IB ChemistryReactivity 1R1.2R1.2.4

Energy Cycle Calculations

Putting it all together. Worked examples combining all three cycle types.

Reactivity 1.2 SL & HL ⏱️ ~5 min revision

Worked Example 1. Formation Cycle

Worked Example 1

Enthalpy of Combustion from Enthalpies of Formation

Find ΔHr⦵ for: \(\text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})\)

Data: \(\Delta H_f^\ominus(\text{CH}_4) = -74\) | \(\Delta H_f^\ominus(\text{CO}_2) = -394\) | \(\Delta H_f^\ominus(\text{H}_2\text{O}) = -286\text{ kJ mol}^{-1}\)

\( \Delta H_r^\ominus = [(-394) + 2(-286)] - [(-74) + 0] \)

\( = -966 - (-74) = \mathbf{-892 \text{ kJ mol}^{-1}} \)

Worked Example 2. Bond Enthalpy

Worked Example 2

Enthalpy of Combustion from Average Bond Enthalpies

Combustion of methane: \(\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\)

Bonds broken: \(4 \times \text{C--H}\) (\(4 \times 414 = 1656\)) + \(2 \times \text{O=O}\) (\(2 \times 498 = 996\)) = 2652 kJ

Bonds formed: \(2 \times \text{C=O}\) (\(2 \times 804 = 1608\)) + \(4 \times \text{O--H}\) (\(4 \times 463 = 1852\)) = 3460 kJ

\( \Delta H = 2652 - 3460 = \mathbf{-808 \text{ kJ mol}^{-1}} \)

(Less accurate than formation data because bond enthalpies are averages compiled across multiple molecules)

Quick Reference. Which Formula?

Given Data Formula
ΔHf⦵ values ΔHr = Σ(products) − Σ(reactants)
ΔHc⦵ values ΔHr = Σ(reactants) − Σ(products)
Bond enthalpies ΔH = Σ(broken) − Σ(formed)
Born-Haber data (HL) ΔHf = ΔHat + IE + EA − ΔHlatt
Solution data (HL) ΔHsol = ΔHlatt + ΣΔHhyd
Exam Checklist

Thermochemical Calculation Sanity Checks

  • Sign check: Ensure exothermic reactions carry a negative sign (−).
  • Units: Always include kJ mol⁻¹.
  • Stoichiometry: Multiply every ΔH by its balanced molar coefficient.
  • State symbols: Write standard state symbols on every intermediate in energy cycles.
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