Learn about Senior Chemistry, Energetics 11, in this comprehensive video by bannanaiscool.
Read the full transcript »

Rob Lederer: Now I'll show you the best way to graph these. So here we have the glucose, and in the end, because there was a release of energy, potential energy because it's bond energy, so we graph this as potential energy and we always right down the unit in kilojoules versus whatever the time is here, we just call it reaction coordinate, that's okay. The lower amount of the energy actually belongs to the products here. So in an exothermic reaction, the products always have less heat content than the reactant does, because this thing turned into this and it released heat. So, we have less energy in the products than the reactants, but in order to connect this line, we need to do something a little more spatial. We need to remind ourselves that energy first had to be added in order to form C and H and O atoms, because we needed to take this chemical and break it down into all of those individual atoms before they can rearrange to form all of these molecules as products. So we needed to add a certain amount of energy to get to this thing called the activating complex. So, we needed to add an activation energy; you seize the activation energy. So that's called an Ea right there, the activation energy. Then the reaction release it, I hope that if you got to straighten out a little bit, the reaction then releases heat to form these products. So we have energy that's released here, energy that's absorbed here, but the difference between the two that is always your delta-H, and the delta-H here of course was negative 68 kilojoules. So that is always - and the delta-H is always difference between where you start and where you stop, but in-between we have a little bit of an activating complex that we have to take into account. Now in that endothermic reaction, the products end up with more potential energy than the reactants have. Remember this has to gain 241.8 kilojoules to turn into that. So we have a higher entropy for the products than we do for the reactants in an endothermic reaction. Okay, but still we need to draw the graph properly because we have to add energy into break bonds and energies released when bonds form. So, we go up to here and we write down the H and O atoms, this is your activated complex again. We had to add activation energy to get to that point. Now, energy is going to be released as well. So even in reactions that are endothermic, energy is still released, but the net amount of energy difference here is a gain in energy between here and here from where you start to where you stop, and of course, that's called the delta-H, the change of energy, which in this case is 241.8 kilojoules to the good, so that's positive. Okay, and that's how you do an endothermic graph. Shortcut method for Hess' law; really it can only be done if you know what the heats of formation are for every compound in the equation. So, it says to you, 'hey, here is your reaction, it's the combustion of methane, but the problem is that, I don't know the delta-H, can you find it for me'? You said, yeah, yeah, give me those equations to add together, I'll manipulate it -- no, you're not given those equations, so then what you're going to do? Well, you find a chart that has heats of formation on them and you take the sum of the heats of formation of the products and you subtract the sum of the heats of formation of the reactants. What you are basically doing there is then taking what are the heats of formation of these compounds because they are forming, and you are taking away from them the heats of formation here because these chemicals really aren't forming, they are decomposing. So, really this formula is, it's the heats of decomposition plus the heats of formation and that of course, gives you the change in the heat. So, take a formation chart and subtract heats of formation of the reactants from the products and that will give you the delta-H. Here is how you do it. So, you take the heats of formation of the products and subtract it f

Advertisement
Advertisement
Advertisement