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)

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.

Note that Fe2O2 isn't correct. Didn't you intend to type Fe2O3

Well, let's take a look at the given thermochemical equations.

We have:

Fe2O3(s) + 3CO(g) ---> 2Fe(s) + 3CO2(g) (ΔH° = -28 kJ)

3Fe2O3(s) + CO(g) ---> 2Fe3O4(s) + CO2(g) (ΔH° = -59 kJ)

Fe3O4(s) + CO(g) ---> 3FeO(s) + CO2(g) (ΔH° = +38 kJ)

Now, to find the ΔH° for the reaction:

FeO(s) + CO(g) ---> Fe(s) + CO2(g)

We can start by manipulating the given equations in order to obtain the desired reaction:

3(Fe3O4(s) + CO(g) ---> 3FeO(s) + CO2(g)) (ΔH° = +38 kJ)

6Fe3O4(s) + 3CO(g) ---> 6FeO(s) + 3CO2(g) (ΔH° = +38 kJ)

Now, we can cancel out the common species on both sides of the equations:

6Fe3O4(s) + 3CO(g) - 3FeO(s) - 3CO2(g) = 6FeO(s) + 3CO2(g) - 3FeO(s) - 3CO2(g)

6Fe3O4(s) + 3CO(g) - 3FeO(s) - 3CO2(g) = 6FeO(s) - FeO(s) + 3CO2(g) - 3CO2(g)

6Fe3O4(s) + 3CO(g) - 3FeO(s) - 3CO2(g) = 5FeO(s)

Now, let's sum up the ΔH° values for the equations used:

ΔH° = 6 * (+38 kJ) + 3 * (-28 kJ) - 3 * (0 kJ) - 5 * (0 kJ)

ΔH° = 228 kJ + (-84 kJ) - 0 kJ - 0 kJ

ΔH° = 144 kJ

So, the ΔH° for the reaction FeO(s) + CO(g) ---> Fe(s) + CO2(g) is 144 kJ.

Now, let's hope these calculations don't turn you into a "furnace" of confusion!

To find the ΔH° for the given reaction:

FeO(s) + CO(g) ---> Fe(s) + CO2(g)

We can use the given thermochemical equations and manipulate them to obtain the desired reaction. Here's how you can do it:

1. Start with the first equation:
Fe2O2(s) + 3CO(g) ---> 2Fe(s) + 3CO2(g)
ΔH° = -28 kJ

2. Multiply the first equation by 1/2 to balance the number of Fe atoms:
(1/2)(Fe2O2(s) + 3CO(g) ---> 2Fe(s) + 3CO2(g))
ΔH° = -14 kJ

3. Reverse the second equation to match the products with the reactants:
2Fe3O4(s) + 2CO2(g) ---> 3Fe2O2(s) + 2CO(g)
ΔH° = +59 kJ

4. Multiply the third equation by 1/2 to balance the number of Fe atoms:
(1/2)(Fe3O4(s) + CO(g) ---> 3FeO(s) + CO2(g))
ΔH° = +19 kJ

Now, we have the manipulated equations:

FeO(s) + 2CO(g) ---> Fe2O2(s) + 2CO2(g)
ΔH° = -14 kJ

2Fe2O2(s) + 2CO2(g) ---> 3FeO(s) + 2CO(g)
ΔH° = +19 kJ

Adding these two equations together cancels out Fe2O2 and CO2:

FeO(s) + 2CO(g) + 2Fe2O2(s) + 2CO2(g) ---> 3FeO(s) + 2CO(g) + 2Fe(s) + 2CO2(g)

The net result is:

2Fe(s) ---> 3FeO(s)
ΔH° = -14 kJ + 19 kJ = +5 kJ

So, the ΔH° for the reaction FeO(s) + CO(g) ---> Fe(s) + CO2(g) is +5 kJ.

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