For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

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### Integrated Rate Law & Reaction Half-Life

- Differential Rate Laws express Rate as a function of concentration
- Integrated Rate Laws express Concentration as a function of Time
- Reactions of orders 0, 1 & 2 have different Integrated Rate Laws
- Each Integrated Rate Law can be put into y = mx + b form
- Each Integrated Rate Law yields a different expression for the Half-Life of a reaction

### Integrated Rate Law & Reaction Half-Life

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Kinetics 0:52
- Integrated Rate Law
- Example 1
- Example 2
- Half-life of a Reaction
- Example 3: Part A
- Example 3: Part B

### AP Chemistry Online Prep Course

### Transcription: Integrated Rate Law & Reaction Half-Life

*Hello, and welcome back to Educator.com; welcome back to AP Chemistry.*0000

*In today's lesson, we are going to be talking about the integrated rate law and reaction half-life.*0004

*Last time, we discussed the differential rate law that says that the rate of a reaction is proportional to the concentration of reactant, to some power.*0009

*We used some concentration-time data to actually work out what that rate law was.*0021

*Now, what we are going to do is: we are going to use calculus; we are actually not going to do the derivations, but using the techniques of calculus, we can take that differential rate law; we can convert it to an integrated rate law; and then, instead of the differential rate law (which expresses rate as a function of concentration of reactant), we are going to express concentration as a function of time.*0027

*Let's jump in and see what we can do!*0049

*OK, so, as we said, the differential rate law expresses rate as a function of reactant concentration.*0053

*So, we had something like this: the rate equals (and there is this interesting symbol)--let's just use di-nitrogen pentoxide, N _{2}O_{5}/Δt.*0059

*And again, this is just a symbol expressing rate; and this negative sign is because, in chemistry, they prefer to have rates being positive.*0073

*Because a reactant is diminishing, the final minus the initial concentration is going to give you a negative value here, so a negative and a negative gives you a positive.*0080

*I wouldn't worry too much about this; what is important here is this thing--this K, times the N _{2}O_{5}; and in this particular case, it would be some value of n, and we actually calculated it before in a previous lesson: n=1.*0088

*So, this is a first-order reaction.*0105

*The decomposition of di-nitrogen pentoxide is a first-order reaction.*0107

*Well, the integrated rate law (this is the differential rate law) expresses concentration as a function of time, which is very, very convenient.*0112

*Given a certain amount of time--10 seconds, 50 seconds, 2 minutes--what is the concentration at that moment?--very, very convenient: as a function of time.*0139

*Now, the differential in differential rate law (or what we call just the rate law--when you hear "the rate law," they mean the differential rate law--they mean this thing)--the "differential" part comes from this symbol, this Δ; in calculus, they become differentials.*0149

*They become something that looks like this: d...let's say dA/dt.*0165

*When you fiddle around with this expression, and then you perform an operation called integration on it (the symbol for which is something like that), that is why it is called the integrated rate law; that is where these names come from.*0170

*If you go on in chemistry, you will understand--you will actually do the derivations; it is actually quite simple, but we will skip it for our purposes.*0181

*OK, so given a differential rate law, we can find an integrated rate law.*0187

*Or, if we have an integrated rate law, we can actually go backwards and find a differential rate law.*0193

*So, since we are going to deal with first-order reaction, let's write a reaction: aA decomposes into products; and the products themselves are not altogether important--what is important is that we have a single reactant on the left side of the arrow.*0198

*So, the rate is equal to (I'm going to skip the symbol; I'm just going to say) K, times the concentration of A, to the first power.*0216

*When we fiddle with this thing (actually, you know what, let me go ahead and write...it is actually pretty important: -ΔA/Δt; we might as well be consistent): K times the concentration of A to the first power; when we integrate this expression, we end up with the following.*0229

*We end up with: the logarithm of the concentration of A equals -K, times t, plus the logarithm of the initial concentration.*0257

*This (oops, try to get red ink here...red ink) is our integrated rate law.*0272

*For a first-order reaction, this is our differential rate law, and this is our integrated rate law, OK?*0281

*This says that the logarithm of the concentration at any given time, t, is equal to minus the rate constant, K (which is the same K here), times the time, plus the logarithm of the initial concentration.*0287

*Now, take a look at this equation here: this here (let's just call it y)...let's call this -K m; this t is your x, plus b.*0303

*So, this equation--the integrated rate law actually takes the form y=mx+b.*0315

*As it turns out, if you have some time and concentration data (which we will do in a minute), if you have a certain time and you are measuring concentrations at different times--if you actually plot, not the time and the concentration data, but if you plot the logarithm of the concentration, versus the time (the time on the x-axis, and the logarithm of the concentration); if you end up with a straight line, that actually tells you that this is a first-order reaction.*0322

*Just given some straight kinetic data (time, concentration), you plot the time on the x-axis and the logarithm of the concentration on the y-axis; if you get a straight line, that tells you that you have a first-order reaction.*0350

*That tells you you could write this and this; it automatically gives you the integrated and the differential rate laws.*0363

*That is what is so great about this particular one.*0370

*It is very practical, because again: at any given time, t, you can measure the concentration of your particular reactant.*0372

*OK, so let's go ahead and do an example, and I think it will make more sense, as always.*0381

*Let's go back to...no, let's keep it red; not a problem.*0388

*Our example will be: The decomposition of di-nitrogen monoxide...di-nitrogen pentoxide; I'm sorry...at a constant temperature (at a constant t) gave the following kinetic data.*0394

*OK, so we have--let's see: t is going to be in seconds, and then, of course, our concentration...*0426

*Now, if you don't mind, I actually tire of doing the whole brackets for concentration; I just like put parentheses for concentration; I'm going to sort of go back and forth between them--I hope you understand what it means.*0433

*When we are talking about kinetics, a parentheses around a species (like di-nitrogen pentoxide) is actually going to mean concentration.*0445

*Normally, it is true: a square bracket is concentration; I shouldn't be so reckless, but I am.*0453

*OK, so the concentration of N _{2}O_{5} (and this is in moles per liter, as always--concentration is in moles per liter when you see it like that)...*0460

*OK, so we are going to do 0 seconds, 50 seconds, 100 seconds, 200 seconds, 300 seconds, and 400.*0470

*The data we have is: we start with an initial concentration of 0.100--four decimal places; 0.0707; 0.0500; 0.0250; 0.0125; and 0.00625.*0480

*This is the kinetic data that we collected.*0505

*At different time intervals, we measured the concentration of reactant left in the flask, and these are the concentrations that we got.*0509

*Now, we want to know what the order of this reaction is, and we want to know what the rate constant is.*0515

*Let's go ahead and start.*0521

*Now, what we are going to do in order to find the order: we said that we are going to plot the logarithm of the concentration, versus the time (y versus x).*0524

*Whenever you see something versus something, it is always y versus x.*0534

*So, this is going to be the y-axis; this is going to be the x-axis...actually, the logarithm of this is going to be the y-axis.*0537

*Let's make a new table and draw a little line here; and this time, I'll do the table in blue.*0543

*It is going to be the same thing; the time is going to be the same thing--it's going to be 0, 50, 100, 200, 300, and 400.*0550

*Now, we are going to take logarithms of these numbers; so, when we take the logarithms of those numbers, we end up with the following data.*0560

*This is going to be ln of the N _{2}O_{5} concentration in moles per liter.*0569

*We end up with -2.3, -2.6, -3, -3.7, -4.4, and -5.*0575

*Now, we are going to plot this.*0589

*We have our y-axis and our x-axis: this is our time; this is going to be our logarithm of the N _{2}O_{5} concentration.*0592

*We will just put some numbers up here: so, -6, -5, -4, -3, -2; so, -6, -4, -2, and we will go with 100, 200, 300, 400; this is 100 seconds (oops, all of these lines are showing up again--this is always happening down at the bottom of the page--it's kind of interesting).*0604

*All right, so 100, 200, 300, and 400 seconds: when we plot this data, we actually end up getting something which is (let's go down to about right there; I guess that is pretty good), believe it or not--we actually do end up with a straight line, with all of these different points at various points.*0633

*Now, I should let you know: this is kinetic data; kinetic data is not always going to line up in an exactly straight line--it's not going to be that the line is going to go through every single point exactly.*0661

*But, it is going to be pretty good; you are going to get a linear correlation.*0672

*So, you should be able to draw a line; in other words, you are not going to get points all over the place.*0676

*You are going to be able to draw a line through many of the points.*0681

*Don't think that is has to be exact; real kinetic data, like real science, is not...doesn't fall into nice, perfect, clean square boxes.*0685

*There is a little bit of deviation; don't let that confuse you.*0697

*The idea is: when you plot this, you get something that is a straight line--pretty much a straight line.*0700

*OK, so because it is a straight line--because this kinetic data gives you a straight line--that implies that you have a first-order reaction.*0706

*That means that your rate law is K times the N _{2}O_{5} to the first power, and that means your integrated rate law is (let's write it down here): the logarithm of the concentration of N_{2}O_{5} equals -K, times t, plus the logarithm of the initial concentration of N_{2}O_{5}.*0722

*That is what is happening here: the initial concentration was 0.100; so we have identified that it is first order.*0757

*We can go ahead and write a differential rate law for it; we can write an integrated rate law for it; everything is good.*0769

*Now, the second part is: how do we find the rate constant?*0776

*OK, well, we have the straight line; the rate constant is (let me go ahead and move to the next page) the slope, the negative slope, of that line.*0779

*So, the slope equals -K.*0791

*So now, all we have to do: we take 2 points on that line--it doesn't matter which 2 points; now, when you are taking points on a line (on a straight line for a graph), you can only use points on that line.*0799

*We just said that real kinetic data doesn't always give you an exactly straight line; so you are going to have a best fit line, and you are going to do a least squares fit on that line.*0813

*Well, when you pick 2 points randomly on there, you don't pick 2 data points, unless those data points actually happen to be on the line.*0825

*If they are not, you pick points on that line.*0833

*There you go: you have your 2 points; you do Δy/Δx; set it equal to -K; and you solve for K.*0836

*When we take a couple of points on the graph that we just did in the previous slide, we end up with the following.*0847

*We end up with: -6.93x10 ^{-3}, and the unit is per second, equals -K.*0853

*Slope equals -K; negative equals negative; so K, the rate constant, equals 6.93x10 ^{-3} per second.*0867

*That is our rate constant.*0879

*What we did: given kinetic data (time and concentration), we plotted time on the x-axis, and the logarithm of the concentration on the y-axis.*0881

*If you get a straight line, it's a first-order reaction; if you don't get a straight line, it is not a first-order reaction; so it works both ways.*0892

*If you do get a straight line, the slope of that straight line is negative of the reaction constant, K.*0902

*OK, so now, let's move on to our second example, or actually a continuation of this first example.*0914

*Using the data and results from the previous example, find the concentration of N _{2}O_{5} at t=150 seconds.*0926

*OK, well, so now we have K; we have the integrated rate law, which is: the logarithm of the concentration is equal to -K, times t, plus the logarithm of the initial concentration.*0958

*What we are looking for is the concentration of N _{2}O_{5} at 150 seconds.*0979

*Well, if we look at our graph, or look at our data table, we have t at 100; we have t at 200.*0984

*We can't just split the difference to find the concentration in between there at 150.*0991

*You will see that in just a minute; but let's just solve our equation.*0996

*We end up with logarithm of the concentration of A (because we are looking for the concentration of A--this is the dependent variable here--this whole thing), equals -K (K was 6.93x10 ^{-3}--let me make this a little more clear), times t (well, at 150 seconds), plus the natural logarithm of the initial concentration (the initial concentration was 0.1000).*1000

*OK, so let's go ahead and do this: equals...well, when you do the math, you end up with: ln of the concentration of A equals -3.3425, and in a logarithmic equation, you solve by exponentiating both sides.*1036

*So, let's go ahead, and I'll just write "exponentiate," which means e to the ln of A equals e to the -3.3425, and you end up (well, the e and the ln go away) with a concentration of A=0.0353 moles per liter.*1059

*There you go: we used the kinetic data to find the integrated rate law, to find the reaction constant...the rate constant; and then we used that equation to find the value of a concentration at any given time (in this case, 150 seconds).*1093

*Now, let's take a look at some values.*1111

*The concentration of N _{2}O_{5} at t=100 was 0.500; that is from the data table.*1119

*The concentration of N _{2}O_{5} at t=200, also from the data table, equals 0.250.*1133

*OK, now, the concentration of N _{2}O_{5} that we found, which was at t=150--it doesn't equal...it's between 100 and 200; it isn't this plus this, divided by 2; it isn't .375; it isn't right down the middle.*1146

*That is the whole idea--that this relationship is logarithmic, so we found this to be equal to 0.0353, not 0.0375.*1165

*This is not right; hold on a second--I think we have some values wrong here; this is .1; this is .05; 0.0500; 0.0250; .0375.*1185

*It is not half of that; you actually have to use the integrated rate law to find the rate--it's this--the concentration--not that.*1204

*Be very, very careful; don't just think you can look at the data and say, "OK, well, half the time between 100 and 200, at 150, is half the concentration between .5 and .25, which is .0375"; that is not the case.*1214

*You have to use the integrated rate law, which gives you .0353; the concentration is actually less--more diminished than you would have expected.*1226

*OK, so let's go back here; now that we have taken care of the integrated rate law, we also want to talk about something called the half-life.*1236

*Let's give a definition.*1247

*The half-life of a reaction is the time required for the concentration of a reactant (in our case, the reactant) to reach 1/2 its original value.*1256

*In other words, when does A equal A _{0}/2?*1296

*When does the concentration of A equal A _{0}/2?--when is half the reactant used up?*1304

*That is sort of an important point; it is a line of demarcation; it gives us a standard--a reference, if you will--a reference point: half-life; it's pretty common.*1309

*We also speak about half-life in radioactive decay: how much time does it take for half of a particular nuclide, radioactive isotope, to completely disappear by half of its original amount?*1318

*So, in kinetics, we talk about the same thing: the half-life of a reaction is the time required for the concentration of a reactant to reach 1/2 of its original value.*1331

*In other words, concentration of A equals 1/2 of A _{0}, the initial concentration.*1339

*OK, well, the symbol is t _{1/2}...half-life; the time.*1345

*Now, let's go ahead and derive our half-life equation.*1359

*So, we have our equation, our integrated rate law: the logarithm of A equal -Kt, plus the logarithm of A _{0}.*1363

*Well, let's see: how shall we do this?*1378

*OK, let me go ahead and bring this over here; so we have ln of A, minus ln of A _{0}, equals -Kt.*1381

*Now, let me rewrite this logarithm thing as (yes, that is fine) logarithm of A, divided by A _{0}, right?*1397

*It is the property of a logarithm; the logarithm of something, minus the logarithm of something else, is the logarithm of that first something divided by the second something.*1408

*Equals -Kt; and now, we will use the fact that A=A _{0}/2, right?*1416

*We will go ahead and put this A _{0}/2 in here.*1424

*We get the logarithm of A _{0}/2, over A_{0}, equals -Kt.*1428

*The A _{0}s cancel; we end up with ln of 1/2 = -Kt.*1439

*We get that t is equal to ln of 1/2, over K.*1447

*So, the half-life--when half of the concentration has completely been depleted, the time that it takes for that to happen--is equal to the logarithm of 1/2, over the rate constant.*1454

*Take a really, really close look.*1469

*We can also write this another way: we can also write it as logarithm of 2 (I'm sorry, this is -K, right?--because the minus sign...) over K, and that just has to do with how we derive the equation.*1471

*Either one of these is fine.*1488

*This one or this one: you can use either one; it is just a question of: logarithm of 1/2 is a negative number, negative over negative is positive, logarithm of 2 is positive, so either one is absolutely fine; it's just that, the way I derived it, we came up with this one.*1490

*Now, notice what is important about this: the time it takes for a first-order reaction to be halfway complete, to be 50% depleted, does not depend on the initial concentration.*1507

*It doesn't depend on concentration at all; it just depends on a number, the logarithm of 1/2, and it depends on the rate constant.*1518

*So, it is a function of the rate constant; for a first-order reaction, the half-life does not depend on concentration; that is very, very important.*1525

*But, we will summarize this towards the end.*1534

*OK, so now that we have that, let's go ahead and do another example.*1537

*OK, a certain first-order reaction has a t _{1/2} equal to 20 minutes--has a half-life of 20 minutes.*1550

*That means it takes 20 minutes for a certain reaction to be halfway complete.*1568

*A: We would like you to calculate the rate constant.*1571

*B: We would like you to find out how much time is required for the reaction to be 75% complete.*1583

*OK, so we have a certain first-order reaction; it has a half-life of 20 minutes.*1610

*We want you to calculate the rate constant, and we want you to tell us how much time is required for the reaction to become 75% complete.*1614

*All right, well, part A: we have: t _{1/2}=...let's just use the ln 2/K.*1623

*We are solving for K here; we know the t _{1/2}, right?*1642

*So, K is equal to ln 2 over the t _{1/2}.*1645

*Well, ln 2 is a certain number, and t _{1/2}; this is just ln 2 over 20 minutes; and what we end up with is 0.0346 per minute; that is the rate constant--nice and simple.*1654

*So, you can find a rate constant from the half-life, or you can find a rate constant from the kinetic data--the slope.*1672

*Either one is fine; it just depends on what you have at your disposal.*1679

*Now, part B: well, 75% complete of the reaction--what does "75% complete" mean?*1684

*That means that your final concentration is only 25% of your original.*1693

*75% means that your final concentration of A is equal to 0.25 A _{0} (that means you only have 25% of the initial concentration left; 75% of it is gone).*1701

*We just plug that in to our integrated rate law.*1714

*So, let's write our integrated rate law again; we write it over and over again, until it becomes second nature.*1720

*-Kt + ln of A _{0} (and you notice, I am getting, actually, very, very lazy; I am not even writing the parentheses anymore).*1726

*We are talking about kinetics; we are talking about rate laws; we are talking about concentrations in moles per liter; I don't want to hammer certain things that are self-evident.*1735

*OK, so now, we said that A=.25 A _{0}, so we write ln of 0.25 A_{0} = -0.0346t + ln, and our initial concentration was...actually, it doesn't even matter; it is just A_{0}.*1746

*We know that it is 75% complete, so we don't actually have to have a number; we just know that this is .25% of that, because it is 75% done.*1776

*OK, so now, we move this over; we put them together; it looks like this.*1784

*The logarithm of 0.25 A _{0}, over A_{0}, equals -0.0346t.*1792

*These A _{0}s cancel; we end up with ln of 0.25, divided by -0.0346, equals t.*1803

*When we solve this on our calculator, we end up with 40 minutes; very nice.*1815

*Now, this another way to actually do this, based on the percentage that they asked.*1824

*They said 75% complete; 75% complete--that means 50% is gone (half-life), and then to 75% means that another half-life (half of the half) is gone.*1829

*So, you can think of it as the first 20 minutes, and then another 20 minutes, because again, the half-life of a first-order reaction is constant.*1842

*It doesn't depend on concentration; it only depends on the K.*1851

*In this case, they say it's 20 minutes; the half-life of any amount is always going to be 20 minutes for this reaction.*1856

*It means 20 minutes must elapse before half of what is left over goes away.*1864

*Then, half of that left over goes away--it takes another 20 minutes; half of that--another 20 minutes.*1869

*But, I think the best thing to do is just to use the equations, instead of sort of reasoning it out that way.*1874

*When you get a little bit more comfortable with it, that is fine; you can look at 75%, and you can say, "Oh, that is 2 half-lives; 2 times 20 is 40."*1879

*Just go ahead and plug it in, and everything will work out fine.*1887

*OK, so we have taken care of the differential rate law for a first-order.*1890

*We have talked about integrated rate law for a first-order.*1894

*We have talked about the half-life of a first-order reaction.*1898

*All of this, basically, comes from standard kinetic data, where you, at different points of time, measure the concentration of the reactant; you plot it, and then the plot will actually tell you whether you are looking at a first-order or a different kind of order reaction.*1902

*So, next lesson, what we are going to do is talk about second-order and zero-order reactions.*1919

*We will just continue with the same theme: new rate laws, new integrated rate laws, new formulas for half-life; and then, we will summarize everything.*1924

*Thank you for joining us here for AP Chemistry, and thank you for joining us at Educator.com.*1931

*We'll see you next time; goodbye.*1936

0 answers

Post by Kaye Lim on April 8, 2017

Greeting sir,

Why is it that when we graph [A] vs. time, we get a curve (which means the depletion rate of A changes at different time), but when we graph ln[A] vs. time, we get a straight line indicating that the depletion rate of ln[A] is constant at all time? Why is it for the same numerical values of A, we get a straight line vs. time just by ln the A values?

How did people find a mathematical way to get A to be linear to time? Did they go through trial-and-error process before they arrived at the conclusion that ln would give a linear graph?

1 answer

Last reply by: Professor Hovasapian

Mon Aug 10, 2015 6:33 AM

Post by Jim Tang on August 10, 2015

should the rate constant be in (mol)/(L-s)?

0 answers

Post by Jim Tang on August 10, 2015

wait, there's no units on -kt. how can you add something without a unit to a unit with mol/L and end up with mol/L. thanks.

1 answer

Last reply by: Professor Hovasapian

Mon Apr 27, 2015 2:35 AM

Post by chitra banarjee on April 25, 2015

Can the k value be negative? And when stating the k value, are units required?

Thank you.

1 answer

Last reply by: Professor Hovasapian

Thu Mar 12, 2015 4:23 AM

Post by Brandon Fell on March 11, 2015

you are an awesome teacher

the enthusiasm is nice

1 answer

Last reply by: Professor Hovasapian

Thu Jan 8, 2015 2:13 AM

Post by Stephen Donovan on January 7, 2015

Isn't ln([A]) = -kt + ln([A0]) equivalent to [A] = [A0]e^(-kt)

1 answer

Last reply by: Professor Hovasapian

Tue Dec 23, 2014 6:36 PM

Post by David Gonzalez on December 22, 2014

I have a second question professor: when does the "T 1/2 = 0.693 / k" come into play? Is that for first order half life problems as well?

1 answer

Last reply by: Professor Hovasapian

Tue Dec 23, 2014 6:42 PM

Post by David Gonzalez on December 22, 2014

Hi Professor Hovasapian, thanks for the great lecture.

It seems that all of my sources are giving me different equations that mean the same thing, so I just wanted to get your opinion on the matter to see which one serves this purpose the best.

The integrated rate law you showed is ln [A] = -kt + [A0]

My question: is that the same as this? ln [A]0 / [A] t = kt

If so, what about this one? ln [A]t / [A]0 = -kt

Do all of these lead to the same answer?

2 answers

Last reply by: Rebecca Bulmer

Wed Jun 5, 2013 12:46 PM

Post by Rebecca Bulmer on June 3, 2013

please explain how you got Ln^2/20 = 0.0346 min ^-1?