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Lecture Comments (22)

0 answers

Post by Jarrah alharbi on September 18 at 09:56:39 PM

is the ----> t1/2=In1/2 / k

or t1/2=In2/k

2 answers

Last reply by: Professor Franklin Ow
Mon Jun 20, 2016 11:35 PM

Post by Parth Shorey on June 20 at 08:43:53 PM

I don't understand why you used the negatives in front of the recants? You said because of the slopes? I still don't understand.

1 answer

Last reply by: Professor Franklin Ow
Thu Jun 16, 2016 3:29 PM

Post by Parth Shorey on June 13 at 08:41:32 PM

Is the Q&A active?

1 answer

Last reply by: Professor Franklin Ow
Thu May 28, 2015 12:28 PM

Post by BRAD POOLE on May 7, 2015

At about the 17 min mark you said that the example was a 2nd order reaction.  How did you come up with this?  Are you just going by whatever your units are for "k"?  I always thought you used the exponents of the reactants to figure out what order it was, then again could be why I can't seem to get these kinetics problems.  

0 answers

Post by Saadman Elman on January 14, 2015

Great lecture as usual!

0 answers

Post by Saadman Elman on January 14, 2015

What you mean by we can write co-effecient as rate order ONLY for elementary steps. What you mean by elementary steps exactly?

2 answers

Last reply by: David Gonzalez
Thu Jul 31, 2014 12:25 PM

Post by David Gonzalez on July 31, 2014

Hi Professor Ow. First of all, great lecture. Although, there is one problem that I have.

In the example (around the 16-minute mark), when determining the order for S2O8, you mentioned that there was a 3x increase in initial rate from 0.015 to 0.044. I'm confused, because 0.015 x 3 is 0.045 - don't these problems need to be exact? Or can they sometimes be slightly off?

Thanks.

1 answer

Last reply by: Professor Franklin Ow
Tue Jun 24, 2014 1:29 PM

Post by brandon joyner on June 24, 2014

Also why for Iodine did you only do the first experiment for the second experiment you went from experiment 3 to 1.

4 answers

Last reply by: brandon joyner
Fri Jun 27, 2014 1:46 PM

Post by brandon joyner on June 24, 2014

For Iodine how is it a triple jump when all you had to do was go up 2?

1 answer

Last reply by: Professor Franklin Ow
Wed May 21, 2014 1:54 AM

Post by Ashley Gwemende on May 20, 2014

Where do I find a lecture of how to read a potential energy diagram ?

Chemical Kinetics

  • Kinetics studies the factors that can influence the rate of a chemical reaction.
  • The method of initial rates and the use of integrated rate laws can help solve for the rate orders and for the rate constant.
  • The slowest step of a reaction mechanism dictates the overall reaction rate.
  • The Arrhrenius equation relates the temperature of a reaction directly to its reaction rate.

Chemical Kinetics

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
  • Lesson Overview 0:06
  • Introduction 1:09
    • Chemical Kinetics and the Rate of a Reaction
    • Factors Influencing Rate
  • Introduction cont'd 2:27
    • How a Reaction Progresses Through Time
    • Rate of Change Equation
  • Rate Laws 7:06
    • Definition of Rate Laws
    • General Form of Rate Laws
  • Rate Laws cont'd 11:07
    • Rate Orders With Respect to Reactant and Concentration
  • Methods of Initial Rates 13:38
    • Methods of Initial Rates
  • Integrated Rate Laws 17:57
    • Integrated Rate Laws
    • Graphically Determine the Rate Constant k
  • Reaction Mechanisms 21:05
    • Step 1: Reversible
    • Step 2: Rate-limiting Step
    • Rate Law for the Reaction
  • Reaction Rates and Temperatures 26:16
    • Reaction Rates and Temperatures
    • The Arrhenius Equation
  • Catalysis 30:31
    • Catalyst
  • Summary 32:02
  • Sample Problem 1: Calculate the Rate Constant and the Time Required for the Reaction to be Completed 32:54
  • Sample Problem 2: Calculate the Energy of Activation and the Order of the Reaction 35:24

Transcription: Chemical Kinetics

Hi, welcome back to Educator.com.0000

Today's lecture from general chemistry is going to be on chemical kinetics.0002

Here is our brief overview of the lesson today.0008

As always we will go ahead and start off with our introduction.0010

Then we are going to get into what is really at the heart of chemical kinetics.0013

That is the rate of chemical reactions; that is how fast does a reaction occur?0017

We are then going to see what are the mathematical equations which actually attempt to quantify the rate of an equation.0021

That is what we call a rate law.0029

Our objective is to then try to derive these equations for any given reaction.0031

There is two sets of experiments that you can do to perform this.0037

Number one is what we call the method of initial rates.0040

Number two is what we call the integrated rate laws.0043

We will then jump into how a reaction proceeds, the step by step process which is0045

called a reaction mechanism, followed by the effect that temperature has on reaction rates.0051

We will go ahead and wrap up the lesson with a brief overview of0059

catalysis followed by our summary and a pair of sample problems.0062

Chemical kinetics basically is the area of chemistry that examines the factors which can influence the rate of a reaction.0070

What do we mean by factors that can influence the rate?0081

There are several--temperature, pressure, the reactant concentration, the addition of a catalyst, and mechanical force.0084

We are going to see that basically as temperature goes up, the rate goes up.0095

We are going to see that in general as pressure goes up, the rate goes up.0104

We are going to see that as the concentration of reactant goes up, the rate goes up.0112

When we add a catalyst, we are going to define what that exactly means.0120

But this also increases the rate.0123

Finally mechanical force, this is what you probably do in lab.0126

All we mean by mechanical force is something as simple as stirring or shaking.0129

Of course this will increase the rate of the reaction.0137

This is what we mean by factors which can influence the rate of a reaction.0142

Let's take a look though how a typical reaction progresses through time.0150

Consider the reaction A plus 2B goes to C.0154

This is telling me that for every one of A, two Bs are required.0158

One C is going to be produced.0165

Remember we are only looking here at the forward direction only.0167

When we go in the forward direction, we say that reactants are consumed which means concentration of reactant go down.0176

The products are made which means that the concentration of product goes up.0190

It looks like because twice the amount of B is going to be consumed as A,0201

the rate of consumption of B is going to be double that of A.0206

It is very typical for us in chemical kinetics to graph this.0221

We can go ahead and graph the concentration of the reactant with respect to time and the concentration of product too.0225

At time zero, I have no product; I only have reactants.0237

Down here right at the origin, this is the concentration of C initial.0240

The way that concentrations change with respect to time is not linear.0247

But instead it is going to be characterized by a simple curve just like that.0251

You notice that at some time t, the concentration of C stops changing.0260

We are going to plateau; it is not going to increase forever.0264

How about the concentration of A and B?0269

The concentration of A let's say is here and the concentration of B.0272

Let's say we had equal amounts.0278

Let me go ahead and draw now a curve that represents the concentration of A changing with respect to time.0280

It is going to be a mirror image of the concentration of C changing because they are both 1:1 ratio.0287

But how is the concentration of B changing?0293

The concentration of B, it is going to drop at double the rate.0296

The concentration of B is going to drop much much much... quicker than the concentration of A.0302

You notice that for all of the reactions, there reaches a point where the curves plateau0315

which means that the concentration of reactants and the concentration of products no longer changes.0325

We are going to talk about this region in another lecture.0337

This is what we call the equilibrium region.0340

In chemical kinetics, what we are interested in is really the start of the reaction, the early part0346

of the reaction where we actually can monitor change in the concentration of reactants and products.0351

That is the representation graphically; how about mathematically?0360

It turns out that the rate of change is going to be equal to the following.0364

The change in the concentration of A, Δ[A]Δt, this is going to be equal to0368

1/2 Δ[B]Δt which is equal to the change in concentration of C Δ[C]Δt.0382

We have to put a negative sign in front of the reactants such that the overall number is0392

going to be positive because Δ[A]Δt is going to be slope of the curve.0401

This is going to be slope of the curve you see.0408

This is going to be the slope of that curve too.0411

This gives us a general equation for relating the rates of change of each reactant and product concentration with respect to time.0416

Now that we have gone through a brief introduction, let's go now into0429

what is at the core of chemical kinetics--that is the rate law.0434

A rate law is going to be a mathematical equation which relates the rate of reaction to the concentration of the reactants.0439

Basically the rate law tells us that the rate of a reaction is proportional to the concentration of reactants typically.0447

When we write out the rate law, all rate laws have the following equation0458

Of the form rate which is going to be usually in molarity per second...0461

is equal to some constant k times the concentration of A raised to0467

some power times the concentration of the B raised to some power, etc.0472

Let's go ahead and define what each of these mean.0478

k is what we call the rate constant.0484

The rate constant is going to be unique to a reaction at a certain temperature.0491

In addition, we are going to see that the rate constant, the units of k vary.0508

We are going to see that very soon.0519

x and y are what we call rate orders.0522

Once again x and y are what we call rate orders.0529

If x is equal to 1, we say reaction is first order with respect to the concentration of A.0537

If x is equal to 2, we say the reaction is second order with respect to the concentration of A.0554

In addition, if the sum of the rate orders x plus y is equal to 0, we say zero order overall.0564

If the sum of the orders is equal to 1, we say first order overall.0577

If x plus y is equal to 2, we say second order overall.0585

This is just some terminology that we want to introduce and clarify.0592

Let's go ahead and get back to the units of k.0598

The units of k will vary.0601

For zero order, the units of k is just going to be reciprocal seconds.0604

For first order, the units of k is going to... excuse me.0619

For zero order.. my apologies.0627

For zero order, the units of k is going to be molarity per second.0629

For first order, the units of k is just reciprocal seconds.0636

For second order overall, the units of k is going to be inverse molarity inverse second.0640

You can easily plug that back into the equation for k and see that the units will cancel.0649

This is the general rate law.0657

Rate is equal to some constant k times the concentration of A raised to0659

the x power times the concentration of B raised to the y power.0662

But let's go ahead now and see what the significance of the rate orders are.0666

Consider the following reaction: A plus 2B goes to C.0673

The rate law is given to us to be k times A squared times the concentration of B.0678

Let's go ahead and study what happens if we change the concentration of one of these reactants.0686

If the concentration of A doubles holding B constant, then we see that the rate is going to increase by 4.0692

It is going to quadruple.0710

If the concentration of A triples holding B constant, we see that the rate is going to increase by a factor of 9.0714

What if the concentration of B doubles holding concentration of A constant?0728

If that happens, the rate is just going to increase by 2.0739

If the concentration of B triples holding A constant, we see that the rate is going to increase by a factor of 3.0746

You see what the significance of the rate order is.0756

The rate order is really proportional to the sensitivity of a reaction rate on0760

the concentration of a specific reactant molecule, on the concentration of a specific reactant.0778

Pretty much we see that the larger the value of these rate orders, the more sensitive0788

your reaction is to a change in concentration of a specific reactant.0795

Basically what the kinetics deals with is focusing on this rate law and solving for x, y, and k.0802

In other words, our goal is to solve for the rate constant and rate orders.0811

We can do this in one of two ways.0820

The first experimental way to derive a rate law is what we call the method of initial rates.0822

In the method of initial rates, you basically do what we just did.0828

You change one reactant concentration holding all else constant and seeing how the rate varies.0834

Let's go ahead and take a look at the following.0857

You are usually given some table of data.0859

The table of data is going to list different concentrations, different molarities of each reactant and the rate that was measured.0865

For example, in experiment number one, this concentration of iodide was found.0873

This concentration of thiosulfate was found; this is the initial rate.0881

In experiment number two, we see that the concentration of thiosulfate was fixed.0886

The concentration of iodide was tripled; we see that the rate also tripled.0894

What does that mean?--basically as the concentration of iodide tripled, the rate tripled.0901

This is a 1:1 correspondence.0913

Anytime you have this 1:1 correspondence, it is a rate order of 1; rate order of 1.0917

Let's now see what the rate order is for S2O82-, thiosulfate.0927

For thiosulfate, we are going to look at experiments two and three.0933

You see that as the concentration of thiosulfate tripled... I'm sorry.0941

We have to do experiments one and three, not two and three.0953

Here as the concentration of thiosulfate tripled, iodide was held constant.0956

What happened to the rate?--the rate tripled.0966

As S2O82- tripled, the rate tripled holding everything else constant.0968

This was also a 1:1 correspondence; the rate order is also 1.0975

Therefore we have the following rate law.0986

The rate of this reaction is equal to some constant k times the concentration of I-0989

raised to the first power and the concentration of S2O82- raised to the first power.0994

The next thing now that we have the rate orders, we can go ahead and now solve for k.1002

We can solve for k by simply plugging in any experiment--one, two, or three--directly into the equation.1011

Solve for k using any experiment number.1019

For example, let's use experiment one; the rate was 0.044 molarity per second.1029

That is going to be equal to the rate constant k times the concentration of I- which is1037

0.125 molar times the concentration of thiosulfate, 0.150 molar, all raised to the first power.1041

Then you can just use your algebra to go ahead and solve for the value of k for this second order overall reaction.1050

Here we are going to get units of inverse molarity inverse seconds.1063

That is how we use the method of initial rates--very straightforward experiment to determine the rate law.1070

The second way of determining a rate law is to use what we call integrated rate laws.1078

Integrated rate laws, they quantify the relationship between the reactant concentration and time; and time.1085

Basically these are all derived mathematically.1095

You should always ask your instructor if you need to know how to derive it or not.1100

For zero order rate law, the rate is equal to the rate constant k.1104

The integrated rate law is the following.1108

The concentration of A at any given time is equal to the initial concentration of A minus kt.1111

For the first order overall, the integrated rate law is natural log of A0 minus natural log of A equal to kt.1116

Finally the second order integrated rate law is 1 over A minus 1 over A0 is equal to kt.1125

The nice thing about these equations is that they are all linear.1133

If several concentrations are determined at different times, you can get the rate constant graphically.1139

For a zero order overall reaction, we are basically going to graph the concentration of A as a function of t.1146

That is going to give us a nice straight line with slope equal to ?k.1156

For first order overall, we can go ahead and graph the natural log of the concentration of A versus t.1162

We are also going to get a straight line whose slope is equal to ?k.1174

For second order overall, we are going to plot 1 over the concentration of A which is equal to time.1179

Here we are going to get a nice straight line with a positive slope equal to k.1189

Once again this is a graphical determination of the rate constant anytime1194

you have data from several different time intervals and measured reactant concentration.1198

Another thing we like to talk about too is the half-life.1210

For zero order, the half-life is equal to the initial concentration of A over 2k.1214

For first order, the half-life is equal to the natural log of 2 over k.1221

For second order, the half-life is equal to 1 over k times the concentration of A0.1228

Once again you should ask your instructor to make sure if you have to know the derivation or not.1235

Basically the half-life is very important because it tells us the time required to reach 1/2 of the initial concentration of A.1241

Again that is what we call integrated rate laws.1261

The next thing we are going to look at is what we call a reaction mechanism.1267

A reaction mechanism basically represents the step by step reactions which when combined give you the overall net reaction.1271

For example, let's say we had the following given, step one.1278

Step one was 2NO gas going on to form N2O2 gas.1285

This tells us that the reaction is occurring both in the forward and reverse directions.1293

This is what we call the reversible reaction; this is usually very very fast.1299

Step number two is going to be O2 gas plus N2O2 gas going on to form 2NO2 gas.1305

You are told that this reaction is very very slow.1318

The overall reaction is going to be 2NO gas plus O2 gas going on to form 2NO2 gas.1323

You notice that N2O2 gets cancelled out.1336

Any item in a chemical reaction mechanism that gets cancelled is what we call an intermediate.1342

That is it is both formed and consumed during the course of a chemical reaction; formed and consumed during a reaction.1351

The reason why we care about what the slow step is is because of the following.1370

The slow step, which is in this case step two, is like the weakest link in your chain.1376

It is like the slowest person on your track and field relay team.1382

The slow step determines, it limits how fast a reaction can go.1387

We call this the rate limiting step.1392

The rate limiting step is equal to the rate of the overall reaction.1399

If we were to write out the rate law for this, we would get the following.1408

The rate is equal to the rate constant k2...1412

This is k2, of the second step.1417

Times the concentration of O2 raised to the first power1419

times the concentration of N2O2 raised to the first power.1422

We can always use the coefficients as the orders if the reaction you are1427

looking at is a part of the mechanism, what we call an elementary reaction.1438

Coefficient is equal to the rate orders for elementary steps only.1444

Otherwise you would have to go through integrated rate laws or method of initial rates1451

to go through the whole process again to find what x and y are.1454

When we look at this, we have a problem.1458

We have N2O2 appearing in this rate law.1460

This is the intermediate.1463

You can never have an intermediate appearing in the rate law.1465

We have to do something about this; what we do is the following.1468

We use what is called a steady state approximation.1473

You utilize the fast equilibrium step where the k1 times the concentration of NO squared1481

equals to k-1 times the concentration of N2O2 where k1 represents the forward.1492

k-1 represents the reverse.1502

What we do then is we solve for the intermediate.1505

Concentration of N2O2 is equal to k1 over k-1 times the concentration of NO2 squared.1507

We then plug this back into our experimental rate law.1518

Rate is equal to k2 times the concentration of O21522

times k1 over k-1 times the concentration of NO squared.1526

You can collect all of the constants together and just call that what we call kobserved.1534

We get left with O2 times the concentration of NO squared.1539

Here we have our final rate law that is going to be the rate law for the overall reaction.1546

It is just by coincidence here that the rate orders are the coefficients.1555

That is not usually the case.1559

But that is how we solve for it where kobserved is equal to k2k1 over k-1.1560

Once again this is how you deal with reaction mechanism problems.1571

Let's now move on to the next topic.1577

This is the relationship between temperature and reaction rate.1579

Basically in general, as temperature increases, so does the reaction rate; just think about this.1583

You know the tea bag is going to brew faster in warm water than cold water.1588

You can see that visually happening right before you.1594

In order for a reaction to occur, reactant molecules must do two things.1597

They must collide; they must collide with sufficient energy.1602

They must collide in the proper orientation; sufficient energy and proper orientation.1606

Basically we can look at a sample here.1613

Let the y-axis be fraction of sample; let the x-axis be temperature.1616

At any given temperature, I am going to have a bell curve distribution of molecules just like that.1628

Let's call this T1.1639

What happens to T2?--what happens when we have a hotter temperature?1641

When I have a hotter temperature, my bell curve is going to shift just like that.1649

I call this T2; T2 is greater than T1.1652

Let's say that in order for the sufficient energy, in order for the1658

reaction to occur, let's go ahead and make that as a dotted line.1663

I am going to call this dotted line EA; EA is equal to activation energy.1669

What this is, it is the minimum energy required for the reaction to proceed, for collisions to occur.1677

Basically anything below EA, anything less than EA, you get zero collisions and no reaction.1691

Anything greater than or equal to EA, you get collisions; therefore a reaction will occur.1704

Basically you see that as you go from T1 to T2,1712

the fractional molecules with an energy greater than EA significantly increases.1717

At T2, larger percent of molecules with an energy greater than or equal to EA.1723

This is why as temperature goes up, so does the rate of a reaction in general.1739

We have a nice equation which can actually quantify this.1749

This is called the Arrhenius equation.1754

The natural log of k1 over k2 equals to EA over R times 1 over T2 minus1756

1 over T1 where R is our universal gas constant in terms of energy, 8.314 joules per K times mole.1766

Temperatures T1 and T2 are kelvin temperatures; what this basically says is the following.1779

If I do a series of reactions at different temperatures and I calculate the1789

rate constant, I can then do graphical determination of the activation energy.1794

If I graph natural log of k1 over k2 as a function of 1 over temperature,1800

I am going to get a nice straight line with slope equal to ?EA over R.1809

Again the Arrhenius equation is very useful because it gives us graphical approximation of the activation energy for a reaction.1815

Again this is the relationship between rate and temperature.1829

Finally the last factor we are going to study is a catalyst.1833

Basically a catalyst's job is to do the following.1837

A catalyst assists reactant molecules to be in the proper orientation for proper collision to occur.1840

What that does is that the activation energy is lower.1847

If this reaction represents without a catalyst, the activation energy is going be basically right here.1853

This is the energy that you must overcome for the reaction to go form products.1867

I can then proceed and draw another curve where this is a catalyst now.1873

With the catalyst, you see that the activation energy EA is much lower.1878

Activation energy catalyst is going to be always much less than the activation energy no catalyst.1884

Because of that, with the lower activation energy, that means a faster reaction.1892

A catalyst again speeds up a reaction by lowering the activation energy.1900

We see that of course the nice about catalysts is that they can be reused over and1907

over again because during the course of a reaction, they are recovered; they are recovered.1913

That is catalysis.1921

Let's now get into our summary before we jump into our sample problems.1923

Kinetic studies the factors that can influence the rate of a chemical reaction.1927

We saw that we can have two main experiments to help us determine1932

the rate law--the method of initial rates and the integrated rate law.1936

We found that in a reaction mechanism, that the slowest step dictates the overall reaction.1940

That is what we call the rate limiting step.1946

Finally we introduced the Arrhenius equation which gives a mathematical relationship between the reaction rate and the temperature.1952

The nice thing about this equation again, this gave us graphical estimate of EA.1963

Let's now get into a pair of sample problems.1975

A certain first order reaction has a half-life of twenty minutes.1977

Calculate the rate constant; that is part A.1980

Part B, how much time is required for this reaction to be 75 percent complete?1984

Let's go ahead; you are told that the reaction is first order.1989

For a first order reaction, T1/2 is equal to the natural log of 2 over the rate constant k.1992

You are told that the rate constant is 20.0 minutes.2001

Here the rate constant is simply going to be equal to the natural log of 2 divided by 20.0 minutes.2007

That gives us our answer in units of reciprocal minutes; that is part A.2014

Let's go ahead and do part B now.2024

How much time is required for this reaction to be 75 percent complete?2026

As soon as you see the word time, you should immediately, immediately, think integrated rate law because2029

method of initial rates does not have time in it; only integrated rate laws does.2036

For the first order integrated rate law, it is the natural log of the initial concentration2041

of A minus the natural log of the concentration of A is equal to kt.2047

We know what k already is because we already solved for that in part A.2056

We are good to go on that.2060

The question is asking for how much time is required.2062

This is what we are trying to solve for.2065

All that matters is is what is the identity of A0 and what is the identity of A?2068

You are told that how much time is required for this reaction to be 75 percent complete?2074

Let's say A0 is going to be 100.2079

If the reaction is 75 percent complete, that means only 25 percent of A is remaining.2083

A is going to be 25.2090

Again we now have enough information to solve for t.2092

We are going to get our final answer of t in units of minutes.2097

How do you know if you have done something wrong?2105

You should always check your answer because if you get a ?t, again that just doesn't make physical sense.2107

You know you have done something mathematically wrong.2119

Always check your answer for a negative time.2121

That is sample problem number one.2125

Let's go ahead and move on to sample problem number two.2127

Consider the following reaction.2130

2N2O5 gas goes on to form 4NO2 gas plus O2 gas.2132

Here we are given several rate constants that were found at several temperatures.2137

The only equation that we know that deals with this is the Arrhenius equation.2145

The natural log of k1 over k2 is equal to2150

EA over R times 1 over T2 minus 1 over T 1.2154

We know that we are going to be using this equation quite easily.2161

Here we can calculate EA from there.2165

The nice thing about this is because this is going to give us a nice straight line,2170

we can use any pair of k and T data points to go ahead and solve for EA.2174

Once again this is going to be using the Arrhenius equation to solve for EA.2193

Our units of EA is going to be in kilojoules per mole.2199

Next one is what is the order of the reaction.2206

This is kind of a trick question; this is something I have asked students before.2208

This is something I have seen asked by other instructors before.2213

The order of the reaction, you don't have to do any work for that.2217

The reason is because they already give you the units for k.2219

The units of the rate constant tell us the rate order.2226

It is only first order where the units of k is reciprocal time; first order overall.2234

Again just watch out for that when you do problems.2248

Again the units of k tell us a great deal of information without doing any work.2251

That is our lecture from general chemistry concerning kinetics.2257

I want to thank you for your time.2262

I will see you next time on Educator.com.2263