R1.2Reactivity 1 — What drives chemical reactions?
Energy cycles in reactions — IB Chemistry SL Revision Notes & Topical Questions
5 min read 11 parts Lesson 2 of 8 1 quiz
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Lesson summary
R1.2 Energy cycles in reactions — IB CHEMISTRY SL notes: Explain bond breaking and forming in energy terms, calculate enthalpy changes from bond enthalpies and apply Hess's law.Bond breaking — requires energy from the surroundings, so it is endothermic and contributes a positive …
- Bond forming
- Equal magnitude
- Net result
- Endothermic case
- Bond strength
- Bond enthalpy
Part 1 of 9Bond breaking and bond forming
Bond breaking
- Bond breaking requires energy from the surroundings, so it is endothermic and contributes a positive term to ΔH.
Bond forming
- Bond forming releases energy to the surroundings, so it is exothermic and contributes a negative term to ΔH.
Equal magnitude
- Equal magnitude the energy released when a bond forms equals the energy absorbed when it breaks, with the opposite sign.
Net result
- Net result if more energy is released forming bonds than absorbed breaking bonds, the reaction is exothermic and the products are more stable.
Endothermic case
- Endothermic case if more energy is needed to break bonds than is released on forming bonds, the products are less stable.
Bond strength
- Bond strength stronger and shorter bonds have larger bond enthalpies, e.g. C≡C > C=C > C–C.
Part 2 of 9Bond enthalpy and average bond enthalpy
Bond enthalpy
Bond enthalpy the enthalpy change needed to break one mole of a specified covalent bond in gaseous molecules, always positive, in kJ mol−1.
Average bond enthalpy
Average bond enthalpy the energy needed to break one mole of a bond in the gaseous state averaged over similar compounds.
Reason for averaging
- Reason for averaging a bond's strength depends on its environment, e.g. successive C–H bonds in methane differ, so an average is used.
Data booklet
- Data booklet values are averages, so calculated ΔH is approximate.
Part 3 of 9Calculating ΔH from bond enthalpies
- Formula ΔH = Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed).
1
Write the balanced equation with state symbols; all species must be gaseous.
2
Draw full displayed formulae and count every bond broken in the reactants and formed in the products.
3
Multiply each bond enthalpy by the number of bonds and by the stoichiometric coefficient.
4
Sum the bonds broken and the bonds formed separately.
5
Calculate ΔH = Σ(broken) − Σ(formed) and state the sign.
Sign convention
- Sign convention bonds broken are positive (endothermic) and bonds formed are negative (exothermic) when added.
Combustion enthalpy
- Combustion enthalpy simplify the equation to one mole of fuel before counting bonds.
Part 4 of 9Why bond enthalpy results are approximate
1Averages
- Averages bond enthalpies are averaged over many compounds, not the actual molecule.
2Gas phase only
- Gas phase only values refer to gaseous species, so liquids such as H2O(l) give a different result because enthalpy of vaporisation is ignored.
3Intermolecular forces
- Intermolecular forces are ignored in liquids and solids.
4Comparison
- Comparison results are less exact than those from Hess cycles with experimental ΔH data.
Part 5 of 9Hess's law
1Statement
Statement the total enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same.
2Basis
- Basis it follows from the law of conservation of energy, since different routes giving different ΔH would let energy be created.
3Use
- Use finds ΔH for reactions that cannot be measured directly, e.g. when reactions are too slow, have side reactions,
- or the elements do not react under standard conditions,
- as for ΔHf of propane.
4Cycle relation
- Cycle relation when the direct route is ΔH2 from elements to products and the indirect route goes via the reactants, ΔHr = ΔH2 − ΔH1.
Part 6 of 9Hess cycles
- Cycle
- a diagram linking reactants and products by two or more routes so an unknown ΔH is found from known values.
1
Write the target enthalpy change at the top of the diagram from left to right.
2
Write the alternative route along the bottom and join the two with arrows.
3
Add the enthalpy data, adjusted for the molar amounts.
4
Write out the calculation along the alternative route.
1Arrow rule
- Arrow rule following an arrow adds the quantity, and going against an arrow subtracts it.
2Clarity
- Clarity put brackets around values and keep the mathematical operator separate from the sign of the enthalpy change.
Part 7 of 9Hess's law using equations
Reverse
- Reverse reverse any equation needed backwards and change the sign of its ΔH.
Scale
- Scale multiply an equation by n to match coefficients and multiply its ΔH by n.
1
Write the target equation with state symbols.
2
List the given equations and their ΔH values.
3
Reverse or scale each equation so the target reactants and products appear on the correct sides.
4
Add the equations and cancel species that appear on both sides.
5
Add the ΔH values to get the answer, with sign and unit.
Part 8 of 9Comparing methods of finding ΔH
| Method | Formula | Accuracy |
|---|---|---|
| Bond enthalpies (SL) | ΔH = Σ(broken) − Σ(formed) | Approximate; gaseous species only |
| Hess cycle with given ΔH values (SL) | Add ΔH values along the alternative route, reversing signs for reversed arrows | Exact if the data are exact |
| Formation data (HL) | ΔH = ΣΔHf(products) − ΣΔHf(reactants) | Exact for standard states |
| Combustion data (HL) | ΔH = ΣΔHc(reactants) − ΣΔHc(products) | Exact for standard states |
- Order
- the combustion formula is reversed compared with the formation formula.
Part 9 of 9More from the specification
1
Standard enthalpy changes per mole
- Standard enthalpy changes per mole calculating formation enthalpy across stoichiometric reactant moles verifies theoretical energy cycles.
2
Balance sheet
- A table of bonds broken (positive) and bonds formed (negative) used to calculate ΔH
3
Hess's Law application
- The overall enthalpy change for a chemical process is independent of the route taken, allowing calculation of unknown reaction enthalpies.
4
Bond dissociation enthalpy
- The average endothermic energy required to break one mole of specific covalent bonds in gaseous atoms or molecules.
5
Standard enthalpy of formation (Delta Hf)
- The enthalpy change when one mole of a compound is formed from its constituent elements in standard states.
6
Enthalpy cycle problem solving
- Setting up algebraic Hess cycles where Delta Hreaction = sum Delta Hf(products) - sum Delta Hf(reactants).
7
Haber process
- Industrial manufacture of ammonia from nitrogen and hydrogen.
Key terms
| Term | Meaning |
|---|---|
| Bond enthalpy | The enthalpy change to break one mole of a covalent bond in gaseous molecules. |
| Average bond enthalpy | The bond enthalpy averaged over a range of different compounds. |
| Hess's law | The total enthalpy change is independent of the route taken. |
| Enthalpy cycle | A diagram linking reactants and products by alternative routes to find an unknown ΔH. |
| Bond breaking | An endothermic process that absorbs energy. |
| Bond forming | An exothermic process that releases energy. |
Avoid these mistakes
1
Adding bond enthalpies of bonds formed with a positive sign — bonds formed are subtracted in ΔH = Σ(broken) − Σ(formed).
2
Missing coefficients in the balanced equation — multiply each bond enthalpy by the number of moles.
3
Forgetting to change the sign of ΔH when reversing an equation — the sign reverses.
4
Using bond enthalpies for liquid species — all species must be gaseous.
Marked practiceQuizzes for this lesson
Last updated:
\n