good afternoon bobcats today we will be discussing hess's law now since enthalpy change or delta h which we call delta h is a state function and does not depend on the pathway we can sum up the enthalpy changes of several chemical equations so we can sum up the delta h or enthalpy change for a series of chemical equations um to find the enthalpy change of an overall reaction so chemically to find the enthalpy change of an overall reaction um regardless remember because it's a state function it's not a pathway we can do this regardless of whether those steps are actually involved in the reaction mechanism so let me state that again so that we understand we get an idea so because enthalpy change or delta h is a state function and it does not depend on a pathway we can take several different chemical equations with delta h values and add them together to get the delta h value of an overall reaction and it doesn't matter whether those steps were actually involved in the reaction or not because remember it's a state function and it doesn't matter about the pathway and in general we can say that the delta h or enthalpy change of a reaction is equal to the enthalpy change of reaction one that was used plus reaction two plus reaction three and so on however many reactions are involved in that uh uh determining the enthalpy change so to give you an idea of what we're talking about let's say i want to know the enthalpy change of this reaction we have nitrogen gas plus two moles of oxygen gas goes to two moles of nitrogen dioxide gas and we want to determine the delta h of this reaction okay so and i'm going to underline it as if so we know that that's my original reaction i'm looking at but we're given these reactions to work with so we have nitrogen gas plus oxygen gas produces two moles of nitrogen monoxide gas and that has a known value delta h value of a positive 180 kilojoules and then we also have this reaction that we can work with two moles of nitrogen monoxide gas plus oxygen gas goes to two moles of nitrogen dioxide gas and this has a known enthalpy change value of negative 112 kilojoules so now what we need to do is we need to add these reactions together so when i do that i'm going to add these two together and when you add reactions together you can cross out the the reactants and products are on this um if i have a reactant on one side and i have a product as a reactant i can cross them out so if you look at this i notice that two nitrile monoxide gas two moles of nitrogen monoxide gas is a reactant we also have two moles of nitrogen oxide gas as a product so i can and you can kind of think of as we're adding these together and so and those are gone they cancel each other it's just like if you have a 2 on each side of an equation you can cancel it or if you have an x on each side of the equation you can cancel it and then we add them up so when i add this we have nitrogen as a product plus and now there's an ox one mole of oxygen here plus one mole of oxygen here so that's two moles of oxygen gas producing two moles of nitrogen dioxide gas and my delta h for this of the reaction is just i add the delta h of 1 plus the delta h of 2 which is 180 plus 1 negative 112 and so i get a positive 160 kilojoules of energy again i just add 180 plus a negative 112 and that gives me a positive 68 kilojoules per mole of kilojoules for this reaction okay now that is simple we didn't have to do anything we didn't have to change or manipulate any of this we just had to cancel what was on the same on both sides just like we do with ions and in spectator ions and so forth but what if we sometimes you have to manipulate or rearrange equations to get the um equation you want uh to find the the enthalpy change four so the next thing we need to look at is this let me do it let's use this page still since i still have this page let me draw a line underneath okay so given reactions remember these are the given reactions so given reactions may have to be manipulated or rearranged to in order to get the desired reaction now when we do this there's a few things that we have to be aware of oops reaction okay desired reaction first you need to decide how to arrange how to rearrange equations so that reactants and products are on the appropriate side next what you have to do is if the equations are reversed the sine of the delta h must also be reversed okay two more things we need to consider before we practice again and let me so first we have to rearrange decide how to rearrange the equation so that the reactance products on the appropriate side if you have to reverse the equation a positive delta h becomes a negative delta h and vice versa the if the equations are multiplied to get correct coefficients then the delta h is also multiplied okay and that also applies for dividing so division is similarly applied okay and then last what you should do is you should check to ensure everything cancels out cancels out to give to give the exact equation you want that includes states of matter and everything so now what we need to do is practice do a couple practices to make sure that we can do this so let's do a practice problem right here so i'm going to use a couple of different colors and show you the process at which i do this so with this practice one we'll start out with two moles of sulfur trioxide gas decomposes to make two moles of sulfur dioxide gas and one mole of oxygen gas o2 gas okay that's o2 and we are looking for the delta h of this reaction what is the delta h of this reaction okay so we're going to have a given so this is what we're looking for given we are given a sulfur solid plus o2 gas goes to sulfur dioxide gas and it has a delta h value and we'll call it reaction 1 delta h value of a negative 296.8 kilojoules and then the other given reaction is sulfur solid plus one and a half or three halves moles of oxygen gas produces sulfur trioxide gas and that has a delta h value of negative 3 95.6 kilojoules okay so now i'm going to have to manipulate these so then i i do a whole nother row of this because it's really hard to keep track of if you if you're not very careful with it so then this is uh how do we want to call this we'll say this is the rearranged equations rearranged okay so in this i notice that i have sulfur dioxide on the product side i need it on the product side because i want it on the product side so i'm not going to flip it but i need two moles of it this the given one only has one mole so i'm going to multiply everything by two all the way across so it's going to be 2 sulfur solids plus 2 oxygen gases produces 2 moles of so2 gas and then i also have to multiply my value i'm not going to put the units in right now we're running out of space now that matches so if you'll notice i've got 2 moles of sulfur dioxide gas on the product side i now have two moles of silver dioxide gas on the product side let's look at the other one i need my sulfur trioxide here it's a product i need it on the reactant side so i have to flip it and not only do i have to flip it but i also have to multiply everything by two because i want two moles and only have one mole so i'm going to flip and write it multiply by two so that's two so3 gas produces 2 moles of sulfur solid plus and 2 times 3 is six and six divided by two is three so that's going to be three moles of o2 gas and in my delta h for this reaction of course i had to multiply everything by two and only that though i had to change the sign so now it becomes a positive it was negative now it's positive okay and so again i've got the correct product i mean reactant and the correct product okay so i'm going to start canceling and i'm just going to kind of do a division line here so you can keep them apart from each other okay now um to cancel and add it up so i'm gonna add it up here so and it's a kind of a plus sign not that you really would show that but i'm just kind of showing you that we're adding it so i've got two sulfurs on the reactant side and then two sulfurs on the product side so those two cancel and then i have two oxygens on the reactant side i have three here so these two are canceled and the three goes to one so we can say we canceled the three so it's just one oxygen and then nothing else can be cancelled so now i write it down so on the if you look at it i've got these are products right this this and this these two are gone so my only product left a reactant sorry my only reactant is so3 gas and it goes to well we've got a product a week product here two so2 gas and then i look down here and i have one mole of o2 gas and does that match the original so i go back up and i'll let 2 so3 2 so3's 2 so2 2 so2 1 oxygen 1 oxygen and now i just for the reaction delta h of the reaction i just add and multiply and add those up so that will come out to be so i have 2 times a negative 2 96.8 and then 2 [Music] times a 395.6 and then i add those two values up and i'll just write it down here so you can see that what we're looking at is the delta h of the reaction is equal to 2 times a negative 296 0.8 plus 2 times a positive 395.6 and if you this this is a negative 593.6 plus a positive 791 [Music] so my delta h of this reaction is 791.2 minus 593.6 and i get a positive 197.6 kilojoules per mole and usually i don't show all of this work i just do it in a calculator and add those two values up and just write it down over here okay so this is the process remember we have the original equation right here this is my original equation that's what i'm looking for i have the two given equations and then i'm going to call this area the rearranging or manipulation equation so i make sure that the this so2 is on the same side so i don't flip it but i have to multiply everything by 2. so i do that i also multiply at 2. this my so3 is on the product but i want it on the reactant so i flip it and then i also need two moles so i multiply everything by two now we're going to do one more practice problem for this one and it's going to be just a little bit more you notice how there was two given reactions here the next problem is going to have three given reactions so let's try to keep track of three given reactions now let's see the original equation is going to be nitro dinitrin monoxide gas plus nitrogen dioxide gas produces three moles of nitrogen monoxide gas and we want to know what is the delta h of this reaction don't know it so we're looking for it so i'm going to just underline it to show you that this is the one that we want to make sure we end up with now here's the given so the given ones we are given uh nitrogen gas plus oxygen gas growth produces two moles of nitrogen monoxide gas and the delta h value for that is equal to and we'll call that reaction one a positive 180.7 kilojoules the next reaction is two moles of n o or nitrogen monoxide gas plus one mole of oxygen gas produces two moles of nitrogen dioxide gas and that has a delta h value and it's a reaction two so a negative one [Music] 13.1 kilojoules and then my third reaction that's given is two moles of nitrogen dioxide gas produces two moles of nitrogen gas plus one mole of oxygen gas [Music] and it has a delta h value of a negative 163.2 kilojoules okay so those are the given ones now i'm gonna have to manipulate them so i'm gonna it's easier if i just do a whole nother step where i manipulate it and so we're going to call these the rearranged or however you want to call it rearranged ones okay so now we've got um if i notice here i've got the nitrile monoxide and the nitrogen monoxide but i'm not i know it's a two and a three but i'm not going to multiply by anything because i also know that i have a nitric monoxide over here and so i may end up getting the number i want without having to multiply this so first i'm just going to rewrite it since i'm not going to flip it it's on the right side the correct side so it's going to be nitrogen gas plus oxygen gas produces two moles of nitrogen monoxide gas and my delta h didn't change for this at this point a positive 180.7 kilojoules now the next reaction let's see how we can manipulate that one i know that i need the see the no2 as a react product here up here no2 is a reactant so i need to flip this not only that though this only has one mole this has two moles so i have to divide everything by two or cut it in half so i'm going to flip and divide by two so when i divide by two and flip it that'll be n o two gas one mole instead of two moles produces sorry um one mole of n o gas plus one half of a mole of o2 gas and the delta h for this and i'm trying to keep my arrows all straight as possible my delta h for this one i had to flip it so it becomes positive and then i divide by two so when i flip and divide by two i get a positive 56.5 kilojoules of energy for that reaction and then the other reaction in o2 well i've got it on the correct side the n2o here the nitrogen dioxide the dinitrogen monoxide gas is a reactant in an original equation it's a reactant so i'm not going to flip it but instead of having two i only want one so i'm going to cut everything in half again so we're going to have n 2o gas produces oh i forgot to line up my arrows sorry about that produces one mole of n2 gas and one half of a mole of o2 gas one and then my delta h or my enthalpy is going to be well i didn't flip it so i keep the same sign but i have to cut it in half so it's going to be a negative 81.6 kilojoules now i'm going to add all of these up you can think of it as adding it up and so i have to cancel everything okay so let's go through and start canceling i have an n2 as a reactant and this into the product and it's only one mole so that cancels i have one mole of o2 here as a reactant and then a half of a mole and a half of a mole makes like a one mole of oxygen as product so this one cancels both of these and then we know that um if we look i don't i can't cancel anything else that's okay that's good everything is good there so now i'm going to write it out as if to see if it matches the original so these are gone this is a product here and this is a product so i'm going to write it as n2o i mean react i said product i meant reacted i'm sorry plus no2 as a reactant produces n2o uh two moles of no here and one mole so those two add up to make three moles of no gas right two moles here one mole here that adds up there's no other reactants let's see if this worked one mole of n2o one mole of n2o one mole of no2 one mole of no2 three moles of no three moles of no and so my delta h of the reaction and all i'm going to do is just add these up and so that comes out to be a 180.7 plus 56.55 minus 81.6 and i'll get a positive 155 but we're gonna round it to seven kilojoules and that's the answer and that's how you go about doing it and so first you have the original equation then you have the given ones with their values and then we have to manipulate or rearrange the equations to match the original and then cancel or you're adding them but by adding you're canceling everything and then just adding these two and then adding those two to get the product and then you add the values i'm just going to kind of draw a dotted line here so you can see that there's a difference there okay and that is hess's law and i will see you next video