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Hess's law and reaction enthalpy change

Video transcript

now that we know a little bit about a heat of formation and enthalpy change and what enthalpy is we can talk a little bit about Hess's law Hess's law and what this tells us is that the energy change of a process is independent of how we get from one state to another and it's it's it and really that's a byproduct of the fact that energy is a state variable whether we're talking about enthalpy or internal energy their state variables and we've talked multiple times that it's independent of how many steps it takes to get there or what path you happen to take but how is that useful to us when we're dealing with everyday reactions so let's say let me just make up some reaction where I have or I have a plus B is yields oh I don't know let's just say this yields C plus D and I wanted to figure out what was the change in enthalpy of this reaction or essentially how much heat is absorbed or released by this reaction I don't know what it is I don't I haven't measured it and all I have are the heats of formation so all I know is how do you go so I know the heat of formation of a so let me call that the heat of formation and remember H isn't for heat even though we kept calling it heat of formation it's actually the change in enthalpy and it's the standard change in enthalpy but the change in enthalpy we know is Heat so it's heat change in enthalpy of formation is same thing as heat of formation this little knot sign tells it's the standard heat of formation we can look up that in a table and let's say that that's some number and then we have our heat of formation of B Delta heat of formation let me call it of B this is heat of formation of a and it's a standard heat of formation and we can look up on a table the heat of formation of C heat of formation of C just change in enthalpy and then the heat of formation for D so all of these things we can look up in the table right and I will do that in a second now what Hess's law tells us is that the change in energy the change in an enthalpy is what we're measuring here the change in enthalpy here is independent of what we're doing so instead of saying this reaction we could say hey let's go from this react and go back let me do it in a different color let's go back to our constituent products so kind of the elemental form of these so you know if this was like carbon dioxide you'd be going back to the carbon and the oxygen molecule so you'd go back to the elemental form and how much energy or what's the change in enthalpy is you go back to the elemental form the heat of formation is what you get from the elemental form to a or the elemental form to B so to get a and B back to the elemental form it's going to be the minus of those you're going to take the reaction in the other direction so it's going to be so that this change is going to take minus Delta the heat of I guess a forming a we give you the minus or the heat it's the heat of deconstructing a you can almost view it and it would also be minus the same thing for B for B and then this is just the elemental form and now we can go from the elemental form back to the products because we have the same atoms here they're just rearranging themselves into two different sets of molecules so now we can go back from the elemental form and go up here and we know what those are we know how much energy it takes to go from the elemental form to C and D that's their heats of formation so Hess's law tells us that Delta H of this reaction Delta H of this reaction the change in enthalpy of this reaction is essentially going to be the sum of what it takes to decompose these guys which is the minus heat of formations of these guys plus what it takes to reform these guys over here so we can just write it as Delta H I'll just write it Delta H of formation for C plus Delta H of formation for D so the heat of formation of these guys - these guys this is what it took you to get to the elemental form so minus Delta heat of formation of a minus Delta heat of formation of B and then you'll have the heat of the reaction and if it's negative we would have released energy and if this number is positive then that means that there's more energy here than on this side so we would have to orab energy for this reaction to happen it would be endothermic so this is all abstract and everything and I've talked about told you about Hess's law let's actually apply it to some problems so let's say I have this reaction right here where I start with ammonia and it's ammonia gas and I'm going to react that with molecular oxygen to yield some nitrogen monoxide four moles of it and some water so what's the heat of this reaction right here so what we do is we just look up the heats of formation of each of these so let's just look them up let's start with the ammonia what's the heat of formation of ammonia and this is it's always given in kilojoules per mole so they'll say to form one mole of ammonia so to form one mole of ammonia let's look up here this is all cut and pasted from Wikipedia and am I starting in the gaseous or the aqueous state well I I think I just see I'm starting the gaseous data I've added that G there so ammonia and the gaseous state has a heat of formation of minus forty five point nine / - so what is that going so minus forty five point nine kilojoules per Joule per mole so minus forty five point nine that's just for one mole of ammonia the heat of formation it's in kilojoules I'll just look them up look them all up right now now what's the heat of formation of oxygen and I'm not going to look it up right now because oxygen is it is in its elemental form so if you see something in the form that it just always takes before you know you do anything to it its heat of formation is zero so if you co2 its heat of formation is zero if you see hydrogen if you ch2 its heat of formation is here if you see carbon by itself heat of formation is here carbon in the solid state heat of formation is zero at standard temperature and pressure now what about nitrogen monoxide let's look that up nitrogen monoxide I have it right here nitrogen monoxide heat of formation it's positive ninety point two nine ninety point two nine ninety point two nine and finally what's the heat of formation of water well let me see water water is water liquid water - two eighty five point eight three - two eighty five point eight three - two eighty five point eight three now you might be tempted to say okay Hess's law says that if we want the Delta H for this reaction we just take this Plus this and subtract that and you would be almost right but you would get the problem wrong because these are the heat of reaction of formation per mole but we notice in this reaction we have four moles of this plus five moles of this reeled four moles of this plus six moles of that so we have to multiply this times the number of moles so here I have to multiply this times four four here and I have to multiply it times four here and have to multiply times six here I don't even worry about multiplying zero times five because it's just going to be zero so now we can apply Hess's law to figure out the Delta H of this reaction so the Delta H of this reaction of this reaction is going to be equal to four times the heat of formation of nitrogen monoxide so four times ninety point two nine plus six times the heat of formation of water so plus I'll switch colors six times minus two eighty five point eight three and just as a side note given that the heat of formation of nitrogen monoxide is positive that means that you have to add heat to a system to get this to a net to its elemental form so it's it has more energy than its elemental form so it won't just happen by itself and water on the other hand it releases energy when you form it from its elemental form so it's in some ways it's more stable but anyway let me so these are the these are the heats of formations of the products and then we want to subtract out the heats of formation of the of the reactants in our reaction so here it's four times forty five point nine I'm gonna make sure it's a minus forty five - 45.9 right that was ammonia had a - 45.9 heat of formation so what do we end up with let me get the calculator out so I have let me make sure let me put it over here I have to be able to read it well I'll just do it off the screen because my screen is getting filled up so I have a let me just do it here for ya four times ninety point two nine plus six times two eighty five point eight three negative is equal to so so far we're at minus thirteen fifty three and does that run sound about right yeah that looks about right and now we want to subtract from that four times minus forty five point nine so we want to subtract so minus four times forty five point nine negative is equal to minus eleven seventy so our Delta H of this reaction is equal to minus eleven seventy kilojoules kilojoules for this reaction and all we did is we took the heat of formation of the products multiply it times the number of moles and subtract it out the heat of formation of the actual reactants there you go let's do one more of these let's say I had some propane I had some propane and I'm going to combust it I'm going to oxidize the probe the propane to yield some carbon dioxide and water well it's the same drill what's the heat of formation of propane look it up here it is amazing how exhaustive these lists really are propane is down here in its liquid state heat of formation - 104.7 so let me write that down - 104.7 heat of formation of what of oxygen is elemental state that's how you always find oxygen so it's just zero heat of formation of carbon dioxide and see carbon dioxide and as a gas - three 93.5 - three 93.5 and water we already figured that out it's minus two eighty five point eight three minus two eighty five point eight three so how much heat is form from we combust one mole of propane right here so let's see we have to take we have to figure out the heat of the products the heat of formation of the products so it's going to be three times this cuz we form three moles of this for every mole we released this much energy and then plus four times this and then subtract out one times this so what do we get we get three times three ninety three point five that's a negative is equal to that plus four times two eighty five point eight three negative is equal to minus 2,300 kilojoules roughly and then we have to subtract out one times this or we could just add 104.7 so let me just do that so plus 104.7 is equal to minus twenty two hundred so here my change my heat of this reaction is equal to minus twenty 219 kilojoules kilojoules as we go in this direction so for every mole of propane that I combust I will actually produce this much energy on the other side because this right here has it has two thousand roughly 2200 less kilojoules than this side right there so I could actually rewrite this reaction where I write all that and I could have added let me actually let me do it I could rewrite this reaction as c3h8 propane plus five oxygens yields three carbon dioxides plus four waters plus 2219 kilojoules that's actually what's released by this reaction it's so thermic this side of the reaction has less heat than this side and that didn't just disappear it got released and this is where it got released now sometimes you'll see a question where they say hey fair enough you figured out the heat of this reaction how much heat is going to be released if I were to hand you I don't know let's say I were to hand you 33 grams 33 grams of propane well then you just start thinking oh well how many moles of propane this is because if I combust one mole of propane I get this much heat so how many moles of propane is 33 grams well how much does one mole weigh the one mole of carbon so one mole of carbon weighs 12 grams one mole of hydrogen weighs one gram so one mole of propane is going to be three times 12 so times three because we have three carbons there and eight hydrogen's so times eight so it's going to be equal to 36 plus eight so it's going to be 44 so this is going to be 44 grams per mole right this is let me write that down 44 grams per mole now if I give you 33 grams how many moles am I giving you well 33 grams times I guess we could say 1 over 44 moles per gram for the whole gram there and then the grams cancel I'm giving you 33 over 40 fourths of a mole or I'm giving you Oh point seven five moles so if one mole produces this much energy 3/4 of a mole is going to produce 3/4 of this so we just multiply that times 0.75 so times 0.75 and you get 16 64 so times 0.75 is equal to 16 64 so if I were to give you one mole of propane and I were to combust it with enough oxygen I'll produce 2200 kilojoules that's released from the cyst so this side of the system has less energy left over but if I were to only give you 33 grams which is 3/4 of a mole then you're going to release 6 roughly 1600 kilojoules anyway hopefully you found that helpful