We're ready to embark on several different methods of determining the delta H for a reaction. We've talked about what enthalpy change is well, how are those values obtained? There are basically four methods we're going to learn about in the remainder of this thermochemistry unit. The first method is via calorimetry. In calorimetry, which is what learning objective number six is about we're going to take experimental data from running in a, actually running a reaction in a specific type of chamber and from that experimental data, we'll be able to calculate the delta H for every reaction. The other methods, will be in the next learning objective we've got the direct method, and in the direct method we're going to learn about delta H of formation or it'll be change of formation then we'll get into the indirect method. In this method, we are going to be using what's called Hess' Law and finally, we are going to learn estimations, and these estimations are determined by way of what we call bond energy. So those latter three will be in our final learning objective of this unit. So let's begin with learning objective six where we're looking calorimetry. In calorimetry, we're going to analyze measurements that are taking from a calorimetric experiment these are measurements of temperature and mass and from that data, we'll be able to calculate the delta H of a reaction so we're using information about the surroundings. To determine the delta H of th, the system which will be our reaction. So calorimetry, is a measurement of heat changes that is what, basically, calorimetry is all about to determine the delta H of a reaction. We can measure the heat change of the surroundings so we're monitoring the surroundings. Whatever happens to the change to their surroundings the system had to provide that change, so it's the reaction of interest that we will be looking at, or determining. And it's always the same value but opposite in sign so we're going to use this little piece of information over, and over in this learning objective number six. So whatever changes, to emphasize it again, whatever the change of the surroundings, the system, which is the reaction that we're interested in, is changed by the same amount but opposite in sign. A way of looking at that with symbols, okay, whatever the system is doing,. And for us that will be the reaction we're studying. The surrounding is doing just the same, but it's opposite. So if this exothermic by a certain amount, then this is endothermic by exactly the same amount. If this is endothermic, then this is exothermic by exactly the same amount. So that is the foundational core to calorimetry what we can do is we could easily monitor the surroundings. We could stick a thermometer in it and know the heat change of the surroundings, and from that, be able to infer what the system or the reaction is doing so, to do this we have various terms and equations that we will utilize. The first term is specific heat there are many different ways that you'll see specific heat abbreviated in literature. We'll use, su, su, we will just use lowercase S. And it's the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. So let's say we had, just for sake of argument, we had a little one gram piece of water, hm, okay? Or, a piece of aluminium, I don't care what it is, just one gram of it It's not very much. And, it's at 25 degrees Celsius, and we put a little flame under it. Bunsen Burner or something and we want to heat it up to 26 degrees Celsius, doesn't take much heat to change one little gram by one degree Celsius, but however much heat it takes to do that is the definition of specific heat. It has units of joules per gram degrees Celsius you might see it in units of joules per gram Kelvin because a change in temperature in the Celsius scale and the Kelvin scale are the same. It doesn't matter, but that's the typical units and I will be focusing and using as examples joules per gram degrees Celsius. The heat capacity is the next thing to define heat capacity is abbreviated with a capital C and it would be the mass of the object times specific heat. So it is amount of heat it would take to raise an object, not necessarily one gram raise a whole object by one degree Celsius. So if we had two grams, so let's take two pieces of this, let's say it's aluminum, and they're both one gram, one gram here, one gram here, and we want to heat them up by one degree Celsius, so we're going to go from 25 to 26 degrees. As we heat it up that, both of those by, it should take twice as much heat, right? So that would be the heat capacity, how much you have times the specific heat of one, which is the math, the amount of heat it takes to raise one gram. So it would have units of Joules per degrees Celsius now, if you were to have not just one, you had not just one gram but you had five grams, okay, so here's the heat capacity but you weren't wanting to raise it by one degram, degree, you were wanting to raise it by two degrees. It's a multiplicative function, it's going to take twice as much energy so let's draw a little picture we're not going to have one gram of aluminium, we're going to have two grams of aluminium we're not going to heat it from 25 degrees Celsius to 26, but we're going to heat it up all the way to 27 degrees Celsius we're going to heat it up twice. The temperature change well it should take twice as much energy to do that so q change, the change in heat of that object, the amount of heat absorbed or lost of that object as it undergoes it's temperature change can be calculated using this equation. Okay. Q we have seen before as our heat change. Delta T would be the change in temperature and it would be final minus initial, because delta always is. Okay? So we have got a question that you can answer here, just using that equation, so see what you come up with for the answer to this, using the previous equation. Well, if you plugged and chugged, as we call it as we put in our numbers, as we should, for this equation and watch our units we would come up with a negative 3.3 kilojoules now, you might've come up with it in terms of joules, but I asked for it in kilojoules. Now, where would you obtain that information well. The q is the mass of the object, the specific heat of the object 139 joules per gram degrees Celsius and a change in temperature, okay It's cooling down to 12 from 77. So it's final 12 minus initial 77, and that would give you a negative sign, and that would be a negative 3.30 joules once it's converted over to joules. Now it's worth mentioning here in terms of the terminology and the language. The q is equal to a negative 3.30 kilojoules or it would have been 3,300 and looks like seven Joules when it came out of the equation. When we ask somebody, what is the amount of heat liberated, we generally say that there would be three hun, oops 3.30 kilojoules of heat liberated. We don't usually put the minus sign in when we're using the language, I liberated 3.30 kilojoules of heat If you were just asked, what is the heat, then it would be a negative 3.3 and that negative sign means it's being given off, it's being liberated so. How much heat is given off, 3.3 kilojoules, how much heat is liberated, 3.3 kilojoules what is the value of heat. It's a negative 3.3 because that's what comes out of your equation, it is a negative sign, and it's exothermic. Okay, we could try another one what's the heat absorbed by 50 grams of copper. When it is heated. So try your hand at that one. So if you got this answer you can advance forward to the next explanation as we talk about the calorimeters. But if you didn't, you might want to watch my explanation here of how 1.25 kilojoules was obtained. q equals mass which was 50 grams. Its specific heat, which is 0.385 joules per gram degree Celsius and change in temperature. 90 degrees Celsius minus 25 degrees Celsius. Now when you plug those values in you will obtain 1,251 joules. And that would be the same as 1.25 kilojoules, and we really only know it to three significant figures. Actually we only know it to two significant figures because our change in temperature we only know to two significant figures So it would probably be even better to call it 1.3 kilojoules. So we're ready to begin looking at the instrumentation, the calorimeters that are used to do Calorimetry. So that's the device here so, the one that we see depicted here is called a constant volume calorimeter. Because, it is the reaction is co, contained within this part of the system. And what we have here, is a highly sealed container, in which we put high pressure of oxygen gas. We put our sample in this little cup here that we want to, to burn. Now, this reaction will get started by by just a spark that we can run down this wire and initiate the reaction and it will combust very rapidly. This thing is sometimes called a bomb calorimeter because of the explosion that takes place inside of there so it has to be real controlled. We don't use a whole bunch of sample in there but this will be an exothermic reaction because combustion reactions are and we generally can use this for reactions in which. The material we put in there is organic material So it'd be carbons and hydrogens and oxygens as your typical scenario. When it is carbon, hydrogens, and oxygens, your products will always be carbon dioxide and water because those are the products of complete combustion of organic substances containing only those three now, you can do other types of reactions with oxygen in this, but I'm giving you those as example product that would occur. So what happens the reaction is exothermic It gives off heat to the surroundings, and the surrounds consists of the water It also consists of the thermometer. The container that is inside of there a lot of empty space which would contain air. And then we would assume that, okay, in the course of this water heating up, we're not going to lose a significant amount of heat beyond. The walls of that calorimeter. If we know the amount that the temperature rose of the system inside of here, we're going to be able to determine the delta H of the reaction. The other type of calorimeter is called a constant pressure calorimeter. And a proc, constant pressure calorimeter it's open to the atmosphere It's not sealed inside of that calorimeter like in the previous picture and so it maintains the pressure of the environment around it, so it's a constant pressure environment and calorimeter. These are often used for aqueous solution reactions, so you can, this is a very simple device. You take a couple styrofoam cups, so it's well-insulated, and you put your reaction inside of the cup, so an aqueous solution is very helpful, you stir it and watch the temperature rise so the reaction is taking place inside of that mixture, inside that aqueous solution. Well again you've got that thermometer in there, you're going to be able to measure the temperature change of that solution. And this would be the surroundings, that's what you'd be measuring and from that you could calculate the heat lost to that surroundings and therefore, the heat given off by the reaction itself. Or if the temperature goes down, the heat absorbed by the reaction so that's the foundation. Of the devices used. But the challenge lies in doing the work so make sure you practice this a lot watch and re-watch if need be, my example problems. because I give you several, here and I give you several extra ones as well. In this unit okay so I'm going to introduce you to this problem and we're going to let you answer this question in just a minute. But let's just see what's going on we have got ourselves cal, coffee cup calorimeter we've got some styrofoam cups that constant pressure. Calorimeter in which we're putting five grams of calcium chloride into 300 milliliters of water. That solution is going to raise in temperature from whatever it was to some other temperature, but they're giving me the change in temperature they're telling me it's raising 3.43 degrees Celsius. If we assume that no heat is lost to the calorimeter, it's not going into the coffee cup parts, not enough to matter anyway. We can measure the change in temperature of that solution and from that be able to calculate the delta H of the reaction so let us look at this first question and you answer it. What's the mass of the surroundings well, if you, well, it didn't pop up, here. If you selected answer number three, you'd be correct. The solution is not the water, the solution is the water plus what you added to it now that calcium chloride is going to break. Part into ions, but its mass is not going to change and that whole thing, that whole solution is raising in the temperature to 3.43 degrees Celsius so let's finish the problem together. When I do calorimetry problems, I start with this simple equation, that the heat of this system is equal to the negative heat of the surroundings. What is my system, It is the reaction so I've got this reaction taking place. What is my surroundings, well that would be the solution. Now if the calorimeter, if the calorimeter didn't have the statement that no significant heat was going into the calorimeter, then we would have the solution plus the calorimeter, okay, because that's the surroundings, the whole thing. But this is telling me that this is essentially zero in terms of it's not enough to make a difference so we don't worry about that number there. Now what is this solution made up of or how would we break it out further? Well, we would take the mass of that solution, the specific heat of that solution, and the change in temperature of that solution, and that would give me how much heat the solution absorbed or lost. So let's put those numbers in we have 306 grams of solution. The specific heat is 4.18 Joules per gram degrees Celsius, and they told me the change in temperature Is 3.43 degrees Celsius It's increasing by that amount so that's a positive value there. So when we plug those values into our calculator, we will obtain a value of a negative 4,391 joules. Now I can convert that to kilojoules and probably should only know it to three signficant figures so that would be a negative 4.39 kilojoules. That is the heat of the reaction now how much reacted was it a whole mole that's what I want to know the delta H for what's the delta H for this reaction and that's for a whole mole. Well no, this is not for a whole mole of calcium chloride this is only for six grams so I'm going to set myself up this little ratio. I know that when six grams of calcium chloride reacts, I will get that amount of heat now in a previous learning objective, we learned how to take this kind of information and relate it back to a balanced equation so let us do that method. I want to know it for one mole so I start with a one mole of calcium chloride that's my starting point I want to know the amount of heat for one mole. Well I know a relation between grams and heat so if I can convert this to grams of calcium chloride I could then use the converter that we just established with the calorimeter. So we go to the periodic table and obtain the molar mass of calcium chloride. It's 110.98. There's that many grams. That's not an eight, grams in a mole. Now we could obviously know how many grams we have. And now we can go from grams of calcium chloride to kilojoules and that's the information we obtained from the calorimeter. This many kilojoules for six grams. So we can multiply and divide as we need to here and obtain a value of a negative 487 kilojoules. So when you are in the 100 of ranges for delta H's reactions that is a typical value, just thinking of magnitude If it a very exothermic type of combustion type of reaction, you would maybe expect it in the 1000. But that's a reasonable number for this solution of a mole of calcium chloride so, it's exothermic, the temperature raised so the negative value makes sense and that is how you work obtaining a delta H for a reaction by extracting the information of calorimetry. Let's going to put some things together here. We're going to want to heat up a cup of water and we want to maybe use a Bunsen Burner that is burning natural gas, which consists of methane. Methane is going to be used by this combustion reaction in order to heat water up and we're given that we're wanting to heat it up from 20 degrees to 100 degrees Celsius. Now when this flame, and let's set up our flame here and a little picture of what's going on, with our little coffee cup above it, okay It's got our water in there. And we've got 200 millilitres of water. So this flame is giving off heat, okay and it's being absorbed into that coffee cup, but it's telling me that only 50% of it's going into that, into the water the rest is going into the surroundings. Whether the surroundings be air or or the mug itself, we know that only half of the heat given off by this flame is going into that water. Now, whatever the system is doing, the surrounding's doing just the opposite well if the system is the reaction,. And the reaction, is the reaction we see up there and the surroundings is the water and the mug and everything else. What this statement right here is telling me is that only 50% of it, only 50% of that, is going into the water. Okay. So that's one way to think about it. The other way we could think about it is, whatever the system is doing, which is our reaction, and then you write reaction there instead. The surroundings is the water plus everything else. Okay. The two together would be all of the heat of that, and half of it belongs to the water, so we could think of it this way whatever the water is doing, if we were to double that, that would get us to the heat of the reaction either way. You could think about that information that only half of, 50%, of the heat given off by the system goes to the water so we're going to start with that messy statement over there on the right, let me clean it up a little bit. [SOUND] And let's expand out the water. The heat of the reaction would be equal the mass of the water, the specific heat of the water, the change in temperature of the water times two. So the mass of the water, so it's 200 millilitres of water the density of water is one gram per millilitre so that's 200 grams of water. Specific heat. Of water is not given up there, but we've used this in previous problems Its value is 4.184 Joules per gram degree Celsius. The change in temperature, it's heating it from 20 up to 100 degrees Celsius, so it's heating it up. 80 degrees, and when we double that, that's the amount of heat the reaction is giving off because only half of it is going into the cup, we double it, we get the heat of the reaction. So the heat of the reaction is equal to a negative 1.34, that's the three significant figures times ten to the fifth Joules. And it looks like, again, with this 80 only having two significant figures, we probably ought to cut out one of those significant figures. This would also be, if we converted it to kilojoules, be valued at a negative 130 kilojoules. Okay so that's the heat, the reaction has to give off in order to heat the water up. But it doesn't ask that, it asks how many moles of methane would have to burn. So lets start with that negative 130 kilojoules and lets look at our thermochemical equation and relate the kilojoules to moles of CH4 and to balance the equation, one mole of CH4 produces 890.4 kilojoules. So when we divide out this value here, we find that it's 1.7 moles. Of CH4 that would have to burn in order to make that react, or that water, heat up to 80 degrees Celsius I mean, 80 degrees up to 100 degrees, right before it starts to boil. So this is the end of learning objective number seven, where we're looking at the surroundings, obtaining information about the surroundings, and knowing about the reaction we've seen a few examples of it I really would recommend you re-watch my examples or you go and do the extra example videos that I have posted on Calorimetry. This is, typically, the most challenging part of the thermochemistry unit for students.
