In this learning objective we'regoing to be learning aboutThermochemical Equations.It's like a chemical equation onlyit gives you the information,associated with delta H for the reaction.So a thermochemical equation, showsthe entropy changes for the reaction, sothere's a delta H associated with it.As well as the mass relationship, soif you were stoichiometry that'sthe balanced equation it is showing you.So here's an example ofa thermochemical equation.Now this is not a chemical reaction,it's a physical change.We're going from water solid,which is ice, to liquid, soit's the melting process.This is an endothermic process.I see that because I have a positive.Value for delta H,but you have to put in heat to breakthose attractions between the molecules.Now, what this equation is stating,is specific amounts.So when one mole of ice meltsto make 1 mole of water.Then 6.01 kilojoules of heat is absorbed.I know it's absorbed,because of the positive value of delta H.We could represent itwith this diagram here.We have water solid you're puttingin the 6.01 kilojoules of heat andit is being converted to H2O liquid.So a thermochemicalequation is giving you,the information this picture is providing.To work with thesethermochemical equations,there's several guidelines that you've gotto keep in mind as you work and use them.The first one is that when you see thosecoefficients of a balanced equation,they're always in terms ofmoles of the substance.So you would never turn, think of themin terms of individual molecules.Or as we did in gas laws think about em interms of leaders you mustthink of em in terms of moles.If you reverse the equation and you're,you're going to flip it around andmake the products the reactants andreactants the products, then you'llchange the sign at the delta H.So if you're going to take a liquid andconvert it to a solid,it is going to be an exothermic process.It is going to give off 6.01 kilojules.So you freeze water by stickingit in the freezer, andthe freezer is a pump to pullthe heat out of the water.The next one should make sense for us.If we double a reaction,we would double the amount of heat.So here we're going back from a solidto a liquid, we're melting it.But, we're melting twice as much, andso the delta H would be twice the 6.01,or it's 12.02.So whatever you do to the coefficients,you will do to the delta H.The other thing that has to happenin a thermal chemical equation,is you would have to always includethe states, whether it be solid, liquid,gas, or aqueous, you always put that.In parenthesis after substance.This equation obviously would makeno sense if you left those off,you've got H2O going H2O.It is the state that makes it make sense,butit's required forall thermochemical equations.So let's examine.A therm, thermochemicalequation involves a reaction.This is a reaction tothe combustion of methane,which is the gas in whatwe call natural gas.What does this equation tell us?It tell us, tells us that if we react2 mols of the methane, the CH4.With two, one mole of that.With 2 mols of the oxygen,we would produce 1 mol of carbon monoxide,we would produce 2 mols of liquid water,and we would give off.It's exothermic, so we would giveoff 890.4 kilojoules of heat.Okay, so what if we hadthe equation balanced this way?Now it's balanced, it's just not balancedwith the lowest possible whole numbers.So it's balanced like this,figure out what the delta H forthis reaction would be.If you said 1780.8 you'd be correct.You are reversing the reaction,and you are doubling it.So if you chose number two, you changedthe sign, but you forgot to double it.If you picked a negative1780.8 you doubled it, butyou forgot to change the sign.Okay, let's continue toexamine this equation,same one we were looking at before.Combustion of CH4.So it tells us that whenone mole of CO2's produced,which is what's in the balanced equationthe question is what is entropy change.Well it be a negative 890.4according to that balanced equation.What if we used a number that'snot a balanced equation?Balanced equation saysthat when two moles,react you would get this much heatwell what if you don't have two moles.What's the enthalpy change?Well anytime you're givena thermal chemical equation,you're giving a relationshipbetween the enthalpy change andthe number of moles of thatsubstance in the balanced equation.What we can do is set up a, conversionfactor for any substance in there.That is related tothe value of the delta H.So this many kilojoules, would beproduced in terms of the substancewe're interested in,when 2 mols of O2 react.Like I said we can do this forany sets that you can put down herein the denominator 1 mol of methane.We could put in the denominator1 mol of carbon dioxide.We can put into the denominator2 mols of water.Since this question is acting forrelationship, of heat forthe oxygen, we're going to usethis first one that I wrote.Don't start with that, start with what'sgiven then we have 2.5 mols of oxygen.And then do your typical dimensionalanalysis, I don't want moles of oxygen.I want kilojoules of heat.And there are a negative 890.4 kilojoulesfor every 2 mols of oxygen that react.That gives to us a value of negative1,113 kilojoules or.To two significant figures.That would be 1.1 times10 to the 4th kilojoules.Now this is a little bit more,than what 2 mols would provide.And you'd expect that.2 mols gives off,as you have a minus sign there andthere, gives off 890.4 kilojoules.This is more.So, I would expect more heat.And indeed, it is more heat.Now here it is written out a littlemore neatly, on your paper.Just want to re-emphasizethat this portion right here,is obtained from the balance equation.It is our conversation factor orour unit factor of conversion,between the amount of a substance andthe amount of heat.That comes from ourthermochemical equation.So let's see if we can work this problem.It's one that you can tryusing a similar technique.But let's point out something.In this problem I didn't start youout with moles of ammonia, okay?Ammonia is NH3.Okay, so this is a substance.We're going to produce this substance.Then I started you out with a mols.I started you with grams.So if you could, see if you can dodimensional analysis starting withgrams of ammonia, andfinishing with energy.Did you fix number three,pick number 3, then excellent job.If you did, then you can mightzoom ahead to the next slide.Otherwise, watch how I've done it.I start with the 1.26 times 10to the 4th grams of ammonia.The first place I want togo is to mols of ammonia,because I know if I canget to mols of ammonia.I can use the thermochemicalequation to convert.The molar mass is 17.03 grams per mole,and I can go from moles of ammonia and usemy balanced equation to get kilojoules.The balanced equation hasa 2 with the ammonia, andwe put a negative 92.6 kilojoules andthat provides with answer number 3.Okay, here's the next questionwe're going to work on.Okay for this problem, we've got somerocket fuel, it's N204 and N2H4 reacting.And it tells me some information.It says that when 10 grams of the N2O4react, so I'll put a 10 underneath here,we know that we are going to produce,or release a 124 kilojoules.So the first question is,what's the sign of delta H?Well, since it is released,it's exothermic.That is a negative value.The next question is,well, if 124 kilojoules isreleased when 10 grams react,if I wanted to figure out the delta H forthis reaction,which I want to go there eventually.Will that delta H, forthat chemical equation as balanced bemore than 124 kilojoules released?Or less than 124 kilojoulesof heat released?Well, to figure this out,think about the balanced equation.Would this amount, in grams bebigger than 10 or less than 10?What is this amount?This amount is 1 mol of N2O4.So, stop and answer the question.Well, if you said more, you're correct.N2O4.Has a molar mass of 92.1.So if I have 1 mol sitting right here,I have 92.01 grams of N2O4.This is far less than that.So 10 grams released 124.If I up it to 92.01 grams,it should be way more than 124.So let's work this part together.I want to know it for 1 mol.So I want to start with that.1 mol of N2O4.I know the relationship that.A negative 124 kilojoules will bereleased for every 10 grams of N2O4.This information here,gives me that relationship there.I cannot use that relationship upthere in the top right-hand corner,unless I convert my moles to grams.So I'll go from mols ofN2O4 to grams of N2O4.I get the molar mass of N2O4 andthat is, like I said, 92.01.That's how many grams are in a mol?Now I can use the relationship wesaw up in the top right-hand corner.For every 10 grams of this reactant,it produces orgives off 124 kilojoules of heat.That will give me a negative 1.14 times10 to the 3rd kilojoules of heat.Over 100, over 1,140 kilojoulesof heat would be given off,when the reaction happens forthe balanced amounts we see up there.So it's exothermic.It's certainly more than 124kilojoules of heat released.Because, I have more than 10 grams.So we've seen lots of example ofusing thermochemical equations, andbeing able to convert, basically,between the amount, of a reactantor product, and the amount of heat,either given off.Or needed for that reaction.So we can convert between those two usingas our tool, a thermochemical equation.The work that we're doing?Right now, it's difficult for studentsto get the handle, get a handle on, sowe've got a few more problemsjust to practice together.In this problem we're startingwith 15 grams of methanol,and we want to know how much heatwould be produced, in kilojoules.So we're going to start with the 15 grams,of methanol.We know that the thermochemicalequation tells me every time,2 mols of this reacts,you're going to produce this much heat.So let's convert the grams to mols.So we need the molar mass.Of the methanol.Which is 32.05.Now that we have that, we can gofrom mols of methanol to kilojoules.And negative 1,452.8 kilojoules isreleased every time 2 mols reacts.And when you multiply anddivide all of this out, you're going toget negative 340 kilojoules of heat.And while I'm thinking about it,I want to talk about the way they'vewritten this equation, orthe way I've written this equation.Sometimes you will see that the delta H,is written as a negative1452.8 kilojoules.Some books use it that way, andsome use it in kilojoules per mol.When you see it written as kilojoulesper mol, what they are saying is.It's kilojoules per mol ratio, of whatwe saw in the balanced equation here.So it's not per mol of CH3OH.It's per 2 mols of CH3OH.It's not per 1 mol of O2.It's per 3 mols of O2.So sometimes you'll see itwritten as kilojoules per mol.I more commonly.We'll just write it as kilojoules,knowing thata thermochemical equation is alwaysbalanced in terms of mols themselves.Okay, now we're going to try todetermine the delta H for this reaction.We don't know it, butwe do know that we get a releasing.Which is the negative 191 kilojoules,of heat every time 100 grams of NO reacts.We want to know it for the balancedequation, and that's for 2 mols of NO.Oops, I didn't write mols,I'll just sneak it in here.Mols of NO.I want to use this relationshipwhich is in terms of grams.So I'm going to go from mols of NO,to grams of NO, andobtain the molar mass of NO.Which is 30.01.Then I can go from grams of NO tokilojoules, which is what's being asked.And I use the relationship that they giveme in the problem, that I'm going to forevery 100 grams of NO that reacts,I'm going to release.A negative 191 kilojules.So this will give mea negative 115 kilojules andthat would be the delta H forthe reaction, because it's the amount ofheat released when 2 mols see whatI have here, when 2 mols reacts.Now, I want you to try another one.In this problem,I'm giving you that 55.5 grams of ammonia.Now, that ammonia is NH3, is reacting.It's producing this amount of heat.It's released, 'kay?And then you're going to determine,what is the delta H for this reaction?Did you pick number 2?Then you are correct.If you picked number,you didn't keep in mind that the energyis released so it's a negative sign.And if you picked a negative 225,you've determined it for 1 mol, andit wants it for 4 mols, so we've got tomake sure that we take that in to account.If you got that right, you have,feel, feel comfortable about it,know that this is the end of ourlearning objective number four,where we're utilizing our thermal chemicalequations to do calculations, andseeing what kind of informationwe can obtain from them.And how they get the values forthe delta H,by knowing a relationshipbetween energy and amount.If you did not get this one right andyou want to see it again,you'll want to continue watching,because I'll lay it all out for you here.So we're trying to determinethe delta H for 4 mols.So I have 4 mols of NH3 andit's exactly 4 mols,I'm not going to limit myself to onesignificant figure in my answer.I'm going to go from molsof NH3 to grams of NH3.I'm going to do this, which was the molarmass of 17.0 for grams per mol.I'm going to do this becausein the balanced equation,they are giving me a relationshipbetween the amount of ammonia in grams,so I can put 55.5 grams of NH3 here.And the amount of heat released,which is a negative 734 kilojoules of heatin this spot, and that is going to give usthe value of a negative 901 kilojoules.