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Hess's Law 19,173 views

Teacher/Instructor Kendal Orenstein
Kendal Orenstein

Rutger's University
M.Ed., Columbia Teachers College

Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.

Hess's Law states that the energy of a chemical reaction is the same regardless of the number of steps needed or the reaction mechanism. Hess's law is directly related to the law of conservation of energy.

Alright sometimes when you're dealing with thermo chemical equations you might not know the delta H of your reaction. So there're different ways you can go about getting the delta H of reaction, you can actually do the reaction itself and a calorimeter and figure out the delta H. Or you can actually use other reactions that might add up to your particular reaction. And what does that mean? Okay there's this law called Hess's law. And Hess's law states that it is possible to add 2 or more thermo chemical equations to produce a final where reaction and sum of the enthalpy changes for the individual reactions is the enthalpy change for final reaction. Now this is definitely a mouthful it's much easier to explain. So let's actually do it, so let's actually come over here and let's see how this reaction. Sulfur plus oxygen gas yields sulfur trioxide and I've no idea what the delta H is okay.

Well there're many, many reference books that tell you what the delta H is for other reactions but this one for some reason I couldn't find or my reference book didn't or I didn't want to do the particular reaction in the lab or something along those lines. So how can I find this, well Hess's law says that I can manipulate this reaction in order to get this overall reaction and that way I can figure out my delta H if I add up the 2 H's. So first I have to make sure that these reactions actually do add up to the final reaction. So let's look at this firstly, I need 2 sulfurs on my reactant side, well here I have sulfur but I only have 1, so I'm going to multiply this whole reaction by 2 so I can get 2 sulfurs on my reactive side. So I'm going to multiply this by 2, so that makes 2 sulfur plus 2 oxygen gases yields 2 sulfur dioxide.

And because I multiplied the reaction by 2 this also must be multiplied by 2 because this negative 297 kilo joules is telling me how much energy is going to be released for 1 mol sulfur but now I'm doing 2 mols of sulfur, so I'm going to multiply this by 2. So my delta H is now going to be negative 594 kilo joules okay great. So now I have my 2 mols of sulfur on my reacting side that's awesome but now I have sulfur dioxide on my product side and I don't it I want sulfur trioxide. So I'm going to look at this reaction here, well here is sulfur dioxide on the product side and sulfur trioxide on the reacting side so I actually want to flip those around so I'm going to do the reverse reaction, so I'm going to say oxygen gas plus 2 mols of sulfur dioxide gas is going to give me 2 mols of sulfur trioxide gas okay.

And because I flipped it this forward direction is going to be an endothermic reaction but if I reversed it, it's going to becoming an exothermic reaction so this is going to be negative. So my delta H is going to be negative 198 kilo joules. Okay so let's make sure this works, so if I were to add these 2 reactions together, note that I have 2 mols of sulfur dioxide here and 2 mols of sulfur dioxide here. I can cross those out because this is a product and this is a reactive so I can just cross those out okay. And now if I add up I can't, nothing else crosses out so I'm going to add everything up. So I've 2 mols of sulfur solid plus 2 mols of oxygen, this should be oxygen too alright O2 gas plus another mol of O2 gas yields 2 mols of SO3 gas okay. So what if I add these together 2 plus 1 is 3 so I can say 3 mols of oxygen gas. So 2 mols of sulfur plus 3 mols of oxygen yields 2 mols of sulfur trioxide which is exactly my original equation is so I added this properly I can check it off and so I want to find my delta H I'm just going to add these guys up. So negative 594 plus negative 198 is going to give me negative 792, so negative 792 kilo joules.

So that's actually now I could know that this is negative 792 kilo joules because Hess's law says that I can do this. Okay this is basically Hess's law in a nutshell you might have several different reactions you're going to have to manipulate but actually once you get the hung of it, it's actually kind of fun and pretty easy. I'm also going to do one last thing, this is the reaction it is the formation reaction meaning we're taking or forming sulfur trioxide from these elements. So in order to do a formation reaction the best way is when you get these three 1 mol. So I'm going to divide the whole thing by 2 because I don't like in formation reaction if you want to learn more about them there's a video on that as well but a formation reaction it has 1 mol of the product so dividing everything by 2 which end up as sulfur solid plus three halves oxygen gas yields sulfur trioxide gas.

And then my delta H is also going to be divided by 2 which is going to be negative 396 kilo joules and we can actually say kilo joules per mol because we have 1 mol there. This is our formation reaction and notice this is fraction in it but in formation reaction it's totally okay we're allowed to have fraction coefficients in our formation reactions because our main important thing is that we have 1 mol of product so this [IB] in action and then we put a little bit of formation delta H formation in there as well. So Hess's law is really, really useful when dealing with reactions where you do not know our delta H so we need to find it pretty easily.