Determine ΔG° for the reaction:
Fe2O3(s) + 3 CO( g) ¡ 2 Fe(s) + ...

Question

Determine ΔG° for the reaction: 
Fe2O3(s) + 3 CO( g) ¡ 2 Fe(s) + 3 CO2( g) 
Use the following reactions with known ΔGrxn ° values: 
2 Fe(s) + 3 2 O2( g) ¡ Fe2O3(s) ΔGrxn ° = -742.2 kJ 
CO( g) + 1 2 O2( g) ¡ CO2( g) ΔGrxn ° = -257.2 kJ

Accepted Answer

Hey everyone. So today we're being asked to calculate the delta G for the following reaction, given the two half reactions above with their delta G or give street energy values. So to do this, we essentially need to do some sort of algebra almost, but we need to essentially add up and manipulate the two half reactions in order to sum up to this final reaction here. So let's just take a look at the first reaction. The first half reaction in the final. The whole reaction, we can see that we have three hydrogen and three water molecules, three hydrogen or three hydrogen gas molecules as reactant and three water molecules as products. So in order to achieve that, the easiest way would actually be to multiply this entire first half reaction by three. This would also mean multiplying its delta G by three as well, we'll come back to this. But let's just take a look at the second reaction as well. We only have 103 in the react inside and we also have no 02 on the product side. In order to achieve this. What we actually need to do is take the entirety of the second reaction, flip it, meaning reversing its delta G and reversing the products and reactant essentially going backwards going to the left. So we flip it and then we would divide by two because we only want 10310 zone for example, or ozone as it is known. So let's go ahead and do this. If we take the first reaction multiplied by three. Well then we get three H two gas plus 3/2 oh two, which looks weird but it'll make sense in just a second, Which gives us three water molecules. and since we've multiplied it by three, that gives us a delta G of the reaction of negative 6 85 85.8 kg jules. I'm leaving out the units here just for the sake of space or actually in hindsight. Let me just move this. Oops, just move this over a little bit. There we go. That'll give us kill jules. And for the second half reaction, we flipped it and divided everything by two. So this will give us 03 gas yields three over 202 gas with a delta G of the reaction, Which has been flipped, which means the signal reverse and then divided by two, Which gives us negative 1 63.2 killer jewels. So with all this. Now, looking at our whole reaction, if we add everything up, oops, if we add everything up, we can see that the 302 here is matched by the or 3/2 oh two on the reactant side as well, which means they will effectively cancel each other out. And our final reaction will be three H two gas plus O. Three gas Heels, three H 20. So since we added it up and it gave us the proper reaction, that means we can also add up the delta cheese to give us the final delta G for the whole reaction. So adding these two up, we get that the final delta G of the reaction Is negative 849 killer jewels. Therefore, using the delta G's From the half reactions, we found that the final Gibbs free energy for the whole reaction is negative 849 killer jewels. I hope this helps. And I look forward to seeing you all in the next one.

Determine ΔG° for the reaction: Fe2O3(s) + 3 CO( g) ¡ 2 Fe(s) + 3 CO2( g) Use the following reactions with known ΔGrxn ° values: 2 Fe(s) + 3 2 O2( g) ¡ Fe2O3(s) ΔGrxn ° = -742.2 kJ CO( g) + 1 2 O2( g) ¡ CO2( g) ΔGrxn ° = -257.2 kJ

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