WEBVTT chemistry/ap-chemistry/hovasapian
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Hello, and welcome back to Educator.com; welcome back to AP Chemistry.
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Today, we are going to start on Part 3 of our discussion of spontaneity, entropy, and free energy.
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Last time, we left off--we introduced this very, very important equation, which was...well, we introduced the notion of free energy, if you remember: so, ΔG=ΔH-TΔS.
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Now, up until this point, we had been talking about entropy, and how the entropy of the universe (which consists of the entropy of the system, plus the entropy of the surroundings)--when that is greater than 0, that defines a spontaneous process, and how, if it is less than 0, then it's spontaneous in reverse, as written for a given process.
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If it is equal to 0, that change in entropy of the universe, then the system is at equilibrium--nothing is going to happen.
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Well, we did a little manipulating; we introduced this thing called free energy, and as it turns out, in order for something to be spontaneous, the free energy should be negative.
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So, just to rewrite here--you can certainly go back to the previous lessons, but just a quick review: for ΔG less than 0 (which is negative), that implies a spontaneous process.
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And, if it's positive--it's greater than 0--that means it's spontaneous in reverse.
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A ΔG, free energy--if it's equal to 0, that means we are at equilibrium.
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So, there is a relationship among the enthalpy of a given system, the entropy of a given system, and the temperature at which this particular reaction is taking place.
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ΔG is dependent on these three factors: the enthalpy (or the heat, ΔH), the temperature, and the ΔS for a given reaction.
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This is our primary equation, at least for this particular lesson--a very, very important equation.
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What we are going to do is: now, we are actually going to start to get into the quantitative aspects: we spent a lot of time discussing entropy, discussing free energy, trying to wrap our mind around this notion; all of that is very, very valuable, but now, we're going to start actually doing some problems.
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We are going to examine the relationship here by examining a process that you actually are very familiar with--the melting of ice, or the freezing of ice.
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Because you are familiar with it, you will understand what the numbers mean, and it will make this equation make more sense in the context of your experience.
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Let's just go ahead and start.
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Example 1: We want to examine the melting of ice at 3 different temperatures--the melting of ice at -20 degrees Celsius, 0 degrees Celsius, and positive 20 degrees Celsius.
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I have a tendency to put a positive sign on things that are positive, as opposed to just sort of leaving them out there.
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The process that we are interested in is...so, as written: H₂O, solid, going to H₂O, liquid--in other words, the melting of ice.
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OK, let's just go ahead, and we're going to set this up as a table, so that we can compare all of the numbers.
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This might end up actually taking...well, you know what, I'll try to do this all on one page; hopefully I can do it here; let's see.
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Let me move this one up a little bit, so that I'll have a little bit more room.
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Our reaction is going to be (let me write my reaction in red): H₂O, solid (ice), going to H₂O, liquid (this is a 2; I know my 2's never look like 2's); now, let me go back to blue.
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We have -20 degrees Celsius--that is going to be one column; we are going to have 0 degrees Celsius--that is another column; and we are going to have 20 degrees Celsius--that is going to be our third column.
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OK, so the first thing we want to write down here is going to be, of course, the temperature, which is in Kelvin.
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And then, we are going to write out the ΔH, which is going to be in kilojoules per mole; and then we are going to write the ΔS; OK, and I'll put a little circle at the top right--that means standard conditions (25 degrees Celsius, one atmosphere pressure for gases, one mole per liter concentration for aqueous solutions...just general standard values; these are the values that you find at the end in those thermodynamic tables).
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OK, ΔS, which is going to be in Joules per mole-Kelvin: notice, of course, one is in kilojoules; one is in Joules; it is very, very important.
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So, when we work this equation, we need to either convert this to kilojoules, or (more often than not) we convert the ΔH to Joules--because our units have to match.
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OK, our next one is going to be ΔS of the surroundings; this ΔS right here--this first ΔS--remember, we said if it doesn't have a subscript, we are talking about the system--that is just a default position.
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But, this one, ΔS surroundings--that is the one that was equal to the -ΔH/T.
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So, ΔS of the universe--that is the sum of the previous two terms; we have the TΔS; and we have the ΔG.
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So again, this is the equation that we are going to use, and we are just going to fill this table again.
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OK, so the temperature--this is going to be 253 degrees; I'm actually going to work my way down.
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253 degrees: the ΔH--well, what I do is: I look up in my thermodynamic tables: the ΔH of a reaction is the sum of the ΔH's of the products, minus the sum of the ΔH's of formation of the reactants.
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I just look them up: H₂O, liquid, minus H₂O, solid; and I end up with 6.03x10³, 6.03x10³; actually, you know what, this is not in kilojoules per mole; I'm sorry, I already changed it to Joules per mole.
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So, again, when you are working with ΔH and ΔS, you should probably work in Joules; you can do kilojoules, but it's probably best to work in Joules.
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OK, the ΔS: the ΔS of the system--you do it the same way: what you do is: you look up in the thermodynamic tables for the entropy of liquid water, and then you look up the entropy--the standard entropy--of solid water (ice).
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You take the products minus the reactants; when you do that, you get 22.1.
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So again, these are all just state functions (the entropy, the free energy, the ΔH), and you have thermodynamic tables, so you can use them to find the Δs.
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OK, the ΔS of the surroundings: this is the one that is interesting; remember, we said that the ΔS of the surroundings (let me write it down here in red--remember that equation, ΔS of the surroundings) equals ΔH, divided by the temperature, Kelvin?
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Because, the ΔS of the surroundings depends on heat flow, and heat is ΔH.
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So, if I take this number, divided by this number, negative sign, I'm going to end up with -23.8.
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OK, ΔS of the universe is equal to the ΔS of the surroundings, plus the ΔS of the system.
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When I add these two, I end up with -1.73; notice, the ΔS is negative--a ΔS that is negative--ΔS of the universe being negative--means that it is spontaneous in the reverse direction.
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Let's do the TΔS; so we'll multiply the T times the ΔS of the system right here; and we end up with 5591.3; and then, when we calculate the ΔG (which is ΔH-TΔS), we end up with 439.
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And again, this is in Joules per mole.
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So notice: ΔG is positive; ΔS universe is negative.
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One is just a different way of looking at the other; that is why we have this equation.
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When ΔG is negative, that means it's spontaneous; when ΔS is positive (of the universe), that means it's spontaneous.
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439 is not negative--it's positive; that means it's spontaneous in the reverse direction; but you know that already--at negative 20 degrees Celsius, your experience tells you that ice doesn't melt into liquid water; but at -20 degrees Celsius, liquid water turns into ice.
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As written--I wrote it as solid to liquid--at -20 degrees Celsius, this ΔG and this ΔS of the universe say that it goes in that direction; it's spontaneous in reverse.
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Now, let's do the 20 degrees Celsius first, before I go to the 0.
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The temperature is 293; the ΔH (again, that is the same; that doesn't change)--that is 6.03x10³; the ΔS is 22.1 (that is still the same).
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Now, when I do the ΔS of the surroundings (which is this equation, -ΔH/T), now T is a lot higher.
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So, what I end up with is (not -23.8, but): I end up with -20.6; and when I add this and that together (this and that together--that is what gives me this), I end up with positive 1.5.
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+1.5: that is my ΔS of the universe.
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That means it is spontaneous as written.
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I do my TΔS; I end up with 6475.3; and then, I end up with a -445 for a standard free energy change.
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This is negative; ΔS of the universe is positive; that means it's spontaneous as written.
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You know from your experience that at 20 degrees Celsius, above the freezing point of water, solid ice turns into liquid water without you doing anything; it just melts at 20 degrees Celsius.
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It is spontaneous--no interruption from anything else; at that temperature, that is what happens.
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Now, we will fill in the 0 column: 0 is 273 Kelvin; this is the same; this is going to be 6.03x10³; this is 22.1; this one is going to be -22.1.
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When I add those together, I get ΔS is equal to 0; when I take TΔS, I end up with (guess what?) 6.03x10³.
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ΔH-TΔS, that minus that: I get 0.
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So, 0: free energy is 0; the system is at equilibrium.
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That means water and ice--liquid water and solid ice--are at equilibrium; there is no net change.
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Back and forth: ice is melting; water is turning into ice; ice is melting; water is turning into ice; it's a cyclic process--there is no net change.
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So, as you see, in this particular case, what you know about ice is corroborated by what we have been talking about--I should say, what we have been talking about in terms of entropy and free energy is corroborated by your experience.
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At -20 degrees Celsius, as written, water will freeze; that is what this says--the free energy is positive; the ΔS of the universe is negative.
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At 20 degrees Celsius, yes, solid ice will turn into liquid water; it's spontaneous as written--I wrote it from left to right, ice to water.
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-445: well, for chemists--again, for chemists--we have this thing...we have this equation where we can calculate ΔH's; we can calculate ΔS's based on thermodynamics, and find the ΔG; for chemists, it's free energy that actually works, because most of the time, under chemical conditions, you are actually working with constant temperature and pressure.
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This is why this whole notion of free energy comes into effect.
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You are probably wondering, "Well, wait a minute: this ΔS actually tells us the same thing"--you are right.
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It does, but free energy, as it turns out, ends up being something that we can actually measure, better than we can entropy; it's easier to deal with.
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That is the reason why we actually use free energy, as opposed to working with entropy itself--the entropy of the universe.
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This is just the free energy of the system--that is the nice thing about this: this is talking about the system.
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This ΔS has to be positive for a spontaneous process; it has to be ΔS of the entire universe, so I have to take the entropy of the universe, the entropy of the surroundings and the entropy of the system, into consideration--whereas, with this ΔG, I only have to worry about the system.
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That is what is nice about it.
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OK, so what can we take away from all this?--so all these things (namely temperature, enthalpy, and entropy) contribute to spontaneity.
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But, temperature has the capacity to shift the direction of spontaneity.
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You just saw that...well, let me see; actually, should I say...yes, that is fine; so, in the example that we just did, you saw that, at higher temperatures, one term dominates; at lower temperatures, the other term dominates.
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So, it controls which direction a reaction is going to be spontaneous in, at any given moment.
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ΔH and ΔS for a particular reaction--they are going to be fixed; you can't do anything about that, but what you can change is the temperature.
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And by changing the temperature, you can actually take a non-spontaneous process and make it spontaneous.
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Or, by lowering the temperature, you can take something that is very, very spontaneous and maybe ease it up a little bit.
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Temperature is your control device, in this case.
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OK, let's do another example: so we have Example 2: At what temperature is the following process spontaneous?
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OK, the process that we are concerned about is: CH₃OH (which is liquid methanol)--at what temperature will it actually become gaseous methanol?
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OK, we want to know: At what temperature is the following process spontaneous?--the process is liquid methanol turning into gaseous methanol.
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The boiling point--that is what they are asking for.
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"Spontaneous process"--meaning, if you don't do anything to it, at what temperature will this happen without any interference, just naturally?
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OK, well, let's calculate what we are going to use: we are going to use ΔG=ΔH-TΔS.
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OK, this is what we are looking for now: in this case, we are looking for temperature.
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Well, what does "spontaneous" mean?--"spontaneous" means that ΔG is going to be less than 0; so we want ΔG (which is equal to ΔH-TΔS) to be less than 0.
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In other words, we want ΔH to be less than TΔS (and I'm going to...well, I'll just leave it as is); so T--we want the temperature to be greater than ΔH/ΔS.
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Well, we can calculate ΔH and ΔS from thermodynamic data in the back; so let's do it.
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Let's see: ΔH, when I calculate it (oh, that's fine; I'll do it down here), equals the products minus the reactants; what you get is -201, minus a -239, equals 38 kilojoules per mole.
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OK, so this is an endothermic process; you knew that anyway.
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A liquid going to a gas--it needs heat, so this is an endothermic process.
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The ΔS...when you look up the entropies in the back, you end up with 240, minus 127, equals 113; this is Joules per mole-Kelvin.
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OK, this is kilojoules; this is Joules; so now, when you put these numbers into here, into here, and into here, we want T greater than ΔH/ΔS; ΔH is in kilojoules per mole; this is in Joules per mole-Kelvin; I need to convert this to Joules (that is the only thing you have to watch out for).
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Temperature has to be greater than...it's going to be 38,000 Joules per mole, over 113 Joules per mole-Kelvin.
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Joules per mole cancels Joules per mole, and you end up with a temperature greater than 336 Kelvin (or 63 degrees Celsius).
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So, as written, this process--if the temperature is greater than 63 degrees Celsius, this will happen.
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At 63 degrees Celsius, liquid methanol will start to become gaseous methanol--just automatically.
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And, in fact, it will keep going that way until all of the liquid methanol is gone.
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At temperatures below 63 degrees Celsius, it's the other way around: gaseous methanol will tend to condense and form liquid methanol.
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At 63 degrees Celsius exactly, there will be an equilibrium between liquid methanol and gaseous methanol.
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There will be no more gas methanol formed; there will be no more liquid actually formed; it will be in a dynamic equilibrium.
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Some will evaporate; some will condense; it will go back and forth, but there will be no net change.
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So, our magic number is 336 Kelvin, 63 degrees Celsius.
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And again, we are just using our basic equation, ΔG=ΔH-TΔS.
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And, in this case, we just solved for T, the key idea being "spontaneous": spontaneous is what allowed me to use this greater than/less than symbol.
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OK, let's see: so again, watch your units: in your thermodynamic tables, ΔH's are in kilojoules per mole.
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I'll talk about this again next lesson, but it can't be overemphasized: ΔH's are in kilojoules per mole; ΔG's are in kilojoules per mole; S's are in Joules per mole-Kelvin.
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Let me write that down: n.b. means *nota bene*--it means make sure you get this one: ΔH is in kilojoules per mole.
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ΔG is in kilojoules per mole.
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S--in thermodynamic data, it isn't ΔS; in thermodynamic data, it's actually just the entropy of that species.
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The entropy is Joules per mole per Kelvin.
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OK, now, let's shift gears just a little bit: now, we said earlier that ΔS of the surroundings is determined by heat flow.
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That is where the equation came from: ΔS of the surroundings is equal to -ΔH, over temperature; it's determined by the heat flowing in and out of the system, or into or out of the surroundings.
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Well, what about ΔS of the system?--well, ΔS of the system--at least in reactions where reactants and products are gaseous--(let me write this a little better) is determined by the number of gas particles.
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The idea is this: the more particles means more entropy--more chaos--more disorder.
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More disorder, more chaos, higher entropy: that is it--that is all it means--but you know this already; I mean, that is the whole idea.
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When we talk about entropy, we are talking about general chaos; if you have 500 gas particles versus 5 gas particles, well, clearly, the 500 gas particles is a more chaotic system: it has a higher entropy.
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That is it; there is nothing here that is not intuitive.
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Now, let's tie this together with a reaction, to show what this actually means.
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Let's take the reaction of nitrogen gas, plus hydrogen gas, to form ammonia gas.
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OK, notice: on this side, the reactant side, we have four particles of gas.
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We have one molecule of nitrogen, three molecules of hydrogen; that is four particles of gas.
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On the right side, we have two particles of ammonia gas; the identity doesn't matter--what matters is the number of particles.
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We have two particles: the entropy of the products is less than the entropy of the reactants.
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OK, the system, in going from this to this, has become less disordered; it has become more ordered; it has become less chaotic.
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It just depends on which word you want to use: order, disorder...it just depends on how you think about it.
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We went from 4 gas particles to 2 gas particles, so our entropy is actually less.
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Now, that means, since the entropy of the products is less, then the entropy of the reactants--the ΔS--is going to be less than 0.
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You know that already--not just quantitatively; you know that because you are going from something that is highly disordered (or more disordered than this), so the change in entropy is actually negative.
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That is why, if you take the entropy of this side, it's going to be a certain number (let's say 50), and at this side, maybe 150.
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It is always products minus reactants: 50-150 is a negative 100.
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Again, when dealing with reactions that have gas particles as reactants and/or products, the entropy of the system is determined by the number of gas particles, on the one side or on the other, of the reaction.
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OK, now, if we actually confirm this by looking up, in a data table, the S of this, the S of this, the S of this, and doing entropy of products minus entropy of reactants, we actually end up getting: ΔS is equal to -199 Joules per mole-Kelvin.
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So, it is actually confirmed when we do the calculation.
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OK, so now, let's do a final example here, Example 3: What are ΔH and ΔG for the above reaction (for the N₂ + 3 H₂ going to 2 NH₃)?
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OK, well, here is our equation; well, we have the ΔS already--that is good; we need that: ΔS equals -199 Joules per mole-Kelvin.
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When we do the ΔH, we are going to look up the ΔH of NH₃ gas, minus the ΔH of this and this, and if you remember, the ΔH's of formation for elements (hydrogen gas and nitrogen gas) is 0, so it's only that number.
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You end up with -92 kilojoules...actually, this is Joules per Kelvin; we have already accounted for the moles.
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We have already accounted for the moles because we are taking 2 times the entropy of this, minus 1 times the entropy of that, plus 3 times the entropy of this.
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We have already canceled the moles, so it's just Joules per Kelvin; my apologies.
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And, this is just kilojoules, not kilojoules per mole.
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OK, so now, we just plug it into the equation, making sure to match our units; we get ΔG=ΔH-TΔS.
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It equals -92,000 Joules, -298 Kelvin, times -199 Joules per Kelvin.
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Kelvin cancels Kelvin; our final answer is going to be in Joules; and what we end up with is -32,700 Joules, or -32.7 kilojoules.
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So, ΔG is -32.7 kilojoules; this is at 25 degrees Celsius, 298 Kelvin.
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So, at 25 degrees Celsius, this reaction (under normal conditions--1 atmosphere pressure) is going to be spontaneous.
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Now again, spontaneous doesn't mean fast; it doesn't mean this actual reaction will actually go; it might be something else.
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This is a thermodynamic discussion: thermodynamics only has to do with energy at the beginning and energy at the end, or energy here and energy there; it just tells us what the relative energies are of products and reactants.
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It is: if I left this reaction alone, eventually, it will happen.
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It doesn't tell me how fast it will happen; that belongs to the domain of kinetics.
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OK, thank you for joining us here at Educator.com to sort of start to round out the discussion of spontaneity, entropy, and free energy.
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We'll see you next time; goodbye.