Question

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

would this be right?

Reaction 1(divide by 2):1/2 Fe_2 O_3 (s)+3/2 CO(g)→ Fe(s)+ 3/2 CO_2 (g)
Reaction 2(reverse & divide by 6):1/3 Fe3O4(s)+1/6 CO2(g) →( 1)/2 Fe2O3(s) +1/6 CO(g)
Reaction 3(reverse & divide by 3):FeO(s)+1/3 CO_2 (g) →1/3 Fe_3 O_4 (s) +1/3 CO(g)
Adding the three reactions together:
1/2 Fe_2 O_3(s) +3/2 CO_((g) )+ 1/3 Fe_3 O_4(s) +1/6 CO_2(g) + FeO_((s) )+1/3 CO_2(g) → Fe_((s) )+3/2 CO_2(g) +( 1)/2 Fe_2 O_3(s) +1/6 CO_((g) )+ 1/3 Fe_3 O_4(s) +1/3 CO_((g) )
since 1/2 Fe_2 O_(3(s)) and 1/3 Fe_3 O_(4(s)) are on both sides they cancel out, so we left with
FeO_((s))+3/2 CO_((g))+1/6 CO_(2(g))+1/3 CO_(2(g))→ Fe_((s))+3/2 CO_(2(g))+1/6 CO_((g))+1/3 CO_((g))
and since the common denominator of 2 and 3 is 6, and by flipping CO_(2(g)) and CO_((g)) we get:
FeO(s)+9/6 CO_((g))-1/6 CO_((g))-2/6 CO_((g))→ Fe(s)+9/6 CO_(2(g))-1/6 CO_(2(g))-2/6 CO_(2(g))
we get:
FeO_((s))+CO_((g))→Fe_((s))+CO_(2(g))

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

Answer 2

Your first steps are correct, I agree with the division, mult. and eq. reverses.
Assuming you also applied the same operations to the enthalpy values, you should be good.