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Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39 kJ Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

Respuesta :

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

This equation we add to the second equation

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

This new equation we will divide by 3

3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) →  CO2(g) + 2FeO(s)

ΔH°  =-3/3 = -1 kJ

Now we will substract this new equation from the first equation

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

The next equation we will divide by 2

2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

CO(g)  + FeO(s)  → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

When ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g) is ΔH° is = -11 k

Calculation of Chemical reaction of Fe(s) + CO2(g).

Data is given as per question

First Part is Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)   ΔH is = −23 kJ

Second Part is 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g)  ΔH is = −39 kJ

Third Part is Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH is = +18 kJ

The balanced equation is:

Then FeO + CO → Fe + CO2

After that we Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

Now To get this equation, Then we need to combine the 3 equations

Then We have to multiply the third equation by 2.

Then 2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH is = +54 kJ

This equation we need add to the second equation

That is 3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

Then 3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

ΔH°  is = 2*18 + (-39) = 36 - 39 =  -3 kJ

After, This new equation we will divide by 3

That is 3Fe2O3(s) + 3CO(g) →  3CO2(g) + 6FeO(s)

Then Fe2O3(s) + CO(g) →  CO2(g) + 2FeO(s)

ΔH°  is = -3/3 = -1 kJ

Now we will Minus this new equation from the first equation

That is Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

Then 2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

ΔH° is = -23kJ +1kJ

ΔH° is = -22 kj

Then The next equation we will divide by 2

Then 2CO(g)  + 2FeO(s)  → 2Fe(s) + 2CO2(g)

After that CO(g)  + FeO(s)  → Fe(s) + CO2(g)

ΔH° is = -22kJ /2

Therefore, ΔH° = -11 kj

Find more information about Chemical reaction here:

https://brainly.com/question/14477259

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