Worked example: formation data with elements and multiplesThe thermite reaction produces molten iron and is used to weld railway lines:2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s) Data: ΔfHθ[Fe2O3(s)] = -824.2, ΔfHθ[Al2O3(s)] = -1675.7 kJ mol-1Step 1. Spot the elements. Al(s) and Fe(s) are elements in their standard states, so their enthalpy changes of formation are zero. That is why they are not in the data.Step 2. Apply products minus reactants, including the zeros:ΔrHθ = [ΔfHθ(Al2O3) + 2(0)] - [2(0) + ΔfHθ(Fe2O3)] Step 3. Substitute:ΔrHθ = -1675.7 - (-824.2) = -851.5 kJ mol-1 This value is per mole of the equation as written, which uses 2 mol of aluminium and 1 mol of Fe2O3.Step 4. Use it. If 0.200 mol of Fe2O3 reacts, the energy released is:0.200 × 851.5 = 170.3 kJ The large negative value explains why the iron is produced as a liquid: the energy released heats the products above the melting point of iron.
Hess's Law Example 9
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Specification: Edexcel Chemistry (8CH0), AS-level, Topic 8: Energetics I. Topic: Enthalpy.
Key terms in this lesson
- enthalpy change of formation
- The enthalpy change when one mole of a compound is formed from its elements in their standard states, under standard conditions.
- standard state
- The physical state of a substance under standard conditions of 100 kPa and 298 K.
More in Enthalpy
- Hess's Law Example 2
- Hess's Law Example 3
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- Hess's Law Example 6
- Hess's Law Example 7
- Hess's Law Example 8
- Hess's Law Example 10
All 21 lessons in Enthalpy · All Edexcel AS-level Chemistry topics