Chem1811 Kinetics of first order reaction-Hydrolysis of ethyl acetate

0:00 good day students welcome to another lab talk for kim 1811 this lab session will be looking at the kinetics of a first order reaction where we're looking at the acid catalyzed hydrolysis of ether acetate on the pseudo first order conditions so ethel acetate or the iupac name ethyl ethanoate or if you want to say ethyl whatever is an ester made from the condensation of a carboxylic acid and an alcohol so that's generally how esters are prepared and the byproduct is water so ethyl acetate or ethernet would be made from ethanoic acid and ethanol

0:47 and you'd form this acetate moiety here now it's a very common this particular is very common and it has application in nail polish remover in paints in varnishes um in cirque circuit board cleaner it's even used as an alternative decaffeinating agent where it was super supercritical carbon dioxide is not available to remove caffeine from coffee teas etc it's a very useful solvent it has a relatively low boiling point and as a consequence as you would have seen last week with vapor pressure measurements its ability to evaporate fairly quickly

1:34 allows it to be used as solvent for certain application that requires hardening on the very quick conditions so it's used like i said in certain pins and varnishes to help the hardening process this esta is of flow toxicity it has a what we consider agreeable order so it's fairly pleasant smelling and it behaves as a very good solvent interestingly this is also a byproduct of fermentation and it contributes to the fruitiness of young wine for example so enough of off of that ester let us look at the theory behind what we're

2:20 doing so we're going to quickly review some basic concepts of chemical kinetics for example we should appreciate that the rates of chemical reaction will vary tremendously depending on the type of reaction some reactions occur slowly some occurs very quickly now how do we define or describe the rate of a reaction we can specify how fast a concentrate the concentration of a reactant or product changes with time so we define the rate of a reaction as a change in the concentration of a specie with respect to time so if we're monitoring a reaction we have a and b going to give us some product c we can either monitor the disappearance

3:07 of the reactants or the appearance or formation of the product and once we are able to measure the concentration of any of these species that hold every time we can describe the rate of the reaction we should also note that if we look at a concentration versus time profile we would see that the products should disappear over time sorry the product should form over time and the reactance should disappear over time so a and b should decrease with time and the product c should increase or grow with time we can describe an instantaneous rate which would be the slope of a tangent to any point on these curves or we can talk about an average rate which looks at how the concentration changes over a specified time range

3:55 so if we're monitoring this the same reaction the rate should be a measurable value and it should be the same irrespective of whichever species we're monitoring so if you do some basic math and look over this time frame here for example you will notice that for any reactant the concentration change will appear to be negative over time which means you get a negative number when you look at the concentration change versus time whereas for a product because this concentration is increasing doing the same math you get a positive number so how do we treat this if we're saying that for the same reaction we should get the same value irrespective of whether we're monitoring reactant or product so what we do is that when we describe the numerical rate in terms of reactants

4:43 we put a minus sign before it because this change is negative and that negative sign before that negative change will make it positive in terms of products we leave it as is because it's already positive all right so negative numbers versus positive number they have different meanings and therefore if we want a measurable value it's usually positive so in order to make the rate positive in terms of reactants we put a minus sign before it and that's quite easy now you should also note that the form of the expression that describes rate will also dictate the units because this concentration change over time whatever your concentration units are and your time units are that's how your

5:29 rate will be reflected so concentration per unit time so if you decided to use mole per kilometer cube or something and your time in year or minutes your units on your rate will be reflected by those concentration units and time units you'll also note that the mole ratio is stoichiometry influences the expression for the rate of reaction as well so the example before we look at one a and one b going to give us some product c what if it is not one to one in this case here you notice b would disappear at twice the rate of a so what you do is to normalize the rate by dividing by the coefficient so if it's a reactant you still have the minus sign

6:14 before it but if the coefficient is a value besides one you divide by that coefficient which means that if b is disappearing at twice the rate of a dividing it by two or multiplying it by a half means you get the same value as if you looked at a so irrespective of the species we're looking at once you normalize the rate you should get the same value for that reaction that we're looking at because you're looking at one reaction and you are choosing to monitor one of the species but for that reaction the rate should be independent of the species you are looking at so we will normalize the rate with respect to each species so that we get one value for that reaction so that's a quick review of some of the

7:01 basic concepts of the rate of a chemical reaction the next concept we'll quickly review is the concept of a rate constant we can state that generally the rate of a chemical reaction should depend on the concentration of at least one reactant caveat here is if it's a zero order then it doesn't depend on the reactant but generally speaking if you use more of a particular reactant you get a faster rate for example if you're going to create a fire if you use a cork full of petroleum versus an entire bottle of petroleum the bottle will generate a large fire more quickly and do more damage more quickly so we can use common sense to tell us that generally the rate is dependent on the concentration also

7:48 so we'd write for example rate is proportional to the concentration of the reactant to raise to some index this index will take care of whether there is linearly or not linearly with that kind with that reactants concentration so if we have a general equation here where reactant a reacts with b and the lowercase letters represents a stoichiometric coefficient or more elusive volume one ratio we can write a general rate law where we can say that the rate will be proportional to these concentration rates to some index so again these in these indices take care of the fact that we don't know if it varies linearly or non-linearly so wherever we have a proportional design we can normally replace that with an equal sign and a constant so that's

8:36 what we're doing here so we can say our rate is simply equal to some constant times the concentration of our reactants raised to some index now how do we measure the rate we spoke about that previously you can look at the disappearance of either reactants or the formation of any product you have but however you decide to measure that rate that's the left-hand side but it's equal to some constant and from the times the concentration of your reactants raised to some index so we'll note that the left-hand side which is the rate is just what you can measure with respect to any specie but the right-hand side is always written in terms of the

9:21 reactants because it's the concentration of erectans that determines the rate of the reaction so this k here is called a rate constant and in simple terms the rate constant is basically a proportionality constant reflecting how the rate is related to the concentration of your reactants that's essentially what your rate constant is now this is what we call for example a rate equation or a rate law and always remember the right hand side of your rate equation or your rate law is written exclusively in terms of your reactants the left-hand side which is the rate which is what you're going to measure can be written in terms of any species in that reaction that you're able to

10:06 monitor the only way we can figure out a rate law is to measure the rate because we need to see how these concentrations affect the rate to figure out a rate law these indices that we call an order with respect to each reactant is in no way related to these stoichiometric coefficients so the lowercase letters here right so your mole ratio those do not determine or that the indices generally speaking the only way to figure out these indices is to do experiments again measure the rate now there are other ways which we can quickly deduce the type of kinetics a reaction follows so in your lecture

10:52 you would have covered the basics in terms of zero order first order second order etc so i'm not going to re revisit those basic concepts what i'm going to look at is how graphically you can quickly deduce the order of a reaction for simple processes that follow simple kinetics so for a zero order reaction where the rate is independent of the concentration of your reactants we see that the rate is just equal to the rate constant and how you measure the rate if you have for example a species disappearing it will be the disappearance of that species with time that gives you the rate and that will be equal to your rate constant if we should play around with this and linearize it by quote-unquote separating variables you would see here that the change in the concentration should

11:39 vary linearly with time because if you solve this differential equation by integrating it of course you would get a function where your concentration varies linearly with time and the slope of that plot is related to your rate constant so the diagnostic platforms in your order process is when you plot the concentration of your reactant with time if you get a straight line it indicates most of the time that your process is following zero order kinetics similarly coming from the form of the expression here if you should plot the rate as a function of the concentration of the reactant you should get a flat line here because your rate should be independent of the concentration of your

12:27 reactant right and that goes in line with your zero order kinetic if we look at first order reaction where your rate varies linearly with the concentration of the reactant we see here if we should do the same process and separate the so-called variables you would see that coming from this expression here because the rate changes as the concentration of the reactant changes if you plot the concentration versus time you get a curvature you're here to talk about an exponential looking plot because as the concentration decrease the rate slows down as depicted by the function here similarly if you should actually take the well if you should integrate this

13:13 differential equation aka solve the differential equation by grouping like terms and separating the variables you would have come across this in a lecture something called a ln plot so the natural log of the concentration versus time should give you a straight line from which we can deduce some information so in your lectures you would have done some differential form of this first order equation uh grouping the like terms taking the integral and at the end of the day between some limits you would see that the natural log of the initial sorry the concentration of your reactant with time should be really linearly with time and the slope would be really give you some information about the rate constant an intercept of that plot would give you

13:59 the in the initial concentration or some information about initial concentration technically it's a natural log of it thus this plot this ln plot is the diagnostic plot for a first order chemical reaction here i've also come across the fact that the half-life for a first order process is independent of the initial concentration which means you can do some forward and some reverse prediction using the rate constant to figure out for example what the concentration would be at some point in time or what it will be in the future for any process that undergoes simple first order kinetics lastly we'll look at second order

14:45 reaction simple second order reaction because that one is easier for you to deduce again starting with our generalized reaction for a and b reacting to give us some product c we would see that our rate would be equal to some rate constant times these concentrations raise the sum index if the index m is two so it will be second order with respect to a and zero order with respect to b or if it is first order in terms of a and b the overall order for this reaction would be second order simple second order so if the concentration of a and b is equal to one if the order with respect to a and b is one we can rewrite it to a similar scenario where if we had it only be in second order in terms of a

15:32 and zero order in terms of b so we have a rate is equal to k times e raised to the power of two so if we write an expression for rate in terms of the disappearance of that reactant and again if you group your like terms separated variables etc etc then you integrate that differential equation you get what they call this reciprocal plot relating the concentration of a reactant as a function of time versus a time now depending on how you approach the expression for rate whether you have a half here if it is a plus a etc you might have a 2 in front of the rate constant irrespective of how it is written

16:19 the 2 would be incorporated in k here as the case might be at the end of the day the important bit is that the reciprocal plot would give us a linear plot from which we can deduce the rate constant from the slope so this reciprocal plot it would be considered as diagnostic tool for a simple second order rate law again notes i said simple you can have other scenarios where the sum of the indices add up to give you two those would still be second order and you can also have a case where it's both first order in terms of a and b but if the concentrations are not the same at the start of the reaction then this solution

17:04 doesn't apply okay now in terms of your second order reaction we would note that the half-life is dependent on the initial concentration so as the concentration of the reactant decreases the half-life changes for a second-order reaction so that should be a quick revision of the basic chemical kinetics that you are taught either keep preliminary chemistry or you would have revised in theory course related to introductory camp so summarizing all of this would see that the rate law will determine the mathematics that we get for the integrated solution so when we solve the differential equation we get a particular solution

17:51 this solution would influence the type of plot that we use to diagnose the type of kinetics and of course the math here also determines the form of the expression for the half-life now at the start we mentioned pseudo-first order reaction condition what is sue the first order reaction condition start with our generalized rate law for this reaction between a and b if we assume that it is first order in terms of a and b so both species are involved in your determining step and we're going to say it's first order in terms of each specie so we're going to use an arrow here where m and n is equal to 1. so we have a simple second-order rate

18:38 expression if for example we at the concentration of one of these reactants so here i'm going to use a being in large axis relative to the other reactant so going off the stoichiometry if one mole of a requires one mole of b if a is much much larger than b at the end of the process would have consumed all b and essentially a remains unchanged for example if a is 10 times the concentration of br greater 100 times if i use 1 out of 100 so let us say is 100 times b and if you take 1 out of 100 it's still roughly 100 and that's the idea here we're at the end of the reaction

19:23 a concentration has changed insignificantly so a is relatively unchanged throughout the reaction so its concentration is considered to be a constant if its concentration is a constant we can incorporate that concentration into the rate constant so instead of having your rate being equal to k times a to the m b to the n it's approximately equal to this combined rate constant in which k and a is combined so we write it as k prime times b raised to the n so this k prime as i stated is a combination of things that are constant

20:08 the rate constant is constant at a particular temperature similarly the concentration of reactant a which is in large axis would also be constant so we combine these two constants into a new constant that we call k prime which is our pseudo first order rate constant it's through the first order because we're saying b is equal to one as stated here so b raised to the 1 mean that is first order in terms of b so if you look at this rate log here rate is equal to k prime times b it looks just like a first order rate equation right but it's the same reaction here is just that we have tweaked the conditions to force it to behave like first order so we call it sue the first

20:55 order because we have forced it to behave like first order it's not intrinsically a first order reaction but we are forcing it to behave like a first order reaction now this is important because it simplifies the mathematics because if as i stated earlier on if the original rate expression is like this the only way we could use that simple reciprocal plot for a second order um read kinetics re read law sorry is if we have to ensure that a and b are exactly the same concentration then we could use the approach shown previously but if a and b are slightly different then we have to use very different mathematics to approach the solution to the problem so by forcing it to behave like first

21:42 order condition then it simplifies the math and eliminates some of the errors that can come about when we attempt to do the experiment so how is this important we're going to be looking at the hydrolysis of the ethiosity so we're doing basically the reverse reaction so to form it we would have condensed the alcohol and the acid now we're going to chop it into to reform the acid and alcohol so we're going to catalyze this process using acid this process could also be catalyzed using base but you will see that the type of experiment were doing it would make more sense to use acid to catalyze the process so because on a molecular level if you were to do the r pushing that you would do in organic chemistry you would see

22:29 that this process is by molecular nature and the rate law would be the rate of disappearance of your ethyl acetate will be equal to your rate constant times the concentration of the ethyl acetate and water right voting to be first order in terms of both but because water will be much much larger concentration than the ether acetate we treat it like a pseudo-first order process where this k combines with a concentration of water which is roughly constant through an experiment so we now have k prime times the concentration of your ethyl acid so this is what we call soda first order condition because practically the concentration of water is so large that

23:16 at the end of the experiments it's basically unchanged so if the process follows um through the first order kinetics it means that we can treat this like a first order reaction where if we do that ln plot of the concentration of the ethyl acidity with time versus time or a slope would give us our pseudo-first order rate constant and we can then resolve that through the first order rate constant if we wanted to because if we know the concentration of water we can then calculate the actual rate constant for the reaction so to keep it simple we're doing it on a soda first order condition now the next task for us is to actually determine the

24:02 concentration of the ethyl acidity as the reaction progressed so we're going to be using something called the aliquot method where we're where we would sample direction mixture quench the reaction meaning stop it from proceeding any further and then we titrate to determine how much of the acid is formed because if we're generating an acidic species we can do an acid-base titration to figure out how much of this is for bearing in mind would have had a fixed amount of the initial acid source given to catalyze the process so let's say we use hcl so if we know how much eighth cell we put in and we know how much of the ethanolic acid is formed based on the total amount of acid present we can

24:49 calculate how much of that is actually due to the ethanoic acid that is formed so it's a very simple process right remember this is a catalyst the hcl that you add is a catalyst which means that its net concentration shouldn't change throughout the reaction all right so remember catalyst speeds up the process without being consumed in the overall process so if you know the exact amount of aids that we put in when we titrated sodium hydroxide the difference between the amount of sodium hydroxide use and the amount of it's easily put in the excess would be due to the ethanoic acid that is produced so we can use

25:35 the amount of ethanoic acid produced to be related to the amount of the ethyl acetate or t that is consumed so experimentally what we do is to titrate aliquots of a fixed volume for example five million milliliters here and you quench that different times before the reaction begins and at various points throughout the course of the reaction then we would quench those with the sodium hydroxide to react both the product form and xx acid that is there so at any point in time the total volume of sodium hydroxide used would be related to the amount of the ethanoic acid form plus the amount of hcl replaced in the flask

26:21 and we can see here that the amount of the ethyl acidity it consumed will be related to the amount of ethanoic acid that is formed so we can use this to figure out how much essentially it remains similarly at infinite time we would have consumed all of the ethyl acidity and produced the ethanoic acid so again we can use that to figure out a bit of information about the reaction process so coming from the infinite volume when the reaction is completely finished and with the volume used at a certain time we can use this to figure out the amount of ethercity that's in the mixture as a function of time now the beauty about the integrated rate

27:06 law or first order kinetics in general you can use in place of concentration any parameter that is directly proportional to the concentration so for example if you're using uv visible absorption spectroscopy you can replace concentration with absorbance if you are using conductance so something that again is affected by concentration we can replace that with concentration so here the difference between the infinite volume and the volume of sodium hydroxide that is used at some time as shown here is directly proportional to the concentration of the ethyl acidity so instead of trying to calculate the exact concentration of the velocity we can use this volume difference

27:53 because it's directly proportional so we can use that volume difference in place of concentration and do the same integrated great law the same plot same everything and the slope would give us a measurement of the rate constant so what we will do is based on the volume measurements from the titration we will then plot the natural log of the infinite volume minus the volume at the different sampling times versus the time in minutes and as you can see here coming from this expression slope of that plot is related to the rate constant so you get a negative slope which is equal to negative k obviously k will be a positive parameter similarly once we determine the rate constant we can calculate a half-life

28:40 for this reaction using the fact that the half-life for any first order process be that true first order so the first order is given as ln 2 upon that rate constant the other thing that is important for us as you need to calculate it is to the to assess how temperature affects the rate constant we're going to be using the arena's interpretation of our temperature effects rate constant and the arenas equation looks at how your rate constant is affected by temperature and something called activation energy as you can see here this is an exponential function and we can linearize it by taking a natural log see that we're using natural number e

29:25 so by taking the natural log of this function here we see that the natural log of the rate constant is equal to the natural log of this pre-exponential factor minus the activation energy over the universal gas constant times one upon a temperature in kelvin so in short if for example you wanted to figure out activation energy for a reaction that follows arrhenius type behavior you would monitor the rate constant as a function of temperature and from the plot you'll be able to determine the activation energy and also the pre-exponential factor you can also use this equation to predict how temperature will affect the rate constant once for example you know

30:11 the activation energy for that reaction so essentially at temperature one you will have a particular rate constant the pre-exponential factor does not change nor does the activation energy change with temperature the only thing that is affected is the rate constant so if you have the rate constant at one temperature if you have another temperature you can ultimately predict the rate constant at that temperature by subtracting the two equations we see the pre-exponential factor would cancel and what we have is that when you subtract logs the same as taking the ratio so we see that ln k2 upon k1 will be equal to because we flip around this portion in a bracket minus sign will go so it becomes e a upon r times one upon

30:57 t one minus one upon t two this is something very easy to derive you can try it on your own if you are interested but at the end of the day knowing the two temperatures knowing the activation energy and knowing one of these rate constants you can calculate the other rate constant at that sorry at that particular temperature so this will be useful in trying to figure out for example what would happen if during the reaction the temperature increased or decreased you can predict how it would influence the rate constant either by looking at a ratio or having one of these rate constants calculating the other value as needs be so that summarizes what we need to do

31:43 for the experiment how to determine the rate constants using that ln plot and the volumes that you determine and then having calculated your rate constant your pseudo first order rate constant you can also predict how the temperature will affect the rate constant while you're doing an experiment give them an activation energy all right so we'll see you guys in the lab take care and all the best oh before i run off so let's summarize the experiment you would pip it 100 milliliters of the hcl particular

32:29 concentration will be given to you into each conical flask and you prepared a certain volume sorry in the conical flask a and you put 20 ml in glass b now we would use ice on water to slow down the process and you would pivot five milliliters of the ethyl acetate into that first flask and you start the stopwatch and you'll be drawing five milliliters from that flask and set in the valves and of course you would record a time which you withdraw the 5ml helicopters then you would titrate this aliquot with a known concentration of sodium

33:14 hydroxide so it would be important to stop the reaction when you withdraw an alligator and that's where the crushed ice water would come into play and you would figure out the volume at these time intervals when you withdraw a certain aliquot and any other direction to figure out what the infinite volume would be as in when the time is infinite would speed up that process instead of waiting a couple days with the speed of the process by heating the mixture and then figuring out what the volume would require that infinite time because infinite time means when the reaction is finished and by heating the flask you would have accelerated the process as you would have seen from the arenas equation rate constant increases with temperature so by heating up the reaction we speed

34:01 up the reaction drive it to completion so that's how we figure out the volume at infinite time you would need to wait for infinite time which would be several days or months and of course you would titrate as before so the manual would explain all of this it's very simple what what you want would be these volumes as accurately as possible to get this difference to the ln clock right so in the lab we'll talk more about this procedure and of course how to acquire the data so i'll see you next time see you in the lab take care and best wishes