 Mae'r bach gweithio gyda'r hammock ar drwsun i gyfer y rhwng Plynedd Llywodraeth. Mae'n gweithio'r ddweud i mi'n gweithio'r mawr cyn staff hefyd i ddod, ac mae'n gweithio'n eich adegau arddangos y vad yma. Yn ymlaen, Mae'r ddweud mewn gweld eich… eich llwyddo yma, eich mas yw'r mathau sy'n mod Why yw'r llwyddo? I'n ddwy meses ychydig gwnaeth. Felly, iawn i'r gweithio'r hwyl yn ei gynnig unig, ond we're going to be looking at differentiation. So differentiation, the idea that you know we can get a speed out and we're also going to be defining rate based around it and then getting units of K are rate constant. So as before each of these do kind of build on each other, we need to get to define rate before we get units of the rate constant. We need to really look at what we're interested in from differentiation before we start defining the rate law and so on. Now the next bit is a bit of revision and it's not mathematically rigorous. So if you are kind of okay with differentiation calculus, you can click one of these links below that I'm going to add on using YouTube's annotations and you can skip ahead past the maths part. It won't be rigorous, it's just going to be getting what are the important parts of differentiation and speed and so on that we are interested in in order to study kinetics. So this diagram I've got here is Superman and he's going to jump over the Empire State Building and the Empire State Building is 443 metres high. So if Superman jumps, it takes him about 10 seconds to get up to the top. You can work this out using something called the kinematic equations if you want to look them up and it takes him about 10 seconds and everyone knows. I'll just make sure that they're written down as an actual unit. 10 seconds, so everyone knows speed is distance over time. So 443 metres divided by 10 is equal to 44.3 metres per second. So that's the speed. Except no it isn't. If you actually look at what would happen, he starts off really fast, then he becomes slow, then gravity takes over and he speeds up again. So there is a change in speed over time here and that change in speed over time is all about what getting gradients and finding differential equations is all about. You can see the beginning of a jump is going to be fast and then slows down and become stationary. Now again if you use the kinematic equations to work this out, this is somewhere closer to 93 metres per second. Here completely stationary at the top of a jump that should be 0 metres per second. So you can see there should be a gradual decrease in speed and finding out that speed is where finding gradients and differentiation comes into play. So what we can see is there's a gradient at the top here, which is practically level at zero and a gradient up here, which is really steep. So that must be the fastest. Now we get the gradient of a particular area, we work out the change in X and Y, divide them together and that gives us the definition of the gradient. Over here we've got another one that's a lot steeper. So what you can see is this value is much higher than this one so the gradient here must be really high. Here they're a bit closer together so the gradient here must be a bit smaller. And the other thing you should notice is that the gradients can be positive or negative. So this change in Y is going up so it must be positive, this change in Y is going down so it must be kind of negative. So this is what you're interested in now. If we turn this into a bit more of an abstract graph over time, height changes in this parabolic fashion here, velocity meanwhile changes linearly. Now here I've switched to using the term velocity over speed because velocity has direction to it. Specifically we're going to be talking about a vertical direction so it is possible to have a negative velocity because it means it's going in the opposite direction. You can't have a negative speed, speed refers to the absolute value or the magnitude of your velocity. Now the rules behind differentiation, I'm not going to cover those are part of the maths course. And we might go into a little bit of it, you don't really need to know it. But if Y is equal to X squared then dy by dx that gradient pattern is 2x. So this tells us that our gradient changes over time and we get a particular gradient at any particular instant. And it goes down quite linearly so that's basically deceleration according to gravity in this particular model. So we are interested in a rate of change and it's defined by a d by dt. That is a change in the thing we want to measure over a particular time. So let's just review a bit of that. Speed is the distance travelled over time on average or a change in any value really. We can do distance in any particular direction or we can substitute different distance for say concentration, which is what we'll do shortly. We know that this speed can actually change over time so we're interested in what is it at any one instance. So we get that from calculating differentiation. So if we have that again, that function of X is equal to say X squared, the differential function prime of X is equal to 2x. We substitute any value in there, we get its speed or its rate of change at any one time. So differentiation, if the distance is, or Y equals to a particular function, then the gradient is a particular function just like that. And we can calculate speed at any one instance. This is what we're interested in. We want the speed at one particular instant in time, not averaged. So let's define rate now. So previously I worked on height and distance. I want to do rate and rate is going to be a change in concentration. Now, how do we denote rate down? I've written it down. Here's rate, v, r, rho. This is because it's written down differently in different textbooks and I've seen it written down in loads of different ways, depending on who's lecturing on it. I'm going to try and keep things a bit simple and just call it rate, but you will see it is the occasionally a few other lecture notes have rho in it. If you read that textbook, he's very helpful. He calls it rate. He just says rate and writes it and then suddenly you turn the page and it switches to r without any explanation. It doesn't matter. This is just how we define the speed of reaction. It is just the speed of the reaction. It's a particular value. The interesting part comes with how do we define it? And it is a differential equation. It won't be booked down by the fact that we've replaced these values with a concentration and a t. It's exactly the same. That is directly analogous to this whole dy by dx thing that you get from differentiation. And we also notice that there are these negatives and positives involved. So in this reaction a goes to b and c. It's going to be a very slow reaction this. If you ever think this is not going to be a particularly stable positive charge. So it's going to be quite slow and the rate that a disappears is defined as negative dA by dt. And then the rate that the other two appear is defined as positive dC by dt. So this is the changing concentrations and we use the negatives and positives to make sure that they balance. Now I'm going to be very kind to everyone and be very explicit by adding these positives in. At least when I remember to, I'm going to try and be as consistent with that as possible. Most of the time they're considered a bit redundant and aren't written in. I'm going to add them in anyway. And the rest we define the rate as proportional to a concentration by a particular factor. So the rate of change changes according to this and we'll get on to a bit more of this in the next screencast when we do something called the initial rates method. Let's have a look at a different version here. We've got two things reacting to form C. Now in this case a is a reactant that's disappearing so it must be the rate is negative to that. B is also disappearing so it's also negative. C appearing so we must be positive. So just think of this in terms of well if our concentrations were changing here then C is appearing. A is disappearing. The gradient of A is always going to be negative. The gradient of C is always going to be positive. So if we take a positive of that and a negative of that we get the same value. So that's why we have a minus and a plus. It keeps things nice and even. And again this is a second order reaction between two things so our care is proportional to two concentrations multiplied together. So let's have a look at it here. A decreases and this is our change over time. So our rate is at any point of the gradient of this. Or the negative of that gradient because the gradient always points down. If we have a more generic concentration or sometimes this is written as a molar fraction so that this would be like 100% concentration and this would be 0% concentration. We'll get on to that a little when we do data processing. One goes down, the other one goes up. So race is also equal to dc by dt. You need to be able to build an equation like this based on a reaction. So let's look at a slightly more convoluted example. That first screencast I did this a little bit where we have different stoichiometry. So there is an implied one here and then there's two and three. Now this is the one where you will have to think what's going on in the equation. What's physically happening? So you kind of say it out loud. When one c appears, three d appears. So d must be appearing three times the rate of c. So obviously that must be quite significantly higher than this. So in order to get this down, we need to multiply this by a fraction or divide it by a number. So what are those numbers? Three, two here and two there. So the stoichiometry is effectively being used to divide by one here. Just to emphasise that it's still true. So this is something that you're going to have to get used to. And you have to know which way around it goes. Don't just try to memorise it in terms of... We have to divide by the stoichiometry. Learn it in terms of what is actually going on. If two of these appear, three of those disappear. One of this disappears, three of those disappear. So the rates must be related by that kind of stoichiometry. So test yourself on these ones. I'm not going to give the answers out, but I'll cover them in lectures in person. So here are some different balanced equations. Define the rate in terms of this. So what you need to know is what's the stoichiometry? What is the identity of this? And then you'll also want to know is it positive or negative. Is it a reactant or a product? Once you figure that out, you can write three answers down. So just to review the rate is the changing concentration over time or DA by DT, which comes from differentiation. That is a gradient of its change over time. So it changes, and we're interested in any one particular instance. So to keep things comparable, if A goes to B, then the reactions that you used up on minus A by DT, there are products that appear or plus DB by DT. And if they're not equal, if two of these go to one, say it out loud, this happens at twice the rate of B, so we must halff that in order to get plus B. Keep that in your head. Now the units of case. So we've defined the rate on this side of the equation. Now we need to define it on this side of the equation. And that equation just says that the speed is proportional to a constant, multiplied by a concentration. Now we've got a sort of equation here that we need units for to give it physical significance. Now I often do say that in physical chemistry you can black most of it just by following the units. And I'm not entirely joking when I say that. If you know the units, you can pretty much follow any equation. So the units here are going to be really important. So let's break the equation up. Now I'm a fan of doing this in mathematics, just highlighting things by different colours. Just to show that the blocks work around. I mean you don't want to get bogged down by the fact that these are just letters and symbols. And this kind of thing is exactly the same as the above. It is just telling us that we're dividing through by one thing of multiplying two things together. So the concentration comes here. The blue bits are the concentration. We have a time here and the rate constant there. So one of the units. Concentration, most per decimetres cubed. We know this. We also do time in seconds, pretty much as standard. You can do it by minutes if you like, or hours if your reaction is slow enough, but converting it to seconds just keeps everything nicely comparable. Otherwise you need to start where you want to multiply it by 60. Oh, I did my rate in days. Do I need to multiply it by 24 or divide it by 24? You don't want to ignore all that. Do it all in seconds. It's much easier. So the thing about units is they must cancel out and balance. So if I have most per decimetre cubed on this side and I have a divide by time on this side and I have most per decimetre cubed on this side, well, they cancel out. So I must have a divide by time on this side as well. So my units of k in this case are seconds to the minus one per time. It's not the same as Hertz, by the way, but it is per time. Let's have a look at the second order one. So this is also the equivalent of saying it's equal to the same concentration raised to the power of two. You've got the same thing. We've got a concentration here, the units of k here, a concentration there, and a divide by time there. So exactly the same as before except, well now this doesn't exactly cancel this out. We've only got a mole to the one here, and that should be a three. I've clearly missed it out. This can't cancel out all of this, so we need to be able to take something else off. If we just crossed out this side onto this side, we'd still be left with a mole per decimetre cubed. We'd still be left with one of those, so we need to cancel that out. So what we do is we add it into the units of k, so now it all cancels. We still always keep this per second, but we have to, with the second order k, keep this. So units need to cancel. So here's a couple of tests yourself one, so one, two, three. Take a look at them. What are the units of k in each case? That you've been given some concentrations. Number four, this is going to be a bit of a tricky one if you're really not used to it, because look at that, that's a half. Now I've said previously these laws are only defined and worked out experimentally, and that means there is nothing in the world that says this needs to be a whole number. This, in fact, could be any number we like. It doesn't need to be rounded at all. So figure out what it means. We'll go through some of the answers in the lecture in person. But if you want to scribble that down, pause the video and try it yourself, please do if you haven't have you. Well, here are all the kind of the answers that are a generic answer, because you will notice a pattern. If we have a first order, second or third, for instance, you will notice this is the pattern. The moles here go up, and then my pen stopped working again. So, come on. Why is this pen not working? Okay, moles per decimate cube per second, and then moles. I've added a zero on these just to kind of imply that they're there. Usually we don't... Oh, the pen's back now. We always get a per second, and you'll notice that moles kind of goes down by one each time. We have moles to the one, goes to moles to the zero, goes to moles to the minus one here, and so on, and the decimusers go up by three each time. We always have a per second. So that's kind of a pattern. You can always remember that a first order reaction, that's the case, units of per second. So, review the units for a moment. DA by DT is equal to K time concentrations, and the units must be the same on each side. So, zero of order. That is when our rate is just equal to K. It is a constant, and that is unit of moles per decimate cube per second, or concentration per time, which is exactly the same as the unit of rate. That's a really easy intuitive unit to get around with. First order, K has units of per second, per time, and second order has units of DM cubed per mole per second. Now, that whole thing of liters per mole doesn't sound very intuitive, but it does have some physical significance when you get into it. It's a kind of an inverse concentration per time. So, let's just review a few of this. Again, maths. We covered a bit of the basics of differentiation. What does DY by DX mean? It means a gradient at any one particular point. Our rate of change in concentration is what our rate is, or DA by DT. We want to put it into maths terms, and then units of K, they must cancel out. So, our first order, K is in per second, our second order, it's in this slightly unintuitive DM cubed per mole per second. So, we're going to start applying this to something called the initial rates method, and that is the third screencast we're going to do. So, we'll see you in the lecture about maybe solving some more of these problems and doing some more initial rates. So, until then, please do crack on and watch the next video.