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Lecture Comments (37)

0 answers

Post by Adris Kowlessar on April 28 at 12:01:38 PM

how did you arrive at the answer for the activation energy i think it is algebriac on my part

1 answer

Last reply by: Professor Hovasapian
Sun Jul 3, 2016 7:35 PM

Post by Jeffrey McNeary on July 2, 2016

The equation you described at 20:20, and the derivation you began at 20:40, the equation being Rate constant = Ae^(activation energy/R*Temperature), has an uncanny resemblance to a lot of the equations in calculus based physics. At least thats how I remember it. Physics was a long time ago...

3 answers

Last reply by: Professor Hovasapian
Fri Jan 8, 2016 3:42 AM

Post by Tammy T on January 2, 2016

Hello prof. Hovasapian,

I have a few questions after watching your great lecture.
-Why is it that K vs T graph won't give a straight line, but lnK and 1/T does?

-I see that it is like the common theme in chemistry that we sort of always trying to mold the 2 variables into linear relationship. Why is it? Is it for the slope m so that we can tell the rate of one variable changing when other variable changes? If so, other type of equation beside linear won't be able to tell us the rate of changing?

-In this lecture, @22:48, when you ln both side of Arrhenius equation to get the equation in y=mx+b form, you said that lnK and 1/T has linear relationship. Is it experimentally determined that the relationship of lnK and 1/T is linear or is it because it is in the linear form y=mx+b so it is linear?

Thank you very much!

4 answers

Last reply by: Derek Marshall
Fri Aug 21, 2015 8:22 AM

Post by Derek Marshall on August 14, 2015

Hi Professor Hovasapian,

Enjoying the AP chemistry lectures. One question I had was when comparing two rate constant values (k1 and k2) and Temperatures (T1 and T2) why is it that you subtract lnk1 from lnk2 instead of setting both equations equal to Ea? I know that this makes lnk1 and lnk2 a ratio and everything works out, but what is the underlying meaning of subtracting them?

Thanks for the help,
Derek Marshall

1 answer

Last reply by: Professor Hovasapian
Mon Aug 10, 2015 6:24 AM

Post by Jim Tang on August 10, 2015

hey raffi!

30:34. not sure if i am mistaken, but isn't (1/s)/(1/K) the same as (K/s)? why is it only in K?

1 answer

Last reply by: Professor Hovasapian
Fri Mar 14, 2014 7:16 PM

Post by Daniel Nguyen on February 23, 2014

How did you get 1.0x10^4 J/mol if you multiplied -1.2x10^-4 K and -8.31 J/mol-K?

0 answers

Post by Tim Zhang on February 18, 2014


1 answer

Last reply by: Professor Hovasapian
Mon Jun 3, 2013 6:36 PM

Post by Rebecca Bulmer on June 3, 2013

Professor Raffi, I have been watching your videos since I started chemistry last fall and they are simply excellent!! I would also like to add that I received a B+ in my first 200 level chemistry course, through distance ed, and i believe your lectures have a great deal to do with that!! Only 4 more chemistry courses to go.... So again, thank you so much!!

5 answers

Last reply by: Willy Hese
Sun May 26, 2013 10:29 PM

Post by Willy Hese on May 26, 2013

Hi Professor,

my maths is really poor

can you help me please?

From this equation - K = Ae^Ea/RT

how do I use the rate, Ea, (T1 and T2) to solve the pre exponential factor? I have done all the calculations from the equation - But do not not know what to do with the figures I have to get A

1 answer

Last reply by: Professor Hovasapian
Tue May 21, 2013 8:25 PM

Post by Nawaphan Jedjomnongkit on May 18, 2013

Thank you for the lecture, so the main idea is that all of the reaction will increase rate when increase temperature right? So what happen to the exothermic reaction when we add the heat or increase temperature it will disturb the equilibrium and the reaction want to relive in the way that reverse the reaction. So in this case will the product increase because of higher rate of reaction or will the product decrease because want to relive the stress of equilibrium?

2 answers

Last reply by: Antie Chen
Mon Apr 29, 2013 7:23 PM

Post by Antie Chen on April 28, 2013

Hello Raffi, in the 13:15, I don't understand the graph. Why the T2 is lower than Ti, and what's the meaning of y axis?

In addition, in the Arrhenius equation, Are the A and the Ea constant for a specific equation?

0 answers

Post by Sdiq Al-atroushi on April 24, 2013

Hello Professor Hovasapian,

I want to thank you for all the hours you have spent teaching us chemistry, I'm a student from New Zealand with no background in chemistry, and currently finding it very hard to learn chemistry in the short period I've been give. The course I'm doing has a chemistry paper with contents near identical to your course's content. Your course has been very helpful along side everything else I'm doing at the university.

Thanks alot,

Sdiq Al-atroushi

5 answers

Last reply by: Professor Hovasapian
Tue Jan 28, 2014 3:06 AM

Post by Odell Glenn on September 26, 2012

The equation for T2...should have the T2 in it

Activation Energy & Arrhenius Equation

  • Reaction Rate is also dependent on Temperature
  • Collision Theory accounts for the rise in rate with rise in temperature.
  • Activation Energy is the Energy Barrier the reactants must overcome to proceed to products.
  • There are 2 forms for the Arrhenius Equation: Exponential and Logarithmic. Each serves its specific function.

Activation Energy & Arrhenius Equation

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 0:53
    • Rate Constant
    • Collision Model
    • Activation Energy
    • Arrhenius Proposed
    • 2 Requirements for a Successful Reaction
    • Rate Constant
    • Arrhenius Equation
    • Example 1
    • Activation Energy & the Values of K
    • Example 2

Transcription: Activation Energy & Arrhenius Equation

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

The last couple of lessons, we have been talking about reaction kinetics.0004

We have talked about the differential rate law; we have talked about the integrated rate law; and today, we are going to close off the discussion of kinetics with a discussion of activation energy and something called the Arrhenius equation.0010

You know from your experience that, when you raise the temperature on a given reaction, that the reaction tends to proceed faster; or, if you drop the temperature, that somehow things seem to go slower.0022

Like, for example, the whole idea of refrigeration is based on that fact: you drop the temperature, and the reactions that cause spoilage actually slow down.0034

So, not only does the rate depend on concentration, but the rate also depends on temperature.0043

The Arrhenius equation actually accounts for that dependence on temperature.0048

Let's just jump right on in.0053

OK, now again, we have seen that the rate depends on concentration; so, if we have some general equation like aA + bB → products, well, we know that the rate is going to be equal to some rate constant, times the product of A raised to some power, times the product of B raised to some power.0056

And, these n and m and n could be integers; they could be numbers; they could be anything.0083

Then again, let's reiterate that the m, the n, and the K are determined experimentally; these are not things that we can read off from the equation, the way we will do later on, when we discuss equilibrium.0090

Well, again, as it turns out, the rate is not only dependent on concentrations, but it is also dependent on the temperature.0101

So, as it turns out, the rate constant itself shows an exponential increase with temperature.0108

This constant of proportionality--yes, it is true that the rate is contingent, is dependent on A and B raised to some power; but in some sense, this K is a measure of that dependence.0132

As it turns out, for the temperature itself, the rate constant shows an exponential increase with temperature.0143

The rate constant depends on the temperature.0149

So now, we are going to introduce something called the collision model, and it is exactly what you think it is.0153

It accounts for this temperature dependence the same way that it accounts for the concentration dependence.0160

Let me just write this collision model; and the collision model basically says (I don't really need to write it down) that, in order for things to react, they basically need to come into close contact with each other.0166

And, since you know that molecules and atoms are sort of flying around, especially in the gas phase, at very, very high speeds--they don't just come close to each other; they collide.0181

And that is what it is; in order for a reaction to take place, they have to collide.0191

Well, so we know that, of course, the kinetic-molecular theory says that, as you raise the temperature, the average velocity of the molecules--of the atoms--of the particles increases.0196

Therefore, you have a higher frequency of collisions.0210

A higher collisions--more things collide; therefore, there is a greater chance of the reaction actually taking place.0213

And, that is all that is really going on.0221

Let's take a look at this.0224

Well, let's see: shall we write...let's write something for the collision model.0231

So, in order for a reaction to proceed, particles of reactants must come into close contact by colliding.0236

Now, I know this is not the most precise, rigorous definition; but it gives us a sense of what is going on, and that is really what is important; we want to understand the chemistry--we want to understand where these equations come from, if it actually makes sense--and this is perfectly fine.0267

OK, now, what is interesting is that, even though the temperature rises, and more particles are colliding--as it turns out, the frequency of collision goes up; but as it turns out, the rate itself seems to be not as high as it would be otherwise.0286

There are some other things going on; so, before we discuss those other things, let's just talk about what this man, Arrhenius, actually proposed.0302

Arrhenius proposed the idea of an activation energy.0312

The activation energy, if we want to define it: it is the threshold energy that the reactants must overcome in order to go to products--or in order for the reaction to actually move forward.0318

And now, we're going to draw a picture, and I think the picture is going to make it a lot more clear.0353

So, this is a standard energy diagram, and on the y-axis we have energy; and this thing is called the reaction coordinate.0358

The reaction coordinate is just a measure of...sort of time moving forward; it isn't time itself--it is saying how far forward the reaction is proceeding.0368

Let's go ahead and put the reactants over here; we'll put the products over here; as it turns out...0378

Now, thermodynamically, notice that the products over here are actually lower than the reactants.0386

Thermodynamically, this reaction is spontaneous; it will actually move forward.0392

But, that doesn't mean that it is going to move forward; there is this energy hill that has to be overcome--it has to get over this hump--in order for it to actually move forward.0397

This is what the activation energy is.0408

The reactants, the molecules...whatever it is--they have to slam into each other; they have to collide with enough energy to actually overcome this barrier.0412

Now, at a given temperature, not all of the velocities of the particles are uniform; it is not like all of the particles have one velocity.0422

There is a distribution of velocities, and only those velocities of the particles that are involved, that have enough kinetic energy for that kinetic energy to be transformed and equal to this activation energy (it's E with an a) will actually move forward.0430

That is why, when you raise the temperature--yes, it is true, you are actually causing more things to move faster and more things to collide; but not everything is going to move at enough speed to actually overcome this threshold barrier.0450

So, let's go ahead and draw a this thing up here--we call it the transition state; we also call it an activated complex--the transition state.0466

So if we were to take something like...let's take a particular reaction; let's take 2 BrNO → Br2 + 2 NO; so here, basically, two molecules of BrNO need to collide, and when they do collide, they end up releasing a bromine molecule and 2 nitrogen monoxide molecules.0478

Well, in order for that to happen--it's true that they need to slam into each other with enough energy; if they actually overcome it--if they actually have enough energy to get over that activation energy and to form a transition state, well, you might think that it actually looks a little bit...something like this.0514

In order for this to happen, a bromine and a nitrogen bond has to break, and a bromine-bromine bond has to form; that is what is happening here.0532

So, we could think of it as BrNO and then another BrNO, and these dots are used for bonds that are breaking and bonds that are forming.0544

This thing would be the transition state; they would slam together, and as this bond is breaking (the BrNO bond), the Br-Br bond is forming.0556

This was sort of what it might look like; now, we don't know exactly what it would look like, but that is what we postulate.0567

We call this thing the transition state; there is a transition from reactants to products.0574

That is all that is going on here.0579

And, we will often see these energy diagrams in any number of contexts.0581

OK, now, let us talk a little bit about this distribution of velocities, so you see what is going on, exactly.0589

Now, we said that, at a given temperature, there is a distribution of velocity; some particles are moving very slowly; some particles are moving very, very fast; in general, they sort of average out, and the distribution looks something like this--like any other distribution.0601

It will be...OK, so this is the energy axis; and now, let's see...this distribution--let's call this at temperature 1; so let's say, at a certain temperature, there is a distribution of speeds; certain things are slow; certain things are fast.0617

Well, as it turns out, if the activation energy is here, that means only those particles have enough energy, have enough speed, to surmount that barrier.0638

All of the others do not: which is why, despite the fact that we raised the temperature, and we have a whole bunch of collisions--as it turns out, the sheer number of collisions that actually lead to a reaction is actually very small.0652

This is why it is very small: because, among the distribution of speeds (very slow, moderately slow...on average, where most molecules fall, and the very fast), it is only the very fast that have enough energy to overcome that activation barrier--the activation energy.0665

That is what this is.0682

Well, if we raise the temperature, we raise, on average, the velocities, and the distribution changes; the distribution looks like this now.0683

So now, notice that this is temperature 2; and the temperature 2 is higher than temperature 1.0694

Well, temperature 2 is higher than temperature 1, which means, on average, you have more things that are fast.0701

Well now, we have basically pushed, on average, all of the particles to a faster speed, but they still occupy a distribution: some are still slower than others; most are in the middle; some are fast0708

But notice, now you have a lot more (I'll do this in red) particles that have enough energy above the activation energy to actually move forward with the reaction, as opposed to (let's do this other one in black) this right here, the original temperature.0721

So, as you raise the temperature, you provide more particles with enough energy to overcome that activation energy barrier.0750

This is a qualitative description of why raising the temperature increases the reaction rate.0762

OK, now, Arrhenius proposed this: he proposed that the number of collisions with enough energy equals some percentage of the total collisions--right?0770

If you have 1,000 collisions, and let's say, of those 1,000 collisions, only 100 of them have enough energy--well, it's a certain percentage.0798

He quantified this; he actually came up with a mathematical formula: so, the total collisions (we won't worry about where these came from or how he derived this) e to the -Ea/RT.0805

So, this ea/RT is the fraction of the total collisions that have enough energy to move forward.0823

And notice: this thing right here, the activation energy, shows up in this exponential.0837

R is the (let me rewrite this, actually: e to the -Ea/RT) gas constant, which is not .08206--we are dealing with energy in Joules, so it is actually 8.31 Joules per mole-Kelvin.0843

T is the temperature in Kelvin; so it's very, very important--you might be given degrees Celsius, but when you use this equation (and we'll sort of make it a little bit better in a minute), the temperature has to be in Kelvin, and R has to be 8.31, not .08206.0868

This Ea--that is the activation energy, the energy that has to be overcome in order for the reaction to move forward--the energy that takes you up to the transition state--the energy that takes you up to the activated complex for it to go over the hump; it is the top of the hump.0888

OK, now, as we said, even those particles with enough energy to overcome the barrier--still, we don't see the kind of rate change that we would expect.0905

Well, that is because there is something else going on.0917

Not only do these particles have to have enough energy in order to overcome the barrier--they actually have to be oriented properly.0920

In other words, they can't just slam into each other this way or that way or that way, and a reaction will take place; there is a specific way that they have to slam into each other in order for a reaction to take place.0928

So, there are two requirements (I'm going to put this down at the bottom here, and I'm going to do this in blue) for a successful reaction.0939

Two requirements for a successful reaction: the first is, of course, that the collision energy has to be greater than or equal to the activation energy, right?--we need to get over the hump.0955

Well, not only do we need to get over the hump, but the molecular orientation (well, I don't want to say "molecular"--well, yes, that is fine) must be such that it allows the reaction to take place.0977

We need an activation energy--the collision energy has to be greater than the Ea--and the orientation of the molecules has to be such that it actually allows for the reaction to take place.1015

That is why the number that we actually see, even though we have a bunch of particles that certainly have enough energy to overcome the barrier--still, only a fraction of those actually makes it past the reaction point.1026

So, we have a bunch of particles; only a fraction of them have a certain velocity, in order to activate, in order to get over the hump.1039

Of the amount that actually have enough to get over the hump, only a fraction of those have the right orientation.1049

So as it turns out, really, it is kind of surprising that reactions proceed the way they do, because really, ultimately, very, very few molecules actually satisfy these things.1057

But again, since we are talking about so many, we actually do see reactions take place.1066

OK, now, I am going to write a mathematical expression that quantifies everything that we have talked about.1072

Now, the rate constant is going to equal z, times p, times e to the (-Ea, divided by RT).1084

Now, let's talk about what these mean.1096

z is a factor that accounts for collision frequency--in other words, how often things collide.1099

p accounts for orientation, and it is always going to be less than 1; but that is fine--we are not really worried about that.1114

And of course, e to the (-Ea/RT)--that accounts for the fraction of collisions having enough energy--in other words, greater than or equal to Ea.1127

These three things: the frequency of collisions, the orientation, and those that actually have enough energy when they collide: all of these things contribute to the rate constant.1161

And, as we said, the rate constant increases exponentially with temperature.1177

All of these things are a function of temperature.1182

This is a constant; this is a constant; all of these things are essentially constants and factors.1184

It is this temperature that changes.1189

Now, we are going to write this as: K (which is the rate constant) equals Ae to the (-Ea/RT).1192

This A is called the frequency factor, and it accounts for these things--it accounts for the frequency of the collisions and the frequency of collisions that actually have the right orientation.1203

OK, this is called...this is the Arrhenius equation, right here.1219

This is the Arrhenius equation.1231

Now, like all things that tend to involve exponentials, let's go ahead and take the logarithm, and see if we can't come up with some linear relationship.1233

When I take the logarithm of both sides, I get the logarithm of K is equal to the logarithm of this whole thing; and the logarithm of a multiplication is the logarithm of one plus the logarithm of the other, so it is going to equal the logarithm of A, plus ln of e to the (-Ea/RT).1240

Well, ln(K)= ln(A), plus--the ln and the e go away, right, so you end up with -Ea/RT.1264

Now, let me rearrange this: I get ln(K)=...I'm going to pull out the -Ea/R, because these are constants, times (oops, no, we want this to be very, very clear) 1/T, plus the logarithm of A.1283

Now, notice what we have: we have (oh, these lines are driving me crazy) RT ln(A); let me actually do it over here.1314

We have y=mx+b.1331

The logarithm of the rate constant--that is the x; let me draw it down here: y=mx+b.1336

When we do y versus x, when we plot the logarithm of the rate constant against 1 over the absolute temperature in Kelvin, what we end up with is a straight line.1350

The slope of that straight line is equal to the negative of the activation energy, over R.1362

So, this gives us a way of finding the activation energy for a given reaction, when we measure the temperature and the rate constant.1368

The y-intercept actually gives us ln(A); well, when we exponentiate that, it gives us a way to find the frequency factor, A.1375

There we go; this is our basic equation that we are going to use when we are dealing with temperature dependence of the rate constant.1386

OK, now, let's see: let's go forward, and let's rewrite the equation again.1397

We have: logarithm of the rate constant equals minus the activation energy, over R, times 1 over the temperature, plus the logarithm of A (which is the frequency factor).1405

Thus, for a reaction which obeys the Arrhenius equation, the logarithm of the rate constant, K, versus 1 over the temperature, gives a straight line with slope equal to -A/R.1422

And again, R equals 8.31 Joules per mole-Kelvin, not the .08206.1464

Now, what is really, really nice is that most rate constants actually do obey the Arrhenius equation.1472

So, because that is the case, it actually ends up lending support to the collision model.1477

We derived this from the collision model; the fact that most things obey this is supporting the collision model--that is supporting evidence; so, our model is actually very, very good.1484

Now, let's go ahead and see if we can do an example; I think it's the best way to proceed.1499

Our example: the following data were obtained for the reaction 2 N2O5 decomposes to 4 NO2, plus O2.1507

The following data was obtained: the temperature, which was in degrees Celsius, and the rate constant, which is in per second.1537

We have 20, 30, 40, 50, and 60; and these are rate constants, OK?--so this is per second, so we are looking at a first-order reaction.1549

2.0x10-5; we have 7.3x10-5; 2.7x10-4; 9.1x10-4; and 2.9x10-3.1562

Notice, as the temperature increases, the rate constant increases; the rate is getting faster.1582

A higher rate constant means a faster rate.1586

OK, now, let's see: what is it that we want to do here?1591

We want to find the activation energy for this reaction.1597

In other words, how much energy do the N2O5 molecules have to have in order for this reaction to proceed properly?1602

OK, well, let's go ahead and graph this data.1610

Now, I'm going to give a rough graph; it's going to be reasonably accurate, but the idea is to see what is going on here.1614

This is going to be 3; this is going to be 3.25; this is going to be 3.50; and then we have 6, 7, 8, 9, 10, 11, 10, 9, 8, 7, 6, so -6, -7, -8, -9, -10, -11.1623

OK, so these are...this axis, the vertical axis, we said, is the logarithm of the rate constant; and this is going to be 1/temperature.1649

So, let's go ahead and what we do is (now, notice: we were given K, and we were given T; what we have to do is) calculate 1/T and ln(K).1667

OK, so let me go ahead and just redraw everything here; let's do temperature in degrees Celsius; let's do temperature in Kelvin; let's do 1/T; and let's do ln(K)--so we can actually see the data, instead of just throwing it out there.1681

We have 20, 30, 40, 50, and 60; then, we have 293, 303, 313, 323, and 333.1701

Now, we have 3.41x10-3; 3.30x10-3; 3.19x10-3; 3.10x10-3; and 3.00x10-3.1718

Then, we have -10.82; -9.53; -8.22; -7.00; -5.84.1739

So, again, our equation is 1/T and ln(K), which is why we took that data that we got, and we calculated the 1/T and the ln(K).1753

Now, this is what we plot: this on the x-axis, this on the y-axis.1764

When we do that, we end up with some line that looks like this.1771

Let's go over here; I'm just going to pick a couple of points: 1, 2, 3, 4, 5--we have 5 points, but I'm just going to pick a couple of them, because it is the point that I want to make that is ultimately important.1776

We go ahead and we draw a line through those; and again, this is kinetic data; everything is not going to lie on a straight line, but you are going to get a linear correlation.1787

You are doing the "best fit" line.1798

When you do that, you pick a couple of points on that line, and you calculate Δy/Δx.1799

Now, Δy/Δx, which, in this case, is equal to Δ of ln(K) (that is the y) over Δ of 1/T.1808

When you do this, you end up with -1.2x10-4 Kelvin.1820

Now, we said -1.2x10-4; that is the slope--well, the slope is equal to the activation energy over R, which is equal to the activation energy (which we are seeking) over 8.31 Joules per mole per Kelvin.1832

When we multiply this through, we end up with an activation energy equal to 1.0x104 Joules per mole; there we go.1856

We were given kinetic data, which consisted of temperature and rate constants; we used the logarithmic version of the Arrhenius equation; we found 1/T and ln(K); we wrote those values down.1872

We plotted that: ln(K) versus 1/T; we got ourselves a straight line; we picked a couple of points on that line; we found the slope; and we know that the slope is equal to negative of the activation energy, over the gas constant.1885

We solved basic algebra, and we ended up being able to find the activation energy of 1.0x104 Joules.1901

That means that the particles--a mole of particles--need to have enough activation, and need to have 1.0x104 Joules (in order for this to actually proceed) per mole.1908

That is all this is; that is all we did.1921

OK, let's see: what else can we do?1926

Let's do another example here.1933

This time, I will go ahead and...actually, before we do the example, let me give you a little preface to the example.1935

Now, instead of kinetic data, the activation energy can also be gotten from the values of the rate constant at only 2 temperatures.1944

In other words, we don't need a whole table of rate constants and their corresponding temperatures...or, temperatures and their corresponding rate constants.1980

As long as we have two temperatures and two rate constants, we can actually find the activation energy.1989

So, let's see how that is done.1994

Now, at temperature T1, we have rate constant K1.1996

The relationship is: the logarithm of K1 is equal to minus the activation energy over R, times 1/T1, plus the ln(A).2009

Again, this is based on the equation: K is equal to A, times e to the (-Ea/RT).2022

This is the Arrhenius equation, and, when we take the logarithm of both sides, we get this version.2032

Well, for any K and T, that is the relationship.2037

And now, let's take another temperature; how about at T2?2042

Well, at T2, we have a K2; we have another rate constant, K2.2047

The relationship is: ln(K2) is equal to -Ea, over R (it doesn't change; it's the same reaction), this time times 1/T1, plus ln(A).2053

And again, A doesn't change; A is the frequency factor--it's a constant for that particular reaction.2066

Well, let me take this equation minus this equation.2072

So, I write: ln(K2)-ln(K1), which equals ln(K2/K1), is equal to...when I take this side and subtract this side, the ln(A)s cancel, and I end up with the following.2077

-Ea/R, times 1/T2, minus 1/T1.2097

Now, in your chemistry books, you will often see this flipped, and you will see the sign change here.2106

All they have done is factor out a -1 from here, in order to get rid of this -1.2112

So, you will also see it as: Ea/R, times 1/T1, minus 1/T2; it's up to you.2117

I personally prefer this, because it doesn't change anything.2128

I think it is important that equations be written exactly as how they were derived, and that signs be left alone, simply so you can see everything, so everything is on the table.2133

When you start to simplify things, yes, you tend to make them look more elegant--and this is generally true of science; we tend to like our equations to look elegant.2143

But, understand something: that in science, and in mathematics, the more elegant something looks--that means the more that is hidden.2153

That is the whole idea: when something looks elegant--when something looks clean and sleek and simple--that means something is hidden.2160

In this particular case, what they have done is: they have hidden the negative sign, and they have flipped this around.2165

That is fine; this is reasonable straightforward--it is not going to confuse anybody too badly.2170

But, I think that if we are going to take something and subtract something else, we should leave things exactly as they are.2174

So, don't let this minus sign confuse you because it looks different than what you see in your books; it's the same equation; they just don't like minus signs--which is generally true of chemists; they tend not to like minus signs; I don't know why.2180

OK, so here we have this equation; if you are given two temperatures and two rate constants, you can calculate the activation energy.2193

Let's do an example.2204

We have the following reaction: we have: methane (CH4), plus 2 moles of diatomic sulfur, forms carbon disulfide, plus 2 H2S gas (hydrogen sulfide gas).2208

OK, now, we take a couple of measurements: temperature in degrees Celsius, and we calculated some rate constants for that.2233

That ended up being...let me see: we did it at 550 degrees Celsius, and we also did it at 625 degrees Celsius; we got 1.1, and we got 6.4; it makes sense--higher rate constant-faster rate; higher temperature-faster rate; so everything is good.2248

We want to find the activation energy.2267

Well, great; we have two constants, and we have two temperatures; let's use our equation that we just derived.2271

And again, you don't have to know that equation; as long as you know the Arrhenius equation, everything else you can derive from there, because you are just taking logarithms and fiddling with things.2282

That is why we are showing you the derivation--to show you that you don't have to memorize the equation; it is where the equation came from and what you can do with it.2290

OK, so let's take the logarithm of K2/K1 (well, you know what, let me write it again), is equal to (-Ea, over R)(1/T2 minus 1/T1).2299

And again, temperature is in Kelvin.2317

This is equal to the logarithm of 6.4 (see this number), over 1.1; is equal to -Ea/8.31,, 1/T2; T2 is 625 degrees; that is 898 it's 1/898, minus...T1 is 550; that is 823 Kelvin.2321

When we solve, when we do this and this, we end up with the following: =1.7609, equals -Ea × -1.22x10-5.2354

We end up with an activation energy of 144,195 Joules, or 144 kilojoules.2372

I am not a big fan of significant figures one way or the other, which is why it looks like this.2386

Numbers are probably going to be a little bit different, as far as you are used to in your book, but it is the process that is important, ultimately--not the significant figures.2393

Later, if you become an analytical chemist, then significant figures will be a bigger issue.2400

That is it; we have used the Arrhenius equation, which, again, says that the rate constant is equal to some constant (called the frequency factor), exponential, minus activation energy over RT.2404

From this one equation, we can do multiple things.2422

This expresses that the rate is dependent on temperature.2427

This is the independent variable; this is the dependent variable.2431

When we are taking kinetic data, we will often have temperature and a rate constant, and we can do things to that, based on how we fiddle with this equation.2435

OK, with the Arrhenius equation, this concludes our discussion of kinetics.2447

I want to thank you for joining us for this discussion, and thank you for joining us here at

We'll see you next time; goodbye.2457