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

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)

2Fe(s)+3/2O2(g)→Fe2O3(s), ΔH = -824.2 kJ
CO(g)+1/2O2(g)→CO2(g), ΔH = -282.7 kJ

Now you take what you know out of the known equations and rearrange till you get your necessary equation like so:

2Fe(s)+3/2O2(g)→Fe2O3(s), ΔH = -824.2 kJ
this first one needs to be flipped, so the Fe2O3 is on the starting material side. So we'll just flip it, but remember to *change* the sign of the delta H:
Fe2O3(s)→2Fe(s)+3/2O2(g), ΔH = 824.2 kJ

CO(g)+1/2O2(g)→CO2(g), ΔH = -282.7 kJ

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

In second one, we have to multiply with three because in reaction 3moles of CO is there, we'll just have to make sure the O2(g), cancels out... on to the next one

So now our equation looks like this:

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

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

Notice the O2(g) will cancel out cause its on both sides of the equation. All you've got to do now is add the delta H and voila: Delta H = -23.9 kJ