8.3 Bond Enthalpy | Calculating Delta H | General Chemistry

bond enthalpy gonna be the topic of this lesson and uh bond enthalpy also called bond dissociation energy uh simply just the energy it takes to break a bond and if we keep track of all the bonds being broken and all the bonds being formed we can actually tally it all up and get an approximation for the delta h of a reaction so this means this actually is the third way that you now know how to calculate the delta h of a reaction and we're going to do our best to make sure that you're not confusing this one with either of the other two my name is chad and welcome to chad's prep where my goal is to take the stress out of learning science now in addition to high school and college science prep we also do mcat dat and oat prep as well you'll find those courses at chadsprep.com now this lesson is part of my new general chemistry playlist it's an entire year of general chemistry i'll be releasing several lessons a week throughout the school year so if you want to be notified every time i post one subscribe to the channel click the bell notification all right so bond enthalpy and another way to calculate delta h of a reaction this is actually a little bit annoying because now you've got three different tools that will all allow you to calculate delta h of reaction students often forget that they've learned three different tools and so when they're given a tool and asked to calculate delta h of reaction they don't even realize that they've learned more than one way and don't realize what the formula should be so uh with hess's law you learned how to add up a bunch of reactions and sometimes you got to reverse them which turns the you know makes the delta h of that reaction negative sometimes you had to double them which would double the delta h but however you manipulate all these reactions they have to add up to exactly the reaction you're looking for and then you can just add up their corresponding delta h's to get the overall delta h that's the worst case scenario then you learned about enthalpy of formation which was way better and it was just simple plug and chug products minus reactants done so well the unfortunate thing is this third way with bond enthalpies is going to be reactants minus products and so you really need to understand was i given enthalpy of formation data or was i given bond enthalpy aka bond dissociation energy data because is it products minus reactants or reactants minus products and we'll explain why they seem a little bit backwards and stuff like that but you absolutely need to be sure of which one you've got here so so it turns out that uh uh these bond enthalpies is simply just the energy it takes to break a bond so it is endothermic process to break a bond and absorb you know requires the absorption of energy it is therefore when you form a bond when two atoms come together it actually releases a bunch of energy and is an exothermic process now students don't often struggle with you know the idea that breaking a bond is going to cost energy but they might think that it actually is going to cost energy when you make a bond no no when you make a bond it actually releases energy two atoms come together it lowers their energy and the excess energy is given off all right so the bond enthalpy is specifically describing the energy it takes to break a bond and so that's uh and we can use those because basically you know in a chemical reaction we just have a whole series of bonds that are broken and formed and if we just kind of keep a tally of how much energy was absorbed in the ones we broke and how much energy was released and the ones that we formed we can get an approximation of delta h now it turns out it really is just a really good approximation it's not going to be exact it turns out that the uh the bond enthalpies we give you here like for breaking a typical ch bond is 413 kilojoules per mole breaking a typical ccl bond is carbon chlorine bond is 328 kilojoules per mole a typical hydrogen chlorine bond is 431 kilojoules per mole and breaking a chlorine chlorine bond is 242 kilojoules per mole so it turns out these are averages so and it turns out i mean the only chlorine chlorine bond that exists is just one in a molecule of cl2 so that one's actually kind of kind of be fixed but you know for a ch bond that can you know show up in a whole different variety of molecules and all these molecules are a little bit different and based on how they're different it will actually alter that number ever so slightly and so it turns out that this number specifically is not very specific it's kind of an average of a bunch of situations so when you actually do use these bond enthalpy values to calculate delta h you're getting a really good approximation but you're probably not getting an exact answer whereas when you did like hess's law or used enthalpy of formation values those you are getting exact values um so that's kind of the deal you should know that going uh going into this up front here so all right so since the bond enthalpy is defined as the energy it takes to break a bond so we can actually uh where do you break bonds you break them in the reactants and that's why we're going to add up all the reactant values but you don't form the bonds that are present in the products you actually i'm sorry you don't break the bonds that are present the products you form them which is the exact opposite of breaking and that's why we subtract the bond enthalpy because the bond enthalpy would be for breaking it so just change the sign and that's what it would make for forming it now notice this means you don't change the sign yourself the subtraction here is going to change the sign for you that's kind of the way this works now it turns out though if we're going to look at all the different bonds we have to be able to draw out the lewis structures for all these lovely compounds and so that's why we couldn't teach you this way back in chapter five we had to first teach you how to draw these lewis dot structures in this chapter so if we take a look at ch4 here and i highly recommend you just draw out your lewis structures so you can kind of keep track and look at all the bonds being broken and formed and if you've got more than one molecule here's just a one to one to one ratio but i might even recommend drawing each separate molecule out all right so this is going to form ch3cl and then plus hcl cool so i'm not focusing too much on actually drawing out the lewis structures i'm giving them to you here but uh get some practice in on that last lesson because this is another reason you need to be really good at drawing lewis structures here so all right so if we take a look at what's going on here it turns out you can do all the bonds in the reactants minus all the bonds and the products and that'll get you a delta h good approximation so however we'll find out that you can actually simplify your calculation in many cases and just do the bonds broken minus the bonds form we'll find out if you do all the reactants minus all the products in an example like this this will be rather redundant and kind of a longer way to calculate this but we'll do it both ways here so if we look at this we can just do all the bonds in the reactants and in this case we got one two three four of these ch bonds and so in this case that means delta h here of the reaction is going to be four times 413 for the ch bonds and then we have a c clcl bond so that's one just one of those and so that's 242 kilojoules per mole and so there's the sum of all the bonds in the reactants and then we'll subtract the sum of all the bonds and the products and again we don't change the sign the subtracting part of this changes the sign for you so just use the the bond enthalpy values right as they are in the chart here and it's just reactants minus products so here we've got three of these ch bonds again so that's going to be 3 times 413 plus now a single carbon chlorine bond so 1 times 328 and then an hcl bond so plus 1 times 431 and we've just got reactants minus products cool and we'll work this out in a little bit so however if you notice we had four ch bonds over here that we're adding in but we still had three of them over here which we're now subtracting off well if we're if three out of these four got added in only be subtracted off from the get-go well then why not just leave it out of the equation and out of the calculation it's just redundant a little longer way to go and that's what doing bonds broken minus bonds formed actually lets you do so if you can recognize quickly which bonds are being broken and which ones are being formed it'll shorten up the calculation for you now if it's too difficult to notice which ones are being broken and formed you can just by all means add up all the bonds that are present in the reactants and all the bonds that are present the products and just do reactants minus products but again really important it is reactants minus products and the reason is because it's in the reactants where we're breaking bonds and if we actually keep track of this so this ch bond is one of the ones that's broken i see that these three ch bonds are still present on the reactant side and the product side and that's how you recognize what's broken and formed is just compare your reactant to your products so react dents to your products and so that's one that's broken but this chlorine chlorine bond isn't present over here either so that one's got to get broken as well and so on the reactant side you find the bonds that are broken and that's why we add in the bond enthalpies of the reactants which are the ones that are broken and again that's what a bond enthalpy is it's the energy it takes to break a bond now on the other side though we're actually forming bonds and we can see again by comparing the products to the reactants what's the new bonds that are present well in this case we've got a new one forming right here that's the ccl bond and a new one forming right here that's the hcl bond and again we're not breaking these bonds we're making them and so notice for a carbon-chlorine bond to break it it would cost kilojoules of energy well when you form it it actually releases 328 kilojoules of energy but again you're not supposed to change the sign yourself that's what subtracting is going to accomplish for us so when we do reactants minus products the sign has already been changed for you one of the biggest mistakes students will make is they'll subtract here but also then change this to negative 4 13 and negative 3 28 negative 3 4 31 don't do that if you're subtracting like the formula says that's changing the sign for you all right now if we just wanted to then do bonds broken minus bonds formed it actually shortens up the calculation quite a bit here notice uh instead we just have the one ch bond and so we just end up with one times 413 and the 1ccl bond so 1 times 242. put that in brackets minus and notice we don't even have to include these three bonds because they're not broken or formed and we just jump right over to here so 1 times 328 plus 1 times 431 so which is definitely a much simpler calculation but again whether it's simpler to recognize what's being broken and formed or simpler just to do all the reactants minus all the products i will leave that in your hands and how comfortable you get by the time your exam rolls around so but if we do this here we're gonna do 413 plus two and i'll do the bottom one here and i lost a parenthesis there so 413 plus 242 minus parentheses 328 plus 431 equals and i'm going to get negative 104 kilojoules here and that is an approximated value of our delta h of reaction now if we had calculated out the delta h of this reaction using hess's law or enthalpies of formation we probably would have got a slightly different value than this that would have been a more accurate number but again this is still a really reasonable approximation for the delta h here and we see being negative that this is an exothermic reaction let's do one more example all right so here's our second example here and this is going to demonstrate a couple of helpful things for us that the last one didn't do and so one we can take a look at a couple of trends here so like if you compare the nh bond to the oh bond 391 versus 463 kilojoules per mole you'll notice that the oh bond takes more energy to break it's a stronger bond so and what you learn is that if you're going across a period like if i was to compare ch to nh to oh to fh so that as the other atom besides hydrogen gets more electronegative the bond gets stronger now it turns out size and electronegativity both play a role in the strength of a bond now turns out size probably has more of an impact but if you're all in the same period they're fairly close in size and so electronegativity ends up being the determining factor now if you were to compare the ones in a in the same group though like if i compared hf to hcl to hbr to hi so then you'd find out that size plays the bigger role and the smallest bond would have been the hf bond which would make it the strongest bond it would have had the highest largest bond enthalpy and so if you're in the same group and you're comparing all the same bond to the same atom like in this case all bonds to hydrogen from each of the halogens size is the determining factor and the smallest or shortest bond uh is the strongest bond it has the largest bond enthalpy but if you're comparing bonds in the same period like a ch c-h-n-h-o-h-f-h electronegativity is the determining factor and the more electronegative atom with that hydrogen in this case would be the stronger bond uh and the larger bond enthalpy so you should also assume that you know typically single bond double bond triple bond like a nitrogen nitrogen single bond is going to be weaker than a nitrogen nitrogen double bond which is going to be weaker still than a nitrogen nitrogen triple bond and so usually the more bonds you have between the same two atoms the larger the bond enthalpy i say usually pretty much always it's going to give you a larger bond enthalpy however it's not like you know a double bond is always going to be double the value of a triple bond i'm sorry double the value of a single bond or that a triple bond would be triple the value of bond enthalpy as a single bond we see that's not the case here like your nitrogen nitrogen single bond is 163 so but your nitrogen triple bond is way more than triple that it turns out and so uh you know that's going to have some important implications when we actually do the calculation here so something we'll definitely have to keep track of so let's take a look at this example now that we've kind of looked at those trends here so once again we've got to draw some lewis structures here and uh once again also since i've got two waters in this case i highly recommend drawing both of them out as i'm about to do here all right you'll find out that i kind of made them with the angles that i did based on some things we'll learn in the next chapter about molecular geometry so but this is how all the bonds would work and uh in this case couple things so one if you try to go the whole broken versus formed thing you'd find out that all the bonds in the reactants are being broken and all the bonds in the products are being formed sort of what's really happening though is that these nitrogens are probably bonded to each other the whole time and we just end up with two additional bonds however with the way bond enthalpies are structured you can't just like well let me just add the extra two bonds here and we know the nitrogen nitrogen single bond so i'll just double that and add those two extra bonds it doesn't work that way when you go from a single to a double or a double to a single or a single to a triple or triple to a single or between the same two atoms you actually have to add or subtract them in for the values for that particular bond you can't just like try to do the extras so in this case if we look at all our bonds in the reactant side here we've got one nitrogen single bond that's going to be 163 and then we've got four of these nitrogen hydrogen bonds which are going to be 391 and then we've got the oxygen auction double bond which is going to be 4.95 and then we'll subtract off all the products we've got a nitrogen triple bond for 941 so and then we've got four of these oxygen hydrogen bonds which are 463 each cool and from here we'll let our calculator do the work for us so we've got 163 plus 4 times 391 plus 495 minus parentheses 941 plus 4 times 463 parentheses and we're going to get negative 571 cool and again there's a good approximation of the delta h for our reaction and one more time one again i drew out both waters a lot of students will make the mistake of drawing only one and instead of getting four of those oh bonds they only account for two of them so be careful with something like that also don't confuse this with enthalpy of formation value and again instead of bond enthalpy if this had said something to the effect of like delta h sub f those are enthalpies of formation and if you're given those values in a table that's when it's products minus reactants and the idea is that in an enthalpy of formation it's for something being formed well what's actually getting formed the reactions to the products well it's the products that are getting formed the reactions are getting unformed and so it's the products that actually get added in but we have to change the sign for the reactants because they're not getting formed they're getting unformed the exact opposite and that's why for enthalpies of formation it's products minus reactants but again with bond enthalpies it's in the reactants that the bonds are being broken and so with the way abundantly defined as the energy it takes to break a bond that's why it's reactants minus products you really need to have a good handle on that you really need to understand all the different ways you have of calculating delta h here these three different ways both uh bond enthalpies as we're doing in this lesson reactants minus products enthalpies of formation products minus reactants or hess's law which is the worst where you got to just like you know add up all the reactions and flip them around and double them and cut them in half whatever it does to add up to the reaction that's desired now if you found this lesson helpful would you consider hitting that like button best thing you can do to make sure youtube shares this lesson with other students as well and if you're looking for practice if you're looking for quizzes and chapter tests and practice final exams check out my general chemistry master course i'll leave a link in the description a free trial is available happy studying

Channel: Chad's Prep

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Keywords: bond enthalpy, bond energy enthalpy of reaction, bond energy calculations, bond enthalpies, bond dissociation energy, bond energy chemistry, bond energies, average bond energy, enthalpy change using bond energies, enthalpy of formation and bond energy, bond enthalpy and enthalpy of reaction, bond enthalpy in thermodynamics, bond enthalpy problems, bond energy calculations questions, enthalpy of reaction from bond energies, bond dissociation energy calculation

Id: G10ZEQNYSQQ

Length: 17min 43sec (1063 seconds)

Published: Thu Oct 28 2021