 This screencast is going to be about complex reactions, so we've done a little bit in the lectures about how you would actually go about doing it. Just this is sort of a recap and some extra detail about it. So if you're struggling to understand why we're doing this, this is going to be quite useful for you, hopefully. And if you think it's quite easy to do, stick around to the end when we do this whole modelling thing. It's an interesting approach to it that I think you might get some use out of. So what we're going to actually cover, we're going to start with reaction complexity. What is a complex reaction in our case? What are we talking about? And then we're going to build up complexity in a few steps. So what I'm going to do is build up the ideas behind the complex reaction bit by bit. The simplest version, equilibria. And then two step reactions, then more than two steps. And then modelling. So we're going to model a far more complex reaction. And that's kind of close to how you would probably do it in research. The two quick definitions I want to cover first. One, this concept of an analytical solution. So an analytical solution is something that is exact. We start by manipulating our maths and our symbols together. And then we have just a simple equation that predicts something. Gossadoni works for a limited number of variables. So a good example of an analytical solution is a first order integrated rate law. So you've come across this equation before. This predicts the concentration of a at time t based only on the starting concentration, the rate constant and at time. Now these, the rate constant and the starting concentration are effectively constant. They are data points that we can find out quite easily. The only variable is time. We just need to take this equation and plug time into it. And we can figure out the concentration of a at any time. 10 seconds from now, 10 minutes from now, 10 hours from now, 10 days from now. Doesn't matter. This will go on forever. In fact, we don't even really need to know the concentration that starts with. We could set that to 100%. And next we have the numerical solution. In this case, we actually do approximation. So we wouldn't be building an equation like this. We would be figuring out what is the concentration a second from now, then two seconds from now, then a second after that. We do it step by step, calculating actual numbers out. Now that has actually an advantage that it works pretty much for everything. We can do numerical solutions for any equation. We like no matter how complicated. If you want an analytical solution, you usually need to have only simplified things. So in this case, we've only got time as a variable. We've only got one concentration and so on. So analytical solutions are really nice, but they are limited in use. So the entire point of this whole complex reaction section is to produce analytical solutions for our rate of reaction or our concentrations. This is what we're aiming for. We've got complicated reactions and we're going to manipulate the symbols that we used to represent them, all the maths, and throw them together to figure something out. What is the rate of reaction? Can we predict it using just time? And when we get on to modelling at the end, that will be the numerical solution. So you don't really need to know any of that, but it's useful anyway. So reaction complexity. So, you know, we can start with a simple reaction. We can make it really complicated. We can make it really, really super complicated and we can tear our hair out thinking about how complicated is this. We can't really do anything about it. But the idea, again, with this topic is to try and simplify it down to much more simple solutions. So here's an example of a reaction that could be complex. I gave you some data associated with this in a workshop. It's a simple elimination. We are eliminating a phenyl ligand and a hydride ligand to form benzene. Those are just coming off and reforming benzene. So that would be simple first order kinetics. If we have time, we have a concentration. It will decay exponentially. That is a terrible exponential curve. I will draw it again. And if we took log of that concentration, positive against time, you would get a straight line. And then our gradient is equal to minus rate constant. Great. This is really easy. Except in reality, the actual reaction we're looking at looks like this. So we can just give you a second to digest what's going on here. There's a lot going on. For instance, what we're actually looking for, everything is in equilibrium. And there are forward and backward reactions everywhere. There are two isomers that this can form. One of these isomers can actually eliminate, not directly, but by undergoing something called migratory insertion. So that ligand goes into there, and then it eliminates a pencil to hide. So there are two pathways that this could actually go back to the starting material. So things get a little bit complex. And we need to take into account all of this at all at different times. We could do some simplifications. We can assume the equilibria don't play too much of a part. We can assume that some rate constants are slow and so on. But this is the kind of thing you need to take into account when we're dealing with complex reactions. So this is really the highest level of how complicated things can get. I mean, there's a little bit more complicated ones out there, but this is probably the most realistic, most complicated example. Anyway, we're going to start with the simplest thing. That is the irreversible reaction. This is the very model of first order kinetics. We've got concentration inside. And we can just call this molar proportion. Percent makes it really easy along the bottom T. And we realize that the products, they go down the reactants. And this is why you don't just do things on autopilot. The products go up. The reactants go down. There we go. It's that way around. And what you'll notice is by the end, there's 100% product, 0% reactants. So this is the kind of thing where maybe this is a gas and it escapes. So there's no way it could come back and reform the starting material. It goes entirely in one direction. Or maybe the equilibrium constant is in the millions or billions. And the energy difference is huge. So this is very simple. And we can model that really, really straightforward. The rate, we define the rate as negative of the reactants, positive of the gradient of the products. And that is equal to K times A. First order kinetics. All that rate depends on is the concentration of this. Nothing else. So the next layer of complexity to add is equilibria. So we covered this basically in the thermodynamic section. Because equilibria are based around some thermodynamic parameters. And equilibria can tell us a little bit about this. So let's look at this exact same graph again. We start with 100%, 0% here for our concentrations. And the products come up. But only to about what? That might be about 70%. And therefore this is 30%. So clearly there is something else going on here. There must be some kind of backwards reaction. Some of this is going backwards in this direction. It's reforming the starting material. And that will then go to, well, not necessarily completion. Only to this 70%. There's still some of the reactant left. So there's a little bit of complexity added there. So here we find what is the rate. We need to figure out what actually changes the concentration of A and B here. So we can find for instance, let's think about what changes the concentration of B. There are two reactions that are going to do that. One, it's formed from A. It goes forward. And that's a first order reaction. What's the rate of change of that? It's equal to K1 times A. It's a first order reaction. And then we've got a backwards reaction. And that's also a first order reaction. It's K minus 1 times the concentration of that. So B changes according to these two. And it appears or disappears. The concentration increases at that rate. It decreases at this rate. And so what we get is this kind of thing. We add them together. Now, in this case, these two equations I've got on screen are exactly the same. I'll just reverse the positive and negative of them. So this is just to underscore the fact that both are perfectly valid. But you need to be aware what would you define the rate as. So you define the rate of reaction as negative and then positive. So we get this one. But if we're actually just straight after the gradient of the reactants, for instance, it's the other way around. So that should be hopefully easy to follow. But don't get booked down if you see negatives or positives reversed at some point. It all depends on how you want to define your rate in the first place. Again, if you get something like A going to B or something like this, you might find factors of two make their way in different places. Same kind of thing. They're all equally valid. Just be careful about how you define it in the first place. So what kind of assumption can we make to make the equilibrium really easy? We can assume that it is at equilibrium. So at equilibrium, there is no change in their concentration. They are equal. They have stabilized. So our rate of change is equal to these two things added together. And finally, we can stick a number on. We can say it's equal to zero. So this is a kind of a recurring theme in the multi-step reaction and equilibrium and so on. Set it to zero. So we're saying there is no change in these concentrations at all. So that means, by extension, the forward and backward reactions are happening at the same speed. k1 times A, that forward reaction, and k minus one times B, that backwards reaction happening at the same speed. So we can stop manipulating these symbols around and get us some new information. And that's what we get down there. So we could take B over to the other side and then k1 to the other side. And we get that equation down there. Storm is really picking up outside it. We get this. So one information to that tell us two concentrations over the top of each other. That is equal to big k, that equilibrium constant. So that's how we start getting thermodynamic parameters from kinetics and relating kinetics to thermodynamics. So let's just kind of review this. We're looking at rates of reaction, rates of proportional concentration, i.e. this, that's what we should know. In equilibrium, there are forward and backwards reactions. So the reaction disappears at one particular rate and it reforms at another rate, backwards reaction. So at equilibrium, our rate is zero. So, well, our overall rate of change, we have to add the two things together. All of the processes that form something we add and we add all the processes that remove this one. So those are two things that's added together. The overall rate is zero in equilibrium and we can start manipulating numbers and rearranging them. So let's go on to two step reactions. So this is the next step up from just equilibrium. It's only a little step in fact because we're only looking at two rate constants again. This time we'll call them K1 and K2. Just to recap the name scheme thing, we can in fact call that KAB to prove that it goes from A to B or KBC and that shows that it goes from B to C. They're just naming schemes. You can use anything you like. We are dealing with three chemicals here but two reactions, so it's not going to be too much more complicated than the equilibrium case. So our rate, we want to define it just in terms of the initial product, initial reactants and by the final product, that's all. So we're only interested in that as far as the rate of reaction is concerned. We don't really care about these intermediates but we do care about the rate of change of everything in between. So DA by DT, how does A change? Well, there's only one reaction that alters the concentration of this and that is the first order reaction where A goes to B. So that must go at minus K1, A. A of reactant will decrease at a particular first order rate. Next, let's go to what does C do? Well, that is produced only by one reaction and that is a first order reaction that depends on the concentration of B. So it's kind of the opposite. Imagine if we just went from A to C not via B at all. These would be what we got. What we'd get out of that. DB by DT, this is slightly more complicated though because we have two reactions that change the concentration of B. One, it grows by a first order reaction and what is that first order reaction? Well, it depends entirely on the concentration of A. So that's K1. A, great. That's pretty much the opposite of this. So this is what we'd expected. That was just a single step reaction that was irreversible. But B is removed by another reaction. K2, multiply by the concentration of B. So we've got this reaction that increases the concentration. That reaction then decreases it. So this is the final result of what we're looking for. Again, two reactions change the concentration of B. So that's a little bit more of a complicated formula. Now, the thing we want to assume is something called the steady-state approximation. And that's to say that the rate of change of B is zero. Again, just like the equilibrium case we are setting something to zero. Really useful for us because that allows us to start moving things around. So if that is equal to zero well, these two rates must be the same. So that tells us that we can rearrange these formulae if dc by dt is equal to this rate constant. Well, we can actually just substitute that in for this. There we go. So if the rate of B doesn't change it's concentration doesn't change. It's rate of change is zero. Then here we have that's an analytical solution. The changing concentration of our final product is only equal to one rate constant and a. Now there are a couple of caveats as to why that works. This has to be really slow and this one has to be really fast. So we'll kind of visualize this in your head. A goes to B really slowly and then B almost instantly reacts to form C. Well, B hardly has any time to accumulate. It doesn't change. It's just hanging around there very, very briefly. So this makes some sense. If it was the other way around, you could kind of imagine what's going to happen. Imagine this happened really quickly. It just disappears to form B but this one is really slow reaction. Well, in which case our B appears really quickly and our rate of C's formation will probably only be k2B. And by the time the steady state approximation in this is the only rate we're interested in. So there's a few things going on. Remember there is not always one size fits all. So what you need to be able to do is happily manipulate these numbers and then predict what happens if one rate constant is a lot higher than the other. Or what happens if they're equal? So let's just review the two step reactions because we're cracking on quite nicely here. We define the rate first. That's always our first step in kinetics defining the rate. And we just ignore the intermediates here. The concentrations, it's not worth bothering about one. We just want to define a rate because our rate of reaction is defined as really just the products or initial reactants changing. But we are interested in what is the rate of change of that intermediate because it tells us something about the other rate constants. So b increases at a rate of k1a, b decreases at the rate of k2 times b and those are added together. Just kind of like in the equilibrium case we are adding things together. And then finally we can simplify things using this steady state approximation. So we assume there's no change in b so it's always going to be equal to zero. Therefore we can say that those are the same. And then again we start rearranging things. We can insert wherever we see k1 multiplied by a we could insert that or vice versa. This is really all about your skill in manipulating the equations. So it is a bit of algebra in mathematics but it is something you need to practice. So this topic really is skills based. It is not something that you can just learn. You really just need to be able to manipulate equations. And then always immediately straightforward or obvious but you can probably get 99% of that with just a few tricks. So now we are going to do multi-step. So I have gone up in complexity again and basically combined equilibrium two-step reactions here. So this we are now looking at three reactions. We are looking at a forward one that is labelled as k1, a backwards one labelled k-1 and a second reaction. So a and b are in equilibrium maybe they are two isomers but only one isomer reacts. So this is kind of similar to that rhodium example I showed you in the beginning two of them are in equilibrium but only one actually reacts. So we have to take into account this. So again our rate is defined as only one. The starting material and the end product is that is fine. So let's start with the easiest one. What is the rate of changes? So this is really really easy. There is only one reaction changes the concentration. See that is positive because it increases k2 b. Really easy. That is the easiest reaction there. So if you are building up a reaction scheme look at the easiest ones. What is the material that only changes according to one reaction? Start with that one then write it down. We now know this is a first order reaction that is true. So let's start with the second easiest one our initial reaction. So that is the second easiest one because there is only two reactions that changes concentration. One that decreases it. That is the first order reaction that it eliminates and one that we will do that at the right very end one that puts it back. So k-1b that is going to increase the concentration k1a that is going to decrease the concentration. So that is our second example. Finally the intermediate. This one is probably the more difficult one because there is now three reactions that changes concentration. One that lowers it. Another one that lowers it and one that increases it. So b can disappear at a particular rate but it goes in two directions. It can't go over to c or it can go back to a and b can only be increased by this one reaction here. So while it's a little bit more complicated the same principle applies exactly. We are adding all the processes that it can be reformed with and then subtracting the processes that remove it. And if we apply the steady state approximation we can set that all to zero. That allows us to start doing it again. Some manipulation of the numbers to make a few other predictions. Can we get rid of the intermediate when it comes to the formation of our product for instance? So we know dc by dt is equal to k2b. Can we get rid of this b by using this approximation? Yeah we can kind of deal with that. So we start with our basic rate law that we've set to zero because we're going to assume that this doesn't change. And then we arrange it. So we bring this to one side and then I'll flip the sign so we have k1 must be equal to this and k1a we can bring b out as a common factor here and then just simply do a little bit of rearranging so we bring that to that side. And there we have it k1 over k-1 plus k2 times the concentration of it is what the concentration of b is. Great so that's really useful. So now if we can say that dc by dt is equal to k2b we've got a formula for what b is based only on one other thing. We can combine that in together. So now our rate of change is a ratio between different rate constants in the reaction which is still just a constant and a. Now at some point you might recognise a little bit of a overlap with what might be called the pseudo first order approximation that is kind of the equivalent of k observed. If we were to plot this the rate is equal to that so it means if we plotted say log a over time we would get a straight line out and that gradient would be equal to k-plots that observed rate constant. Now that is not necessarily saying it as a first order reaction exactly that k-obs would be equal to all of those rate constants put together. So there is a limited amount of data that we can actually begin to extract out of this reaction but the numbers might break down a little bit differently. So let's just kind of review that now. It's been quite a long and arduous screencast I think. We are looking at multiple rates of change. So for instance the equilibria means we can do forward and backwards reactions so I think label the rate constants. I am dealing principally with k1 k-1 for a backwards reaction and so on. We can still do kab kba kbc that kind of thing. Maybe experiment with that if that makes more sense to you use that and we can use the steady state approximation. So if we are talking about two-step and multi-step this is the one we are interested in. The intermediates don't change and now if we assume that it means we can ignore them because if we look at how those equations will all be manipulated together we are finding that if that intermediate here is equal to the change of the intermediate is equal to zero we can start plugging different values in and it turns out we don't need to know that concentration. So most of the time as far as you are all concerned that's what you are trying to do. We are trying to get an analytical solution to what is the rate of the reaction and the way we approach that is by assuming certain rates don't change. It gives us a little chink into the equations that lets us rearrange them. So complex reactions. We are looking for analytical solutions to predict a rate of formation and our main approximations are the equilibrium one and steady state. So we can approach those in different ways. Then we want to do mathematical manipulations in order to get things. This is of course really important. There is not one thing you have to memorize as a procedure here. Yes, if you want to get a rate constant, yes, plot log over you get a gradient. That's a procedure. Dealing with complex reactions is a skill. It is manipulating equations. You might find yourself going down on a completely wrong track. You might not be able to figure out how things cancel. You are building analytical solutions here. They are a little bit tricky to deal with. So unfortunately that is just how it works. There is no one way of memorizing. There are a couple of tricks. Equilibria, steady state and so on. But you just have to learn some maths and some algebra to manipulate things and that's it. So that's kind of it for complex reactions. That's all you really need to know. The rest is just practice. What I want to go on to now is modelling. So this is a little bit of an extension. If you are struggling to kind of get into your head how we would do this in reality this might still be useful for you. If you are finding the complex reactions kind of easy. This will be a nice extension. If you are finding complex reactions really hard to a degree this is such a simpler then we are going to do a numerical solution. So I introduced that concept at the beginning that we have analytical solutions and numerical solutions. This is the numerical side. So what we can kind of see is this is kind of a real issue reaction that is two step and what were we assuming before? We were assuming the dB by dt equals zero. Now if you look at this graph that's clearly not the case at all. This is zero well there and here. So that is roughly to zero and very for only a couple of seconds in the action and by the time it proxmits to zero the reaction is pretty much done. So that assumption half the time is actually quite useless to us it's not really brilliant. But there are too many variables in here to make an exact analytical solution. So we are going to model it step by step. So where do we start with this? One thing is to start with kind of the integrated rate law because this gives us a little bit of a hint of how we are going to do this. So you should be used to at least integrating a rate law like this. You go from time zero to time t. That's how we deal with an analytical solution. We are only interested in the initial conditions and then we can plug any value of time into it to get a concentration out so we solve that. But there is a more generic way of doing this. Zero and t are the only kind of limits that we can stick into this integration. We can actually stick in these. This is t and t plus delta t. In fact this is the more generic version. This is that's t and that is t plus delta t but the delta t is much bigger and t is zero. So this one at the bottom here is the much more generic solution. So it tells us we can actually kind of integrate it just in tiny little steps. We can see what's the rate of change over maybe a second and then the next second. Then a second after that. Or maybe we can do even in smaller discrete chunks. A tenth of a second, a hundredth of a second or so on. So this is kind of the starting point of how we want to figure this out. We want to break the reaction down into chunks of time because we're doing a numerical solution. So this is going to be a model or a simulation. What happens from one second to the next? So that is sort of the starting hint of how we're going to do this. So if we integrate this, we can find the concentration of a based on whatever it was at the beginning. Here we find what's the concentration in a second's time based on what was well now. Or what's the concentration now based on a second ago. Either way it's the same. These values are entirely arbitrary. This is just the most generic version of the one you used to see. So now let's start with what these rate laws are. So we covered these rate laws. Now what happens to these equations if suddenly we start integrating into small chunks of time? Well, these actually become meaningful. We can change this dt to literally one second. So this rate of change tells us how far does this change in one second? Okay, so maybe it's time well maybe 17 seconds into a reaction. This is at 75% concentration. 18 seconds. It's at 68% concentration and so on. That is literally a step change of reaction. We've got delta t there and then concentration. So we can actually stick actual values in there. So this is going to be an approximation show and I'll get onto consequence of using the approximation in a moment. But they are physically meaningful values. So if that means this is an actual change in value over a particular set of time it gets us something like this. Here's the concentration we started at. At that change in concentration and that gets us what's going to be. So maybe 10 seconds into the reaction there's 0.5 moles. 11 seconds into the reaction there's 0.6 moles. Then dA by d. 1 second is equal to 0.1 moles. So you can actually stick actual numbers into this and work it out. Now because of this first-order equation we can actually substitute that speed for something a bit more precise. So actually, look at this, this actually becomes really really simple. The concentration at any particular time is equal to the concentration that it was before minus the rate law. That's really easy, that's very linear basic equation. It's a little bit complex to get there but it's really easy equation. Now that means we can actually apply it. So let's just assume that this is 100. So I'm not going to stick units on this, I just have a nice round number to illustrate this. This is a schematic. So our rate of change is equal to k1 times a and our k1 I'm going to put it as 0.1 per second. So I'm now going to multiply that by the time. We're going to advance this forward by 1 second. So that's really just kind of our change is going to be that times 1. So we're going to just ignore that for now. So 1.1 times the concentration 10. So a tenth of this is going to react. So we actually want 100 minus 10 and that's 90 and then that will move to B. We've done something like this. So these are the numbers. This we've literally changed to 90 and then moved to 10 here. Okay, so what's changing the concentration to be? We've now got this to change. Well, we've got a rate that increases concentration at any rate that decreases it. Previously we couldn't change this because this concentration was 0. That was equal to 0. So it meant nothing. But now k2 is here. So we're going to add what? 0.1 times 90 at that and we're going to subtract that 0.05 times 10. So fill that in as necessary. So I've pretty done this. Remember we've jumped this is going down by 10% each time and adding to here and then this number is at the same time going down as it transfers to C. So we are doing this step by step by calculating that out and by treating this changing concentration as an actual number. Now here's the caveat to this. So you could type these equations into a spreadsheet, no problem and do it or you can do it by hand. Certainly easier in a spreadsheet because then you can drag down and do hundreds of data points at once. Here's some manner caveats to it. Notice here I've taken my delta T as one second. That's fine but here's delta T at 5 seconds. So I'm actually I've gone ahead and done this in 20 separate steps and my concentration predicted to be 12.16. Here I've taken delta T as chunks of 5 seconds and it goes to 100, 50, 25, 12.5, 6.2. That's a huge disagreement. So when I say that numerical solutions are an approximation this is why you are actually deviating a little bit from the exact analytical solution each time. This is much more widely applicable certainly and you can minimise the effect of this certainly but it is going to be there if you try this method. So it's deviating quite a bit from this. In fact let's change this again. What you might see is that we've got one second, half a second and we do a step change so this requires twice as many steps now. We're subtracting this times one second and subtracting that time to 0.5 seconds we get slightly different but the agreement is a little similar at least it begins with a 12 but it is very different. Now the reason for that is because the analytical solution that first order rate integrated rate will reach delta t as if it was 0 that's what we do when we integrate things we are treating this kind of steps the steps as kind of infinitesimal in size. So it's a little bit mathematically if that's even a word and you can see there's a little difference we go to 13 which we go forward. So what we can see is there's actually a progress here progression here at 5 seconds we estimate that it's 6.25 if we simulated one second intervals 12.16 we estimated half a second interval was 12.85 and if we assume that 0 seconds this is our analytical solution it's 13.5 but actually gradually increases to this number so the shorter that delta t is in our simulation the closer to the real value we get so if you there is an additional maths video added on about e and exponential solutions you should be able to spot kind of a similarity here that exponentials equation the one with e in it is where the limit of the step changed goes towards 0 so that whole progression is useful. So anyway let's just assume that this is meaningful quantity and we're going to pick a time scale of one second so our reaction lasts an hour one second is pretty good that gives us a couple hundred data points to simulate then we build the model and then every change changes by d by dt so we take time times that rate of change times dt and we substitute it in for the rate law so we can see that it will change by this all that and for really complicated reactions we could just make that as big as we like you know k3a minus k4 ab or something like that these things could get really complicated all we're doing is bolting on more rate laws so that's kind of where it's simple we're bolting on the rate laws really so that could be as huge as we like and we go through and simulate it and we can get the rate constants just by manipulating the numbers and our simulation getting it to match data so that's probably another video's worth of material but this is modelling it's one way of approaching it as a numerical solution but as I said earlier what we're really interested in in this topic what you're expected to do in an exam is still just the analytical solutions from here