Question

Is this right?

for
Using Hess's law, calculate the ΔH value for the following reaction:?

FeO (s) + CO (g) → Fe (s) + CO₂

Use these three reactions:

1. Fe₂O₃ (s) + 3CO (g) → 2Fe (s) + 3CO₂ (g) ΔH = -25 kJ
2. 3Fe₂O₃ (s) + CO (g) → 2Fe₃O₄ (s) + CO₂ (g) ΔH = -47.0 kJ
3. Fe₃O₄ (s) + CO (g) → 3FeO (s) + CO₂ (g) ΔH = +38.0 kJ

Answer 1

Reaction 1:Fe2O3(s) +3CO(g) -> 2Fe(s) +3CO2(g) (divide by 2)
=1/2Fe2O3(s)+3/2CO(g) -> Fe(s)+ 3/2CO2(g)
Reaction 2:3Fe2O3(s)+CO(g) ->2Fe3O4(s) +CO2(g) (reverse and divide by 6)
=1/3Fe3O4(s)+1/6CO2(g) -> 1/2Fe2O3(s) +1/6CO(g)
Reaction 3:Fe3O4(s)+CO(g) -> 3FeO(s) + CO2(g) (reverse and divide by 3)
=FeO(s)+1/3CO2(g) ->1/3Fe3O4(s) +1/3CO(g)
Adding the three reactions together:
1/2Fe2O3(s)+3/2CO(g)+ 1/3Fe3O4(s)+1/6CO2(g)+ FeO(s)+1/3CO2(g)-> Fe(s)+ 3/2CO2(g)+ 1/2Fe2O3(s) +1/6CO(g)+ 1/3Fe3O4(s) +1/3CO(g)
since 1/2Fe2O3(s) and 1/3Fe3O4(s) are on both sides they cancel out, so we left with
FeO(s)+3/2CO(g)+1/6CO2(g)+1/3CO2(g)-> Fe(s)+ 3/2CO2(g)+1/6CO(g)+1/3CO(g)
and since the common denominator of 2 and 3 is 6, we get:
FeO(s)+9/6CO(g)+1/6CO2(g)+2/6CO2(g) -> Fe(s)+9/6CO2(g)+1/6CO(g)+2/6CO(g)
by flipping the CO2(g) and CO(g)we get:
FeO(s)+9/6CO(g)-1/6CO(g)-2/6CO(g) ->Fe(s)+9/6CO2(g)-1/6CO2(g)-2/6CO2(g)
we are left with:
FeO(s)+CO(g) -> Fe(s)+CO2(g)

Reaction 1:∆H=(-25.0kJ)/2=-12.5
Reaction 2:∆H=47kJ/6=7.833333333 (It^' s+because we flipped the reaction)
Reaction 3:∆H=(-38)/3=-12.66666667(we reversed the reaction)
add the reactions ∆H=-12.5+7.833333333+(-12.66666667)=-17.33333334 or-17.3kJ

Answer 2

Wow that's very long. Are you trying to balance the chemical equation?

Answer 3

Omg, my brain can literally not handle this right now.

Answer 4

what's that? @abb0t

Answer 5

how can i make it shorter?

Answer 6

what do you do?

Answer 7

well, calculate delta H for the target Feo(s)+Co(g) ->Fe(s)+CO2(g) using the 3 reactions