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

what are the óptions. pléase list them

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.