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Lecture Comments (13)

0 answers

Post by Soo bin Hwang on February 27, 2017

Dear Professor Hovasapian,

  Hello, I have learned the slope= -k in the "integrated rate law and reaction half-life" lecture. So for the example of second order, would
k= -0.06145 since m= 0.06145? But then t(1/2)= -1627 seconds. Can reaction time have negative value?

Thank you!

1 answer

Last reply by: Professor Hovasapian
Mon Feb 8, 2016 2:11 AM

Post by Gowrish Vaka on January 31, 2016

Greetings Professor Hovasapian,

The explanation for half-life (t1/2) of a second-order reaction on a University website problem said that the the formula for half life was as follows:

t1/2 = 1/(a*k) * 1/[A0]

It said that the variable "a" was the reaction order, in this case 2. I kept inputting my answer as 140 y, but the explanation stated that the answer was 70 y. I followed the problem's explanation exactly, using the given value of k and the given value for [A0], yet my answer was double the "correct value" for half-life since I didn't divide by 2 (a).

Do you know anything about this mysterious "a" variable in their given equation for the half-life of a second-order rxn? Or did the University make a mistake in their explanation?

Thank you

1 answer

Last reply by: Professor Hovasapian
Mon Aug 10, 2015 6:15 AM

Post by Jim Tang on August 10, 2015

without calculating any values, i feel the rate constant does vary (albeit slightly) when you choose diff. points to find your dy/dx. is this a big deal?

2 answers

Last reply by: Professor Hovasapian
Tue Apr 7, 2015 11:02 PM

Post by Lyngage Tan on April 7, 2015

hello professor im using a non programmable calculator scientific on solving this problems. i got the following values for example 1. K= 0.06126 and T1/2 = 1632 can this values also be considered as correct.

2 answers

Last reply by: peter alabi
Thu May 28, 2015 5:56 AM

Post by Datevig Daghlian on February 18, 2015

Dear Professor Hovasapian,

    As always, thank you very much for your lecture! these lectures are so very helpful! I was just wondering if it was possible to determine if a reaction is first order or second order without plotting the data (in other words, is it possible to formulaicly solve this?). Thank you very much and may God bless you!

Thank You,
George Daghlian

1 answer

Last reply by: Professor Hovasapian
Sun Jun 2, 2013 2:48 PM

Post by Kate Bevan on May 31, 2013

I'd just like to say how helpful I've found all these videos. It's amazing what a difference a good teacher can make!

Second Order & Zero-Order Rate Laws

  • Given Time and Concentration data, if a plot of Reciprocal Concentration (y) vs. Time (x) gives a straight line, then the reaction is 2nd order. The slope of this line is the Rate constant (k) for the reaction.
  • Given Time and Concentration data, if a direct plot of Concentration (y) vs. Time (x) gives a straight line, then the reaction is Zero-Order. The slope of this line is the negative of k, the rate constant.

Second Order & Zero-Order Rate Laws

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 0:22
    • Second Order
    • Example 1
    • Zero-Order
    • Summary for the Kinetics Associated with the Reaction

Transcription: Second Order & Zero-Order Rate Laws

Hello, and welcome back to; welcome back to AP Chemistry.0000

Today, we are going to continue our discussion of integrated rate laws.0004

Last lesson, we introduced the integrated rate law for a first-order reaction, and also the half-life formula for a first-order reaction.0007

Today, we are going to talk about second-order reactions and zero-order rate laws.0015

So, let's just jump right on in.0019

OK, so again, we are talking about (in order to simplify matters for ourselves, just so we get a good sense of the kinetics) a single reactant that decomposes to products.0024

That is what we have been doing this entire time.0035

Our fundamental reaction that we have been working with is the following products.0037

OK, now, the rate law for a second-order is the following--this is the differential rate law; and again, when we say "rate law," they mean differential rate law.0043

When they mean integrated rate law, they will specifically say "integrated rate law."0055

So, the rate is equal to -Delta;A/Δt; this rate symbol, which is the differential part in differential rate law, is equal to some constant, K, times the concentration of A to the second power.0060

That is the differential rate law for a second-order reaction.0079

Now, when we integrate this particular function, we end up with the following.0084

Let me write; this is the differential rate law; now, the integrated, which relates concentration as a function of time.0091

It is going to be 1 over the concentration of A, is equal to Kt, plus 1 over the initial concentration.0112

For a second-order reaction, your integrated rate law says 1 over the concentration of A is equal to the rate constant, times the time, plus 1 over the initial concentration of A.0124

So again, in this particular case, we can set it up as y=mx+b, but this time, the y is 1 over the concentration--the data that we take.0138

That is what is important; and the slope of the line that we get is the rate constant.0151

Notice, this time it is a positive slope--positive K instead of negative K.0158

What we are going to do is: we are presented with some standard raw kinetic data; we are going to find...we have the time; we have concentration; we are going to also change those concentrations to 1/concentration, and we are also going to do logarithm of concentration, because now, we have to check; now, we have 2 different types of equation, 2 different types of reaction orders.0161

Now, we have to check to see whether it is first-order or second-order.0185

We have to have 2 columns: one with the logarithm of A, one with the reciprocal of the concentration of A.0188

And again, you will see that in just a minute, when we do the example.0195

So, now, this is the integrated rate law; let's go ahead and work out the half-life for a second-order reaction.0199

And again, remember: half-life means that the concentration of A is equal to the initial concentration, over 2; it is when half of it is gone.0209

We will go ahead and write; that implies using this equation with those values: 1 over A0 over 2 equals Kt, plus 1 over A0; this ends up being 2 over A0, minus 1 over A0.0219

I flip this and move this over; it equals Kt; 2 over A0 minus 1 over A0 is 1 over A0, equals Kt.0247

Now, I will divide through by K, and I get that t of 1/2 is equal to 1/K, times the concentration (the initial concentration).0257

This is the formula for the half-life of a second-order reaction.0269

Notice, in this particular (you know what, this is not too clear; let me write this out; I need to erase this, and let me write it bigger over to the side; I think it will be a little more clear; let me get rid of this thing;, we get t is equal to 1/K, times the initial concentration)...0274

Now notice: in this case, the half-life of a second-order reaction not only depends on K, but it also depends on the initial concentration.0302

That was not the case with the first-order reaction; a first-order reaction does not depend on the concentration--it just depends on K; it's constant.0310

This is not constant; so the half-life--if you start with a certain amount, half of it is going to take a certain amount of time to get rid of.0317

Well, now that you have gotten through half of it, now that is a new initial amount.0325

In order to release half of that, that half-life is going to change; it is actually going to end up being longer.0328

So now, it depends on two things: the rate constant and the concentration.0334

There you go: we have that equation, which is the integrated rate law; we have this equation, which is the half-life; and, of course, we have this equation, which is the differential rate law.0339

These are the things that matter in kinetics: the differential rate law, the integrated rate law, and the half-life formula.0354

Pretty much everything else can be worked out from this information, and all of this, of course, comes from raw kinetic data: time, concentration.0360

OK, now, let's go ahead and do an example, because that is the best way that these things work.0368

Example: we have butadiene; butadiene dimerizes (and dimerization means that two molecules of something stick together); butadiene dimerization was studied, and the following kinetic data were obtained.0377

OK, so let's go ahead and write out the formula.0417

2 C4H6 turns into C8H12.0420

OK, so now, we have our time value; I'm going to go ahead and write it as one big plot.0429

Our time value; we have our concentration of C4H6, which is our reactant.0437

Now, what we want to do here is: we want to find the rate constant; we want to find the order of the reaction; and we want to find the half-life (I'm sorry about that; I should have actually told you what it is that we are actually going to be doing here).0446

So, the first thing we want to do is: we want to find the order of this particular reaction, based on the data that we are given.0459

Then, we want to find out what the rate constant is; and then, we want to find out what the half-life of the reaction is, based on this data.0467

So, t (let me go ahead and write this down) is 0, 1000 seconds, 1800, 2800, 3600, 4400, 5200, and 6200.0475

And what we have is 0.0100, 0.0625,, that is not right; that is 0.0100; so this is going to be 0.00625--the butadiene is diminishing--0.00476, 0.00370, 0.00313, 0.00270, and we are almost done--no worries--0.00241; and 0.00208.0498

OK, given this raw data, find the order of the reaction; find the rate constant; and find the half-life of this reaction, based on the fact that we started with .0100 moles per liter of this butadiene.0552

All right, so we need to check: is it first-order or second-order?0569

We need to plot both the logarithm of C4H6 versus time, and the reciprocal of C4H6 versus time, to see which one of these gives us a straight line.0574

If this gives us a straight line--the logarithm--it is first-order; if this one gives us a straight line, it's second-order.0591

That is how we do it.0597

OK, so here is what the data looks like.0599

Let's see; let's do logarithm first.0602

We get -4.605, -5.075, -5.348, -5.599, -5.767, -5.915, -6.028, and our last one is -6.175; reciprocals are a lot easier.0604

What we have is 100; 160; 210; 270; 320; 370; 415; and 481.0639

OK, so now we are going to plot (I'll do it in blue): we are going to plot this versus time, and we are going to plot this versus time, to see which one gives us a straight line.0654

Let's go ahead and take a look at what these plots look like.0669

I'm going to go ahead and put them next to each other; so, I'm not going to give you too much detail on these graphs--you can actually go ahead and plot them yourself on a piece of graph paper, or using a software like Excel or something like that.0671

What you are going to end up with, as it turns out--let's see here, this is going to be 200...yes; so let's do the logarithm of the butadiene versus time here; and let's do the 1 over the reciprocal of butadiene (which I'll just call A) over time here.0687

When I do this, the logarithm graph is something like this--definitely not a straight line.0714

When I do the reciprocal, I get a straight line; so I will have you confirm this, but this is exactly what happens.0721

So, because it is the reciprocal that gives us a straight line, it is a second-order reaction.0729

Second-order: so, we can go ahead and write the differential rate law: C4H6/Δt equals the constant times C4H6 to the second power.0737

We know it now; it is a second power; we derived it from the graph now, so we have an order of 2.0754

Now, the next thing we want to know is: what is the rate constant?0763

OK, well, the rate constant--the best thing to do is to go ahead and (let's see, how shall we do the rate constant)...let's just go ahead and use the integrated rate law.0767

We know that the second order is: 1, over the concentration of A, equals -K (actually, you know what?--what am I saying?--let me just...write it out, so we have it)...equals positive K times t, plus 1 over A0.0780

This is y=mx+b.0808

We just graphed the reciprocal over that; we got a straight line, so now, what we want to do is: we want to pick two points on that straight line, take the Δy over the Δx of those two points...0811

And again, the points need to be on the line; if the data points that we have in the original data, when we calculated it, are on the line, that is fine--you can use those data points.0825

But, if they are not, make sure that you are using points that are on the line.0835

When we calculate Δy/Δx from this graph, based on the kinetic data that we derived, we end up with the following.0838

Let's see: let's use a couple of points, actually; so let's use Δy over Δx; it it turns out, a couple of the data points do fall on the line.0848

I pick the first and the last; so it's going to be 481 minus 100, over 6200 minus 0, which gives me 0.06145, and the unit is going to be liters per mole-second.0862

I know the units for the rate constants don't really make sense; they tend to change, depending on what order it is; it is actually the number that matters.0882

This is our rate constant; and now, we want to find t1/2.0889

OK, well, t1/2 of a second-order reaction is 1/K, times the initial concentration.0897

It is equal to 1 over 0.06145, and the initial concentration was 0.0100; so, when we do that, we get 1627 seconds.0909

There you go: raw data: time and concentration; we calculated the logarithm of concentration; we calculated the reciprocal of the concentration; we plotted ln of concentration versus time for one graph; we plotted reciprocal of concentration versus time for a second graph.0926

We saw which one of those graphs gave us a straight line; in this particular case, it was the reciprocal versus time that gave us a straight line: second-order.0947

Because it is second order, I go ahead, and I can write the differential rate law, if I need it.0955

We need the rate constant; so I just found a couple of points on that straight line; I took the slope, y2-y1 over x2-x1.0961

I ended up with the rate constant; and then, because I have the half-life formula, and I have the rate constant which I found, and I have the initial concentration (which is the initial concentration from raw data), I was able to calculate the half-life of this reaction--very, very straightforward.0969

Nothing altogether too complicated: again, differential rate law, integrated rate law, half-life--those three things are what is important, as far as the kinetics of a reaction are concerned.0985

OK, that was second-order reaction.0996

Let's talk about a zero-order reaction.0999

All right, so a zero-order reaction is exactly what you think it is; it says that the rate (well, let me actually write it in the top here)...1001

Zero-order reaction: well, the rate (which is equal to -Δ of some reactant/Δt) equals KA to the 0 power, which is K times 1, which is K.1011

In other words, the rate itself is constant; the rate doesn't change.1032

The integrated rate law: when we integrate this particular function, you end up with the following.1039

You end up with the concentration of A (you know what, I am not even going to write these anymore; they are driving me crazy), equals -Kt, plus A0.1045

The concentration at any given time is equal to minus the rate constant, times time, plus the initial concentration.1060

Or, it equals the initial concentration, minus rate constant, times time; it's diminishment.1067

Again, y=mx+b.1072

So now, if you had some raw kinetic data, and you just plotted the concentration versus the time, and you got a straight line, that is a zero-order reaction; there you go.1077

Now, half-life: A=A0/2; so let's put it in here.1088

A0/2=-K times t, plus A0.1097

A0/2 minus A0=-Kt.1111


Therefore, t1/2 equals A0 (these lines--they always show up in the most inopportune moments), divided by 2K.1125

Let's do it in red.1143

We have this; we have this; and we have this.1145

For a zero-order reaction, the differential rate law says that the rate is equal to K, the rate constant; it is a constant.1155

The integrated rate law says that the concentration at any time A is equal to -Kt, plus A0.1163

The half-life formula says that the initial concentration, divided by twice the rate constant, will give you the amount of time it takes for half of whatever you started with to disappear.1169

Notice, in this case: again, it depends on the initial concentration, and it depends on K.1181

Now, you might be wondering what a zero-order reaction is; where would you run across a zero-order reaction?1187

When would you have a reaction that doesn't actually depend on the concentration?1192

Zero-order reactions show up in places where...usually in things that involve catalysis; and, extended to the biological realm, that means enzymes.1197

So, for example, if you have--let's say just 10 enzymes, catalyzing a reaction; well, if all of those 10 enzymes are busy (meaning if they are full, doing what they are supposed to do), it doesn't matter how much more reactant you actually put in there.1210

It is called a substrate--the thing that the enzyme grabs onto and fills with; so it doesn't matter--once those 10 enzymes are full, you can't make the reaction go any faster, because it doesn't matter how much concentration--how much more of the reactant--you actually put in; the reaction rate is controlled by how many enzymes are actually doing the work and how fast they are doing it.1229

So, under catalysis conditions, the reaction rate actually depends on the catalyst, not so much the concentration of the reactant.1252

Now, this isn't always the case; we're just saying that, when you run across a zero-order reaction, that more often than not, it is going to be in a catalysis situation.1260

That is just a qualitative bit of information that you should know, if it happens to come up.1268

OK, so now, let us take a global view of what it is that we have done with zero-, first-, and second-order reactions.1273

Let's summarize what we have, and it will give us a good, nice one-page summary of reaction kinetics and how to deal with them.1280

OK, so this is going to be a summary for the kinetics associated with the reaction aA going to products.1288

So again, this entire time we have been talking about a single reactant decomposing into products.1314

It diminishes, and it forms something.1320

How can we figure out the rate?1322

All right, so here, let's go ahead and make ourselves a little table; and we will go ahead and put a line here, and we will say first-order, second-order, zero-order reaction.1324

OK, the differential rate law for a first-order reaction is: the rate is equal to K, times the concentration of A to the first power.1345

The rate is equal to K, times the concentration of A to the second power.1360

And, the rate equals just plain old K.1366

That is the differential rate law.1369

OK, our integrated rate law: we have, for a first-order reaction--we have: ln of A is equal to -Kt, plus (oops, I always do an h for ln; I don't know why) ln of A0; again, concentrations, kinetics--we are talking about concentrations.1371

Now, the rate law for a second-order is reciprocal: 1 over A is equal to Kt, plus 1 over A0.1395

And, this one is: A is equal to -Kt, plus A0.1408

These are the integrated rate laws.1414

OK, a plot giving a straight line is going to be ln of A versus t, y versus x.1416

So, the plot of ln(A) for a first-order reaction--the plot of ln(A) versus t gives a straight line.1440

Here, it is 1/A versus t that gives a straight line.1445

Here, it is A versus t that gives a straight line.1450

And again, it is just the left-hand side of the integrated rate law: ln(A), 1/ check the data.1456

Slope: well, the slope is -K, K, -K of the particular line, if you happen to get one.1464

In other words, the slope is that; so you can find K; that is how you find K.1478

And, half-life: the t1/2 is equal to ln(2) over K; notice, for a first-order reaction, it is kind of interesting; it does not depend on concentration at all; it is constant.1483

The half-life for a second-order reaction is equal to 1/K times the concentration of A0.1503

It depends on the initial concentration.1512

And, t1/2 of this one equals the initial concentration over 2K.1515

So, this is everything that we have done: first-order, second-order, zero-order reaction rates.1521

When we are given kinetic data, which involves time and concentration, we have to plot the logarithm of the concentration versus time, the concentration versus time, and 1 over the concentration versus time.1528

Whichever one of these graphs gives us a straight line, that is the order of the reaction.1542

When we have the order of the reaction (let's say, for example--boom!--we come up with second-order), we know the differential rate law; we can write the integrated rate law; we can find K from the slope; and we have the half-life formula.1547

We can tell you what the concentration is at any time; we can tell you what the concentration is going to be at any percentage that we want along the way, instead of the time--all of this information right here.1561

This is a summary of standard kinetic operating procedure.1573

We'll go ahead and leave it at that; next time, we will talk about temperature dependence of reaction rates, and talk about something called the Arrhenius equation, and activation energy, and things like that.1581

But until then, thank you for joining us for AP Chemistry and kinetics, and thank you for joining us at

Take care; goodbye.1599