy (S) measures the number of ways energy and particles can be arranged; gases have far more than liquids, and liquids more than solids. Entropy values are given in J K⁻¹ mol⁻¹.
ΔSsystem = ΣS(products) − ΣS(reactants), multiplying each standard molar entropy by its coefficient in the equation.
Worked example: N₂ + 3H₂ → 2NH₃ with S = 191.6, 130.6 and 192.3. ΔSsystem = (2 × 192.3) − (191.6 + 3 × 130.6) = 384.6 − 583.4 = −198.8 J K⁻¹ mol⁻¹. It is negative because 4 mol of gas become 2 mol.
Forgetting to multiply each S value by its coefficient, or subtracting the wrong way round (it is products − reactants).
Section 2
Entropy change of the surroundings
When a reaction releases heat to the surroundings their entropy rises; when it takes heat in, their entropy falls.
ΔSsurroundings = −ΔH/T
ΔH is in kJ mol⁻¹ but entropy is in J K⁻¹ mol⁻¹, so multiply ΔH by 1000 first. T is in kelvin.
Example: ΔH = +25.7 kJ mol⁻¹ at 298 K gives ΔSsurroundings = −25 700 ÷ 298 = −86.2 J K⁻¹ mol⁻¹. An exothermic reaction (ΔH negative) gives a positive ΔSsurroundings.
Leaving ΔH in kJ. This gives an answer 1000 times too small.
Section 3
Total entropy change and feasibility
ΔStotal = ΔSsystem + ΔSsurroundings
A reaction is feasible (thermodynamically able to happen) if ΔStotal is positive (or zero at the limit).
The balance matters, so one term may be negative:
- An endothermic reaction can be feasible if a large positive ΔSsystem outweighs the negative ΔSsurroundings, as when ammonium nitrate dissolves in water.
- An exothermic reaction with a negative ΔSsystem can be feasible if the positive ΔSsurroundings is larger.
Example: ΔSsystem = +108.5 and ΔSsurroundings = −86.2 give ΔStotal = +22.3 J K⁻¹ mol⁻¹, so it is feasible.
Section 4
Temperature and the feasibility temperature
Because ΔSsurroundings = −ΔH/T, a higher temperature reduces the magnitude of ΔSsurroundings.
- For an endothermic reaction the negative ΔSsurroundings becomes smaller, so ΔStotal rises and the reaction becomes feasible at high enough temperature.
- For an exothermic reaction the positive ΔSsurroundings becomes smaller, so ΔStotal falls (this matters if ΔSsystem is negative).
The minimum feasible temperature is where ΔStotal = 0, so ΔSsurroundings = −ΔSsystem and T = ΔH/ΔSsystem.
Example: CaCO₃ → CaO + CO₂, ΔH = +178 kJ mol⁻¹, ΔSsystem = +160.4 J K⁻¹ mol⁻¹: T = 178 000 ÷ 160.4 = 1110 K.
Section 5
Thermodynamic and kinetic stability
A positive ΔStotal says a reaction can happen, not that it will happen at a measurable rate. Rate depends on activation energy.
A substance that is thermodynamically unstable (feasible reaction) can be kinetically stable if the activation energy is so high that, at that temperature, almost no molecules have enough energy to react. Methane and oxygen do not react at 298 K despite a huge positive ΔStotal until a spark supplies the activation energy.
The Gibbs energy equation ΔG = ΔH − TΔSsystem may also be used (ΔG < 0 means feasible) but is not required for this specification. It is just ΔStotal multiplied by −T.
When asked why a feasible reaction does not happen, say there is a high activation energy so the rate is negligible. Do not say it is not feasible.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Total entropy change and feasibility
- Instant cold packs contain solid ammonium nitrate and a sealed pouch of water. When the pouch is burst, the solid dissolves: NH₄NO₃(s) + aq → NH₄⁺(aq) + NO₃⁻(aq). The enthalpy change of solution is +25.7 kJ mol⁻¹ and the entropy change of the system is +108.5 J K⁻¹ mol⁻¹. The pack is used at 298 K and becomes noticeably cold as the solid dissolves.Calculate the total entropy change for the dissolving at 298 K and state what it shows about feasibility.2 marks
- In the Haber process, nitrogen and hydrogen react to make ammonia: N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.0 kJ mol⁻¹. Standard molar entropies at 298 K are: N₂(g) 191.6, H₂(g) 130.6 and NH₃(g) 192.3 J K⁻¹ mol⁻¹.Explain, with a calculation of ΔSsurroundings at 298 K, why this reaction is feasible even though ΔSsystem is negative.2 marks
- Limestone is heated in a kiln to make quicklime: CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹. Standard molar entropies at 298 K are: CaCO₃(s) 92.9, CaO(s) 39.7 and CO₂(g) 213.6 J K⁻¹ mol⁻¹. Assume that ΔH and the entropy values do not change with temperature.Calculate ΔSsystem and ΔSsurroundings at 298 K for this reaction, and use them to show that the decomposition is not feasible at 298 K.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).