What is entropy — disorder and spreading out — Chemistry

Before this

This leads to

Gibbs free energy
Entropy: a statistical interpretation · coming soon
The second law in physics · coming soon

A reaction that runs uphill

Squeeze an instant cold pack, the kind used on a sports injury. A bag of water bursts inside it, and solid ammonium nitrate dissolves. Within seconds the pack is cold. The calorimetry lecture measured this change: $Δ H = + 25.7$ kJ/mol. It takes heat in from its surroundings.

Everything in the Enthalpy topic suggests that reactions "go downhill", giving out heat and ending at lower enthalpy. Most do. But this one goes uphill in enthalpy, and it still happens by itself, quickly, with nobody pushing it. Ice melting in a warm room does the same: it takes in heat, and it still happens.

The calorimetry lecture ended with a promise: enthalpy alone cannot tell you whether a change will happen, and the next topic, entropy, explains why. Here is that explanation. It is one of the deepest ideas in science, and it starts with counting.

Why things spread out

Open a bottle of perfume in a corner of a room. Soon the smell is everywhere. You never see the opposite: perfume spread through a room never gathers itself back into the bottle. No force pushes the molecules out. They simply move about at random, as the diffusion lecture showed. So why does the change only go one way?

Count the possibilities. Imagine a box divided into a left half and a right half, with just 4 gas particles in it. Each particle can be on either side, so there are $2 \times 2 \times 2 \times 2 = 16$ possible arrangements. Only 1 of these has all 4 particles on the left. So if the particles move at random, you find them all on the left only 1 time in 16.

Now try more particles in the explorer.

Explore — counting the arrangements

A box holds N particles, each moving at random between a left half and a right half. Each bar shows how many arrangements put that number of particles on the left. Green bars are the arrangements that are close to half and half. Increase N and watch the "all on the left" arrangement become extremely rare.

Number of particles, N: 10

Try:

For N particles moving at random between two halves of a box, there are $2^{N}$ arrangements, and only one of them has every particle on the left.

N	Arrangements	Chance all are on the left	Arrangements with 40–60% on the left
4	16	1 in 16	37.5%
10	1024	1 in 1024	65.6%
40	≈ 1.1 × 10¹²	1 in about 1.1 × 10¹²	84.6%

What should you have seen? With 4 particles, finding them all on one side is not unusual. With 40, it happens once in about a million million tries. And the spread-out arrangements, close to half and half, become most of all the arrangements. A real sample of gas has about $6 \times 1 0^{23}$ particles in each mole. The chance that they all gather in one half is so small that it never happens, not in the whole age of the universe.

That is it. That is the whole idea. Nothing pushes the perfume out of the bottle. Spread-out arrangements simply outnumber gathered ones so enormously that random motion almost always ends in one of them.

What entropy measures

Entropy, $S$ , is a measure of the number of ways the particles of a system, and their energy, can be arranged. The more ways there are, the higher the entropy.

You will often read that entropy is "disorder". That everyday word is a fair first picture, but it can mislead. A messy desk does not have more entropy in the chemist's sense. Think instead about spreading: particles spreading through space, and energy spreading among the particles. Both give more possible arrangements.

The second kind of spreading matters as much as the first. Energy comes in small packets, and particles can share it out in many ways. The more energy there is, and the more particles there are to share it, the more ways it can be shared. This is why heating any substance raises its entropy.

The Austrian physicist Ludwig Boltzmann gave this idea its mathematical form in 1877. In the form later written by Max Planck, it is $S = k ln W$ , where $W$ is the number of arrangements and $k$ is a constant now named after Boltzmann. This equation is carved on Boltzmann's gravestone in Vienna. You do not need it for calculations at this level, but notice what it says: entropy is counting.

Every substance has a measured standard molar entropy, S^⦵, at 298 K. Its unit is J/K/mol: joules, not kilojoules. Unlike enthalpies of formation, entropies of elements are not zero. A perfect crystal at absolute zero would have only one possible arrangement, so its entropy would be zero, and every real substance at 298 K has more.

Substance at 298 K	S⦵ in J/K/mol
C(s, diamond)	2.4
C(s, graphite)	5.7
H₂O(l)	70.0
NaCl(s)	72.1
I₂(s)	116.1
H₂(g)	130.7
H₂O(g)	188.8
O₂(g)	205.2
CO₂(g)	213.8
I₂(g)	260.7

Read the table for patterns. Diamond, a rigid network, has the lowest entropy of all. Water vapour has almost three times the entropy of liquid water. Iodine gas has more than twice the entropy of solid iodine. Gases sit at the top.

The same number of particles as a solid, a liquid and a gas. In a solid, each particle vibrates about a fixed place. In a liquid, particles touch but move past each other. In a gas, they move freely through the whole container, so there are far more ways to arrange them, and far more ways to share their energy.

Predicting the sign of ΔS

For a physical or chemical change, $Δ S = S (final) - S (initial)$ . You can often predict its sign without any data, from four ideas, each with its reason.

Change of state. Solid → liquid → gas each increase $S$ , because the particles gain freedom to move and places to be. The biggest jump is liquid → gas.
Moles of gas. Gas particles carry most of the entropy. So if a reaction makes more moles of gas than it uses, $Δ S$ is positive. If it uses more than it makes, $Δ S$ is negative.
Dissolving. Usually positive, because the ions or molecules of a solid spread through the liquid. But not always: small, highly charged ions hold water molecules tightly around them, which lowers the water's entropy. So check data before you are sure.
Temperature. Heating any substance raises its entropy, because there is more energy to share among the particles.

Use the gas count first. It almost always decides the sign.

Sort — which way does entropy go?

Put each change in a box. Count the moles of gas first. Read the reason after each choice.

Eight changes, sorted by the sign of ΔS, with the value from data where it helps.

Entropy increases: water evaporating (+118.8 J/K/mol); iodine subliming (+144.6); CaCO₃(s) → CaO(s) + CO₂(g) (+160.2); NaCl dissolving in water (+43.4).
Entropy decreases: N₂(g) + 3H₂(g) → 2NH₃(g) (−198.1); 2SO₂(g) + O₂(g) → 2SO₃(g) (−188.0); steam condensing (−118.8).
Small, need data: H₂(g) + Cl₂(g) → 2HCl(g): 2 moles of gas on each side; data give +20.0 J/K/mol.

Calculating ΔS from data

Entropies are absolute values, so you can add them up directly:

Δ S ⦵ = Σ S ⦵ (products) - Σ S ⦵ (reactants)

Multiply each value by its number in the balanced equation. And remember that elements have real entropies. You cannot leave them out, as you can with enthalpies of formation.

Worked Example 1 — Burning hydrogen

Find ΔS^⦵ for H₂(g) + ½O₂(g) → H₂O(l).

Products: $70.0$ . Reactants: $130.7 + \frac{1}{2} \times 205.2 = 130.7 + 102.6 = 233.3$ .

ΔS^⦵ = 70.0 − 233.3 = −163.3 J/K/mol.

Check the sign: one and a half moles of gas become a liquid, so entropy must fall. ✓

Worked Example 2 — Heating limestone

Find ΔS^⦵ for CaCO₃(s) → CaO(s) + CO₂(g). Use S^⦵ values of CaCO₃(s) 91.7, CaO(s) 38.1 and CO₂(g) 213.8 J/K/mol.

ΔS^⦵ = (38.1 + 213.8) − 91.7 = +160.2 J/K/mol.

Check the sign: a gas is made from a solid. ✓ Notice that almost all of the products' entropy belongs to the CO₂.

The second law: count the surroundings too

Here is a puzzle. Water freezes in a freezer, and freezing lowers the water's entropy. So how can it happen by itself, if spreading out always wins?

The answer is that the water is not alone. When water freezes, it gives out heat to its surroundings. That heat gives the particles of the surroundings more energy to share, so their entropy rises. The rule of nature, the second law of thermodynamics, is about the total:

The second law of thermodynamics

A change happens by itself only if the total entropy of the system and its surroundings increases: $Δ S_{total} = Δ S_{system} + Δ S_{surroundings} > 0$ .

How much does the surroundings' entropy change? It depends on the heat it receives and on its temperature:

Δ S_{surroundings} = - \frac{Δ H}{T}

Read the formula with reasons. The minus sign is there because the surroundings gain the heat the system loses, as in calorimetry. $Δ H$ must be in J/mol, because entropy is in joules. And $T$ is in kelvin, underneath, for a reason you can feel: a gift of energy makes a big difference to a cold, quiet set of particles. The same gift makes much less difference to particles that already have plenty.

Worked Example 3 — Why ice melts above 0 °C and not below

For melting ice, $Δ H = + 6010$ J/mol and $Δ S_{system} = + 22.0$ J/K/mol. Assume these stay the same for a few degrees either side of 0 °C.

At 25 °C (298 K): $Δ S_{surr} = - 6010 \div 298 \approx - 20.2$ . Total: $22.0 - 20.2 = + 1.8$ J/K/mol. Positive, so ice melts.

At 0 °C (273.15 K): $Δ S_{surr} = - 6010 \div 273.15 \approx - 22.0$ . Total: about 0. Neither direction wins, so ice and water stay together.

At −10 °C (263 K): $Δ S_{surr} = - 6010 \div 263 \approx - 22.9$ . Total: $22.0 - 22.9 = - 0.9$ J/K/mol. Negative, so ice does not melt. The reverse change, freezing, has a total of +0.9, so water freezes.

Look at what just happened. The melting point was not put in. It came out of the arithmetic, as the temperature where the two entropy changes balance.

Stop and think

A student says: "An exothermic reaction gives out heat, so it always happens by itself." Is that right?

Show the answer

No. Giving out heat raises the entropy of the surroundings, which helps. But if the system's own entropy falls by more, the total still falls, and the change does not happen. Water freezing gives out heat at +10 °C too, but there the water's entropy falls by 22.0 while the surroundings gain only about 21.2, so the total is negative. Being exothermic helps. It does not decide on its own.

The promise kept: the cold pack

Now return to ammonium nitrate dissolving: NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq).

Worked Example 4 — Why an endothermic change happens

Data (J/K/mol): NH₄NO₃(s) 151.1, NH₄⁺(aq) 113.4, NO₃⁻(aq) 146.4. $Δ H = + 25.7$ kJ/mol. Take $T = 298$ K.

System: $Δ S = (113.4 + 146.4) - 151.1 = + 108.7$ J/K/mol. The ions spread out from the crystal through the water.

Surroundings: first convert, $Δ H = + 25700$ J/mol. Then $Δ S_{surr} = - 25700 \div 298 \approx - 86.2$ J/K/mol. They lose heat, so their entropy falls.

Total: $108.7 - 86.2 = + 22.5$ J/K/mol. Positive, so the salt dissolves by itself, even though it takes heat in.

Check the units: if you forgot to convert, you would get $- 25.7 \div 298 \approx - 0.086$ , a meaningless total of 108.6. Always put joules with joules.

So the cold pack works because the spreading of the ions creates more entropy than the cooling of the surroundings destroys. The next lecture combines these two entropy changes into one quantity, the Gibbs free energy, which makes this test quicker to use.

Three classic mistakes

Trap 1 — Mixing joules and kilojoules

Entropy is in J/K/mol, but $Δ H$ is usually given in kJ/mol. Before you combine them, convert $Δ H$ to J/mol (× 1000), or convert $Δ S$ to kJ/K/mol (÷ 1000). Mixing them gives an answer that is wrong by a factor of 1000.

Trap 2 — Leaving out the elements

Enthalpies of formation of elements are zero, but entropies are not. In H₂ + ½O₂ → H₂O, the reactants' entropies (130.7 and 102.6) are the larger part of the calculation. Leaving them out gives +70.0 instead of −163.3.

Trap 3 — "The system's entropy must increase"

The second law is about the total entropy of the system and the surroundings. The system's entropy can fall, as when water freezes, if the surroundings gain more. Always ask about both.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Predict the sign of $Δ S$ for each change, with a reason. (a) CO₂(s) → CO₂(g). (b) 2H₂(g) + O₂(g) → 2H₂O(l). (c) 2NO₂(g) → N₂O₄(g). (d) Sugar dissolving in water.

Hint 1
Count the moles of gas on each side first.

Hint 2
For (d), think about the molecules spreading from the crystal into the water.

Full answer
(a) Positive: solid to gas. (b) Negative: 3 moles of gas become a liquid. (c) Negative: 2 moles of gas become 1. (d) Positive: the sugar molecules spread from an ordered crystal through the solution. Check (b) with the gas count: 3 → 0, a large fall. ✓
Find ΔS^⦵ for 2SO₂(g) + O₂(g) → 2SO₃(g). Data (J/K/mol): SO₂(g) 248.2, O₂(g) 205.2, SO₃(g) 256.8.

Hint 1
Multiply each value by its number in the equation.

Hint 2
Products: $2 \times 256.8$ . Reactants: $2 \times 248.2 + 205.2$ .

Full answer
Products: $513.6$ . Reactants: $496.4 + 205.2 = 701.6$ . ΔS^⦵ = 513.6 − 701.6 = −188.0 J/K/mol. Check the sign: 3 moles of gas become 2, so entropy falls. ✓
For N₂(g) + 3H₂(g) → 2NH₃(g), ΔS^⦵ = −198.1 J/K/mol and ΔH^⦵ = −91.8 kJ/mol. (a) Find $Δ S_{total}$ at 298 K. (b) The reaction is "allowed" at 298 K, yet nitrogen and hydrogen can sit mixed for years without reacting. Explain.

Hint 1
Convert $Δ H$ to J/mol, then use $Δ S_{surr} = - Δ H / T$ .

Hint 2
Entropy says whether a change can happen by itself. Does it say how fast?

Full answer
(a) $Δ S_{surr} = + 91800 \div 298 \approx + 308.1$ J/K/mol. Total: $- 198.1 + 308.1 = + 110.0$ J/K/mol, positive. (b) A positive total means the reaction can happen by itself. It says nothing about speed. The strong N≡N triple bond (946 kJ/mol) gives the reaction a high activation energy, so at 298 K it is far too slow to notice. That is why industry uses a catalyst and a higher temperature. Check: $308.1 - 198.1 = 110.0$ . ✓
Steam condenses: H₂O(g) → H₂O(l). Use the table above for the entropies, and ΔH_f^⦵ values of −241.8 kJ/mol for H₂O(g) and −285.8 kJ/mol for H₂O(l). (a) Find $Δ S_{system}$ , $Δ H$ , $Δ S_{surr}$ at 298 K, and $Δ S_{total}$ . (b) Does steam condense at 25 °C? Which side "wins", and why?

Hint 1
$Δ S_{system} = 70.0 - 188.8$ . $Δ H = - 285.8 - (- 241.8)$ .

Hint 2
Convert $Δ H$ to J/mol before dividing by 298.

Full answer
(a) $Δ S_{system} = - 118.8$ J/K/mol. $Δ H = - 44.0$ kJ/mol $= - 44000$ J/mol. $Δ S_{surr} = + 44000 \div 298 \approx + 147.7$ J/K/mol. Total: $- 118.8 + 147.7 = + 28.9$ J/K/mol. (b) Yes. The water's own entropy falls a lot, but the heat it gives out raises the entropy of the cool surroundings by even more. The surroundings win at 25 °C. At a high enough temperature the surroundings' gain would shrink, and the gas would win, which is why water boils. Check: $147.7 - 118.8 = 28.9$ . ✓

The bigger idea

Entropy explains why time seems to have a direction. A film of perfume spreading looks normal. Run it backwards and it looks impossible, even though every single collision in it obeys the laws of motion just as well in reverse. The backward film is not forbidden. It is just so unlikely that it never happens.

Heat engines, refrigerators, living cells and stars all work inside this rule. A refrigerator makes its inside colder, lowering that entropy, only by putting out more heat at the back, raising the entropy of the kitchen by more. Your body builds ordered proteins only by releasing heat and simpler molecules. Life does not break the second law. It pays for its order by increasing the entropy around it.

Key concepts in this lesson

Entropy, S, measures the number of ways the particles of a system and their energy can be arranged, and its unit is J/K/mol.
Changes tend towards higher entropy because spread-out arrangements vastly outnumber gathered ones, not because any force pushes them.
Entropy rises from solid to liquid to gas, with more moles of gas, usually on dissolving, and on heating. The change in moles of gas is the quickest guide to the sign of ΔS.
ΔS⦵ = ΣS⦵(products) − ΣS⦵(reactants), with every substance included, because elements have real entropies.
The second law says a change happens by itself only if the total entropy increases, where ΔS(surroundings) = −ΔH/T, so an endothermic change can happen when the system gains enough entropy.