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Yuan Han 12S7F Chemistry Tut 5 Qn 9

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Question (a) Iron is commonly produced in blast furnace from the ore haematite, which contains Iron (III) oxide, Fe 2 O 3. Construct a Born-Haber cycle for the formation of Fe 2 O 3 (s) and use the data provided, together with relevant data from the Data Booklet to calculate enthalpy change of formation of Fe 2 O 3 (s).

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Data provided H atom of Fe = +416kJ/mol Lattice Energy of Fe 2 O 3 (s) = -1.51 x 10 4 kJ/mol 1 st electron affinity of oxygen = -141kJ/mol 2 nd electron affinity of oxygen = +844kJ/mol From Data Booklet: 1 st I.E. of Fe = +762kJ/mol 2 nd I.E. of Fe = +1560kJ/mol 3 rd I.E. of Fe = +2960kJ/mol Bond Energy of O=O = +496kJ/mol

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Born-Haber Cycle 2Fe(s) + 1O 2 (g) 2Fe(g) + 1O 2 (g) 2Fe(g) + 3O(g) 2Fe 3+ (g) + 3O(g) 2Fe 3+ (g) + 3O - (g) 2Fe 3+ (g) + 3O 2- (g) Fe 2 O 3 (s) +416 x 2 H f 1 x (+496) (1 st + 2 nd + 3 rd IE of Fe) x 2 0 Energy / kJ mol -1 -141 x 3 +844 x 3 -1.51 x 10 4

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Electron Affinity of Oxygen 1 st electron affinity of oxygen = -141 kJ/mol 2 nd 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.

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Electron Affinity of Oxygen Why is the 2 nd electron affinity of oxygen so endothermic? 1. Definition of 2 nd electron affinity of oxygen. Therefore, energy is needed to overcome the interelectronic repulsion. 2. The electron is forced into a small, electron-dense region of space.

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Solution (a) H f = 2 x(+416) 1 x (+496) 2 x (762+1560+2960) 3 x (-141) 3 x (+844) (-1.51 x 10 4 ) = -851 kJ/mol

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

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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). H rxn o for Reaction I = +493 kJ/mol H f o of CO 2 (g)= -394 kJ/mol

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Energy Cycle 2Fe 2 O 3 (s) + 3C(s)2Fe(s) + 3CO(g) + Fe 2 O 3 (s) 4Fe(s) + 3CO 2 (g)4Fe(s) + 3C(s) + 3O 2 (g) H rxn o +493kJ/mol 2 x (-851)kJ/mol 3 x (-394)kJ/mol

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Solution (b) -394 x 3 = -851 x 2 + 493 + H rxn o H rxn o = -1209 (-1182) = -27.0 kJ/mol

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Thank you