Equilibrium Constant Kc calculation - Exam Question Grade 12 Chemical Equilibrium

Added: 2026-05-14

[00:00:00]Let's look at another KC exam question, equilibrium constant exam question. I've labeled this one medium hard. I wouldn't really say that, oh, it's super difficult, super super difficult. I would say medium hard as in this is sort of what you could expect if they're trying to make it a little bit challenging in your exam. So, let's get into it. The question says methanol, which you should recognize, that's an alcohol, is manufactured, we mix carbon monoxide hydrogen, they give me the pressure. This is not relevant in this question. The mixture is passed over zinc oxide that acts as a catalyst.

[00:00:33]Again, not relevant in this question, and they give me the temperature. Then they give me the chemical equation, and they say, initially X moles of carbon monoxide and eight moles of hydrogen are placed in a container. They give me the volume of the container, and again the temperature. When equilibrium is established, then they give me the concentration of hydrogen, this one, in the container, and they give me the value of the equilibrium constant.

[00:00:57]Remember, equilibrium constant is KC.

[00:01:01]It's the KC value, and it's given it has no unit. Define the term chemical equilibrium, and that we know we need to memorize. It is a dynamic equilibrium, and this is the important part, when the rate of the forward reaction equals the rate of the reverse reaction. You need to say rate twice. Rate of forward equals rate of reverse. Okay, let's get into the calculation. Determine the initial amount X of CO, and X is in moles. So, the initial moles of carbon monoxide that was placed in the container. Nine marks. So, grade 12s, or people that are rewriting, whoever's watching this video, when you get to question six in your prelims or your finals, even your June exam sometimes, you know it's a chemical equilibrium question, and you see a nine mark question, and somewhere they mention equilibrium constant, you know you have to draw what we call the KC table, what I call the mice cube table. Some teachers call it the rice table, some teachers call it the ice table, it doesn't matter as long as you know what everything stands for. As you know, I've done this in my previous videos, the more simple videos. So, if you've missed that, please go watch that first. So, I draw the skeleton, the outline of my table, and then what I also do, because this gets you one mark, is you write out the KC expression for this particular reaction. Remember, you can't just write a general KC expression like this. I've seen it a lot in exams where students think writing KC equals concentration of products divided by concentration of reactants, like those words, they think that gets them a mark. It doesn't. What will get you a mark is doing the specific KC reaction. So, remember, only gases and aqueous solutions goes into the table and into my KC expression. These are all gases, so we're all good. We can put them in. Products will always go at the top. My product is C, you square brackets, C3OH, and that's my only product. Then these two are my reactants. So, CO in square brackets, and then H2 to the power of two. The reason why I'm putting it to the power of two is because the coefficient here is two. The coefficient of this one and this one is a one, which is why I don't need to put the one in there, but you know that it's one. And remember, you're multiplying these two at the bottom together. Okay, then let's leave that for now. Let's go look at the question and see what information I have. So, I It says here initially, X moles of carbon monoxide and 8 moles of hydrogen are placed in the container.

[00:03:25]So, I in your table is initial moles.

[00:03:29]Remember, this is a mole table. So, these three rows is moles. So, initially, X moles of carbon monoxide and 8 moles of hydrogen are placed in a container, given. One thing that I'm You must always fill out in the beginning.

[00:03:43]Some teachers leave this out of their table completely. Some teachers call it the R, it's the mole ratio, and that comes from these numbers here. So, one to two to one. Okay, so we fill that in. Then, what I also like to do is where the arrow is separating the reactants and the products, I like to put a double line here.

[00:04:01]Just so that I know everything on the left-hand side of this double line is a reactant, everything on the right-hand side is a product. And because they didn't say otherwise in the question, we know that when we start off a reaction and some people don't understand this, but when a reaction starts off, you initially don't have any products.

[00:04:17]There's nothing. You don't have a product, you only have reactants. So, my initial of my product is going to be zero.

[00:04:24]Super, super, super important. Back to the question. Other things that they give you over here is they give me, if you read here, it says the concentration of hydrogen at equilibrium is 0.225.

[00:04:38]Now, please, this is where a lot of students go wrong. They see the word at equilibrium. When equilibrium is established, and then they see this number and they go, "Ooh, ooh, okay.

[00:04:47]0.225, I'm going to put that in the E row." No. It says here, "Concentration at equilibrium of hydrogen." So, this row is concentration. So, that is 0.

[00:04:59]Now, I've forgotten it already. 225, okay? Concentration at equilibrium. The last thing they give me is the value of the equilibrium constant, and you can see that that is 6.22.

[00:05:10]That is your Kc value, 6.22.

[00:05:14]And technically, everybody, what you should also know is that what goes into your Kc expression is your concentration at equilibrium, this last row of the table.

[00:05:23]So, this 0.225, I can actually already put that in the table. We can use round brackets now.

[00:05:30]0.225, and we're going to square it cuz that's what my formula says. Okay, so let's just keep sight of what the question wants. It wants me to calculate the initial moles of CO that was placed in the container. So, I'm looking for X in my table over here. This is what I'm looking for. So, now the best thing to do is to fill out the table by starting with the column that has the most information. So, I hope you can see that it's clearly this column that has the most information. So, as soon as you have equilibrium moles, you can get concentration because we use this C is equal to n over v. So, if you're looking for concentration, you go C, you take whatever's here and you divide it by the volume. But, if you're working backwards, you can also work backwards.

[00:06:15]So, in other words, they give you the concentration and you want moles at equilibrium. So, basically, you manipulate this formula. You want moles at equilibrium.

[00:06:26]This one over here.

[00:06:28]You say C take V over C times V. So, if I multiply this concentration here by the volume, and remember the volume is given in the question, it's 10. If you multiply this by 10, if you say 0.225 * 10, you're using this formula. Just write it somewhere on your page. Then, what you get is your moles at equilibrium, 2.25.

[00:06:53]Okay? Then, what's important to notice is that, and think about this carefully, reactants, everything on the left-hand side of this double line over here is a reactant. Reactants get used up. So, think about it. We start with eight. We end up with 2.25.

[00:07:09]So, what was my change? How did what change happened? We subtracted 5.75.

[00:07:18]And work down on the table to check your answer. 8 5.75 is 2.25. That's what's left over at equilibrium. Okay, that is our moles at equilibrium.

[00:07:31]Then, next step. So, we filled this column completely. Then, as I said in my explanation video on this table, the C and the M column, they are best friends forever. They go together. And what that means is that the change row happens in a mole ratio. So, what that means is Do you see here is a two to one ratio. So, it's like a normal, you know, when we do normally do stoichiometry, we go carbon monoxide to hydrogen, and we say it's a one to two ratio, okay? One to two ratio. We have the change of hydrogen. So, what we have is 5,75, and we're looking for the change of carbon monoxide. So, a quick way to think about the ratios to get from here to here, you divide by two. So, to get here, you would divide by two. That means that the change for carbon monoxide, because you have to use the ratio, is 2,875.

[00:08:26]And again, we're going to put a minus there because we use up reactants, okay?

[00:08:31]Reactants get used up, which is why they get minuses. On the other side of the double line, this is a product. Products get produced. So, we're going to put a plus here for the change. So, the change row is where we put the pluses and the minuses. Usually, it's products get a plus, reactants get a minus, okay?

[00:08:46]Depends on the question, but most of the time. Now again, this is a two to one ratio. So, you divide by two going this way. So again, you would divide by two to get this change. So, it would be the same, 2,875.

[00:09:00]Then, we work down on the table. So, for the products, in the beginning, we start with zero product. Then, we produce 2,875.

[00:09:10]So, at equilibrium moles, we have 2,875.

[00:09:15]Okay? Then, what we can do is Remember I told you, if you have equilibrium moles, you can get concentration at equilibrium. And how we get from moles to concentration is we take our moles and we divide it by the volume. So, we're dividing it by 10, basically. We're basically saying 2,875 / 10 cuz the volume is 10, and you get 0,2875.

[00:09:42]Okay, you just move the decimal one place backwards. You're dividing by 10.

[00:09:46]Then, what you do next for the question is up to you. Oopsie, there goes my line of my table. How you approach the rest of the question depends on how you feel about mathematics, okay? So, what a lot of my students prefer to do is they look at what they're what they're trying to get at. They're trying to get at X.

[00:10:06]Okay? Yes, you could technically say that at equilibrium, you start with X and you subtract 2,875.

[00:10:14]That's the moles that you're going to have left over at equilibrium. And then you take that equilibrium amount to get it to concentration. Remember, you take moles at equilibrium divided by volume.

[00:10:23]So, your moles at equilibrium is that expression.

[00:10:26]You divide it by your volume, which is 10. So, some of my students like to create expressions in this row of the table, this column of the table. But what my other students prefer, I'll do it both ways. What my other students prefer is this. You leave that open for now, and what you can see is the following. Remember I told you, let's use a different color, that what goes into the Kc expression is this row over here, the concentrations at equilibrium.

[00:10:50]So, what you can see is that I've already substituted, this is for hydrogen, 0,225. I've substituted it in the place of hydrogen.

[00:10:57]Then what I can do is I can take this, which is for this one, methanol, okay? I can take that and I can substitute it in the place of methanol, 0,2875.

[00:11:09]You can use round brackets. So, that goes in there because it's in the place of methanol, and I do not yet know what this is. So, I'm going to leave it as concentration of CO.

[00:11:21]And again, this depends on your mathematical abilities or preferences.

[00:11:25]Some of my students take, remember in the previous example, when my previous way of doing it, I worked up this expression to represent, so let's use a different color. I worked out this expression over here to represent concentration of CO. So, in the place of CO, what some of them like to do is to put that expression in there, like that.

[00:11:45]But, some of my students say, "Uh-uh, ma'am, the maths they No, I don't want to see a fraction in a fraction. That's Nope, can't do it." If your brain can work with this, great.

[00:11:54]But, I'm going to start showing you how you would do it this way first. So, first thing we would need to do is solve for CO or concentration of CO. So, to show you the maths here, the easiest way I can explain it to my students is if you're looking for a variable that is at the bottom of the fraction, this thing that I circled swaps places with this thing, so we'll get this like that, and then you can work this out. And then, what you do to get the concentration of CO by itself is you take your right-hand side and you divide it by this. So, you say 0,046 {blah} {blah} {blah} divided by this.

[00:12:32]And what I end up getting is 0,9130244929.

[00:12:40]I'm not going to round off because I'm not at the end of my question. So, that would be the ugly fraction. Remember, if you are working with the stuff like this in exams, keep it to at least six decimal places, okay? Don't round off.

[00:12:51]So, this I had to have a lot of space here, so I did it on a separate page, but let me just show you. This concentration of this, there we go. That goes in here in my table. And now, we can work backwards in the table. So, this is going to be 0,913024.

[00:13:09]Now, we work backwards to get up to my X. Now, when we work backwards, you need to think of like inverse operations. So, to get from concentration to moles at equilibrium, we already spoke about that.

[00:13:20]If I want to get from concentration to moles, you say C * V is equal to N, okay? That's how you get moles at equilibrium. So, you take your concentration and you times it by your volume, which is 10. So, we're going to get 9.13024.

[00:13:35]Then, when we work backwards, you see that my table says minus 2.875.

[00:13:41]When you work backwards in your table, you must plus. So, you go now you take this and you plus 2.875.

[00:13:48]And what I get when I plus those two, 2.875, is my answer for x. And my answer for x is 12.005, which you can round off to 12.01.

[00:14:02]So, therefore, x is equal to 12.01 moles. Cuz remember, x is moles. Now, if you don't like working the way I did over here, another way that you can do it is as follows. If you don't like putting it back in the table and working back in the table, another way you can do it is like this. So, you can sit with the maths and how it looks like that, manipulate the equation, and isolate x.

[00:14:28]And if you isolate x, you will get exactly the same answer, 12.01.

[00:14:33]But, if you don't want to sit with something like this, you want a little bit of a middle ground, what you can do is the following.

[00:14:39]Remember we said that the concentration of CO we worked out over here to be that number, and we also did the following. We said we can get an expression over here. So, if you start with x and you use up 2.875, then at equilibrium you're left with x minus 2.875.

[00:14:55]And then, you to get from moles at equilibrium to concentration, from moles to concentration, you divide by volume.

[00:15:02]So, you take that and you divide it by volume. Then, what you can do is this expression represents concentration of CO at equilibrium. And this is also the concentration of CO at equilibrium. So, what you can then do is the following.

[00:15:19]You make your little expression equal to the concentration. And then you do a little math calculation and you end up with the exact same answer, 12.005 or 12.01 moles.

[00:15:32]There we go. I know that was long. I hope that it was helpful. Please check out the rest of the playlist for more help on this stuff and more of chemical equilibrium. I'll see you in another video very soon. Bye, everyone.