Asked by Lenny
In the recovery of iron from iron ore, the reduction of the ore is actually accomplished by reactions involving carbon monoxide. Use the following thermochemical equations,
Fe2O2(s) + 3CO(g) ---> 2Fe(s) +3CO2(g)
ΔH° = -28kJ
3Fe2O2(s) + CO(g) ---> 2Fe3O4(s)+ CO2(g)
ΔH° = -59kJ
Fe3O4(s) + CO(g) ---> 3FeO(s) + CO2(g)
ΔH° = +38kJ
What is ΔH° for the reaction
FeO(s) + CO(g) ---> Fe(s) +CO2(g)
Fe2O2(s) + 3CO(g) ---> 2Fe(s) +3CO2(g)
ΔH° = -28kJ
3Fe2O2(s) + CO(g) ---> 2Fe3O4(s)+ CO2(g)
ΔH° = -59kJ
Fe3O4(s) + CO(g) ---> 3FeO(s) + CO2(g)
ΔH° = +38kJ
What is ΔH° for the reaction
FeO(s) + CO(g) ---> Fe(s) +CO2(g)
Answers
Answered by
DrBob222
Note that Fe2O2 isn't correct. Didn't you intend to type Fe2O3
Answered by
Lenny
Correction:
3Fe2O3(s) + CO(g) ---> 2Fe3O4(s)+ CO2(g)
ΔH° = -59kJ
Is this right?
FeO(s) + CO(g) ---> Fe(s) +CO2(g)
ΔH° = -28/2
ΔH° = -14
3Fe2O3(s) + CO(g) ---> 2Fe3O4(s)+ CO2(g)
ΔH° = -59kJ
Is this right?
FeO(s) + CO(g) ---> Fe(s) +CO2(g)
ΔH° = -28/2
ΔH° = -14
Answered by
DrBob222
No, I don't get that answer. I only went through it once so I could have made an error. Here is what I did.
The reverse of twice equation 3 + the reverse of equation 2 + 3x equation 1 and that gave me
6FeO + 6CO ==> 6Fe + 6CO2 and -101 kJ/mol. That divided by 6 = about -17 kJ/mol.
The reverse of twice equation 3 + the reverse of equation 2 + 3x equation 1 and that gave me
6FeO + 6CO ==> 6Fe + 6CO2 and -101 kJ/mol. That divided by 6 = about -17 kJ/mol.
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