IB Chemistry SLTopic 4 — Energy CyclesPaper 1 & 2Core skill~14 min read
Applying Hess’s Law
Hess’s law is one idea, but in the exam it arrives in three costumes: formation data, combustion data, and a pile of equations to rearrange. Learn to recognise which one you have been handed and the rest is arithmetic.
📚 What you need to know
Given formation data: ΔH = ΣΔHf(products) – ΣΔHf(reactants).
Given combustion data: ΔH = ΣΔHc(reactants) – ΣΔHc(products).
The two are the opposite way round because formation arrows point up out of the elements and combustion arrows point down into CO2 and H2O.
ΔHf of any element in its standard state is zero.
Every value must be multiplied by the coefficient of that substance in the equation.
Rearranging equations: reverse → change the sign, multiply → multiply ΔH, then add the equations and their ΔH values.
Two definitions you have to be exact about
Standard enthalpy of formation, ΔHf
the enthalpy change when one mole of a compound forms from its elements in their standard states
Standard enthalpy of combustion, ΔHc
the enthalpy change when one mole of a substance burns completely in excess oxygen
One consequence falls straight out of the first definition: forming an element from itself is not a change at all, so ΔHf of an element in its standard state is zero. O2(g), Fe(s) and C(graphite) all contribute nothing. That is not a special exception to memorise; it is just what the definition says.
Both definitions are per one mole, which is why half-integer coefficients turn up so often. ΔHc for ethanol is written for 1 mol of ethanol burning, however awkward that makes the oxygen.
Method 1 — formation data
Formation arrows always point away from the elements, so the route to the products goes down one arrow the wrong way and up the other.
Put the elements at the bottom, because every substance in the equation can be built from them. Now trace the route: from the reactants you go down against their formation arrow, then up with the products’ formation arrow. Against then with means subtract then add:
Using enthalpies of formation
ΔH = ΣΔHf(products) – ΣΔHf(reactants)
WORKED EXAMPLE
Calculate ΔH for the blast furnace reaction Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g), given ΔHf: Fe2O3(s) = –824, CO(g) = –111, CO2(g) = –394 kJ mol–1.
Step 1 — productsFe is an element, so its ΔH f is zero.2(0) + 3(−394) = −1182Step 2 — reactants(−824) + 3(−111) = −824 − 333 = −1157Step 3 — products minus reactants−1182 − (−1157) = −1182 + 1157ΔH = −25 kJ mol⁻¹Only just exothermic, which is exactly why a blast furnace has to be kept ferociously hot by other means.
Method 2 — combustion data
Combustion arrows always point towards carbon dioxide and water, so this cycle is the formation one turned upside down.
Now the common destination is the burnt products, and both arrows point down into them. Trace the route again: down with the reactants’ combustion arrow, then up against the products’ one. With then against means add then subtract — the mirror image of Method 1:
Using enthalpies of combustion
ΔH = ΣΔHc(reactants) – ΣΔHc(products)
Do not memorise the two formulas as unrelated facts — you will mix them up under pressure. Memorise which way the arrows point, draw the cycle, and read the formula off it in ten seconds.
WORKED EXAMPLE
Calculate the enthalpy of formation of methane, C(s) + 2H2(g) → CH4(g), given ΔHc: C(s) = –394, H2(g) = –286, CH4(g) = –891 kJ mol–1.
Step 1 — reactants, burnt(−394) + 2(−286) = −394 − 572 = −966Step 2 — products, burnt1 × (−891) = −891Step 3 — reactants minus products−966 − (−891) = −966 + 891ΔH f = −75 kJ mol⁻¹This is a reaction nobody can perform — carbon and hydrogen simply do not combine. Three combustions you CAN perform deliver the answer anyway.
Notice the pattern: combustion data is how enthalpies of formation get measured for organic compounds, and formation data is then used to predict enthalpies of reaction. The two cycles feed each other.
Method 3 — rearranging equations
Sometimes you are simply given a list of equations and asked to combine them. No cycle is needed: adjust each equation until adding them up produces the target, then add the enthalpy changes the same way.
What you do to the equation
What happens to its ΔH
Why
Reverse it
Change the sign
Energy released going one way is absorbed coming back
Multiply by 2
Multiply ΔH by 2
Twice the substance, twice the energy
Halve it
Halve ΔH
Same reasoning, other direction
Add two equations
Add their ΔH values
Two consecutive steps of one route
🧩 The method
Write the target equation and keep it in front of you.
Take each given equation and ask: does this substance need to be on the other side? If so, reverse it and flip the sign.
Does it need more moles? Multiply the equation and the ΔH by the same factor.
Add the adjusted equations together and cancel anything appearing on both sides.
Check the result is exactly the target, then add the adjusted ΔH values.
Step 1 — the target needs 2C on the leftEquation (1) has 1 C on the left. Double it.2C + 2O₂ → 2CO₂ 2 × (−394) = −788Step 2 — the target needs 2H₂ on the left2H₂ + O₂ → 2H₂O 2 × (−286) = −572Step 3 — the target needs C₂H₄ on the RIGHTIn (3) it is on the left, so reverse the equation and flip the sign.2CO₂ + 2H₂O → C₂H₄ + 3O₂ +1411Step 4 — add and cancel2CO₂, 2H₂O and 3O₂ all appear on both sides and disappear, leaving 2C + 2H₂ → C₂H₄.−788 + (−572) + 1411 = +51ΔH = +51 kJ mol⁻¹Positive, and it should be: ethene is an endothermic compound, less stable than the elements it is made from.
Which method?
What the question gives you
Use
The giveaway
A table of ΔHf values
Method 1
Elements listed as zero, or compounds only
A table of ΔHc values
Method 2
Every substance listed is something that burns
Two or three full equations with ΔH
Method 3
No common substance to build a tidy cycle around
A table of average bond enthalpies
Bond enthalpies
Values are per bond, not per substance
💡 Exam tip
Multiply by the coefficients. 3CO2 means three times that formation enthalpy, and it is the most common single slip in the topic.
Elements have ΔHf = 0. Write the zero into your working so the examiner can see you knew.
Brackets around every value. –1182 – (–1157) is a different thing from –1182 – 1157.
If you cannot remember which formula is which, draw the cycle and read it off. Formation points up, combustion points down.
Check the sign is plausible. Combustion answers should be strongly negative; if yours is positive, something has been subtracted the wrong way.
⚠️ Common mix-up
Swapping the two formulas. Formation is products – reactants; combustion is reactants – products.
Including O2 in a formation calculation as though it had a value. It is an element; it contributes zero.
Reversing an equation and forgetting to reverse the sign.
Scaling ΔH by the wrong factor after multiplying an equation through.
Forgetting that ΔHc and ΔHf are per mole of one particular substance, so the target equation may need scaling too.
That completes Energy Cycles. You can now find an enthalpy change three ways without ever running the reaction: from the bonds inside the molecules, from a cycle built out of formation or combustion data, or by rearranging equations you were handed. Up next: Energy from Fuels, where these numbers stop being exam questions and start deciding what we burn.
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