hello bobcats in this video we'll be just asking how to calculate the enthalpy of a reaction using standard enthalpy of formations so to in here to determine the delta h or enthalpy of a reaction involves i'm sorry involves so determined to determine the delta h of reaction involves the use of standard standard meaning these values have already been determined molar heats of formation of formation so we're forming a compound so it's the formation of a compound and we use delta h this stands for standard temperature and pressure and we'll go over that in just a second and s stands for formation of a substance so what are we talking about when we use these standard molar heats of formation well the standard heat of formation occurs when one mole now a substance is formed from its constituent elements okay in their standard states now a standard state is the um is the most stable [Music] state at which that compound exists at 25 degrees and one atmosphere so the standard state is how the compound exists at 25 degrees celsius and one atmosphere okay and that's what that little zero means it's that standard state okay now an example would be if we know that we have water as uh h hydrogen gas plus oxygen gas going to [Music] h2o gas producing h2o gas and if i look at this or h2o liquid i'm sorry it should be liquid liquid liquid h2o liquid and normally to balance this you would write a two here and a two here so we would have to remember we want one mole i wrote it right up here i underline it i'll even square it one mole of the pr of the compound of the substance so in order to have one mole here i had to cut everything in half so that's where the half a mole comes from well this reaction has the delta h of formation has already been determined it's got a value of a negative 200 and 85.8 kilojoules per mole of that reaction so when we are looking at other reactions that have water liquid water involved in it we can use this delta h to determine the delta h of the reaction this delta h formation and there's a huge list of these so and you will be given either a list or the list will be written on your problem but if you'll see here usually textbooks have an appendix called thermodynamic quantities and those would be the delta h of formation the gibbs free energy of formation and enthalpy of formation we just care about the delta h so the energy involved in formation of let's say calcium carbonate is a negative 1207.1 kilojoules per mole or i can say the energy involved in producing um carbon tetrachloride gas down here would be a negative 106.7 kilojoules per mole if i wanted to do it in liquid form it would be 139 negative 139 0.3 kilojoules per mole if we look at the formation of water let's see if we can find the formation of water where do we have water water should be on here well real quick the formation of carbon dioxide gas would be a negative 393.5 um i'm just i thought water was on the first page maybe it's not and that's the problem with some of these if you've got to find his water so the water liquid is negative 285 water as a gas is negative 241.82 so you got to make sure you have the right state the right state of matter and so you have the values and then we just plug the values into an equation i'm going to show you in just a minute but it's important to understand this idea that we're looking for the formation of a compound from its elements a thing to understand too is that the standard heat of formation for elements is zero for a pure element is zero and it's because the reason it's zero is because l is already formed you don't form an element so you're not going to use energy to form an element so they have zero value but remember we're looking at forming compounds so and the standard conditions of course are 25 degrees and one atmosphere now the deal is uh we got to make sure i guarantee you'll see something like this we need to make sure that we have the correct problem or equation for the standard enthalpy of formation so a lot of times you'll see examples like this i want to go through it because it's so common i want to make sure you understand this let's say which of the following i should do this in a different color and i'm going to change my colors real quick so that we can do it and i know i started but i'm just going to put a red over this so you can see that i'm really going to switch over to red apologize for the color issue it's formation which of the following see i even messed up here which of the following reactions represent a standard enthalpy of formation reaction for to represent a standard enthalpy road formation reaction for no2 gas so let's look at this i'm going to write four reactions here we're going to decide which one's the correct one to represent the standard enthalpy of formation reaction so let's say we have n nitrogen gas plus two moles of o gas to produce n o two gas then we have this reaction to look at no gas plus one half o2 gas produces no2 gas okay or we have n2 gas plus two moles of o2 gas will produce two moles of no2 gas and the last one one half of a mole of n2 gas plus one mole of o2 gas goes to one mole of no2 gas okay so let me explain which which ones are right which ones are wrong and why and then pick the right one so as i go through this remember it's the formation of one mole of a substance from its constituent elements in their standard states well nitrogen doesn't exist as just n as in its standard state it's n2 and this would be o2 so this one is incorrect because the elements are not written correctly here we have nitrogen monoxide plus half oxygen goes to no2 remember it's from its constituent elements this is a compound so this is incorrect based on the fact that it's got a compound in the formation so remember we're not we're not using compounds to form it we're using elements to form this compound here into that's correct o2 is correct and it's balanced correctly but remember it's all per one mole of substance and we have two moles of substance here so that's incorrect so the last one one half of a mole of nitrogen plus a mole of oxygen gives you one mole of no2 this is the correct one because we have one mole of the compound and we have the correct states of and the correct elements that are forming this one mole of this compound so you have to be able to identify uh the correct reactions that represent a standard enthalpy of formation reaction now the next part of this is the equation that we use to determine the enthalpy for of a reaction so to to determine the delta h of a reaction we we use the following equation so in this equation we say that the delta h of a reaction is equal to the sum of the moles times the delta h of formation of the products minus the sum of the moles times the delta h of formation of the reactants now remember these delta h of formation values come from that table and so let's put this is the equation and where most students make a mistake is right here it's always products minus reactants not the other way around it's not reactants minus products and these n values come from the coefficients in the reaction that tell you how many moles of that element or product compound is made or used so let's do a practice problem let's use green so practice problem determine enthalpy of reaction for the following of the following okay here's my reaction we're going to take propane that's cah c3h8 gas plus 5 moles of o2 gas to produce three moles of co2 gas plus four moles of h2o liquid okay and so my reaction what we're going to do is the delta h of the reaction is equal to the sum of the products so i'm going to put a bracket here and we have our products are carbon dioxide and water so i'm going to write 3 moles of carbon dioxide and the value for the carbon dioxide that we determined from this chart was lower co2 carbon carbon dioxide carbon dioxide is a negative 393.5 kilojoules for carbon dioxide so that's going to be a negative 393 point five kilojoules per mole okay the moles cancel plus we're going to sum it the other product which is going to be four moles times and water we said just a minute ago was a negative 285.8 kilojoules per mole okay and then that's all my product so there's my sum and then i'm going to subtract and then i'm going to do the reactants here so i have one mole of the propane and the value for propane c3h8 if we have that that's going to be right here i know you probably probably can't see it on the the the um video but uh trust me that's the value it's a negative 193.85 and so we're around to negative 190 103.9 negative 103.9 so that has a value of a negative 103.9 kilojoules per mole plus remember it's the sum of all the reactants here and so that's 5 moles of oxygen and remember oxygen is an element and so that's going to be 0 kilojoules per mole and there's the sum of that now let's go ahead and so when i sum these two up this times this that's going to be a negative one thousand and one hundred and eighty five kilojoules plus a negative when i multiply these two the four times this a negative 1143.2 kilojoules and then these minus this one times this plus five times zero so that's going to be minus a negative 103.9 kilojoules of energy and so in the end this minus an a becomes plus a positive so it all ends up with a negative two to one nine point eight kilojoules of energy and then and because it's delta h of reaction it's really you can say kilojoules per i know we cancel out the moles here but remember it's understood as kilojoules per mole of the reaction based on how it's written up here oh i don't know why i put a square around that i'm going to put a square around the answer okay so that's how we do this we're going to do one more practice um and you don't have to show all the units because i just wanted to show you the units how we cancel some of this the information there let's do one more practice problem and then we'll be done with this video and in this case i've already got it typed up so i'm just going to pull the typed up one and this is usually how it's presented on an exam or quiz that you've got the information for your your reactants and products available and so we want to know what the delta h for this reaction is so the delta h for this reaction remember is the sum of the moles times the delta h will form ah formation of the products which products minus the sum of moles times the delta h of formation of the reactants okay now so all we do is we take the values and we put them in so the next thing is i've got okay here i've got 2 so i'm going to go 2 times the value of this and that's going to be a negative 2 4 2 negative 124.7 kilojoules plus 13 uh i did it wrong did you guys catch me and this is the same mistake that students make and i made the exact same mistake that students tend to make so we got cross through this because this is a reactant it's always products first products first so we got to do the products first so let's try this again and if i'm showing you how we it's so easy to make that mistake so you're aware of it and will not make that mistake again let's do the product so now we're going to take eight times carbon dioxide which is a negative 393.5 and we don't need to put the units in this time because we know that they're kilojoules per mole and we cancel the moles and then we have 10 times the value for the water which is a negative 285.8 kilojoules per mole now that's the sum of the products minus and i'm going to write products on here so we make sure we do this correctly minus the sum of the reactants and so that's going to be 2 times the value for the butane negative 124.7 plus 13 well the value for o2 is zero and this is the reactants and so our value and i'm just i'm not going to do each one individually i'm just going to show the work here let me show the products and reactants so we've got 8 8 times 393.5 so this is going to be a negative 3 1 4 8 plus 10 times that is a negative 2 hundred and fifty eight minus two times that so that's two times a negative one twenty four point seven so that's a negative two forty nine point four plus zero and so we get three one four eight plus two eight five eight so you don't have to show all this work i'm just showing it so you can follow along and see how it's how we get it so there is the negative 6006 kilojoules and then minus a negative 249.4 kilojoules and so minus a negative becomes plus a positive and so we really have six zero zero six negative six zero zero six negative six zero zero six plus uh two forty nine four enter and so this is going to be a negative 5756.6 kilojoules per mole of reaction and that's my delta h of the reaction using the enthalpy of formation standard enthalpy of formation so if you see the word standard enthalpy of formation this is the type of work you'll do and we'll stop this video at this point and again this is standard enthalpy of formation problems