Hello my name is Chris Harris and I'm from Allery Chemistry and welcome to this video on OCRA lattice enthalpy. So the whole point of this video is to go through some of the the main points of lattice enthalpy specifically for OCRA. So it's a revision video.
Everything you see on here is dedicated and specifically to OCRA. And this isn't the only video. There is a range of videos that covers the full exam board for OCRA for year one and year two.
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The slides can be useful for if you've got your tablet, if you've got your phone, or if you've got anything like that where you can revise on the go. And they can be purchased separately as a PowerPoint presentation. If you click on the link in the description, description box and you can get a hold of them there um but like i say the whole video these videos are purely designed to go through ocr ocra lattice enthalpy in particular and they're particularly designed if i could spit it out and they're particularly designed to go through the through the content so for that reason they do adhere to the specification and this is the specification for this topic which is looks quite a scary topic with the born here bicycles etc there's a lot of a lot of information in there um and there's a lot of that you need to remember but we'll go through it anyway and it's a really quite a succinct summary for this topic particularly for this example it's a perfect if you study in OCR okay so let's get started so we're going to look at the types of enthalpy change okay now you will have seen enthalpy in year one chemistry and so this one's looking at the different types that's that you may have seen before and some which you won't have done and this will all become apparent shortly so the first one is the enthalpy change of formation now you may have seen this from Hess's cycles yes they're horrendous aren't they but no they're not they're actually alright if you know how to them if you're not sure then have a look at the video I've done loads of videos on Hess's cycle within alloy chemistry so go and have a look if you're not too sure and so anyway enthalpy of formation which is Delta FH which is this bit here Delta means change in F is formation H is enthalpy and so the enthalpy change when one mole of compound is formed from its elements in their standard states under standard conditions now Now what you'll see is a lot of these definitions have this at the end. Standard states, standard conditions.
Make sure you add that bit. Definitions are vital. You've got to remember your definitions.
The ones that I've included on here are the ones that you need to remember. There are other ones, but the exam board are likely to give you them. But these ones you definitely need to remember.
Okay, so enthalpy change of formation. We'll use these later on. These are just literally definitions. Lattice enthalpy of formation. Now, you might not have come across this one, but we give it, you'll notice a pattern here.
We have lattice, not lettuce, lattice in the middle. middle delta which is changing enthalpy obviously is there this is the enthalpy change when one mole of a solid ionic compound is formed from its gaseous ions under standard conditions okay and we've got an example here again with definitions like this the best things to do in context and so we will see where these fit in later so don't worry too much about it we do need to know the the definitions but don't worry about where it fits in for the time being at least. So the next one is ionization.
Now you would have seen this one in year one about the ionization of substances, the removal of an electron. So this is the enthalpy change when one mole of gaseous ions, gaseous one plus ions, are made. from one mole of gaseous atoms okay so effectively we're removing an electron from from an atom and it's the amount of energy so removing one mole of electrons from that now the reason why this one mole and standard states is relevant is because These are factors which can affect the amount of energy required to conduct the action that we're describing.
So if we had one and a half moles for one thing and two moles for another, then obviously the two moles is going to take more energy than one and a half moles. So all these different elements here, these caveats, I suppose, are in there to really emphasize that the standard conditions must be met. Okay, so let's look at ionic bonding.
Because this is going to come into play quite a lot when we look at enthalpy and energy and the structure of substances and the energy required to break them or put them together. So this is why this is important. So the size of the charge on the ion affects the strength of the ionic bond.
So the bigger the charge on an ion, the stronger the electrostatic attraction between the ions. So that makes sense. So the 2 plus charge is going to have a bigger...
bigger electrostatic attraction with a two minus than a one plus and a one minus so it's about the size of the charge which has a which has an impact like we say more energy is required to overcome these forces so they have higher melting and boiling points and this is classic of things like sodium chloride is a is a as an ionic compound i don't know if you've tried to melt sodium chloride in a pan but it is incredibly difficult to do that because it's an ionic It requires so much energy to break the ions, the sodium and chloride ions, apart from each other, that it has a really high melting point. So here's potassium chloride. So this is also table salt.
So K plus and Cl minus has a melting point of 770. So it has a 1 plus and 1 minus. Calcium oxide is made up of Ca2 plus and O2 minus. So we have a 2 positive and a 2 negative charge. This is a melting point of 2572, so it's significantly bigger. So the effect of that is the main effect is the size of the charge on the respective ions.
Okay, so the size of the ion. So the ionic radii is basically the radius. So it's the size of the atom. And that does affect the strength of the ionic bond as well. So the smaller the ion, the stronger the electrostatic attraction between the ions.
You've got to get used to using this terminology as well, electrostatic attraction. So these are key terms that you must put into your answer. Just get used to integrating them with your answer. Try and think, when you're putting a word down, try and think, is there a scientific word that I can use to describe that?
It sounds very geeky, but you've got to make sure you include these words because the exam board gives you marks for using key words. Talking like a scientist. So the smaller ions, they can pack together much more effectively, so they really can squeeze in. And so that means there's more energy required to overcome these stronger forces that are effectively holding them together And the melting and boiling points are significantly higher Um as well as a result. So a classic example is sodium chloride which um Na plus and cr minus has a mountain point of 801 But if we take potassium chloride now potassium is bigger.
It's a bigger iron then the melting point is actually lower at 770 We can't pack these irons in closely enough and so therefore the melting point is reduced as a consequence of that. So generally the smaller the ion and the higher the charge the stronger the electrostatic attraction and hence the higher the melting point and so we say what they've got is a high charge density. So if we've got a two plus charge in a smaller ion there's a higher charge density in there because we've got the same charge but in a smaller area. So that's quite important.
There's a lot of keywords here. And really what I've got to emphasize is key terminology in this, such as charge density, such as radii, such as electrostatic attraction, such as charge, so things like that. Okay, so this is where I said we're going to apply some of that terminology that we've seen, the definitions, and we're going to apply it into the Born-Haber cycle.
Now the Born-Haber cycles look really, really complicated. But the key thing is to be methodical. Okay, and the definitions really learn your definitions through doing born-haber cycle so you're doing two things at once trying to learn definitions is is probably one of those boring things you could do and also it's really difficult to actually learn definitions you know off scratch unless you're literally drumming it into your head over and over and over again it becomes a bit monotonous and boring and you end up just not doing it so the best way of doing it is actually by doing a born-haber cycle which is slightly more interesting because there's lines and arrows etc you But also, when you're drawing out your Born-Hebber, just think of the definition.
Look at the definition and look at the step that you're doing. And it really does become a little bit more easier to remember. Okay, so Born-Hebber cycles are useful to calculate lattice enthalpies.
So this is because you can't calculate this directly from experiments. And so it is structured in a particular way. Okay so let's look at the first bit.
So the basic thing with the Born-Haber cycle is right at the bottom of the cycle we always have the ionic solid. Okay so in this case we're going to basically break down the formation of lithium chloride here or well lithium chloride is the salt. But the actual intricacies of how we make lithium chloride or how lithium chloride is broken down is actually in various different steps. And so what we're doing here is we're showing how this can be structured. So the first one, as you can see here, we've got lithium and half chlorine going to lithium chloride solid.
So we've got lithium solid, because that's how lithium normally exists, and chlorine as a gas, because that's how chlorine normally exists. But chlorine always goes around as a pair. Okay, so we must, but we only need one for lithium chloride.
So we must put half in front of the Cl2 because we only need one chlorine, not two of them. Okay, so only forming one lithium chloride. Now this reaction shows us the enthalpy of formation. Okay, so we're forming, now if you can remember this from our definition, we're forming a solid compound here, which is lithium chloride, from elements in their standard states.
okay so this is what states these elements would normally existed at room temperature and pressure okay this step is always exothermic okay exothermic means negative so delta delta uh delta h it's negative value so that means when we draw our arrows on our ball inhibitor cycle we always draw exothermic um processes as a downwards arrow in other words energy If you imagine this is energy along the side here, and you might see that in the exam actually, I haven't put it on here, but energy is effectively being lost. It's decreasing, it's negative, it's going down. So all exothermic processes are negative, and all exothermic, sorry, all downwards arrows are negative, and all exothermic processes are negative. Okay, so let's look at another step. So we're going to go the other way.
Okay, so let's say, for example, if we want to form lithium chloride, we can take these elements, lithium and chlorine. And we can form lithium chloride through a formation reaction. But we can also go another way as well.
And this is going an opposite way. So there's always two routes. It's a bit like, so my closest town is Newcastle.
So, closest town, it's the city. So my closest city is Newcastle. So I can get to Newcastle a few ways. I can either drive down the A1 and get to Newcastle that way.
I can go down the A19. So I can go two different ways. And it's the same with chemical reactions.
You could go one way, which is, for example... lithium lithium plus half cl2 to lithium chloride or it can go the way which i'm going now with the blue arrows so this way what were the first step we do in a born herba cycle is um enthalpy of atomization so what we're doing is we're taking lithium sorry taking chlorine sorry and we're atomizing chlorine now atomization is always an endothermic reaction okay so that means heat energy is required to do this step so what we're doing is we're effectively Breaking the bond between the chlorine atoms and turning it into an atom so we call that atomization So this is an endothermic process the arrows going up because it's positive Okay, the higher up the scale we go the more energy is required. So basically this is going up Okay, so now we now need to atomize lithium because we've got it in a solid Ideally, we'd like to get it into the same state because it's easier to handle now.
This is theoretical Of course it is, it's theoretical, but what we're trying to do is work out the actual energy required to do this. So, L3 of atomisation of lithium, this is an endothermic process. So, atomisation also means converting things that are not in a gaseous state into a gaseous state. So, effectively, we're breaking apart the structure of lithium, solid, and turning it into lithium gas.
And that requires energy, it's an endothermic process. Okay, the next step... is to form ions because we need to form lithium chloride and lithium chloride is an is an ionic substance so we've got to make sure that we form ions to do this so to form ions we're going up to form lithium plus cl and electrons so this is ionization energy it's the first ionization energy for lithium and to remove an electron from lithium requires energy again so this is an endothermic process so we're forming lithium gas chlorine and electron Now you might be starting to see something similar here where we are effectively forming lithium chloride from ions in the gaseous state. So remember that definition from the first one. So this is the lattice enthalpy formation.
So what we're trying to do here is form ions in the gaseous state but we have to do it step by step. So what do you think the next step is going to be? Well what we have to do is create a negative charge in chlorine.
So this time we're going to add the electron that we've removed from lithium to chlorine. Now, do you think that will be an endothermic or an exothermic process? Well that one's going to be exothermic and the reason why is we're adding an electron to chlorine.
Remember chlorine has seven electrons in its outer shell if we're looking at on that on that on the Bohr's model scale and it has seven electrons in the outer shell it is desperate for another electron so it's not going to pull up a fuss to to receive an electron and in fact it's it's inviting the electron to be part of the atom and so energy is released when we add an electron to chlorine so this is electron affinity we call it. So affinity just means to attach or to if you affinitize with something, it means you're associated with it. So this is electron affinity and it's an exothermic process for this one here because we're adding an electron to an atom. OK, and so the final step, look what we've got here. We've got ions in the gaseous state.
So now we can go for the final process of using these ions to form lithium chloride. And so we call this the lattice enthalpy of formation of. lithium chloride here so that was one of the definitions that you have to remember is the lattice enthalpy of formation and you can see we're forming from the ions in the gaseous state to lithium chloride okay so what the exam board will want you to do because you might think well so what what the exam board will want you to do is is construct maybe it's part of a cycle they might have put some of it there or maybe it's not even construct one at all and just use figures and use numbers to actually calculate the different parts of a cycle so for example they might say right here's the information here we've calculated the enthalpy um the enthalpy um values of of um of something so for example we've given you the um all the figures here but not for lattice enthalpy we want you to work out this bit then you have to use this cycle to calculate it that's really what the exam board are looking for okay so it is very when you're using it um it is very much like a Hesse's cycle okay so you remember with Hesse's cycle when you go with an arrow you keep the sign the same when you go against an arrow depending on the direction that you're going you flip the sign the opposite way so for example if it's positive going forwards if you go against that arrow you go negative okay the value stays the same it's just the sign in front of it changes so what I encourage you to do if you're ever a little bit lost because this is just like a giant Hesse's cycle effectively it's just a Hester cycle is broken down into more bits the principles the same okay so that makes it a bit easier but if you're a little bit rusty on Hester cycle right encourage you to do is to go and have a look at the video and that have made in year one and the year one video for OCR that looks into Hester cycle as well I've got some whiteboard videos as well with Hester cycle so go and have a look at them as well okay so let's look at the calculation side because I said the treat them same as Hester cycle Have a look and see what you think and if you're ever a bit stuck like say go back and have a look So we calculate that lattice at lattice.
I'm going to stop saying lattice lattice enthalpies Um by using the cycle in the same way as hess's cycle So go with the arrow keep the sign the same and go against the arrow and you change the sign That's basically what I said just before So let's have a look at the same cycle that we looked at before that we just constructed Obviously, we've broken it down to the different bits and we know what them parts mean so the enthalpy change of, sorry, the enthalpy change and the enthalpy, these are the different parts of the cycle. But the difference is what we haven't done is we have labelled them here, but we haven't labelled them on the side here. So this is, first of all, this is why I think it's useful to do the cycle first, you're born here, but, and then learn your definitions from there. Because actually, you can see where these fit by looking at a cycle like this, and then subconsciously, well, not kind of subconsciously, you then remember the definitions.
So we've got enthalpy of formation of lithium chloride. First ionization, enthalpy of atomization of lithium and of chlorine and electron affinity. And we've got the values here.
Okay, so you'll be given these values. You don't need to worry about it. Okay, so let's look at the first one. So enthalpy of formation is minus 409. Okay, and remember, so just before we carry on, what we're calculating is the lattice enthalpy.
Okay, so remember lattice enthalpy. If you can... If you can remember from the previous slide, but lattice enthalpy is this bit here. This is your lattice enthalpy, so we're forming it here.
So we'll want to work out this value here. So what we're going to do is we're going to take these figures here and basically just fill them in on there. And then we're going to use the cycle to calculate that. So enthalpy of formation is here. First, so this bit here is...
your atomization because what we're doing is we're atomizing chlorine to chlorine okay so we take that figure there which is one two one now the reason why it's one two one and this is a really really important part is the data they've given you is the atomization of chlorine cl2 we are only atomizing to get one atom of chlorine so we halve the value there it is vitally important that when you get this data that you check to see what that data is actually asking you to do so this is asking you to um this is the enthalpy of atomization of chlorine it's 242 but we're only after one chlorine so it's one two one really important to check the data it might say and through atomization um and it might already have worked out as chlorine one atom of chlorine so really look at it carefully in this case we have to have it okay so the atomization of lithium um is 161 because we're turning lithium solid to lithium gas And then we've got our first ionization energy, which is 519. And then our electron affinity, which is minus 364. Okay, so we've used the data there. And really, you do need to know what is what, which is which. So we can put this data in. If you don't, then you're going to struggle.
But at least by looking at it, you can see how it's structured. So we need to calculate the lattice enthalpy formation of lithium. Okay, lithium chloride, sorry.
So that is delta lattice H. Okay, so what we're going to do is we're going to use it in a similar way to Hess's cycle. And so like I say, there's two ways of getting there.
Okay, but imagine it as a bit like a roadblock. So go back to my previous example about my nearest city was Newcastle. So I can either go down the A1 or I can go down the A19. Now the A1 is a nightmare to get down.
So sometimes the A19 is easier, but you're coming from the wrong side. So let's assume that the... a1 is closed and i have to go down the a19 i'm taking an alternative route it's longer to get down the a19 um it's shorter to go down the a1 um but it's the same with hess's cycle i could go two ways now imagine one of them route is one of them routes is closed so i want to work out going from here to here okay this is where i want to go but what we've got to imagine is because we don't know the value of that we imagine that that's a roadblock so that's like the a1 closed So what we've got to do is go down the A19, or we're going to go the opposite way.
So we're going to go this way, round here, round here, and we're going to get to the same place. It's just going to take us a little bit longer. But what we're going to do is we're going to go via certain spots.
And every time we go a stage, we've got to take into account the number. So we're going to start from here, and we're going to go against the arrow. Now we know this is minus 364. Because we're going against the arrow, then we're going to flip that to plus 364. So there it is. Okay, so it's plus 364. Then we're going to go down on this side here because we're going against the arrow here because this is the direction of travel. Then we're going to go 519. So this is going to be minus 519. There we are, minus 519. We're then going to go against the arrow for the next one.
So it's minus 161. There it is. We're then going to go against the arrow for the next one. So it's minus 121. So there's a lot of minuses here.
And then the last step, We're going with the arrow for once for this one. So this is going to be minus 409. So we keep it as minus 409. Okay, so there it is there. So we've eventually arrived at our final destination.
We've arrived there. We've just taken a longer route to get there. So the actual value is going to be minus 846 kilojoules per mole.
So it's still an exothermic reaction. But what you can do, and this is the great thing about it, is we can start anywhere on the cycle. and we use the same arrow rules as we've done there and we should always always always get a value of zero okay so this is effectively a bit like if you do maths this is like vectors okay um if you add them all up everything should add to zero effectively or even physics if you do displacement in physics um then you'll know because that's a vector quantity and then you'll know that it always adds up to zero so it's effectively like um um it's a bit like me going to newcastle and then coming back to morepeth so i live in morepeth um coming back to morepeth um and um in fact it's about 16 miles from morepeth to newcastle and then 16 miles back i've traveled um 32 miles in total but my total displacement is zero i haven't actually really gone anywhere i've gone to newcastle i've come back so um it's the same with the cycle as well so everything adds up to zero you haven't actually moved anywhere you've gone to the same you've got the same step So, effectively, this is, we add all these up, and all of these numbers here, providing we adhere to the rules, should add up to zero.
So, try it. If it doesn't add up to zero, something's gone wrong. So, just go back and check and have a look.
It's always worth the effort, just to make sure that you've got it right. If you do get zero, you either... it more than likely it's going to be right if you do get zero and if you don't get zero um and it's wrong then you probably the most and luckiest person the luckiest person should i say in the world because to get out of all the numbers that you could get by fluke zero um is very difficult so always check so if you put all them numbers in um just put all those together and they should add up to zero okay and also just check the figure this is minus eight four six you That makes sense because this is a negative value, remember, because it's an exothermic.
So we know it must be negative. It can't be positive because it's always exothermic. So always check that.
Okay, so let's imagine we take another cycle and we're going to calculate a different part of the cycle. So there we calculate lattice enthalpy. I will stop saying lattice, I promise.
Lattice enthalpy. And then what we're going to do is we're going to... work out a different part of that cycle.
So remember go against the arrow we change the sign. Okay so here's the formation of sodium chloride this time so it's a different cycle and we've got our data as normal so we're going to work through that cycle again. So we're going to start with our enthalpy of formation this time we're going to work out we're going to work out something different here but we're going to work out what that is in a moment but what we're going to do is putting all these data in to see what's missing. So we've got our enthalpy of formation We've got our atomization of chlorine again.
We have it because we only want one chlorine atom We need to calculate the atomization and three here because we don't have data for that and we've got our ionization energy We have that and we have our smart our electron affinity. Sorry. We've got electron affinity And we actually have all that ascent to be here. So we've got everything here, but we need to work out this bit here Okay, so remember this is the route we want to take we want to go from here to here because that's the bit we don't know but this road has been blocked so this is like the a19 is blocked so we have to go down the a1 instead so we're going to go from here to here but if this is blocked the only way we're starting from here so if i put it on there there you go starting from here that's where we're starting from if we can't go this way directly we have to go indirectly so we have to go backwards backwards up up and then down Okay to get to the same place, so let's have a go so we're going to go down So we're going against the arrow so it's minus 1 2 1 we're going with the arrow here, so that's minus 411 we're going against the arrow on this bit here, so that's gonna be plus 787 there it is We're going against the arrow here, so that's plus 3 6 4 there. We are okay, and then we're going Against the arrow here, so that's minus 496 and then eventually we get to where we need to be we get to our final point so then if we put all of that into a calculator we should get our final value of plus one two three kilojoules per mole check to see if that's right that is positive okay so that's a arrow pointing up so it should be positive we're getting negative that's the first thing first checkpoint that's not right then all you have to do is take all these numbers here just chuck them into the calculator and they should add up to zero It's magic, okay?
And remember, if it doesn't, just go back and check. It's worth, it is absolutely worth the effort. It gives you that confidence that if you do get that right, then you know you've definitely got it right. Because the chances are, if you get something zero and you've got it wrong, it's very unlikely.
Okay, so here's another Born-Haber cycle, okay? Or Born-Haber cycle, however you want to pronounce it. Tomato, tomato. Right, so it's an extended Born-Haber cycle, but it's for ionic compounds made up of...
two double ions so this one's going to be magnesium two plus and o2 minus okay now this looks a little bit more complicated but just bear with me you follow exactly the same rules we're just adding a few extra steps in here because we have to ionize twice and we have to electron affinitize twice because we're forming um a two minus and a two plus that's the only difference okay this bit looks a bit scary but i'll explain okay it's easy marks okay it's a real real good tip the examiners love this okay so here we go so notice we've got the second ionization step like i say so this is needed because we need to create that mg2 plus ion and this is an endothermic process okay heat energy is required to remove that second electron so we've got that extra step there and likewise we've also got a second electron affinity step as well because we're producing an o2 minus iron okay we're not just producing like like before where we looked at cl minus and that was it we just needed to add one electron with this one we have to add two electrons okay now this process is also endothermic that's strange because you've got a first electron affinity is is exothermic but the second electron affinity is endothermic and can you think why can you ever think well this is because energy is needed okay you have you're trying to add remember when i said last time you have an electron that you're trying to add to an atom that electron oxygen really does want that electron okay because um it's got a shortage of electrons so it's going to really take that electron and for the first electron affinity so that's going to be exothermic it's more than happy to take it however When you're trying to add another electron to something that's already negative, because you've already got an O-, it's not so keen, because you've got electron repulsion. You're trying to put two negative charges together, and that isn't favourable. So it's not massively endothermic, but it's a little bit endothermic. You need to give it a little bit of an incentive, a little bit of a push, to take that second electron and form O2-. So you've got repulsive forces, and this is electron repulsion.
electron repulsion effectively between the electron and the negative ion so just watch out for that but you know it's easy to spot that will only happen if you're adding two electrons to the same atom okay so enthalpy change of solution so this is going to be quite important because we make solutions all the time okay that is fundamental in chemistry and yes there's an energy change when you form a solution so um This is what we call enthalpy change of solution, and this is when one mole of an ionic substance, and you do need to know this one as well, 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. So for a substance to dissolve, it must meet two main criteria. So the substance bonds must break, okay? That's the first thing in this example here.
They must break. That's an endothermic process. You need to put heat energy in. to break something don't you that's that's that's that's common sense effectively and new bonds are formed between the solvent and the substance so for something to dissolve what happens is the ionic substance breaks apart and it forms new bonds with the solvent that's dissolved in that could be water for example and when it forms them new bonds that's an exothermic process so it's endothermic to break exothermic to form so just remember that Okay, so let's have a look here.
So we've got the ionic lattice in solid form. So there it is there. And we've got in this case, our solvent is going to be water. So there's our water here. Okay, so substance is broken into create free moving iron.
So that's the first step. Remember, substance bonds must break. Okay, whatever the substance is. So this is the first step. So the substance is broken into free-moving ions when we add it to the water.
Then what happens is the water then surrounds these ions. It's almost like trying to break up something that is massive with loads of different ions. And these waters then swarm the individual ions and form individual interactions with them. So the bonds formed between ions and water, these ions are what we call hydrated. So we have hydrated ions here.
because effectively they're surrounded by water. And this is effectively dissolved. We don't have a solid structure anymore. We have ions.
The ions are still there. This is like dissolving salt in water. You don't have solid salt when you dissolve it in water. But this is, obviously there comes a point where it becomes saturated.
You can't just keep adding loads of salt. Eventually it becomes saturated. But this is why we're saying the enthalpy change when one mole of an ionic substance is dissolved in the minimum amount of solvent. In other words, at saturation.
Okay, so most ionic compounds dissolve in polar solvents like water. And this is because you have a delta positive on the hydrogen is attracted to the negative ions. There it is.
And the delta negative on the oxygen is attracted to the positive ions and the structure breaks down. So water is a really powerful solvent. When you think to break structure using heat, like here's the salt in the pan again, it's really difficult.
But water has this immense power. To break up a structure which takes an immense amount of heat to break. That's mind-blowing that in my mind anyway.
So the water molecules they surround the ions. We call that hydration. Remember that because we're going to come on to that in a moment. And what happens is new bonds are formed.
And these new bonds formed must be the same strength or greater than those that are broken. That's an important criteria. If they're not then the substance is unlikely to dissolve. Soluble substances tend to have exothermic entities for that reason. So when you dissolve something, generally we have an exothermic process that's happening overall.
But this is why some substances are soluble and some aren't. For example... You have substances which may be massively insoluble, or partially insoluble, they don't dissolve very well, and some that don't. So this is quite important.
Okay, so this is where we're going to use the enthalpy change of solution. I want you to remember that structure. So if I just go back, I want you to remember that process there, because we're going to use this. And just have that in your mind, try and visualise that in your mind as we go through these structures here.
That's really important. Okay, so what we're going to do is we're going to calculate it. So we're going to take these processes here and calculate it.
So the enthalpy change of solution, okay, this is one of the parts that we need to know, can be calculated by using the lattice dissociation enthalpy and the enthalpy of hydration. So remember we've seen that word hydration previously. So we can use a cycle to help work this out, okay, and you've got to remember how this is structured.
Okay, so you've got to remember we've got lithium chloride here. There it is, so lithium chloride forming Li plus aqueous and Cl minus aqueous. So this is what we call the enthalpy of solution, and this is what we're going to try and work out.
So this is similar to a Hesse cycle. So the first thing we need to do is we need to break that up. So remember that diagram.
So we started with our salt here, solid compound. We're going to go through enthalpy of dissociation, which basically means we're breaking the ions up. No water's involved here, technically.
We're just breaking them up. So how much energy is it required to dissociate them ions? So that's Li plus Cl minus gas. Okay.
And then enthalpy of hydration. So this is where the water then forms an interaction with the ions. Okay. So we're breaking up. We're going via a two-step process.
So the enthalpy of hydration is when one mole of aqueous ions is made from one mole of gaseous ions. Okay. So this is the process here.
So effectively it's breaking it up. Then. Forming bonds with with the individual lines and this these two processes here combined equals The enthalpy of solution. Okay, so we have the same the same Well, the sum of them two it gives you the enthalpy of solution So we assume that we do the following like we say you break the solid lattice into its gaseous ions first That's lattice dissociation and then we dissolve their minds in water.
So this is the entropy of hydration and so remember when we're doing this when we go with the arrow we keep the sign the same when we go against the arrow we change the sign and you will be given data to work this out so let's have a look so here this is going to be plus eight four six because we're going with the arrow this is like the again the the journey to newcastle i'm going to newcastle quite a lot um so it's like the journey to newcastle we need to work out this bit here we can't go this way because this is the one we want to work out so it's like a roadblock So we must go here, take a diversion to get to the same place. So we're going with the arrow here, plus 8, 4, 6. And then we're going with the arrow here again, which is, we've got two different steps here. So we've got minus 5, 1, 9 for lithium and minus 3, 6, 4 for Cl-. So the total is minus 8, 8, 3. And you will be given this data in your exam, so don't worry about it. And so, therefore, what we do is we work out what the enthalpy of solution is, which is minus.
37 kilojoules per mole. So we do plus minus and that gives us the total amount. So this is negative. It's slightly exothermic which is what we kind of expect as well.
Okay so let's look at the enthalpy change of hydration. So there are two things that can affect the enthalpy change of hydration. That's charge and the size of the ion. Okay so this is kind of going back a little bit to what we were mentioning before. so ions with a higher charge these attract water molecules much more strongly so this is the interaction remember between water molecules and the ions so these attract them much more strongly more energy is released when the bond is made which means they have a more exothermic enthalpy of hydration the larger the charge the greater the enthalpy of hydration okay so if something's got a like a calcium for example calcium 2 plus and na1 plus calcium is going to have a larger entity of hydration than sodium.
The size of the ion has an impact as well. Smaller ions have a higher charge density than larger ions. So ions which are much smaller with the same charge, there's more of a charge per square area, let's say, square millimeter, nanometer, whatever it is. So they attract water molecules more strongly and hence there's a more exothermic entropy of hydration here. So the smaller the ion, the greater the entropy of hydration.
Okay, very similar to what we looked at before with ions forming an ionic compound. This is an interaction between water and the individual ions. Okay, so what we've got here, we've got a diagram on here just showing some of these ions. So the diagram on the left shows that they are smaller and have larger charge. Ions that are smaller and have a larger charge have a higher charge density.
And there's a stronger interaction. So here, you can see here, we've got a 2 plus charge and a 1 plus charge. This is smaller, this is bigger.
So you can see here, there's a stronger interaction with water here. than there is with the 1 plus charge. So you've got a bigger exothermic reaction going on here.
Very similar to, like I said, from the ionic one, the ionic forming the ionic structure before. Okay, and that's it. So that was a look at lattice enthalpy for OCR.
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