[Music] [Music] hello my name is Chris Harris and I'm from a Lurie chemistry and welcome to this video on AQA thermodynamics so in this video we're going to specifically look at the a QA topic of thermodynamics R naught C and this video is a it's dedicated to this example so if you are studying AQA chemistry at a level then this is perfect for you there may be some example you might see some things on some resources online that may have generic information then you might wonder in whether you know whether it's it actually is relevant for your exam board or whether you need to know it or not well this video has everything that you need to know and is designed in accordance with the specification to make sure that it has everything you need to know and nothing more and there is a full range of AQA with revision videos on a Lurie chemistry youtube channel there's also some white board tutorials on there and also some exam technique videos there's a lot of stuff on there and that will help 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like I say this is dedicated to AQA and has and meets all the specifications points that you can see in here so the main topics that were going to look at are things like born-haber cycle z' and gives free energy and entropy as well so we're going to look at quite a bit of stuff in this video so I hope you find it useful ok so let's get started so this they say this topic is particularly because thermodynamics it's got a particularly look at entropy changes and enthalpy and Gibbs free etcetera and we're also going to look at born-haber cycle x' which as you may have seen in year one are very similar to Hess's cycles unfortunately and also people don't don't like HESA cycles but they're not too bad really you just need to practice at them and know the rules really and I'll try and make it as simple as I can anything you're doing here try and simplify because chemistry's not an easy subject to study so but what we're gonna do is look at a born here this cycle which is basically just an extended cycle but the cycle how extended has a cycle but the cycle has loads of different steps involved and different processes involved so and then processes need definitions and you need to know your definitions like the back of your hand ok so you need to try and you need to try and memorize these you need to go through them and I think the best way though of learning these is actually just doing the cycles and because it fits in much more readily rather than just learning it as individual quotes and individual definitions integrate it within your cycle because you need to know your cycles as well so why not put both of them in one go and then you'll actually know it because once you get familiar with you're born here you're born here besides you'll be familiarize yourself with the steps and then you'll know exactly what the definitions will be from that so that's probably a better way of trying to remember these because there is a lot of definitions but I'm going to go through the definitions first and then once we've gone through the definitions and then going to go through the born-haber cycle and look at how we can integrate them definitions within the cycle ok so don't panic when you see a lot of definitions on here AQA like their definitions put it the hook ok here's the first one there so enthalpy change the formation which you would have seen before from year ones there's nothing new for this one so it's the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions and so an example would be for example two lots of carbon hydrogen will form your ether in molecule there so that's just enthalpy of formation very straightforward to highlight some of the key points there sirs one mole of a compound and it's got to be formed from standard states as well so solid and gas as you can see there okay so lattice enthalpy of formation so this is the enthalpy change when one mole of a solid ionic compound is formed from its gaseous ions under standard conditions so you can see here we are we have our gaseous ions here and that's going to form our solid compound which in this case is calcium chloride but you'll see these later you see all these in the born-haber cycle later another definition is the lattice enthalpy of dissociation very straightforward this is just the direct opposite of lattice entropy of formation so dissociation means to break so it's just the opposite so the enthalpy change with one mole of a solid ionic compound is dissociated into its gaseous ions under standard conditions all of these are under standard conditions so you can see a cacl2 source calcium chloride will form calcium 2 plus and 2 CL minus enthalpy change of dissociation so the enthalpy change when one mole of the bonds of the same type of a molecule in the gaseous state is broken so you can see here that we've got f2 gas go into 2f gas so this is called the entropy change of dissociation in case you'll see that later as well and the enthalpy change of first ionization so now there's the first ionization energy you would have seen this from year 1 chemistry as well so this shouldn't be new so the enthalpy change when one mole of gaseous 1 plus ions are made from one mole of gaseous atoms so we've gone from sodium gas to sodium plus gas so we're moving an electron remember okay enthalpy change of atomization so this is where the enthalpy change of one mole of gaseous atoms is made from an element in its standard state okay so for example we're going from 1/2 F to 2 F is an example of atomization okay so ionization second ionization energy so the enthalpy change when one mole of gaseous two-plus ions are made from one mole of gaseous one plus ions so you've got to remember that one that's basically just going from you've already ionized your atom you just taken another electron from it typically found as you can see there in Group two elements because they have two electrons that they have to shell to valent electrons and so therefore we can remove both of them to form a two plus I'm okay and first electron affinity so this is just the opposite of ionization so it is the enthalpy change when one mole of gaseous one minus ions are made from one mole of gaseous atoms so you can see here what we're doing is we're going from a gas this is an oxygen gas - OH - so you can see all we're doing is adding an electron to our oxygen and so therefore that it's electron affinity okay and then we've also got the second electron affinity as well so this is the entropy change when one mole of gaseous - minus ions are formed from one mole of gaseous one minus ion so effectively all that means is we already have a negatively charged ion and we're going to add another electron to something that is negatively charged already so that will form a 2 minus ion okay so all these definitions you're gonna see you're gonna see in the cycle so don't worry too much about the definitions learning them as of this screen like I say the best way to learn them is by just doing the cycles and doing loads and loads of cycles and then put your definitions in there and then you see it a lot more easily I think we're born here but it's it's just trying to make sure you know the process and obviously the definitions alongside that so let's have a look at the born-haber cycle in so born hero cycles are useful to calculate lattice enthalpies so this is because we can't calculate them directly from experiments and so here it is structures so let me start with a born-haber cycle what we do is we start with the bottom line and the bottom line is your lithium chloride so here we've got licl solid so in other words we're looking for the born-haber cycle for the production of lithium chloride which is a solid ionic compound and the line immediately above it onto the left-hand side slightly to the left a soft Center is going to be elements in the standard States so this is lithium solid and chlorine gas so these are the elements required to form lithium chloride so of course this is your enthalpy of formation but we'll look at the definitions in a moment okay so they go so entropy of formation so we'll go through the rest of them as well on the cycle so that's the first step so the best way to look at born here besides is a bit like HESA cycles when you've looked to HESA cycles is to have different routes so there's two ways in which you can form lithium chloride okay one way is doing that way which you would see on the screen there so that is the enthalpy of formation so that's just taken elements in their standard states and forming your solid ionic lattice which is lithium chloride so that's pretty straightforward alternative but we can do instead of going directly to form lithium chloride via a formation process what we can do is we can break it down and we can actually form ions in the gaseous state and then form lithium chloride okay so there's two ways either elements in the standard States we can form it there or we can form it from ions in the gaseous state now to form ions in the gaseous state we need to do a few steps before we actually get to the ions and the cases then you'll see this icon pointing the cursor around here because there you'll see the cycle as it's formed from that so there's two ways so it's a little bit like I mean I live I live north of Newcastle at sound called morpeth and there's two ways in which I can get to Newcastle I can either go down the year one or I can go down the year nineteen now the ear one is the more direct route it's quicker to get Newcastle via the a1 then it is via the 19 and bots if the ear one is blocked I'm gonna have to go down the Year 19 because I've got like a good hour other little country roots but and the ear 19 is the major it's where I have to go down that way now it's the same with these cycles we're trying to get to form one particular product we're trying to form the same products but there's different ways in which you can do it so all born here but does is just show you the different ways in which you can form lithium chloride one form elements in the standard state and one from ions in the gaseous state so let's have a look going upward so let's look at the alternative route instead of going directly look at how we get there so the next thing we need to do is form lithium lithium solid and chlorine gas so what we're going to try and form is ions in the gaseous state but we need to get it into that state first so the first thing is the enthalpy of atomization of chlorine so we're going from 1/2 cl2 gas to CL gas so remember we're breaking the bond between chlorine the two chlorine atoms and forming a single chlorine acid we only need one because we're formerly employed okay so the next step is once we've done that we then need to we then need to convert lithium solid into lithium gas and so we need to authorize this and you notice that all these are endothermic so we need to put energy in to do these types of reactions so to break the bond to convert from a solid to a gas we need to actually we need to actually convert them as well and these have got to be endothermic there's an ice cream van ago and by lovely and it's not even that nice either right okay so and the first ionization sorry about that the window is shutters also right so the next step is the first ionization energy of lithium so what we need to do is remove an electron to form lithium plus okay so because this is ionization this is the removal of an electron so because we're moving electron we need to put energy in so that's an endothermic process so first ionization energy the removal of an electron and so we formed our lithium ion as a gas so we've got the lithium in the right state so remember there's another way of forming ions from the gaseous state and then chlorine is remained as a gas so chlorine atom is the gas and then it's plus an electron so all these are the theoretical steps of course subsume in reality you know we don't hold these separately you know for a long period of time this is just the the methodical process the energy steps required okay so we just need to get chlorine to be an iron now there to be because it's already in the gaseous state so we need to get it into an iron and then we can then form our lithium chloride so here it is this is obviously the electron affinity so we are adding an electron to chlorine and that is an exothermic yes because obviously adding an electron to write to complete a full shell because member chlorine has seven electrons valence electrons so gaining one electron will give it a full shell that's quite good so it's more stable so it will give out energy it's not going to put up any fuss at all so this is an exothermic process okay so you can see here we've now got to the state where we've got our ions in the gaseous state so now we can then form lithium chloride and so this is the lattice enthalpy of formation so we're forming our lithium chloride from ions in the gaseous state so you've got two methods we either go by the lattice enthalpy of formation so we're going from the elements in their standard States forming it or we can go via all these steps here to form lithium chloride as well so there's multiple steps here and the key thing is about putting figures in there and trying to work it out okay so like I say it's a bit like has a cycle so the key thing if you remember from HESA cycle is that the enthalpy change of a reaction is the same no matter what route is taken so it's independent of the route taken so in other words the amount of energy required to go via this step is the same as the amount of energy required to go via these steps here we're still ending up in the same place so it's a bit like like I say it's like getting to Newcastle I can either go down the a1 which is the direct route or if that's closed then I'll have to go in directly down the air 19 and then back into Newcastle again okay so let's have a look at the born-haber cycle but put some let's put some values into this okay so we can calculate lattice enthalpies okay that was the last section of it seemed just the last hour on that diagram by using the cycle in the same way as has a cycle so what we mean by that is if we go with the arrow we keep the sign the same if we go against the arrow we change the sign now what you've got to be familiar with is what endothermic and what's exothermic on them born here besides and I'll put another one up on here in a minute but as you go further up that cycle the more energy it is okay so all of our oils are endothermic down arrows are exothermic remember exothermic is a negative enthalpy change endothermic is a positive and and the enthalpy change okay so let's have a look so here we go we've brought our born here beside back so this is our cycle that we had from the last slide so it's exactly the same except it's all pre completed so everything's done for switches which is brilliant okay difference is what we've got now is a table and this table contains information so we've got the lattice and look at the enthalpy of formation of lithium chloride minus 409 you've got the first ionization of lithium okay which is five one nine where the optimisations of lithium and chlorine and we've got the first electron affinity of chlorine which is down here okay so got all the figures in there so now what we're going to do is put the figures in and see what we've got so we need to work out remember the lastest enthalpy so it's this one here this is what we're trying to work out so what we're going to do is put all our figures in that we that we know into the cycle to work out this bit here okay so the first one maximization of chlorine so we do 242 divided by two now the reason why we divide this by two is because the data we've been given here which is two for two which is this one here is the enthalpy of atomization of chlorine that's of CL two but obviously we're only doing that for 1/2 CL 2 so we've got to divide it by 2 here so just read the data really carefully what you've been given and then once you've been given that data then work out if you need to half it or not and you basically look at what you've written and your cycle here so this is half CL 2 so we need to divide by 2 so I'm not doing it for the 4 1 mole of chlorine we're doing it for half a mole okay so the other one is converting lithium solid to lift the gas so this is this is 161 as you can see on there and and then we've got lithium and chlorine lithium and chlorine forming the will form and lithium sorry forming a anion so we're removing an electron this is an endothermic process so you can see it's 5 4 9 okay so first ionization is 5 1 9 so you notice all the exothermic SAR negative or your endothermic sar positive ok then we're going to look at the next one which is electron affinity which is mine three six four so we've got that so only adding one electron so that's that's fine and the bit that we need to work out is our lattice enthalpy of formation of lithium chloride okay so we're going to use this to calculate it and this is exactly the same as has two cycles so if you're if you're comfortable with has a cycle then you should be fine with this if you're not don't worry I don't think many people are and what we're going to do is then we're going to go through this step by step and show you how it's work out so it's a good reminder of how to use these cycles so remember we need wherever you need to work out at there's loads of ways to do this but I like to do it this way so and I find it quite methodical they've seen me other videos as well you'll see that I like to do things that are in a methodical way and break it down so obviously for the purposes so you can understand but also I just find it a bit easier if you've done something wrong you can go back and check and make sure it's right so we need to work out this bit okay now imagine this is a road let's say I'm gone from morpeth to Newcastle okay so but this road is brought that's slightly a 1 so close that road so I need to find an alternative route us they want to get a Newcastle but I need to find alternative route same with has a cycle so I want to find this this road is blocked so I want to get from here to here so we're going to push start and an end this is where we're going from we're going to start from here and go here okay this roads blocked so we need to go an alternative route we need to go backwards up here okay anything you want to work out just say this road is blocked and you just go the opposite way around the cycle okay so I'm going to start from here remember that rule go against the arrow or change the sign go with the arrow we'll keep it the same so we're going against the arrow here so we've got the start here we're gonna have to go up and then down here to get to our final product so this is minus 364 so we're going to go up that's plus three six volt we change the sign we're then gonna go down here against the arrow so it's going to be minus five one nine and then we're going against the arrow again that's plot minus one six one and we're going to go against the arrow again that's minus one to one and then with the our here so we keep that the same that's minus 409 and you literally just put that in the calculator it's straight forward when you think of it and to put all that calculator and we get our lattice enthalpy value of minus eight four six kilojoules per mole now what I would do there's two ways to check this first thing is this is the lattice enthalpy of formation of lithium chloride so you can see here that we've got their irons in the gaseous state and we're going to form lithium chloride solid this is clearly an exothermic reaction because we're bond forming we're going from ions to a solid compound so check to make sure that your figure is showing as an exothermic value in other words it should be negative okay if it isn't something's gone wrong so go back and just check this as well mean if you're methodical you'll be able to spot your mistake and also another way just like Hester cycle is that if you took all these start anywhere on the cycle whatsoever anywhere so we start from here and start anywhere and go around the cycle back to where you started from okay the total value should always add up to zero go remember these hours that the rules so go with the arrow keep the sign keep the sign the same against the arrow and we change the sign so try it put all your figures in there with the arrow keep the sign the same if we go against the arrow you change the sign okay so then once you've once you've worked that out and if you get an answer of zero and then you've probably got that right more than likely and it's a really good way of checking because it gives you a little bit of confidence in the exam and the least got that that question correct okay so a big fan of checking things because you know there's got to be a reason or purpose for doing something rather than just doing it for the sake of it okay so we can look we can calculate other parts of the cycle as well obviously we've just calculated they're the lattice enthalpy of formation and but we can actually calculate any part of the cycle so let's have a look so here we've got a different cycle here I've gots the formation of sodium chloride same type of steps different type of data okay so we've got all the all the information here okay so what we're going to try and work out is what data is missing so what I would do is put all the information in all the data that you've got just put it in the cycle so let's do that now so the first one and to be a formation - form on optimization of chlorine we need to calculate this which is the atomization of sodium we need to do the ionization as well this is the ionization so we've been given that data so it's the ionization of sodium we've been given the electron affinity of chlorine I don't see we've got our lattice enthalpy as well - seven eight seven so have a think just from that previous cycle that we've looked at we need to try and find a start and an end point so remember this is the bit we want to calculate want to calculate that are all there so remember we need to start from here and end here okay because that's the direction we want to go in but when we starting from there and end in there there we go let's put it there start and and then remember roadblocked okay so this is like the a1 it's blocked so I need to go down here 19 so I'm starting here so I need to go it's blocked so I need to take the slip Lord off on the year 19 and then swing back round into Newcastle so I'm starting here so I'm gonna go down down up up and then down and then then I'll get that figure there and that will tell us this value so let's go so against - 1 - 1 because we're going against the arrow we're going with the hour here so it keeps set up - for 1 1 we're going against the arrow here plus 7 8 7 we're going against the arrow here plus 3 6 4 and by the way before go any further and what do you think the answer should be do you think it should be positive or do you think it should be negative so have a think before a reveal the answer ok so the next one is minus 496 okay and so we put all that in our calculator and we get plus 1 2 3 so if you said positive you've got it right the reason why it's positive is because it's an endothermic reaction we're going up the our was going up so there's an endothermic reaction and it's a positive value if you got a negative value something hasn't gone right and then again check it again put all them numbers every single one start any way you want in the cycle and for all the numbers in until you come back to where you were before and you should get 0 it should always add up to 0 if it's not something's gone wrong ok so just check right so we're going to look at an extended born here the cycle this looks horrendous than it but it's not it's exactly the same type of stuff so the only difference is we've got an extra ionization step because we're forming that should say magnesium oxide not lithium chloride I'll get that changed so this should say magnesium oxide which is down here so we have you can see we've got an mg 2 plus and OH two - so we're going we're added an extra step here so mg plus 2 for mg 2 plus and we've got this funny step here as well going on but we'll have a they'll have a look at that in a moment so you can see we've got the second ionization step because we need to form magnesium two-plus but you see we've got a second ionization a second electron affinity here now the first electron affinity is exothermic because we're adding an electron to a nectar to a neutral atom and were forming a negatively charged ion however when we try and add another electron to oxygen to that it's an endothermic process and the reason why is because we're actually adding an electron to something a negatively charged electron to an ion that's already negatively charged now there that just isn't going to happen you know we're trying to have a negative a negative reactant with a negative I'm trying to add a negative on there that's going to repel now that means you need to put a bit of energy in to put that electron in there so that's why it's an endothermic process so let's get that confirmation there so you've got your second electron affinity step okay it's an endothermic process I'd like to say the reason why is because we've got repulsive forces between a negative electron and a negative ion so because they repel each other normally you need to put a bit of energy in to actually get this reaction to go okay and like again just to confirm that this bottom one should be mg or not licl so that's just a typo but I'll get all that changed anyway if you are if you are going to get the slides I'll get that changed okay so let's look at theoretical and experimental lattice enthalpies okay so the theoretical and experimental values of lattice enthalpies can be different depending on how purely ionic the pound is okay and that sums a bit stranger thing well hang on I was told that everything is either covalent ionic or metallic and well yes that is true you normally when you're describing that you're describing and what its main attributes are so for example and you would normally clasp a metal and nonmetal as an ionic compound so which is true that's fine but just a little bit of a curveball is that some ionic compounds actually have a little bit of covalent character in there as well and so what we're going to do is identify something as purely ionic or does it have a little bit of covalent character and there's mainly ionic but it might have some covalency and some covalent characters in there as well characteristics so let's have a look so theoretical lattice enthalpies can be calculated from data assuming a perfectly ionic model sounds intriguing so a perfectly ionic model is just the one where you have ions that are perfectly spherical okay so there's no distortion to them at all and the charge is evenly distributed so we call them plink charges in other words the charge is equally distributed amongst them so Spears so that is a perfect ionic model and we know that because we know that they're my M Spears will then be attracted to each other so electrostatic attractions between them oppositely charged ions okay it's common to carry out an experiment to work out the lattice enthalpy only to find the number is different to the theoretical value so you do a practical and you get a value and then you check it in the data book and it comes out different and anything I've done something wrong you haven't this is the reason why so this tells us that the compound they're being experimented on doesn't follow a perfectly ionic model and actually has some covalent characteristics DQ isn't it right so most of the time the positive ion distorts and the charge distribution on the negative ion as you can see here and we can say that the positive ion polarizes the negative ion so instead of having a perfectly spherical ion we actually have one that's slightly distorted and some of the electrons are being lured over to that positive iron there and so the more polarization we get the more covalent character there will be so it's it's that I'm distortion of the negative iron there that's towards that positive island that is actually creating some kind of covalent characters remember converting characters about sharing electrons so these are not just obviously are obviously charged ions just just being attracted to each other and this is also some distortion in their minds as well so acting a little bit like a covalent bond okay so the latter sent to be values they can tell us how much a substance is purely ionic okay so let's have a look so we see we've got our table here okay and we've got a table of ionic compounds here's who got all of our net compounds lithium compounds with different halogens and we've got magnesium compounds with different halogens as well bonded to it so in all examples the experimental value shows a higher lattice enthalpy than the purely ionic theoretical value okay so we can get a slightly higher value experimentally so this shows that we actually have some form of covalent character in their molecules so and this is obviously caused by them larger distortions by the by the negative ion like we've seen in the previous the previous slide so this can be seen clearly and between a1 plus which is lithium and a2 plus sign and which is magnesium and all we have to do is just compare the percentage difference in lattice enthalpy so you can see here this is the this is the difference the deviation in between the two so you can see experimentally for lithium chloride it's minus 8/5 - but in theory it should be a two four five so it's a little bit higher experimentally so that's a deviation of not 0.8 but you can see the more the more polarized the bigger the charge and obviously the larger the the larger the the eye on the bigger the distortion here that's nineteen point seven and this is basically what we're going to look at here is a summary so basically the lithium chloride this one here is the most perfectly ionic model so your data should closely match the theoretical data and but the bigger the difference in lattice enthalpy okay and the more polarization you have and the greater the covalent character so you can see here with magnesium iodide it's a much bigger positive charge so that's going to distort that negative ion which is quite large for iodide a lot more and so therefore we've got a lot more covalent character in that type of compound and so therefore the difference in lattice enthalpy is going to be much greater 19.7% difference that's quite a big difference so effectively you could say it's like a percentage of covalent see how covalent it is it's still ionic overall of course but just those covalent character so make sure you aware of that because there will be asking for data and that will be asking you to comment on what's theoretical and experimental values in the difference between them okay so we're going to look at something called an enthalpy change of solution so what we've looked at so far are ionic compounds being formed from gaseous ions and they've in there from gaseous ions or from elements in their standard states so it looks a lot about how we make these ionic compounds and the theoretical difference between each ionic compound because they're covalent character now we're going to look at okay we take these ionic compounds and what about their energies when we throw them into water okay and they dissolve and we call this enthalpy change of solutions so we need to know a little bit about the behavior about why the salts dissolve and what energy do they release or take in when they do that so this is where we're going to discuss around these points here so we're going to look at another type of enthalpy change which is enthalpy change of solution and this is the enthalpy change when one mole of an ionic substance is dissolved in the minimum amount of solvent to ensure no further enthalpy change is observed upon further dilution Wow so all that means is we're going to take a solid right we got your anak compound we're gonna add that to a beaker and we're gonna add small amount of water okay we're going to keep adding water until we've just dissolved our ionic compound just about - in other words we've got a very saturated solution remaining leftover okay so this is what we're doing and we're gonna measure the energy change when we do that simple as that okay so so remember for something to dissolve we need two things okay so the first thing is the substance bonds must break so that ionic you must break that up first that's the first step that's an endothermic process because we need heat energy to actually break the ionic compound apart and the second thing is new bonds have gotta be formed between the solvent and the substance so the solvent and surrounds the individual ions that have been broken up and that's an exothermic process because we've got interactions happening between the water or the solvent whatever the solvent is and the and the ions as well so let's have a look here so this is our ionic lattice in solid form unbroken fine okay and we've got a water molecule here which is separate so Wolffe imagine we've got salt in a container and water in a beaker in the case you'll get them separate so then what happens when we add that salt to water what happens the first thing is the ions are broken up okay so that needs energy energies need to this an endothermic process so substance is broken into three moving ions and then what happens is the water then jumps in the opportunity to surround these ions and keep them apart so the water so the bonds formed between the ions and water and the ions are hydrated so we see they're hydrated ions so you can see here that we've got the negative Delta negative oxygens on the water orient it themselves the water molecule orientates to point towards the positive ion and obviously Delta positive hydrogens then point towards the negative ion there so this is effectively dissolved so most ionic compounds like I say they dissolve in polar solvents okay because you've got that polarity there so the Delta positive hydrogen is attract to the negative ions and Delta negative oxygen stripes the positive ions and the structure starts to break down and so then the water molecules surround the ions and this is obviously what we call hydration and so for this to happen this is very important because not all annex substances are soluble most are but some are not so for this to happen the new bonds that are formed between and so the new bonds formed here in this process must be the same strength or greater than those required to break it into the ions in the first place if it's not then the substance very unlikely to dissolve and so soluble substances tend to have exothermic enthalpies for this reason because remember this process here is endothermic this one's exothermic so if this is greater than the energy required to break it in the first place then obviously then your overall gonna get an exothermic reaction so that's why when you dissolve any other compound and water generally warms up okay so what we can do is the enthalpy change of solution can be calculated and we can note we can do this by knowing the lattice dissociation enthalpy and the enthalpy of hydration which is what we've just looked at there so we've looked at the end to be of hydration we looked at how we can actually how it can actually do that so what we can do is we can use a cycle so this is very similar to a Hesse a cycle it's just a special type of cycle dedicated to and to be changes in solution okay so it's not not the same ones that you would have seen in any year one chemistry when you've looked at at the enthalpy of combustion for example your combustion cycles and your formation cycles so this one is looking at hydration but we treat it in the same way okay so we use the cycle to help work out so remember how its set out you've got to remember this so you can see we've got our solid here lifting employed and enthalpy of solution is turning that solid its aqueous ions okay so we got lithium plus and chloride minus ions now we can go and via this method here so we can say right we're going to take our ions remember this here is going from a solid to ions aqueous iron so this is this is your hydrated ions but what we have to do remember from the previous slide is we have to take these ions we have to break them up into their separate ions first these are gaseous ions and then what we do is with until engage the signs and form our a crease ions which is the hydrated ones and I put the little diagrams on there just so you can see what each part of the cycle actually represents so remember the enthalpy of hydration here is the enthalpy change of one mole of a co-signs is made from one mole of gaseous ions so you've got to remember that okay so what we've got to assume is that we do the following so we break the solid lattice up into its gaseous ions first we call this lattice dissociation so remember that and then we dissolve these gaseous ions in water and we call that an enthalpy of hydration okay so this is the assumption we make when we're dissolving something in solution we break it up first then we hydrate it okay that's the step that we're taking so remember well if we go with the arrow we keep the sign the same and if we go against the our what we change the sign and you will be given data in the data book to calculate that so what we're going to do here is we're going to put some data in that we've been given and we're going to put that into our cycle and we're going to work out the enthalpy of solution so the first one here so enthalpy of solution so remember this is exactly the same as a bond here but has a cycle so we want to work out this bit here so imagine that rods blocked so we're starting from here we want to work out when it gets here so we need to go with the arrow here so that's minus a plus 846 sorry it's an endothermic process we put energy in and this is going to be exothermic and actually hydrating them so let's have a look and there it is okay so exothermic for lithium exothermic for chloride ions and that's going to give us a total of minus eight eight three so then if we put that in the calculator we should get an enthalpy of solution of minus 37 kilojoules per mole so remember when we said that most most of the time for something to dissolve and this bit here so the formation or the interaction between water in the ions these interactions must be stronger or the same as and the amount of energy then or higher sorry are the same as then the amount of energy required to break this up so generally we get exothermic reactions it's not massively exothermic but it's still exothermic there's still a negative negative value okay so just treat that exactly the same as a Hess's cycle as you can see it's just a very special cycle dedicated to the enthalpy change of solution or when you dissolve a nine a compound and water okay so we're going to look at another pot we're going to move away from the enthalpy side a little bit okay we're going to introduce a new type of energy you know just when you thought you've got your head round enthalpy we're gonna bring in another one called entropy okay entropy is a really strange type of energy it's it's the entropy of order yes that's right and we're going to talk about chaos we're gonna talk about all that and disorder I'm gonna be using quite kind of strange language like that in chemistry so but it is and it's a measure of disorder but you'll see it kind of makes sense when we go through it so entropy is the number of ways energy can be shared out between particles okay so we can have disorder and order like I say but the more disorder there is the higher the level of entropy okay so you can see and when we're looking at disorder we're looking at randomness so how random things are so if you just imagine it like tickets that sales like your bedroom and it's easier to allow your bedroom to get into a state of disorder in other words to allow it to get messy than it is to keep it ordered so when you think about it it's easier to keep it in that way but if you had to tidy it up it'll take more effort to tidy it up than it does to make a mess doesn't it and generally everything in the world tends towards more disorder okay that's just how things happen everything becomes more disordered because it's the lowest energy so in order to correct things or put things back into some kind of disorder back into this back into order sorry we have to put energy in to do that because everything and tends towards a disorderly fashion because it's lower in energy okay so if you leave your bedroom and you leave it for a month and not touch it eventually it's you know and not not you know don't be bothered about putting things away or putting it back into any kind of order it's easier to do that isn't it and then you think well I have to get this tidy doctors it's getting a bit of a mess you know you can have to put energy in so it's the same with atoms and if you've seen any other videos as well I always say the atoms particles molecules protons neutrons blood body bar these type of substances are incredibly lazy and they want to be in a state where it's the lowest energy possible that's just as a general rule and there's no difference here so let's have a look at these different models here and yes I have put up the diagrams of a solid liquid and a gas it might have been something you might have seen in year 7 but it's it's very useful to explaining entropy so solids have the lowest level of disorder okay the particles arranged neatly in rows and they vibrate on the spot if you can remember that and then liquids are a little bit more disordered as you can see the particles we've got a little bit more space between them just a tiny bit there's still quite densely packed though and gases i've got the greatest amount of disorder there's loads of space between the particles there so we've got increasing disorder and entropy okay so the number of particles it's not just the arrangement of particles it's the number of particles as well also affect central change so if a reaction is in the same state and but more moles are produced and entropy also increases and this is because there's more ways in which energy can be distributed so if we have a look at this example here we've got dinitrogen tetroxide producing nitrogen dioxide shows increasing entropy so we've got more moles of gas are produced so you can see here we've got one mole of n2o4 on the left here and we've got two moles on the right and tropically that's a favorable process because we've gone from one mole to two moles there's more ways in which you can arrange two moles of no2 than there is obviously of one mole of n2o4 so that's another way in which we can increase entropy okay so a reaction can be spontaneous or feasible if it's install quickly unfavorable okay which is endothermic no math seems a little bit weird to say there isn't many you remember enthalpically favorable reactions are exothermic ones because the products of a lower energy than the reactants endothermic reactions are rare and because because the products have in a higher energy State but just because enthalpy okay it's not favorable doesn't mean the reaction won't go and the main reason why that is for example a classic examples for synthesis is an endothermic process and but the reason why this is endothermic reason why endothermic reactions can go feasibly is because entropically it could be favorable so let's have a look so a reaction will tend towards more disorder and hence increase entropy soaps that remember us what we said before okay so increase in entropy is energetically favorable okay that's a good thing the more disorder you have the more favorable that reaction is going to be and some reactions that are enthalpically unfavorable such as endothermic ones can still be spontaneously and could still spontaneously react if changes in entropy overcome the changes in enthalpy so in other words if our entropic forces are greater than our enthalpic forces then the reaction can still go ahead because entropy is effectively overriding the enthalpy values so let's look at an example here so we've got this is of an endothermic reaction I love this reaction you might have seen this one so this is a reaction between hydrated barium hydroxide and ammonium chloride and it shows how enthalpy changes are over and this shows how entropy changes overcome you enthalpy changes in this reaction so this reaction is a pretty cool one literally it's a cool reaction you have you might have seen this if you if you go to school or college or if you're working independently but if you go to school of college you might have seen it where you have a wooden block and you have a beaker and you mix these two solids spurn remind these solids you put two solids in the beacon you mix them up and what you do is you put blob of water between the beaker and the block of wood so you put it in between and you sit there because on top of the wood and then you mix these two solids together and the temperature drop is incredible you get approximately a tenth at 20 degrees Celsius drop in temperature and this is enough to freeze the water between the beaker and the block and effectively you can lift the beaker up and the block is actually stuck to the beaker you get this strong smell of ammonia coming from it because it does produce ammonia and so it's a really cold reaction but it's it's just weird because you mix them two solids no liquid at all just mix two solids together and you get this reaction so go and have a look if you if you haven't seen it go and have a look online sure that'd be so many would have done it and you'll see the the temperature change is really erratic so enthalpically that wouldn't go it's an endothermic reaction our products higher in energy than our reactants okay so normally that wouldn't go but and we can see here here's our overall reaction the dot it hitched oil just says it's just the water of crystallisation so it's just we've got eight waters surrounding our barium hydroxide so the summary points what we can deduce from this is that this reaction is very endothermic so in farbe eclis it's not favorable so we've said that already and it has a value of plus one six four kilojoules per mole so definitely endothermic the three moles on the left-hand side of the reaction you see here we've got three moles on the left and we have thirteen moles on the right so definitely entropically favorable who's got more moles on the right than we do on the left so there's more ways in which you can rearrange them atoms and then the other thing as well as we're starting with two solids but we're making a gas and a liquid as product and so we've got more disorder as a consequence because remember gases and liquids are more disordered than solids so there's another factor that actually entropically this reaction is is a favorable is a favorable reaction so you can see here and the difference between something which is blatantly obviously endothermic and not feasible actually does happen it do you don't need anything you just mix these two powders together and just give them a stir and they go so but this is why cuz entropy is actually allowing this to happen okay so entropy change Delta s yes they give it an s a value of s I'm sure somebody will be able to tell me why it's given the letter S and but you get used to this in chemistry they give letters for things that actually doesn't really have much much logical sense to call it but proving out letters on margin but anyway entropy change Delta s can be calculated between reactants and products okay so Delta s your entropy change is the entropy of products minus reactants and we always measure entropy in joules per Kelvin per mole okay very different to enthalpy which is kilojoules per mole okay so this is joules per Kelvin per mole so entropy values are given as a standard entropy so all of these are under standard conditions this allows us to compare figures and results and conditions are we must use one mole of substance when were measuring it must be a 100 kilo Pascal's of pressure and a room temperature as a temperature of 298 Kelvin so these are standard there's nothing different here and for entropy there's just standards conditions which you would apply for anything in chemistry ok so let's have a look at an example calculation so the reaction between M hydrated barium hydroxide and ammonium chloride so let's calculate the change in entropy so we can see here there's the reaction that we've got already but we're going to look at the entropic changes now you'll be given data to do with entropy so you can see here we've got our entropic values or entropy values for barium hydroxide ammonium chloride these are the two reactants and there will still produce an ammonia water and barium chloride as a product so we've got all our entropy values there now all we have to do is work out the total entropy of the products so we literally put all the figures in here so you've got to lots of ammonia we've got 10 lots of water and we've got a molecule of barium chloride and so we add all that up and get 1208 joules per Kelvin per mole and then we do exactly the same for our reactants so we've got one molecule of barium hydroxide which is M which is there so it includes all I've C the water as well and then we've got two molecules of ammonium chloride and then we form we get a value of 617 now we put all that into our total entropy calculation and we get plus five nine one joules per Kelvin per mole now that is a positive value so that means it is entropically favorable so anything that's positive for entropy is a good sign it's phase it's a feasible reaction a favorable reaction okay with enthalpy remember a negative value is favorable and this will become important when we look at hypothetical situations later on so at a negative enthalpy value is favourable a positive entropy value is favorable okay so what we're going to do now is we're going to link entropy and enthalpy together and luckily there's an equation that allows us to take into account both factors because both have a big seg in whether in terms whether a reaction is feasible or not and this is called Gibbs free energy so Gibbs free energy Delta G at last we've used a proper letter for this the Delta G tells us if a reaction is feasible or not so Delta G equals Delta H minus T Delta s you need to know this equation you're going to use it quite a bit in this topic so Delta H is enthalpy so that's a change in joules per mole if you're given else being kilojoules per mole you need to convert to joules per mole so watch them units T is temperature in Kelvin Delta s is entropy change that's joules per Kelvin per mole and so the units of Delta G or Gibbs free energy is joules per mole see and see why we need to convert our enthalpy from kilojoules per mole to joules per mole okay so the basic rule here a reaction is feasible in theory if Delta G is negative or zero okay so Delta G needs to be negative or zero for a reaction to be feasible what we've got to be careful though is that even if a reaction is calculated to be feasible sorry you may not observe a reaction occurring and this is due to the activation energy probably been too high or the rate of reaction is actually been other rate reaction is slow so a classic example of that is rust so if you leave a and I in nail out m in just in the atmosphere so you need water and oxygen to allow the iron to turn to iron oxide which is rust and that is definitely a feasible reaction because we see it everywhere you know that's a natural process however we don't just leave a nail I won't just see it react instantly it's a very slow reaction to the low rate of reactions it's slow so calculating the free energy Delta G for the reaction between and barium hydroxide and ammonium plied at 298 Kelvin so let's have a look and see how we can work this out so we know the following from the previous slide so we know that M entropy is plus 591 joules per Kelvin per mole and we know that enthalpy is plus 164 kilojoules per mole or remember we'll convert that into joules per mole and is 164 thousand joules per mole so then we put it into this equation so Delta G so Delta H is one hundred and sixty-four thousand minus 298 because that's our temperature remember there it is there two 9s and then five nine one is our entropy value because that's what works out just before so then that's going to give us a delta g of minus twelve thousand one hundred eighteen joules per mole so this is a negative value Delta G's negative here so that means that this reaction is feasible at this temperature okay so anything it's negative or zero is feasible 0 is just feasible so if the Dubs t is zero means it is just feasible at that point okay so temperature may have an effect on reaction feasibility okay that's very important that seems quite obvious actually because if you sometimes you need to heat things up to actually get them to react so that's quite an obvious point so what we're gonna do here is we're going to look at different scenarios when Delta H and Delta s is changed has changed and we're going to look at the Delta G value as well which has changed so let's have a look so we're going to remember this basic room very important that for a reaction to be feasible Delta G must be negative or must be zero okay anything above that and it's not feasible so what we've got is we've got a range of different scenarios in this table and we're going to look at each one of them and this is a theoretical square I've got any numbers here no figures we're going to look at use numbers as an example but we've got no figures here now the exam board will expect you to be able to comment on reaction feasibility even if you don't have any numbers okay to actually calculate it and this is the way in which we're going to do it so we can see Delta G Delta H minus T Delta s that's the reaction remember the easiest way to actually work these theoretical scenarios out is by splitting that equation into two so we take the Delta H part which is this bit here and we take the T Delta s part which is this bit here so we've got an enthalpy part and we've got an entropy part which is here so if we split that down the middle and effectively what we're trying to do is imagine and we've got a C source and we've got this minus this so if we subtract by a larger number than here then basic maths will tell you you know if you have C if you have 2 minus 2 let's see if these were equal then that would be 0 okay but let's see if you had 2 minus 3 then that's going to be minus 1 that's going to be a feasible reaction okay but let's say if you had let's say you had 3 minus 2 so the Delta H bit was bigger so that's 3 minus 2 then that's going to be obviously that's going to be a positive values can be positive 1 so split it up into halves okay and you'll see what I mean so let's have a look at the first one there that's it we're going around there at the top so look at the first row so even if T was at its lowest so if we stress this to 0 Kelvin Delta G would always would always be negative um so that should say negative B rather die ok so it should always be negative so exothermic reactions with a positive entropy will always be feasible whatever the temperature okay so there than nice reactions because they will go at whatever temperature you put them in so Delta H is negative Delta s is positive always negative Delta G so yes it's feasible at any temperature so let's look at that second well look at the next one so even if T was at its lowest at 0 Kelvin so if we put these into the into the data there Delta G would always be positive always okay so endothermic reactions with a negative entropy would never be feasible at whatever temperature so they would never actually react ok so just think remember look at it see it has two parts even if Delta G even if this bit was 0 okay there was no value there if Delta s here was bigger than Delta here it's that's always sorry if Delta F sorry was was less than this value here which always will be because it's negative in this scenario then Delta G will always be positive ok so just have a have a look at it okay so here's another one so let's look at the third scenario here so if we look at the third row so freezing water is an exothermic process okay so it gives out heat energy when we when we freeze water and but we're going to a more ordered arrangement so we're going from a liquid to a solid okay so that's not good entropically that's not a good process so freezing only happens below a certain temperature we know that could we see that all the time so we know that obviously if we have water and we freeze that of so we've got the liquid water with to freeze that it needs to pee at zero degrees or below doesn't it that's obvious this is all to do with enthalpy and entropy so this is where it becomes really kind of real almost so let's prove it okay so let's put some numbers in just to show it as an example okay so Delta H is minus let's see it in this example here there's nothing to do with the water or freezing of water but let's imagine Delta H is minus 100,000 and Delta s is minus 100 joules per mole okay so let's say we've got both of them are negative okay so both got negative values so let's put them in let's do it at 500 Kelvin so we do it at 500 Kelvin put the numbers into the into that equation and we get Delta G is minus 50,000 to say certain low temperature but at a higher temperature if we put a higher temperature in here then you can see that the Delta G actually becomes positive and it's not feasible so there is a point in between here when the reaction just becomes feasible so reactions like this we have a negative and negative it's only negative at lower temperatures it's feasible like this and at higher temperatures it isn't okay so let's look at the last one okay so this last one here so let's look at the fourth row the decomposition of sodium hydrogen carbonate is an endothermic process okay so that's not an thought okay that's not a favorable process but we are going to a more random arrangement because we're going from a solid to a gas so this is your solid here your bicarbonate which is in here and we when we make an cinder toffee or Crunchy's for example and what they do is they exactly I'd love to know dramatic crunchy but and they use a baking powder based of the user sodium bicarbonate here and what they do is eight when it's heated or baked it turns into a gas it reacts and turns into a gas and we get the gas that's trapped within the within the toffee within the sugar and then when it sets you get bubbles of the gas where it was trapped before and that's obviously crunchy there so you can see it's gone from a solid to a gas that's entropically favorable because you go into a more chaotic or random arrangement there so this can only happen these type reactions can only happen above a certain temperature so let's look at the same figures again let's stress test it so remember at 500 degrees Kelvin for this type of reaction so we've got a positive so an endothermic reaction so it's positive and a positive entropy so if we put these figures in at 500 Kelvin it's not feasible because you get a Delta G at plus 550 thousands and but above a certain temperature so this is 1,500 it is feasible because we're getting a negative value so you can see this one is above a certain temperature it will work so the temperature what temperature is it at does it actually become feasible now this is valuable to chemists because what chemists want to do is produce a product but they want to use as little energy as they possibly can to produce that why heat it more than you need to to waste the money and his bath for the environment as well because we're burning fossil fuels more than likely burning fossil fuels to produce that energy so let's have a look so our reaction is just feasible when Delta G is zero so we can rearrange the equation to work out the temperature when Delta G is zero so this is the rearranged equation so T equals Delta H over Delta s now notice the Delta G has been emitted we've just removed it because as a value of zero it doesn't need to appear in the equation so because that's the when Delta G is zero that's when it's just feasible and that's what we want to work out the temperature for so that's why it's not in there so hasn't it's not a mistake for that one okay so we need to calculate the minimum temperature required for sodium hydrogen carbonate to decompose let's look at this question here so we've got sodium hydrogen carbonate here decompose into sodium carbonate water and carbon dioxide enthalpy value is 129 it's positive it's an endothermic reaction we've got our entropy figures here for each of the substances there it's the first thing we need to do is work out Delta s ok because that's part of our equation so Delta s remember is products minus reactants we other our entropy changes for our products so this is this bit here and entropy changes for our reactants which is here and then we subtract them away from each other and we get 335 joules per Kelvin per mole this entropy value is positive that's a favorable process ok enthalpically it's not entropy and tropically it is so what we want to know is and we need to know what temperature does this reaction become feasible ok so we know it will be feasible because we've got a positive enthalpy and a positive entropy so it's going to be feasible above a certain temperature but we need to know what temperature that is so at least we know we can heat it to that so it's very simple we assume that Delta G is 0 and we need to convert enthalpy remember to joules per mole so that's what we've done there 129,000 divided by 335 and our temperature is 385 kelvin so remember it's got to be in Kelvin so that's the temperature the minimum temperature what's required to actually get this reaction to go so fairly straightforward ok so that's it so that's the end of the video on thermodynamics so there's a lot of information there a lot of calculation but I hope it was helpful anyway I hope it all there it'll help you too yeah it's getting you get your answers right at least and like I say there's a full range of AQA videos from year 1 to year 2 full around your whiteboard tutorials and exam practice as well so all for free all I ask is you just hit the subscribe button that is more than enough as long as people keep subscribing and watching I'll keep making the videos and remember these videos are available these slides are veiled to purchase for your own use if you just click on the link below in the description box and you'll be able to get ahold of them there and right that's it bye bye