Absolute Entropies and ΔS°

To put numbers on entropy changes, chemists tabulate standard molar entropies, $S^\circ$. You combine them the way you combined enthalpies of formation, with one important difference.

Key idea

The third law: a perfect crystal at $0\ \text{K}$ has $S = 0$. So every substance, element or compound, has a positive absolute entropy $S^\circ$ at $298\ \text{K}$. For a reaction, $$\Delta S^\circ = \sum nS^\circ(\text{products}) - \sum nS^\circ(\text{reactants})$$

Why it works

A perfect crystal at absolute zero has only one possible arrangement: $W = 1$, so $S = k\ln 1 = 0$. That gives entropy a true zero point. Warming a substance from there adds microstates, so $S^\circ$ at $298\ \text{K}$ is measured from heat capacities as an absolute value. Enthalpy has no such zero, which is why $\Delta H^\circ_f$ of an element is defined as zero; $S^\circ$ of an element is not zero at all.

Standard molar entropies at 298 K. Note the elements: none is zero, and gases are far above solids.
Substance	$S^\circ$ (J/(mol·K))
$\mathrm{H_2(g)}$	130.7
$\mathrm{N_2(g)}$	191.6
$\mathrm{O_2(g)}$	205.2
$\mathrm{NH_3(g)}$	192.8
$\mathrm{H_2O(l)}$	70.0
$\mathrm{H_2O(g)}$	188.8
$\mathrm{CO_2(g)}$	213.8
$\mathrm{CaCO_3(s)}$	92.9
$\mathrm{CaO(s)}$	38.1

Worked examples

Worked example: Making ammonia

Find $\Delta S^\circ$ for $\mathrm{N_2(g) + 3H_2(g) \to 2NH_3(g)}$.

Products: $2(192.8) = 385.6\ \text{J/K}$.
Reactants: $$191.6 + 3(130.7) = 191.6 + 392.1 = 583.7\ \text{J/K}$$
$\Delta S^\circ = 385.6 - 583.7 = -198.1\ \text{J/K}$. Negative, as expected from 4 mol of gas becoming 2.

Answer: $-198.1\ \text{J/K}$.

Worked example: Decomposing limestone

Find $\Delta S^\circ$ for $\mathrm{CaCO_3(s) \to CaO(s) + CO_2(g)}$.

$\Delta S^\circ = (38.1 + 213.8) - 92.9 = +159.0\ \text{J/K}$.
Positive: a gas forms.

Answer: $+159.0\ \text{J/K}$.

Why gases have large entropies: on the same lattice, a gas has many more arrangements than a liquid or solid.

Three trends make the table easy to read. Gases have much larger $S^\circ$ than liquids, and liquids than solids. Within one phase, larger and more complex molecules have larger entropies (more ways to vibrate and rotate): $S^\circ$ rises from $\mathrm{CH_4}$ (186.3) to $\mathrm{C_2H_6}$ (229.2) to $\mathrm{C_3H_8}$ (270.3). And hard, tightly bonded solids such as diamond (2.4) have very low entropies, while soft solids such as lead (64.8) have higher ones.

Tables of $S^\circ$, $\Delta H^\circ_f$ and $\Delta G^\circ_f$ appear side by side in the back of most textbooks; the values here come from those tables.

Common mistake

Setting the standard entropy of an element to zero, as you do for $\Delta H^\circ_f$. Only a perfect crystal at 0 K has zero entropy. Check: in the ammonia example, leaving out $\mathrm{N_2}$ and $\mathrm{H_2}$ would give $+385.6\ \text{J/K}$, the wrong sign entirely.

See why absolute zero can be approached but never reached in the Science Lab's Absolute Zero and the Third Law chapter.

Absolute Entropies and ΔS°

Worked examples

Practice

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Sources and further reading