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Thermodynamics is more than just enthalpy. Many reactions that are endothermic nevertheless happen spontaneously - a piece of ammonium nitrate dissolved in water chills the beaker, yet the process proceeds without any input of energy. Enthalpy alone cannot predict which reactions are thermodynamically favourable. We need another variable: entropy.
Entropy (S) is a measure of the disorder or spread of energy in a system. More rigorously, it is proportional to the number of ways the particles and their energies can be arranged while still giving the same overall macroscopic state. The more arrangements (microstates) available, the higher the entropy.
A pack of playing cards, neatly ordered from Ace of Spades to King of Hearts, has only one "sorted" arrangement. Throw the cards in the air and the number of disordered arrangements is vast. Tossed cards have high entropy; the sorted deck has low entropy.
At the particulate level:
The clearest guide to entropy is the physical state of a substance.
| State | Particle arrangement | Entropy |
|---|---|---|
| Solid | regular, fixed | lowest |
| Liquid | disordered, free to move | intermediate |
| Gas | random, high KE, large volume | highest |
graph LR
A[Solid<br/>low S] -->|melting| B[Liquid<br/>medium S]
B -->|vaporising| C[Gas<br/>high S]
C -->|much higher than liquid| D[Gases dominate<br/>the entropy term]
Vaporisation gives a particularly large entropy increase because the volume available to each particle jumps from a few cubic nanometres (liquid) to a few cubic centimetres (gas at 1 bar, 298 K) - an increase of about 1000-fold in accessible space.
Rules of thumb for predicting the sign of an entropy change:
Counter-examples exist. Hydrating a highly charged ion can DECREASE the entropy of water because the orderly water shell around the ion is more structured than bulk liquid water.
(a) 2H2(g) + O2(g) -> 2H2O(l)
Moles of gas: 3 -> 0. Entropy decreases sharply. ΔS° is very negative (about -327 J K^-1 mol^-1).
(b) CaCO3(s) -> CaO(s) + CO2(g)
Moles of gas: 0 -> 1. Entropy increases. ΔS° is positive (about +161 J K^-1 mol^-1).
(c) NH4NO3(s) -> NH4+(aq) + NO3-(aq)
Solid dissolving to give aqueous ions. Entropy increases. ΔS° is positive (about +108 J K^-1 mol^-1), and this is why the process is spontaneous even though it is endothermic.
(d) N2(g) + 3H2(g) -> 2NH3(g)
Moles of gas: 4 -> 2. Entropy decreases. ΔS° is negative (about -198 J K^-1 mol^-1). Despite this, the Haber reaction can proceed because it is strongly exothermic (see lesson 7 for the full treatment).
The standard entropy change of a reaction is calculated from the standard entropies of the reactants and products:
ΔS°_reaction = ΣS°_products - ΣS°_reactants
Just as with enthalpy, you take "products minus reactants". The difference is that standard entropies are tabulated as absolute values (referenced to S = 0 at 0 K), not relative values. A pure substance has a well-defined standard entropy.
Units: J K^-1 mol^-1 (joules, not kilojoules)
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