Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

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

📚 IB Chemistry SL Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
0%read
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

MethodFormulaAccuracy
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 arrowsExact 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

TermMeaning
Bond enthalpyThe enthalpy change to break one mole of a covalent bond in gaseous molecules.
Average bond enthalpyThe bond enthalpy averaged over a range of different compounds.
Hess's lawThe total enthalpy change is independent of the route taken.
Enthalpy cycleA diagram linking reactants and products by alternative routes to find an unknown ΔH.
Bond breakingAn endothermic process that absorbs energy.
Bond formingAn 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.

Last updated:

\n