[Music] [Music] hello there my name is Chris Harris and I'm from Allawi chemistry and welcome to this video on ocr-a this is for module 5 and it's the how fast topic so in this video we're going to look at some revision notes really for OCR a v' and for this for this topic in particular so everything you see on here will give you the basic outline of of the information that you need for OCR a for the how fast topic and it doesn't have for example things like exam technique and practice questions etc so what you need to do is try and work through that and practice these questions because that is just as important it's one thing knowing the content it's another thing actually using the content as well and all the slides you see here are available to purchase and if you click on the link in the in the comments box below you can get them there the great value and it means you can use them on your phone on your tablet and you can access them as and when you require and and move it forward and make notes from it and as well so they're all available so we just click on the link there and also all the videos on here all for free so they are accessible by anybody and you can do whatever you want with you know with the videos with regards to your revision so all I ask is that you subscribe to the channel so if you click on the the button on the screen there to subscribe and that would be that would be fantastic and but there are a full range of videos on Alawis chemistry youtube channel and that covers all them in all the wayon exam boards they are these ones in particular are specifically tailored to the exam board so you know that this contains everything you do need and nothing more and that keeps it a little bit concise but it is there to supplement your your revision it's not there as an isolated tool but it should it should give you a little bit of help anyway and so I can say all of it is is specifically to the specification and these are the points that I've used to make sure that it does meet that that criteria force yeah okay so let's start by looking at how can rate to be measured in it so the change in pH of a reaction so when we're when we're looking at a reaction we've got to make sure that we'd be able to measure its progress over time and there's numerous ways in which you can do this so pH is one of them now this is useful if your reaction is actually changing pH from reactants to products so a pH and reaction may change over time because H+ ions have been used up for example and a pH meter can be used to measure the pH of a substance at regular intervals so that can be quite useful and then what we can do is we can use that to calculate the H+ concentration so the amount of mass loss so you might have seen this already so this is particularly if we go way back into GCS here you might have seen this in particular but this is useful for any reaction that produces a gas you basically put the reaction on the top pan balance allow the reaction to to work itself aware and you get the gas that's produced the gas escapes and that's a loss obviously the the beaker will get well there'll be a reduction in mass because the the gas has been lost so that's a good way particularly if reactions are producing gases obviously we can use mole calculations and you Tholian you'll not get away from all calculations but you can use more calculations to calculate the number of moles of a gas and that's that's been released so in other words reactants that have left another one is actually instead of measuring the mass losses measure the volume of gas that's been produced this might be useful if the gas in particular might be quite toxic or not very pleasant to have in the atmosphere you know you want to try and contain it in it in the vessel or even measure a volume that's produced itself just the mass loss so another way of measuring it like say is using a gas syringe and that's measured over a period of time so every say a minute or 30 seconds depending how quickly reactions going then you've you know then you take the the volume of gas that's collected at that time you plot that on a graph you've got to be careful you've actually got to set this up correctly because sometimes particularly I've been in the lab where you've set a reaction away and actually the reagents have either too concentrated or there's too much of it and you end up producing far too much gas there the syringe can hold that it it shoots the the plunger off the other side so to make sure you set it up properly that's all I would say okay and and what we can do with these with this information is we can use them call an ideal gas equation and we can work out the number of moles of gas produced and use a molar ratio in the equation to work that out so you'll see later on the slide there's loads of ways in which we can measure rate of reaction using a graphic cetera okay so if a reaction changes color M is another one then we can use something called a color limiter a color limiter you might have these in your labs your college labs or school labs they're basically an instrument that measures absorptivity of light in a colored sample so this is really useful if you're producing something that's perhaps changing color or change in a shade of color so the more concentrate the sample is the darker it is and the more light is absorbed as well and there's a there's a picture there as you can save a color imita particular useful in environmental sciences you can measure concentrations of say water their water quality if it's murky or if it's quite clear okay so the common reaction that we can use where we can use a color meter is between propanone and iodine and the reaction goes like this so we have iodine which is i2 which is here you can see there and that's propanone which is a ketone and that's gonna form and and I order compound here which is this one here and you form I I died I died I know this is I minus and h plus if I can get me get my tongue n so you can see the the main thing here is the color change so we've got iodine which is a classic classic brownie like a brownie color when it's dissolved in solution I things a funny one actually it comes those are different colors it's actually a grey solids but when it's dissolved in solution it can you can't form a brown color so in this case it's a brown solution colorless is you is your propanol and we're going to form our products and all our products are colorless so what we can do is we can monitor the progress of this reaction by monitoring how how much light is absorbed by the colored compound we know there's only a color at the products and the reactants sorry so if the solution that we've got is colorless then we know we don't have any products left so we can monitor how that brown color fades into something which is colorless so what we do is we can plot a calibration curve and this is basically a benchmark it allows us to measure something against something so for example in this case where a calibration curve and it's made up of a range of known different concentrations of the iodine solution and we measure the absorbance of each one of them and plot them on a graph so for example we have a concentration here of iodine which is this bit here of zero so the absorbance is zero because we don't have any so a concentration of nought points let's say that's not point three you think they'll know it's a concentration of about two or two-and-a-half and the concentration of the absorbance increases as well and you can see here that as we increase the concentration of iodine the absorbance increases as well because it's getting darker in color so we use this to then measure a sample that we don't know the concentration of and so this one so we use the concentration like so we use the calibration to measure out the the absorbance of something that we don't know so for example if we take an unknown sample and it has a concentration of let's say seven and a half we can go up and then we can go along and see what the absorbance is or do it the other way around which is probably the most the most likely one is you get your you get your non sample you put it in your absorbance you drop and color limiter you measure the absorbance of that whatever it meant whatever it says on the screen and and you go along and you find out the concentration that's the way that you do it that's that way okay right so so let's look at rate rate is to do with how fast the reaction goes and of course this is the premise of this of this video is looking at that so when we when we monitor the rate of reaction we take regular results at regular intervals and we plot them on a graph like this now in this example we're going to look at the volume of gas produced so this is a reaction where we can use that gas syringe that we've seen before and we're going to measure that every set I think here we're measuring it regularly so nearly every minute so a month over half a minute sorry so we're monitoring the the the progress of this reaction and we're just plotting the results of when we've actually taken that what we've then done is then drawn a line of best fit and you should know what one of them is it's just trying to make sure that the the line that you draw doesn't go through all the points they might do but it's unlikely and what you're trying to get through as many of the points as you can and that's the line of best fit that's very important that we draw that so the gradient and you might have seen this in maths as well particularly in GCSE maths but if you do a level math you certainly will have seen it gradient is the change in Y over the change in X and your Y axis is this one the x axis is along the bottom here so the bigger the section of the graph that we do and then this is a basic skill at anything when you're drawing graphs the bigger the section of the graph use the better the gradient we're going to get so we're not drawing a graph that's squashed up in the bottom you might have heard your math teacher science teacher talk about this putting the squashing your graph into the one tiny part of paper is no good because it's really difficult to read and you're more likely to get higher errors so it's always better to try and as a general rule try and draw it with at least half the paper half the paper of your graph paper and and when you're trying to work out when you've drawn your graph we then try to work out the gradient when we're trying to work that I we again adopt the same principle and we try and take the biggest section of that line as we possibly can for the same reasons so here we are so Joyner picks a sample here so we've got the gradient of this section so here the length here is 6 because that's the difference between 1 and 7 and here the length is 4 so now we've got the change in Y and the change in X for this section of the graph so the gradient is four point six - naught point six which is which is the gradient of the change in Y and the gradient is seven and one so this gives us the sum of the one is the one at the bottom here so this gives us the change and so the gradient here is naught point six seven centimeters cubed per minute now how do we know it's these units for the key thing is with units is always y over X so Y here is centimeters cubed there it is and time which is the x axis is minutes so it's minutes so really really easy all you have to do if you have a scratching editing in what you know what what's the units do I have here look at your graph look what's on your wire access put that first we look what's on your x-axis and put that second and anything that's on the X is always over something so it's centimeters per minute and but obviously in science you would have seen this before anyway way we use minus 1 to represent per so it's always written like that so if you ever get a graph and they give you a really really strange units just literally write it on there look one of us from the y-axis put it there one of us from the X put it there okay straightforward okay so a little bit more difficult if your graph is curved now that's a straightforward one if it so if it's a line graph to destroy triangle and there you go but if it's a curve it's not quite so straightforward because it's changing rates changing along that profile so with curves what we have to do is draw tangent that's a diagonal line that measures at the gradient at a very specific point on that curve and what we want to try and do just like the last time is they draw that line as big as we possibly can to fill that graph because the bigger the graph though the bigger the gradient line the more the less errors we have and the more accurate our results going to be so the gradient they could say is across a specific point so in this example we're going to work out the rate of reaction at three minutes so all we do is you find three minutes and we draw our tangent in there and you can see yeah there's three three minutes there we draw up and that is just where it hits the the curve okay it's really difficult to get it exact don't worry you just need to be as as best as you can as close as you can principles the same change in Y over change in X so we find out what what the M values are so we're going to use this gradient here it's easier to try and get whole numbers rather than decimals it just makes it a little bit easier than that so the change in Y is not point eight and the change in X is four so then put all that into our equation and we get a gradient of nought point two grams per minute notice have changed the units here again so we've got a massive reaction vessel this is not a gas produced so G which is grams per minute which is time that's in minutes there okay so let's look at how we can calculate the initial rate from a graph now this is going to become more useful later on when we look at the initial rate of a reaction because we're going to use it to work out orders and rate of reaction but I'll come on to that later so let's just look at the initial rate but just bear in mind that you we will be using this later on okay so this is why I put it here so the initial rate can be found out using the gradient of a tangent okay so the initial rate is the rate right at the start of the reaction hence the word initial okay and just like the slides before slope we've seen before the we take the gradient of the tangent but with the initial rate we take it at zero minutes so it's right at the start so the tangent at zero minutes is going to meet there okay so it literally adopt the same principle so that's at zero minutes and so therefore in this case the gradient is going to be the change in Y over the change in X and you will get a little bit of grace for this one as well it's not exact with the example the lot so it must be this no Miles's really unless they set it up so that we can only have one answer it is really difficult to get an exact figure so there is a little bit of tolerance here so don't worry too much about the figures but here is one point seven and the change here which is which is one so if we put all that in we get a gradient of one point seven grams per minute so exactly the same principle change in Y over change in X okay so you can see there's a lot of these graphs here so is it is useful to talk about how to set up the graphs and how to get the best from your graph as well because you take time you know that you don't just throw one of these out in a couple of minutes you know you spend a lot of time with them so you know make sure you get the right results from them okay so let's look at clock experiments now these are interesting because you get some all sorts of different clock experiments and some of the best ones are really colorful so they're really good good practicals and they're actually they're really good because they effectively time themselves hence the word clock so um a clock reaction tells us how long a reaction how long it takes for a reaction to occur so a clock reaction is used to simplify the initial rate method okay so it makes it a lot a lot more straightforward so most reactions and you might have seen this as well most reactions like iodine clock and I'll go through iodine clock and a little bit more see you do need to know about that specific clock reaction and it can be monitored by sitting the reaction vessel on some paper with a cross on it now you might have seen this at GCSE as well some examples do do this they call it disappearing cross experiment but what we do is we time how long it takes until we can't see the cross through the beaker and there is a color change and this is known as an endpoint okay so basically we set our reaction away and it literally it'll say for example it'd be colorless to start off with which which it would be so you look down the top of the flask here you look down there and you can see underneath that because you put a bit of paper underneath you can see a black cross and then if you just wait and wait and wait and wait and eventually it'll just change color it'll go cloudy and it'll be fairly rapid it's not instant it's not like split-second it will but it'll change fairly quickly and the moment that changes and you can't see that cross you stop your stopwatch and that's basically the time and we can alter concentration we'll see that in a minute and but we can use column imagery as well we can do as a calibration graph have you seen so you don't have to use the disappearing cross but you will have seen this type of reaction but effectively the quicker the clock reaction the faster the initial rate of reaction is okay so we make three assumptions for this and it's very important that we need to know what these assumptions are because rate is dependent on new risk factors and if we start and alter or change one of them you end up in a bit of a in a bit of a mess and you get a reaction that doesn't actually do as it should do as you all know what it should do when you start it you know what you're looking for and sometimes it's easier scientists too because you think well our should be seeing this and when you don't see it in the experiment you think you've done something wrong these don't get stressed over it but quite often practicals never go right okay you've just got a you know you're looking to your practical and you've got somebody else next to you who's got a different result than somebody else has got a different you know that's just science you've just got to keep repeating it to know if it's an anomalous result but anyway the three assumptions were making is the temperature of the reaction must main constant so we can't do it on a really hot day in June and then in January and have a different temperature in each room because you're gonna get a different result on you the other one is the concentration of the actions it doesn't change significantly during the two during the time period of reaction so when we're when we're allowing this reaction to proceed the concentrations of the reactions don't change obviously they do a little bit in reality but we assume in this case and that because the the changes are quite insignificant that we don't have any any change there and the reaction has not proceeded too far when the endpoint is seen so what we don't want to do is we don't want the the assumption that we make is that reaction doesn't hasn't gone beyond the the point at which we want to work out so all we can go off is the endpoint but and that's all and that's a fair assumption to make otherwise you would never know you know you can't see the atoms you know they were that small so we have to use indicators to help us okay so on the basis of these assumptions what we can say is that the rate of reaction remains constant during the time of the period that we're measuring okay on them but and they're fair assumptions to make it's not a it's not something where it's gonna have a huge distortion and new salts so M so yeah so it's a fair assumption to make so as a result what we can say is that the rate of their clock reaction is a good estimate of the initial rate of the reaction okay because we assume that the rate is the same all the way through so if we can work out the rate of this reaction then we can work out from a practical side we can work out the initial rate and you'll see again this rate will be used later so let's look at that specific because also I want you to know specifically about this clock experiment this is an iodine clock experiment and the iodine clock experiment uses sodium thiosulfate which you'll see later on and and basically that is reacting with the iodine that's produced in this reaction okay so we can use this to work out concentration of something okay so all we have to do in this case is we add sodium thiosulfate and starch cuz starts will react with iodine you'll see that in a minute starch as the indicator and we add that to excess hydrogen peroxide okay so we add all of that into into our into our system and we have our iodine in there as well okay I died iron so the same okay so the sodium thiosulfate what this does is it reacts immediately with the iodine that's produced so we set this reaction away and the moment I Dean's produced it reacts with the sodium thiosulfate and it disappears that's it so it's reacted okay now there is a time so we here's the reaction here so you've got your sodium thiosulfate reacting with the iodine to produce I - and so that's 406 to minus ions here okay so you've got your thiosulfate ions here so that's right and with the iodine and that's that's fine however when the sodium thiosulfate runs out and this is dependent on how much you add in the concentration etc so this is where you change it and then what happens is the iodine reacts with the starch because remember we have it starts in there and what happens is iodine when iodine reacts with the starch you get this jet black color so it is blatantly obvious that there is a color change or that your thiosulfate ions have ran out so that is really obvious that's ran out because you get this very stark black on it and it's a very quick change very quick as well okay so if we vary the concentration of iodine and or hydrogen peroxide and keeping everything else constant this will result in the time taken for the blue-black color to appear and to change so that will change depending on the concentration of iodine that we add in there because we've got lots of iodine in there or the the amount of peroxide as well okay and then we can work out the order of reaction from that and we'll look at orders in a minute okay so the rate of the rate equation and you might have seen this already if you're looking at this as a from a revision perspective but if you're looking at this from new then there is something called a rate equation and the rate equation is something which we can use in formation from practical results to workout the rate of reaction okay so that's all it is so it links rate with concentrations of substances so the rate equation is given this standard formula which is rate equals K and you'll notice here it's got concentration of a and the concentration of B so there we are so rate remembers in moles per diems moles per diem cubed per second okay it's what we call a rate constant and the unit's vary and you you will see the units of K vary as well later bombs I'll show that in the later slide B is the concentration of a substance so this is whatever you're adding in of course and so this could be for example a plus B gives C plus D so B is one of the reactants in this case and the little letters above are what is known as orders of reaction and we'll look at look at them later so orders of reaction so in order M is the power to which the concentration is raised to in the rate equation and it basically tells us the concentration how the concentration of the substance affects rates so for example you might find that if you add chemical B and if we double B it might double the rate of reaction you might find that if we double B it does absolutely nothing it doesn't actually change the rate of reaction and you might find that if we double B we get quadruple the rate of reaction and actually if we if we increase B we significantly increase the rate of reaction so that's quite a that's quite a significant parts to chemistry because if we understand that as a chemist that's really powerful because it means we can we're not wasting our times adding more reagents and it's doing nothing to the rate of reaction so this is a quite useful but the powers tell us that so a zero order basically tells us that something has no effect on rate so that's what I was saying before so if we double a rate doesn't change we call that zero order first order is if the changes in concentration has a proportional change on rate in other words if we double a we double the rate or frequently we quadruple the rate so it has a proportional effect second order is where the changes in concentration is a square proportional change on rate so it's an inversely proportional rate so for example if a doubles then the rate quadruple so it has a squared proportional change on that so you'll see this this is gonna be you're gonna see loads of examples to do with orders here and initial rate I mean you know this is gonna be really important so make sure you understand what an order is and how that how that order links practically into concentrations and rate okay but you'll see loads of examples will show you loads of examples later as well one of the most important thing those orders can only be determined by experiment you cannot ever ever ever look at a reaction and just decide it's how fast it's going to go or what it's going to do and you know without doing a practical to work that out for yourself so practical experimentation is critical here okay so the rate equation links rate with concentrations of substances okay so rate constant rate constant is a number that allows us to equate rate and concentration together okay so it's that it's a it's a figure that we use to basically allow the one side of an equation and the other side of the equation to to balance so we call it a constant and you might if you don't maths you may have seen this in a level of a maths if you do do a little math don't worry if you don't need to do a little math to do chemistry but if you do you might have seen this before and but I'm gonna assume that you haven't so the rate constant is only fixed at a particular temperature in this case now it's strange because the word constant make may make you think that actually it's the same it doesn't change but this one does but it is dependent on temperature that's just chemistry fear so if the temperature changes so does the rate constant so K increases when temperature increases okay that's important so the larger the value of K the faster the rate of reaction try and remember that as best as you can okay because that's going to be vitally important particularly in things like multiple choice questions so the larger the value of K the faster the rate of reaction so let's have a look so the explanation so if we increase the temperature and the particles then have more kinetic energy so they move around a lot more and they collide more often and this increases the rate okay because they're bumping into each other a lot more excuse me right so here's our rate equation but the concentrations of the substances remain constant so they don't change okay so we're not having but not have more a or more beige is because we increase the temperature we're just getting a and B moving around with a lot more energy okay so to make this equation balanced then the value of K must increase as well because a and B isn't changing all we're doing is increase in the temperature so this is the reason why K varies even though we call it a constant it actually changes in value because one side of the equation is actually maybe increasing but this side here isn't that doesn't have any impact so we need something in the middle to bridge the gap basically the value here it must be the value there must be the same that's what that sign means it's equal as much not as like scales or something so this must be the same value okay so let's look at how we can calculate rate then there's quite a bit of calculation in this so you're gonna have very hot calculator buttons by the end of this but rate can be calculated using the rate equation so let's look at an example so a reaction was carried out in a lab and here we've got and nitrogen monoxide sorry no reacting with oxygen to form nitrogen dioxide which is over here and it was found that this that the reaction is second order with respect to no.2 nitrogen monoxide and first order with respect to oxygen and so we have to calculate the rate of reaction given the concentration of N or is three times ten to the minus three moles per diem cubed and or two is one times ten to the minus three mostly in cute so that's the units on there and the rate constant K is seven hundred thousand mole to the minus 2d m6 s to the minus 1 remember these units can change you don't you're not bugged you certainly not expect to remember the units of K because it changes and so you know you'll know that this changes so don't worry if the units of K are changing because that's normal okay so the first thing we need to do is write our rate expression so we know rate equals K the reagents we've got here is an O and ot so the reason why we put number two here is because we are told that n o is second order and we know that it's first order in respect to O two okay so we've got two and one there so what we do is substitute the numbers in there so we know that K is 700,000 and it's always worth making sure that the units are the same so these are obviously moles of DM cube because that's concentration and so that's fine so the units of the vine so seven hundred thousand times by concentration of N or which is three times 10 to the minus 3 squared times by 1 times 10 to the minus 3 now because that's the power of one we just it's the same pickup so the same value okay so the rate here is six point three times by 10 to the minus 3 moles per diem cube per second so it's as straightforward as that but you've just got to look out for the M for the orders okay but this rate equation is going to come up quite a lot more later on so keep it in your mind and be prepared to rearrange it as well okay so this is the rear-engine part actually so this is the rate constant so rate constant is calculated using the rate of equation two so let's look at a slightly different reaction so the following reaction was carried out in a lab and so this is nitrogen monoxide reacting with carbon monoxide and oxygen and that's producing nitrogen dioxide and carbon dioxide so we're going to calculate the rate constant which is K for this reaction given that the order with respect to nitrogen monoxide is second-order as you can see there and zero order with respect to carbon monoxide and o2 the concentration of all the Acton's is for saying it says 1.5 times so 10 to the minus 3 moles DN cubed and the rate of reaction is one point one seven times by 10 to the minus three so we've got a lot of information there so the first thing they could say is you write down our rate expression so our rate expression is rate equals K n o and squared notice that we haven't included carbon monoxide or oxygen that's because there's zero order and if there's zero order they have no effect on rate so what's the point and putting them in the rate equation so that's why they're not there I haven't forgotten about them it's not a mistake that's deliberate okay so we rearrange it to make cave the subject so in this case it's going to be K equals rate divided by zero squared okay we substitute the numbers in so the numbers are going to be K which is one point one set of K equals one point one seven times 10 to the minus three which is your rate divided by the concentration of errno squared and then we should get K is 520 now there's a bit of a blank space there this is because we've got to work out the units okay okay so work out the unit's we just substitute them into the rate equation okay so what this means is that here's our real equation here and we know that this bit here this bit is the rate and rate has the units of moles dim cube per second so we put that in there so instead of putting numbers in we're putting units in concentration as moles per diem cubed but because it's squared we need to write the unit's twice so we're going to put moles per diem cubed here and most diem cute here so this is how you work out the units of K remember it does change so the next thing we cancel them out so we've got a moles per diem cubed and a moles per diem cubed on the top so we're left with an s to the minus 1 on the top and so what we need to do is nudge the moles DM cubed remaining on the bottom up and we invert it so we literally just put minus in front of or anywhere where it's positive and positive anywhere in front of a minus so for example mole to the minus one so that was mole there was a positive one there diem cubes that's DM minus cube so when we flip it up it goes to a positive and then s to the minus one remains the same because it's already on the top line okay so let's look at some rate graphs so rate concentration graphs can actually help us to identify the order these are derived from practicals so when you do a practical you'd probably plot the graph depending on what the axes are and that can help us to establish the order now in this case we would have to work out rates first and then offset against concentration which you would have changed in the practical so a rate concentration graph is created by knowing the rate and the rate is found by taking the gradient of variant points on a concentration time graph so you've got to remember the difference between them so in this case we've got to work out rate first to be able to use these graphs but you work out rate from the graphs that we've seen before so don't get the two mixed up so let's have a look so the rate on a straight line graph is constant okay so the rate concentration graph shows a horizontal line and changing the concentration does not change the rate and we call that a zero order reaction so for example if we've got concentration over time and you might have plotted this from an experiment from that we can work out the rate of reaction and we know that if we take the rate of this gradient and plot that against the concentration of one of your reagents then we get a flat line so a rate concentration graph flat line is zero order so you've got to remember that okay so let's look at a graph we get a graph that looks a little bit more like this now instead of a straight line now if we take the rate at different points along now remember we've seen that before about how to do that and the rate on a shallow graph changes in equal amounts and so the rate concentration graphs shows a straight diagonal line and we can see that there as a change in the concentration changes the rate equally and so this is first-order so if we increase a the rate increases by the same amount and this is why we get a straight line graph for this it's first-order okay and another type of graph so the final one is the second rate so this is where we get a rate of reaction that's really accelerating rapidly at the start and significantly slows down towards the answer it's a very rapid start for these type of reactions if we get this classic much steeper curve similar to that one but steeper and so if we again do the same thing rate against concentration we get a curved upwards graph like this and upwards curve graph so this is a classic sign of a second-order reaction and so this is where and the change in the concentration changes the rate squared and this is what we'll call a second-order reaction you are expected to be able to look at these graphs and identify which one 0 first and second don't get them mixed up between the two I've put them side-by-side so you can see the difference between them so once concentration time the other ones rate concentration you only work out these graphs from these ones here these are the ones that you're going to be drawn from your practicals ok these ones are derived from them graphs so there's a big difference so make sure you don't get too confused because they look similar ok half-life any of you who do physics would probably know what half-life is and if you don't well then it's a fairly straightforward concept a very logical concept so half-life is the time it takes behalf of the reactant to be used up so the half-life graph can be used to calculate a rate constant which is K which you've seen before so here we are we have a graph here now this is a graph showing the rate of reaction or the rates profile from the concentration of hydrogen peroxide to time in seconds and basically we're measuring how much hydrogen peroxide there is in the reaction over time ok so this is a concentration time graph and it's shown a first-order reaction for the decomposition of h2o2 ok so hydrogen peroxide so here what we're going to do is we're going to measure our half-lives okay so remember half-life is the time it takes for half of the reactant to be used up so if we pick it at this point and we're going to drop it down to here so how long does it take for half of that so that's just about two and a half and it's dropped to the half point here how long has it taken for it to do that so in this case that's the first half-life the first half-life here is 200 seconds so it's gone from 2.6 to 1.3 so it's dropped by heart so the next one is how long will it take to go from 1.3 to 1/2 of 1.3 ok so here it is here so from one point three 2.65 is another 200 seconds so it's dropped by the same amount so a first-order reaction and the half-life is independent of concentration so each and half-life will actually be the same length and in this example that confirms up because each of the half-lives as you can see that is the same it's exactly the same length and we can use this equation right that uses the halfway values that we've just worked out there and K is the rate constant is the natural log of 2 natural log 2 divided by the half-life which is given the symbol T 1/2 with the subscripts at the bottom the units are s to the minus 1 because to do with the time so the rate constant in this example here is calculated as the following now you'll see on your calculator if you got a scientific calculator which you should do I don't know how earth you do in chemistry we don't have a scientific calculator but on there there'll be an Ln button and that's the button that you need to use so just have a look at that first but if you press that button we do Ln 2 which is three point four seven sorry Ln two divided by two hundred seconds because that is the half-life that's how long each half-life was and that gives us the value of K which is three point four seven times by 10 to the minus 3 s to the minus one okay so that's per second make sure you get lots of practice this it is fellow fairly straightforward the Ln bits the natural log might you know might freak you out a little bit when you look at it but it's it's really straightforward the calculator does all that for you so don't worry about it okay so initial rates is we've seen this before already we know how to work it out and we know that the iodine clock and clock experiments are really good ways of trying to work out the initial rate of her reaction so initial rate is a really good way of working out the rate equation for a reaction so we've looked at the rate equation there but this is if we we want to work out what that rate equation is and then use it to do whatever we want to do with it so let's say we want to work out the rate equation for a plus B plus C will produce D plus C okay and we can only work out the rate from the experiment okay we can't just look at that reaction I don't know I have no idea well as you know cuz I made it but if that was completely new to me I have no idea what the rate of that reaction is and I've deliberately put in generic letters to take away any form of chemical identity and so this just proves that it's irrelevant what the reaction is it's about it's about the datum for that reason in an exam they will have to give you the data because you can't just look at an equation worked out so the first thing we need to do is we repeat the experiment several times but we're changing the concentrations of a B and C one at a time in each experiment because what we want to do is want to find out the effects of changing the concentration of each one of these reactants to the rate and K and C and C so changing the changing the amounts of each of them reactants see what effect that has on rate because some of them might have no effect or none of them might tube so it might be the fact that a has a double a you double the rate B might have no effect at all and neither will see so you might have a way but I don't know by looking at that if that's the case so what we need to do is work out the initial rate for each experiment and so we calculate this using the graph shown before now we've seen that before okay so we just want to know what the effect is of a B and C so record the concentrations of reactants used for each experiment and their initial rates on a table okay and for this we can work out the orders with respect to each reactant and then on the back of that write a rate equation and so you'll see how this all fits in but the key thing here is about being methodical with your with your processes here so we we change a and keep being C the same we then change B and keep a and C the same we then change C and we keep it in be the same okay so it's about being methodical with this so here we are so here is an experiment for a following reaction we're now going to bring in some chemicals here okay for this one so this is nitrogen monoxide reacting with chlorine to form nocl okay so we're going to look at the the data for this so we're going to work out the rate equation and the initial rate and for experiment four okay so we've been given some data here so the experiments been repeated four times okay and we've got the respective concentrations of N or and cl2 and with each one of them experiments and then what we've done is worked out the initial rate of some of them but we don't know the initial rate of experiment four so that's what they've asked us to work out so first we need to work out the order with respect to n all now ideally we need to compare experiments where N or is changing and cl2 is constant okay so let's have a look here so go to experiment one and experiment three so you see here that we've got the amount of cl2 remaining the same in experiment one and three so you can see it's both naught point one but the amount of N or is changing so that's going from point two two point eight so that's important when you're looking at these tables is to try and keep it simple try and keep it really straightforward look for things that aren't changing look for things that are which is what you want to work out so experiment one and three n always quadrupled okay and the initial rate is quadrupled as well we can see that because it's gone from point six three to two point five eight so the order with respect to n all is first-order some strip forward dead simple okay so we know that one so there's one bit one bit ticked off okay second and we need to work out the order with respect to CL two because that's part of the equation as well and ideally again we need to compare the experiments where chlorine is changing but I know isn't so let's have a look so we've got experiment one and experiment 2 so experiment one and two have n oh the same this time so they both got no point two but chlorine is changing so it's gone from point one to point three so chlorine has traveled so it's multiplied by 3 and the initial rate has traveled as well so the only thing which is cause that rate to change has got to be chlorine because we haven't done anything to n o it hasn't changed so it's actually first order with respect to chlorine as well and so the rate equation here is rate equals a n o + CL - okay so that's fairly straightforward because it's first-order both of them are in there both have an impact okay and so like I see the other thing what to do as well is to work out the the initial rate because there's a bit of data missing so one of Turner's have brought back the rate equation that we just calculated before so that's that's just there so what we need to do is we need to calculate K okay that's really important if in doubt work out K because that's going to get you a lot it's going to get you a lot of places so we'll just pick up any experiment with all the data complete k is the same value for all experiments in this table because we're using the same reagents and we're humans at the same temperature okay so let's look at experiment one for example okay so rate equals K nocl - so rearrange that k equals rate over n o + CL - so k equals the rate is nought point six three in this experiment here and we've got a concentration of ni of nought point two which is there and a concentration of nought point one which is there so that gives us the value of K to be thirty one point five moles to the minus one DM GM cubes s to the mine as well so second we use the rate equation and the value of K that we've just worked out to calculate the rate in experiment four so the race is KN o + CL - that's the rate equation and the rate is calculated by using K which is thirty one point five that we've just worked out there naught point five for the rate the concentration of eddore because that's the value in experiment for a nought point five four CL - because that's the value written just before and then we get a value rate of seven point eight eight moles per diem cube per second so that is the value that goes in there and it's as simple as that okay this is a bit tricky now sometimes that will give you things like this so just be just be vigilant for this but sometimes rate tables don't have a substance where the concentration remains constant and this is an absolute pain but I'm gonna try and show a method and there may be various different methods me personally I think this is the simplest one because I think it's Alec to see things really clearly and broken down so if I make a mistake I can work backwards and see where I've gone wrong so this is my method but there are other ways in which you can do it it tries to simplify it a little bit easier so we've got a reaction between an A and B and the experiment was repeated three times varying the concentrations of a and B give the rate equation for this reaction using the data below okay so we've got an experiment here and we've got experiment a and B again we've anonymized the chemicals because it doesn't make any difference and we've got exponent one two and three and we've got the different reagents here and we've got the initial rates there so first of all what we need to do is work out the order with respect to a now ideally we need to compare experiments where a is changing and B is constant okay so here we are so here's the two here so that's fine because we know that actually B remains the same so that's staying constant and age changing so that's fairly straightforward because we can work that out so we can see here that a is traveled and the initial rate is increased by nine okay so the order with respect to a is second-order because it's increased disproportionately okay so second what we need to do is work out the order with respect to B so this is tricky as a changes in each experiment so that whoever's done this really hasn't thought about the end user and you would never set this up by the way practically this is just you know you just wouldn't so this is really just to test your mathematical skills here so so what we're gonna do here is we're gonna add another column at the end of this table and what we're going to do is we're going to show what the rate would be if we only changed a okay so looking at experiment two and three which of these two here that we've highlighted we see that a doubles so here it is okay so you can see that a doubles so the rate must quadruple as we know it's second-order we've already worked that bit out with respect to a so we know that in theory if we ignore be completely so just take B out completely I'm completely looking at it in theory the rate should be 15 point 1 times 10 to the minus 3 this is a theoretical change ok now what we can see then what we're going to do is compare the difference so this is spot the difference so finally what we're looking for is there is a difference in the rates okay so we know that actually in theory if only a had an impact that would be the rate of reaction but unfortunately they're not going to make it easy other unfortunately that isn't now with them two numbers were the same then we know that B has no effect and there's no impact but these numbers aren't the same so B must have some kind of influence here so what we could say here is that the rate is actually half half the amount into what it is theoretically some fifteen point one to seven point five six so we can see it's about half and this is caused by having B so we say that it's first order with respect to B okay so that's a really tricky example because you really have to if they're add an extra column in again this is my method that's what I think I think is a little bit simpler because I can see I can visibly see that the difference between them if there is one but it's a way in which you can spot that so the rate equation here is rate equals K a squared and B okay with these types of questions it's practice it's keep practicing keep practicing and keep practicing and just try it and look for changes look for patterns look for similarities you know make sure that you're you're getting the answers right there okay so let's look at something called rate determining step so rate determining step is the slowest step and a multi step reaction that's quite important because we don't have reactions that just proceed in one way okay they've they break down in multiple ways so some of these steps though in in a reaction at least one of them is going to be a very slow step and some of them are going to be quick and basically that one which is the rate determining step is going to be the one that determines the overall rate of this multi step and process so here we are we're going to look at some baking examples here so when we're baking a Victoria sponge we we bake it in a multi-step we don't just start with ingredients and end up with a cake at the end of it we've got to do things to it we've got to mix things together we're gonna bake things etc so there's a there's various steps here some steps are quicker than others but it's a process in which you're turning raw material into something useful which is this case a Victoria sponge so step one we gather the ingredients takes about ten minutes step two we mix the ingredients it takes about five minutes and take long for that step three bake in the oven and cool afterwards which is an hour and thirty because it takes a long time to do that and the the last one is just with icing and then add your filling which is which is five minutes so you can see here we've got a multi-step process here some steps taken longer than others so we can see here the cake takes one hour and 50 minutes and total to make and the rate-determining step is step three okay so that's the total that's the total length of time but there's one step that's really taken at most the time there so if we're to reduce the overall time significantly what we've got to do is speed up step three okay now we can do that by for example putting the cake in the fridge and that will speed up the rate at which it cools down we can use a fan oven which will which me which means the cake will take a little bit slightly quicker - slightly quicker - to cook so in chemical reactions we have exactly the same process except to speed up a step we use a catalyst or change the temperature so we're doing slightly different things but the principle is exactly the same so the rate equation remember the reactants that appear in the rate equation affect the rate of reaction because we've seen there where we emitted some of the reactants because they have no effect on rate so only the ones in the rate equation will affect the rate of reaction so it's as simple as that these reactants or substances derived from them must appear in the rate determining step because these are the only ones which have an impact on the rate so substances that are not in the rate equation will not be in the rate determining step and so you've got to be mindful as well it's not always reactants that appear in the rate equation catalysts can as well just because they're not used up in the reaction doesn't mean they have an influence on rate because it calls to do okay so the rate equation can be found from a multi-step reaction so we're going to look at this particular process here so we're going to look at the overall rate equation on the overall rate for the following generic steps so we've got step one which is a plus B forms to see which is a fast reaction step two which is to see forming D which is a slow reaction and we've got D / Z forming F plus G is a fast reaction so the overall equation here is a plus B plus a gives a plus G okay so doing mob again this generic so the first we know that the rate determining step is step two because that's the slowest one okay so that's the one if we were chemist and we want to look at a way in which we can make this quicker we need to look at things which can speed up that bit in particular one of the reactants is C so this must appear in the rate determining step because in the rate equation because that's the rate determining step what do you know this is a two next to it this is quite clever because this actually tells us the power is two so in other words its second order with respect to C so that's pretty clever so it's all linking together now so C is an intermediate which is formed from or derived from a and B and so for this reason a and B must also be in the rate equation two so there is so you see there's only one of a and one of B so they're both first order with respect to a and B in terms of reaction so from that we can then deduce the rate equation is rate equals K a B C squared okay so first order respect to a first order respect to B and second order with respect to C so C is not a reactant though and this is really important but it's fine that's okay so C is not a reactant it doesn't appear in the overall equation sees a catalyst and this reaction is an example of auto catalysis and you'll see this later on in chemistry so and but C is a catalyst and remember it is allowed to appear in the rate equation it doesn't have to be in the main equation there's no sign of C in there but there's a classic thing if it appears in the rate equation and it's not in your overall equation here then that must be a catalyst so there's a bit of a hint okay so we can also find it from the rate equation as well which is the rate determining step so we deduce the rate equation for the following generic reaction step so we've got step 1 which is n or + n or forms n 2 or 2 and n 2 or 2 + or - forms 2 + or - ok so the overall rate equation we've been given this already rate equals K + o squared Oh - so we know that this reaction is second order with respect to n or and first order with respect to O 2 so the first thing to notice here M is the ratio of molecules that were looking for okay so the first thing is we're looking for - an or and 1 or 2 because we remember the powers tell us what the number is in front of it in terms of an equation so looking at the reaction steps there isn't one step with this ratio in okay so notice there isn't a step here that has - lots of N or 2 + or - and neither in step two either so that's going to prove a little bit difficult so what we need to do is we need to start from step one and we start scoring off the Molly this is a bit like chemical bingo okay we need to start from step one and score off the molecules or atoms needed to match the ratio in the Reard equation and we stop when we've accounted for them all okay so are you ready eyes down right it's the first one step one and we have to and or and we score them off but we still have all two to get so we continue to step two so you can see here in the first reaction we have N or and N or so tic we've got that but we haven't got Otto yet so we need to keep going so then we go on to step two there we are okay so we score them off and then we go to step two and we have the aw - molecule in this step in or - and so we've now accounted for all the molecules in the rate equation so we stop we've we've scored them all off we've got all the countless for them all so there it is so what does that mean then it chemistry terms so it means that the rate determining step is step two now there is only two steps in this if it has four or five or six steps then it's wherever whichever stop whichever steps sorry is the one where you finally score all of the elements in your rate equation then that is the rate determining step it's as simple as that it looks horrendously complicated but it isn't it's fairly straightforward as you can see okay okay so the rate mechanism and can be found from the rate determining step as well so the following reaction can occur via two potential mechanisms okay so there's the reaction there so mechanism one is via this web via this method it's just a single a single step or mechanism to it either goes by two steps so this is two possible routes that this reaction could take so the reaction is found to have the following rate equation so k ch3 CBR okay so this is the rate equation so which mechanism is going to be correct so this is quite clever as well so the strong bond the strong CBR bond is broken and this is why it's the slow process in this example here so you can see and we've got to we have to break that CBR bonds and that's really that's really difficult to break that is quite a strong bond okay so also just an another note there's a good chance of a collision as well if there's enough OAH minus ions hence why this step is fast here so there's loads of our minus sign because this is positive and this is negative so this is a fast step because it's very likely that these reactions will will happen so the rate equation tells us that only ch3 3cbr can appear in that rate determining step mechanism one chose ch3 3cbr plus o h minus mechanism to only shows the ch3 3cbr so that only shows that in mechanism 2 so this suggests that the reaction is likely to proceed by a mechanism to as the rate determining step matches the ratio in that rate equation ok so that matchstick appellee says we think it's gonna be more likely to proceed via a mechanism to than mechanism one okay right so let's look at this which is the Arrhenius equation makes a pirate chemistry our Arrhenius the Arrhenius surgeon move on tree Harvey this equation links the activation energy temperature and the rate constant K and this looks absolutely horrendous but it's fairly straightforward if you follow a method and if you can account for everything okay so don't panic too much or try and simplify it as much as I can told you this is the rate this is the Arrhenius equation and it has the following components so we have a rate constant which we've seen before so don't worry too much about that that's just K the Arrhenius constant which is which is this letter here which is given a capital A and that is constant this time and E is the exponential this is the exponential now there is it you're gonna use in this button a lot told you it's now so look out for the e^x button on the calculator that's what you're looking for when when putting in these these figures here EA is the activation energy which is in joules and you would have seen that before gas constant which is constant which is 8.31 joules per Kelvin per mole and T is temperature so there's a lot it looks scary but most of it you've seen before so the first important point to do with a rhiness equation is the activation energy gets smaller and so as the activation energy gets smaller sorry and the rate constant K gets bigger okay that's really important that's important point number one and and the main reason is that the as the activation energy drops the rate of reaction increases okay because if you think about eunji profile diagrams when your activation g reduces or it's slightly lower you don't need as much energy to get particles to react successfully so therefore that that's really why it by that explains that so we've got more particles with enough energy when they collide the second important point is that as we increase the temperature okay the rate constant K increases okay so we increase the temperature the rate constant K increases and the reason why is is that when the temperature increases the particles have more kinetic energy so they're moving around a lot more the more late collide and with at least the minimum amount of energy needed to react which is the activation energy and as a result the rate of reaction increases so gets quicker okay so now we're going to use that calculation using the Arrhenius equation so the Arenas equation is great for calculating the activation energy or the rate constants as we've seen and but now we're going to use that equation and put it in good use so for example calculate the activation energy in kilojoules per mole of a reaction at 330 Kelvin and a rate constant of one point three zero times by 10 to the minus four per seconds assume that the arena's constant is four point five five times a tenth the mind at times ten to thirteen and that the gas constant is eight point three one so first to write up the arena's equation so that's the equation there and we can make it easier to use by taking the natural log of both sides and this gets rid of that 'part or at least if we get rid of that there's one bit that we don't need to worry about so we take the natural log on both sides so we just put Ln K and then put Ln all of that governs on the side there and that gives us something which is a little bit tidier and then we rearrange it so we start to move EA to one side of the equation so we've got EA divided by RT equals Ln a minus Ln K okay so we just rearranged that we shift that across to one side and get the aid of try and get that to one side at least then we multiplied both sides by RT to get EA on its own so here we're going to multiply both sides by RT so that means we can get rid of that bit at least so we get EA on its own but then we have to add RT onto that side so this is the fiddly bit if you can remember this method here and know it's tricky it's really hard there's a lot of unusual things here if you remember this bit it literally strips out all the complicated stuff later on when you put in your numbers in because when it gets really tricky trying to put numbers in something like that is horrendous so try and simplify it get it down to somewhere where you can handle it where you can manage it but it's this initial bit here that's tricky once you've stripped all that out then you can then simplify it's a lot easier so we've got our equation okay it's nicely simplified so all we do is we substitute the numbers in and we'll you'll need the Ln button okay so you have seen that already and with your half-life equation so the Ln button on your calculator so here we go so put ei equals Ln a so got Ln and a is four point five five times seven thirteen that's your rhiness equation minus Ln and then this is K we've got the rate equation which is here the rate constant which is there multiply that by RT so R is your gas constant T is the temperature remember temperature must be in Kelvin if it's in degrees Celsius convert to Kelvin can't stress that enough right so we put all the figures in so this is just simplified it so I'm breaking it down into different steps and so EA is going to be M 11 one hundred and ten thousand seven hundred eighty one joules per mole or 111 kilojoules per mole okay so there is a video and I've gone through a much much greater detail the video to do with that and I'll I'll see if I can try and put that in the video as well so you can you can have a look at it basically talk through the detailed breakdown of this calculation a lot more on me on my whiteboard okay so our rhiness plots still on our eanes plots so what we can do is we can use our enos graphs so plots and we can use that as well so we can use the Arrhenius equation which we've seen and plot the graph of Ln K over 1 over T ok and then we draw a line of best fit and the gradient represents minus EA over R ok that's what the gradient represents don't worry too much about this it's more about the application ok and we can then use that to work the activation energy and the arena's constant so here we go we've got a graph here so we've got 1 over T over Ln K that would be unusual because like upside down but the principle still the same so so we're going to work out the activation energy from this so to work out the activation energy we need to work out the gradient of the line first okay so the gradient is minus 1 point 1 divided by naught point 1 times 10 to the minus 3 so the reason why is remember the gradient is changing Y over X so changing Y is this is y-axis still just because it's upside down still the y-axis is minus 1 point 1 and the x-axis could remember this is times 10 to the minus 3 really pertinent with units so it's naught point 1 times by 10 to the minus 3 so that gives us a gradient of minus 11 thousand so minus EA over R equals 11 minus 11 thousand that's not your activation and you get rid of our so we move EA to one side that's your activation energy so the EA is the activation energy is minus minus 11 times not 8.31 and the reason why it's it's it's minus because of other minus in front of the EA and it's it's a negative graph as well as you can see and so therefore the EA is ninety one ninety one thousand four hundred and ten joules per mole or ninety one point four kilojoules well again a super trend there's a Dunham's a video where I actually talk through it on the whiteboard which makes it a little bit easier to display rather than on these slides but these are just purely for revision and so they have a lot more information on them videos there if you just try and see if I can get a link put up on here so you can click on that and access a brief cards then have a look on the on the channel and you'll see it it's the one with the white background so there's black ones which of these ones the black thumbnails the ones are the white thumbnails or the ones where the whiteboard okay so let's look at working out the arena's constants as well so we can work out the arena's constant as well as the activation energy so all we do is you substitute the value of the gradient and the coordinates of any point on the line into the equation okay so here's the equation which is a line k equals minus EA over RT plus Ln a so we've natural log both sides because it makes it so much easier to work out getting rid of that exponential bit in the in the equation okay so we substitute into the equation using the coordinates circled here okay so we've got and one of the points we've got three point one times ten to the minus three which is up here that's one over T and the Ln K is minus four okay so this is the math C bit and we can rearrange this formula into the form of y equals MX plus six is a straight line graph and that's the formula for that so that's effectively Ln K is the Y bit okay the M bit is the minus Yi over our X is one over T because that's your x axis which is there and Ln a is is represents C so we have that straight line graph so we put the numbers in so four is Ln K which is this bit equals and then we've got minus EA over R okay which is not put one times ten to the minus three we'll work that out last time that's the gradient three point one times ten to the minus three is your x value which is the one over T and then plus Ln a which we don't know okay that's what want to try and work out so simplifying it so they'll go so just worked out some of the parts there and got to this part again strip it back even further so we've multiplied the 11 minus eleven thousand by the three point one and that gets us this value here so we're just breaking it down rearrange it so Ln a is minus four plus thirty four point one Ln a is thirty point one and so therefore a is basically the exponential button that's where you would use that so you press the e button on your calculator put in the number three thirty point one and we should get the value of a there's a lot of work there but thankfully get a lot of marks what I would encourage you to do is to practice it and look at the the whiteboard videos that I've got there and which go into it a little bit more detail I'll try and explain it using white board sometimes a lot easier than doing things like this that's why I have the two two videos out there and that's it that was a tough topic um so that's everything to do the how fast part of OCR and please subscribe to my channel that's that would be fantastic if you can do that so all there for free all ask is just that you subscribe just just to show you support for it all these like I say all these powerpoints are available to purchase we click on the link in the subscription in the subscription box the description box if you click on that link and you'll be able to purchase them from there great for a revision and there are loads more videos and like this for all the major examples on a Larry chemistry ok that's it bye bye