Absolute Entropy and Entropy Change: Study Guide (Chemistry 9.2)

From predictions to numbers

In Topic 9.1 you predicted the sign of ΔS from what the particles were doing: changing phase, gaining space, gaining energy, or changing the number of moles of gas. That reasoning tells you the direction of the change, but not its size. It also fails when the moles of gas are equal on both sides.

This topic gives you the numbers. Chemists have measured the entropy of thousands of substances and listed them in tables. From those tables you can calculate ΔS for any reaction or physical change whose substances are listed.

Absolute entropy and standard molar entropy

Enthalpy has no natural zero, so in Topic 6.8 you measured it relative to the elements. Entropy is different: it has a true starting point.

• Background (the third law of thermodynamics): a perfect crystal at absolute zero, 0 K, has zero entropy. Its particles are perfectly ordered and all in their lowest energy state, so there is only one way to arrange them and their energy.
• As a substance is warmed from 0 K, its energy becomes more dispersed, and it melts and boils on the way. Each step adds entropy.
• So every substance at 298 K has a positive absolute entropy. Because it is measured from a true zero, it is called absolute.

The standard molar entropy, S°, is the absolute entropy of one mole of a substance at standard pressure (1 bar). Tables usually list values at 298 K. The unit is J K⁻¹ mol⁻¹ (joules per kelvin per mole). Notice: joules, not kilojoules.

Two points that trip students up:

1. The symbol is S°, not ΔS°. It is an amount the substance has, not a change.
2. Elements are not zero. O₂(g), H₂(g) and C(graphite) all have positive S° values. Do not carry the “elements are zero” rule over from enthalpies of formation.

Patterns in the table

What the chart shows, and how it links to Topic 9.1:

• Gases have much larger S° values than liquids and solids. Compare H₂O(l), 70.0, with H₂O(g), 188.8. This is why moles of gas dominate ΔS for reactions.
• For one substance, S°(gas) > S°(liquid) > S°(solid). Matter is more dispersed in each later phase.
• Background: among substances in the same phase, molecules with more atoms usually have larger S°. Propane, C₃H₈, has more ways to store and spread out energy than H₂: it has more atoms and bonds, so more ways to vibrate, bend and rotate.
• Background: hard, strongly bonded solids such as graphite have very low S°.

Calculating ΔS° for a process

The entropy change for a process is the total entropy after it minus the total entropy before it:

ΔS°reaction = ΣnS°(products) − ΣnS°(reactants)

where n is the coefficient of each species in the balanced equation. This is the same “products minus reactants” pattern you used with enthalpies of formation in Topic 6.8.

Here are the values used on this page. All are standard molar entropies at 298 K, rounded to one decimal place.

Substance	S° (J K⁻¹ mol⁻¹)	Substance	S° (J K⁻¹ mol⁻¹)
C(graphite)	5.7	H₂(g)	130.7
CO(g)	197.7	O₂(g)	205.2
CO₂(g)	213.8	H₂O(l)	70.0
C₃H₈(g)	270.3	H₂O(g)	188.8
CH₃OH(l)	126.8	CH₃OH(g)	239.9

A reliable routine:

1. Write the balanced equation with state symbols. H₂O(l) and H₂O(g) have different S° values; using the wrong one is a common error.
2. Predict the sign first from the moles of gas (Topic 9.1). This gives you a check.
3. Multiply each S° by its coefficient and add each side separately.
4. Subtract: products minus reactants.
5. Write the unit: J K⁻¹ mol⁻¹, meaning per mole of reaction as written.
6. Compare with your prediction. If the signs disagree, look for a missed coefficient or a wrong state.

What “per mole of reaction” means. The value belongs to the equation exactly as written. Double every coefficient and ΔS° doubles. Reverse the equation and ΔS° changes sign. For an actual sample, find how many moles of reaction occur and multiply.

Worked example 1: combustion of propane

Question. Calculate ΔS° for the combustion of propane at 298 K:

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l)

Then (b) give the answer in kJ K⁻¹ mol⁻¹ and (c) find the entropy change when 11.0 g of propane burns completely.

1. Predict. Moles of gas: 1 + 5 = 6 on the left; 3 on the right (the water is liquid). Fewer moles of gas, so ΔS° should be negative and fairly large.
2. Products. 3(213.8) + 4(70.0) = 641.4 + 280.0 = 921.4 J K⁻¹ mol⁻¹.
3. Reactants. 270.3 + 5(205.2) = 270.3 + 1026.0 = 1296.3 J K⁻¹ mol⁻¹. O₂ is an element, but its S° is not zero, so it must be included.
4. Subtract. ΔS° = 921.4 − 1296.3 = −374.9 J K⁻¹ mol⁻¹.
5. Check. Negative, as predicted.
6. (b) Divide by 1000: −374.9 J K⁻¹ mol⁻¹ = −0.3749 kJ K⁻¹ mol⁻¹. You will need this form in Topic 9.3, where ΔS° is combined with ΔH° in kilojoules.
7. (c) M(C₃H₈) = 3(12.01) + 8(1.008) = 44.094 g mol⁻¹, so n = 11.0 g ÷ 44.094 g mol⁻¹ = 0.24947 mol. The equation has 1 mol of propane, so this is 0.24947 mol of reaction. ΔS = 0.24947 mol × (−374.9 J K⁻¹ mol⁻¹) = −93.53 J K⁻¹.

Answer. (a) −374.9 J K⁻¹ mol⁻¹; (b) −0.3749 kJ K⁻¹ mol⁻¹; (c) −93.5 J K⁻¹ for the 11.0 g sample (3 significant figures). Note that the unit in (c) has no “mol⁻¹”: it is the change for that particular sample.

Why the state matters. If you used H₂O(g) (188.8) instead of H₂O(l), you would get +100.3 J K⁻¹ mol⁻¹, a different sign. With water as a gas there are 7 mol of gas products against 6 mol of gas reactants, so a positive value would then be correct. Read the state symbols.

Worked example 2: a physical process

Question. Calculate the standard entropy change for the vaporisation of methanol at 298 K, and explain the sign and size.

CH₃OH(l) → CH₃OH(g)

1. Predict. Liquid to gas: matter becomes far more dispersed, so ΔS° should be large and positive.
2. Calculate. ΔS° = S°(CH₃OH, g) − S°(CH₃OH, l) = 239.9 − 126.8 = +113.1 J K⁻¹ mol⁻¹.
3. Explain. In the liquid the molecules are held close together by hydrogen bonds and other intermolecular forces. In the gas they are far apart and move freely through a much larger volume, so the matter is much more dispersed.

Compare. The same calculation for water gives 188.8 − 70.0 = +118.8 J K⁻¹ mol⁻¹. Both vaporisations are large and positive, which matches the Topic 9.1 rule that liquid → gas is a big entropy step. The equation ΔS° = ΣnS°(products) − ΣnS°(reactants) works for physical changes as well as chemical reactions.

Worked example 3: working backwards to an unknown S°

Question. For the reaction

CO(g) + 2 H₂(g) → CH₃OH(l)

ΔS° = −332.3 J K⁻¹ mol⁻¹. Use this and the values for H₂(g) and CH₃OH(l) to find S° for CO(g).

1. Write the expression. ΔS° = S°(CH₃OH, l) − [S°(CO) + 2 S°(H₂)].
2. Substitute. −332.3 = 126.8 − [S°(CO) + 2(130.7)] = 126.8 − S°(CO) − 261.4.
3. Rearrange. S°(CO) = 126.8 − 261.4 + 332.3 = 197.7 J K⁻¹ mol⁻¹.
4. Check. It matches the table value. It is also sensible: a gas, with a value in the same range as other small gas molecules (130 to 240 J K⁻¹ mol⁻¹ in Figure 1).

Watch the sign. The given ΔS° is negative. Substituting it as +332.3 would give 126.8 − 261.4 − 332.3, a negative S°, which is impossible at 298 K. A negative absolute entropy always means a sign slip.

Common misconceptions

• “Elements have S° = 0.” That rule belongs to ΔH°f. Every substance, element or compound, has a positive S° at 298 K. Leaving out O₂ is the single most common error in combustion questions.
• Forgetting coefficients. In Worked example 1, 5 O₂ contributes 5 × 205.2 = 1026.0, not 205.2.
• Reactants minus products. Always products minus reactants. Your sign check from moles of gas will catch this.
• Using the wrong state. H₂O(l) and H₂O(g) differ by almost 119 J K⁻¹ mol⁻¹. One wrong state can flip the sign.
• Mixing J and kJ. S° is in J K⁻¹ mol⁻¹; ΔH° is in kJ mol⁻¹. Convert ΔS° to kJ (divide by 1000) before combining the two in Topic 9.3.
• Writing S° when you mean ΔS°. S° is what one mole of a substance has; ΔS° is the change for a process.
• Treating “per mole” as per mole of one substance. ΔS° is per mole of reaction as written. Scale by moles of reaction for a real sample.

Where this leads

Next, Topic 9.3: Gibbs Free Energy and Thermodynamic Favorability combines ΔS° with ΔH° to decide whether a process is thermodynamically favourable, which is where the J-to-kJ conversion matters. Try the practice questions now, then use the revision notes and the checklist to consolidate.