okay so this is the uh second of the crash course videos but the first of the chemistry crash course videos for people with a non-science background we're going to be looking at chemistry today for people who do have a bit of chemistry knowledge bear with me the first bit's going to be very much introductory from the ground up there's a few concepts in chemistry that are very very difficult to grasp and then we're going to move on to reaction rates and equilibria as well because that seems to be quite a high frequency topic in the test that being said this is very much a theory based video we're not really going to be going through gamsat style questions although i will explain some of the ways that they might assess it as well and it's really about understanding the principles so that it's a quick way of trying to get up to speed with what you need it may not be everything that you need and you may still need to go and do some wider reading um if you're from a non-science background but hopefully this gets the ball rolling as well okay cool so as a bit of a checklist of what we're going to go through um so we're going to start with the mold concept then we're going to go through stoichiometry solids liquids and gases as well very briefly and then reaction rates in equilibria and we're going to go through this in quite a bit of detail we're going to look at kinetic theory and how that helps explain it so we're very much looking at the qualitative side of things le chatelier's principle which is quite important as well to just reasoning and understanding equilibria questions different types of changes and how that impacts extensive reaction or the position of something called the equilibrium and then we're also going to look quickly at rate laws that is somewhat of a quantitative method the way that they assess this knowledge and then finally we'll look at equilibrium constants and reaction quotients as well because again they often explain them in the question but they're good to just understand how they work so that you can just get straight into it really really quickly all right so uh first one the mole concept right so let's just move that out of the way so basically in chemistry you're obviously trying to quantify very very small amounts of things but we're talking about the number of molecules the number of atoms in something so you're generally actually going to be dealing with very very large numbers of particles and it would be very very tedious to be working constantly with exponential values the entire time when we're doing all of the algebra which we're going to be calling stoichiometry in chemistry so instead what we do is we create a new unit called the mole right and so in the same way that if i was to measure the distance from say melbourne to sydney i wouldn't measure it in centimeters because then i would end up dealing with i don't know exactly the distance but you know let's say 627 kilometers but then that would be this many centimeters and that's a really really tedious number to be dealing with if i have to keep using it and referring to it so instead if we bundle up and this is exactly what we've done with units right we bundle up say obviously this is not to scale 100 centimeters and we go okay let's call that one meter we create a new unit for it and even that's not good enough because then it becomes 627 000 meters it's still a very kind of cumbersome uh value to be working with so instead we are well what if we bundle it up even further and we say well every 100 000 centimeters or every thousand meters let's call that one kilometer and all of a sudden now we're talking about the same distance but we can use a much simpler number to 627 bundles of distance and in this case 627 kilometers so we do the same thing in chemistry the difference is because we're dealing with very very very large numbers we're going to say that every 6.02 by 10 to the 23 particles whether that's atoms or molecules or whatever it is that we're concerned with we're going to call that one mole like this and so this number here is uh technically avogadro's number you don't need to know that for the test they're not going to test your knowledge of the history of sciences or anything like that it's a good number to know though again they'll probably give that to you in the question a lot of the time you don't actually need to even know it because they're not really concerned with that but it's a good understanding of what a mole is the first time you learn the mole concept it can be very confusing to kind of adjust to so a mole is really just a bundle of particles many many particles so that now we're talking about a large large number of particles in chemistry we can just talk about one or two moles or even half a mole and that kind of thing and it brings the numbers back into a reasonable kind of realm so we could be talking about atoms specifically so we might have for example a carbon atom and so therefore if we have one of those well then how many moles do we have we have one out of as a fraction 6.02 by 10 to the 23. that's obviously a terrible example because that then still stays in a very very large number but imagine instead if we were talking about quite a large amount like say i had carbon in my hand i would literally have billions of them in my hand even though it would be a very small volume of it or a very small mass of it i might have something like 30 times 10 to the 23. and so if i want to know how many mole i have i'm really asking how many lots of avogadro's number do i have here so if i divide this by 6.02 by 10 to the 23 i can kind of much more quickly calculate how many mol i have and it'll be the same amount that i'm talking about but a much more reasonable number so again we can see index laws allow us to cancel these two off and we end up with 30 divided by effectively six let's just simplify that and say it's basically five mole and you'll notice as well a bit of the notation even though the technical word is with the e um we shorten it in the same way that we write kilometers as km or centimeters as cm or uh ft for feet if you're using the imperial system or anything like that then we do the same thing with mole except it's kind of pointless we just drop the e off and it goes from four letter word to a three letter word it apparently saves a whole lot of time so this is effectively what we mean by the mole and so really we use the mole to calculate just about everything in chemistry it's the only thing that's actually helpful because we can no longer can compare masses of different things because every single compound has a different weight to it because each atom is heavier or lighter than the next so if we're talking about something like water we'll take simple things h2o we have two lots of hydrogen and one lot of oxygen and the weight of hydrogen from the periodic table is effectively one so we've got two lots of one and over here the weight of an oxygen atom is uh 16. and so putting those two together we get 18 overall that is a different weight to something like say co2 although this still has three particles it has much heavier particles and so this has a carbon which has a weight of 12 and this has two lots of oxygen which each again have a weight of 16. so we end up with 32 and we end up with 44. so we can't really compare the masses of two two compounds because they have different weights to them which then leads me to these numbers so these numbers we're getting from the periodic table and these are the mass numbers they are effectively a weighted average of the mass of one mole of that atom it's not the weight of a single atom technically it's of a batch or a bundle of them so that we can then compare their weights overall so the idea is that one mole of h2o technically weighs 18 grams grams being the standard unit but as well as that one mole of just hydrogen weighs one gram and one mole of just oxygen atoms should weigh 16 grams and this is where these mass these ones here these mass numbers are actually coming from and they're all given in the periodic table you do not get given a periodic table in the test so they usually give you the mass numbers though in the questions now i've noticed so it's very very unlikely that you need to memorize that if you wanted to really go overkill i guess you could memorize the first 20 it's not it's doable but uh it's really not a memory test so they generally give you the mass numbers to one decimal place in the actual thing and remember as well it's a reasoning test it's not about your ability to do very very finite calculations it's about estimating and moving quickly so you should be able to get away with relying on the questions to give you that information if you're not familiar with those masses already and so the next thing then is we come into a bit of the stoichiometry so stoichiometry is just a god-awful word to say the algebra of chemistry itself and there's a few different ways that we can look at it we'll start with the mass stoichiometry which relates to solids and so there's a primary formula that we need which is n is equal to little m over big m sometimes a little subscript m as well and the idea is this here is the number of mole often referred to as the amount if you see that word in chemistry always means n this is measured in mole which we were just talking about the molecular mass is the capital n and this is from calculated from the periodic table so if you're talking about a pure atom then it's just the number that's given in the periodic table if you're talking about a molecule you have to add up all of the weights contributed by each of the atoms within that bundle and then this is measured in grams per mole because as we said it was the weight of one mole right so of that item of that compound so grams per one mole and finally little m is reserved for the actual mass measured in grams of the sample that you're dealing with or the thing that you have overall the key distinction to make is the difference between amount which is n and mass which is little m so we use this to calculate the number of moles we can do it because this is effectively a rate of mass per per mole so if we're really working out well how many mole do we have well how much does each mole weigh right if we have 30 grams of something and something weighs 5 grams per mole then it means we have 6 lots of that the second one that we have is for liquids or volume stoichiometry so liquids and solutions and this one is number of moles is equal to concentration multiplied by volume so again this is just the amount in mole nothing's changing there c is the concentration and concentration is measured in moles per liter because liters is the standard unit of volume and this is often written as capital m a little bit confusing because so is this it's got little subscript m the difference is this is a variable name whereas this is a unit overall so wherever you see capital m used as a unit following a number this is referring to the moles per liter or sometimes called the molarity as well right can also be written as moles per liter like this can also be written as they often do this in gamsat moles per liter like this using index laws writing the inverse of l instead all equivalent then finally we have the volume of the sample and that's always measured in liters you're often dealing with very small samples so you have to often convert mils into liters dividing by a thousand each time and then the third one is i might do this in a different color because it does kind of separate itself a little bit but to do with gases technically you can calculate concentrations of gases as well because they have an amount and they have a volume but this one we will use the gas law and you might have seen me talk about this one in a previous video i think in the maths one i use it as an example this is pv equals nrt i know there's a lot of information to take on at once but you can rewatch it kind of take notes and that kind of thing and go through it bit by bit p stands for the pressure and standard unit is to use kilopascals which is again just another unit of of uh of pressure then volume is measured in liters as normal so they don't change the letters from formula to formula which is nice and again is that amount in mole r is actually a gas constant usually given as 8.31 and again that will give you that in the question if they use this they very i don't think i've ever seen them assume that anyone has to know that number again it's a good number to know but just speeds it up a little bit but i don't think you would actually need to memorize that the formula though i would probably memorize there's a good chance that they may take that formula away from you and then t is the temperature and although we're very used to celsius in science we always use kelvin and the conversion for kelvin is going to be whatever it is in degrees celsius add 273 like that so the idea is that zero kelvin is absolute zero the lowest possible temperature um that is that is possible and then it's represented as negative uh 273 degrees so the idea is that wherever you've got degree celsius um you add 273 to it and you'll convert it to kelvin so the most common being room temperature 25 degrees plus 273 and you get 298 which we always basically say is around about 300 just to clean up the calculations a little bit more and there we go so those are the three if we do look at some examples of just the straight up calculations like i said it's very unlikely that you would get a question that only relies on calculation it'd probably look more so at the mathematical relationship but just so that we can understand each of them say we had for example three grams of water a water sample and we wanted to know the number of moles of it that we had we can start by treating it as a mass and say well the amount is going to be 3 grams divided by the molar mass of water so this here is the little m which is going in here and then water which is h2o and again don't worry about memorizing chemical formulae for most water and carbon dioxide and oxygen hydrogen gas i'd probably commit those to memory just in case anything else they pretty much give it to you in the question so you can use the the actual formula to rebuild the molecular mass and we'll go through that in more detail in the organic chemistry section where i look a little bit more at the actual chemical formula in the makeup of them so water we've got two lots of hydrogen which have each a weight of one and then one lot of oxygen which has a weight of 16 which will give us 18 grams per mole so there we go and then this gives us technically 1 6 of a mole so around about 0.17 mol that is the amount that we have in 3 grams of water i can't really use water for the next one for concentration because normally things are dissolved in water so let's say we had a different example let's say i dissolved 10 grams of sodium chloride in 100 grams of water and we know that 100 grams of water would be 100 mils so immediately we can go well that's 0.1 liters converting straight to standard units and again we can get the number of mole i like to do brackets to see what i'm looking at is going to be 10 grams like that divided by the molar mass of sodium chloride so sodium has a molar mass of 23 and then chlorine has a molar mass of 35.5 which is 10 over and this is all far too specific but 58.5 and let's actually just do estimation instead of calculator work right let's just again here we go call that basically 60 which is basically 1 6 which is basically 0.17 because i'm dividing by a slightly smaller denominator that tells me that i can probably round up a little bit but given that really is 0.1666 that's already slightly rounded up so i know that's around about it that's how many mole i have if i wanted to then get the concentration of uh say sodium chloride although again yes this is a salt yes it dissolves we'll look at that more in another video for now i'll just simplify this to just say the concentration of sodium chloride the salt and that would be n over v so the number of moles of stuff that we have divided across because solutions form an equal distribution of mole we will divide that across 0.1 liters which dividing by 0.1 is the same as timesing by 10 so we'll get 1.7 and that would be moles per liter so even though we only have 0.17 moles of sodium chloride in 0.1 of a liter that is equivalent scaling it up to a liter we would then have 1.7 moles across a full liter of solution and so we could write that as 1.7 moles per liter or capital m as well and there we have the concentration and then finally i'll go through this one so the gas law if we had a gas let's say we had co2 in a three liter tank at let's see 200 kelvin and one atmosphere of pressure and let's throw in a little bit more of a conversion they would probably give you this in the actual question though one atmosphere in kilopascals is 101 point kilopascals and we're going to work out the let's do the concentration so we can do it in full so from that we can do it in two ways we can do it the long way which would just be to get the number of moles so i can divide pv by rt and then dump in all my numbers so 101.3 times the volume which was three liters already in standard units r is 8.31 that's the constant and the temperature which was already given in kelvin 200 like that and then from there i'm just going to simplify this to 100 3 8 and 200 and then from there i can cancel out the two zeros and then i've got three over eight times two so three over 16 like that which is basically three over 15 which is a whole lot nicer which is one-fifth which is around about 0.2 mole and then for the concentration that is n over v so 0.2 over and the volume was 3 liters so dividing this i would get 0.067 like that and that would then be in moles per liter the shortcut though is you would have noticed that the volume basically was used cancelled out and then came right back so really what you could do is you could rearrange the gas law knowing that concentration is n over v is we could just rearrange just for n over v so what i can do here i'll do this on the side pv equals nrt is i could divide the volume this way so i'll leave behind p and then i can divide the rt back across this way and i actually then can solve for the concentration as n over v as the ratio itself now i can just put these numbers in and it makes the calculation a lot quicker and i should jump straight to this number so pressure was we were calling it 100 the rate we were calling effectively 8 even though it's 8.31 and the temperature we're calling 200 and again we can do here uh there there and then from here we've got one over 16 like that and then from here i could say well maybe 1 over 15 or that probably is a little bit tricky to do actually i'd probably just go old school just do short division and 16 into zero doesn't 16 to 1 doesn't work do it like this point uh then add on a zero doesn't work add another zero 16 into that goes around about six times to make 96 remainder four uh and then round about two so obviously there's a slight rounding error here because i've used different methods but they're both in the same ballpark and if you were doing it with a calculator would give the exact same answer all right um moving along then so that is all of your key kind of stoichiometry one thing to keep in mind is that the number of mole is effectively like your link or your bridge across uh different states different compounds everything because things are only going to interact in ratios of their particles themselves so if we were to draw out an actual reaction let's say we've got let's do a really simple like sodium chloride and it's going to react with magnesium hydroxide and again ignore the fact about how we've put that together again we'll go through that probably in another video if it's needed but let's actually color code some of this stuff here so we've got all of our pieces and we're looking at how is it going to interact the idea is that they're going to interact and change in the reaction based on the number of particles right and what goes with what since we're not really dealing with the number of particles we're using the moles as a measure of the number of particles the moles is actually what we care about overall so we want to know how many moles of one thing react with some number of moles of another thing so if i was to do this the product here should be magnesium let's actually keep the color coding and so we get this and so because of the actual uh charges which again we'll go through in another one we'll just just follow along with it for this one um with the magnesium we need two chlorides there but for the sodium we only need one of these hydroxides or these ohs so what we actually need in that case is we can't suddenly say that there were two chlorine atoms out of nowhere when there was only one over here before the reaction started and so this is really about how we balance equations and so we'll see two lots of sodium chloride needed here but then in order to have two sodiums which will then turn into one we need two of these good news is that that will mean that we have two lots of the hydroxides which then balances this over here so now we should have the same number of particles on the left as we do on the right or at least the same number of atoms on the left as we do on the right and so really what these numbers mean is this means that two mole of sodium chloride reacts with just the one just like in normal algebra coefficients are not written if they're one with one mole of magnesium hydroxide and that's really important because we can't say that the same amount of one is going to be used as the other right and so we're going to track the number of moles in all of our chemistry and that is why this is going to be the bridging the bridging number all right i'll remove all of that actually because we don't really want to focus too much on that anyway and let's move to the second thing so that's your stoichiometry then reaction rates and equilibrium so this is a little bit more um i guess a little bit more qualitative looking at different chemical reactions we obviously first know that there are going to be rates of reaction and we measure these based on the changes in concentration of something so if we have a reaction like compound a plus compound b reacting to form compound c and compound d like this so we'll get all of the the tedious details out of the way what we can do here is we can measure the rate at which this reaction is going forwards based on the rate at which either these two things on the left here are being consumed or the rate at which these two on the right the products are being produced the way that we're going to measure that is we're going to measure the change in concentration and that is going to be in moles per liter and we're going to measure how quickly that happens per second so time generally measured in seconds and this will be our units of our rate measurement now so obviously if it's higher that it means it's going forwards much more quickly if it's slower than it'll be a lower number overall you wouldn't be expected to just look at an equation and just go oh i know what the rate is obviously it's a dynamic process generally speaking rates occur very very quickly at the beginning and they slow down as they approach completion for most situations and in the case of equilibria which is systems where the reaction can go forwards and backwards and needs to find some happy medium where the rates of the forward and the back reaction are equal then what we can do is we can use rate laws for them right so this is kind of the whole roadmap of kind of where we're where we're headed with all this so the first thing is to understanding what actually causes a rate to change we have to look at something called kinetic theory or something it's called particle theory as well and it's really quite simple it just basically looks at particles in a box or in a container it's acting like ping pong balls all floating around and bashing into each other the key thing is they need to they need to clash or they need to collide so collision of particles with enough energy to overcome some barrier kind of like an energy price or cost in order for the reaction to occur and we call this the activation energy which i might go into in a little bit more detail in a kind of thermodynamics chemistry video as well it's generally speaking not going to be all that helpful to the gamsat for the most part so i'll just mention it here we just call that an activation energy luckily it has a very nice name the energy to activate a reaction um so the collision has to have that occur and this is going to become more likely if the particles are say moving faster right so if we draw a box like this and then we just have a bunch of particles like this and i'll draw some purple ones and some green ones as our reactants like this and let's say for the moment they are a liquid so they're able to move but they don't move all that quickly so we'll put little directions on them and they're moving in completely random directions at the moment like this and the same thing these are going to be going everywhere as well and you can imagine kind of like the old screensaver on a dvd player or something they're going to bounce off the walls and just kind of move in some particular pattern until they crash into one of the other reactants and hopefully then form a reaction the more that happens every second the higher the rate of reaction is going to be so what will actually increase the number of successful collisions per second one thing is going to be increasing the amount of stuff right so if you put more reactant in even if it's just one of the reactants or both that will increase the likelihood of a collision because then you have a more crowded room right think of it as if you're at a party although that's very out of date these days um is the likelihood of randomly bumping into someone and then having a conversation with them goes up when there are more people in the room right let's say everyone's just kind of like aimlessly walking around bouncing off the walls and then waiting until they bump into someone and then striking up a conversation so increasing the amount of stuff that means increasing the concentration or adding to it in some way the second thing that you can do is you can raise the temperature the reason why that will occur is because if you raise the temperature then it means that there is more energy in the system and more of that energy will go towards moving those particles and become kinetic energy and kinetic energy is moving energy so the more kinetic energy they have the higher their velocity the higher they're moving or the faster they're moving and so therefore even if you've got a fixed amount of particles if they're moving more quickly it means that they're going to crash into each other more quickly and therefore you should have more collisions per second and therefore the rate goes up and then finally the other thing you can do is you can change the volume of the container and this effectively changes the concentration anyway because if concentration is n over v and your number of moles is fixed and you decrease the volume decreasing the denominator of any fraction increases the overall value of the fraction so you can also decrease the volume so imagine then the walls were to shrink in let's see if i can actually do this like this now they're kind of squashed in a little bit further which means you're more likely now to bump into someone because it's a more crowded room at that point and so all three of these factors will increase the rate the fourth one is adding an enzyme or adding a catalyst i should say and catalysts the idea is they lower the activation energy that energy barrier or price which means that more of the particles should be able to overcome that barrier now with the energy they have and therefore able to contribute to successful collisions per second and then raise the rate as well so no they're not going to like straight up ask you questions like which of the following contributes to an increase in rate of reaction it's very unlikely because that is a straight up theory question they're more likely to describe a particular reaction process and make mention of temperature pressure volume and that kind of thing and then ask a question that relates to how that may impact the the rate is it ambiguous are there some things that are helping push it up and some things that are helping push it down is everything helping push it up and having that reasoning ability is really what's going to get you through on it so it's much more qualitative than anything else the next thing then is looking a little bit more at equilibrium the important thing is to understand that equilibrium is really a dynamic equilibrium meaning things are still moving so although things are seemingly stationary it just means that the rates of the forward and the back reaction are equal so a simple example would be if we took a container and then we put water in it and we let it sit still initially it might be kind of sloshing around in there and we might see it kind of doing this right and so you can imagine water might net be moving one way as it tries to equilibrate but then eventually under gravity everything will hold constant it doesn't mean the water is no longer moving those individual water particles are still randomly moving around just like we had up here as well it's just that the rate of the for of the leftward movement and the rightward movement is equal all right so that would be an analogy for dynamic equilibrium what we're really talking about is that if we bring back that equilibrium set up and we've got these arrows for the both the forward and the back reaction the rate of the forward reaction is equal to the rate of the back reaction and so what that means is that every second you have a fixed amount let's say like three reactions of this go forward in that same time frame three went back and replaced them or replenished them and so overall you have no net change in um the the concentrations of the two everything starts to balance out so if you were to graph that just to get a visual on it because this is something that they may give you is you might have graphs of concentration and then you'll have measurements so maybe a and b were like this and a decreased over time let me do that again and then it kind of flattened out b also decreased and then kind of flattened out and then maybe had c and d which were a little bit lower and they had to go up a little bit but eventually everything and so everything kind of flattens out like this it means then that we've achieved equilibrium kind of around about here where everything's kind of flattened out and the concentrations aren't changing reactions are still happening it's just that the forward and the back reactions are occurring at equal rate so this is referred to as equilibrium and we have this kind of arbitrary concept of uh the equilibrium position where this actually occurs as sitting kind of left or right right favoring one side or the other of the reaction or favoring one group of chemicals and so the way that we measure this is with reaction quotients so if we again have a plus b in equilibrium with c and d then we have a reaction quotient which is a mathematical calculation and we usually call this q and the idea is if we had coefficients let's put coefficients on each of these let's say this was little a this was little b little c little d like that the reaction quotient is the concentration of compound c to the power of its coefficient multiplied by the concentration of d to the power of its coefficient divided by the same thing of everything on the left side as well and we by standard always do this particular order this would be the q for the forward reaction and so then down here we'll have concentration of a in moles per liter still to the power of its coefficient multiplied by the concentration of b to the power of its coefficient i'll point out as well the square brackets is shorthand for saying concentration so you don't have to do c of a because that could get very confusing if then you're putting things to the power of a or if that was c and then to the power c it could get very very confusing and so this number will help us understand where the reaction is at and then what we can do is when it's finally reached equilibrium and nothing's changing anymore then what we can do is do the same calculation and then we call it k which is our reaction or our equilibrium constant i should say so equilibrium constant and then we use the word or the letter k it's really just the same calculation as q it just identifies that now this is when it is at its equilibrium so we do the exact same thing but we put a little e or eq for equilibrium and we're using those concentrations whereas every other calculation is when it's not yet achieved equilibrium it's at some point in time as it tries to achieve that so a equilibrium and b equilibrium if we wanted to get the q value for the back reaction all you'd do is just flip the denominator in the numerator it's also equivalent to q to the negative one you can just invert it and that will give you the q for the back reaction as well so that's more of a mathematical relationship and so if we take an arbitrary number line right and this is going from they're always positive numbers so from zero technically to infinity uh then and i'm gonna put one in the middle because you'll find some are much much less than one some of them are much much more than one as well and so it's obviously not to scale in any particular way let's say that k for a particular reaction because every reaction has a different k value sits here and if we do the calculations of q at some point in time so using the concentrations of a b c and d at some point in a reaction's progress we might get the q to be sitting here and what that means is that it is currently left of equilibrium it's to the left of the equilibrium position and what that means is that the forward reaction would still be getting favored it's still trying to progress and make its way towards k if however we did it the other way uh or let's say we did another calculation and we found that the q value was greater than k that would mean that it's sitting to the right of equilibrium and that would mean that it's currently favoring the back reaction meaning the rate of the back reaction is higher than the rate of the forward reaction so it's net going backwards overall once q is equal to k that means you're at equilibrium and there is no longer a net movement one way or the other the forward and the back reaction are equal and that position is no longer budging unless you make some changes to it and so then that's where we come into less shadowless principle which explains what happens [Music] when you actually change an equilibrium uh system once it's already achieved equilibrium so there are three different things that we can do so we can change the amount i.e change the concentration the second thing that we can do is we can change the pressure and then the third thing that we can do is we can change the temperature and all of these can have impacts for different different reactions so in the case of changes in amounts and concentration this affects every single reaction and all it means is just think of it as a seesaw like this actually now the seesaw is a bad example i might go with think of it as a hill right so let's say you've got flat ground like this and you've got a ball sitting here and this is currently at equilibrium right so k and q is equal to k what happens is if you were to add so say this is a and b reacting to make c and d and it can go backwards if you were to add a you're effectively raising the left side and so now the hill rolls so where's the ball going to roll it's going to roll this way so what that means is the new equilibrium position is favoring the right side it's moving further this way and what you'll find is that when they equilibrate you'll actually have the ball sitting further to the right of where it used to be so this was k1 now it's over here k2 so what that means is that an increase in the amount has favored the right side it's favored the forward reaction and it has moved the position of equilibrium to the right is the way that we describe it and what that means is that when the system reaches equilibrium it will have a higher concentration or a higher amount of c and d than it did at its old equilibrium and you can do every other combination so you could lower you could remove stuff from the left hand side and that would then make the ball roll left and that's going to favor the left side right and shift the equilibrium to the left you could raise the right that would be equivalent to lowering the left you could also lower the right that would be equivalent to raising the left in terms of the movement of the position of equilibrium i'll also point out that it does change the overall absolute amount of stuff that you have in there so the amount of mole and the concentrations a simple way to think about it i'll do this here hopefully this works is if this is the equilibrium position right and then if i were to add stuff on this particular side and i raise it like that and then now what happens is the reaction goes forwards and it starts creating more stuff on this side and this side gets consumed then this side is going to lower a little bit again and this side is going to raise but it will be still at a higher level than it used to be which should be my finger hopefully so the idea is that even though you raised a little bit and it lowered it's still equilibriated at a higher amount absolute amount or concentration so the equilibrium position is shifted the amount at equilibrium i.e the yield is also greater as well likewise if you reduce it a little bit on one side it'll equilibrate at a lower yield then second is we've got pressure so this only applies to gases these are the only things that can be impacted by pressure liquids are semi-compressible solids not at all and so this one's a whole lot easier if we have a plus b creating c and d and vice versa back and forth all you have to do is the shuttle is would explain that it'll try to alleviate the change in pressure so if you increase pressure it's going to favor the side or the reaction progress that decreases pressure again right so it's very much homeostatic so say we had over here three two one and two that would mean that here we have three particles and two particles making five particles on the left whereas over here we only have one lot of c and two lots of d so that is three lots of particles total so if we raise the pressure the best way to alleviate that is to reduce the amount of stuff you have and that would favor in this case the forward reaction again it's like removing people from the room almost to alleviate the pressure so raising pressure will favor the reaction that decreases it and that will mean moving towards the side of the reaction that has less stuff on it and likewise decrease pressure it'll try to resist that and increase the pressure and then the third thing is change in temperature so this one only occurs well it occurs if or we can explain it if we understand the thermodynamics of the reaction so we have exothermic reactions which try to increase temp and we have endothermic reactions sometimes called endergonic as well and these ones decrease temp i might put in the full exothermic and endothermic like that if they give it to you you'll see a little delta h again we'll get to that in a thermo video but delta h should be negative for exothermic and positive for endothermic and if you have a reaction pathway the idea is that one side will be exothermic the other side the reverse should be endothermic and vice versa and so all that happens here is same concept if you raise temp it's going to favor the side that lowers temp right so if we increase the temperature the reaction that lowers temperature is going to be endothermic because what this actually means is it consumes energy from the environment and cools down the environment steels steals energy from the environment itself so if you put more energy into the environment by raising the temp then it will try to consume it and take it from the environment to cool it back down so that would favor the endothermic direction likewise if we lower temperature that will then favor the exothermic direction which tries to raise temperature back up and so if we zoom out a little bit looking at le chatelier's principle although you don't need to know it by definition what it really explains is that every equilibrium system will attempt to partially resist any changes made to it right so if you try to raise temp it'll try to respond to decreased temp if you try to add more stuff it'll try to decrease the amount of stuff it won't be able to do it fully because again it equilibrates back to a new equilibrium position cool and then finally jumping around a little bit i did want to get through equilibrium and k values because then it somewhat leans on the final thing which is rate laws so jumping back to rates a little bit here they do have this in the practice material i haven't seen it in a recent one so it may be a little bit outdated now but it's good to know because i still see it as a potential for coming up rate laws is treat it very much like a maths topic overall the idea is that a rate is going to be equal to some constant multiplied by the concentrations of your reactants let's just go with a and b raised to the power of some exponent like this and this is going to explain the relationship between the rate and the concentration of that reactant so again let's say we have a plus b c plus d and we're looking at the rate of the forward reaction we're using the two reactants in this particular case you don't do what you do in the reaction quotients and equilibrium constants where you use the coefficients these can be different numbers we have to mathematically kind of understand what that actually is so say that we run three experiments so we run experiment one experiment two experiment three and we alter the concentrations of our reactants and we measure what happens to the rate so we have concentration of a concentration of b and then the rate as well and we can completely ignore the rate constant it'll just be some fixed number that's either given or you can mathematically solve it we're not going to look at that in this one so say we have 0.1 moles per liter of a and 0.2 moles per liter of b and the rate comes out as 0.04 moles per liter per second like that then we do a second experiment we do 0.1 moles per liter again but this one we do 0.4 moles per liter and all of a sudden it goes to 0.08 moles per liter per second then we finally do another one where we do 0.2 moles per liter but we return this one to 0.2 moles per liter and then we find that and in this case it's going to go up to 0.16 moles per liter per second like that and we're going to try and work out all right what is what is x what is y and what is therefore the the rate laws between them so all you have to do is analyze look at two individual pairs so let's take pair experiment one and two we can see that a was held constant that's good it means that the independent variable the thing that we're altering is b and you can see what's happened is we've multiplied by two then what we do is we look at what happened to the rate the rate has multiplied by two so the relationship between these is linear because it means that if we double b we get a doubling of the rate so what that actually means is that y is equal to one right so so far now we know that the rate is k times a to the power of its exponent which we don't know times b to the power of one right so this is what's referred to as a first order relationship if we then look at a second pairing we could compare experiment one where a is constant here sorry where a is changed with experiment three because the reason why we're doing that is we want to keep b constant now we want to hold all other variables constant so we hold point two and point two constant and so this time we doubled a what happened to the rate from 1 to 3 it multiplied by 4 here and so 2 to the power of x should be equal to four we can just do i didn't do this in the previous one because it's a little bit simpler but we just do the proportionality again as we've done in previous videos and so here we go well that would be 2 to the power 2 is equal to 4. so that means that x is equal to 2 and that's referred to as a second order rate and so what that means is we can put it all together the rate of the reaction is some rate constant k which is not the same as an equilibrium constant so this is a lower case k multiplied by the concentration of a to the power of two multiplied by the concentration of b to the power of one and then if we needed to we could use that to calculate a rate or to solve the k value algebraically as well and there we go so that is i think pretty much everything that i wanted to go through we've gone through so the mole concept just a bundle of particles 6.02 by 10 to the 23. you don't need to know that number the stoichiometry i would uh memorize those three formulas uh although the gas law may be given the other two i don't think will often be given they pretty much assume that you know how those two work then the reaction rates and equilibrium so understanding kinetic theory that's a big one and also understanding the shadow is principle they may explain le chatelier's principle it's not that you need to absolutely memorize it i can't say for certain but it's a relatively simple concept to to understand without too much diving into deep chemistry or anything and it would absolutely come in handy with being able to reason with these chemistry questions much more quickly and then finally you've got the rate laws and the equilibrium constants and reaction quotients which is a little bit more mathematical than than it is qualitative the final thing as well is just looking at with equilibria um is looking at the difference between yield and rate a lot of people get these wrong so yield is the amount of stuff you get right regardless of how long it took to get there rate is the rate or the speed at which you can you know have the reaction go forwards regardless of how much you actually get at the end and you'll find there's often a trade-off between these two in most reaction progress of most reaction pathways although you're able to raise temperature to say increase rate sometimes that means that that's going to decrease yield because it's favoring the back reaction and that kind of thing they would most likely set you up with a particular one that's why i haven't gone through specific chemicals or anything in most of the examples in this one here this is very much just about a theoretical understanding so that it kind of puts everything in one place for a lot of this inorganic chemistry what we'll do in the next one i think i'll i'll try to balance between chemistry and physics i'll probably do another physics one and then another chemistry one at around about the same time later this week hopefully but we'll be looking next in chemistry at the organic chemistry breaking down things like the naming structures how to read different structural formulae and also the structural representations uh and then moving into some of the heavier organic chemistry pretty quickly which i know a lot of people are probably waiting on as well other than that uh i think that's everything and there's been a long enough video to begin with so i will see the next one [Music] you