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Chemistry Tut 5 Qn 9

Chemistry Tut 5 Qn 9. Yuan Han 12S7F. Question (a). Iron is commonly produced in blast furnace from the ore haematite, which contains Iron (III) oxide, Fe 2 O 3 .

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Chemistry Tut 5 Qn 9

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  1. Chemistry Tut 5 Qn 9 Yuan Han 12S7F

  2. Question (a) • Iron is commonly produced in blast furnace from the ore haematite, which contains Iron (III) oxide, Fe2O3. • Construct a Born-Haber cycle for the formation of Fe2O3(s) and use the data provided, together with relevant data from the Data Booklet to calculate enthalpy change of formation of Fe2O3(s).

  3. Data provided • ∆Hatom of Fe = +416kJ/mol • Lattice Energy of Fe2O3(s) = -1.51 x 104 kJ/mol • 1st electron affinity of oxygen = -141kJ/mol • 2nd electron affinity of oxygen = +844kJ/mol From Data Booklet: • 1st I.E. of Fe = +762kJ/mol • 2nd I.E. of Fe = +1560kJ/mol • 3rd I.E. of Fe = +2960kJ/mol • Bond Energy of O=O = +496kJ/mol

  4. Born-Haber Cycle Energy / kJ mol-1 2Fe3+(g) + 3O2-(g) 2Fe3+(g) + 3O(g) +844 x 3 (1st + 2nd + 3rd IE of Fe) x 2 2Fe3+(g) + 3O-(g) 2Fe(g) + 3O(g) -141 x 3 1½ x (+496) 2Fe(g) + 1½O2(g) -1.51 x 104 2Fe(s) + 1½O2(g) +416 x 2 0 ∆Hf Fe2O3(s)

  5. Electron Affinity of Oxygen • 1st electron affinity of oxygen = -141 kJ/mol • 2nd electron affinity of oxygen = +844 kJ/mol Why is electron affinity usually negative? Potential Energy is released when an electron is brought close to the nucleus.

  6. Electron Affinity of Oxygen Why is the 2nd electron affinity of oxygen so endothermic? • Definition of 2nd electron affinity of oxygen.Therefore, energy is needed to overcome the interelectronic repulsion. • The electron is forced into a small, electron-dense region of space.

  7. Solution (a) • ∆Hf = • 2 x(+416) • 1½ x (+496) • 2 x (762+1560+2960) • 3 x (-141) • 3 x (+844) • (-1.51 x 104) • = -851 kJ/mol

  8. Question (b) • At the bottom f the blast furnace where the temperature may be above 1500oC, reduction is principally by carbon. Reaction I: Fe2O3(s) + 3C(s) → 2Fe(s) + 3CO(g) • At the top of the furnace where temperature is lower at 800oC, the principal reducing agent is carbon monoxide. Reaction II: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

  9. Question (b) • By means of a suitable energy cycles or any other method, calculate the standard enthalpy change of reaction for Reaction II using the following enthalpy change data and the answer in (a). • ∆Hrxno for Reaction I = +493 kJ/mol • ∆Hfo of CO2(g)= -394 kJ/mol

  10. Energy Cycle +493kJ/mol 2Fe2O3(s) + 3C(s) 2Fe(s) + 3CO(g) + Fe2O3(s) 2 x (-851)kJ/mol ∆Hrxno 4Fe(s) + 3C(s) + 3O2(g) 4Fe(s) + 3CO2(g) 3 x (-394)kJ/mol

  11. Solution (b) • -394 x 3 = -851 x 2 + 493 + ∆Hrxno • ∆Hrxno = -1209 – (-1182) = -27.0 kJ/mol

  12. Thank you

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