WEBVTT chemistry/ap-chemistry/hovasapian
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Welcome back to Educator.com; welcome back to AP Chemistry.
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Today, we're going to continue our discussion of thermochemistry.
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We're going to talk about enthalpy and Hess's Law.
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I have to, before I begin...I wanted to discuss--just take a couple of minutes to talk about--thermodynamics and thermochemistry, and a lot of the terms that are sort of being thrown around, and some of the equations.
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Back in the early...well, not early days; in the turn of the century, there is a saying about thermodynamics; a very famous thermodynamicist says this; he said, "None of us really understands thermodynamics; we just get used to it."
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Now, of course, that is not completely true; we understand it; but to a large extent, a lot of it really is true.
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Thermodynamics is a very, very strange thing; heat behaves in very difficult ways--very unusual ways.
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These terms, like enthalpy and heat and energy and work, and pressure-volume...I know that a lot of that is very, very odd; it's difficult to wrap your mind around; it's difficult to sort of get a good, intuitive feeling for what is going on.
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My recommendation for dealing with that is: don't really worry about it too much.
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A lot of the comfort that comes from dealing with thermochemistry and thermodynamics, aside from some of the stuff like today's (it's actually not that bad)...a lot of it just comes from familiarity.
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You will be seeing it over and over and over again, and you will be doing the problems over and over again, so that you will get more of a sense.
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So, if you don't understand it the way you do other things, I wouldn't worry about it too much; that is just the nature of thermodynamics.
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Again, just to throw something out there to make you put your mind at ease: it's like that for everybody.
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With that, let's go ahead and get started.
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OK, so let's start with a definition.
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We're going to define enthalpy: enthalpy (which we use the symbol H for, and you will understand in a minute, when we actually equate it to heat) is equal to the energy of a system, plus the pressure of the system, times the volume of the system.
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So, let's say I had some container of gas; it's going to have a certain energy that is associated with it; it's going to have a certain pressure that is associated with it and a certain volume that is associated with it.
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By definition, that is the enthalpy; if I take the energy, plus the pressure, times the volume, I get the enthalpy.
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Now, like most thermodynamic properties, we don't really know what absolute enthalpies are; the only thing we can actually measure (which is science: in science, we measure things) are changes in enthalpy.
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So, ultimately, we are going to be concerned with ΔH; and, in fact, all of thermodynamics is going to be concerned with the Δ of some properties.
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Later on in the course, we will talk about ΔS; we will talk about ΔG, which is free energy entropy; and H is enthalpy.
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We are concerned more with Δ, and, a little bit later in this lesson, we will show why the Δ is important...in any case, just so you know...
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The definition of enthalpy is the energy plus the pressure times the volume of the system.
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Now, let's recall what we did last time: we said that the change in energy of a system is equal to q, plus the work, which was -PΔV.
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Now, let me rearrange this; let me bring this -ΔV over to this side; q equals the change in energy, plus PΔV--now, let me just set that aside for a minute.
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Now, let me come over here, and let me take my definition of H as equal to E, plus PV, and now I'll do ΔH.
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Well, ΔH, which is final minus initial, equals Δ of this, which is the ΔE plus Δ of PV.
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Well, in this particular case, if we keep the pressure constant, that means we can pull this out of the Δ; we end up having ΔH=ΔE+PΔV.
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Notice what we have: we have that q, which is the heat that is transferred, is equal to the change in energy plus the pressure times the change in volume (constant pressure).
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And we have that ΔH, just based on the definition, equals ΔE, plus pressure, times Δvolume.
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These are the same; q equals ΔH at constant pressure.
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So, this is a constant pressure situation here, OK?--at constant pressure (which is pretty much what we are doing--what we do most of our chemistry in, at a constant atmospheric pressure), q equals ΔH.
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In other words, the enthalpy is nothing more than the heat that is transferred--the energy that flows as heat.
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In other words, if a certain reaction takes place, and it gives off a certain amount of heat or it takes in a certain amount of heat, that heat is equal to the enthalpy.
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So, I can talk about the enthalpy by just referring to the heat; all I have to do is deal with the heat--it happens to be the same as the enthalpy, under constant pressure conditions.
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That is why constant pressure is really, really nice, because under those conditions, I don't have to deal with enthalpy; I just deal with the heat; they are equivalent.
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Heat is a very easy thing to measure; you just run a reaction, and you measure how hot something gets or how cold something gets; that is your enthalpy.
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It is a thermodynamic quantity that is related to heat; under constant pressure conditions, it is the heat--it is equivalent to the heat; so that is really nice.
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This is the important thing to know.
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Heat and enthalpy are equivalent in magnitude under constant pressure conditions.
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OK, so when we talk about the enthalpy of a reaction, ΔH < font size="-6" > rxn < /font > , well, that is equivalent to the heat of the reaction.
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We will often talk about the heat of the reaction.
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OK, now, ΔH of the reaction is equal to...actually, you know what, no; I'm not going to...I'll save this for the next time; there is no need to confuse us with any information that we don't need right away.
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OK, so let's just do a quick example to get a feel for what is going on--a little bit of stoichiometry.
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Example 1: When 1 mole of C₆H < font size="-6" > 12 < /font > O₆, which is glucose, is fermented to ethanol at a constant pressure, 67 kilojoules of heat is released.
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The system releases heat; heat is flowing out of the system; it is negative.
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How much heat is released when 7.6 grams of glucose is fermented?
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1 mole of glucose is fermented to ethanol at constant pressure; 67 kilojoules of heat is released--exothermic.
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The enthalpy is negative; now heat and enthalpy is the same--constant pressure.
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How much heat is released when 7.6 grams of glucose is fermented?
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OK, so let's write out what this looks like to get a sense of how we sort of start these problems, and what it looks like, notationally, for a chemist.
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So, C₆H < font size="-6" > 12 < /font > O₆ is fermented to 2 C₂H₅OH (ethanol--this is regular drinking alcohol), plus carbon dioxide gas.
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And we often write the ΔH over here as -67 kilojoules; so we see that heat is released; ΔH is negative; it is an exothermic process.
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That is what it means: exothermic--ΔH is negative; ΔH is negative--it is exothermic; it is giving off heat, which means, in addition to this product and this product, one of the other products is that much heat.
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That is why we write it; imagine heat as that third product that also comes out of the reaction.
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This heat comes from the bonds in the carbon, hydrogen, and oxygen; that is where it is coming from.
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OK, let's draw a little energy diagram, so you see what is going on here.
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C₆H < font size="-6" > 12 < /font > O₆: this is H, enthalpy (heat at constant pressure), and this is just the reaction coordinate.
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The reaction coordinate just means that the reaction is proceeding in that direction.
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Well, there are some...here we have the C₂H₅OH, ethanol, plus our CO₂.
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Now, what this means--this -67 kilojoules--as it turns out, thermodynamically, the reactants--there is more heat in these bonds; when going from glucose to ethanol as CO₂, the amount of energy in these bonds is actually 67 kilojoules less.
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So, this difference right here, from this point to this point, is the 67 kilojoules.
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Because, again, energy cannot be created or destroyed, the energy in these bonds is returned into the energy of these bonds.
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But now, I have an excess amount of energy; what am I going to do with it?--well, the reaction just releases it as heat.
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That is what this says; it is going to a lower heat--this excess heat is what is given off.
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That is why it is -67; this is what it actually looks like.
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The products are thermodynamically at a lower energy.
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OK, so let's go ahead and calculate: well, C₆H < font size="-6" > 12 < /font > O₆ is 180 grams per mole; we have 7.6 grams of it, times 1 mole--180 grams; that gives us 0.0422 mol, and I hope that I did my arithmetic correctly.
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Well, I have 0.0422 mol, and it is telling me that it releases 67 kilojoules per mole (that is what the problem says).
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It is a simple arithmetic problem.
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-2.8 kilojoules of heat is released.
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2,800 Joules of heat is released with 7.6 grams of glucose.
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7.6 grams of glucose is not very much; it's a handful--not even a handful; it is 2,800 Joules; that is a lot of heat.
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There is a lot of heat in those bonds; that is why the human body metabolizes glucose--it breaks it down, not into alcohol and CO₂--it breaks it down completely into carbon dioxide and water, and all of the energy that is released--the body uses that energy to produce a molecule called adenosine triphosphate, and it is the adenosine triphosphate that runs the body.
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That is our energy currency, and it all comes from the energy that is stored in the bonds of C₆H < font size="-6" > 12 < /font > O₆.
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Well, just 7.6 grams produces 2,800 kilojoules of energy!
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You can imagine the amount of glucose we actually take in, in the form of carbohydrates and other things; the body requires a lot of energy to run.
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All right, now let's talk about something called a state function.
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Let me...all right, let's define a state function.
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A state function is a property (you could call it a state property; I don't know why they call it a state function, but it is a property) that does not depend on the path taken to achieve that state.
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OK, so let's say I have something here and something there; these are two states; I can get from this state to this state--I can either go this way, directly, or I can go this way and come back; I can go this way, this way, this way, this way, this way, this way.
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Now, as it turns out, there are certain properties that are not state functions, like heat and work.
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So, for example, if I went from here to here, I would have to do a certain amount of work.
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Clearly, if I went from here to here to here to here to here to here, I am doing more work.
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But, as it turns out, energy (which, as we said, is equal to heat plus work, neither of which is a state function)--as it turns out, energy is a state function.
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As it turns out, it doesn't matter how I get to the final state--all that matters is where I started and where I ended up.
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That is why we are concerned with ΔS: ΔH, ΔG--those are state functions; enthalpy is another state function.
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So, we said that enthalpy is equal to the change in energy, plus the pressure, times the change in volume, right?
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Well, this is a state function; pressure is a state function--it doesn't matter how I get there; at a certain point, and at the pressure that I start and end up with, it is just a certain pressure; it doesn't matter how I get there.
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Volume: it doesn't matter how I go from 3 liters to 5 liters; I can go up to 18 liters, drop down to .1 liter, and then go up to 5 liters; I have still just gone from 3 to 5--the net effect is the same.
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So, ΔH is also a state function.
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It is a state function because it is the sum of two state functions.
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Energy is a state function despite the fact that neither of these is a state function.
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This is also quite extraordinary, that that is the case.
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OK, so as far as chemistry is concerned, now: chemistry--if we start with certain reactants, and we want to end up with certain products, well, as far as the enthalpy is concerned, it doesn't matter how I get there.
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I can get there in 2 steps; 15 steps; 147 steps.
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Now, yes, there are areas of chemistry where we are concerned about the steps, but as far as a thermodynamicist is concerned, all he cares about is the enthalpy at the beginning and the enthalpy at the end.
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It is a state function; it doesn't matter how you get there; all that matters is that you get there.
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It is the two states that matter; that is all that matters.
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Because of that, we can actually take a reaction that we are interested in, and perform it, and if we want to find the enthalpy of that reaction, well, if we have enthalpies of other reactions that we can use to get to our final reaction, we get our final enthalpy.
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OK, so now, this is the idea of Hess's Law.
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And, rather than talking about it or writing about it, the best thing to do is just do a problem, and of course, it will make sense.
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So, let me write out Hess's Law here.
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Well, I won't write it out; we'll just do an example, and it will make sense.
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OK, we want to find the ΔH for the reaction of sulfur, plus oxygen gas, going to sulfur dioxide gas.
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In other words, there is a certain heat of reaction associated with this; either it absorbs heat to create SO₂, or in the process of creating SO₂, it releases heat.
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I want to find the enthalpy--the heat of the reaction.
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Well, how do I do that?
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OK, so we want to find ΔH for the reaction; well, as it turns out, we just happen to know that we have a couple of reactions at our disposal.
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We know that, if I take sulfur plus three-halves oxygen gas, in the process of creating sulfur trioxide gas, we happen to know that the ΔH of that is -395.2 kilojoules.
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We also happen to know that sulfur dioxide gas, 2 moles of that, plus oxygen gas, goes to 2 sulfur trioxide gas, and we happen to know that the ΔH of that equals -198.2 kilojoules.
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So, we have this reaction that we know the ΔH for; we have this reaction that we know the ΔH for.
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Hess's Law says it doesn't matter how we get to our final reaction, if we can come up (oops, no, we don't want these stray lines here) with a way of manipulating these equations (switching them, multiplying them by coefficients, a lot like you do for linear equations in linear algebra or algebra courses that you have taken).
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If we can fiddle around with them and add the equations to come up with a final equation--this one that we want--well, we will just add the ΔHs, and we will get the final enthalpy of the reaction, because again, ΔH is a state function; it doesn't matter how you get there, as long as you get there.
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So, let's see what we are going to do.
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So, how can I fiddle with these equations in order that, when I add them vertically, I end up with this equation?
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OK, well, let's see; let's reverse...let's see, what can I do?
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I am trying to create SO₂ gas; so notice that SO₂ here is on the right-hand side, but in these equations, SO₃ is on the right-hand side.
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The only equation that has SO₂ in it is over here; I want to reverse it, and there is one SO₂ here, but there are 2 SO₂, so I am going to flip equation 2, and I'm going to divide it by 2.
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OK, so let me...actually, let me rewrite everything again, so that we have it on one page.
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We have S + O₂ going to SO₂; that is the reaction that we want; and we are given S + 3/2 O₂ goes to SO₃; ΔH equals -395.2 kilojoules.
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Our second equation (we'll call that #1) is 2 sulfur dioxide gas, plus O₂, goes to 2 SO₃, and the ΔH for that is -198.2 kilojoules.
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OK, so we said we want this equation from these equations.
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We are going to flip this equation, #2; so, we are going to reverse #2 and divide it by 2.
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When we reverse it and divide it by 2, this 2 SO₃ comes to the left, and it ends up becoming just SO₃; this 2 SO₂ ends up on the right, but becomes SO₂; and this O₂ also ends up on the right, but it becomes 1/2 O₂.
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Now, what happens to the ΔH?--well, exactly what you think.
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If you flip a reaction--if you flip an equation--you change the sign of ΔH; if you divide an equation by 2, you divide the ΔH by 2.
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So now, the ΔH is no longer -198; it is going to equal +99.1 kilojoules.
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And now, we leave the...here, our other equation is S + 3/2 O₂ goes to SO₃, so S is on the left--there is one S on the left--so let's leave that one alone.
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We have S + 3/2 O₂ going to SO₃; that ΔH stays the same: -395.2 kilojoules.
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Now, we just add straight down.
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Everything that is on both sides--if there is something on the left and something on the right, they cancel.
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SO₃ on the left; SO₃ on the right; it cancels.
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S comes down; that is taken care of.
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I have 3/2 O₂ on the left; I have 1/2 O₂ on the right; 3/2 minus 1/2 is equal to...well, there you go: O₂ (two halves)...+ O₂.
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O₂ is taken care of; now, the only thing left is the SO₂.
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There you go; I have the final equation that I wanted by messing around with the equations for which I did have information.
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Now, all I do is: I just add the ΔHs.
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When I add the ΔHs: +99.1, minus 395.2; we get -296.1 kilojoules.
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I used reactions that I knew, manipulated the equations, made the appropriate changes to the enthalpy, and I added straight down, added straight down; now I have the enthalpy for this reaction.
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This is Hess's Law; I can use reactions that I do know to find a reaction that I want.
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OK, let's do another example here.
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Let's do this on a new page.
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Calculate the ΔH for the synthesis (which means formation) of di-nitrogen pentoxide gas, N₂O₅, from its elements.
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OK, elements--so, the reaction that we want is nitrogen gas, plus oxygen gas, goes to N₂O₅ gas.
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Now, we have to balance this: so, we have 5 and 2, so we'll put a 5 here; we'll put a 2 here.
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We'll put a 2 here; now it's balanced.
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This is the equation that we want; OK, now here are the equations that we have at our disposal.
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Equation #1: we have H₂ + 1/2 O₂ goes to H₂O; the ΔH of that is equal to -285.8 kilojoules.
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Our second equation is: N₂O₅ + H₂O goes to 2 HNO₃, which is nitric acid; the ΔH of that is -76.6; exothermic, exothermic.
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3: we have 1/2 N₂ + 3/2 O₂ + 1/2 H₂ goes to HNO₃; the ΔH of this is equal to -174.1 kilojoules.
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So, our task is to take these three equations, manipulate them by multiplying by coefficients, reversing them, and then adding them straight down and arranging them in such a way so that, when they add, they add to the final equation.
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OK, so let's see what we have; what can we do?
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I notice I have N₂O₅ on the right, and here I have N₂O₅ on the left, so let me just leave that one alone for now.
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Let me see: I have 2 N₂ on the left, 5 O₂; I have 2 N₂, and the only equation here that has N₂ in it is this one, so I'm going to multiply equation #3 by 4.
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So, multiply #3 by 4, and what I end up with, when I multiply the third equation by 4--I get 2 N₂ + 6 O₂ + 2 H₂ goes to 4 HNO₃, and the ΔH also gets multiplied by 4.
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Negative...so this is multiplied by 4 minus 696.4 kilojoules; so whatever I do to the equation, I do the same thing to the enthalpy.
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OK, now let's see what I have; HNO₃...I have 4 HNO₃, but I don't have any HNO₃ here, so I need 4 HNO₃s on the left.
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I have 2 HNO₃s here, so I'm going to flip this equation and multiply it by w.
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So, we'll flip 2 and multiply by 2.
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When I flip this and multiply by 2, I get 4 HNO₃ goes to 2 N₂O₅ + 2 H₂O, and now the ΔH for this...I have flipped it, and I have multiplied it by 2, so now this is a positive 153.2 kilojoules.
00:29:01.000 --> 00:29:15.000
Now, #1--let me see: #1 equation: H₂; I have...what do I have?...I have 2 H₂s on the left; I have an H₂ here; I need to cancel the H₂s.
00:29:15.000 --> 00:29:25.000
I have 2 H₂Os on the right; I have an H₂O on the right here; so I have used equation 3 and used equation 2; I need to get this on the left, and I need to multiply it by 2, so I'll do the same thing.
00:29:25.000 --> 00:29:36.000
I'll flip #1 and multiply by 2, and I end up with the following equation.
00:29:36.000 --> 00:29:57.000
I get 2 H₂O goes to 2 H₂ + O₂; and again, the ΔH is going to be +571.6 kilojoules.
00:29:57.000 --> 00:30:15.000
And now, I should end up with what I have; so let me see here; let me...we are concerned with this equation, this equation, and this equation; so let's see what cancels.
00:30:15.000 --> 00:30:24.000
H₂O; 2 H₂O, 2 H₂O; 2 H₂, 2 H₂, right?--this is on the left of the arrow; this is on the right of the arrow.
00:30:24.000 --> 00:30:30.000
They are on top of each other, but it is where they are on as far as the arrows are concerned.
00:30:30.000 --> 00:30:41.000
I have 4 HNO₃, 4 HNO₃; 2 N₂, so I'll bring that down; that takes care of the 2 N₂.
00:30:41.000 --> 00:30:53.000
I have 6 oxygens on the left; I have 1 oxygen on the right; so, 6 minus the 1 leaves 5 oxygens on the left.
00:30:53.000 --> 00:31:06.000
So, that is taken care of and that is taken care of; now, the only thing left is the 2 N₂O₅, which is on the right, so I am not adding it.
00:31:06.000 --> 00:31:20.000
That is on the right-hand side of the arrow; so I get 2 N₂O₅, which is exactly what we want: 2 N₂ + 5 O₂ goes to 2 N₂O₅, exactly what we found.
00:31:20.000 --> 00:31:33.000
Now, let's just add the ΔHs straight down, and when we add them, we end up with 28.4; again, you might want to check my arithmetic; I'm notorious for bad arithmetic.
00:31:33.000 --> 00:31:46.000
So, this reaction: nitrogen gas plus oxygen gas to form di-nitrogen pentoxide: it is positive 28.4 kilojoules, so this is endothermic.
00:31:46.000 --> 00:31:51.000
That means, in order for this reaction to go forward, I actually have to add heat to it.
00:31:51.000 --> 00:32:04.000
Or, if I leave it alone, and there are other circumstances where there is something happening--there is enough heat that it can pull from the surroundings--it will pull that heat from the surroundings in order to make this go.
00:32:04.000 --> 00:32:06.000
So, there you have it--Hess's Law.
00:32:06.000 --> 00:32:27.000
We want to find the enthalpy, the heat of a given reaction; well, if we have other reactions at our disposal that we know the heat for, and we can rearrange those equations in a certain way--we can fiddle with the enthalpies appropriately--add the equations; we will get the equation that we are looking for, and we will get the enthalpy of that equation.
00:32:27.000 --> 00:32:33.000
OK, thank you for joining us today at Educator.com for the discussion of enthalpy.
00:32:33.000 --> 00:32:34.000
We will see you next time; goodbye.