A Hess' Law Problem

Question

Given the following data:

Fe2 O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔH = -28 kJ

3Fe2 O3(s) + CO(g) →2Fe3 O4(s) + CO2(g) ΔH = -59 kJ

Fe3 O4(s) + CO(g) →3FeO(s) + CO2(g) ΔH = +38 kJ

Calculate the ΔH for the reaction

FeO(s) + CO(g) → Fe(s) + CO2(g)

Answer 1

1. Label the top three equations Eq1, Eq2, and Eq3.
2. Find the first reactant of the goal equation in Eq1, Eq2, and/or Eq3.
3. The first reactant of the goal equation is FeO. I will look at Eq1, Eq2, and Eq3 to see which one contains FeO (it doesn't matter if it's a reactant or product right now).
4. Begin to manipulate the equations if necessary by flipping or multiplying the equations by coefficients.
5. Equation 3 contains FeO as a product. To make it a reactant, I will have to flip the equation. When I flip the equation, the enthalpy sign will change to the opposite sign.
6. Fe3 O4(s) + CO(g) →3FeO(s) + CO2(g) ΔH = +38 kJ
7. After the flip: 3FeO(s) + CO2(g) → Fe3O4(s) + CO(g) ΔH = -- 38 kJ
8. If I start looking at how to cancel out the CO2, I'll notice that all three equations have CO2 as a product. Since that makes it hard to tell which equation to use, I'll pick something else to focus on.
9. Fe3O4(s) is a product. I will see which equation has Fe3O4(s), which is in equation 2 as a product.
10. I need to cancel out the Fe3O4 from the equation I'm working with. This means that I must flip equation 2 so that the Fe3O4s are on opposite sides.
11. Eq 2: 3Fe2 O3(s) + CO(g) →2Fe3 O4(s) + CO2(g) ΔH = -59 kJ
12. Flipped: 2Fe3 O4(s) + CO2(g) --> 3Fe2 O3(s) + CO(g) ΔH = +59 kJ
13. Flipped Eq2 has a coefficient of 2 in front of the Fe3O4, but Flipped Eq3 does not. In order for them to completely cancel, I will multiply Flipped Eq3 by 2. When I multiply it by 2, I will multiply *everything* by 2, including the enthalpies.
14. Fl Eq3: 3FeO(s) + CO2(g) → Fe3O4(s) + CO(g) ΔH = -- 38 kJ
15. After x2: 6FeO(s) + 2CO2(g) --> 2Fe3O4s) + 2CO(g) ΔH = --76 kJ
16. Now I add Fl*2 Eq3 and fl Eq2 together.
17. ***Substances that are on the same side of the arrow get ADDED together. Substances that are on the opposite sides of the arrow get SUBTRACTED**
18. Fl*2 Eq3: 6FeO(s) + 2CO2(g) --> 2Fe3O4(s) + 2CO(g) ΔH = --76 kJ
19. Fl Eq2: 2Fe3 O4(s) + CO2(g) --> 3Fe2O3(s) + CO(g) ΔH = +59 kJ
20. Eq2+Eq3: 6FeO + 3CO2 --> 3Fe2O3(s) + 3CO(g) ΔH =-17 kJ
21. The only equation left is equation 1. In my new equation (Eq2+Eq3), I have 3 Fe2O3 as a product. Equation 1 has Fe2O3 as a reactant. This is good because they will cancel each other out (since one is a reactant and one is a product)! But, I need to multiply Eq1 by 3 in order to completely have the Fe2O3 cancel out.
22. Eq 1: Fe2 O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔH = -28 kJ
23. Eq1*3: 3Fe2O3(s) + 9CO(g) --> 6Fe(s) + 9CO2(g) ΔH = -84 kJ
24. I can now add Eq1*3 with Eq2+Eq3.
25. Eq2+Eq3: 6FeO + 3CO2 --> 3Fe2O3(s) + 3CO(g) ΔH =-17 kJ
26. Eq1*3: 3Fe2O3(s) + 9CO(g) --> 6Fe(s) + 9CO2(g) ΔH = -84 kJ
27. **For the CO, notice how the COs are on opposite sides, so they will be subtracted. The top equation only has 3CO as a product, but the bottom equation has 9 as a reactant. This means that the 6 remaining COs will be as reactants. In other words, if you're subtracting two coefficients from opposite sides, the final answer will go on the side that had the bigger coefficient.
28. The same thing happens for the CO2. Since the 9CO2s are products, this means that when I do 9-3 for the CO2, the remainder I get will stay as a product (because 9 was the bigger coefficient and the 9 was on the product side).
29. 6FeO(s) + 6CO(g) --> 6Fe(s) + 6CO2(g) ΔH = -101 kJ
30. The coefficients are all 6, but my "goal" equation has implied coefficients of 1. This means that I will need to multiply everything by 1/6th. (There's not really such a thing as "dividing" for Hess's Law, but multiplying by a fraction *is the same thing* as "dividing')
31. FeO(s) + CO(g) --> Fe(s) + CO2(g) ΔH = -17 kJ (rounded up from -16.8333)