those views some of you I think mostly month in calculus you'll learn about it using calculus to solve some real problems and so here is actually all this integrating law it come from the integration of the of the kinetics of the rate law I will not explain how to do that but you needed to find the from Khan Academy whatever place they will explain here how they come up those equations the first order reaction we have a rate which is equal to this and that should be rate constant times the first order of the reactant this equation if you use the differential the first derivative becomes this then you separate you separate time with concentration into two different parts and they integrate you have an equation looks like this and the integrated rate law it has several forms I'll just write down one form write down one natural log of the reactant a and time T that equals natural log of the concentration react a at time zero okay the initial concentration minus K times T this equation looks very nice and neat it's just natural log so that's the first order reaction second order reaction the second order reaction we do the similar integration but we have a different form inverse of the concentration of a at time T divided by the inverse of the concentration at time zero plus K times T so that's a second-order reaction and you can see these two mathematically if the first ones Y that T become X you have different curves on the on the x and y graph so that's actually how we find that order by crossing which one give me the linear then that shows you showed me the order okay zero order is a be the easiest and we have this zero order reaction it is okay we have a a time T concentration initial concentration - Kate time T so we'll have a three different forms would have straight even form of the reaction how would I use the integrated form by graphing the three ways further for the first order reaction as you can see from the equation this equation looks different from what I write but it's the same thing keep in mind so you could have familiar ways here natural log a over B it equal natural a minus natural log of B so what are you quizzing I wrote earlier it is same equation that says okay and this form will be native Katie this portion you need to memorize they're not so much easy way out we had memorized the for this chapter the integrated rate laws are three of them and the equation will be giving the test but still to use it get a very familiar need to memorize this equation so here's how to memorize this is the initial concentration this is the time T concentration remember the ae0 initial concentration always higher than the later moment just because a reactant reactant will decrease a decrease then the eighty less will be less than zero and by doing the a zero over eighty you get larger than one this importantly more than one it would be a natural wrong he was positive number and so you have that anyway this reaction in usual concentration constant and if you graph the concentration change versus time natural log concentration waste time when you do the graph you this is why that use the time as X then you have a graph like this for the first-order this graph will give you a straight line okay you streamline what exact streamline next Thursday I was show you didn't do the lap time show you how to get streamline we're going to learn how to I want to go over how to use Excel to get a stream on anyway when you do the graph use natural the concentration as Y and X what we get it is the the y intercept the whiners that it is natural log of the initial concentration here and I do this graph the slope is actually net here in mega stone the next hope it come from okay this is a slope for T and this slope is negative so basically you look into the slope you can find the rate constant or the rate constant equals negative slope by doing graph you can find that all of the reaction this is the first one what about second-order reactions second-order reaction I wrote down earlier in that in that slide and that's so mean this this little equation either if you rearrange use this as Y then we have that so this would be the Y inverse the concentration as Y and the X is T then the K will be the slope and it's actually positive slope and the y-intercept is the inverse concentration so this graph looks different from the first graph because we are inverting concentration even though initial concentration a0 is the highest largest as a0 any concentration decrease the universe the concentration decrease even worse content comes whenever you increase so you can see this kind of slope the slope is positive and it's in the streamline give you a straight line on the inverse concentration graph so this is why time is tax either linear and by the way if a first-order reaction you could graph this inverse concentration you will not get a straight line so by graph three ways and compare which give you the best linear graph that will shows you the order of the reaction it is zero or first or second the second order reaction graph this way you get a a linear now it looks the last one okay if it's zero order your order of the easiest zero order you graph simply this concentration with time and you can see the concentration decrease straight line and the slope again is negative slope again the negative K y-intercept is the initial concentration obviously so we have them it should be negative K T plus 2 concentration I'll write this way just because normally we write the equation y equals a times X plus B and a the slope B is the y-intercept so that'll be match with so that's the third way to to graph and if the third away this way to graph graph concentration waste time I'm give you a straight line that means the reaction is dere order and you can see the rate is constant because rate is a slope the K is slope and it's also rate constant all right so we learned the three ways to graph that will help us find out all the reaction and you can see the real calculations there are many work can be done with the three integrated rate law and this afternoon after the either go or lab which was very short and we'll get it back to here and do some more progress on some I show they show you how to solve those problems in the this section when we wrap up the determine the order the rate law of the reaction that is basically the first half of the kinetics chapter has a more mass more mass than the second half of this chapter take some practice and let me pause the video before the I focus on more on the first-order reactions the first-order reaction as you've seen earlier one equation have so many forms it's almost like a transformer just more round and complete become a different look even appearance but it all come from the same equation depends on the what kind of question you solved the first order rate law equation the first order equation have two forms why does the natural law and natural log reform the natural log form so we'll never see the this kind of questions you see the first order the most because it is harder so this is the natural log of form the logarithm form the exponent so form it is basically I do the exponential use these to an exponent and I will get a different equation and so these two equations based basically the same thing you can anybody even more it doesn't matter what how many believe for the 11 having questions give you a question asked about a time or rate constant you this equation is much better because you'll be given all the others you're given the concentration or you'll be given the ratio between these two and you'll be given one of these two to solve the other one we the national Agra form is easy to solve time or read constant and with this form it'll be easy to solve the concentration either this or that and given the others even okay even the time it just or given one of these to solve the other one so these two equations remember these two equations all the other manipulations it just takes a little more practice if you even ask kind of rusty algebra to rust just more process but a basic form is I do not a give you example for the first order for the zero order or second order because those equation manipulate is just much easier than this you can do it you can figure out yourself the Spurs holder just needed more practice and so this question we have we have a temperature we have this chemical decompose first-order so there we go you can see the first order you're going to use one of these two equations even though they're the same thing and so we can see the K is given rate constant K is given and so we have first question so it's not quite a few questions based on this reaction and so here we we have initial concentration so this is a you need a concentration the reactant a the constant feeling 0 10 0 that's the 802.1 is the concentrated after such a time so the question asked for our concentration and so between these two equation you're going to use this equation because this equation to solve the concentration is just very easy if you have time K and so forth and we do have those so this question asked concentration after the point zero one second this is basically asking for concentration at time T he is point zero one zero but anyway I still write down the general formula it's a T we need for software this software is we start on this equation and you just it should be the AG e equals you guys know how to manipulate this is very easy so I just go ahead I will not show you more steps you can use this or use a 0 times e exponential negative KT it doesn't matter so you can either manipulate this equation to get that and so the rest is just a substitute with all the numbers given you have a 0 which is 2 point 0 0 keep it unit because exponential have no units okay you need to keep the concentration unit and this one no unit in the end have M and this is part exponential the K is given 81 87 you should have painted into the unit because sometimes the question give you one place have a one unit a different place have a different unit and they may not cancel so make sure use the use a unit here this s minus one means inverse of the inverse of second okay or and now a plug in time the time is point zero one zero second and so you can see the s is minus the 1 times s you could cancel and so the rest is just use your calculator you should be need to get a number the number should keep about two three sig figs okay we should apply the rules we just learned the rules we do these financial the rules should be the number one decimal place give you number sigfigs this is the number of decimal place in the exponent it should have becomes a number sig figs in the decimal part in my scientific notation so let's do one more step and try to practice this rule to point your and divided by that that should be exponential the Year point eight seven and two point zero zero and exponential points exponential what I'm getting is two point three eight three eight seven two point sweetie seven but here since I have a two decimal place I should keep only two sig figs in my answer for this to a decimal place become two sig figs so Daniels I put two points two point two four have no units because this part the you negative cancel and the exponent have no units so baited two hundred zero zero divided by two point four and that should give me mine by announcer and final to keep on a two-state fix because keep two sig figs and the answer is in the power point I have the monster point V over either and so that's how we solve this problem with the use the integrated rate law to solve problems so depends on which one you saw you can you have to liquation to pick this is one of the national law and the other one is the exponential form all right I see one more a couple more examples I think there's quite a few more examples so first order reaction rate constant search what a fraction of this have decomposed after three point zero seconds this question is a quite a bit of twisted so what fraction have decomposed and this question do not give you the concentration fraction is supposedly means the part divided by the whole that's what called fraction because part is the part is part of the whole thing and so here the asks about a how much decomposed the eighth zero means initial concentration all right the 80 it is concentrate on reaction at time T this is a Romanian reactant this is starting reactant so the fracking decompose it should be how much lost so basically a time T your concentration smaller than the beginning the difference between these two the initial minus final this is how much he composed it is deep the molarity is a compound decomposed it's still not fraction but the homo smaller decompose divided by the initial the whole it will give me the fraction of the decomposed so this divided by the whole concentration that should be the fraction of decomposed and there's another way to understand this if you look the 80 over a a zero this means how much fraction remained this is the initial concentration and this is at a time T concentration or reactant remain so that should be basically shows you how much fraction remains the question asked for constant a fraction decompose which means all the other part the reactant that it lost it's a fashion decompose and the whole part is a 1 so 1 minus this it is 1 minus the fraction remained it is fraction decomposed decompose a fraction anyway anyway this question a little bit twisted but the I haven't the users kind of putting my test equation this example I did on purpose make them go trickier just thinking more make sense you guys [Music] this is how much I remained fraction remaining then the out of 100% you subtract the part remaining it is how much you lost decomposed so basically this question the tricky part is a figure out of what this question asking we have a whole bunch of equations and the seam that are not relevant to this the key to figure out was question asking as typical word problems okay you when you learn the math in high school elementary school called algebra if you have the foreman or try to solve a formula is easy if you remember those equations but if you give you word problems it's always hard because you want to translate the word problem into the foreigner and then you can solve it this question the the harder part is get to understand what they are trying what are they are what they're trying to ask and then the rest just apply the equations and that's the what is the shooting for we just need a fun this equation where we actually do not see them but we see this equation you can remember my last slide we have this equation the exponential form we have this basically since we know the rate constant K and we have this T here three second what can you need you to solve this and then I just need to invert it upside down and use a 1 to subtract that subtract this part from one I get an answer and so this question basically it becomes solve for this and they're realized we have this exponential form I just use that here substitute this is the a 0 over 80 and here 80 over a zero it's upside down so what you need to put that answer down the in one over e katie is actually equal native katie you don't need to do this because at this point you can actually is just plugging in numbers and get that fine answer but if you want to work more universe of the exponent it is the negative exponent and so the rest will be easy okay once you solve the Oh get over the initial barrier to overcome to understand what's the asking solving 3d just 1 minus e K is one point zero second inverse so it's the inverse of second and times the time which is three seconds three point zero second so these two a good cancel you get a e 3.0 and now just wait a minute I'm sorry there's native here got a carried away we have a net him don't forget I almost forgot I almost made them misty cause you know how I realize doesn't it states because once beginning here every night e to the third power just gonna be a more larger than what I would have get a native numbers because we because I lost the initial sign exponential of three that's a big negative bigger number more than one I would again Nick in there you can try to calculate this that will not make sense okay so don't forget negative and so the rest we can one more time apply the rules and let's see how many sig fig we should have the exponent the number of decimals it should give me number of decimals it should have give me how many stay fixed this is a one decimal so my answer should be only one sig figs are here so let's turn around and get here one minus e to the neck is straight so if I plug in my calculator t1 sig figs oh oh I see yeah my color is weird I have the use of parentheses no I don't like this is the actually the programmable calculator I rarely use it so okay what I get is point point zero four nine seven okay point a zero four nine eight but since here's only one decimal I should keep only one sig figs I want to keep it here so I just around around 2.0 fire so final answer is point nine five no units because the one wanted mean is 100% and this part is whatever person they remain and so my final answer is at point nine five which means 95% of the compound cyclobutane has decomposed have answer a key for that there we go so that's how we use the equations for for this okay let's see there's a few more examples any questions on this this question little bit twisted so just make sure Yanis on the the ratio between this especially the 80 over zero is how much fraction remained so 80s is the concentration of reactant at time T and eight heroes initial concentration I don't see we have I have this kind of question on quiz so don't worry about it this kind of question we have some other questions more you see the homework questions all right so what happened steel I got stuck with the same reaction I had a certain temperature decompose the first-order reaction after 10 second 10.0 second twenty point zero percent cyclobutane Oh No here we use a how much person decomposed again we use that and we have that equation that equation suppose of a concentration matter here we see no concentration not any concentration being mentioned we only see the how much percent just like when you saw any algebra problem you can see the you know the equation and you see the the information on some of the variables not a given in that case use algebra make sure you use algebra solve a problem even though you don't know the concentration and so the same equation I have two forms just let's write down the what does it mean twenty reason decompose so that means let's sing it will have a you know I don't know how much easier is and then we have eighty at time T which means ten second the time is ten second time in ten second that we know the needs of concentrate I don't know but even we we keep the initial concentration as a zero what should be the eighty at time 10 seconds the question asks a question management 20% is cyclobutane decompose this is how much reactant remained after 10 seconds each time zero yes that's a key the remaining reactant is eighty percent eighty point zero percent of the Orang mode so you can just use 80.000 so that's a key even though I don't have numbers you can assume the a is a zero and then use that a person use it whatever given the ratio to proportion the percentage to write down a formula to represent 80 because a first-order reaction it's actually the ratio between the reactant the nice reaction and the final reaction is a ratio of 10 mattress to concentration extra snow that important so once again that that are making a life much easier because we have those equations the question asked about the rate constant can you remember the I mentioned earlier which the equation is solved the rate constant K easier we have a natural log reform we have the exponential form it is actually natural log form to solve kzz ok if you remember those equation that will be life it will be much easier but it's very easy to confused which one where should I put a 0 where so I put a tee okay there's a way to memorize this if you try to just memorize that the way it is it's gonna be hard if you understand it'll be much easier a 0 and a 80 which one should be bigger between a zero int a 0 supposed to be larger than 80 because a zere's is react and edit beginning it's always lager so because that and a K is always positive time can be never popped negative times always pause it's only one direction so this is the whole thing with a positive to make it a positive this ratio has been larger than one larger than one that means I have to put whatever bigger on top and the smaller the bottom so that's how we memorize this equation and so this equation it will make solving the rate constant easy you just need to send to manipulate K equals natural log and / T at time and so there though you can see the ratio only racial matters and if you substitute the eighty was this substitute eighteen with this your problem solved that makes sense because the concentration it's the concentration since first organs independent of the rate right actually concentration depends on rate first order it is the reason just because the rate actually is a proportional linearly proportional concentration but if you consider the rate and integration every equation you're getting just results it's kind of unique for its first order only the first order the racial matters the other order reaction zero and second the concentration matters the concentration required to solve any time or K you need a concentration but only for first-order the ratio it actually matters the specific concentration not necessary so that's the way it is and this question is a more algebraic it's more like not so much chemistry but it's a whole chapter not so much chemistry there's a bit about convene later but so far I seem that I was just math it is in many ways it is not but this is not a mathematical initial concentration large and final concentration it is because for any chemical reactions reactant should decrease and products for the increase so anyway let's plug in numbers a zero we don't know about leave the way it is leave that zero and eight he that's how I figure out we can substitute a tee with zips on a zero zero because this will end up canceled very nicely and the national autumn is divided by time which is 10.0 all the rest will be Z unit okay what are the unit the the ratio will make the unit cancel and when you take in the natural log this will not add a unit so the base is the whole thing have no units and time is in second second yes your final answer the unit will be inverse of the asks so that's the uniform for the first order by the way the first order rate constant unit it is inverse on time that's just because it's first order and so the rest just use a calculator that'll be easy part and sigfigs say fixed the when you do the natural log they're going in okay we need to watch for sig figs so here is the natural log of the 1.25 go in at one point 1/8 it's not 1.25 what is that it is 1.25 I should keep a 3 sig figs because the point a zero zero so I should have this doc divided by 10 points you know okay when you take a natural log the roux is number of sig figs and this is natural log but we use the same rule for the log of 10 tamago to a log of 10 rule we mentioned early in the afternoon he's a number of sig figs give me number decimal place in my child so here are the 3 sig figs that's very down the three things and when you take a log I should have a street decimal place a log of a 1.25 so let me do the log of 1.25 natural log I went one point two five my answer is a point two to three this is three sig figs I should have can use three decimals in my house or some point two to three I round to point to the street and divide by 10 and so my final 10.0.0.0 to two three [Music] and that's how we find the rate constant based on the time and based on the how much ratio decomposed the actual homework I don't have this kind of a very twisted question I have the question even give you how much decomposed you not give you the ratio of the concentration and have more thinking that involved but in the end I still have the answer done so that's basically use the first-order equation to solve question like this in the homework I only ask you to still work on the first-order reactions if you take a look at our PDF file the homework assignment the whole packet the later questions it has some we'd asked about the zero order or second order and that give you some varieties the diversity on the problems that even work out but it's very important I will ask mainly in the quiz or test and especially the test I only asked about a first-order reaction it's just more interesting and also has more manipulations on the numbers and also first order reaction the unit for the rate constant it's actually easier than zero the second order first order reaction the rate constant as a unit is inverse of time whatever time you use second or minute or hour is a Hubert's of the time they become the unit for your rate constant and the best practice here so we'll have just a little bit and the we've gone through a lot in the afternoon we're wrapping up this chapter so just get it back to the we learn about the three integrative a loss the street integrated rate law zeroes first and second saying order how do we use that to find out an order of reaction okay if there's a reaction has only one reactant we can easily use the graphing basically when you do the experiment you'll be collecting you'll be collecting the concentration one way or the other you collect concentration of the reactant at a different time and then you can do the graph you graph three ways you can grab the just concentrating always time treat this is why you this is the y and this is X times X you do a graph even just growing off give you a straight line a linear that means a to be zero order okay even call them freedoms at this time if this one if the concentration which time do not give you straight line then you try the next one you take a natural log of each of the concentration okay and you get a whole bunch of numbers these are based in the natural log of those concentrations now you use this graph again use this as X prime of X natural log of concentration as to why you take another shot you try one more time see if you've got a dinner if the natural log of the concentration at waste time even linear that it means that means the reaction is first order is the first order and so anyway the oh there's not one more let's say the this graph he knocking me linear nope nonlinear this part let's say which the with actual data IIT not a given linear way this graph either so that means it's not a zero order and it's not first order we have one more thing to try let's try one more time I even versus those concentrations in words it comes what slide is this on 30 Street do you have this one can you see this light by the way there's no reason I can't find it yeah we only see the three graphs not this table oh sorry this is the same actually it is the same file see what happened oh I see if you you haven't run the full screen and they will see the table first okay you have the run of one the full screen first they should unless you're using PDF the PDF out then you won't see the table yeah if you're opening the PowerPoint put us to de tous full screen you should be able to see that see this table first because these are same slide slide number no change just line numbers are sorry same okay no we just don't have the table oh the table is actually under the the Streamy graphs yeah by the way the textbook if you check the textbook see how something similar yeah I wish I could use a text the OpenStack textbook PowerPoint but so far I have not get to that yeah I still have to use my own its base the same same kind of material thank you yeah so beta to try three graphs and the reason because we have with those three equations all the way back yeah there we go so remember this we have those three equations right down it's either natural aqueous time given linear means the first order okay and if the universe concentration weeks time gave you a linear it means it's a second order and then if the concentration itself concentrated it's tough which time give a linear that is sorry the proof evidence that the reaction is zero-order so we do whatever I said the three waves graph is all come from these three sites come on these three sites the reason because it's come from the algebra the three order reaction they behave in the graph very differently you can either tell what order reaction based on which cross give you the linear okay by the way the non of data can give you all three gravity that's impossible mathematic that's impossible so if the reaction on one of these three it won't give one of the three graph a straight line yeah and the other will be curved so that's I must be speed back to the rest you can see the data you see here I need to the graph that's a ground for looks like any way I can or we undo the my scribbling and it's still here so basically this shows you the concentration which time is nominee all right so the three graph first one is the just concentrated playing concentration you graph at least time it steadily decline well not really steadily actually the ended beginning decrease very quick later is slowed down to decline because in the university we're a high slope high rates and so that I will not give me linear means is not there order and then you will try the second graph the natural law and natural always time give you beautiful very nice streamlining that means the reaction is a first order yeah and then of course you want to take one more just in case nothing unusual happen and so you do the inverse concentration with time well too bad it's not linear that means it's not second-order so it depends on which graph give you a straight line okay if the first graph gave me a linear that means concentration waste times is straight line that means is a zero order if this one giving any that means the first order and finally the inverse concentration if the inverse concentration with time giving an error that means is a second order and so that's theta to be how we use integrated rate law to determine order of the reaction this kind of question is hard for me to put in the test I'll just ask you the matching questions plainly it basically that you got a three kind of graph and you know which one's the linear means what what kind of order reaction because this kind of question to do in the test you need to really use a ruler use a graphing paper graph paper do that it's but in the lab report in mind you might have a this kind of question in your lab report and the good for homework next time I show you how to use Excel to the graph this graph you can either to quickly do with Excel without struggling the feeling dart living line linking the link all the dots and so forth so that's basically use integrated rate law to find the reaction rate I'm almost done for the day the last part half life let's go over this and it will push stop today getting to an almost last minute ID not planted masses long so the half-life what exactly half life they have my means when the concentration increased to half which means the when the eighty equals half of the a zero take time it is we call half-life and we have a way to represent half-life is T 1/2 subscript T 1/2 and the half-life status that's where this the for the all first order is very unique first order we have an equation even remember the equation T equals natural log of the a 0 over 80 [Music] divided by K that's the first order reaction and then the half-life the half-life means 18 equals half of the a zero and so that apart of give me to you right okay that's where it come from okay so T 1/2 is natural log of 2 divided by K this is actually a new equation over time we'll just use the way this we don't really ask you how to prove it question asked let's say you know the half-life what about ready to constant they can use this equation to solve you can write down like T decay time T 1/2 equals natural log of 2 so you can solve anything yourself ok or solve a half-life if with the answer given the half writing is easier have their own depends on rate constant and if the rate constant is very high which means very faster reaction high rate constant is a faster reaction then the half-life will be very short it'll make sense when the reaction very fast is sooner to consume the reactant it's faster is reduce the concentration to 1/2 so that's pretty much it for the first half of the chapter I don't see any more stuff okay there's more ground and I gotta explain this and so this is when you have a reactant I like the decomposition of dinitrogen pentoxide we have this and it's a first-order reaction so basically we find out every 24 minutes the concentration decreases to half okay the first start round full concentrate in point zero six after the first half-life concentration is 1/2 so this is time it is half-life because it is time for the concentration decrease to half what about another 24 minutes another half-life another half-life the remaining reactant nigel pentoxide it would be 1/4 of the initial because star from the way to look at this star from point zero three zero from here going in through one half-life one more one half-life even lose exactly half of this which is point zero one five so basically the same half-life it become half of the end of the first half life and this compare the orange no it is one quarter so basic the amount of remaining reactant we have a equation to use the ratio between the remaining product and the compare with the initial this in equals the number of one half-raised the power of the number of half-lives the number half-life is baited a time divided by half-life this equation no need to memorize but it actually very easy to use the time divided by half-life means how many half-lives so if the time is twice as a half-life then the ratio is two and so the remaining amount of our reactant is 1/2 square become 1/4 if you get a one more half five let's start from point zero one five another half life then the amount will be half over the point zero one five one zero zero seven five this is one eighth just be one eighth of the very beginning point zero six zero zero so that's where this is from coming from the more half-life the more longer time it passes they are less fraction of the reactant still remains so half-life the the basically one half-life you lose half to half lives become quarter three half-lives becomes one ace it's almost like a cutting pizza and by the way almost dinnertime it's time to talk about pizza so what are the half-life you basically count of time and you cut in Pisa it's almost peanut time so let's talk about FINA