Hess cycles from combustion data
Key idea: A combustion cycle connects both sides to the same combustion products. Subtract the products' total combustion enthalpy from the reactants' total, matching coefficients and states.
Offline pages
Saved on this device only. Connect at least once every 30 days to confirm Premium. Sign-out or a change of account removes saved copies.
Loading guide controls…
Guided lessons · coming soon
Guided lesson
Guided steps
- Choose what must match at the combustion routes’ endpoint.
- Follow both combustion routes and derive the subtraction order.
- Calculate both combustion totals and the reverse reaction enthalpy.
- Find ethanol’s formation enthalpy using the combustion data.
- Draw the ethanol cycle and explain the oxygen accounting.
- Find methane’s formation enthalpy using the complete reaction equations.
- Draw the methane cycle and explain the hydrogen coefficient.
On this page
You stopped at:
ContinueCombustion releases heat that can be measured by calorimetry. A useful combustion enthalpy needs complete combustion to known products in stated physical states. A Hess cycle turns them into enthalpy changes that cannot be measured, such as the formation of a compound from its elements. In a formation cycle the routes start from the same elements; in a combustion cycle they finish at the same combustion products.
Check your understanding 1 Choose the matching endpoint
For C₂H₄(g) + H₂(g) → C₂H₆(g), what must be identical after completely burning either side in oxygen?
Show answer
Both sides contain two carbon atoms and six hydrogen atoms, so both routes must end at 2CO₂(g) + 3H₂O(l). A route to water vapour would need an additional state-change step.
Combustion cycles
Take the hydrogenation of ethene:
The combustion enthalpies are ethene -1411, hydrogen -286 and ethane -1560 kJ mol⁻¹. Burning either side completely gives the same products, 2CO₂(g) + 3H₂O(l). Put the target equation across the top and the combustion products underneath, then draw an arrow down from each side. Each side needs 7/2 O₂ to burn, so the oxygen cancels.
From the reactants there are two routes to the combustion products: burn them directly, or form ethane and then burn it. The two must be equal:
Rearranged, the route equation gives a general rule for combustion data:
This is the reverse of the order for formation data, because the arrows now point to a shared destination instead of from a shared origin. Using “products minus reactants” here would give + 137 kJ mol⁻¹, with the wrong sign. If you are unsure which order to use, write the route equation.
Check your understanding 2 Use the model for the reverse reaction
Using the ethene, hydrogen and ethane combustion values above, calculate Δ H for C₂H₆(g) → C₂H₄(g) + H₂(g). Enter the combustion total for each side, then the signed reaction enthalpy.
Show the steps Optional working
Show answer
-1560 = Δ H + [-1411 + (-286)], so Δ H = +137 kJ mol⁻¹. Reversing the target reaction reverses its enthalpy sign, while all the combustion arrows still point towards the common destination.
Try it yourself 1
Find an enthalpy that was not measured
Problem
Calculate the standard enthalpy of formation of liquid ethanol for:
Use these rounded standard combustion enthalpies at 298 K. All combustion routes produce CO₂(g) and H₂O(l) where applicable.
| Substance | Δ Hc⦵/kJ mol⁻¹ |
|---|---|
| C(graphite) | -394 |
| H₂(g) | -286 |
| C₂H₅OH(l) | -1368 |
Construct a combustion cycle or write equivalent route equations. Explain why no separate combustion enthalpy is needed for the oxygen in the target equation.
Try this before viewing the solution
Show the steps Optional working
Hints
Hint 1
match the final products
How many moles of carbon dioxide and water form from each side?Hint 2
balance oxygen on both routes
Add the same amount of oxygen to both sides of the target equation, so each route has enough oxygen for complete combustion.Show solution step by step
Match the combustion endpoints
Method
Both routes end at 2CO₂(g) + 3H₂O(l).Reason
Two carbon atoms and six hydrogen atoms are present on each side of the target equation; the final products must also match in physical state.Working
Answer
Add 3O₂(g) to both upper states. The left state then contains 7/2 moles of oxygen with the elements; the right contains one mole of ethanol and three moles of oxygen.Account for amounts burned
Method
Burn two moles of carbon and three moles of hydrogen on the direct route.Reason
Each combustion value is per mole of the named substance, so the coefficients multiply the enthalpy changes.Working
Answer
The total is 2(-394) + 3(-286), which is -1646 kJ for the stated amounts.Equate routes and rearrange
Method
Find the enthalpy change for forming one mole of ethanol.Reason
The direct route equals ethanol formation followed by combustion of that ethanol.Working
Answer
-1646 = Δ Hf⦵ + (-1368), so Δ Hf⦵ = -1646-(-1368). This gives -278 kJ mol⁻¹.Explain the oxygen accounting
Method
No separate oxygen combustion term is added.Reason
Oxygen is the oxidant, not another substance being burned. Adding the same oxygen amount to both upper states leaves the target enthalpy change unchanged.Working
Answer
The direct combustion route uses 7/2 moles of oxygen. Forming ethanol first uses 1/2 mole, then burning it uses three moles: the total is again 7/2.
Quick check
Check your understanding 3 Construct and explain the ethanol cycle
Construct the ethanol cycle on paper. Record the common destination, route equation and why oxygen has no separate combustion term.
Show answer
Check your destination is 2CO₂(g) + 3H₂O(l). Add 3O₂(g) to both upper states; the elemental side then contains 7/2O₂(g). The direct route is 2(-394) + 3(-286) = -1646. The other route is Δ Hf-1368. Equate them to get Δ Hf = -278 kJ mol⁻¹. Oxygen is the oxidant: forming ethanol uses half a mole and burning it uses three, giving the same total oxygen as the direct route.
Check your understanding 4 The formation enthalpy of methane
Construct a combustion cycle and use the data below to find Δ H for: C(graphite) + 2H₂(g) → CH₄(g)
Data:
-
C(graphite) + O₂(g) → CO₂(g) Δ H = -394 kJ mol⁻¹
-
2H₂(g) + O₂(g) → 2H₂O(l) Δ H = -572 kJ mol⁻¹
-
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) Δ H = -890 kJ mol⁻¹
Show the steps Optional working
Show answer
The combined elemental combustion route and the methane combustion route end at CO₂(g) + 2H₂O(l), so that is the common destination.
Burning the elements: add the first two equations, -394 + (-572) = -966 kJ mol⁻¹.
Forming methane and then burning it: Δ H + (-890).
The two routes are equal: -966 = Δ H + (-890), so Δ H = -76 kJ mol⁻¹.
Check your understanding 5 Compare complete equations with per-mole data
Sketch the methane cycle on paper and record its route equation. Explain why the given -572 kJ mol⁻¹ hydrogen value is used once.
Show answer
The common destination is CO₂(g) + 2H₂O(l). The given hydrogen equation already burns two moles of hydrogen, so -394-572 = Δ H-890. Doubling -572 would burn four moles of hydrogen and give the wrong endpoint. The correct result is -76 kJ mol⁻¹.
Common mistakes
- Using “products minus reactants” with combustion data.
- Forgetting to multiply a combustion enthalpy by the number of moles burned.
- Including a combustion enthalpy for oxygen. Oxygen does not burn.
- Using routes that end at different states, such as H₂O(g) on one side and H₂O(l) on the other.
Spotted an error? Report a correction
Syllabus and review details
- GCE A-Level H2 Chemistry 2027 · 9476-2027
9476 (2027), complete syllabus
Last reviewed: