WEBVTT chemistry/ap-chemistry/hovasapian
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Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.
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Today, we are going to continue our discussion of spontaneity, entropy, and free energy.
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Last lesson, we spent a fair amount of time just discussing entropy, discussing what things are; we started off with a review of general thermodynamic principles.
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But, it was entropy that we were concerned with, and we closed off by discussing the notion of spontaneity--that a spontaneous process is when the entropy of the universe is greater than 0.
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If it happens to be less than 0, then it is spontaneous in the other direction.
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So, let's just jump in, continue our discussion of entropy, and we're going to try to do it in the context of something that is familiar to us--something that we experience every day: the melting of ice.
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We have an intuitive sense of what is going on when ice melts, or when water freezes; so we are going to use that to sort of wrap our minds around this notion of entropy, and how the entropy of the universe is broken up, and made up of two part--the entropy of the system and the entropy of the surroundings--and what that means.
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And then, towards the end, we are going to introduce this thing called free energy.
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Free energy is a way of discussing entropy; it's an accounting tool, if you will--it's a thermodynamicist's way of accounting for entropy without dealing with it directly.
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So, let's just jump right on in and see what we can do.
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Remember, we said that if the ΔS of the universe is bigger than 0, then we have a spontaneous process.
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In other words, it will happen without our interference; it will just naturally tend toward that direction, as written ("as written," meaning if you wrote a chemical reaction, and it turns out that the entropy increases, then it will happen).
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OK, so now, let's rewrite our equation: so ΔS of the universe is equal to ΔS of the system, plus ΔS of the surroundings.
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OK, so each term--the system and the surroundings--the change in entropy of each of those, because they are separate entities--contributes to the tendency for a given process (I'll write "reaction") to proceed spontaneously.
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So there we go: whenever I'm considering a particular process, a particular reaction, I have two things going on: I have the change in entropy of the system itself, which is the reaction itself; but I also have the change in entropy of the surroundings--the environment in which the reaction actually takes place.
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Both of those entropies contribute to the ΔS universe; it's the ΔS universe that we are interested in.
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So, we have to account for both of these things.
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So, before we actually do the example, let me just write a convention that we are going to use; and this convention makes absolutely perfect sense--in fact, I probably don't have to write it, but I do want to be as complete as possible.
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It is something that you know, just from your experience.
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Adding energy to a system to a system (or adding energy to anything)--adding energy to something (I'll just say "something"--I won't say "system," because we speak about system and surroundings; you can actually add energy to the surroundings also, right?--an exothermic process is heat leaving a system, going to the surroundings, so you are adding energy to the surroundings) raises its entropy, because it causes (let's see...) an increase in the random motion of the molecules (or atoms, as the case may be--I'll put "molecules and atoms").
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So, remember what we said when we defined what entropy is: entropy--think of entropy as chaos.
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You know that, if I add energy to a system (like heat it up), the molecules--the atoms--they start to move faster; they are bouncing around more vigorously.
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Well, that is more chaotic; it is not just sitting there sort of doing its thing--now, you have sort of supercharged the system, and now it's more chaotic.
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So, adding energy to a system increases the entropy; that is just something I wanted to throw out there, because it's going to be an issue when we discuss what we discuss next.
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But it's natural to think about that; I mean, you know that, just naturally speaking.
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OK, so let's examine a system that we are familiar with (let me go back to...you know what...that is fine; let me go back to blue); so, let's examine a system we are familiar with, or "with which we are familiar," for those of you who prefer correct grammar.
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OK, H₂O liquid going to H₂O gas--liquid water turning into water vapor: OK, we know that at lower temperatures (and by "lower temperatures," I mean temperature greater than 0, less than 100--somewhere above the freezing point, but below the boiling point--that is what I mean by "lower"), it is spontaneous to the left.
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In other words, when the temperature is below 100 degrees and above 0 degrees, water vapor spontaneously turns into liquid water.
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That is the whole idea; it is below its boiling point--below its boiling point, it wants to be liquid water.
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Therefore, at those temperatures between 0 and 100, if there is any water vapor around, it will more than likely condense--well, not "more than likely"--it does; it naturally condenses; it's spontaneous to the left.
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That is what it means, "to the left"--it means it goes from gas to liquid--"spontaneous to the left."
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Now, at T above 100 degrees Celsius (which is boiling point), now it is spontaneous to the right.
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In other words, liquid goes to water vapor; it's boiling: to the right.
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In other words, if I took a beaker of water, and if I all of a sudden stuck it in an oven that is at 200 degrees Celsius--well, the water is going to be gone in about 5 minutes, because liquid water will spontaneously turn into water vapor.
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That is the whole idea: I haven't done anything--I have stuck it into a set of circumstances where the temperature is high, and now, spontaneously, the water will evaporate.
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OK, it's spontaneous to the right.
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All right, so (now I'll do this in red) clearly, temperature has something has something to do (I'll actually write it out--temp) with whether ΔS of the universe is (I won't say "greater than 0"--I'll say) positive or negative.
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So, at the lower temperatures, it's spontaneous to the left; in other words, as written, H₂O liquid to H₂O gas--it's spontaneous this way, which means that the entropy is actually going to be less than 0.
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In other words, it's not spontaneous as written; perhaps it would be better if I eliminate this one.
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As written, at lower temperatures, it is spontaneous this way--which means that the change in entropy is going to be less than 0.
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As written--"spontaneous as written"--means that ΔS is greater than 0; that is for temperatures above 100.
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This is how I have written it; OK--I hope that makes sense.
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OK, so clearly, temperature has something to do with whether the ΔS of the universe is positive or negative; the question is how.
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OK, how is this possible?
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Well, we know that the ΔS of the universe is equal to the ΔS of the system, plus the ΔS of the surroundings.
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OK, since it is temperature-dependent (this is temperature-dependent; we know that it is from our experience), that means either this is temperature-dependent or that is temperature-dependent.
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One of these is--maybe both; well, let's take a look.
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ΔS of the system (let's deal with this one first) is equal to the entropy of the product, minus the entropy of the reactants (right?--that is what ΔS is for a reaction).
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Recall, our reaction is: H₂O liquid to H₂O gas; that is what is happening.
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So, in this reaction, this is the product; this is the reactant.
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Well, what do you think has a higher entropy--a mole of water gas or a mole of water liquid?
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Which is more disordered--which is more random--which is more chaotic?
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Well, you know this already: a gas that is bouncing around, all crazy, is more chaotic than the liquid, which is sitting there, all nice and peaceful.
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One mole of liquid water at 25 degrees Celsius is about 18 milliliters (18 grams, 18 milliliters--1 gram per milliliter, so the volume it occupies is 18 milliliters).
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At 100 degrees Celsius, one mole of water occupies a volume of about 31 liters; 18 milliliters versus 31 liters--there are a lot more possibilities for all of the water molecules to be occupying places all over those 31 milliliters than contained here.
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This is ordered; this is disordered; therefore, the S of the products, the entropy of the products, is higher than the entropy of the reactants.
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Therefore, ΔS of the system is positive; it is greater than 0, because the product entropy is bigger than the reactant entropy--entropy of water vapor is bigger than the entropy of liquid water.
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Therefore, the ΔS of the system is positive; it is greater than 0.
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OK, here is a statement: In fact, in general, ΔS of a system is normally a fixed quantity.
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The system in one state and the system in another state (in other words, the products and the reactants)--they have sort of a fixed entropic value--a fixed entropy value.
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There is not much that can be said about it; it's a fixed quantity.
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OK, we want to discuss (let me rewrite this, actually--"it's a fixed quantity"...)--whenever you are dealing with the entropy of a given system, we are always going to be talking about some sort of a reaction: in this case, this is the reaction.
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The change in entropy of the system is going to be a fixed quantity.
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If I look at a table of thermodynamic data (the ones that I used for ΔHs, for enthalpies), there is also a column for entropy values--not ΔS values, entropy values.
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I can find the entropy of all the products, subtract the entropy of the reactants, and I get my change in entropy of my system.
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So, I can always calculate the change in entropy, if I happen to have a nice, complete table of thermodynamic data.
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But, it is usually a fixed quantity: this quantity, as it turns out, is not temperature-dependent--it is the next quantity.
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Now, I won't exactly--we won't--go into why this is the case, but this is the one that is going to be temperature-dependent.
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OK, so we won't discuss why here; you may discuss it in your subsequent courses, in an actual formal course in thermodynamics, where you spend the whole quarter or semester in thermodynamics; you may discuss it there.
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But know that ΔS of the surroundings (in other words, this one) is what is variable, and depends on the flow of heat at a given temperature.
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As it turns out, it's the ΔS of the surroundings that is the one that is temperature-dependent.
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It depends on heat flow--how heat is flowing into and out of the system (or, in this case, into and out of the surroundings).
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OK, now we said before that a thermodynamic quantity (thermodynamic property) has two components: a sign and a magnitude.
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It has two components: it has a sign, which indicates direction (for example, enthalpy--if enthalpy is negative, that means heat is leaving the system; if enthalpy is positive, that means it's coming into the system; so a sign tells us which direction this property is flowing), and 2) it has a magnitude.
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Magnitude doesn't need to be explained; it is a magnitude.
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OK, now, the sign of ΔS (surroundings) depends on the direction of heat flow.
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Here is how it depends: exothermic, from the system's point of view--an exothermic reaction releases heat from the system into the surroundings.
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The ΔS of the surroundings is greater than 0; that makes sense--here is why.
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If heat is leaving (so let's say here is my boundary; this is my system; this is my surroundings)--an exothermic process: heat is leaving the system and going into the surroundings; well, we said that when we put energy into something, it increases the entropy of that something.
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Well, heat is leaving the system (exothermic) and going into the surroundings; energy is going into the surroundings; the entropy goes up.
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The entropy is greater than 0; there is nothing new here.
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Endothermic: an endothermic process--the ΔS of the surroundings is less than 0.
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OK, an endothermic process means heat is coming into the system.
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Heat is leaving the surroundings; if you drop the energy of the surroundings, you drop its entropy; entropy is less than 0.
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That takes care of the sign.
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The magnitude: the magnitude of ΔS of your surroundings--that is what depends on temperature.
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It depends on temperature, and here is the definition: the change in entropy of the surroundings is equal to negative ΔH, divided by absolute temperature (temperature in Kelvin).
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ΔH is the change in enthalpy of the system, OK?
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Any time you don't see a subscript, we are talking about the system; any time we talk about surroundings, we will say "surroundings."
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But, if you don't see a subscript (let me put that in blue--if you *don't* see a subscript) on a thermodynamic property, it's the system's point of view.
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If you don't see a subscript, it is the system's point of view.
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So, the magnitude of the change in entropy of the surroundings is equal to the negative of the enthalpy, divided by the temperature.
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Well, an exothermic process--the enthalpy is negative; negative, negative makes it positive; that is what this says.
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If the enthalpy is positive (an endothermic process), negative means that it is less than 0.
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OK, so let's go back to our H₂O: so we have H₂O, liquid, going to H₂O gas.
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The ΔS of the system, we said, is positive; and the ΔS of the surroundings (so let me write the ΔH for this--the ΔH for this is positive 44 kilojoules)--well, ΔS of 44 kilojoules--negative 44 over T (we are not worried about T just yet)--this is negative.
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Now, you see: this is equal to ΔS of the universe; when the ΔS of the universe is greater than 0, the process is spontaneous as written.
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When the ΔS of the universe is less than 0, the process is spontaneous in the reverse direction.
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One of these--this is equal to this plus that; there is a positive term, and there is a negative term; depending on what these are...well, ΔS of the surroundings is negative; not only that, it is also temperature-dependent.
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It is equal to -ΔH over temperature.
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We see that, at lower temperatures, if this number is low, this quantity here is large.
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Because it is large, it tends to dominate this sum; therefore, even though this is positive, the magnitude of this will be bigger than this magnitude, and you will end up with a negative entropy at lower...which means it's going to be spontaneous in that direction.
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However, as you raise the temperature and raise the temperature and raise the temperature, once you get to 100 degrees Celsius or higher, once this temperature rises, notice, this is an inverse relationship; the temperature is in the denominator.
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As the denominator gets bigger (the temperature rises), the magnitude gets smaller; therefore, the effect of the negative sign diminishes.
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Therefore, it stays positive, and now it becomes spontaneous in this direction.
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I hope that makes sense--let me say it again.
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The entropy of the universe of this particular reaction consists of the surroundings and the system.
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The system, in this particular case, is positive, because the products minus the reactants gives you a more chaotic system (less chaotic system: the change--products minus reactants--is positive).
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The ΔS of the surroundings is equal to -ΔH /T; ΔH is positive; therefore, this value is negative; but the temperature is in the denominator.
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As the temperature rises (as the denominator gets bigger), this whole fraction gets smaller; therefore, the effect of the negative sign diminishes.
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It comes to a point where this completely dominates, and you end up with a ΔS which is greater than 0, positive, which means make it spontaneous as written.
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That is why, above 100 degrees Celsius, liquid turns into water vapor gas.
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Below 100 degrees Celsius, water vapor gas turns into liquid.
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It's pretty extraordinary.
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OK, let me actually draw it out: if we mark this as 0, well, at 25 degrees (you know what, let me make this a little bit smaller here, because I want a little bit more room)...so let's just say: At 25 degrees Celsius, this is lower temperature (between 0...).
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If this is our 0, well, our ΔS of the system goes here, and then the ΔS of the surroundings is bigger; so we end up with a negative value.
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This plus this is the ΔS of the universe; we end up negative, so it's spontaneous in the reverse direction: gas goes to liquid (or liquid doesn't go to gas--however you want to look at it).
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At, let's say, 110 degrees Celsius (we start at 0), well, the ΔS of the system is going to be the same; and then, the ΔS of the surroundings is going to take us here; we're going to end up positive, and it's going to be spontaneous as written, from left to right--liquid water will become water vapor.
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That is what is happening here; so, what is important is: ΔS surrounding equals -ΔH over T; this is the first of our great thermodynamic equations.
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The change in entropy of the surroundings is equal to the negative of the enthalpy of the system, divided by the absolute temperature at which the particular reaction is being run.
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That is it--a profoundly important equation.
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Heat, divided by temperature; heat divided by temperature; Joules over Kelvin, Joules over Kelvin; that is entropy.
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OK, let's do an example.
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Example 1: OK, antimony is found sometimes as a sulfide ore.
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In other words, it's antimony sulfide, as opposed to just pure antimony; we have to find a way to pull out the antimony--separate it from the sulfur.
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OK, and it is Sb₂S₃.
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Now, pure antimony is recovered by reducing the antimony ion in this ore to pure antimony metal, with iron metal (we use iron to reduce it).
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Iron gets oxidized, and it becomes iron sulfide; well, here is the reaction.
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Sb₂S₃, plus 3 moles of iron atoms, produces 2 moles of pure antimony metal, plus iron (2) sulfide, FeS.
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Let's erase some of these random lines here.
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The (oh, this is driving me crazy) ΔH for this is -125 kilojoules.
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This is an exothermic reaction; as written, it releases heat.
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The system is exothermic; OK.
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So, we want you to calculate (just a little bit of practice, getting used to using the equation) ΔS of the surroundings at 25 degrees Celsius.
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Well, OK; we know what ΔS of surroundings is: ΔS of our surroundings is equal to -ΔH, divided by T, is equal to negative, *negative*, 125 kilojoules, divided by 25 degrees Celsius.
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Absolute temperature--meaning in Kelvin: so 298 K; we end up with 0.419 kilojoules per Kelvin, or (the more standard unit of entropy is Joules per Kelvin) 419 Joules per Kelvin.
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There you go: so, that is what we wanted to sort of get to--our first real equation dealing with entropy that we were actually able to quantify in terms of something that we have dealt with before.
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So, hopefully, the discussion and the process that we actually got to made sense; but this is our first primary equation.
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OK, so now, we are going to introduce something called the Gibbs free energy, or free energy.
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Let's just go ahead and start that discussion; we'll do some definitions, and then we'll leave it for now, and then we'll continue on with an example in the next lesson.
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So, for a system, define free energy.
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OK, free energy is symbolized by a G; it equal H minus TS; the free energy of a system is equal to the enthalpy of that system, minus the product of the temperature of that system times the entropy of that system.
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Well, again, we are interested in Δs; so, at a constant temperature, we have ΔG=ΔH-TΔS.
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I cannot, cannot, *cannot* overemphasize the importance of this equation.
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The free energy of the system, the change in the free energy in the system (and we will talk about what we mean in just a minute) is equal to the change in enthalpy of the system, minus the temperature at which the thing is taking place times the change in entropy.
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OK, now let's do a little bit of mathematical manipulation; so let me rewrite it: I will write: ΔG=ΔH-TΔS.
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Well, we said that, at constant temperature and pressure, the ΔS of the surroundings is equal to -ΔH/T, right?
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Let's move this around: I have -TΔS < font size="-6" > surr < /font > =ΔH; let me take this, which is ΔH and stick it in there; I get ΔG is equal to -T, times ΔS of the surroundings.
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-T times ΔS (and remember what we said--no subscript--it means the system--so I'll just go ahead and write "system"); that means ΔG is equal to -T, times ΔS of the surroundings, minus ΔS of the system; now let me divide by negative T; I get -ΔG/T (it doesn't matter where I put the negative sign; I always prefer to put it on the numerator, just to be consistent) equals ΔS surroundings (I'm sorry, plus, because I pulled the negative out).
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Well, what is ΔS < font size="-6" > surr < /font > , plus the ΔS of the system?--it's the ΔS of the universe.
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ΔS of the universe--let's take a look at what we have just done here.
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I have defined this thing called free energy; in a little bit of mathematical manipulation, I was able to come up with an expression that involves temperature for the ΔS of the universe.
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We said that the ΔS of the universe--if it's greater than 0, we have a spontaneous process; the only way that it is greater than 0 is if ΔG is negative (because negative, negative makes it positive).
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Let me say that again: "ΔS of the universe is greater than 0" implies that the process is spontaneous as written.
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In order for ΔS of the universe to be 0, that means -ΔG/T has to be greater than 0, right?
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That means -ΔG has to be greater than 0; that means ΔG has to be less than 0.
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So now, we have a way of discussing spontaneity under conditions of constant temperature and pressure (which is most chemistry--it takes place under constant temperature and pressure).
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Now I don't have to worry about entropy; instead of talking about entropy, I can just talk about free energy.
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Free energy is a form of entropy; the universe--ΔS of the universe--I didn't have to break it up; so now, I have free energy, temperature, ΔS of the universe.
00:34:47.000 --> 00:35:04.000
So now...if you remember, from the thermodynamic tables at the back of your book, they have three columns: ΔH (enthalpy) in kilojoules per mole; ΔG (which is also kilojoules per mole), and S (entropy, which is Joules per mole-Kelvin).
00:35:04.000 --> 00:35:11.000
So now, if I want to talk about whether something is spontaneous, like a reaction, I don't have to worry about its entropy.
00:35:11.000 --> 00:35:16.000
I can if I want to--it's not a problem--or I can just talk about free energy.
00:35:16.000 --> 00:35:22.000
And again, the important equation is...well, actually, I'll get to it in just a second.
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Let me just finish this part up--let me write it down--so: Under conditions of constant temperature and pressure, which is what matters, which is...actually, I should say which are...the conditions of interest to chemists (not just chemists, though), ΔS of the universe is equal to -ΔG/T.
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ΔS universe positive implies that ΔG is negative, implies spontaneous.
00:36:49.000 --> 00:37:00.000
There you go; and let me write the equation: ΔG=ΔH-TΔS.
00:37:00.000 --> 00:37:05.000
I'm going to make a little bit of a broad statement here: it's the most important thermodynamic equation for chemists.
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Certainly, you can argue this--but it's pretty important.
00:37:09.000 --> 00:37:21.000
"The most important thermodynamic equation for chemists"; OK.
00:37:21.000 --> 00:37:49.000
So, for a reaction (rxn), the enthalpy, the entropy, and temperature all contribute to whether the reaction is spontaneous as written.
00:37:49.000 --> 00:37:59.000
That what this is all about--"is spontaneous as written."
00:37:59.000 --> 00:38:22.000
The enthalpy (which I can get from the table of data); the temperature at which I am running the particular reaction; the ΔS (which I can get from a table of thermodynamic data)--if I take the ΔH, minus T (absolute, Kelvin) ΔS, and I get a ΔG; if that ΔG is less than 0, the reaction is spontaneous as written.
00:38:22.000 --> 00:38:29.000
If the reaction's ΔG is positive, that means it's spontaneous in the other direction as written.
00:38:29.000 --> 00:38:34.000
If the ΔG equals 0, that means the system is at equilibrium.
00:38:34.000 --> 00:38:40.000
If the ΔG is at 0, that means the system is at equilibrium--very, very profoundly important.
00:38:40.000 --> 00:38:45.000
OK, I'm going to go ahead and put little circles on top here.
00:38:45.000 --> 00:38:58.000
If you remember what those circles mean, they mean "at standard temperature and pressure, and standard conditions" (meaning 1 atmosphere, 1 mole per liter for solutions, 25 degrees Celsius, things like that).
00:38:58.000 --> 00:39:07.000
These are the values that are written in the table of thermodynamic data; so I think, let's go ahead and...it's nice to just go ahead and put these in.
00:39:07.000 --> 00:39:11.000
So again, this is a profoundly, profoundly, profoundly important equation.
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If you don't take anything away from your study of chemistry, at least take this away from it, because it will come in handy, and you will impress a lot of people later on in your graduate studies if you can actually draw out this equation and know what it's about.
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It says that a spontaneous process depends on three things at constant temperature and pressure.
00:39:30.000 --> 00:39:44.000
It depends on the (well, "under conditions of constant temperature and pressure"--mostly it's under conditions of constant pressure) enthalpy, the entropy, and the conditions under which you are actually running that particular process.
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OK, thank you for joining us here at Educator.com for a further discussion of spontaneity, entropy, and free energy.
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We will see you next time, and we will start working some problems in thermodynamics.
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Take care; goodbye.