IB Chemistry HLTopic 4 — Energy CyclesPaper 1 & 2Core skill~11 min read
Enthalpy Changes from Combustion Data
Combustion is the easiest enthalpy change in the world to measure — set fire to something and watch the thermometer. That is why so much combustion data exists, and why chemists lean on it to work out enthalpy changes they could never measure directly. The cycle looks like the formation one turned upside down, and the formula flips with it.
📘 What you need to know
Standard enthalpy of combustion, ΔHc: the enthalpy change when one mole of a substance burns completely in oxygen under standard conditions.
In this cycle the combustion products go at the bottom and every arrow points down.
The result: ΔHreaction = ΣΔHc(reactants) − ΣΔHc(products).
Notice the order is the opposite way round to the formation version — because the arrows point the other way.
Anything that cannot burn (water, carbon dioxide, oxygen) has ΔHc = 0.
Combustion values are always negative, so expect a lot of double negatives.
What counts as “complete” combustion
Standard enthalpy of combustion
The enthalpy change when one mole of a substance burns completely in excess oxygen under standard conditions
“Completely” is doing real work in that sentence. For an organic compound it means:
every carbon atom ends up as CO2(g), never CO or soot
every hydrogen atom ends up as H2O(l) under standard conditions
the oxygen is in excess, so nothing is left half-burnt
And as with formation, it is one mole of the substance being burnt. That again forces fractional coefficients on the oxygen — ethanol burns as C2H5OH + 3O2, but methanol needs 1½O2.
Why so much combustion data exists. Combustion enthalpies can be measured to real precision with a bomb calorimeter, because the reaction is fast, complete and gives out a lot of heat. Formation enthalpies, by contrast, usually have to be calculated from combustion data using exactly the cycle on this page.
The cycle: combustion products at the bottom
Burning always destroys a compound and pushes it down to CO2 and H2O. So every arrow in a combustion cycle points down, into the combustion products. That single geometric fact is what flips the formula around.
Compare this with the formation cycle. Same triangle, arrows reversed, and the two terms in the formula swap places. Nothing has been memorised — it is read straight off the picture.
The combustion equation
ΔHreaction = ΣΔHc(reactants) − ΣΔHc(products)
Two formulas, one difference: formation is products first, combustion is reactants first. If you can only hold one in your head, hold the arrows instead. Formation arrows go up out of the elements, combustion arrows go down into the ashes — and the term you subtract is always the one you have to travel backwards.
Things that cannot burn
Students often panic when a reactant has no combustion value in the table. There is usually nothing missing. Water, carbon dioxide and oxygen are already fully oxidised or non-combustible, so their enthalpy of combustion is zero. They are already at the bottom of the cycle.
Substance
ΔHc
Why
C(graphite)
−394 kJ mol−1
Burns to CO2, and this is also ΔHf[CO2]
H2(g)
−286 kJ mol−1
Burns to H2O(l), and this is also ΔHf[H2O(l)]
H2O(l)
0
Cannot burn — it is already the product of burning hydrogen
CO2(g)
0
Cannot burn — carbon is already fully oxidised
O2(g)
0
Oxygen is what things burn in, not what burns
A useful coincidence worth spotting. The combustion enthalpy of carbon is the same number as the formation enthalpy of CO2, because both describe C(s) + O2 → CO2(g). The same is true of hydrogen and water. That is why the two cycles so often produce identical answers.
Working out a formation enthalpy from combustion data
The left-hand total is 2 × (−394) for the carbon plus 3 × (−286) for the hydrogen. Both sides burn to exactly the same CO2 and H2O, which is what makes the comparison legal.
WORKED EXAMPLE
Enthalpy of formation of ethane
Calculate ΔHf for 2C(s) + 3H2(g) → C2H6(g).
Substance
C(s)
H2(g)
C2H6(g)
ΔHc / kJ mol−1
−394
−286
−1560
Step 1: total the reactants, with coefficients2 × (−394) = −7883 × (−286) = −858ΣΔHc(reactants) = −1646 kJStep 2: total the productsΣΔHc(products) = −1560 kJStep 3: reactants minus productsΔH = (−1646) − (−1560)ΔH = −1646 + 1560ΔHf[C₂H₆] = −86 kJ mol−¹a small negative number, which is typical for a simple alkane
WORKED EXAMPLE
A reactant that does not burn
Ethene is converted to ethanol industrially by adding water. Calculate ΔH for C2H4(g) + H2O(l) → C2H5OH(l), given ΔHc = −1411 for ethene and −1367 kJ mol−1 for ethanol.
Step 1: spot that water contributes nothingH₂O(l) cannot burn, so its ΔHc is zero. It is not missing from the question.Step 2: total the reactants(−1411) + 0 = −1411 kJStep 3: total the products−1367 kJStep 4: reactants minus productsΔH = (−1411) − (−1367)ΔH = −44 kJ mol−¹mildly exothermic — which is exactly why industry runs this reaction hot to shift the equilibrium the other way
WORKED EXAMPLE
When the answer comes out positive
Calculate ΔHf for benzene, 6C(s) + 3H2(g) → C6H6(l), given ΔHc = −394 for C(s), −286 for H2(g) and −3268 kJ mol−1 for benzene.
Step 1: reactants6 × (−394) = −23643 × (−286) = −858total = −3222 kJStep 2: products−3268 kJStep 3: reactants minus productsΔH = (−3222) − (−3268)ΔHf[C₆H₆] = +46 kJ mol−¹positive! benzene sits higher in energy than its own elements
A positive ΔHf does not mean benzene is unstable in the everyday sense — it sits in a bottle quite happily. It means the formation from graphite and hydrogen is uphill. Thermodynamic stability and kinetic stability are different things, and examiners like to test whether you know that.
Comparing the two cycles side by side
Feature
Formation data
Combustion data
What sits at the bottom
The elements
CO2 and H2O
Arrow direction
Upwards, out of the elements
Downwards, into the burnt products
Which arrow you reverse
The reactants arrow
The products arrow
Formula
Σ(products) − Σ(reactants)
Σ(reactants) − Σ(products)
What has a value of zero
Elements in standard states
Water, CO2 and O2
💡 Exam tip
Decide which data type you have before you draw anything. Formation data means elements at the bottom; combustion data means CO2 and water at the bottom.
Balance the cycle by atoms. If the left of your bottom row has 6 carbons and the right has 2, something has gone wrong upstream.
Write the zeros in explicitly for water and oxygen. It shows the examiner you knew rather than forgot.
Every combustion value is negative, so nearly every step involves subtracting a negative. Use brackets religiously.
You may leave the oxygen off the diagram for clarity, as long as your calculation still accounts for it. Say so if you do.
If the question hands you ΔHc values and asks for ΔHf, it is this page. If it hands you ΔHf and asks for ΔHc, it is the previous page. Both are one subtraction.
⚠ Common mix-up
Using the formation formula with combustion data. Products minus reactants here gives you the right number with the wrong sign, every time.
Thinking water’s missing combustion value is a printing error. It is zero because water cannot burn.
Forgetting the coefficients. Six moles of carbon is 6 × (−394), a difference of nearly 2000 kJ if you miss it.
Producing CO instead of CO2 when writing a combustion equation. Complete combustion means fully oxidised.
Writing H2O(g) in a standard combustion equation. Under standard conditions the water is liquid.
Assuming a positive ΔHf means the compound cannot exist. Benzene, ethyne and nitrogen dioxide all have positive formation enthalpies and are perfectly real.
Up next: Born-Haber Cycles — the same Hess logic applied to ionic solids, where the quantity you are chasing cannot be measured at all.
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