 Right so this screencast is all about thermodynamics. I'm currently coming to you from a laptop that's been literally held together with gaffer tip but we'll see what holds up. What we're going to cover today are hydraulic kinetics and thermodynamics so we did a little bit of this in the lectures. So this is going to recap most of that and kind of show the the derivation again for how to link various things together. So I'm going to recap what's the difference between kinetics and thermodynamics. What are those involved? And then connecting them together. So activation in g2 delta h and then I'm going to put it onto the slide but we're going to do the Arrhenius factor to delta s. I've given you the summary of that but I didn't go through it in details and then extend to something called transition state theory. So that's going to be at the end. It's technically outside the scope of the course but it should set the ground work for what you will do later on involved in physical chemistry. So now equilibrium versus kinetics. What's the difference between the big k and the little k? So hopefully this is going to be revision but at least sets the scene for what we're going to be talking about later. So what you should be aware of is the difference between diamond and graphite. What you might not know, you may know is that diamond is actually a higher energy structure than graphite. By considerable amount 1.9 kilojoules per mole is the delta h value between them. I remember rightly. It's in kilojoules per mole. It's delta g value is somewhere in the region of about three kilojoules per mole. So obviously one is going to be more stable than the other and if this was just a chemical equilibrium we could plug this into various equations to get us our rate constants. So if we start with three kilojoules per mole we want to insert that here rearrange to get k. This e to an energy over RT type function is what all the physical chemistry seems to revolve around this sort of function and we can evaluate that to get 0.2979. Now we can then rearrange that and if this was a chemical equilibrium we would expect 77 percent graphite and 23 percent diamond. Evidently that doesn't happen. Diamonds are pretty much solid. They don't react. Why isn't that the case? For instance so this is why we deal with kinetics. So what you will find is actually the activation energy to go from diamond to graphite or vice versa is somewhere in the region of 540 kilojoules per mole. At least the figure that I can find out that's the at least the activation enthalpy ish. This is reaction. So those are two energy values that these would have to overcome in order to into conversion. So as a result one might be thermodynamically stable but it can't transfer there the kinetic energy barriers way too high. So this cannot happen past enough and we'll go on to quantifying how not very quick in a moment. Just to watch it be worth watching if I remember to put a link to this is that you can actually get diamond to overcome reaction barriers if you set it on fire. This is a screenshot from the 2012 Christmas lectures where they set a diamond on fire in in concentration of oxygen. So yes they are reactive you can get them to react but you do need quite a bit of energy to get over that activation barrier. It's thermodynamically unstable but it's so kinetically inert it takes a lot of energy to actually get it over an activation barrier and to react. One kind of analogy for this that I found that's actually quite interesting so you're interested in teaching and later on this might be something that's useful. The idea of knocking something over if you actually had a box up on end like this obviously it's more stable if it is lying down like that. The trouble is in order to tip it over that central gravity the center of mass has to go up so it's almost like this is your starting material that's the end but it won't just spontaneously tumble over you've got to get it over some kind of activation barrier so if you want to knock something over you've actually got to do the kind of picking it up by that kind of distance so you can actually work out how much energy that takes. If you're really really interested you can figure out how much it is to tip a cow over and that's why cow tipping is another myth but anyway activation energy to enthalpy this is the important thing that you probably need to be aware of. So what I'm going to do is just throw the answer at you immediately what we're finding is that Delta H here the difference between all reactant and product in terms of their internal energy that enthalpy is equivalent of the difference between the activation energy so there are two reactions one goes forward it has this activation energy one goes backwards it has this activation energy the difference between them is then Delta H of the reaction so how are we going to deal with this well let's start with the very basics an irreversible reaction our rate is equal to the rate of change of the product the reactance of the rate of change of the product and that can be controlled by a rate law of dA by dt is equal to negative kA that's our starting point for any kind of kinetics but in equilibrium things are a little bit different on definition of the rate is exactly the same the rate of reaction will go at a particular speed and A will convert to B the same rate but they're controlled by two reactions this time so A will grow by whatever this reaction is but it will decrease by whatever that reaction is so it will be consumed by one reaction when we form by another so at this point we can say the rate of change is equal to that plus this or we can reverse it we said dB by dt is equal to positive of k1 because B grows by that reaction minus k negative one times B because of that is now decreasing concentration so remember these are just one set of ways of denoting different rate constants I use one and minus one for a forward and backwards reaction you can actually get k and someone right forward and k backward you might even see kA, B, kB, A or someone might be weak and imagine doing k1 and k2 they are all the same as only you're kind of consistent and you're happy that you know which is referring to what you can use any numbering scheme or labeling scheme you like so at equilibrium you should know as chemists the reaction is not changing there's not a change in concentration how microscopically yes the molecule might be converting from one to the other but microscopically in the lab we can't measure a concentration change so these must add up to zero the rate of change of A and the rate of change of B is both zero that means these must add or subtract to cancel each other out and that leads to this conclusion k1A is equal to k minus one B so this is one of those approximations we can start using to get information from it so we assume everything's at equilibrium these rates are now the same and we can then rearrange this rearrange this equation bring that over to this side this over to that side and what we get is a ratio of our concentrations equals a ratio of rate constants now you should also be aware from thermodynamics that a ratio of concentrations is equal to an equilibrium constant and this is the exact same equilibrium constant that appears in this delta g equation here this is one of the core equations of physical chemistry this is one of the ones that you should be able to rattle off pretty much instantaneously if i say delta g equals to you should say that's all the delta h minus t delta estimate both of the same both are quite equally important so that gives us a hint of what we're going to do next to this equation we've got k here we want to get it into log form so we can do this k to log k we've now got equal to that part enough we take a log of our rate constants the log rule say that they become subtractive so log of k over k the minus one becomes this so that's just logarithmic rules it's a little bit of mathematical rearrangement but this is a another equation that gives us k so previously we started with a ratio of rate constants is equal to k here the difference in the logs of them is equal to log k it's just a mathematical transformation this becomes very very useful to us all right so after that this is a bit of an obscure step you might not necessarily think about it but we're going to differentiate it with respect to temperature check the delta g equation for a moment t is in there so delta g is a function of the rate of the equilibrium constant and temperature so it's not entirely out of nowhere but you might not necessarily think to do this in this stage so we've just differentiated with respect to t that gives us this equation and we will start substituting in these different components so this is one to remember we're going to take this and figure out what does this mean and substitute it and what does that mean and substitute it so let's start with this one let's have some written blanked out but if we start with delta g minus r t log k rearrange it put them to this side and then we substitute delta g for delta h minus t delta s so we could in fact swap this entire factor here for this one and then it's just a case of breaking it up into two components delta h over r t and t delta s over r t and then well we've got negative here so we need to reverse this so we've got that there so this is the sort of manipulation that you need to get used to doing as well if we have a delta g function at some point you should be able to substitute in delta h and delta s and vice versa as well if we have a delta h equals something you should be able to break it up into delta g and delta s again it's going to be really useful especially in some tutorial questions or some quite complicated ones right from here we just start cancelling things down there's the t and the delta s there the t delta s cancels out there uses a much more simplified version here and then we're going to differentiate with respect to temperature because previously we differentiated log k with respect to temperature we've got log k on this side makes sense that we want to do this to both sides so here's my d by dt that's an operator and we apply that to both sides of an equation in order for this to work so we want to differentiate that with respect to t and that with respect to t okay so how does this actually work well this well we can just leave this as is this we can actually do something with we can actually evaluate that function quite nicely so we want to differentiate this with respect to t what you'll find is that is the equivalent of t to power zero so well we'll multiply that by zero it cancels out uh this one i'll draw it up over here that's equal to delta h over r t to the minus one so what we do is differentiate that as before we want that to be t to the minus two then we times that by minus one uh so that becomes minus delta over r and that is what we insert over here so what we find is d log k by dt is equal to delta h over r t squared and so that's if you're still not quite comfortable with it that's what we use minus two they're equal to just to find us right so from that we've gone from delta t delta g equals minus r t log k to over that expression okay so we're going to leave that one there now go back to our actual rates again and we're going to look at the Arrhenius equations so that's what is the rate constant equal to so the rate constant is a function of two things one is temperature one is an activation energy and we also have that pre-exponential factor that we're not as interested in at the moment but it is also component and by extension that also applies to an equilibrium we have a rate going forward and a rate coming back we have exactly the same equation on both sides difference being I've just labeled them differently so a forward reaction and a backwards reaction will have a different pre-exponential factor it will also have a different activation energy so go back to that very first diagram I showed there are two different activation energies going forward and backwards so we've got two rate constants related to two activation energies great now we can take logs of that as well because this gets the equation at least linear but remember previously we had some logs in a previous equation so we take a log of that our a comes down to here and when we time something right to the activation energy of our rt that part comes down okay so that's again happens for both ways forwards and backwards and now we can go back to this sort of equation what you can see is this was kind of what we were we got a little earlier and we could if we wanted to substitute those in although we're going to right now we're going to do something a bit different which is differentiated with respect to temperature so what do how do we differentiate this with respect to temperature there's our dy dt operator well we can just leave these as is for now and what's that going to be that's t to the minus one just as before that's going to go to t to the minus two uh at least negative this becomes zero because there's no t in it this comes down to here that's the exact same rules as before and we do that twice to get this and that function out of it so what we've got is two functions from our radius expression one function that we did when we were trying to rearrange this expression and differentiate it by t and we had this one earlier as well and now we're going to just substitute those in and what do we get from it we get this here so we've substituted those all in we've converted this to here we've converted each of these log k values into into these activation energy terms and what you should notice is these are t squares all canceled and we end up with this rather boring looking equation that looks exactly the same as what our intuition said the difference in activation energies is equal to the reaction enthalpy great it is really nice when the maths actually works out so there's our activation energies here's the delta h the difference between them uh the activation energies is delta h fabulous uh so all that long convoluted maths does prove something that you might be able to show with a diagram but it's actually kind of the other way around the diagram comes from this set of calculations just the diagram's really easy for you to understand so you usually get that presented first so anyway now the pre-exponential factor to entropy i've given you this in the lecture notes but in a kind of condensed form so we'll go through it now kind of step by step so just in case you're a bit lost with it we're starting with this as before we start with the urinius equation and we linearize it because we're interested in logs and we also want that relationship that we had before so you know previously i said we could just substitute these straight in without doing any um deriving with respect to temperature here so let's go ahead and do that each of these k's are equal to that so we can substitute in for all of this here okay so we've done one we've done another this is just some rearrangement now so we'll remove the brackets out first because that just helps you ensure that we've got two values there and then rearranging so i want to collect these terms on one side and these terms on one side so now we've got something that looks at least a little tidier you can see for instance that has over rt in common so that can be rewritten as this for instance which is usually useful for us uh so let's have a look what we want to do is then do that but also pay attention to these log rules okay so we are subtracting two logs that's equivalent of doing dividing numbers it's just one of these rules you need to get used to in order to make faith the numbers so actually the log of the ratio of these pre-experimental factors here is what that simplifies to we can also take that one over rt out so we're collecting these together as well now that section there once we've cancelled out those negatives and so on looks very similar to what we had previously that is what was equal to delta h so you can actually then you do this begins substituting that in so we have delta h here but you also can notice that this section now looks very similar to that previous equation so we've had this one we've rearranged it and then we can substitute delta g for delta h and delta s cancel out the t's obviously want to cancel those ones out but that's not entirely a formal proof we can't just say one equation looks very similar so what we need to do is just substitute that log k for the delta s and delta h so we want to take this and we want to shove it in the place of that so that's what we've cut down here so we've just rearranged them and now we're going to replace that with a delta h and what you find is here's the delta h over rt here's delta h over rt they're going to cancel and we end up with delta s equals log of the ratio of these pre-experimental factors so once again a little bit of convoluted mathematics but it's there hopefully you can follow it if not the oh well so that's a slightly less intuitive result than this one is so just to review the two things we're looking at delta h this is our intuitive result it's the difference between the activation energies this is the less intuitive result less needed for an exam certainly delta rt over r is equal to a log of these two factors so we could get delta s out of our Rhenius data if we do the right kinetics together in reality you'd probably use something like the Eyring equation to get delta s and delta h out of this because it's getting forward and backwards equilibria data is a little bit more difficult in reality but still that is the main relationship you need to know so the whole that's previous derivation anyway now this is transition states so this is I don't want to spend too long on it it just lay the ground work for something you might come across a bit later called transition states theory so this is how we relate kinetics to thermodynamics a little bit more qualitatively I don't want to put too many huge derivations into it so this is what we're looking at here we've got our activation energy in our reaction profile so one can go from reactant to product delta h separating them the controls the equilibrium so these tell us what k is equal to the small k that tells us what the forward direction is this tells us big k the equilibrium and so on and we could also in principle swap these over out for delta g and if it's delta g we can get the equilibrium constant out of it and what we're interested in is actually this transition state here because we're going to set up an equilibrium between the reactant and not the product but whatever is up here because a transition state is still a chemical it will still be a physical entity and it will have energy associated with it so if a reacting has to convert to a certain form that's really stretched and strained and high energy it's got to take an engine and then it will relax back down to the product now the interesting part about this is that once a molecule hits this transition state uh that's level it can actually in fact it has a 50 probability going that way and a 50 percent probability if one kind of works off going that way so that's a new caveat you need to know we get it up to a transition state bubbles a little bit is it going to fall down back to the reactant what's it going to fall down towards the product it's a 50-50 chance so what we're so what we're going to do next is set up a equilibrium between that and the transition state because we know the energy difference between this one up here the transition state and the ground state of the reactant we know how many molecules in theory have enough energy to get up to here not too dissimilar to the Maxwell Boltzmann distribution and the back send the Boltzmann factor and all of that but this is a slightly more rigorous way of getting to maths a bit so let's look at the diamond example again that activation energy is about 540 kilojoules per mole and I'm just going to assume delta g is roughly about that it is not going to make as much of a difference as you might think so if delta g is roughly that we can actually work out what k is what is the equilibrium constant between say diamond and its transition state towards graphite well you can actually plug that into the equation and you get 10 to the minus 88 that is a hugely small equilibrium constant that tells you everything that is going to everything you need to know about the speed of the reaction everything is going to be on this side of the equilibrium between diamond and the transition state so we've got diamond on one side transition state on the other the equilibrium very very much lies in that direction so do the maths about 1 in 10 to the 88 atoms almost to the point where to simply imagine what the exact numbers are that is huge and to put it in perspective this is the estimate for how many atoms are in the entire universe so if 1 in 10 to the 88 diamond atoms are going to be carbon atoms in diamond are going to be at its transition state then if the entire universe was made a diamond you still wouldn't find one that's quite an impressive stability there so this is not going to climb over that activation in the new barrier spontaneously at all certainly not at room temperature you have to really pump it up to tens of thousands of degrees to make it spontaneously change to graphite so that is why even though it is quite highly quite significantly less stable less stable than graphite it won't convert spontaneously nowhere can these atoms get to that transition state at least to a first approximation with transition state theory but let's try it with a slightly different delta g so imagine we have a delta g of activation of 100 kilojoules per mole so activation energy here the difference between the ground state that reactant and our transition state is 100 kilojoules per mole in delta g we can then put that back into our equation to find out what is k we rearrange that and that ends up at 1.17 times 10 to the minus 14 at 375 again that sounds really low but that's a mole of materials is on the scale of 1 times 10 to the 23 so actually a quite an appreciable number if it's 1 10 to the 14 of these would in fact react so at any one point in time something with just an activation energy of this quite a number should be capable of reacting so that's roughly how the transition state theory begins to work we try to figure out how many molecules can reach the transition state what is the equilibrium constant and can we convert that into rate data so when you get into transition state theory properly this is what you're going to be end up doing in a little bit more rigor there are a few other caveats that we don't we have time to cover right now but this is setting the scene for what we're going to be doing climate transition state and dropping down let's just review it again so equilibria so states are separated by energy so the higher energy states are less populated this is true of all chemistry um you're after something with low energy because more thing will fill that state so you can do this in more transition state there's an equilibrium between a reactant and it's transition state but that energy is a lot higher so whereas with a normal delta g of you know only a couple of kilojoules per mole or something like that you might be expecting you know an equilibrium constantly you can measure so few of here you're going to be talking about 10 to the 1 10 to the 14 molecules only a few molecules at any one time we're going to have the right energy so therefore we can kind of start working on what should affect the rate based on activation energies so more populated transition state or a low energy transition state means the reaction is faster so it shouldn't escape your attention that that is what a catalyst does it lowers the energy of the transition state so it's more likely the molecules will have that energy things will move faster then a higher energy transition state will make the reaction slower so just like we're going from diamond to carbon the transition state if i was to draw a half decent looking reaction profile is absolutely staggeringly huge we go from diamond to graphite that looks like it's a really low energy jump but it's got to get up to here first in fact to scale i think if you go back to that diagram the energy difference on the diagram will only be a couple of centimeters the transition state is about 10 meters above you to scale on that diagram so this is huge hundreds upon hundreds of kilojoules per mole you're not going to see that reaction occur so that's the basics of transition state theory before we're just covering most of the derivations so you don't need to learn those derivations but you do need to be aware of how you would go about doing that kind of thing and so that really is it for dealing with thermodynamics and kinetics