Calculate enthalpy changes from Hess cycles
Use formation, combustion and bond-energy data in Hess cycles to calculate otherwise inaccessible reaction enthalpies.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Report signs, states, units and significant figures consistently. Standard conditions use 298 K and 101 kPa in this syllabus. Written practice is self-checked
Before you begin
- Rearrange a formula, balance a chemical equation and use temperature differences.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is ΔHf for an element in its standard state?
What you will learn
- Determine enthalpy changes using formation and combustion cycles
- Use bond-energy data as an alternative route
- Match the calculation to the target equation
Formation and combustion data define useful cycle routes
For an equation, ΔHr = ΣΔHf(products) − ΣΔHf(reactants), with each value multiplied by its balanced coefficient. Elements in their standard states have ΔHf = 0.
A combustion cycle can connect reactants and products through common combustion products such as CO₂ and H₂O. Add the reactants' combustion enthalpies and reverse the products' combustion path as needed; follow an energy cycle and check the resulting equation.
Bond-energy data can provide a Hess route
For gas-phase molecules, the sum of bond energies for bonds broken minus the sum for bonds formed estimates ΔHr. This is an energy route based on gaseous atoms and molecules; it may differ from experimental enthalpy because some supplied bond energies are averages.
Identify the requested reaction, multiply for every bond in the balanced equation, and check the final units and sign. Formation, combustion and bond-energy routes should all reduce to the same target equation if they use consistent data and states.
Pause and explain
Which formula uses standard enthalpies of formation?
Worked example
Find ΔH for C(s, graphite) + ½O₂(g) → CO(g) from ΔHc(C) = −393.5 and ΔHc(CO) = −283.0 kJ mol⁻¹.
Show the worked solution
- Carbon combustion gives C + O₂ → CO₂, ΔH=−393.5.
- Reverse CO combustion: CO₂ → CO + ½O₂, ΔH=+283.0.
- Add the routes and cancel CO₂ and half of O₂.
Answer ΔH = −110.5 kJ mol⁻¹.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Use ΔHf(C₂H₆)=−84.7, ΔHf(C₂H₄)=+52.3 and ΔHf(H₂)=0 kJ mol⁻¹ to find ΔH for C₂H₄ + H₂ → C₂H₆.
Give me a hint
Use products minus reactants, with the equation coefficients.
Compare my reasoning
- Products: −84.7 kJ mol⁻¹.
- Reactants: +52.3 + 0 = +52.3 kJ mol⁻¹.
- Subtract reactants from products.
ΔH = −84.7 − 52.3 = −137.0 kJ mol⁻¹.
Look for these in your work
- Formation values matched to equation
- Element reference value used
- Products-minus-reactants calculation correct
Use ΔHc(C)=−393.5 and ΔHc(CO)=−283.0 kJ mol⁻¹ to find C + ½O₂ → CO.
Give me a hint
Connect both substances to CO₂ and reverse the CO combustion step.
Compare my reasoning
- Combust carbon to CO₂: −393.5.
- Reverse CO combustion: +283.0.
- Add the enthalpies.
ΔH = −110.5 kJ mol⁻¹.
Look for these in your work
- Common product selected
- Reversed step sign correct
- Unknown enthalpy calculated
Estimate ΔH for H₂(g)+Cl₂(g)→2HCl(g) using H–H 436, Cl–Cl 242 and H–Cl 431 kJ mol⁻¹. State one limitation.
Give me a hint
Use bonds broken minus bonds formed and label average-data estimates appropriately.
Compare my reasoning
- Broken = 436 + 242 = 678 kJ mol⁻¹.
- Formed = 2×431 = 862 kJ mol⁻¹.
- ΔH ≈ 678 − 862; note that average bond values can make it an estimate.
ΔH ≈ −184 kJ mol⁻¹ for the equation as written. Bond-energy results are estimates when average values are used.
Look for these in your work
- Bond counts correct
- Signs and subtraction correct
- Estimate limitation stated
Common mistakes
- Forgetting coefficients in a formation-enthalpy sum
- Assigning non-zero formation enthalpy to an element in its standard state
- Reversing a combustion step without reversing its sign
- Treating bond-energy estimates as exact experimental data
What can you explain now?
What is ΔHf for an element in its standard state?
Compare with the explanation
Zero
Formation from the element in its own standard state is the reference.
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Make it stick
How do formation, combustion and bond-energy data connect to the same target reaction enthalpy?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.