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Exam level7 past questions

Enthalpy Changes and Bond Energies

Learn to calculate enthalpy changes for reactions using bond enthalpies, enthalpy of formation, and Hess's law.

Why this shows up in the exam

You will be tested on your ability to use enthalpy data to solve for unknowns in reaction energetics.

How NEET tests this

Numerical · 5 QsDirect recallApplication

Learn the idea

Enthalpy change of a reaction is obtained by adding the enthalpy changes of steps that lead from reactants to products; using bond‑energy data you subtract the total energy of bonds formed from the total energy of bonds broken, and using formation enthalpies you apply Hess’s law: ΔH_rxn = ΣΔH_f(products) – ΣΔH_f(reactants).

🧠 Memory hook: Break bonds → pay energy (+); make bonds → get energy back (‑); net ΔH = pay – get back.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Bond enthalpy = average energy required to break one mole of a specific bond in the gas phase (positive value).
  • ΔH_f° of an element in its standard state = 0 kJ mol⁻¹.
  • Hess’s law: total enthalpy change is independent of the pathway; you may add or subtract any equations as long as you keep stoichiometry correct.
  • ΔH_rxn = Σ(bonds broken) – Σ(bonds formed) when using bond‑energy values.
  • When using ΔH_f values, ΔH_rxn = ΣΔH_f(products) – ΣΔH_f(reactants).
  • All enthalpy values must be in the same unit (kJ mol⁻¹ or kcal mol⁻¹) before calculation.

How to approach it

  1. 1Write the balanced overall reaction you need.
  2. 2Choose the data set given (bond energies or ΔH_f).
  3. 3If using bond energies: list all bonds broken, sum their enthalpies (positive); list all bonds formed, sum their enthalpies (positive) and subtract the latter from the former.
  4. 4If using ΔH_f: sum ΔH_f of each product (multiply by its coefficient) and subtract the sum for reactants; remember elements in standard state contribute zero.
  5. 5Report the sign exactly as obtained; a negative result means exothermic, positive means endothermic.

Worked example — watch it click

Consider the following reactions : (i) H⁺(aq) + OH⁻(aq) = H₂O(l) ΔH = -X₁ kJ mol⁻¹ (ii) H₂(g) + 1/2O₂(g) = H₂O(l), ΔH = -X₂ kJ mol⁻¹ (iii) CO₂(g) + H₂(g) = CO(g) + H₂O(l), ΔH = -X₃ kJ mol⁻¹ (iv) C₂H₂(g) + 5/2O₂(g) = 2CO₂(g) + H₂O(l) ΔH = +X₄ kJ mol⁻¹ Enthalpy of formation of H₂O(l) is

  • A)+X₃ kJ mol⁻¹
  • B)-X₄ kJ mol⁻¹
  • C)+X₁ kJ mol⁻¹
  • ✅-X₂ kJ mol⁻¹

The concept behind this problem

The example asks for the standard enthalpy of formation of liquid water; reaction (ii) is exactly the definition of ΔH_f°[H₂O(l)], so the answer is –X₂ kJ mol⁻¹.

Step by step

  1. 1Enthalpy of formation is the enthalpy change when 1 mole of compound is formed from its elements in their standard states.
  2. 2Reaction (ii) H₂(g) + ½O₂(g) = H₂O(l), ΔH = -X₂ kJ mol⁻¹ directly represents the formation of H₂O(l) from its elements H₂ and O₂ in standard states.
  3. 3Therefore, ΔH_f[H₂O(l)] = -X₂ kJ mol⁻¹

Watch out

Students often look for a combination of the other reactions, but the formation enthalpy is read directly from reaction (ii) without any manipulation.

Common slip-ups that cost marks

  • •Mixing up the sign: breaking bonds costs energy (add), forming bonds releases energy (subtract).
  • •Treating average bond‑energy values as exact for a specific molecule – they give an estimate, not a precise value.
  • •Forgetting that ΔH_f of elements in their standard states is zero, leading to extra terms in the calculation.

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 7NEET 2007

Given that bond energies of H – H and Cl – Cl are 430 kJ mol⁻¹ and 240 kJ mol⁻¹ respectively and ΔHₓ for HCl is -90 kJ mol⁻¹, bond enthalpy of HCl is

Push further

More challenging

14 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 14

Given the bond dissociation enthalpies: H-H = 436 kJ/mol, F-F = 158 kJ/mol, and H-F = 568 kJ/mol. Calculate the standard enthalpy of formation of HF(g).

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