Hess's Law

this video is about Hess's law now this video to be able to use Hess's law to calculate enthalpy changes using combustion to divide the users' law to calculate changes using the formation of these and new tests Calculator changes Lord using bond enthalpies now notice that you are using Hess's law to calculate the enthalpies using these types of enthalpies so it is about the type of enthalpy you are using rather than what you are trying to calculate and that becomes important and when you look at the methods of how we use them, then I think the best way to explain to you how to do this is to use some past exam questions to show you, so the first one I'm going to show you is where we use enthalpies of combustion to calculate the reaction. of something now in this question it says that the following equation shows the formation of beuter 1/3 die in c4 1/6 anyway this is the equation for the formation of c4h6 and it gives you some dates that are here to calculate that standard enthalpy of formation and the first step is to realize that each of these reactions are enthalpies of combustion.

Now generally they are the enthalpy. Check the enthalpies you should use to calculate the reaction of P are given in a table and in a table the combustion of P would be You have to symbolize this so if there is a table that is the first place you would look, see what enthalpy changes you are using and if you are using enthalpies of combustion, then you would be given this symbol, but here you have been given the reactions that These enthalpy changes relate to and you can see that each of these are four piece combustion, so in the Higher you have carbon reacting with oxygen to produce carbon dioxide, so this is the combustion of carbon and that entropy change is the standard. enthalpy of combustion of carbon, this is for hot hydrogen to produce water and this is for c4h6, so we are using combustion 3, so the first thing we need to do is write the reaction at the end for the enthalpy change you are dealing with to calculate.

So that's the top reaction there, so 4 c + 3 h 2 is going to make c4h6. The next step is basically right at the bottom what would burn these two things and burn this and if it burns, provide either 3h2 completely oxygen or burnt. c4h6 completely oxygen, you do the same thing and what it would be is for carbon dioxides and three waters, so the enthalpies of combustion are where we are producing this carbon dioxide and water from these things, so on the arrows we go to draw in our cycle. they will all point down first, so if you are using combustion for peas or on your arrows at first they will point down, then you will put the numbers from your question on your arrows so that there are four carbons, see you all on this first day. four times minus 3 9 4 which is equal to minus 1 5 7 6 on the next arrow you will have three times the hydrogen 1, so 3 times minus 286 and that is equal to minus 8 5 8 and the last one will be 4 c4h6, that is minus 2 5 4 - there is that number now, right now all your arrows are pointing down and what you want is a path from your reactants to your products through this carbon dioxide and water, so you want these first two hours go down and then you want to travel back to your c4h6, so what you have to do is swap this arrow, you have to reverse this arrow to make it go up again, so you swap this air to make it go up and by doing so you will change the sign of the 2 5 4 - 2 a + basically, if you swap mara, you have to reverse the sign so it's the opposite of what it was and now all you do is add the 3 as it is on your arrows so your answer is like this.

Delta H will be equal to -1 5 7 6 - 8 5 8 + 2 5 4 2 which will equal + 108 kilojoules per mole. Now in the lesson I have told you that what you write here doesn't matter and no No as long as you get the answer right, so most 108 Virgil of all, even if you just wrote that here without any of your work, you will get 3 points , but if you don't do it right then your job becomes important and if you don't write the right thing here so balanced combustion products won't get the marks for your method so if you are sure you can do this correctly , there's no need to worry about making sure you have the right thing here. but if not, make sure you have the correct combustion products if you are using enthalpies of combustion in your cycle, so in the next question you will see that we have another question that says that methoxy methane burns completely an arrow in an equation there is, it says to use the standard formation abuse given in the table below to calculate a value for stella through the combustion of mimi fox anything, the most common mistake is to see or calculate an enthalpy of combustion and follow the process I showed you on the previous slide, but like I said, the first step for this question is to look at what they're using and you can see that instead of burning the peas, they're now using training pups that have the following symbol delta hf, so again your first step is to draw, write the equation for the reaction you are using, whatever you are trying to solve, sorry, you have ch3 or CH 3 + 3 or 2 going to Make 2 co2 + 3 H 2 O again at the end of that cycle, we're going to write down what that substance is going to form, so we're going to have carbon plus hydrogen plus oxygen, we just want to balance it out, so if we use our right hand. size balance if there's going to be two carbons, there's six hydrogens, so you're going to need three h2 and there's four five six seven oxygens, you're going to need three and a half oxygen, the same as 7 over 2, so what are we doing this time? using a formation of the pieces, these formations represent the reaction in which you are making these things from these things, so if you are using a trapezoid formation, this time your arrows will go up to your reactants and products first, whereas if you are using Combustion of the piece now goes down first, if you are using information about the piece they will go up first, that's the only real difference when solving them, so now you put the numbers on the arrow so you see that 4 1 CH 3 or CH 3 which is minus 185 so you put minus 185 on this arrow oxygen has zero formation and that's because it's an element so it has all the elements that are in their standard states have zero formation by definition and that's basically because you're creating an element from an element that didn't consume any energy because if you want, it's already made, but if they ask you why it's hero, you just say by definition source/zero on that arrow, we have two carbons.

Each carbon dioxide has a P formation of minus 394, so we have to put two x minus three and nine four here, which equals minus 788 and this in the spine, Larry, you have three times water, so this once you got three times. 286, which is equal to negative 8/5 8 now, at the moment when all the arrows are pointing up, there is no root from the reactants to the products, oh, we have to cut a switch, two of these areas around we have a work to determine what we have to do. go from these reactants down here and back to the products, so we want our path to go down and then back up.

That's the same for all Hess's law questions, so you want to go down and then back up, so it's these first two arrows that we need to change this time, so we change this air to point down and in doing so we need to make sure we change our negative sign at 185 to positive and we're going to change this one down as well. but that doesn't really matter because they are positive or negative, they are the same thing, so all we do is add these three numbers and again we have the answer, so Delta H will be equal to 180 5 minus 7 8 8 minus 8 5 8 , which will give you minus 1 4 6 1 kilojoules per mole, there is an exothermic reaction, you know this because it has a negative enthalpy change, which makes sense because you are burning something, so again you will get all three marks if you get this number correctly . minus 1 4 6 1 but if you don't get that number, you will only get method marks if you balance this lower bit correctly in this cycle, so another way to calculate reaction enthalpy values ​​is using bond enthalpies, so again Here we have a table with some average bond enthalpies of some bonds and remember that the bond enthalpy is basically the amount of energy that you need to put in to break one mole of this bond and it takes that value over your average of 11 different ones. compounds and we have an equation here, so we use the same cycle method, so first write the equation, the chemical equation, it has half m2 plus 1 and 1/2 f2, which makes em f3 and when you link, the Enthalpies basically mean breaking these bonds. so if we break these bonds, we're going to form n atoms, F atoms and that's it, sorry, we just want to balance that out, so 1/2 n 2 is going to form 2 n atoms and 1 and 1/2 f2 is going to form a make 3 F atoms so just determine where the arrows will go first there the arrows will go down first because with the bond enthalpies we were creating these larger atoms by breaking the bonds up here so for the bond enthalpies the air arrows they also go down like they did for combustion for peace, then you open up the numbers and you get 1/2 n2, so there is your number n2, so it's not 9 4 5, so it will be 1/2 multiplied by 9 4 5 , which is equivalent to four hundred seventy-two point five.

I have one and a half one point five times one five nine for fluorine which is equal to two three eight point five and finally what nf3 and in nf3 you have three NF bonds now actually I always say in lesson two draw all your reactants and products with all the bonds shown so you can remember to count all the bonds there are and in one molecule out of three you have three bonds left you need two times this 2 7 8 times 3 3 times 2 7 8 and that equals 8 3 4 now you have to think about which one you are going to move or which arrows you are going to talk about remember that you have to go down from the reactants to the product, so it will be this arrow that we swap so that this arrow goes up instead of down, we change the sign in our 8 3 4 to minus a 3 4 and then our Delta H will be the added ones, so it will be four seven two point five plus two three eight point five Oh, someone else didn't write, I write plus two three eight point five negative eight three four and that's equal to negative one two three kilojoules per mole, so another question that you might be asked when using Hess's law and bond enthalpies is to work Calculate the average bond enthalpy of one of the bonds using the p reaction and the mean bond enthalpies and you can do this using a cycle again, so I'll do it again so you have h2 Pissarro to make h2o. so this time I'm going to use the advice I gave you in the lessons and write all the links so that you have h2, so that only one H joined to one H plus half of O 2 is going to form h2o and only one of them and from all of this, if you break the bonds you will make with h plus oh, and remember your arrows with middle bonds in which the enthalpies go down first, so on these arrows we can write the numbers that equal those in p, so to a link oo We have 496, so it's going to be 1/2 multiplied by 496 and that equals 2 times 8.

You have to look at the links in this, so 2 multiplied by 4 6 3, which equals 9 2 6 and on this side for h h1, that's where I try to figure it out, so I'll leave that as this will become negative 9 to 6, which means that these three numbers 2 4 2 is equal to x plus 2 4 8 minus 9 a 6 so alone, which means that X will be equal to plus 4 3 6 / joules per mole. Now if you are trying to calculate a bond enthalpy, they will almost always be positive because you are breaking a bond and therefore that takes energy. For this to happen, if you end up with a negative bond enthalpy, then you've probably done it wrong.

One of the things you may have done wrong is that you have to remember that when your arrows go down, your X values ​​will be Positive, but if you were to change a time, then if your of the X, it would become negative, so you should make sure to change the signs if necessary for like your numbers and remember that your bond enthalpies are positive for arrows going down and will be negative for arrows going up again, so just to recap the most important points, remember that Hess's law and cycles depend on the enthalpies that you are using, not what you are trying to calculate the enthalpies that you are using, you will look at the table that they give you or you will look at the reactions that they can give you, like in the first example I reviewed, and you will determine if their combustion formation of enthalpies for peas or bond enthalpies if they are combustion for peas h delta c will be their symbolif your formation the piece H Delta F will be your symbol and if your bond enthalpies will only say average bond enthalpy if given Your reactions instead of a table, then remember combustion from PISA, where you are burning a mole of something in oxygen to convert it completely into oxygen.

If you're forming something, you're forming the product from its constituent elements, they should be fairly easy to detect. if you are using commercial each, your arrows go down first, you are using formation, peas, your arrows go up first and if you use medium bond enthalpies, your arrows go down first and then you have to swap the appropriate arrows so that the first reactive arrows go down and the product arrows go up to move from reactants to products through the middle step of your cycle. If you have any more questions about Hess's law, be sure to send me an email and I'll give you a happy antral