Predicting the sign of ΔS gets you one mark. Putting a number on it gets you three or four, and it is one subtraction with data straight from the booklet. The only thing standing between you and full marks is the balancing numbers, so we will nail those.
This equation is not in the data booklet. You have to know it.
The S° values are in the booklet, section 13, for substances at 298 K and 100 kPa.
Every value must be multiplied by its balancing number from the equation.
The answer is in J K−1 mol−1, and it can be positive or negative.
Unlike enthalpy, elements do not have S° = 0. Every substance has a real, positive entropy value.
Always sanity check the sign against the moles of gas before you move on.
The equation, and why it looks familiar
You already did this shape of sum for enthalpies of formation: add up what you end with, subtract what you started with. Entropy works the same way, because entropy is a property of the substances themselves.
Learn this one
ΔS°reaction = ΣS°(products) − ΣS°(reactants)
The Σ just means “add up all of them”. The little ° means standard conditions: 298 K and 100 kPa, with everything in its normal state at those conditions.
Products minus reactants. Get that order the wrong way round and every sign in your answer flips, which usually costs the final mark as well as the number.
🧩 The method, step by step
Write the balanced equation with state symbols. If the question gives it, copy it out anyway.
List the S° values for every substance. Booklet section 13.
Multiply each value by its balancing number from the equation.
Add up the products. Add up the reactants. Two totals, written down separately.
Subtract: products − reactants.
Check the sign makes sense by counting moles of gas. If they disagree, you have made an arithmetic slip.
Step 4 feels like a waste of time until the day you lose two marks to a mis-typed bracket. Write both totals down. Examiners give method marks for exactly that line, even if your final number is wrong.
Balancing numbers: the mark most people drop
A balancing number is not decoration. If the equation says 2SO3, you have two moles of SO3, so you have twice the entropy. The multiplication is not optional.
Write the multiplication out in full, like the two middle lines here. It takes five seconds and it is where the method marks live.
Values you will keep meeting
These are the ones that turn up again and again. You will always be given them or be able to look them up, but knowing roughly how big they are helps you spot a silly answer.
Substance
S° / J K−1 mol−1
Worth noticing
H2O(l)
70
Liquid water is surprisingly low
H2O(g)
189
Same substance, more than double
H2(g)
131
Small light molecule, so lower than most gases
O2(g)
205
A typical simple gas
CO2(g)
214
More atoms, more ways to vibrate
C(graphite)
6
A very rigid solid, so almost nothing to arrange
MgO(s)
27
Strong ionic lattice, tightly held
Look at carbon.S° for graphite is only 6 J K−1 mol−1. That is why burning carbon, C(s) + O2(g) → CO2(g), has ΔS° of just +3: one mole of gas becomes one mole of gas, and the solid barely contributes anything. Gas count tied, tiny answer — exactly as the quick check predicts.
Worked examples
WORKED EXAMPLE
Calculate ΔS° for MgCO3(s) → MgO(s) + CO2(g). Use S°: MgCO3(s) 66, MgO(s) 27, CO2(g) 214 J K−1 mol−1.
Write the equation you will use
ΔS° = ΣS°(products) − ΣS°(reactants)
Total the products27 + 214 = 241Total the reactants66Subtract241 − 66 = +175ΔS° = +175 J K⁻¹ mol⁻¹Check: no gas on the left, one mole on the right. Positive was the only sensible answer.
WORKED EXAMPLE
Calculate ΔS° for 2H2(g) + O2(g) → 2H2O(l). Use S°: H2(g) 131, O2(g) 205, H2O(l) 70.
Total the products, remembering the 22 × 70 = 140Total the reactants, remembering the 2(2 × 131) + 205 = 262 + 205 = 467Subtract140 − 467 = −327ΔS° = −327 J K⁻¹ mol⁻¹Three moles of gas turn into a liquid. A big negative number is no surprise at all.
WORKED EXAMPLE
Calculate ΔS° for N2(g) + 3H2(g) → 2NH3(g). Use S°: N2(g) 192, H2(g) 131, NH3(g) 193.
Products2 × 193 = 386Reactants — the 3 matters here192 + (3 × 131) = 192 + 393 = 585Subtract386 − 585 = −199ΔS° = −199 J K⁻¹ mol⁻¹Hang on to this number. You will use it again on the next page to work out whether the Haber process is spontaneous.
💡 Exam tip
Write “products − reactants” at the top of your working before you touch the numbers. It is one line and it stops the classic sign error.
Show the multiplications. “2 × 193 = 386” earns method marks even if you fat-finger the calculator later.
Keep the units in J here. Do not convert to kJ yet — that only happens when entropy meets enthalpy in the Gibbs equation.
Watch the state symbols in the data table. H2O(l) and H2O(g) are 119 apart, and questions pick whichever one you were not expecting.
Include the sign in your answer. “175” is not the same answer as “+175” when the question asks you to comment on it.
Round sensibly. The booklet values are whole numbers, so your answer should be too.
⚠ Common mix-up
Doing reactants minus products. The most common error in this whole topic, and it flips your sign.
Forgetting the balancing numbers. Especially the 3 in 3H2 and the 2 in 2NH3.
Setting elements to zero. That habit comes from ΔHf°. It is wrong for entropy: O2(g) is 205, not 0.
Converting to kJ too early. Standard entropies are in joules and stay in joules on this page.
Picking the wrong state from the table. Read the equation, then read the state symbol in the data list.
Never checking the sign. Ten seconds counting moles of gas catches most slips.
You can now put a number on the spreading out. But a number on its own does not tell you whether a reaction will actually go. For that you need to weigh entropy against enthalpy. Up next: Gibbs Free Energy.
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