Right did everybody get the email I send out this morning. A little reminders about things that might be useful as you prepare for your first exam which is two weeks And we discussed briefly calling your attention to a little quirk of the calendar that you may want to pay attention to because they don't emphasize that challenge that faces here. Now in this we have already done what we're going to do with the material for exam. Number one gaffers one through four Dojo where to draw the lighting notes last time. So obviously one thing you're going to be doing in the course of the next couple of weeks is reviewing all that stuff and making sure you're ready for your exam on October third. But the other thing you need to do is sort of keep up with what we're doing in class. Because I'll point out to you that your second exam is on October 24th only three weeks after your first exam. So you're going to want to be doing two things at once. You're going to be preparing for your first exam but also continue to come to class. Take notes What's unique after videos if necessary things like that. And I apologize for the October 24th but think of it this way. If we had bumped back another week it would have been. Hello. It's bad enough that your final exam is going to be on Friday the 13th which I'm really want Last time we were talking about Chapter five. The time of chapter five is thermochemistry which has to do with the interaction between energy typically in the form of chemical reactions. They may have mentioned last time exothermic and endothermic reactions. Exothermic And he comes out endothermic means heat comes in. And we were talking about how this plays out in the context of forming or breaking chemical bonds. We're going to look at a specific example in a moment but suffice it to say that if you break the chemical bonds that generally requires putting energies games so that's an endothermic process. You form a chemical bond you usually get energy out. So that's an exothermic functions. So that's where we were at the end of last class. And what I'd like to begin today is show you a specific example that will illustrate some of these points The reaction called the thermite reaction. It's a very straightforward reaction in terms of the chemical equation. It's just the reaction between elemental aluminum and ferric oxide or iron three oxide which the form aluminum oxide. And I. However there's a reason it's called the thermite. What's this little l and parentheses mean liquid Hired is not normally liquid. This reaction gives off so much heat that the iron that forms generally comes out in the form of molten iron. The thermite reaction has commonly been used for welding. Especially if you're welding big pieces of metal together like say you're putting down railroad tracks. And you want to Well a couple of big pieces of metal together typically what's done is to put some of the thermite mixture at the juncture. The two pieces of Bell said off the reactions. Iron forms very quickly cooled very quickly solidifies. What else those two pieces of metal together they used to do this An in-class demonstration and then I got tired of torching tabletop. So he quit. But you can probably Google a thermite reaction and see some more impressive videos of it than what I would be able to do for you here in this room. Then what I want you to do is imagine this reaction occurring in a couple of steps. Get a picture in your mind of what the atoms look like. And here's what I mean by two steps. In step one. We're going to take apart the ferric oxide into the iron atoms and the oxygen atoms that it's made from. And then in step two we're going to take the oxygen atoms that we got from step one and combine that with aluminum which is the other React in here to form the other product which is aluminum oxide we form the iron here step one. So it's sort of like taking apart the Legos are the Tinker Toys and then reassembling them to form which is pretty much what Dalton said takes place in a chemical reaction. You just rearrange the atoms to make new stuff. Okay step what are we forming or breaking chemical bonds break So based on what we said on the previous slide is that Steph exothermic or endothermic. Endothermic is correct. In general when you break bonds when you take atoms that are attracted to each other and move them farther apart. You have to put energy in to make that happen. So step one should be endothermic or energy consuming. That step two we forming or breaking forming modes. So exothermic or endothermic. If you bring in aluminum and oxygen atoms together which are attracted to each other in the first place and allow them to form bonds. That process gives off energy. So step two is exothermic. We said before that the thermite reaction gives off so much he heard that the iron is formed as ultimately. So therefore the overall reaction must be exothermic. What does that suggest to you about the relative amount of energy involved in step one versus step two which step involves a larger amount of energy. Could you so much Yeah. Well if the overall process is exothermic would it make sense that step one consumes more energy than Step two gives off. Now. So the right answer would be step two. However that does make an assumption. It assumes that whatever amounts of energy are involved here are additive. And you'll be able to show later on that that does turn out to be a valid assumption. For now is everybody comfortable with these concepts and is everybody done with this one. Sorry a fair chunk of what we're gonna do in the first part of today's class is defined some terms. But then later I will show you how to use these concepts in a more quantitative fashion. In both my lecture notes and the textbook you will see the symbol Q used to represent heat. So q with a subscript R represents the heat of a reaction. And that is simply the amount of heat or the amount of energy consumed all released by a particular chemical reaction. There's a related term that means pretty much the same thing. And it's something called enthalpy The title of section 5.3 in your textbook is simply enthalpy. And a lot of the concepts we're going to be talking about today have to do with this concept called enthalpy. For the moment we will simply say that what enthalpy is is the potential energy of a system at constant pressure. Now you may recall that last time we were talking about chemical energy. We said that chemical energy is just one form of potential energy where the he consumed or released by a chemical reaction comes from Is changes in chemical energy come about as a result of forming or breaking chemical bonds. We'll talk later about what this constant pressure thing means for the moment and it will simply say this Q subscript R the heat of reaction is equal to delta H subscript R. That triangle in front of the age is a capital Greek letter delta which generally means change and whatever follows. And the capital H is the symbol for enthalpy. Some people think that the reason enthalpy gotten civil eight because it represents heat Not this one this equation is saying is that where the heat of a reaction comes from is the change in enthalpy or the change in potential energy that the compounds undergo as bonds form and break during a chemical reaction. So these two symbols are commonly used interchangeably. Your textbook mostly uses delta H. I use some of H and the lecture notes whatever gains pretty much the same thing. By agreement between scientists the value of delta H is a positive number for an endothermic reaction And a negative number for an exothermic reaction. So if you look at the value of delta H for reaction you can tell right away whether the reaction is exothermic or endothermic. By simply observing the sign of delta H they would like more time with this slide. Okay Now another concept that has mentioned in section 5.3 in the textbook is what a state function. It's the reason it's called a state function is and it's a quantity that depends only on the current state of whatever it is you're measuring. It doesn't depend on what it was like a few minutes ago. And it doesn't depend on what pathway it took to get from where it was to where it is just what's happening then Here's another way we can define a state function. If x is a state function then this statement will be true. The change in x delta x is equal to the final value of x minus the initial value of x. Let me give you an example. Temperature is a state function because any change in temperature you measure is simply the final temperature minus the initial temperature. So for example Suppose you are heating up a beaker of water in the laboratory using a Bunsen burner. The war starts out at a temperature of 25 degrees Celsius. By the time you're done heating it up it's it's 75 degrees Celsius. What's the change in temperature 50 degrees suppose you started at 25 degrees warm it up to 90 degrees then COO at 75 degrees. What's the change in temperature for the overall process if the degree is to start at 25 degrees cool it down to five degrees and then warm it up to 75 degrees. What's delta-t 50 degrees doesn't matter how you get there. If you started a year and in here the change is just subtracting the two. So temperature is an example of a state function. Now let me give you an example. I'm sorry question. It says in the lecture that state function depends only on the initial and final states and you'll have to show me where amplifiers. They couldn't get back. Let me give you an example of something that's not a state function that we were talking about last time. Work work is not a state function and here's why work depends on the pathway take. Suppose I want to take this pen and lifted up off the table top until it's here. Let me show you two possible pathways to accomplish. Here's pathway. That's pathway. Pathway B. That's bathroom. Which pathway is more work. To be cores The reason work is not a state function is the work depends on the pathway you take. To get from point a to point B. One of the reasons that people who do that sort of thing for a living like talk about bullying is an enthalpy is also an example of a state. That is to say that delta H the change in enthalpy is just the final value of the enthalpy minus the initial value. But if you want to think about this in the context of chemical reactions that here's what this means Any chemical reaction. The initial state is simply the reactance. The final state is simply the products. So you could conceivably find out the change in enthalpy for a reaction. If you know the enthalpy of the products and the enthalpy of the reactants and just subtract them. And we'll find out later that it's actually pretty feasible to do that. So that's part of what we'll be doing later on today. Questions about any of the terminology in bone What are the emitter so this y they may need more time here. When people first started measuring things like values and delta H they found that there were several variables that they needed to control And one of them was the weather. If you did these measurements on different days with different weather conditions you might get different results. So what was agreed upon was a set of standard conditions under which these measurements would be made. The standard conditions are a temperature of 25 degrees Celsius and at atmospheric pressure of one standard atmosphere in the units that we use in this country for weather reports 77 degrees Fahrenheit 29.92 inches of mercury barometric pressure For the benefit of anybody watching the video at some later time today's September 192019. I'm going outside the couple hours. Is it still is utterly lovely day as it was two hours ago. Got warmer. Probably still pretty close to the standard conditions. Somebody can check the weather forecast or the weather summary at the end of the day and see if that's that's the way it is. But a reasonably nice spring light day. That's the standard conditions. The standard heat of reaction It's simply that e to the reaction and measured and under the standard conditions. On the previous slide we use the symbol delta h subscript r. Here we're using delta H subscript or superscript 0. The superscript 0 refers to the standard conditions. So this symbol should be read as the change in enthalpy of a reaction under the standard conditions. A thermochemical equation Simply if an equation of a chemical reaction that also tells you what the value of the standard heat of reaction is. Here's an example. This is the equation of the reaction for the rusting of iron. Ferric oxide or iron three oxide is just blessed. What happens when iron reacts with oxygen and here's the balanced equation. Delta H zeros of are the standard heat of reaction for that red. Negative a 196.5 kilocalories exothermic or endothermic reaction. Exothermic reaction because delta H has a negative side. Terminology make sense. While people are finishing with this slide let me just point out a couple of things here. First of all a little subscript test suite style a little subscript g means gas subscript S smooth silence. No surprises there right Iron is normally a solid oxygen normally a gas rust. Islam. Also notice that we are forming one mole of rust in this reaction. And we're forming it from that two elements iron and oxygen that make this stuff. Yes isn't that nothing else equation because of this part Well not really because the three-halves coefficient here okay you can have half of a molecule that you can have half of a mole ends up what this is really saying is two moles of iron and 1.5 moles of oxygen form f2 O3 in that context. One mole of that video O3 it's balanced way. Well maybe more time. Okay. All right There's a reason I was pointing out those things. Here's part of what those reasons though. The standard state of a substance is simply the normal physical state solid liquid or gas that we expect that summons to beam under the standard conditions 25 degree Celsius one atmosphere pressure. So no surprise irons normally a silent oxygens normally gans water is normally liquid. These are the standard states of those substances. I pointed out that the reaction shown on the previous slide involve forming one mole of the compound from the elements that make it up. Because that's the definition of something called the standard heat of formation. The standard heat of formation is simply the standard via the reaction for the reaction in which one mole of a compound that is for the elements that make it up with all of the substances involved being in their standard states The symbol for standard heat of formation is this delta h 0 the subscript f where the f stands for formation. And the significance of standard heats of formation. Because you can look them up in tables in your textbook. While you're reading we find we're told him about this. On page 263 in your textbook. It talks about standard enthalpies of formation. Same thing as standard heats of formation It calls your attention to Appendix G in the back of the textbook which lists standard heats of formation for a whole bunch of different frequencies. So the point is as you work problems in chapter five you may from time to time have to look up standard heats of formation in Appendix G in your textbook. But the point is these are not numbers. We expect you to have stored somewhere in your head. You can look them up in tables in your book. Obviously in an exam situation we would also give you those numbers as well Mm-hm why are we talking about for most of the rest of our time together today. Is this database to 65 makes mention of something The aim of this a lot is Hess's Law of Heat summation. We said a moment ago that question I was asking you before about the thermite reaction. Depends on the assumption that the amount of heat involved in each step could be added together. Turns out that's a valid assumption. That witnesses Law says is if you can imagine that a reaction occurs in a series of steps which is why I asked you to imagine the thermite reaction occurring in step one. Step two. You can imagine that the reaction occurs in a series of steps than delta H for the reaction is just the sum of the values of delta H for the individual Stems. In other words in equation form delta H for the overall reaction is equal to delta H for step one plus delta H for step two plus delta H for step three. However the step jam and it doesn't have to be the pathway and the reaction actually goes. Because enthalpy is a state function doesn't depend on the pathway. So you can make up any pathway that works for you And here's one pathway that we can imagine any chemical reaction taking place by start with reactants break them down into the elements the rate from then reassemble those elements to form the products. Again sort of like taking apart the Legos and then putting them back together some different way. However we can define the second half of this in terms of standard heats of formation. Because the definition of the standard heat of formation is forming one mole of the compound from the elements that make it up. So the second half of this is just the standard heat of formation of the products. And you can look up those numbers in the appendix to the Mac your textbook. If we're going from the elements to the reactants we could do the same thing. But we're not we're going from the reactants to the elements. However let me ask you to think about something for a moment. Suppose we have some reaction doesn't matter what it is X going to Y. The value of delta H negative a 123 kilojoules. What would you guess would be the value of delta H for the reverse reaction. In which y gets converted into x gets very positive. That's the point. If you reverse the reaction you simply change the sign of delta H. If you get a 123 kilojoules out you go in this direction You put a 123 kilo joules back in and you go in the opposite direction. So actually we can express the first part of this in terms of standard heats of formation. If we were going from the elements to the reactants it would just be the standard heat of formation of the reactants or not we're going in the opposite direction. Five. We just change the sign negative standard heat of formation of Hess's Law says that we can add those two things. And if we do that we get an equation that is most commonly written this way. Delta H 0 sub R the standard enthalpy change of the reaction is equal to delta h 0 sub f a standard heat of formation of the products minus delta H sub f the standard heat of formation of the reactants. What I want to do when you're done with this slide is showing you a couple of examples of how you can solve delta H problems using this equation at the bottom. And then we'll go back and look at some examples of using the more general expression written in blue is the middle of the slide All the problems I'm going to show you appear the lecture notes. So for those of you who have your lecture notes with you if you want to save results in writing you can just refer to the various pages and the like everybody. Okay first problem we're going to show you is at the top of page 45 minutes. Now I should point out those textbooks list standard heats of formation and kilo joules per mole. The lecture notes I tend to do it in kilo calories per mole that really make any difference whether you work in kilojoules Calories as long as you're consistent you can always convert one to the other. Using the conversion factor for 0.14 joules. Saying there's one cower. Here's a reaction that you've seen before work sort of. Normally when we think about hydrochloric acid reacting with sodium hydroxide we think about the reaction happening in aqueous solution. But actually hydrogen chloride as a gas. What we normally call hydrochloric acid. It is hydrogen chloride dissolved in water. Likewise sodium hydroxide normally a solid you can dissolve in water make a solution out of it. But the point is if you mix these two things together you get salt and what. Now the question is what is delta H for this reaction to the standard enthalpy change the react. Given these values of the standard pizza formation which you can look up in the appendix in your textbook. How would you go about solving this problem bearing in mind the equation at the very bottom of the previous one. Yes around waiting for the formation of the products minus omega. Okay so the fundamental idea is products minus reactants. Now I don't want to insult anybody In intelligence but I have seen people make this mistake. So I'm just going to ask in this reaction one of the products. The salt in the water it's the stuff on the right side. The reactants or the HCl and the NAOH stuff roadside. Okay so that's really all there is to it. Products minus reactants. And we have the numbers for everything. So For the products sodium fluoride negative 98.23 kilocalories per mole. Water negative 68.32 kilocalories per mole. Add those together. Subtract the reactants which are HCl. Negative 22.06 kilocalories per mole. And sodium hydroxide negative 101.99 kilocalories per mole. What do calculators do they lie to you about saying things easily they lie to you by giving you too many sig figs If you punch this up on the calculator you will get to few sig. Figs calculators will round this off and say negative 42.5. But you are allowed the hundreds column here because all you're doing is adding and subtracting numbers all of which are known after the hundreds column. So you are allowed the hundreds column here which means your OLAP for seeing things in the answer. Delta h 0 for this reaction is negative 42.5 kilocalories per mole. Exothermic or endothermic reaction. For both answers Exothermic is negative sign means exothermic reaction. Most acid base reactions tend to be exothermic. That should not be a surprise. Any questions about this brow. Hopefully the arithmetic is pretty straightforward. Main thing to remember products minus reactants writes I slept. When you're done with this slide we'll look at another example also on page 45 in the lectures. All right I'm going to refer back to that slide just a moment when outlets out of this This is the reaction that takes place when you use a Bunsen burner in the laboratory. And ask that burns at a Bunsen burner is mostly methane CH4. And like any other hydrocarbon when it burns in the presence of enough oxygen. The products are carbon dioxide and water. Plus the value of delta H for this reaction. Given these standard heats of formation. Again we don't expect you to know these numbers. You can look them up in the tables in your book But let me point out a couple of things before we started working on this breadth. First of all in the previous problem we gave you the value of a water standard heat of formation negative 68.32 kilocalories per mole. In this problem it is negative 57.80 kilo calories per mole. Why is that yeah although we will take that into consideration later. Yes you said. In the previous example water was a liquid In this example water is a gas. One thing to be careful when you're using the tables in the back of your textbook. Some compounds might be listed more than once depending on what physical state they happen to be. Using those tables and looking up numbers. Make sure you not only get the right number for the right compound but also the right number for the right physical state. One other thing to point out. The last time we gave you for standard heats of formation. This time only give you three Say anything about what the standard heat of formation of oxygen is. What do you think about the definition and the standard heat of formation of oxygen. You might be able to figure out what the standard heat of formation of all. The way down. It'll give me a hint. Water is a compound CO2 Zika bad things have gone bad. What's oxygen it's an elegant. Definition of the standard heat of formation is the amount of heat involved in forming one mole of a compound from the elements that make it up. In this case there's only one element that makes it up and that's oxygen. It takes 0 energy to convert oxygen into itself. Standard heat of formation of oxygen is 0. When you look at those tables in the back of your book you will see that there's a lot of substances in there that have standard heats of formation of 01 all those sentences have in common is that they're all elements. So compounds will have some non-zero standard heat of formation for a element in its standard state standard heat of formation Okay so how do we go about solving this problem yes I do see the products minus the same thing. Products minus reactants. Coming back. Yet one thing we have The intention do here that we didn't have to worry about in the previous problem is how many moles of everything we've got. In the previous problem all the coefficients happened to be what I didn't matter. But notice these things are given in kilo calories or kilo joules per mole. So if two moles are involved you have to multiply by two. And in general whenever you have any coefficient other than one you have to multiply by whatever the coefficient is. So as I go about solving this problem we take the number for water which is negative 57.8 kilocalories per mole and multiply it by two Because there are two moles of water for carbon dioxide negative 94.05 to ten quickly we're multiplying it by one gives us kilocalories per units. And from that thing same thing negative 17.89 technically times one mole. We're subtracting that. And technically we're multiplying oxygen by two but oxygen the zeros and that the men. In this case you're allowed five sig figs because you're still allowed the hundreds column. Turns out negative 191.76 kilocalories Exothermic or in exothermic Of course. This is the reaction that takes place when a Bunsen burner is lit. Wanna stick your hand but Berners-Lee. Wise person. Most people would say no. Why too hot or too cold too hot the reaction feels hot to the touch. It's because it's giving off. Which is why it's an exothermic reaction. Endothermic reactions would feel cold or the touchscreen yesterday Questions about this. So those are two examples of products minus reactants. But the rest of today I want to work out a few examples of the more general form of Hess's law up delta H equals delta H for step one plus delta H for step two etcetera. We will awake until everybody is done with this one. And it looks like everybody is On page 46 and the lecture notes for this. Lets consider the reaction between acetylene C2H2 and hydrogen to form ethane C2H6. And the question is what is delta H for this reaction given these two thermochemical equations. You said before that when a thermochemical equation is simply an equation of a reaction that also tells me the value of delta H for that reaction. So take a look at these two thermochemical equations. And then see if you can figure out a plausible strategy for finding the value of delta H for the top equation up here. Good You just add these two. And as it turns out in this particular case if you just add these two numbers together you will get the right answer. That's why it's called Hess's Law of Heat. Summation. Summation means add them together. But there's a reason it works in this case. Let's look at what happens when we add the two chemical equations to get what I mean by add the two chemical equation is bring everything from both equations down and make one big equation out of it. So on the left side C2H2 comes down. H2 comes down. Other H2 comes down. C2h4 comes down. The right side C2H4 comes down. C2h6 comes now. Now I've got everything in one big chemical equation looking like that. Couple things to notice. Normally when we have one mole of hydrogen and then another mole of hydrogen. We don't write it as H2 plus H2. We want a coefficient in front of it. Since that's two moles of hydrogen you would normally just write that as two Also notice that we have C2H4 on the left side and on the right side. For purposes of what we're trying to figure out which is the amount of heat involved in this chemical reaction. We can cancel those two guys out. Same logic as before. It takes 0 h0 to convert C2H4 into itself excuse me. So just in general if we have similar terms on opposite sides of the era they cancel out If you have similar terms on the same side of the arrow you can combine them and put a coefficient. Rather. If you do those things you get this equation. And this equation looks just like this equation. Where we said we wanted to know the value of delta H for. What Hess's Law eat summation says is that if you can add the chemical equations together and get the chemical equation that you want. Then you can add the values of delta H together and get the value of delta H for the overall reaction So at this point we go ahead and add those two numbers together. Take the five sig figs worldwide and find that delta H for this reaction is negative 74.4 kilocalories. Does the logic makes sense I'm assuming the arithmetic makes sense. You're only adding two numbers together. But I'm hoping the logic behind it makes sense. Because the next problem we're going to look at is slightly more complicated Wait till people are done with this slide before we show you that what do you get here the next problem is on page 47. This is the reaction between nitrogen and oxygen form dinitrogen pentoxide N2O5. And the question is what is the value of delta H for this reaction given these two thermochemical equations which I have labeled a and B Now if we add these two chemical equations together do we get this chemical equation no we do not. So now what do we do now it turns out we'll go ahead and read the name and multi-level payments Well you're on the right track. You don't want to multiply by four but we'll get to that. Point is you are allowed to multiply or divide a thermochemical equations by any quantity you want as long as whatever you multiply or divide the chemical equation by You do the same thing to the value of built case which divide the second equation by two. Ok. And what would you do the first way you take the NO2 is included. What let's let's go back to this part for a moment. Why did you say divide by two there yeah like N2O5. Okay. In other words the question to ask ourselves at this point is what does the equation B have in common with this equation that we want to know something about and the answer is N2O5. But Equation B has two moles of it. What do we want one mole so we can divide this by two. And that means we have to divide everything by including the value of adult age. Let's go back to equation a football was equation a have in common with the equation up top here which I'm going to call the target equation from now on. Yep anyone 5B. Okay. But what do we have to do two equation a to make it look more like the target language. Right because in the end beginning Well the point is what equation a has in common with the target equation this nitrogen into that equation a only has half a mole of it. You want a hole. So multiply this by two and multiply that by two as well. In other words the strategy here is going to be multiply equation to divide equation b by two. And when you do that here are the two new thermochemical equations that result. The point is the red equation is equation a times du and its value of delta H W b is the blue equation down here except we divide an equation v by two and gotten this value for delta H. If we now add these two chemical equations together do we get our targeted way yes we do. Here's what moles of NO2 you on the left side cancels two moles of NO2 on the right side to moles of O2 and another half a mole of O2 is 2.5 moles of O2 on the left side. That's the same thing as five over two. Turns out if we add a chemical equations together we do in fact get our target equation. And therefore Hess's Law says that we are now allowed at these two numbers together and get this number which is delta H for the overall reaction Exothermic or endothermic or above answers. Earned both answers louder. Year. Positive side. Anytime any reactor that has a positive value of delta H endothermic degree. I'll just point out one thing while people are finishing up with this line. This reaction involves the formation of one mole of N2O5 from the elements nitrogen and oxygen that make it up That was kind of our definition the standard heat of formation. So it can also be argued that this answer positive 3.6 also represents the standard heat of formation for N2O5. There will be some problems in the textbook that we'll ask you to calculate a standard heat of formation for such and such a compare. And it might be helpful when you're working on problems like that. To remember what standard heat of formation means. It means for one mole of a compound from the elements that make it Standard heats of formation are usually given kilocalories kilojoules per mole which is why I wrote that this way. Everybody following all iss. One more problem to show you soon as people are done with this slide we'll get to the surprise of virtually no one at this point. It's the problem on page 48 in the lecture notes. Everybody. Okay. Here we go. To our task You see determine the standard heat of formation for octane. Octane gasoline has the formula CA age 18. Given these three thermochemical equations labelled a B and C. Now in the previous two problems we had some target equation that we were shooting for. In this case we don't. But you're being asked to calculate a standard heat of formation You know what standard heat of formation means. You can conceivably write your own target equation. What standard heat of formation me that you're forming one mole of the compound the elements that make it up. In this case the compound is octane. Ca. Changing what I make one mole of that from the elements and I hope it's obvious. The elements that make up octane are carbon and hydrogen. Carbon can be thought of as occurring is just individual atoms Hydrogen hers in the form of diatomic molecules with a formula each two. How would you write a balanced equation to show the conversion of C and H two into C a HAT. Do they figure this out yes you want to find the equation here. We're just trying to finish writing this equation. Tell him you're on the right track. When they say 99 okay. Point is what the nine Applying there is a hydrogen because you need a total of eight di nitrogens. And if they come in groups of two you need nine molecules of H2 and make that app. Okay how many carbon atoms do you need a balanced equation for the formation of one mole of octane from carbon and hydrogen looks like this. So the point is if you're trying to figure out a standard heat of formation start by writing an equation like this one which are forming one mole of the compound from the elements that make it up and make sure it's a balanced equation. The point is now that we have a time equation to look at. Now we can think about what we have to do two equations a B and C to make this work out. And you already took care of part a. We want to multiply this one by not. You saw this line over here. Okay that's fine. I'm looking at this nine right here and saying okay what equation a has in common with our targeting One is H2 but at the target equation we need nine moles of it. That's why we have to multiply equation eight by nine. Or we have to do to equation B. Yeah we'll multiply it by y. Equation B has carbon. Eight moles of carbon Mumbai equation be bindings. For a epidemic waned C to make it look more like the targeting wait. Yeah. Turn it backwards. Because u1 results. We have CHA theme which is also at the target and the target equations on the right side the equation c it's on the left side. So we can reverse it. And if we do what does that do to the value of delta H endothermic well okay it changes the sign. So in that case yes it would become positive. So this times nine which means that number times nine times eight which means that number times a this times negative one switch around which means change the sign of delta H And if you do those things here's what happens. Here is nine times equation a. And nine times the value of delta H. Here's eight times equation be and eight times the value of delta H. Here's the reverse that equation c changes sign from negative to positive. If you add these chemical equations together here's what happens Nine molecules of water on the right side cancels nine molecules of water on the left side. Eight molecules of CO2 on the right side cancels eight molecules of CO2 on the let say. Nine halves is the same thing as 4.5 moles of oxygen up here. Plus eight more moles of oxygen is 12.5 moles of oxygen on the left side. 12.5 same thing as 25 over two all the cancels and all that remains is our targeting weight Hess's law says that if you're going to have chemical equations together and get the chemical equation you. Then you're allowed to add these numbers together and find the value of delta H for the overall reaction. That you do that in this case you get negative 53.57. So the standard heat of formation of octane is negative 53.57 kilocalories per mole. Problems like these Or discussed in gory detail in Section 5.3 title empathy in the textbook. And specifically problems like the ones we were just doing or in the Hess's Law part which begins on page 265 and carries on the page to 70. And I'll just say one last thing as we wrap up later this semester you'll be doing a laboratory experiment involving calorimetry. You might have been wondering this whole time where we get these numbers. The answer is calorimetry So we'll talk about calorimetry experiments on Tuesday. I begin. Thank you. Ultimately we are. Yes well. Thank you. Liver wasn't ready. I don't know. Because I don't use English for lab. That's me. I said okay well January theorist or forgotten There might be things like what style of a file if you have any questions about anything he tells me. Okay where the limit definition of the function works and the correct. Spelling. Change temperature depend only essential temperatures occur because the original initial state final state here this way The initial value of weight by adding up the house. Can I practice what I'm saying is if the current temperature of say five degrees and this has nothing to do with the fact that ten minutes exotic might have been 15. That's what you can do is figure out the change in temperature. If I subtract my legend Yes. Yes because I felt like the main point is if you're going to find definitely didn't find temperature state function by this equation. Diff vital went essential things gradually and that's only going to isolate the matrix. Okay I also just thank you right click a bringing back. Back. Back. Back today. Nobody else sounds. Okay. That's why I question. Okay. So further equations these are those are the numbers you get after you multiply ways by Edna since we're canceling out and I'm left with this equation why do we still use them one-off reports the canceling out only has to do with chemical equations. What you're trying to do and it carefully outward and convince yourself that if you bring them all the chemical terms it make one equation out of it. This is all. Okay that's your targeting wages. That's what you want. So now this law say you're allowed to add together these numbers try adding together the original numbers it wouldn't work because the chemical equations no it doesn't
chem103-070-20190919-153001.mp4
From Dana Chatellier September 19, 2019
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