 Oh, welcome to our fifth chemical kinetics lecture as you can probably tell from the teacher I'm recording these straight after each other just in case you thought I was you know spreading this out and I'm kind of doing them in bits and pieces now previously we looked at something called the initial rates method that's if you got three kinetic runs and you took three rate constants from some experiments you could solve for your rate law and your rate constant K now I'm going to look at something called first-order reactions so those are the really simple ones and a very neat little approximation called the pseudo first order approximation and those allow us to get a rate constant for any first-order reaction from a single experiment really quite quickly and easily and we're going to determine K from that now there are some other ways of doing this via Excel but you remember if you watched the whole of the last video I give a little bit of a tutorial on Excel I'm going to do that this one separately because there are two methods you can do it graphically you can do it by simulation again it's not going to be examined but you can watch that video anyway but this one is going to be the explanation the next video is going to be kind of the procedure for how to do it so what we're going to cover in this one first-order reactions what are they now we know that they are basically one thing goes to another okay what's the equation so like and so on and a pseudo first-order reaction as you can probably guess from the word first-order it's going to be very similar to a first-order but thanks to that word pseudo why is it slightly different then we're going to look at some actual data and how it would work and then determine K our rate constant graphically so it's a we're going to be able to draw a graph and get a K value out of it so you've seen some of my intro videos I sketched graph out like this and said there was a gradient is equal to K and so on this is the part we are now going to cover so that's so if pseudo first-order react well not sorry no pseudo first this is an actual first-order reaction this will basic dissociation pretty much follows first-order kinetics this is where our rate is dependent on only one variable and that is the concentration of the starting material so our rate our change of AB is proportional to oh this should be B I'm really sorry that should be B I will change it so just the rate of the starting material disappearing equals that so we have a nice exponential to K curve that we could measure say spectroscopically and here's a reaction example so to peroxide molecule breaking down to water and oxygen once again you know microscopic symbolic microscopic think about all the different ways we can actually look at kinetics and which different perspectives we can see it from all the exact same thing going on so this is a first-order reaction so let's have a look at something goes from one thing to the other let's have a look at what it would look like over time you see one decay that's the reactants products appear B that's a I've written this is more fraction so this would be like 100% of a going to zero percent of B here so at the very end and the rate is proportional to a single factor and we can define that kinetically via our differential equations now this is something here we'll cover this in a bit more detail later it is called the integrated rate law so how to get this is well it's integration so you are comfortable with calculus and differentiation integration you should be able to get from this equation to this one no problem but in short the integrated rate law is basically what predicts these two curves that you can see what you can see here is a is equal the concentration a at a particular time is equal to the concentration of a at time zero times an exponential to minus KT okay that looks a bit weird but that is why this decay happens very gradually in a nice exponential fashion because it follows that law now if you see let's rewrite it out as a different way if you might see an integrated rate law like this means pretty much exactly the same thing that's a T basically instead of saying initial concentration we're just going to pick an arbitrary time and then it arbitrary time plus a slight gap so if we were going to do the simulation method where we try to simulate race constants we might use this method a bit more it preaches pretty much the same result as long as you're consistent in fact this is kind of the same thing as this we're just saying that T is in this case we're saying T is zero and the delta time is or whatever the time is produces the same results give a take so an obvious first order reaction is something like peroxide decomposing so in this kind of video hopefully the screen captures gets the gif here you've probably seen at the elephants toothpaste that's pretty much a first order reaction it is peroxide breaking down in this case catalytically but pretty much of its own accord according to just the concentration a second order reaction so we come across this sort of thing before this is the sign of thing like inucleophilic addition so this is an elementary step this is a single step it is the simplest chemical reaction so this is not obviously the end product of a reaction this is just kind of a single step so this means we can do things like k a b to determine the rate so again we're only interested in elementary steps if we've tracked an entire reaction as in what does this then proceed on to we have to determine it experimentally but for the most part we can say this is second order two things are coming together so therefore the rate is dependent on two concentrations so here we have a OH coming in targets a ketone group and there's again I should pay attention to my own diagrams here this should be a seal there and it's nucleophilic addition so that's the second order reaction so how does that work out as well the rate is proportional to two things and we can define that as the speed that the two reactants disappear or the speed that C appears now this is a slightly different plot because I'm actually now doing this not as a mole fraction I'm doing it as absolute moles so the blue line here say that that's two moles we start with one mole of that and one mole of this and that will obviously because this combination go to one mole of that so we're talking about discrete chemical entities when we're talking about moles hence why we do concentration in moles in kinetics so obviously by the end of the reaction we've got zero moles of that left in one mole of this so here we go so it doesn't quite swap one to one in this case but the rate of each of these individually this is not one of those well we could do it as a squared if you want but for now we're keeping them distinct but what if we had a lot of be we'd actually have a plot that looked like this this wouldn't in fact go to zero this would stay quite a bit above zero because we can only do one more of this plus one more of that to one more of C so what if we started with about 10 moles of a and then we added one more of B or what we'd get one more of C out and then nine moles of a leftover there's nothing left for it to react with so as a result one of the concentrations isn't changing very much so and in this case well I've actually ever in the round in this case of the diagram it's B that's unused so B is in an excess so you're used to this in chemistry something is in excess so what's the consequence of that well have a look at this sort of scheme here we've got the exact same you could feel a condition was before but we've got a lot of waitress left over so one two three four six of them one two three four five six seven of them over here seven of them over here so we've got a lot of excess and these aren't going to react at all so let's look at the kind of the mathematical consequences of this when we look at the rate so we're actually six numbers in here these are you can see that's a vast excess so about 0.03 moles per decimeter cubed as a starting number that's a really minuscule amount compared to say 0.5 moles per decimeter cubed of the other of B here equals a so we can then multiply this out to get a rate and the rate will be k times this value k times this value k times that value so that means the rate is decreasing as the concentration of a drops and obviously the concentration of B is dropping as well but only slightly so no point no one moles of this react so one point no point no one moles of this disappears each time so you can see each of these are dropping by minus no point no one but obviously this is dropped by a third this is dropped by just over a percent and so I'm going to make an approximation in the next screen I'm not going to know what the concentration of B is I'm going to assume it's no point no point nine five each time I'm going to assume B doesn't change because it is in such excess I'm just going to take that value as the value of the initial concentration so if I set up a reaction that has B in it I know how much I put in at the beginning I know how much AI put in at the beginning and I'm going to assume B doesn't change I get these values out so I've actually calculated these values these are exactly right you can check the math if you want and what you will find is compared to the previous values notice they hardly changed there in fact only within a couple of percent so obviously it gets as the reaction proceeds a bit further they do start diverging a bit this is within two percent if with this kept going it would probably or if you know if a wasn't a bit more excess and was allowed to proceed to the end obviously this would begin to this approximation would definitely stop holding so at the beginning of a reaction when I think it's in excess is still in excess the pseudo first-order approximation applies that is we can assume the concentration doesn't change so what is the consequence of that consequences that is that we can simplify our rate equals k a B equation to something else k is constant B is constant we can change that into something called k observed now this is where a bit of notation comes gets a bit confusing now depending on which lecture you talk to or which textbook you read k the observed rate constant the pseudo first-order rate constant does change I prefer k obs because this is literally the observed short for observed my handwriting is terrible so is the observed rate constant if we got some data out of this particular reaction that's the rate constant we will be able to determine quite easily it's literally seen but sometimes k one sometimes k prime sometimes k something else k obs is the one I'm going to use but if you are reading all the textbooks and all the sources get used to seeing slightly different versions so let's look at some actual data if a goes to be we're just going to jump back to covering a first-order reaction now the so this is going a going to be or it's a plus something that's in excess going to be so maybe a catalyst maybe another reactant but either way the pseudo first-order approximation will apply so we can treat it like a first-order reaction so it will have first-order kinetics this line as we see will behave as if it is a first-order equation and that it runs according to first-order kinetics now in real life real life you're not likely to get these smooth perfect lines you're more likely to get a little jaggedy nonsense number like this so you can see it does still have that curve down but see these don't fit very well maybe one of these are slightly off we've got enough data points but if you wanted to plot that out it would look a bit jagged so be aware that there is going to be some real data that is always a bit iffy now we are going to cover something about errors and solving this in the lecture series later but for now be aware that your data can be a bit jagged so we're going to plot and do some best fits more than anything else so what actually controls this equation we're going to assume that it's first-order or pseudo first-order it runs according to that this is the integrated rate equation so the integrated rate law comes from a rate law we just apply some integration to it and then we take it out of the log form to this so this is a sort of an equation that predicts and models how the concentration will change and it becomes quite useful because if we take a log of it we actually get a linear equation so this is remember physical chemists like y equals plus c there we go minus mx or y equals mx plus c other way to linear equation and as we can see these four data points here form pretty much a straight line these two don't they're a bit off why would they be off well if it's a pseudo first-order reaction remember as the concentration would be genuinely does decrease it starts to diverge away from that first-order behavior you can't tell just by looking at a curve all rate look quite curves look a bit just like exponential decay curves if human eye really can't eyeball that very well we turn it into a straight line however we can very much see that those do not fit so those last things at the end I've also converted that into seconds remember keep it in seconds to make it useful we can see that those last two data points we can get away with so we want first you know handful four of data points and then let's see what that looks like so if we plot those we get a straight line graph and if we run a straight line through it we get no point 0.79 that's the gradient so there's the gradient out k is equal to 0.079 it's a first-order or pseudo first-order reaction so I will leave you to figure out what those units are yourself I'm gonna tell you you should be able to figure that one out and that's it so let's kind of review all that there's a quite a bit to take in so first-order reactions they are really simple to plot rate is equal to k times a concentration so we can solve for k really easily so we did a bit that towards the end a pseudo first-order reaction is well obviously the real rate is equal is proportional to real k times a times b but suitably large it be the height of concentration then the rate is actually we can k be is equal to k obs and solve for that because that is more or less constant or at least it is constant at the beginning of a reaction for maybe the first first 80% of the reaction usually the last 20% you just wanted to discard in practice so obviously a pseudo first-order reaction is as simple as a first-order reaction to solve you just take a log of the concentration plotted against time get the gradient and so the data for these two types of reactions we collect concentrations versus time so that's our graph and then we plot log a versus time and we actually see that it's a straight line and the gradient is minus k so that is a y equals mx plus c graph really really simple get physical chemistry love it when you can put things into a straight line so that's how you would solve a great constant for a first-order or pseudo first-order reaction I do go over through that again I didn't think in first time I'm gonna go through and do this in Excel in another video shortly so that will cover two different methods one