hello and welcome to this second video on enthropy changes this is just dealing with specifics of hess's law so hess's law is that the enthropy change of a reaction is independent of the route taken if reaction goes from reactants to products in one step or if it goes in many steps the eny change of that reaction is the same it's just the sum of the eny changes of those reactions in that multi-step process and in exams you be asked three main types of H law questions in all of them you're asked to find out the enthropy change of a reaction you're not given the enthropy change of that reaction you're given the enthropy change of other reactions that you have to make a hess's law cycle from the three main types of data that you can be given are enthropy changes of combustion enthropy changes of formation or just some other data that somehow you need to link together to get from reactants to products without going straight there the majority of the questions that come up are either combustion data or formation data occasionally you get one and they call that unfamiliar data those are the really hard questions and for those who really need to understand what you're doing in the process in order to work out what you need to do and in what order you need to put those reactions okay so let's assume this is the eny change we're trying to work out the eny change of that reaction this is the hydration eth to make ethanol there's a few bits of data we could be given we could be given the enthropy change of formation of each of these three chemicals or we could be given the entry change of combustion of the two organic chemicals and I'm going to do the combustion one so they'll give you information they'll give you data and the data they need to give you for combustion is the combustion of ethine combustion of Ethan okay so these data are for the entty change of combustion of ethine it's minus 1,411 and for the entty change of combustion of ethanol minus 1371 K per Mo what we need to do now is combine these data with this equation to work out this unknown quantity with the enty change of reaction with this reaction hydration reaction and the way you do that is to burn everything on the left hand side and to burn everything on the right hand side and put in these data to show what those entty changes would be now hopefully it's obvious if we burn this side and this side we get the same products because atoms on this side have to be the same as the atoms on that side you be completely combust the same ATS then we have to get the same products and those products are 2 CO2 and 3 H2O what I need to put now are the numbers which correspond to these changes on the arrows so that I can use them more easily so this is the entty change of combustion of eth so put Inus 1,411 and this Arrow represents the ENT change of combustion of ethanol so I need to put in this number now remember the answer if we want to know what this changes and H's la says it doesn't matter which way we go that change is always the same so if I go from here to there then the enty change will be the same as if I just went straight there now the trick is if you go with the arrow then the eny change stays the same if you're going in the opposite direction of the arrow then you get the opposite eny change so in this case minus 1411 plus 1371 will give me my answer ofus 40 and minus 40 is the enthropy change of that reaction hydration of ethine to make ethanol doing K's law calculations with combustion data you just combust both sides you write down all the numbers corresponding to the substances that you burnt on this side all the numbers which correspond to the chemicals you're burning on this side and then do these all as minus as you've written them down all of these are positives and then equals do eny changes with formation is actually quite similar so we do one of those now in all of these there's one equation that you don't know the eny change of you're trying to calculate in this case it's the addition of hydrochloric acid to propan to make chloropropane so same as before we're going to need some data this time I'm going to do it with formation data I need to know the enpy change of formation of propan change of formation of hydr chloride and then be change of formation to chloropropane and those are plus 20 for propan - 92 for hydrogen chloride and- 116 for two chloropropane and again we need to combine those data with this equation so we need to form everything on the left hand side in the same atoms as we form stuff on the right hand side remember formation reactions make one mole of a compound from its elements under standard conditions with everything in it standard States okay so both of these are made from three carbon atoms seven hydrogen atoms and a chlorine atom those as elements in their standard states are three carbons is solid in graphite 7 over two hydrogen molecules gives you seven atoms of hydrogen and half a chlorine molecule be the gases in the standard States so same as the last one we need to fill in these arrows with the information which is shown by so on this arrow on the left we're making propan and we're making hydrogen chloride from the elements and we have the information the data you need is plus 20 for propan - 92 for hydrogen chloride and on the right hand side we're just forming two chloro propane that's -6 and in the same way as we did in the last one again if you're going with the arrow so it's this way you still use minus 16 if you're going in the opposite direction of the arrow then you use the opposite side and the same way as the last one we go down and then up to get to the opposite side so I'm going to do minus 20 plus 92 - 116 gives the answer - 44 and so you can see the steps for the formation combustion are quite similar with combustion the arrows go down to the products of a combustion reaction and in a formation reaction the arrows go up that's just because combustion of a compound creates products and so the arrows go in the same direction they would if you were burning these chemicals and in formation they are making the chemicals so forming up and the last type is with unfamiliar data so it's not combustion or formation data it's other reactions some of those reactions could be formational combustion reactions but there could be other reactions which aren't either formation or combustion and what we need to do is very similar to this you need to find a way of getting from the reactants to the products in in a different way so in these cases we go down to the elements and up to the products and in the combustion we went down to the combustion products and then back up to the product of the reaction now with unfamiliar data often there's more steps I'll show you one of those there okay so for this reaction which is nitrogen dioxide reacting with water to make nitric acid and nitrogen monoxide what we need to do is combine these three equations with this equation to somehow get from the left hand side side to the right hand side and the way I'd suggest doing that is by finding something which only appears once in these equations and then linking it to this main equation it's almost like you're playing a game of mix and match trying to get from one side to the other so the first one I'll start off with is the middle one because it makes nitric acid and I've got to make nitric acid so I take that middle equation half it because I'm making two instead of four and then place it here okay so you can see here got the nitric acid being made by nitrogen 52 oxygen and water so that's these all halfed making half of that and then finally because the nitr monoxide doesn't take place in that reaction it doesn't change the enthropy change of this reaction would be half of the eny change of that reaction because it's got half of the reactants making half the products so that's minus 128 K per so now the problem is we got to get from here to here now in this nitrogen diox only appears once here I'm getting rid of nitrogen dioxide so the reaction I'm going to use is the top reaction making nitrogen dioxide on this side this one's making two nitrogen dioxide whereas I have three so I'm going to times that reaction by one and a half so three nitrogen monoxides react with one and a half oxygen to make three nitrogen dioxides the same as the nitrogen monoxide on this side doesn't take place in that reaction the water doesn't take place in this reaction so I've used two of the these equations I've got one left I've got this which is almost a complete cycle but not quite hopefully the difference between this side and this side will also be difference in that reaction so I'm just looking for the differences so here there's water on both of them so that doesn't change on this side there's three nitr monoxides and on this side there's one so my left hand side has two more nitr monoxides and on this side I've got three over two oxygen whereas here I've got five over two oxygen so this side has one oxygen molecule more it's also got a nitrogen molecu so if you look at the actual difference here I've got nitrogen and oxygen and two fuel nitrogen monoxides this side I don't have an oxygen I don't have a nitrogen but I have two extra nitrogen oxid hopefully you can see that's this reaction I react this nitrogen with one of those oxygen I made two extra nitrium monoxides and then the whole thing balances and because it's in the same symetric amounts as that reaction I just put in exactly the same number as that reaction notice the direction of the arrow it goes from nitrogen and oxygen to nitrogen and oxide to get to the unknown same as before I start off with the reactants I go along the arrows if they're going in the same direction I add them if they're going in the opposite direction I subtract them so this would be minus- 174 minus 183 128 and that gives us the answer ofus 137 K per Mo which would be this unknown and that's how I use unfamiliar data to calculate an unknown enthropy change I treat it like a puzzle but I know that these three because those are the three that the exam Bo gave me those three reactions must combine in some way to get from the left to the right and so I find the ones which only imp once like nitric acid and like nitrogen dioxide I start there then once I'd used two of them I knew the last one was going to be the way that these two link together I just made sure that that was correct and I worked out which way the hour should go okay so I hope that helped in your hess's law questions and I hope you join me in the next one goodbye