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 enthalpy of reaction (ΔHrxn) of the reaction Fe₂O₃(s) + 3CO → 2Fe(s) + 3CO₂(g) calculated from the given enthalpies of the 2 reactions is -23.9 kJ.  

We need to find the enthalpy of reaction (ΔHrxn) of the following reaction:

Fe₂O₃(s) + 3CO → 2Fe(s) + 3CO₂(g)    (1)

We have the enthalpies of the reactions:

2Fe(s) + 3/2O₂(g) → Fe₂O₃(s)  ΔH₁ = -824.2 kJ    (2)

CO(g) + 1/2O₂(g) → CO₂(g)  ΔH₂ = -282.7 kJ     (3)

To calculate the ΔHrxn of equation (1) we must modify equations (2) and (3).

First, we need to invert equation (2) so the Fe₂O₃(s) be in the reactants rather than in the products. After doing this, the sign of ΔH₁ will change:

Fe₂O₃(s) → 2Fe(s) + 3/2O₂(g)    ΔH₁ = 824.2 kJ    (4)

Now we have the Fe₂O₃(s) in the reactants side as in reaction (1).

Second, we need to add a coefficient of 3 before the CO and CO₂, so we need to multiply equation (3) by 3.

3*[CO(g) + 1/2O₂(g) → CO₂(g)]    

3CO(g) + 3/2O₂(g) → 3CO₂(g)    

This also applies for ΔH₂ = -282.7 kJ:

Hence, reaction (3) is now:

3CO(g) + 3/2O₂(g) → 3CO₂(g)  ΔH₂ = -848.1 kJ   (5)

Now we have the coefficients of 3 before CO and CO₂ as in equation (1).

Finally, by adding equations (4) and (5) we can get equation (1)

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

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)   (1)

We get the reaction (1). The enthalpy of reaction is given by the sum of ΔH₁ (eq 4) and ΔH₂ (eq 5).

Therefore, the ΔHrxn for the reaction (1) is -23.9 kJ.

You can find more about the enthalpy of reaction here:

I hope it helps you!

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