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0:01
- Equilibrium and equilibrium constants are gonna be the topics in this lesson. My name is Chad and welcome to Chad's Prep, where my goal is to take the stress out of learning science. In addition to high school and college science prep, we also do MCAT, DAT, and OAT prep as well. You'll find those courses at chadsprep.com. I'll leave links in the description. Now, this lesson's part of my new general chemistry playlist. I'm releasing several lessons a week throughout the school year.
0:23
So, if you wanna be notified every time I post a new one, subscribe to the channel, click the bell notification. All right, so, we're first gonna start off with some definitions here and the first is simply the definition of equilibrium. In chemistry, we talk about what's known as a dynamic equilibrium. The important part here is that when a reaction reaches equilibrium, the concentrations of reactants and products are no longer changing and this is important. Back in first semester gen chem,
0:48
we talked about limiting reagent problems and things of this sort. The assumption we made with those limiting reagent calculations is often a reaction that went to completion until one of the reactants ran out. Well, it turns out that's not often all of, it's not always gonna be true, it turns out. Most reactions don't actually go to completion and often or even close to completion. They're gonna reach some intermediate, what we call equilibrium. And once they reach equilibrium,
1:12
the concentrations of reactants and products are no longer changing. However, it's not because the reaction has stopped, it's just that the reaction is going in both the forward and reverse directions at exactly the same rate. And that way, every time we convert reactants into products, we're converting products back into reactants at exactly the same speed. That way, the overall concentrations aren't changing. That's what we mean by a dynamic equilibrium. A static equilibrium would decide
1:40
something that's completely stopped and that's not what we have typically in chemistry. We have, excuse me, this dynamic equilibrium, again, where the forward rate equals the reverse rate and that's the definition of equilibrium you wanna take home. Now, keep in mind again, this means the concentrations aren't changing anymore. However, it does not mean that they are equal and this is a huge misconception. At equilibrium, the concentrations of reactants
2:03
and products do not have to be equal at all. For some reactions, they end up with a lot more products when they reach equilibrium. For other reactions, they end up with a lot more reactants when they reach equilibrium. And still yet for other reactions, you're gonna end up with a fair amount of both. But there's nothing that says they have to, in principle, be equal for any system that has reached equilibrium. For some small number, that might be true,
2:22
but for the vast majority, it's just simply not the case. So again, it means that the forward rate of the reaction equals the reverse rate of the reaction, but it does not mean that the reactant concentrations and product concentrations are equal. Now, in addition, you should know that we're gonna talk about what's called an equilibrium constant. So in this equilibrium constant, we're gonna symbolize with the letter K, and it's gonna be a capital K. And I point that out in this case
2:49
because in the last chapter on kinetics, we used a lowercase k to represent the rate constant. And we saw that with a rate constant, they were affected by temperature and they were affected by when you add a catalyst, 'cause that lowers the activation energy. So, it's affected by both temperature and activation energy. Well, this equilibrium constant, the only thing that can change it is a change in temperature. If you add a catalyst, it doesn't change.
3:12
So if you'd change the activation energy, it doesn't change. But only a change in temperature is actually gonna lead to a change in the numerical value of an equilibrium constant. So, we'll find out a little bit more about what these means. But almost always, when they give you an equilibrium constant, they're gonna specify what temperature it applies to because if you change that temperature, it's probably gonna change the value.
3:31
All right, so, it turns out this equilibrium constant is equal to the ratio of the concentrations of products to the concentrations of reactants, raised to the power of their coefficients. And you should know that typically. only aqueous and gaseous species show up. No solids, no liquids. Now technically, liquids might show up if they are not the solvent. So, if you've got some liquids in combination but none of them is actually the solvent, well, it's when the liquid is a solvent
3:56
that it's not gonna appear. So on the odd case, and in this often happens more like in a particular laboratory experiment you might do, you might see some liquids. But in the lecture, you're probably not likely to see liquids showing up in these equilibrium constant expressions. And so in this case, we're gonna say, it's the ratio of the concentration of the products, CO, raised to the power of their coefficients. So, it's got a coefficient two. So, product is on top in the numerator here
4:20
and these brackets mean molar concentration relative to the concentrations of the reactants. And notice, we're leaving carbon out of this 'cause carbon is a solid and it turns out that
4:31
solids have an activity of one, we say. And so technically, if you put its activity down here, if you take an advanced thermodynamics course, you'll actually find out that instead of concentrations, it's really activities and solids have an activity of one and anything multiplied or divided by one is itself and it doesn't affect the equation. So, that's ultimately why solids and liquids are gonna stay out of here. They have an activity of one in most cases and so they just stay out.
4:57
Well, in our cases, we're not talking about activities. We're simplifying this in gen chem. We're just going straight to concentrations of either aqueous or gaseous species. Now, in this case, we've got, again, the concentration of CO squared all over the concentration of O2. However, for reaction involving gases like we've got here, you might also use partial pressures instead. And so, you might also give this ratio a little bit different and say, partial pressure of CO squared
5:25
all over the partial pressure of O2. Now, I've got a problem because partial pressures and molar concentrations are not necessarily equal for gases. So, if you come and take a look at the ideal gas law for a second, so say, PV equals nRT. You'd see that we can relate pressure to molarity here in kind of a unique way. So, pressure is gonna equal n over V times RT. If I solve for P here and divide through by the volume, moles over the leaders is molarity.
5:58
And so we see that pressure and molarity are not equal. There's this extra factor of RT that's going to be important as we'll see in a little bit. And so, it turns out with gases then, you have two different ways of expressing the equilibrium constant. And because pressures and molarities are not equal, then the equilibrium constant values themselves won't necessarily be equal and so we've got to actually define two types of equilibrium constants. If you express it for gases
6:20
in terms of molar concentrations, we'll call it a Kc value. And if you do it in terms of partial pressures, we'll call it a Kp value and these are kind of the proper designations and we'll find out a little bit that Kp and Kc under certain cases might be equal in value, but most of the time they won't. But we'll see how to kind of, if you're given one, figure out the other here in a little bit. Now in general chemistry, it's quite common to generically just use the term Keq.
6:47
And based on context, you got to kind of figure out, are they talking about Kc with molar concentrations which is probably most of the time. And if it's a reaction involving only aqueous species, it would have to be a Kc, they don't have Kps.
7:00
But technically, this is generic and technically it's kind of improper. If you were taking an advanced thermo class or advanced physical chemistry class, you wouldn't see this showing up there. They're either gonna use Kc or Kp. They're never going to use Keq. So, it's kind of improper term but we use it in general chemistry all the time, just so you know. But I'm gonna refrain from using it. I'm gonna try and do my best to be proper. I just wanted to give you a heads up
7:21
in case you see it in your class or in your textbook.
7:26
Okay, so, these equilibrium constants, again, are constants. The only thing that can change them is a change in temperature. Now, one thing to note, we're actually got to get a little tiny bit more specific than this. So, it turns out these concentrations have to be measured at equilibrium and I'm writing this in red because we never specify this. I'm adding this myself so you know that this is the concentration of CO for a reaction that has reached equilibrium
7:51
and this is the concentration of O2 for the reaction that has reached equilibrium. But that's kind of implied when you set this ratio equal to K as it turns out. So we don't actually write the eq there, it's just implied. So, what if the reaction hasn't reached equilibrium? Well, then this ratio
8:08
wouldn't actually equal your K value yet. We'll find out in the next lesson, there's something called the reaction quotient that looks at the same exact ratio for a system that doesn't necessarily have to be at equilibrium. So, I just wanna allude to that for the future.
8:25
Cool, so, big thing about these equilibrium constants, they're super handy because by a single numerical value, you can figure out, is this a reaction that's gonna end up with a lot of products or not so much? If you work in an industrial factory and stuff like this and you wanna make a lot of a certain particular chemical, well, then the hope is that it's reaction where you can make a lot of that chemical. So, let's take a look at what I mean here.
8:51
So, it turns out you're gonna have one of really three scenarios. You're either gonna have an equilibrium constant that is a lot greater than one, an equilibrium constant that is a lot smaller than one, or you're gonna have an equilibrium constant
9:06
that's somewhere in between. And when we say in between, oftentimes we actually set the outer boundaries for what we mean by in between. And so for an equilibrium constant that's close to one, that's really kind of nebulous and you're gonna be like, that's not very close to one, Chad. Well, it's just kind of the boundary. Sorry, but if it's in between 0.001, 10 to the -3, and 1,000, 10 to the third, that's what we're gonna say is K that is close to one.
9:29
And depending on where an equilibrium constant lies, if it's in this range, this ranger or this range, it's gonna tell you something important about this reaction. Now, in the case where you're equilibrium constant is a huge number like, let's say, it's equal to 1,000,000,000, 10 to the ninth. Well, in that case, that means you're gonna have a very large numerator and a very small denominator. In comparison, your numerator is gonna be way bigger than your denominator.
9:50
Well, that means you're gonna have way more products than reactants when that reaction reaches equilibrium. And to describe that, we say that that reaction favors
10:01
the products at equilibrium. And that'd be the ideal scenario if you're trying to make this particular chemical in your industrial factory. If it's the product of the reaction, then you'd be really convenient if it has a very large equilibrium constant because when that reaction reaches equilibrium, you end up with a lot of that product. Now, on the other hand here, if your equilibrium constant is way smaller than one, like let's say instead of 1 billion, it's 0.0000001, a really tiny number.
10:29
Well, what that's gonna indicate is that you have a very large denominator now, a lot of reactants and a very small numerator in comparison, a small amount of products. And so, because you have lots of reactants and not very many products at equilibrium, we say that that reaction favors the reactants.
10:51
Now, just because the reaction favors the reactants at equilibrium doesn't mean you can't make a lot of the product in that reaction, but it probably is gonna be a little more difficult and it might be a little more expensive to manufacture and things of this sort. But we'll find out in the next lesson, there are ways to still shift the equilibrium as we'll see in the next lesson and actually still make a fair amount of that product. Now, your last scenario here is again
11:11
for an equilibrium constant that's somewhere close to one, and again, close to one means bigger than 0.001, smaller than a thousand. And in this case, it doesn't really favor the reactants or the products we say, but you're gonna have a fair amount of both. Now, these are kind of somewhat arbitrary where we set these boundaries but they're kind of given good liberty to what approximately one or close to one means. And the reason they do that is 'cause
11:33
the number of products and reactants you might have,
11:37
don't have to be that similar in some cases. So if you notice here, like in the numerator, we've got a squared term, in the dominator, we don't. And so let's say that both these concentrations were equal. Let's say that, I had 10 molar O2 and 10 molar CO for some reaction that reached equilibrium at some temperature or something like that. Well, if these are both equal to 10 molar, they're equal. However, notice your equilibrium constant is gonna be 10 squared over 10.
12:00
So, 10 squared is a hundred, divided by 10 equals 10. And so even though you have equal amounts or equal concentrations of reactants and products, that doesn't actually indicate an equal and constant that's exactly equal to one and that's the difference of having a square on top and only having it to the first power on bottom. And you have other cases where it's not just a difference between a square and a first power, but maybe a cube or to the fourth power or something like that.
12:22
And so you can end up with big differences or big deviations of your equilibrium constant away from one but still have roughly equal amounts of reactants and products. And so that's why they give huge Liberty to saying approximately one or close to one. Well, again, bigger than 0.001, smaller than a thousand. Cool, so now if I give you an equilibrium constant, and if I say, like actually, if you look at this reaction here, kind of normal room temperature here, this is a combustion reaction.
12:50
It is very spontaneous, we'll learn later in the semester and it goes almost entirely to completion like most combustion reactions do. Now, this is only partial combustion but it's still highly, highly favorable and it goes virtually to completion. If you actually looked up at equilibrium constant at a certain temperature like say, 298K, you probably find it's like 10 to the 40th or 10 to the 50th. It is a ginormous number.
13:13
And when you see a number that is that big, that it heavily favors the products. In fact, if it's that big, like 10 to the 40th, 10 to the 50th, it virtually goes to completion. So, it's just an FYI. Now, on the other hand, if you have an equilibrium constant that's at the other end of the spectrum, like let's say it's 10 to the negative 10. Well, if you have an equilibrium constant on power of 10 to the negative 10, then it's gonna heavily favor the reactants.
13:36
You're gonna have way more denominator than numerator for that reaction to equilibrium. And then finally, if I said, you're equal and constant is equal to like 13, well, 13 is definitely in this range and you're probably gonna have a fair amount of both reactants and products and you might be like, well, maybe more products, but not necessarily really depending on how many squares and stuff you have in your numerator and dominator. But it is in this range if I said it had a value of 13
14:00
and so it would have both reactants and products at equilibrium. So, that's what an equilibrium constant ultimately allows us to do. From a single number, we know something about how much product we can make from that reaction. Now let's take a look at Kp and Kc one more time. So, we said that again, these don't necessarily have to have the same value. So, and again, we only have Kcs and Kps for reactions that involve gases. If you only got aqueous species, you only gonna talk about Kcs.
14:29
There is no Kps for those. So, it's only for reactions that involve gases that you can have these Kps as well. And again, Kc and Kp don't have to be numerically equal, but they might be. So, let's go back and take a look one more time.
14:43
So we rearranged the ideal gas a little bit ago and said that, pressure equals molarity times RT. And so, if you were to take and substitute for every one of these pressures here, the molarity, the molar concentration times RT, well, you'd end up with a molarity on the bottom just like here and a molarity squared on top just like here, but you'd end up with an extra factor of RT on the bottom and an extra factor of RT all squared on top, and they're not gonna cancel.
15:09
So, you've got that extra factor of RT you've got to factor in and there's an equation for this. So and the equation just goes, Kp equals Kc
15:19
times RT to the Delta n power. And Delta n here refers to the change in the number of moles of gas only. Only gas.
15:29
And so notice, if that change in the number of moles of gas is zero, well, anything to the zero power is one and anything times one is itself. And so if Delta n is zero, I.e. you have the same number of moles of gas on the reactant side as on the product side, for that case and that case only would Kp equal Kc. But if Delta n doesn't equal zero, then Kp and Kc won't be equal. And so you might get a question on the test that just says, for which of the following reactions would Kp equal Kc?
15:56
You don't have to do any calculation here. You just gotta look and say, okay, for which of the provided reactions is there an equal number of moles of gas on the reactant side and on the product side. And again, leave out the solids, leave out the liquids, leave out the aqueous, Delta n here is only the change in the number of moles of gas. So like in our reaction here that we've been looking at, leave out the solids, but we have one mole of gas on the reactant side,
16:18
two moles of gas on the product side, Kp is not gonna equal Kc. And in fact, that's the next thing we're to look at, is a calculation involving this reaction 'cause that's the other thing you might have to do. You might actually have to do the math here and you might be given Kc and asked to calculate Kp or given Kp and asked to calculate Kc. Either one of those could show up and that's what we're gonna take a look at here. And so, question on the study guide here says,
16:41
if for this reaction, which has Kc equal 1200,
16:51
and it says, at 400K. And so in this case, it turns out again that the one thing that can change the value of an equilibrium constant is temperature. And so oftentimes when you're given an equilibrium constant, you're almost always gonna be given the temperature at which that equilibrium constant applies. 'Cause if the equilibrium constant Kc here is 1200 at 400K, at 300K, it's gonna be a different number. At 200K, it's gonna be a different number again and stuff like that. And so almost always,
17:16
when they give you an equilibrium constant, they're gonna specify what temperature it applies to
17:21
and just keep in mind that the temperature is not necessarily gonna be used in every calculation you see. It's just proper to give you a temperature for which that equilibrate constant applies. But in this case, we're definitely gonna use that temperature as part of this. So, we're given Kc, we're being asked to find Kp. Let's do some plugging and chugging here and so we've got Kp equals 1200. Now, we haven't talked about units yet, but these equilibrium constants are unit less, it turns out.
17:49
And you might be like, wait a minute, Chad, that's not possible 'cause you got the molarity here squared over molarity here. Molarity squared divided by molarity is actually molarity, Chad not no units. I know and that's the problem here. Technically, I told you that we write in the little eqs here, but they're not really there. It's implied if you set it equal to K that you're measuring these concentrations at equilibrium. Well, the other thing that we often don't write in
18:16
and we'd be a little more explicit about in Pchem class, something like that is that technically, we're always gonna divide every concentration by unit molarity, one molar. And so here would be the concentration of CO over one molar all squared and you have molarity over molarity and the units cancel. Here, you have molarity over molarity and the units cancel. We don't write this in, but it turns out it's also implied. It actually is technically a part of what an equilibrium constant is
18:41
but you'll not likely to see it in a gen chem class. But for that reason, equilibrium constants never have units. Notice, it's not the same as rate constants. Rate constants, all we had units and it all depended on what order the overall reaction was and stuff. But equilibrium constants are always going to be unit less. At least that's proper.
19:02
All right, so, we got an equilibrium constant of 1200
19:07
and in this case we've got this RT and in this case, your R is gonna be the 0.08206 liter atmospheres per mole Kelvin.
19:20
This is the one you're gonna end up using. Now, if you look here, this was the value we used when we did PV equals nRT ideal gas calculations. Although for much of this second semester, you're gonna be using the one involving energy, more proper and you know SI energy units, the Joule, the 8.314 Joules per mole kelvin. When you're dealing with energy, you're almost always gonna be using that value. But anything that goes back to PV equals nRT, in fact that's derived from PV equals nRT
19:46
is still gonna be using this value. In all likelihood because of you're gonna be using both this semester, you're probably gonna be provided with both and you've got to know when you're gonna use one versus the other. Well, again, anything involving gas is back in chapter 10 that we saw or in this one place, in this formula right here. This is the one we're gonna be using with the liter atmospheres per mole Kelvin. But any other contexts it's likely to come up in in general chemistry
20:09
is probably gonna come up in terms of energy like we saw with activation energy in equation last chapter and that was customary to use the 8.314 Joules per mole kelvin. So, I just wanna make sure you realize which value of R your gonna use. Now, in this case, it's the only one I'm providing you with but on an exam or a homework company, you got to know that, oh, this is the one place in second semester you're likely to have to go back and use the value we used back with PV equals nRT stuff.
20:34
All right, so here we've got this
20:37
0.08206 liter atmospheres per mole Kelvin
20:43
and then temperature in this case, again, 400K given in question,
20:49
and then we got this Delta n. So, we're gonna put this in brackets. And then the Delta n again, is the change in moles of gas and a change in anything is always final minus initial. And so, the final number of moles of gas we end up with is two and the initial number of moles of gas is just one. And again, we're ignoring the solids. And so, final minus initial, two minus one is going to be positive one. And so this is gonna be raised to the positive one power and that just signifies that
21:16
we're increasing by one mole of gas and going from one mole of gas to two moles of gas. That's an increase of one mole of gas total and that's what actually that Delta n signifies. So when it's an increase, then you have more on the product side, it's gonna be positive. And when you have a decrease, there's more on the reactant side. Then it's gonna end up being a negative Delta n in that case. All right, so, and from here, it's just plugging and chugging. So, let's pull out our calculators.
21:38
And so in this case, we're gonna take 1200 times 0.08206 times 400, and we don't have to square to anything. It's just to the one power. And so we're gonna end up with a Kp value in this case, 39388.8 and I'll probably just round that to
21:55
a couple of sig figs and round it to 39,000.
22:04
- And again, also unit less. So, the Kc value is 1200. The Kp value is 39,000. Notice they're not that similar in this case but that's pretty typical. But notice both of these, you would have probably recognized as meaning the reaction's gonna favor the products. They're both bigger than a thousand, so quite a bit bigger than one. All right, the last thing we're gonna do in this lesson is we're gonna look at one reaction. We've been given the value of the equilibrium constant
22:28
for is equal to 16 in this case. And then we're gonna be given some related reactions and asked to give a numerical value for their Kc values, their equilibrium constants as well. And so, if you look these reactions can't just be random. They've gotta be related to the first one we've been given the Kc value for it. And so if you look at this top reaction, you look at the one right below it, how are these related? Well, if you look, the second reaction
22:50
is just the exact reverse of the first reaction and there's gonna be a relationship we can derive from that. Now, a lot of students see oh, it's reversed. I guess if Kc here is equal to 16, maybe here it's equal to negative 16. Well, equilibrium constants are never negative so we can just scratch that right off the bat. If they're just concentrations of products of reactants, the only way they can come out negative is if you have a negative concentration which is impossible.
23:13
So, no such thing as negative equilibrium constants. So, we'll come back to that in a minute. If you look at the next one here, how does that relate to the first one? Well, it's the same thing as the first reaction with everything cut in half. So it turns out again, you might be like, well, if I carried this reaction out in a beaker, whether I wrote it like this or wrote it like this, it'd be the same reaction. So, I've got SO2 mixing with O2 to form SO3 just like we do here.
23:36
However, if we change this to equiometry, we treat this as a different reaction. And so you might realize, earlier I said that the only thing that will change the value of an equilibrium constant is temperature. And what I mean by that though is, in such a case like this, if you change this to equiometry, you have changed the reaction. I realized if it's happening and occurring in a beaker, it's the same reaction. But if you change this to equiometry, we view it as a different reaction
23:59
and it will change the value of the equilibrium constant as we'll see. But in this case, you find out that we've cut all the coefficients in half and before you think, well, that might cut the equilibrium constant in half, maybe it's equal to eight. Well, you'll find out that's wrong again as well, but based on our definition of equilibrium constant, we'll see why that is and we'll get a nice little pattern for it. Then finally for this last one,
24:18
it relates to the first one and the fact that it's reversed, you're gonna go in the reverse direction and cut everything in half. We'll see how that plays out. Now, to help us do this, I'm gonna put the equilibrium constant expressions for every one of these reactions up on the board to make this a little easier to see. All right, so we're gonna start with this first one here or the second one, I guess really technically. And if you look at the expressions here,
24:40
for the first one products over reactants is SO3 squared over SO2 squared times O2. Whereas for this one which is again, the exact reverse reaction to the first one, we've got written out here, so again, products over reactants now it looks like so. And if you look at the expressions, so you'll see that this is the exact inverse of what we had up here. And so if this one is equal to 16 and notice that 16 is the same thing as saying 16/1. Well, if you flip this over,
25:11
then you're gonna get an equilibrium constant now of one over 16 and you can put that in your calculator and find out what that is in decimal form or something like that. But it's 1/16. And so what we've just learned now is that reversing the reaction just based on how equilibrium constant expressions here are defined, products of reactants raised to the power of the coefficients, reversing the reaction inverts the value of the equilibrium constant. So for the first reaction, Kc was equal to 16.
25:35
For this second one, that's the reverse reaction. It's now a 1/16.
25:40
Okay, so let's take a look at the next one here. Let's get a Kc expression up for this one as well. Okay, so for this next one here, relating it back to the first one here again, it's in the same direction, but all the coefficients are cut in half, twos become one, ones become 1/2, twos become one. And with all the coefficients cut in half, don't forget that in the equilibrium constant expression, your coefficients show up as the powers. And so if you're cutting all the coefficients in half,
26:05
you're cutting all the exponents in half in your equilibrium constant expression. And cutting your exponents in half is not the same thing as cutting the number in half. And so our Kc for the first reaction 16, the Kc here is not gonna equal eight. You don't just divide it by two. You have to take the one half power when you're cutting exponents in half and the one half power means square root. And so in this case, your Kc is actually not gonna equal 16/2,
26:29
it's gonna equal to square root of 16, which is four. And so cutting your coefficients in half square root, cutting them by 1/3 of their value, cube root, doubling them, square the equilibrium constant value, tripling them, cube the equilibrium constant value. That's kind of how that works. And again, you can figure it out if you just do a simple comparison of the equilibrium constant expressions, but that's kind of time consuming. And on the exam, you wanna get to the point
26:55
where you can look at the relationships, is it forward? Is it reversed? Is it doubled? Is it halved? Those kinds of things and be able to figure out what the equilibrium constant is gonna be without even having to write these expressions out. But in a pinch, if you still need to write these expressions out to figure out how they're related mathematically, great. It'll still lead you to the right answer. So, let's take a look at this last one here.
27:15
In this last one, we not only reverse the reaction, we also cut it in half. All right, so there's our expression. And if, again we related this back to the first one, we'd see that not only is it reversed, but also all the equilibrium constants, oh sorry, all the coefficients are cut in half which again means square root. So reversing it means take the inverse, cutting all the coefficients in half means take the square root. Now, we've got to do both.
27:40
We've got to inverse it and take the square root and it doesn't matter which one you do first. They're linear operations, so order doesn't matter it turns out. And so in this case, if we start with 16, I can invert it and get 1/16 and then take the square root and get 1/4.
27:59
So I noticed if you did it the other order, it'd be the same thing. If you said you want to take the square root first, square root of 16 is four and then inverting it would get your 1/4.
28:08
Cool, but these are the kinds of relationships again, you might be on the hook for, make a great exam question. I just wanted to show you some patterns here. You wanna get good at this for the exam again, to save yourself some time. But if at the end of the day you need to write these out, it will still lead you to the right direction, just a little more time consuming.
28:25
Now, if you found this lesson helpful, that thumbs up button goes a long way to letting YouTube know that other students should see this lesson as well. And if you're looking for practice problems on equilibrium, check out my general chemistry master course. I'll be sure to leave a link in the description, free trial is available. Happy studying.
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