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0:01
- Gibbs free energy is gonna be the topic of this lesson. My name is Chad, and welcome to Chad's Prep, where my goal is to take the stress out of learning science. Now, in addition to high school and college science prep, we also offer MCAT DAT and OT prep as well. I'll leave a link in the description for where you can find those courses. Now, this lesson's part of my new general chemistry prep list. I'll be releasing the several lessons a week throughout the rest of this school year,
0:20
so if you wanna notified every time I post one, subscribe to the channel, click the bell notification. So Gibbs free energy, so symbolized by the letter G, and more commonly, we're gonna talk about the change in Gibbs free energy, Delta G, so a Delta G is simply defined in terms of quantities we've already looked at, the enthalpy change and the entropy change, so, and this is one of the more famous equations in thermodynamics here, so Delta G equals Delta H minus T Delta S.
0:45
Now, before we get there, and technically, that is kind of the definition (chuckles) of Delta G, it turns out. It's a mathematical definition. Mr. Gibbs here, in coming up with his definition, it really was just a mathematical definition. Now, what you've gotta realize though, is that when we talk about Gibbs free energy, don't think of free like it doesn't cost anything. It's not that kind of free. It's like, are you free for lunch on Friday? Are you available?
1:07
So when you think of Gibbs free energy, you wanna think of energy that is available to do work, energy that is free to do work. So we like talking about the interaction between the system and the surroundings. The system is whatever you're looking at, often a chemical reaction, and the surroundings is everything else in the entire universe. And so it turns out, the system can't create its own energy. If the system ends up with more free energy, in this case,
1:32
Gibbs free energy, it's because it got it from somewhere. It's 'cause the surroundings gave it to it. So however, the system can use up its free energy. Again, this is the energy available to do work. And if system uses up its free energy, it's using it up to accomplish something. And so in this case, it turns out when a system uses its energy, its free energy up to accomplish something, that, by necessity, means Delta G is going to be a negative number, less than zero. So, and in this case,
1:59
that's when a reaction is spontaneous. So the system can choose to use up its own energy, but can't create its own energy. And so when Delta G is positive, that is a non-spontaneous process. The system can't create its own energy. The only thing it can do is get energy from the surroundings. The surroundings can give it that free energy, but it can't just create it on its own. And so big thing here is that when Delta G is a negative number, that is the kind of reaction
2:23
that is spontaneous, one that actually occurs. The system is using up some of its free energy to do work. So, but when Delta G is positive, that's a non spontaneous reaction. And then finally, when Delta G equals zero, that is a system at equilibrium. So one quick thing here, we'll talk about this more later, but contrast Delta G versus Delta G standard, that little circle means standard, and that means talking about the Delta G under a certain specific set of standard conditions.
2:50
And those standard conditions mean that all reactants and all products, if they're aqueous, have a one molar concentration, or all reactants and all products, if they're gaseous, have a one atmosphere partial pressure. Now, technically, standard conditions doesn't imply a temperature, but most of the time, like, when you look up, you know, thermodynamic properties, say, in the back of your textbook, they're gonna be recorded as at standard values at 298 Kelvin, i.e., 25 degree Celsius.
3:17
And so because standard values for these thermodynamic quantities are most commonly reported at 25 degrees Celsius, a lot of students come away with the impression that the temperature is part of those standard conditions, but I'll remind you, it's technically not. You can have standard conditions at any temperature. The big things that tells you what the concentrations of your reactants and products are, again, for aqueous species, all one molar concentrations, for gaseous species,
3:41
all one atmosphere partial pressures. And so in this, the last three lessons, we're really gonna be dealing with Delta G in this thermodynamics chapter, and so I just really wanna make sure you distinguish between these two. That'll be important later on, and sometimes as professors, we don't do a good job of it either. So for right now, we're talking about plain old Delta G here. So, and so we're saying here's that when Delta G is negative, it means that the reaction is spontaneous
4:04
under whatever conditions you happen to be under that this value applies, when it's positive, non-spontaneous. When it's at equal to zero, you're at equilibrium. Okay, so where did Mr. Gibbs come up with this? Well, how, you know, how did he come up with the idea that Delta G being negative is a spontaneous reaction? Well, we learned from the second law of thermodynamics what the requirement for a spontaneous reaction was. We learned that Delta S system,
4:28
plus Delta S surroundings has gotta be positive. So the Delta S of the entire universe is positive for a spontaneous process. That's what our second law of thermodynamics come from. So, and that's actually where Gibbs began. He started with this lovely expression, and said, "Well, you know, that's where I'll start too, then." So, but he took this a little bit different route here. So, and by the way, when you see system here, so we technically don't have to label that one.
4:56
If you ever see Delta G, Delta H, or Delta S without a label, that's automatically for the system, which is often a reaction, or something along these lines. But if you're gonna say Delta S surroundings, or Delta S universe, you better label those. So, but the one that's not labeled, that's the Delta S of the system. Well, it turns out this would be a rather inconvenient way to figure out if a reaction is gonna be spontaneous, 'cause measuring the Delta S of the system is one thing,
5:18
but measuring the Delta S of the entire rest of the universe, the that's kind of problematic. So he said, "Well, you know, it turns out we could measure that indirectly though, by just looking at the system." So if you recall the last lesson, we learned one of the definitions of Delta S was that it was Q reversible over T, Q reversible over temperature. And we also learned earlier in this, early in the course, way back in chapter five, that enthalpy was the same thing
5:46
as heat for a process carried out at constant pressure. And so as a result, we can kind of come up with an expression for Delta S that it's equal to Delta H over T when done at constant temperature, as well as constant pressure, as it turns out. So what we're gonna do is make some substitutions here. So in this case, for Delta S of the surroundings, we'll keep the Delta S of the system, but again, this is not super convenient to try and measure it for the whole surroundings.
6:11
So what we're gonna do is make some substitutions here. We're gonna put plus the Delta H of the surroundings over the temperature. So we just substitute Delta H over T for that Delta S value of the surroundings. And again, this has gotta be greater than zero. Well, we can take this a step further. So between the system of surroundings, we're just exchanging enthalpy, and so in this case, if enthalpy, which we've most closely associate with heat, you know, leaves the system,
6:40
that's a negative value for Delta H, and for the surroundings, it would gain the exact same amount, the exact same value, but positive. And so it turns out the Delta H of the system and surroundings are equal in magnitude, just having the opposite sign. And so you could substitute in, you could say that the Delta H of the surroundings is equal to the negative of the Delta H of the system, which again, we don't actually have to include that label.
7:05
And so we can rearrange this just a little bit further, and go Delta S plus, we'll put a negative sign in there, negative Delta H of the system. I'll erase that in a second, just FYI, is greater than zero. And so once again, we don't actually need to label the system, so we'll take it off. So, but the interesting part here is that now I've got an equivalent expression to the second law of thermodynamics at the requirement of a spontaneous reaction
7:33
that Delta S minus the Delta H over T has gotta be a positive number. Well, let's rearrange this just a little bit. So what we're gonna do is we're gonna multiply through by negative T, and so we're gonna get negative T Delta S plus Delta H. And when I multiply zero by negative T, I still get zero. So, oh, but I did mess up one thing. If you multiply or divide by a negative, it changes the sign here, so it becomes a less than sign instead of a greater than sign.
8:03
And so we end up with this expression right here. So, and if we rewrite this just the opposite way,
8:10
Delta H minus T Delta S is less than zero, and again, we derive this from the second law of thermodynamics. This is a requirement for a reaction to be spontaneous, something that's true for any spontaneous reaction process. And so Mr. Gibbs thought this was rather convenient, so that now, instead of having to measure anything involving the surroundings, I can just measure the enthalpy change of the system, and the enthalpy change of the system,
8:35
and not have to measure anything in the surroundings at all, and come up with a requirement for a reaction to be spontaneous. And it turns out when this lovely quantity is a negative number less than zero, that's when a reaction spontaneous. He said, "Well, you know what? I'm just gonna call this combination of things, Delta G." And that's where Mr. Gibbs came up with this idea. So the derivation of this is not super important. For those of you that are interested,
8:56
that's where it kind of comes from, but you do have to know this equation. Delta G is equal to Delta H minus T Delta S, and sometimes I like to be facetious, and shock some students. I'll be like, "Hey, I talked to your professor the other day, and you guys need to get higher test scores." So, and I, turns out, I didn't talk to their professor. I didn't know their scores, but they need to get higher test scores. Just a way of remembering that equation, 'cause that equation
9:19
for Gibbs free energy is super important. So now we know what Gibbs free energy is. It's a mathematical equation. We know where it comes from, but we know what it means. When Delta G is negative, a reaction is spontaneous, when Delta G is positive, a reaction is non-spontaneous, and when Delta G equals zero, the reaction's at equilibrium. One thing to note, so if Delta G is negative in one direction, so then the opposite direction of the reaction Delta G would be the exact same number,
9:45
but positive. And so another way of saying Delta G is positive means a reaction is non-spontaneous, what also you could just say, "Well, the reverse reaction is the one that's spontaneous." So you could say it that way as well, just an FYI. So now we've gotta talk about some relationships between Delta G, Delta H, and Delta S, and some conditions for when a reaction's gonna be spontaneous or not. And so we're gonna rely back on this equation first, but before we look at this mathematically,
10:09
it might be helpful to kind of talk about what the universe wants in a chemical reaction. Turns out the universe wants a couple of things. It wants your, well, one, we wanna note, free energy needs to go down. Well, it turns out from enthalpy's perspective, your best bet to have your overall free energy going down is to have your enthalpy go down as well, to give off heat, so to speak. And so exothermic reactions tend to be much more likely to be spontaneous than endothermic reactions,
10:36
but that's a only part of it. So that's one of the things the universe wants. It wants Delta H to be negative, wants the reaction to be exothermic, and give off heat. Now, on the other side of the coin, it turns out that we learned from the second law of thermodynamics that one of the big things the universe wants is, so for a spontaneous process, the entropy change of the universe is gonna be positive. It's gonna increase, so in terms of entropy.
10:58
Well, great way to get a good start on that is just if the entropy of the system is increasing as well. And so the second thing the universe wants, in addition to having an exothermic reaction, and a negative Delta H, is it wants a positive Delta S for that reaction, a positive Delta S of the system. So, and it turns out if you give the universe both of those things, that reaction will always be spontaneous. If you don't give the, you know, the universe, either of those things, like let's say,
11:23
instead of an exothermic reaction, you have an endothermic reaction, and instead of Delta S of the reaction being positive, Delta S of the reaction is negative? Well, for an endothermic reaction, where Delta S of the reaction is negative, that reaction's never going to be spontaneous. So, and then finally, if you give the universe one out of the two things it wants, either it's exothermic, but Delta S is negative, or it's endothermic, but Delta S is positive, well, it turns out,
11:46
then that reaction will be conditionally spontaneous. It will be spontaneous under certain conditions, but you gotta give the universe at least one of the two things it wants. So, and again, the universe wants things to be lower in energy, and so things that give off heat more likely to be spontaneous, so exothermic reactions. So, and if you looked around my office here right now, you'd find out that the second law of thermodynamics is definitely in effect.
12:10
It's more likely to, on its own, get messy, rather than to get clean, and the universe likes more disorder and things of this sort. And so, again, second thing the universe wants is a Delta S that is positive for the reaction for the system. All right, so let's take a look at this mathematically now, and before we get too far into this mathematically, we gotta make sure your math skills are up to speed. So my question for you is, what do you get
12:36
when you add a positive number to another positive number? Yeah, you get another positive number, an even bigger positive number. Now let's not go, you know, assume too much here. What do you get when you add a negative number to another negative number? Great, you get an even more negative number. Now the difficult one, what happens when you add a positive number to a negative number? Well, it depends. You notice, I didn't say multiply. We're not multiplying here, we're just adding.
13:02
If I add a positive number to a negative number, well, the result come could come out positive, negative, or zero, depending on if one's bigger than the other. If the positive number's bigger, well, then adding 'em together gets positive. If the negative number was a higher magnitude, then it's overall gonna come out negative, and if they were exactly equal in magnitude, then they'd add up to exactly zero. These are the math skills you need. And so when we take a look at this,
13:23
this Delta G equals Delta H minus T Delta S, we're gonna look at this as A plus B,
13:34
where A is simply Delta H, but B is not simply Delta S. It's this entire second term, including the negative sign. It's negative T Delta S. And what you should realize about this second term here is that it's always gonna have the opposite sign of whatever Delta S is, 'cause the temperature here, it turns out, has to be in Kelvin. And there's no such thing as a negative temperature in Kelvin, it stops at zero, so on the Kelvin scale. So, but you've got this negative sign over here,
14:00
and so if Delta S is positive, then this whole term is gonna end up negative, due to that negative sign right there. And if Delta S is negative, well, a negative times a negative would be positive, and this whole second term's gonna end up positive. And so now we can evaluate based on the signs for Delta H and Delta S, we can look and say, "Well, when would Delta G come out negative?" Because again, that's when the reaction would be spontaneous.
14:24
So let's give the universe both the things it wants. And the first thing the universe wants, it wants an exothermic reaction, and it wants the entropy change of the system of the reaction to be positive, again, as a contributing factor towards entropy of the entire universe being positive. All right, well, if we look at this then, this means that A, our first term is negative. So, and then B, if Delta S is positive, well, again, a negative times a positive times a positive means
14:52
that B is gonna come out negative as well. And if you add a negative number to a negative number, you're gonna get a negative number, no matter what. When we say no matter what, what we really mean is no matter what value of the temperature you plug in. Again, you can't plug in a negative number for temperature, 'cause it's on the Kelvin scale. So T is either gonna be zero, or it's gonna be a positive number, but it can't be negative. And so no matter what possible temperature you plug in here,
15:17
you're just still adding two negative numbers. Now it turns out at low temperatures, this would be a small negative number, and at higher temperatures, this ends up being a large negative number, but it's a negative number either way, and a negative plus a negative is going to be negative period, no matter what temperature we talk about. And so when Delta H is negative, and Delta S is positive, when you give the universe both of the things at once, that reaction is gonna be spontaneous,
15:44
we say, at all temperatures.
15:49
That's kind of how that works. Now, what happens if you don't give the universe anything that it wants? So in this case, what if we say that Delta H is positive, and Delta S is negative? So if we go back and take a look at this situation now,
16:05
we said earlier, you gotta give the universe at least one of the things that it wants. If you don't give it either one, that has no chance of being spontaneous, and it turns out this is the not gonna be spontaneous at any temperature, as we'll see. So, but in this case, Delta H is positive, so A here is a positive number, and with Delta S being negative, that means that negative T Delta S, that whole second term, well, again, a negative times a negative is a positive,
16:32
and I don't care if we're doing this at low temperatures, and the second term is a small positive number, or at high temperatures, and the second term is a large positive number, I'm adding two positive numbers in either case, and Delta G is gonna come out positive, no matter what. And if Delta G is positive, that reaction is non-spontaneous.
16:57
So a couple different ways you could phrase this as well, you could say it's non spontaneous at all temperatures, you could say that it's spontaneous at no temperatures, not usually the way we say it, but you could also say that it's the reverse reaction that would be spontaneous at all temperatures as well. All right, finally, some conditional things. So let's say that Delta H and Delta S are both positive, or both negative, and in both these cases, you're giving the universe one
17:21
out of the two things it wants, which means we're gonna see some conditional,
17:27
some conditional spontaneity in these cases. So let's take a look at this for a sec mathematically. So if Delta H is positive, then our first term here is positive. And if Delta S is positive, it means our second term here is going to be negative. Now, if my goal is to get a spontaneous reaction, then which one of these terms do I want to dominate? Well, again, I want Delta G to come out negative, so I want the negative term to dominate. So, and it turns out that Delta H and Delta S,
17:56
as long as you don't change the temperature too drastically, the values for Delta H and Delta S don't fluctuate a lot. So the big place this equation's gonna change then is by changing the value of the temperature you plug in, again in Kelvin. And so in this case, I can make this a very large negative number by putting in a very large temperature. I can make this a very small negative number by plugging in a small temperature. And so the key is which one has got a better chance
18:22
of having Delta G come out negative? Well, again, I want the negative number to be bigger than the positive number, and the negative one's the one with temperature, so let's make this a large temperature. And so it turns out when Delta H and Delta S are both positive, we like to say spontaneous
18:44
at high temperatures. Now, we have to be careful. What do we mean by high temperatures? Well, it just means a over some threshold temperature, just over some threshold temperature, and that may be something you consider high, it may be something that you don't consider high, but it's not about you or me. It's about the reaction of interest. So let's say we took a look at, say,
19:04
H2O solid going to H2O liquid, and H2O liquid going to H2O gas.
19:15
Now, if you look at this, and we'll start with the second one first, in going from a liquid to a gas, we're gaining moles of gas. And we learned earlier in the chapter that if you increase in the number of moles of gas, it's definitely gonna have a positive Delta S for the reaction or process. So in this case, that's definitely got a positive Delta S, but again, to boil something, to go from liquid to gas definitely takes heat, and so it's an endothermic process.
19:37
And so we've got exactly one of these scenarios, where we've got an endothermic process that results in an increase in entropy for the system, Delta H and Delta S are both positive. And you very well know that to be spontaneous, to boil water, at least at one atmosphere pressure, you've gotta have a minimum temperature of 100 degrees Celsius. It's gotta be over 100. Anything less than 100 degrees Celsius, it's not gonna be spontaneous. Anything over 100 degrees Celsius, it will be spontaneous.
20:02
And so in this case, when we say spontaneous at high temperatures, we mean spontaneous above some threshold temperature, in this case, 100 degrees Celsius, or 373 Kelvin. Same thing here, with going from solid water ice to liquid water, so in this case, it's gonna be all about zero degrees Celsius, right? Which maybe you don't consider high, and I don't consider it high, living in Arizona, but it's not about you or me. So in this case, it's all about just being
20:29
above some threshold temperature, and so in this case, for the process of melting, for in this case, water, you've gotta be above zero degrees Celsius, not below. Any temperature above that would be considered a high temperature for the melting of water, and any temperature below zero degrees Celsius would be considered a low temperature for melting of water, so that's kind of how that works. So I just wanna make sure when you see the word high here, you don't think
20:53
about what you personally consider hot or cold.
20:57
All right, if we go one step further, and by the way, one thing to note here, I wanna clarify that high temperature. If you were asked to calculate at what temperature it is, well, again, the key is, you know, if we looked at like, say again, the boiling of water, so if you plug in the Delta H for vaporization, and the Delta S for vaporization of water, well, it turns out if you plug in any value of T higher than 373 Kelvin, again, it's gotta be put in Kelvin, so higher than 100 degrees Celsius,
21:27
which is 373 Kelvin, then Delta G's gonna come out negative. If you plug in any temperature lower than 373 Kelvin in for T, then Delta G would come out positive. And if you plugged in exactly 373 Kelvin, that's when Delta G would come out to zero. And so oftentimes you're asked to find that threshold temperature. And so to find that threshold temperature, you just set Delta G equal to zero, and then plug in the value for Delta H and Delta S, and then solve for that temperature
21:53
that comes out in Kelvin. Now, we'll see some examples of later, you just gotta be careful on your units, because Delta H is usually given in kilojoules per mole, but Delta Ss are usually a lot smaller than Delta Hs, and those are usually given in joules per mole Kelvin. And so with this being in kilojoules per mole, but this only in joules per mole Kelvin, you either gotta make 'em both joules, or both kilojoules before you start doing your calculation. All right, so last scenario here,
22:19
where Delta H and Delta S are both negative,
22:24
so if Delta H is negative, well, that means definitely our first term is simply negative, but if Delta S is negative, again, a negative times a negative means your second term is positive. And again, that second term here, that a whole negative T Delta S is always gonna have the opposite sign of whatever just plain old Delta S has. And so in this case, again, we're adding two numbers, one's negative, one's positive. If my goal is to come out with an overall negative number,
22:48
that way Delta a G comes up negative, and the reaction is spontaneous, well, then I want the negative term to have the larger magnitude. Well, again, I can't really change the value of Delta H in any significant way, but the second term has temperature in it, and in this case, I don't wanna make it as big as possible, I wanna make it as small as possible. That way the positive term will be smaller than the negative term. And so in this case, I wanna limit the size of it
23:08
by lowering the value of T I plug in. And so here, we'd say that the reaction is spontaneous
23:20
at low temperatures. And again, it's not what you or I would necessarily consider low, it just means below some threshold temperature. And so instead of melting ice, what if I wanted to freeze liquid water? Well, if freezing liquid water is spontaneous below zeros degrees Celsius, it's non spontaneous above zero degrees Celsius. That's kind of how it works, so low in this case would just mean below zero. Well, what if I wanted steam to condense into liquid water?
23:44
Well, again, steam will condense into liquid water if you're below 100 degrees Celsius. So would 90 degree Celsius considered a low temperature in this case? Yes. Now, do you personally consider 90 degrees Celsius a low temperature? No. So again, it's all about, is it below the threshold temperature for the given reaction you're looking at? Cool, you've gotta know these four situations, and when that reaction's gonna be spontaneous, either at all, temps, no temps, high temps, or low temps.
24:11
Now, a lot of students simply are just gonna memorize this chart, and if you want to go that route, fantastic. So, whereas some students will actually work out the math every time in such fashion, and I personally do it that way myself. Truth be told, I kind of have memorized now as well. So, but I personally like that way. So, but I usually get about half students working out the math, half the students just memorizing this chart. Whichever one makes more sense for you personally,
24:38
go with it, but at the end of the day, you are definitely on the hook for knowing, given the sign of Delta H and Delta S, under what can conditions that reaction is gonna be spontaneous in terms of temperature. So the first problem we're gonna look at is you're given a reaction here, N two plus three H two, going to two NH three. You're given the Delta H for this reaction, and then you are asked whether or not this reaction is spontaneous at all temperatures, no temperatures,
25:01
high temperatures, or low temperatures. And so obviously you're just supposed to match it up with one of these or scenarios. Well, right off the bat, you're told Delta H is negative. It's an exothermic reaction. You know that for sure you're giving the universe one of the two things it wants. The question is what is Delta S? Because unfortunately, Delta S was not provided. However, in the last lesson, we learned how to predict the sign for Delta S, and the single most important factor was
25:28
change in moles of gas. And we're definitely going from one, two, three, four, these are both gases, so four moles of gas, ending up as two moles of gas. That's an overall decrease in the number of moles of gas, and therefore a definite decrease in entropy. That tells us that Delta S is negative. And so, as a result, we've got Delta H and Delta S both being negative, and that reaction's definitely only gonna be spontaneous at low temperatures. And again, if you wanted to look at this
25:56
kind of mathematically, you'd go back here and say, "Okay, if Delta H and Delta S are both negative, well, then my first term is negative, but a negative times a negative means my second term is positive, and if I want the negative term to win, but Delta G comes out negative, then I wanna limit the size of the second term, which means make that temperature small." And so that's where we get this idea of it being spontaneous at low temperatures.
26:20
All right, that is the first example problem. Let's do a little bit of math here as well. Okay, so first thing I wanna look at is phase changes with regard to Delta G, Delta H, and Delta S, and so it turns out, when you are at a phase change temperature, like the boiling point, or the melting point, AKA melting point same thing as the freezing point, FYI, it turns out you have two phases in equilibrium. At the boiling point, you have liquid and gas in equilibrium.
26:44
At the freezing or melting point, you have solid and liquid in equilibrium. And so having them in equilibrium means that Delta G equals zero at the phase change temperature. Well, if we look at what that implies then, we could say that zero equals Delta H minus T Delta S.
27:04
So, but again, in this case, let's say, we're doing this for this problem. We're gonna, we're given the Delta H of vaporization, and the Delta S of vaporization. That means the temperature at which this is gonna be true is gonna be the boiling temperature. And then we can rearrange this, and notice, we can just take this Delta H vaporization on this side, and on this side, we'll add T times the Delta S vaporization, and then we'll divide through by that boiling temperature,
27:32
and get the Delta S vaporization equals the Delta H of vaporization over the boiling temperature. So, and that's where this kind of comes from, so, and you can do it again for any phase change. So at any phase change, Delta G equals zero, and we can derive expressions like so. So, and as a result, you've now got two equations. They've got three variables, and if I give you two of them, you should be able to calculate the third, which is what we'll do in this next problem.
27:56
It says for water, Delta H vaporization is 40.68 kilojoules per mole, Delta S vaporization is 109 joules per Kelvin mole, what is the boiling temperature of water? So we're wanting to get this boiling temperature. So in this case, we can rearrange this, and get that the boiling temperature, so we'll move this up, and then we'll divide through by the Delta S vaporization, but it's gonna equal the Delta H of vaporization, divided by the Delta S of vaporization,
28:26
and again, both these values were provided. Now, we gotta be a little bit careful. Delta H vaporization was given us 40.68 kilojoules per mole.
28:36
The Delta S was given as 109 joules per Kelvin mole.
28:45
So whether it's joules per mole per Kelvin, or joules per Kelvin mole, same diff. So, and then we can calculate that, and we'll see that that's where the Kelvin part is gonna come from. So we need all the other units to cancel, though. And the mole's part that's great, but kilojoules and joules, those aren't gonna cancel each other out. And so here's, again, one of those places you gotta be careful. In fact, Delta G and Delta H are most commonly reported
29:06
in kilojoules, because they're significantly larger than Delta S, which is only reported in joules per Kelvin. So it's kinda like if I said, "Hey, how tall are you?" And you said, you know, "Six foot tall." And I said, "No, no, no, how many miles tall are you?" And you just kind of looked at me funny, because you probably have no idea how many miles tall you are, some, you know, .05 or something miles tall, or maybe probably smaller than that, actually. So, but we don't know, we don't,
29:32
we would never measure our height in miles. That would be the most inconvenient unit ever. Well, same thing, it turns out, for Delta S. Delta S, we wouldn't use kilojoules, 'cause here, instead of being 109, it would've been .109. And so it's much easier to, instead of talking about fractions, to talk about whole numbers, and that's why for Delta S, which is usually much smaller than Delta G and Delta H, we'll use joules per mole Kelvin instead of kilojoules.
29:56
But it does mean that when you're starting to do some calculations, you need to check your units. Put everything in kilojoules, or everything in joules. It's your choice. It'll work out the same either way, but you need to go there. So we need to change some things here. And personally, I like changing kilojoules to joules, instead of, instead of 46, 40.68, we're gonna have 40,680.
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Let's make that eight look a little better.
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And now that's in joules, rather than kilojoules, and now we're ready to do a little bit of math. Let me grab my calculator.
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All right, so 40,680 divided by 109, it's gonna give us 373.2 Kelvin,
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which would be right around 100 degrees Celsius. Now, obviously, you could have been like, "Chad, I already knew the bowling temperature of 100 degrees Celsius. Why would I have ever done it this way?" That is a great question. Why would you have ever done it this way? You should have been like, "I know that one." So, but aren't they're gonna give this to, you know, kind of a calculation for something other than water? And I'm guessing that you haven't memorized
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the boiling temperatures of every known liquid, right? So, but I wanted to work one out for one that you did know, just to show you how it worked out. And again, the big things here is that when you solve for your temperature, it's gonna come out in Kelvin, so your Delta H and Delta S have to either both be in joules, or both in kilojoules for this to work out properly. So, but oftentimes, a professor loves to get you this far. Not only could you have made a mistake in not making
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your units match, but then once you get the right answer, they might put 373 degrees Celsius as one of the answer choices on the test, but it's not 373 Celsius, it's 373.2 Kelvin. The correct answer in Celsius would have been subtract off 273 and been 100 degrees Celsius. So be careful, a couple different ways professors can kind of give you some good detractors to throw you off on just such a calculation. Now, if you found this lesson helpful,
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the thumbs up button and a comment are the best things you can do to support the channel. And if you're looking for practice for your general chemistry course, take a look at my general chemistry master course. It includes over 1200 practice questions. I'll leave a link in the description. Free trial is available. Happy studying.
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