 this is our kind of revision screencast so I'm just going to go through some of the questions that have been asked and maybe the topics that you probably want to revise a bit more so after doing that I asked you a bunch of questions via Socrative whether you knew them very well or you had no idea or you thought you were a bit unsure about them so what I've done is ranked all those scores and given them a score so the higher this value the less you seem to know about it so things like this you can describe how temperature effects rate everyone said that they could do that just perfectly fine fantastic everything that's a little bit higher I'm going to go through so I'm selecting basically anything that's called higher than 10 on the bright side though if you all answered that you didn't understand any of it this would add up to about 1500 as it stands as you only adds up to 136 so you're actually 99% of the way well 90% ish the way there so that's not bad so the topics I'm gonna cover all these and hopefully which direction would they be going they'll be here I'm gonna stick those YouTube annotations on those so if you want to just skip to one of these things you can click on those and then if you want click the back to start one there so that should stay up there so it's gonna be a screen has you can navigate around if you're using YouTube just quite nice also I'm gonna probably focus on these three a bit more because this is these are the only three where more people said you really weren't sure next to I could do this with some revision so there's gonna be a little bit more on those anyway I'm gonna go through them in order now you don't need to take it with all in order of course those annotations are there if you want to click through at least they should work and hopefully they do anyway so on with it first pseudo first order approximation so couple people have a few issues about how to apply this or when to apply it it's certainly a bit of a weird thing to come across the first time you find it and the basics of this in a second order reaction that is our rate law it's proportional to two concentrations but we can simplify this down if sorry that case should not be there to just a first order reaction but this is only true when one of the reagents in excess so in this case be one is in excess it's concentrations doesn't change if this was say five moles and a was no point one moles for instance just actually write it down the starting concentration of B is five moles a is on the order of 0.1 moles then the end concentration of this would be zero and four point nine okay that means as far as we're concerned that is pretty much state constant which is fine so that means we can simplify things down to here where this k obs the observed rate constant is equal to k times concentration of B so that tells us that you know if we were to do this reaction and we plotted log A versus T and we get a straight line a couple of caveats to this B will decrease towards the end of an action so we might actually start to see a little bit of a curve so if you did your reaction this is the T and this is log A you might start seeing things go down and then they'll curve we'll go away from linearity so in that case it would be perfectly possible to get rid of those data points and just plot a straight line through there great and that gets you k that is your rate constant and you can feed that back into this equation to get your real rate constant because presumably you know what B is it averages out to whatever your starting concentration is so mostly you want to use it to simplify some of your reactions and you only use it when one is in quite a bit of excess so it's just a little bit of a mathematical trick and it just happens to work out experimentally it's it's a pretty nice approximation so this Arrhenius equation and simple collision theory so we've come across the Arrhenius equation before it's the one that tells us that the rate constant is basically a function of temperature the rate constant changes according to temperature but there are a few things that fit into it and so how do we relate that to what we can pick up from simple collision theory so this should be very straightforward simple collision theory tells us that molecules must collide but normally must they collide they must collide in the right orientation and with the right energy so that means for every collision we have there we've got to modify that a little bit so z zero is the collision rate and that's the collision rate independent of any concentration so we had the whole Z A B thing that was the cross section times the speed times bar density so if we ignore the density for a moment that is z zero that is the rate of collisions it's a theoretical rate constant as well so if all collisions were successful the reaction would proceed with z zero equals k so that's a interesting fact to remember it is a theoretical rate constant unfortunately that theory that all collisions react doesn't hold true okay it's not brilliant theory so we need to modify it for a few things namely energy do they literally have the right energy to react if they don't that's kind of independent of the molecule speed although it is related if they have the right energy they react and if they have the right orientation they react so remember you have a big long chain molecule with a functional group at one end and a long carbon chain at the other you know if a collision happens over there it's not going to react so you need to have a rights orientation and the right position so we have this steric factor here so that brings us back to the A in the Arrhenius expression you should be able to see that these are very very similar so when I'm asking you to relate the Arrhenius expression and simple collision theory this is what we're asking you to spot the connection between there's a pre exponential factor and then exponential term and notice these are very very similar in form and that's no accident so basically says that a our pre exponential factor in the Arrhenius equation that we can get from data is equal to this collision rate times the steric factor because these two are related so in other words it basically means the number of successfully orientated collisions so that's great and we also have this other factor here and the important thing to realise is that that uses the Boltzmann constant so this is talking about individual molecules this simple collision theory assumes that we're only looking at the molecules the Arrhenius expression that's derived empirically we're working with huge quantities of molecules so we use the gas constant that's all in per mole this is in per molecule or molecules anyway that's what we need when we want to connect these two together notice the equations are very similar and then see what the pass okay so experimental methods I think more people click that they weren't quite sure about this one probably because we didn't spend much time in the lectures doing it that's sort of intentional there's a lot of derivations that are part of the course that I've shipped onto the screencasts they are worth going through you are expected to know them or at least you are expected to be able to follow those manipulations quite well you're not expected to memorise them for instance but you do need to follow them so go through those screencasts and follow the derivations at some point in the revision period probably but here's just a quick way of going through it I'm going to recap them you've got to think of the properties of your reaction and link them to the properties of the method you want to follow them with so let's just go through these in situ reactions of these in situ methods sorry first of all UV viz okay that's an excitation of electrons you'd be very used to that from the labs that's where you should get really broad peaks that look like this and that's your wavelength usually in nanometers and then absorbance now you require for that to really work nicely the molecule should be absorbing white things like transition metal compounds really really good for that they have a really strong charge transfer bands or DD transitions depending on how deep you've gone into an organic chemistry so far but those are really good for organic reactions so look at these molecules these are dyes I think specifically these are hair dyes and look at all these alternating two ones everywhere low in pairs that can do on it so that is one whole huge congiated system similar here I think that's an azo dye with that group in the middle all conjugated all conjugated all of these alternating double bonds that are really extensive or a good indication that your molecule has massively good chance of absorbing light they're called chromophores they absorb light very very effectively so if you've got a molecule that looks like that massive congiated system UV it will be easily detectable by UV and sometimes if you react it you're going to break that chromophore up so that it no longer absorbs light so a lot of reactions involving dyes have a tendency to break that chromophore really obvious ones being the crystal violet crystal green type solutions you have this double bond here with a with phenyl rings I'm drawing it terribly here and if you add OH to this double bond here you would break that in half and then it goes colorless so a lot of dyes reactions like that were really good to follow with UV infrared this excites with vibrations so you should be very familiar with that you obviously OH bands are really visible CH ones quite typical very characteristic RCO bands as well so if your reaction gets rid of one of those or produces one of those infrared is a good bet to use so you probably couldn't see seen in organic chemistry so far nucleophilic substitution reactions like this or nucleophilic addition reactions that's very easily detectable by infrared it's a completely different functional group here so you have that huge change in functional group infrared it's a really quick fast method so as the peak the corresponds to that bond disappears that's your reaction an MR this is another kind of fallback which is really good for following kinetics providing they are slow an NMR scan takes at least a minute to do so that's one minute per data point and you want 10 or 15 data points to get a decent trace so if your reaction is faster than about 15 minutes don't bother but you don't need to necessarily have UV active molecule you don't necessarily need to have a major change in functional group you can really follow almost anything by NMR anything with hydrogen or whether any NMR device to open it you can follow with it again slow but if you've got the material then you can see it and you can get time on the spectrometer for long enough it's a really good bet to use all right so the other in situ methods so these are the ones that the screencasts have a lot of derivations with so I've only briefly covered them in the actual face-to-face lectures conductivity and pressure now these are useful because conductivity is proportional to concentration at least once you've taken into account the final and initial conductivity of the equation slightly complicated but I'm not going to go over it this time pressure well number pv equals nrt so you know your pressure is directly proportional to the number of moles so therefore directly proportional to concentration so you can use pressure directly so if your reaction produces ions now things can eliminate cl like well come here up in time they eliminate or consume things like this if they are changing their charges at some point conductivity is good unfortunately conductivity is proportional to the total ion concentration so it's if you have h plus and cl being liberated that is what's proportional to the conductivity can't quite hand write that would be compared to okay so you would have to take into account the fact that that might be twice the concentration of your actual product so you produce one product in one of these each you then have to divide your conductivity in half so that's the one that's worth paying attention to pressure in other words is on the other hand is pretty much direction directly proportional to the number of moles a little bit convoluted mass to get there but it's you know the punch line is pretty decent so if you're your reaction produces gas or consumes gas followed by pressure really really useful and then there's going to fast reactions so all of those previous ones need to be done on a lab time scale where you've got time to fiddle about with things and take readings usually reactions tend to happen a bit faster than that usually a couple of seconds and they're done or your milliseconds and they're done and that's where you use flow methods and flush potatoes so flow methods if your reactions are over a couple of seconds you can go back and use those again I'm not going to describe those again you can look those up in a text but what are the flow methods you've got continuous flow and stopped flow do you just shove those into a surge engine if you'll want you to find out more about them and flush patolis is the one for reactions that are really fast you can spot intermediate to flush patolis because you are initiating the reaction with a pulse of light and then you're seeing what changes there are on a nanosecond time scale down to that kind of level you can do quite reasonably because you know very precisely when the reaction starts because you've flushed it with a UV initiation it does require a photochemical reaction to initiate unfortunately so if your actions just thermal you can't quite use that method but usually a good pulse of light will set up most reactions and you can follow them through kinetics that way so this is not canonical this is just the kind of questions you will want to ask yourself so does that react you over several minutes we'll know if it's really quick how quick if it's almost instant flash for tall assist that's a good bet that's for you if it reacts just over a couple of seconds maybe stop flow is what you want but if it reacts quite slowly have a look what changes are there just the function group change yes infrared great there's usually good pretty clear change in the infrared there's absorbance change that's specifically UV vis absorbance so yes use UV to follow it does a producer consume ions do conductivity does a producer concern to consume gas pressure developed by following by pressure and if your answer is no to all of those you might need to just go back to non-insertion methods and quench the reaction to follow it through something else usually reacting it into something that you can detect but you will need some time to do that so go back over the quenching if you want to understand that but this is the kind of the idea of what your thought process should look at it's not so much something that you need to memorize this floor chart this is just the process of what you should be thinking about what are the properties the reaction do they match up with the method I want to use to follow it if it happens if your reaction is over and done within two seconds don't use magnetic resonance to follow it because you need at least a minute for a data point if the number of molecules on the left and the right of the same the pressure isn't going to change so you can't follow it by pressure of volume changes okay there's that kind of thinking so you don't want to get confused right so integrated by laws I'm thrown up a few videos on campus about this so I don't want to spend too long on it so I'm just going to give the general method for how you want to go about doing this so if I remember I'm going to stick one of those annotations there so you should be able to click that and go to that video it covers the second order integrated rate law so in general what you need to do is take your rate law it's going to be changing concentration over time is equal to K times some concentration and then you want to rearrange that so dA by dt equals K we can bring that dt to the other side and that means we are integrating one side with respect to A and the other side with respect to T okay that is how we figure out what we want to integrate we integrate one side with respect to the concentration the other side with respect to time doesn't get as much useful information that alone so we want to use the definite integral so 0 and T 0 and T some people will very explicitly say that we are actually going to do this between a0 and aT both are you know valid enough really so you want to label the concentration as time t and the initial concentration at time 0 and you when you do the definite integral for KT you'll get KT minus K0 okay that cancels out so most of these integrated rate laws end up with KT on the right what happens on the left is a little bit changes depending on what it is so I've found a couple of derivations for when you've got KAB and also KA cubed I've stuck those up on canvas but these are not ones that you need to worry about those are kind of very much extension if you want to follow the maths involved with it you're only expected to you know the first and second order one so this is them so the first order one that well when you rearrange that you end up with one over A and you want to integrate that with respect to A the concentration well that's your special case where you end up with log A now that log A produces a linear graph if we want to get what A is equal to you need A to the power of something so that's what we got here so both versions of the integrated rate law one is linear one actually predicts the concentration and this is why when you plot log of your concentration versus time you get a straight line with a gradient of k so let's just leave that up there for a moment the second order reaction when you rearrange that you're gonna get one over A squared DA or if you want to put it a different way A to the minus 2 DA so the things that you might want to spot with that is make sure your minuses work out because if you integrate this you should get A to the minus one but then you divide by minus one so that should end up this kind of form so you check that your minuses are right and so that depends whether you've put a minus there or not and this 2 as well comes depending on whether you use rate equals 1 over 2 DA by DT so things get a little bit more complicated here that you might need to be aware of so depending on where you look you will find slightly different versions of this formula so you just need to be very careful about how you define the rate in the first place if you want to define it as this half DA by DT you'll find this two in it if you use B the product DT is equal to k of the reactant squared you'll get a positive instead of a minus at this point so a few things start changing they're not always a straight that's the most straightforward version doesn't look it because you have to use logs and exponentials but there is usually only one version of that the rest start taking multiple versions depending on what you define things as so you need to be able to take into account that they should be straightforward they're all equally valid as long as you're consistent basically it's thermodynamics and equilibrium kinetics so we cover kind of a long convoluted derivation about this but I think it's a lot simpler than you might think though so the derivation was quite complex but you don't need to know that as such you should be aware of it and you should be able to follow the maths but you won't be expected to regurgitate it in an exam because regurgitating a long derivation like that is a bit pointless to be fair but you are expected to know the punchline to it and the basics of it so where do we start well let's have a look at the reversible case DA by DT is equal to minus kA that's the first order kinetics but in a reversible reaction there are two processes that create that that alter the concentration so if we have a goes to be or goes back again we have potentially two rate laws we have this DA by DT is equal to minus kA yeah that's just what we expect we also have the reverse reaction k by DT equal to positive k so that would be k going forward and that's k going backwards and make sure that you keep these rate constants labeled properly so really this is your gentle introduction to complex reactions where you get more than one rate constant and your concentration of a product or reactant might depend on more than one factor so we've got a forward and a backward reaction and we combine those by adding them together and you get this now the only major thing that you need to know about equilibria is what happens in equilibrium well your concentrations aren't changing so if your concentrations aren't changing we can do that we can set it to zero and that lets us do this little rearrangement here so those are equal to zero they must be equal to each other so that's the main leap of logic that you need to be able to make so that is the first introduction to kind of complex reactions as well so I've added another video on complex reactions it builds it up from this going through so do watch that one as well so this is the main leap of logic so this is also your introduction to how to do the complex ones the other final punch line is you need to know this diagram so we did a long convoluted derivation that proved this the difference between the activation energies is equal to delta H of the reaction pretty much this should be intuitive to you if you've seen this diagram you know that this value here is equilibria big K this value here these values is kinetics little K so this is the punch line of thermodynamics and equilibrium kinetics once you this is all the derivation you might need to know this is the punch line to it and that's really what you're expected to know you should be able to follow the maths but you don't necessarily to regurgitate it in an exam it's very useful if you can follow it because you might be asked to follow far more complicated stuff in the future and certainly if you get used to manipulating numbers it's definitely a massive benefit to you right so the steady-state approximation are this is the next jump into kind of complex reaction so the steady-state approximation does have a few similarities with equilibrium namely we're setting certain rates to zero and using that to get new information so what does the steady-state approximation well we have a two-step reaction a goes to be goes to see and well it's very similar in many respects to the equilibrium because well the concentration of this depends on two rate constants so we have k1 first and k2 second and that means we've got two factors controlling the rate of change of the intermediate so the steady-state approximation says that rate does not change it also says that the concentration of B over all is roughly zero but that's possibly less useful to you than knowing that the rate doesn't change very much so if this is all equal to zero then k1 8 is equal to k2 B that means we can straight up substitute that for this and get us that final equation so that is the steady-state approximation it just basically says the concentration the intermediate doesn't change so the production of that final product is basically first-order kinetics with respect to a that is the steady-state approximation there is very little to it it's one of the things that are very useful to us as an approximation it kind of does break down in reality but it's very useful for certain things that are fast so when would this apply and a goes to be goes to see well why would the concentration of this be very constant or very low well obviously if it appears and then react instantly and this reaction was very fast then we're not going to see much change in the concentration of B because as soon as it's formed it's a negligible concentration but it reacts again so if this is a low figure here and this is really the other ones really fast we can apply the steady-state approximation if it's the other way around we can't apply it if k1 was actually quite fast then B would have time this was high and this was low then B would have a chance to accumulate it wouldn't be reacting as soon as it appears so we wouldn't be able to say the disconsentration doesn't change we'd have to use a slightly more complicated approach to figuring out the rate of formation of C manipulation rate laws and equations I think this may just throw some people because you wouldn't quite know what does it mean and fortunately the short answer is it doesn't really mean one individual procedure you want to learn how to manipulate rate laws and do complicated reactions there is no one size fits all approach to this you have got to be able to use a couple of tricks and solve some problems so I'm going to go through some of them now was like examples of how you would approach it but they're not going to be completely worked out I'm going to give you the tricks for what you need to do and now from an exam perspective it's going to depend entirely what you're asked it's going to depend on what set of reaction conditions are and what you then ask to calculate from it so there should be enough hints thrown into the exams to guide you through this you're not expected to do something really crazy and convoluted and it will just say do this 15 marks you're not going to be asked to do that it will be broken up so this is all really about learning of the skill in manipulating equations what do you want to what tricks can you use so let's just have a look at this one for instance we're going to do two separate two reactions so this is similar to our two-step reaction that we applied the steady state approximation to I shouldn't say that clearly have not checked that slide over so he's a multi-step reaction we've got number one we got number two so now what information can we get well what's the rate of change of a that's starting material well that is minus k B right what is the rate of change of our final product DT well that is also a one-step reaction K-1 K2 so that is K2 C B so this is probably where we're going to start with so you need to be used to building these rails up well this one actually because a is reformed by k minus one illustration of C so now what do we want to assume well we can take various assumptions what we can assume is say DC by DT equals zero great so that's like the steady state approximation so if DC by TT that would be equal to well it's gonna be a bit more complicated so we're gonna do K1 A B that's the formation of it minus because it's been removed minus one concentration of C and then see well be disappearing again so that's K2 C so we can start manipulating that because we now know it's equal to zero and say various things for instance we can say K1 A B is equal to K minus one C minus K2 C we can take those C's out this gets a little bit convoluted K minus one minus K2 K1 A B missed something here that should be B and then we can rearrange so we can say that the concentration of C is equal to K1 A B that hand over K minus one minus K2 B this is just kind of ficking that forward this is just kind of an example of the manipulations you have to do again it exam perspective it depends on what question you asked so if I was to give you this reaction and say tell me an expression for the concentration of C this is how you would do it you would assume that it's really changed some change so you can work it out I alternatively we could start with a completely different assumption I think I've deleted the wrong scrub down the wrong thing we could use another assumption for instance we could assume that that is A and B and C they're all had equilibrium in which case the concentration of A and B and C don't change to be honest that would be less sensible because A is obviously reacting to D we could notice that C is intermediate so it doesn't change so we could also ask what's actually the rate of change of B well two B's get consumed at this point so A plus two B equals C so we'd be wanting to define the rate of reaction as minus DA by DT minus a half DB by DT equals to D plus a touch DC by DT all of this sort of thing could be asked so there like I said there is not a one-size-fits-all it's just case I'm looking at the reactions and then what's being asked of you so let's just go through another version here so this is something like a kind of a weird complex reaction A plus B goes to they combine to form AB then this hits B and then you get B2 so this actually if you look at what's happening here those sort of cancel so what's actually happening as an overall reaction is B what to be is going to be to and then AB is an intermediate areas like a catalyst so a catalyst coming together to associate with something and then dissociating this is kind of a typical reaction so we could work out some other things to do with this the rate dT is one over half plus 2 by DT okay that's the sort of information that could be asked we could also assume that DAB by DT is equal to something how would work that out you can also say that DA by DT because that is being produced and consumed at the same point that is equal to zero oh so that's a chink in the armor that we can actually go in and look at this reaction it's not manipulating numbers so let's say what what is that equal to so DA by DT that is it's removed via a second-order reaction it's K1 AB it's added back again backwards reaction and that's just dependent on the concentration of AB and it's also generated again by a K2 and that's AB B great so that's something so we can actually say that that's all equal to zero and we could even say that the rate of change at that's zero because it's a steady state approximation so we can start applying things to this we could say that the steady state approximation says that zero and that the concentration of it should be roughly equal to zero we can look at this equation and cancel out of that because that should be about zero and that zero so what we can actually see is that the rate of change of that is in fact going to be a mass K1 AB which which should be equal to zero and so on so we can say things about the concentration changes and so on anyway this is all about trial and error see what you can come up with and it depends kind of on what you've been asked so the various things can get quite complicated sure but it's kind of a little bit about creativity reading the question and learning some tricks so the main tricks you need to think about is that your intermediates look at the intermediates their rate of change should be zero we can tell you to do that and see what you can get from it catalysts their change should be okay in this case that should be equal to zero as well so this case that concentration be it doesn't change and so on and you also might want to ask what happens to all of these equations when certain rate of constants are higher than the others so we might say that this is a multi-step reaction what happens when K2 is very very much greater than K1 K-1 what happens with that what happens to the equation what happens if K2 is in fact much much smaller than K1 okay minus what happens when K1 is very very much greater than K-1 well and it's probably not reasonable to call it an equilibrium is it so you can be asked that kind of thing so this again this is the more advanced end of what you could be asked an exam or any kind of kinetics question whatsoever it requires a bit of creativity it just requires manipulating the equations around and to be honest don't panic about getting a completely right answer usually if you're given an exact typical exam question we'll actually tell you to justify a rate that just says it's k-obs of a over b or something like that and once you get to a convoluted set of rate constants multiplied by by that you're done the rest is just that is it for this revision lecture I'm sure we'll be plenty more questions about this last topic in general there's just unfortunately very few ways are approaching it is all about manipulating things to try and solve some problems so that's it for steady state thing. Maybe I could re-record this with some better slides I don't know