Answer:
ΔH° = -11 kj
Explanation:
Step 1: Data given
1) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ
2) 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39 kJ
3) Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ
Step 2: The balanced equation
FeO + CO → Fe + CO2
Step 3: Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).
To get this equation, we need to combine the 3 equations
We have to multiply the third equation by 2.
2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ
<u>This equation we add to the second equation</u>
3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2
(g)
3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)
ΔH° = 2*18 + (-39) = 36 - 39 = -3 kJ
<u>This new equation we will divide by 3 </u>
3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)
Fe2O3(s) + CO(g) → CO2(g) + 2FeO(s)
ΔH° =-3/3 = -1 kJ
<u>Now we will substract this new equation from the first equation</u>
Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)
2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)
ΔH° = -23kJ +1kJ
ΔH° = -22 kj
<u>The next equation we will divide by 2</u>
2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)
CO(g) + FeO(s) → Fe(s) + CO2(g)
ΔH° = -22kJ /2
ΔH° = -11 kj
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