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:

[tex] \Delta H_{2} = 3*(-282.7) kJ = -848.1 kJ [/tex]

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).

[tex] \Delta H_{rxn} = \Delta H_{1} + \Delta H_{2} = 824.2 kJ + (-848.1 kJ) = -23.9 kJ [/tex]

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

You can find more about the enthalpy of reaction here:

https://brainly.com/question/1657608?referrer=searchResults

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I hope it helps you!