Calculate ΔHrxn for the following reaction: Fe2O3 (s) + 3CO (g) →2Fe (s) + 3CO2 (g) Use the following reactions and given ΔH′s. 2Fe (s) + 3/2O2 (g) →Fe2O3 (s), ΔH = - 824.2 kJ CO (g) + 1/2O2 (g) →CO2 (g), ΔH = - 282.7 kJ

The answer to your question is: ΔHrxm = - 23.9 kJ

Explanation:

dа ta:

2Fe (s) + 3/2O2 (g) →Fe2O3 (s), ΔH = - 824.2 kJ (1)

CO (g) + 1/2O2 (g) →CO2 (g) ΔH = - 282.7 kJ (2)

Reaction:

Fe2O3 (s) + 3CO (g) →2Fe (s) + 3CO2 (g)

We invert (1) and change the sign of ΔH

Fe2O3 (s) → 2Fe (s) + 3/2O2 (g) ΔH = 824.2 kJ

We multiply (2) by 3

3 (CO (g) + 1/2O2 (g) →CO2 (g) ΔH = - 282.7 kJ) (2)

3CO (g) + 3/2O2 (g) →3CO2 (g) ΔH = - 848.1 kJ

We add (1) and (2)

Fe2O3 (s) → 2Fe (s) + 3/2O2 (g) ΔH = 824.2 kJ

3CO (g) + 3/2O2 (g) →3CO2 (g) ΔH = - 848.1 kJ

Fe2O3 (s) + 3CO (g) + 3/2O2 (g) → 2Fe (s) + 3/2O2 + 3CO2 (g)

Simplify

Fe2O3 (s) + 3CO (g) →2Fe (s) + 3CO2 (g) and ΔHrxm = - 23.9 kJ