Entropy

A2 · 12 min

Some endothermic processes happen on their own: ice melts on a warm day, ammonium nitrate dissolves while cooling its solution, and calcium carbonate decomposes in a hot kiln. Enthalpy alone cannot explain this. The missing idea is entropy, a measure of how many ways the particles and their energy can be arranged. Paper 4 asks you to define entropy, predict and explain the sign of entropy changes, and calculate $\Delta S^{\ominus}$ from standard entropies. These feed straight into Gibbs free energy.

What entropy measures

Definition

Entropy, $S$ , is the number of possible arrangements of the particles and their energy in a given system.

Think about the particles in a substance and the energy they carry.

In a solid, particles vibrate about fixed positions. There are relatively few ways to arrange them, so entropy is low.
In a liquid, particles move past one another. There are many more possible arrangements, so entropy is higher.
In a gas, particles move freely and randomly through the whole container. The number of possible arrangements is enormous, so entropy is highest.

Energy can be arranged in different ways too. Particles store energy in quantised packets (translational, rotational and vibrational energy). The more energy a system has, and the more ways it can be shared between particles, the more arrangements are possible and the higher the entropy.

A system tends to move towards the state with the greatest number of arrangements, simply because that state is overwhelmingly the most probable. This is why gases spread out and why a drop of ink disperses in water: you never see the reverse.

Units and values

Standard molar entropies, $S^{\ominus}$ , are measured in $\text{J K}^{-1}\ \text{mol}^{-1}$ (note joules, not kilojoules). Every substance has a positive entropy at room temperature, including elements. (Contrast enthalpy of formation, which is zero for elements.)

Substance	State	$S^{\ominus}$ / J K⁻¹ mol⁻¹
C (diamond)	s	2.4
C (graphite)	s	5.7
MgO	s	26.9
Mg	s	32.7
NaCl	s	72.1
$\ce{H2O}$	l	69.9
$\ce{H2}$	g	130.6
$\ce{H2O}$	g	188.7
$\ce{N2}$	g	191.6
$\ce{O2}$	g	205.0
$\ce{CO2}$	g	213.6

Patterns to notice: gases have far higher entropies than solids; diamond (a rigid, very ordered lattice) has a lower entropy than graphite; heavier, more complex gas molecules generally have higher entropy than simpler ones, because they have more ways to store energy.

Predicting the sign of an entropy change

You are expected to predict and explain the sign of $\Delta S$ in three types of situation. In each case, the explanation must mention the number of possible arrangements of particles or energy.

Changes of state

y = 0.3x + 25(1 + tanh(3(x - 30)))/2 + 55(1 + tanh(3(x - 70)))/2 + 3

The sketch shows the entropy of a substance against temperature. Entropy rises gradually as the solid warms, jumps at the melting point (here at 30 on the temperature axis), rises gradually through the liquid range, then jumps by a much larger amount at the boiling point (here at 70).

Change	Sign of $\Delta S$	Reason
melting (s → l)	positive	particles gain freedom to move; more arrangements
boiling (l → g)	large, positive	particles become free and widely spaced; vastly more arrangements
sublimation (s → g)	large, positive	as above
freezing, condensing	negative	fewer arrangements
dissolving a solid in water	usually positive	the ordered lattice breaks up and ions or molecules spread through the solution

The jump on boiling is much bigger than on melting because the gas occupies a far larger volume, so the number of positional arrangements increases enormously.

Tip

Dissolving is "usually" positive. For very small, highly charged ions, the water molecules become tightly ordered around each ion, which can make the overall entropy change of solution small or even negative. You will not be asked to predict these exceptions, but do not write "dissolving always increases entropy".

Temperature change

Raising the temperature of a substance increases its entropy (positive $\Delta S$ ). The particles have more energy, so there are more ways of distributing that energy among them (a wider spread of energies), and the particles move more, giving more positional arrangements. Cooling has the opposite effect.

Reactions that change the number of gas molecules

Because gases dominate entropy, the most reliable way to predict the sign of $\Delta S$ for a reaction is to count moles of gas.

More moles of gas in the products than the reactants: $\Delta S$ is positive.
Fewer moles of gas: $\Delta S$ is negative.
No change in gas moles: $\Delta S$ is small, and you cannot predict its sign reliably without data.

\ce{CaCO3(s) -> CaO(s) + CO2(g)} \qquad 0 \to 1 \text{ mol gas: } \Delta S > 0

\ce{N2(g) + 3H2(g) -> 2NH3(g)} \qquad 4 \to 2 \text{ mol gas: } \Delta S < 0

Predicting the sign of ΔS for a reaction

Check state symbols carefully. Count the moles of gas on each side.
If the moles of gas change, that decides the sign. Explain it as "an increase (or decrease) in the number of gas molecules, so more (or fewer) possible arrangements".
If gas moles do not change, look for other state changes: solid to liquid or solution increases entropy; solution to precipitate decreases it.
If no clear change exists, say that $\Delta S$ is likely to be small.

Calculating ΔS⦵ for a reaction

Standard entropies are absolute values, so you can calculate the entropy change of a reaction directly, much like using enthalpies of formation.

Key result

\Delta S^{\ominus} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})

Multiply each $S^{\ominus}$ by its stoichiometric coefficient. Elements have non-zero entropies and must be included. Units: $\text{J K}^{-1}\ \text{mol}^{-1}$ .

The syllabus only requires $\Delta S^{\ominus}$ of the reacting system. You do not need the entropy change of the surroundings or the total entropy change; Gibbs free energy takes care of the surroundings for you.

Worked examples

Predicting signs

Predict, with a reason, the sign of $\Delta S$ for each change.

(a) $\ce{H2O(l) -> H2O(g)}$ (b) $\ce{2SO2(g) + O2(g) -> 2SO3(g)}$ (c) $\ce{Ag^+(aq) + Cl^-(aq) -> AgCl(s)}$ (d) $\ce{H2(g) + Cl2(g) -> 2HCl(g)}$ (e) heating a sample of copper from $300\ \text{K}$ to $400\ \text{K}$

Solution

(a) Positive. A liquid becomes a gas; gas particles are free to move throughout the container, so there are many more possible arrangements.

(b) Negative. Three moles of gas become two, so there are fewer possible arrangements of particles.

(d) Small; sign cannot be predicted reliably from the equation, because there are two moles of gas on each side. (Data give $+20\ \text{J K}^{-1}\ \text{mol}^{-1}$ .)

(e) Positive. The atoms have more energy, so there are more ways of distributing the energy between them, and they vibrate more.

Thermal decomposition of calcium carbonate

Calculate $\Delta S^{\ominus}$ for $\ce{CaCO3(s) -> CaO(s) + CO2(g)}$ . $S^{\ominus}$ / J K⁻¹ mol⁻¹: $\ce{CaCO3}$ 92.9; $\ce{CaO}$ 39.7; $\ce{CO2}$ 213.6.

Solution

\Delta S^{\ominus} = (39.7 + 213.6) - 92.9 = +160.4\ \text{J K}^{-1}\ \text{mol}^{-1}

Positive, as predicted: a gas is produced from a solid.

The Haber process

Calculate $\Delta S^{\ominus}$ for $\ce{N2(g) + 3H2(g) -> 2NH3(g)}$ . $S^{\ominus}$ / J K⁻¹ mol⁻¹: $\ce{N2}$ 191.6; $\ce{H2}$ 130.6; $\ce{NH3}$ 192.3.

Solution

\Delta S^{\ominus} = 2(192.3) - [191.6 + 3(130.6)] = 384.6 - 583.4 = -198.8\ \text{J K}^{-1}\ \text{mol}^{-1}

Negative, because four moles of gas become two.

Combustion with a liquid product

Calculate $\Delta S^{\ominus}$ for $\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$ . $S^{\ominus}$ / J K⁻¹ mol⁻¹: $\ce{CH4}$ 186.2; $\ce{O2}$ 205.0; $\ce{CO2}$ 213.6; $\ce{H2O(l)}$ 69.9.

Solution

\sum S^{\ominus}(\text{products}) = 213.6 + 2(69.9) = 353.4

\sum S^{\ominus}(\text{reactants}) = 186.2 + 2(205.0) = 596.2

\Delta S^{\ominus} = 353.4 - 596.2 = -242.8\ \text{J K}^{-1}\ \text{mol}^{-1}

Three moles of gas become one mole of gas and two moles of liquid, so a large decrease is expected.

Working back to an unknown entropy (exam-hard)

For $\ce{H2(g) + Cl2(g) -> 2HCl(g)}$ , $\Delta S^{\ominus} = +20.0\ \text{J K}^{-1}\ \text{mol}^{-1}$ . $S^{\ominus}(\ce{H2}) = 130.6$ and $S^{\ominus}(\ce{HCl}) = 186.8\ \text{J K}^{-1}\ \text{mol}^{-1}$ .

(a) Calculate $S^{\ominus}(\ce{Cl2})$ . (b) Explain why $S^{\ominus}(\ce{Cl2})$ is larger than $S^{\ominus}(\ce{H2})$ , although both are diatomic gases.

Solution

(a)

20.0 = 2(186.8) - [130.6 + S^{\ominus}(\ce{Cl2})]

S^{\ominus}(\ce{Cl2}) = 373.6 - 130.6 - 20.0 = 223.0\ \text{J K}^{-1}\ \text{mol}^{-1}

(b) $\ce{Cl2}$ molecules are much heavier and larger, with a longer, weaker bond. They have many more closely spaced energy levels for rotation and vibration, so there are many more ways of arranging the energy among the molecules, giving a higher entropy.

Common mistakes

Units. Entropy is in $\text{J K}^{-1}\ \text{mol}^{-1}$ . Gibbs calculations use $\text{kJ}$ for enthalpy, so you must divide $\Delta S$ by 1000 there.
Setting elements to zero. That rule is for $\Delta H^{\ominus}_{\text{f}}$ , not entropy. $S^{\ominus}(\ce{O2}) = 205.0$ , not zero.
Ignoring coefficients. In the Haber process, $3 \times S^{\ominus}(\ce{H2})$ .
Vague explanations. "It becomes more disordered" scores less than "there are more possible arrangements of the particles (and their energy)". Use the syllabus language.
Forgetting state symbols. $\ce{H2O(l)}$ (69.9) and $\ce{H2O(g)}$ (188.7) have very different entropies.

Exam technique

Exam tip

The definition "the number of possible arrangements of the particles and their energy in a given system" is worth a mark on its own; learn it exactly.
For "predict and explain" questions, give the sign and a reason tied to the change in the number of gas molecules or the state of matter. One word is not enough.
In calculations, set out $\sum S^{\ominus}(\text{products})$ and $\sum S^{\ominus}(\text{reactants})$ separately; partial credit is available if one sum is wrong.
Give answers to 3 or 4 significant figures with sign and units, for example $+160.4\ \text{J K}^{-1}\ \text{mol}^{-1}$ (or $+0.160\ \text{kJ K}^{-1}\ \text{mol}^{-1}$ ).
Sketch-graph questions on $S$ against $T$ : gradual rise in solid, vertical jump at melting, gradual rise in liquid, much larger vertical jump at boiling, gradual rise in gas.

Summary

Entropy is the number of possible arrangements of the particles and their energy in a system; units J K⁻¹ mol⁻¹.
$S$ : gas ≫ liquid > solid. All substances, including elements, have positive $S^{\ominus}$ .
$\Delta S$ is positive for melting, boiling, sublimation, usually for dissolving, and when temperature rises.
For reactions, an increase in moles of gas gives positive $\Delta S$ ; a decrease gives negative $\Delta S$ .
$\Delta S^{\ominus} = \sum S^{\ominus}(\text{products}) - \sum S^{\ominus}(\text{reactants})$ , including coefficients and elements.

Practice questions

Question

Define the term entropy.
State and explain the sign of $\Delta S$ for: (a) iodine subliming; (b) $\ce{2Mg(s) + O2(g) -> 2MgO(s)}$ ; (c) dissolving sodium chloride in water; (d) $\ce{N2O4(g) -> 2NO2(g)}$ .
Explain why the entropy change on boiling water is much larger than the entropy change on melting ice.
Calculate $\Delta S^{\ominus}$ for $\ce{2SO2(g) + O2(g) -> 2SO3(g)}$ . $S^{\ominus}$ / J K⁻¹ mol⁻¹: $\ce{SO2}$ 248.1; $\ce{O2}$ 205.0; $\ce{SO3}$ 256.7.
Calculate $\Delta S^{\ominus}$ for $\ce{C2H4(g) + H2(g) -> C2H6(g)}$ . $S^{\ominus}$ : $\ce{C2H4}$ 219.5; $\ce{H2}$ 130.6; $\ce{C2H6}$ 229.5.
Calculate $\Delta S^{\ominus}$ for $\ce{Na(s) + 1/2Cl2(g) -> NaCl(s)}$ . $S^{\ominus}$ : Na 51.2; $\ce{Cl2}$ 223.0; NaCl 72.1. Comment on the sign.
Calculate the entropy change when one mole of water boils at its boiling point, using $S^{\ominus}(\ce{H2O(l)}) = 69.9$ and $S^{\ominus}(\ce{H2O(g)}) = 188.7\ \text{J K}^{-1}\ \text{mol}^{-1}$ .
Diamond has $S^{\ominus} = 2.4$ and graphite $5.7\ \text{J K}^{-1}\ \text{mol}^{-1}$ . Calculate $\Delta S^{\ominus}$ for graphite → diamond, and explain the difference in entropy in terms of structure.
Calculate $\Delta S^{\ominus}$ for $\ce{2NaHCO3(s) -> Na2CO3(s) + H2O(g) + CO2(g)}$ . $S^{\ominus}$ : $\ce{NaHCO3}$ 101.7; $\ce{Na2CO3}$ 135.0; $\ce{H2O(g)}$ 188.7; $\ce{CO2}$ 213.6. Explain why the answer is large and positive.
For $\ce{2NO2(g) -> N2O4(g)}$ , $\Delta S^{\ominus} = -175.8\ \text{J K}^{-1}\ \text{mol}^{-1}$ and $S^{\ominus}(\ce{NO2}) = 240.0\ \text{J K}^{-1}\ \text{mol}^{-1}$ . Calculate $S^{\ominus}(\ce{N2O4})$ , and explain why $S^{\ominus}(\ce{N2O4})$ is greater than $S^{\ominus}(\ce{NO2})$ even though the reaction has a negative entropy change.

Answers

The number of possible arrangements of the particles and their energy in a given system.
(a) Positive: solid becomes gas, so the particles have far more possible arrangements. (b) Negative: one mole of gas is used up and only solid forms. (c) Positive: the ordered lattice breaks up and ions spread through the solution, giving more arrangements. (d) Positive: one mole of gas becomes two moles of gas.
In a gas, molecules are far apart and move freely through the whole volume, so the number of positional arrangements is vastly greater than in a liquid. Melting only gives particles freedom to move past each other while still close together, a much smaller increase in the number of arrangements.
$\Delta S^{\ominus} = 2(256.7) - [2(248.1) + 205.0] = 513.4 - 701.2 = -187.8\ \text{J K}^{-1}\ \text{mol}^{-1}$ .
$\Delta S^{\ominus} = 229.5 - (219.5 + 130.6) = -120.6\ \text{J K}^{-1}\ \text{mol}^{-1}$ .
$\Delta S^{\ominus} = 72.1 - [51.2 + \tfrac{1}{2}(223.0)] = 72.1 - 162.7 = -90.6\ \text{J K}^{-1}\ \text{mol}^{-1}$ . Negative, because half a mole of gas is converted into an ordered solid.
$\Delta S^{\ominus} = 188.7 - 69.9 = +118.8\ \text{J K}^{-1}\ \text{mol}^{-1}$ .
$\Delta S^{\ominus} = 2.4 - 5.7 = -3.3\ \text{J K}^{-1}\ \text{mol}^{-1}$ . Diamond is a rigid three-dimensional network with every atom strongly bonded to four others, so its atoms vibrate less and there are fewer ways of arranging energy. Graphite has weakly held layers that can vibrate and slide, allowing more arrangements of energy.
$\Delta S^{\ominus} = (135.0 + 188.7 + 213.6) - 2(101.7) = 537.3 - 203.4 = +333.9\ \text{J K}^{-1}\ \text{mol}^{-1}$ . Two moles of gas are produced from solids only, giving a very large increase in the number of possible arrangements.
$-175.8 = S^{\ominus}(\ce{N2O4}) - 2(240.0)$ , so $S^{\ominus}(\ce{N2O4}) = 480.0 - 175.8 = 304.2\ \text{J K}^{-1}\ \text{mol}^{-1}$ . One $\ce{N2O4}$ molecule is larger with more atoms, so it has more ways to store energy (more vibrations and rotations) than one $\ce{NO2}$ molecule. But the reaction turns two moles of gas into one, and the loss of a mole of independently moving gas molecules outweighs this, so $\Delta S$ is negative.

What entropy measures

Units and values

Predicting the sign of an entropy change

Changes of state

Temperature change

Reactions that change the number of gas molecules

Calculating ΔS⦵ for a reaction

Worked examples

Exam technique

Summary

Practice questions

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