AP Chemistry Reaction Rates and Rate Laws

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

We just finished talking about thermochemistry, and now we are going to move on to discuss a series of topics that are going to form, really, the most important part of chemistry.0003

Not that what we talked about wasn't important--I shouldn't say important; I should say the real core of what it is that you are going to be using in your future work.0015

We are going to be discussing--the first thing we are going to be starting with is kinetics.0025

We are going to talk about kinetics; after that, we are going to talk about equilibrium; after that, we are going to talk about acid-bases and some more aspects of acid-base chemistry.0030

And then, we are going to talk about thermodynamics in greater detail--not just enthalpy, but we are going to talk about free energy and entropy, and then we are going to round it out with a discussion electrochemistry.0040

Afterward, we will actually come back and discuss some of the topics like bonding and solutions and things like that.0052

But, I wanted to get to the heart and soul of what chemistry is--the things that you are going to be using--because we want to spend a fair amount of time with it and get really, really comfortable with it, over and over and over again.0060

Our first jump into this is chemical kinetics.0073

Chemical kinetics is the study of how fast something happens.0078

As it turns out with chemistry, you have thermodynamics, which tells you whether something can happen; and then you have chemical kinetics, which is to tell you how fast something happens.0083

You might have this wonderful, wonderful reaction that works thermodynamically, and it is the greatest thing you have ever seen, but it is just so slow that maybe it comes to completion after four thousand years.0096

There are plenty of reactions like that...so both are important: you have to know that a reaction actually works--will go forward without too much effort on your part--and that it will actually go forward in a reasonable amount of time, because there is the practical issue.0106

So, let's just jump in, and today we are going to talk about reaction rates, and we're going to introduce this idea of rate laws.0119

Much of our discussion today is just going to be getting you familiar with the context--that is what it is going to be about today.0126

OK, let's get started.0136

Let us start with a definition, like we often do.0139

No, I am not going to write the word "definition"; I am just going to write the rate.0147

Well, you all know what a rate is; it is the change in something over the change in time.0153

In this particular case, let's write ΔA over ΔT, where A, in the square brackets, is the concentration of a particular species (it could be any species--molecule, atom, whatever we happen to be on...ion...more often than not, it is going to be something in solution).0160

So, it's the change in concentration; so Δ is the concentration in moles per liter: molarity--pretty standard: M, and you have also seen it as m with a line on it; that is my particular symbol for it.0183

The rate is the final concentration of something, minus the initial concentration of something, divided by the time it took for it to go from the initial concentration to the final concentration.0198

Let's just take, as an example, a particular reaction: we'll write the balanced reaction, 2 moles of nitrogen dioxide decompose into 2 moles of nitrogen monoxide, plus a mole of oxygen gas.0210

Of course, all of these are gases, so let's go ahead and write that, just to be absolutely complete.0227

Now, I'm going to give a table: it's going to be a timetable, a table of time increments.0233

Then, I'm going to give you the concentrations of this, this, and this.0239

Then, we are going to see what is going on here.0242

Let's draw this out; so we have time t: 0, 50, 100, 150, 200, 250, 300, 350, and 400.0245

Now, we have NO₂; and again, square brackets always means concentration in chemistry, in moles per liter.0265

NO₂, concentration of NO, concentration of O₂...so, 0.0100, 0.0079 (it's going to take me a couple of minutes to write this; I apologize), .0065 (I just want to give you as broad a picture as possible), 0.0055, .0048, and then these are 0 and 0...1, 2, 3, 4, 5; you know what, let's just stop with 5 of them.0274

0.0021, .0011, 0.0035, .0018, 0.0045, 0.0023, .0052, 0.0026, and it goes on; but we have a pretty good idea here, so let's talk about what is happening here.0312

This is a particular reaction that was run, and it is a decomposition of nitrogen dioxide to nitrogen monoxide plus oxygen gas.0347

At different time intervals (at 50 seconds, then 100 seconds, then 150, and so on), I actually measure the concentration.0354

Well, at time 0, where we started, I started off with (let me use red) 0.0100 moles per liter of the NO₂.0362

Notice, this is 0 and this is 0; in other words, nothing has happened yet.0373

This is right at the beginning of the reaction.0377

So, 50 seconds later, I measure the concentration in the flask; I notice that I have 0.0079 moles per liter; there is 0.0021, 0.0011 of the oxygen.0379

That is all this is saying; I am just tabulating different concentrations of all three species at different times--50-second intervals.0392

100 seconds, 150...at 200, I have .0048--just about half; .0052, .0026.0399

Something you should notice here: notice that this column tends to be half of this column.0410

Well, it makes sense; if you go up here and you look at the coefficient, this coefficient is 1; this coefficient is 2.0416

It is telling me that, for every 1 mole of oxygen produced, 2 moles of nitrogen monoxide is produced.0422

These numbers confirm the stoichiometry.0429

This is a standard sort of kinetic data.0432

You are measuring concentrations, and you are taking time.0438

Well, let's go ahead and see what the graph of this looks like.0441

OK, so let me move over here; now, let's go ahead and draw this graph: it will be something like this.0445

This is the time axis, and this is the concentration axis; I'll just put brackets for concentration.0453

Let's see, let's do 1, 2, 3, 4...let's do 0.0100 here, and let's...so these are in increments of...0.0025, 0.0050, 0.0075; that is right--I know how to do math.0464

Then we have 1, 2, 3, 4, 5, 6, 7, and 8; OK, so now, we are going to take the data that we have, and we are going to graph it.0491

One of the graphs that we have looks like this: it goes down to about something like that; we have another one that goes up to about like that.0507

And then, we have another one which is about halfway--something like that.0527

OK, so this right here--this is the NO₂.0533

It starts off at 0.0100 concentration, and as time passes, it decreases; the concentration goes down, down, down, down, down.0539

Now, notice what is interesting to notice about this: this is not a straight line; this is a curve--it starts at a certain thing, but it actually starts to slow down--it starts to stabilize and become...OK, so that is interesting.0547

This up here--this graph--is for the NO.0558

As the reaction starts, there is none of it; the concentration increases as time goes by: boom, boom, boom, boom, boom, boom, boom.0564

Because the stoichiometric coefficient is 2 NO₂, 2 NO, this line and this line are actually just copies of each other.0573

As this one drops, that one goes up, and the crossover is going to be completely symmetric.0581

Now, this line over here--this graph--that is the O₂.0587

Notice, it is half of the NO, which works with the stoichiometry; for every 1 mole of O₂, 2 moles of NO is actually produced.0591

This gives a graphical representation of what we have.0601

Now, let's just find an average rate from, let's say, from t0 to t50; so, average rate that equals Δ...in this particular case, let's do ΔNO₂ over Δt.0604

Let's take from 0 to 50--the rate from 0 to 50; it is going to be...Δ is final minus initial, so it is going to be 0.0079 (after 50 seconds, that is the concentration of the NO₂--it has gone down), minus the initial, which is 0.0100.0630

The time difference is 50 minus 0; what you end up with is -4.2x10^-5 moles per liter per second.0657

So, remember these numbers right here: this is concentration--it's moles per liter.0672

That means that the nitrogen dioxide is actually decreasing, on average, between 0 and 50 seconds--the first 50 seconds of the reaction, it is decreasing at 4.2x10^-5 moles per liter per second.0676

That is what this negative sign tells me--it tells me that it is decreasing; it is confirming this.0689

You also know this from the slope; the slope along that curve is negative.0695

OK, now, in kinetics, it is customary to express all rates as positive.0700

So, this negative 4.2x10^-5--we will usually express it as a positive rate, so we will actually just negate it, so it will be something like this.0710

Let me go back to blue here.0721

The rate is equal to -ΔNO₂/Δt, equals -(-4.2x10^-5) moles per liter per second.0725

Now, again, this is just a customary thing; when you get a negative rate, you know that something is decreasing; by now, chances are that you have had calculus or you are in calculus--you have a sense of what a rate is.0744

A negative rate means something is decreasing; a positive rate means that something is increasing.0760

Again, I myself don't particularly care for using positive rates--I like to keep things...I like to use the negative and positive signs, but it is customary, so I just thought that you should know that.0764

Symbolically, it looks like this.0773

This is the symbol that you will see: -ΔNO₂ when something is a reactant that is actually decreasing as you move forward in a reaction.0776

OK, now let's see some average rates for different time intervals.0786

Let's do 0 to 50, 50 to 100, 100 to 150; so, let's put together another table here: the time interval and the average rate (which is -ΔNO₂/Δt); we are just going to calculate rates for different time intervals.0791

OK, and again, this is just a symbol; that is why this negative is here.0812

From 0 to 50, we already calculated as 4.2x10^-4; again, we are keeping it positive.0815

From 50 to 100, when we use the data, we end up with 2.8x10^-4.0825

So again, we take the concentration at 100 minus the concentration at 50, divided by 50, because 50 seconds passed.0835

That is what we are doing; and then, from 100 to 150, you have 2.0x10^-4.0843

This is going to be in (by the way, the units are) moles per liter per second.0852

If we go from 150 to 200, and then we will go from 200 to 250; we will get 1.4x10^-4, and we will get 1.0x10^-4.0858

So, at different times along the reaction (from 50 to the next 50, next 50, next 50), notice: the rate of the reactions, the rate of concentration decline--it is not constant.0871

That is confirmed from the graph, of course; if it were constant, you would have seen a straight line, but the graphs don't actually make a straight line--they actually slope down a little bit, so...different values.0886

OK, so not the same--the rate actually decreases.0899

Now, these are average rates: averages...sometimes we want the instantaneous rate at a given time.0904

Instantaneous rates: for an instantaneous rate, what you do is: you take the graph; you draw a tangent line at a particular point that you are interested in; and then you take two points on that line, and you calculate the slope.0913

The derivative is what you are calculating, but it is just the slope of the tangent line; so those of you who have had calculus--it is just--you know that that line...the slope represents the derivative of that graph, if you were to have an equation for it.0936

More often than not, you will not, because you are dealing with raw data.0951

You are going to take the data, make the graph, draw the tangent yourself, and then estimate.0954

So, for an instantaneous rate, we form the tangent line to the graph at (t, f(t)), or (t, concentration)--that is what f(t) is--it's the concentration--so maybe I should write that.0961

Time t, and the concentration of whatever species you have to be dealing with...then, calculate the slope.0991

That will give you an instantaneous rate: calculate the slope.0999

OK, so let's do something like that.1007

Let's take our graph, and we have a particular curve, and then we are going to draw a tangent line to that; let's say we choose 100--that is, we want to know what the instantaneous rate is; what is the rate of the reaction at t=100?1011

OK, well, you draw the tangent line; you pick a couple of points on that tangent line; you see where they hit the y-axis; and then, of course, you see where they hit the x-axis.1034

You take Δy/Δx (in other words, that is x₂; that is x₁; y₂, y₁); you just take y₂-y₁/x₂-x₁ from points that you pick from the line that you drew (not other random points).1050

The point that it passes through--the tangent--is at the point that you are interested in; so, you will get (when we do this calculation based on the data that we have earlier--this is just a small portion of the other graph around 100)...the rate ends up being 2.4x10^-5.1072

And again, it is going to be moles per liter per second.1094

So, at 100 seconds, that is the rate of decrease of nitrogen dioxide, based on the particular data that we collected.1097

OK, well, notice: we did the reactants; we had NO₂ going to NO + O₂; balance it.1106

Well, what if I wanted to express the rate with respect to the products, either nitrogen monoxide or oxygen gas?1118

I can do that, and it ends up looking something like this.1126

Rate of NO₂ consumption: well, I know that that equals the rate of NO production; it is also equal to twice the rate of O₂ production.1130

That is what the stoichiometry tells me--that the rate of this is twice as fast as this, because the ratio is 2:1.1158

Well, the symbols are -ΔNO₂/Δt; this is a symbol for the rate.1165

Again, we put a negative sign here so that everything is positive, because this Δ is going to be negative; so negative, negative makes this positive.1173

Well, that is equal to ΔNO/Δt; that is equal to twice the ΔO₂/Δt.1182

That is it; it doesn't really matter; more often than not, we tend to work with reactants--the depletion of the reactants.1193

If you wanted to, you can express it as the increase of the products, but again, most of the time, we'll be talking about (it's just sort of standard, at this point, to talk about) the depletion of reactants.1202

OK, now, let's talk about what it is that is actually going on here, and how it is that we can derive some sort of a rate law, some sort of a mathematical expression.1217

This is a symbol, and we have some data; how do we take some of this data and convert it into an actual, mathematical formula that expresses just how fast the rate is changing?1228

These right here give me rates of change of concentrations.1241

OK, if we take the reaction 2 NO₂ → NO + O₂, well, when you start the reaction, there is no nitrogen monoxide, and there is no oxygen gas.1248

So, the reaction is just going to fall forward as fast as it can.1265

Well, in chemistry, many reactions...some go to completion all the way, but at some point, you actually do reach a point called equilibrium (which we will talk about later in more detail), but understand that this reaction is actually reversible.1268

At some point, enough nitrogen monoxide has formed, and enough oxygen has formed, so that now the reverse reaction starts to take place.1282

Well, so now you have the forward reaction of the decomposition of nitrogen dioxide, and you have the backward reaction, the reverse reaction, of the formation of nitrogen dioxide.1290

Things can get kind of complicated here.1300

So, what we do when we run these kinetic experiments is: we try to measure concentrations as close to t=0 as possible.1303

In other words, start the reaction, and immediately take all of the measurements that you need, in order to find the rate of the reaction, before this reaction has had a chance to get to a point where the reverse reaction becomes significant--where it starts to change the actual depletion of the nitrogen dioxide.1314

If I measure the rate of the depletion of nitrogen dioxide right at the beginning of that calculation, I don't account for how fast...I don't have to worry about when the reaction has run a little bit, and now the reverse reaction is going to affect the rate at which this depletes.1333

It changes things; it is very, very complex to deal with that mathematically and physically.1353

What we try to do is: we try to measure reaction rates right at the beginning of reactions.1358

When we do that, as it turns out, under these conditions, the rate is actually proportional to the concentration of the reactants, raised to certain powers.1366

So, in the case of NO₂, as it turns out, the rate of the reaction--the actual rate of the reaction (this is just a rate of consumption and depletion; that is not the rate of the reaction) actually ends up being proportional to...raised to some power.1384

The rate is equal to some constant, K, times the concentration of nitrogen dioxide, raised to some power.1404

Now, this K and this n have to be determined experimentally.1416

I can't just read it off from the equation.1422

Let me say this again; let me move to another page and write this again.1425

Now, as it turns out, the rate of a reaction is proportional to some power of the reactant concentrations.1429

If I have some species, A, going to B + C, the rate of the reaction, which is symbolized as -ΔA/Δt, is equal to some constant, K, times the concentration of A raised to some power.1465

This right here is what we call a differential rate law.1486

Now, what if you had something that had 2 reactants, A + B, going to...I don't know, C + D.1491

Well, now the rate law (and you can choose any one of these to actually symbolize it; let's just go with A again): -ΔA/Δt (this is just the symbol for the rate) is equal to some constant, K, times A to some power, times B to some power--because now, both reactants are involved.1498

Again, if you have 3, you will have 3 terms; it is just like that.1528

So again, the rate of a reaction is proportional to some power of the reactants' concentration.1532

Depending on whether you have one reactant or two reactants...if you have 1 reactant, you have 1 term like this; if you have 2 reactants, you have 2 terms like this...1537

This is called a rate law; specifically, it is called a differential rate law.1549

Now, again, this K, this n, and this m--they have to be determined experimentally; we have to run the reactions, and we have to work this out.1554

We will actually do this in the next lesson; we will actually talk about how to derive this rate law; it is actually quite nice--quite easy.1562

OK, now, let's write: these are differential rate laws.1571

We also have something called integrated rate laws, which we will also talk about in, not the next lesson, but the lesson afterward.1591

These are called differential rate laws.1596

It is when the rate of a reaction is expressed as some constant, times the concentration of the reactant, raised to some power.1600

If there is 1 reactant, that is 1 thing; if there are 2 reactants, you include those; if there are 3 reactants, you include all of those.1609

You would have Aⁿ, B^m, C^s power...something like that, times some constant.1615

What is important is: the K, the n, and the m are determined experimentally.1621

You can't just read them off from the stoichiometry.1625

These coefficients in the actual equation have nothing to do with these numbers.1628

OK, now, let's see: let's actually go through a sort of basic example of determining a particular rate law for a particular reaction.1634

Let's start with a reaction: 2 N₂O₅ → 4 NO₂ + O₂.1647

OK, now let me just put in carbon tetrachloride; so this happens in carbon tetrachloride solution.1661

I'm actually going to...you know what, let me actually write the equation again up here.1670

2 N₂O₅ → 4 NO₂ + O₂, in CCl₄.1679

OK, so now we have some time measurements, and we have some concentration measurements: N₂O₅...1692

0, 200, 400, 600, 800, 1000, 1200, 1400; we have 1.00, 0.88, 0.78, 0.69, 0.61 (not 71), 0.54, 0.48, 0.43.1702

OK, so we have this set of data that we collected at different time increments; we measure the concentration of the N₂O₅, and sure enough, it is decreasing.1750

Now, when we graph this (I'll just do the graph right here), we are going to end up with something like...if this is 1, start off with 1; let's go ahead and put 0.2 down here, and let's put 400, 800, 1200, 1600...so, it's going to look something like this.1763

This data is just plotted on a graph: 400, 800, 1200, 1600--nice and simple; nothing particularly complicated.1791

This is time; this is concentration of the species; OK.1805

So now, what we want to do is: we end up actually...so now, I'm going to leave this here, and I'm going to move to the next page...when I take that graph, and I pick a couple of points, I can actually measure the concentration of N₂O₅ at particular points.1810

When I do that, I have the following (I can actually measure the rate, the instantaneous rate; remember, you draw tangent lines at a given point; you measure the rate at that point; you measure the rate someplace else)...1836

So, at a concentration of 0.90 moles per liter, the rate ends up being 5.4x10^-4.1857

0.45: 2.7x10^-4; now, notice what has happened here.1871

At a certain concentration, I measured the rate of the reaction with a tangent line.1884

At another concentration, I measured the rate; but notice the relationship between these two--I did this specifically.1891

This is half of that; well, this is half of that; what happens here--the correlation--this also happens to be half of that.1899

So, when you have something like that, where when the concentration is halved, the rate is also halved (it won't always happen like this; this just happens to be this particular data that we did)--we arranged it so we can check the rate at concentration that is easily comparable.1911

.9; .45; this is twice that; this is half of that; as it turns out, when I halve the concentration, the reaction rate is also cut in half; that is a linear relationship--that is the whole idea of it.1946

If you double something, and the thing that you are measuring also doubles, that is a linear relationship.1963

If you double something, and the thing that you are measuring goes up by a factor of 4, that means it is a quadratic relationship.1970

2, 4; 3, 9; here, you have halved it or doubled it, depending on which direction--how you want to look at it.1977

Here, you have doubled something; the difference has doubled; so it is a 1:1 relationship--that means it is a linear relationship: linear means an exponent of 1.1987

So, our rate law is: this is the symbol when we write the rate; in fact, oftentimes, you don't really need this symbol; I know that this symbol actually confuses a lot of students--it doesn't need to; I don't particularly care for it very much.1998

It is just the symbol--it means rate; for all practical purposes, you can actually leave this off and just write "rate =."2018

We said that the rate equals some constant (K), times the concentration of the reactant (N₂O₅), raised to some power (n).2026

Well, here, because the relationship is linear and I read it straight off, when a concentration is halved, the rate is halved; or when a concentration is doubled, the rate is doubled.2037

That is a linear relationship; that means n equals 1.2049

So, I don't need this; so this is the rate, equals some constant (K), times N₂O₅.2055

That means the rate of this reaction, based on the data, is actually some constant (K), raised to the concentration of the di-nitrogen pentoxide, raised to the first power.2071

This is the differential rate law for this particular reaction.2083

The most important thing to take away from this is that, under conditions where we actually measure rates, near points where the reaction is just starting, where we have a concentration and then another concentration: we have the rate at one concentration and the rate at the other concentration.2101

When we see how the concentration changes, and then we compare it to how the actual rate changes, that gives us the exponent of the particular concentration term of the reactant.2128

That n is called the order--the order of the reaction.2140

This is a first-order reaction; so the decomposition of di-nitrogen pentoxide is a first-order reaction.2148

Now, I'm going to go ahead and stop this lesson here; I just wanted to give you sort of a brief run-down about how this works.2155

Next lesson, we are actually going to get deeper into this, and we are going to calculate differential rate laws; we are going to calculate the exponents; and we are going to calculate the rate constants more systematically.2161

But, I just wanted you to see and get a sense of what this looks like and how things operate.2173

Thank you for joining us at Educator.com; we'll see you next time for a greater discussion of kinetics; goodbye.2178