Bond breaking, bond making and enthalpy estimates
Count every bond in the balanced reaction, then subtract energy released from energy required.
Section 5.1 · Exothermic and endothermic reactions
Extended only (Supplement) · Chemistry 0620, 2026–2028.
Before you begin
- Balance equations and count bonds in simple molecular structures.
- Use signed enthalpy changes and multiply a supplied bond energy by bond count.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How many O–H bonds are formed in two H₂O molecules?
What you will learn
- Explain why breaking bonds takes energy and making bonds releases it.
- Estimate enthalpy change from supplied bond energies and balanced coefficients.
Separate the two contributions
Breaking bonds requires an energy input and is endothermic. Making bonds releases energy and is exothermic. A reaction may break some reactant bonds and form new product bonds; whether it is exothermic overall depends on which total is larger. It is incorrect to say fuel releases energy simply because its bonds are broken: breaking them is the energy-requiring contribution.
Estimate ΔH by adding the supplied energies of all bonds broken and subtracting the total supplied energies of all bonds formed. If bond formation releases more energy than breaking requires, the difference is negative. If breaking requires more than formation releases, it is positive. Use the balanced equation's coefficients to count molecules and bonds before doing the arithmetic.
Count structures, not just formula letters
For H₂+Cl₂→2HCl, break one H–H and one Cl–Cl bond and form two H–Cl bonds. For 2H₂+O₂→2H₂O, break two H–H bonds and one O=O bond, then form four O–H bonds. A double bond's supplied energy is the energy for that double bond, not twice a separate single-bond value.
These calculations use average bond-energy data and give estimates, typically for gaseous molecular species. Use the supplied table consistently; phase changes or different data can make a measured value differ. State the answer per balanced reaction quantity used in the calculation. Doubling the whole equation doubles its enthalpy change, whereas changing only a molecular subscript changes the substances and is not a valid way to rescale it.
Pause and explain
Why can a reaction be exothermic even though bonds must be broken?
Worked example
Estimate ΔH for H₂+Cl₂→2HCl using supplied bond energies H–H=436, Cl–Cl=242 and H–Cl=431 kJ/mol of bonds.
Show the worked solution
- Break one H–H and one Cl–Cl bond: energy required is 436+242=678 kJ per mole of reaction as written.
- Form two H–Cl bonds: energy released is 2×431=862 kJ per mole of reaction.
- Estimated ΔH=678−862=−184 kJ/mol of reaction; the negative sign identifies an exothermic change.
Answer Estimated ΔH=−184 kJ/mol of reaction as written.
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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.
Estimate ΔH for 2H₂+O₂→2H₂O using H–H=436, O=O=498 and O–H=463 kJ/mol of bonds. Use the gaseous-molecule estimate model.
Give me a hint
Count two H–H, one O=O and four O–H bonds.
Compare my reasoning
- Energy required is 2×436+498=1370 kJ/mol of reaction.
- Energy released is 4×463=1852 kJ/mol of reaction.
- Estimated ΔH=1370−1852=−482 kJ/mol of the reaction as written.
−482 kJ/mol of reaction as written; exothermic.
Look for these in your work
- Coefficients and two bonds per water molecule are included.
- The estimate's sign and reaction quantity are stated.
Estimate ΔH for CH₄+2O₂→CO₂+2H₂O using C–H=413, O=O=498, C=O in CO₂=805 and O–H=463 kJ/mol of bonds. Use the gaseous-molecule estimate model.
Give me a hint
Break four C–H and two O=O bonds; form two C=O and four O–H bonds.
Compare my reasoning
- Required energy is 4×413+2×498=2648 kJ/mol of reaction.
- Released energy is 2×805+4×463=3462 kJ/mol of reaction.
- Estimated ΔH=2648−3462=−814 kJ/mol of reaction as written.
−814 kJ/mol of reaction as written; exothermic.
Look for these in your work
- The full balanced reaction fixes bond counts.
- The supplied CO₂ double-bond energy is used directly.
A bond-energy estimate gives −184 kJ for one mole of H₂+Cl₂→2HCl as written. Find the estimate for the reverse reaction and for twice the original reaction amount. Explain why a measured value might differ from the estimate.
Give me a hint
Reverse the sign when reversing the reaction; scale the amount separately.
Compare my reasoning
- The reverse reaction 2HCl→H₂+Cl₂ has the opposite change, +184 kJ for the corresponding amount.
- Twice the original reaction amount gives 2×(−184)=−368 kJ.
- Average bond energies and differences in physical states can make the estimate differ from measured data.
Reverse: +184 kJ; double original amount: −368 kJ; average data and states can affect agreement.
Look for these in your work
- Reaction direction and amount are handled separately.
- An estimate is not advertised as an exact experimental result.
Common mistakes
- Calling bond breaking exothermic.
- Subtracting breaking energy from formation energy and reversing the sign.
- Counting one O–H bond per water molecule rather than two.
- Ignoring balanced coefficients or treating a double-bond table entry as a single-bond value.
What can you explain now?
A reaction requires 500 kJ to break bonds and releases 620 kJ when new bonds form for the stated reaction amount. Find the overall energy change and classification.
Compare with the explanation
−120 kJ; exothermic.
Required minus released is 500−620=−120 kJ for the stated amount; energy is transferred out overall.
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Make it stick
Count bonds for a new balanced molecular equation before opening a data table. Calculate required and released totals separately, then explain the sign and limits of the resulting estimate.
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.