Q9 Chem Tut 5 - 12S7F-note

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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, 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).
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
Born-Haber Cycle
Energy
/ kJ mol-1
2Fe(g) + 3O(g)
2Fe3+(g) + 3O2-(g)
2Fe3+(g) + 3O(g)
(1st + 2nd + 3rd IE
of Fe) x 2
2Fe(g) + 1½O2(g)
0
2Fe(s) + 1½O2(g)
Fe2O3(s)
+844 x 3
2Fe3+(g) + 3O-(g)
-141 x 3
1½ x (+496)
-1.51 x 104
+416 x 2
∆Hf
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.
Electron Affinity of Oxygen
Why is the 2nd electron affinity of oxygen so endothermic?
1. Definition of 2nd 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.
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
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)
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
Energy Cycle
2Fe2O3(s) + 3C(s)
+493kJ/mol
2 x (-851)kJ/mol
4Fe(s) + 3C(s) +
3O2(g)
2Fe(s) + 3CO(g)
+ Fe2O3(s)
∆Hrxno
3 x (-394)kJ/mol
4Fe(s) + 3CO2(g)
Solution (b)
 -394 x 3 = -851 x 2 + 493 + ∆Hrxno
 ∆Hrxno = -1209 – (-1182)
= -27.0 kJ/mol
 Thank you

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