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Lecture Comments (17)

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?

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

Transcription: Integrated Rate Law & Reaction Half-Life

Hello, and welcome back to; 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, N2O5/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 N2O5; 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 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 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, let's keep it red; not a problem.0388

Our example will be: The decomposition of di-nitrogen monoxide...di-nitrogen pentoxide; I'm 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 N2O5 (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 N2O5 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 N2O5 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 N2O5 to the first power, and that means your integrated rate law is (let's write it down here): the logarithm of the concentration of N2O5 equals -K, times t, plus the logarithm of the initial concentration of N2O5.0722

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 N2O5 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 N2O5 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 N2O5 at t=100 was 0.500; that is from the data table.1119

The concentration of N2O5 at t=200, also from the data table, equals 0.250.1133

OK, now, the concentration of N2O5 that we found, which was at t=150--it doesn't'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 A0/2?1296

When does the concentration of A equal A0/2?--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 A0, the initial concentration.1339

OK, well, the symbol is t1/2...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 A0.1363

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 A0, equals -Kt.1381

Now, let me rewrite this logarithm thing as (yes, that is fine) logarithm of A, divided by A0, 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=A0/2, right?1416

We will go ahead and put this A0/2 in here.1424

We get the logarithm of A0/2, over A0, equals -Kt.1428

The A0s 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 t1/2 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: t1/2=...let's just use the ln 2/K.1623

We are solving for K here; we know the t1/2, right?1642

So, K is equal to ln 2 over the t1/2.1645

Well, ln 2 is a certain number, and t1/2; 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 A0 (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 A0 (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 A0, so we write ln of 0.25 A0 = -0.0346t + ln, and our initial concentration was...actually, it doesn't even matter; it is just A0.1746

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 A0, over A0, equals -0.0346t.1792

These A0s 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

We'll see you next time; goodbye.1936