Question

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

Answer 1

Answer:

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

Explanation:

Data:

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