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### Method of Initial Rates

• We run the reaction several times, starting with different Initial Concentrations, then measure Initial Rates.
• We then compare experiments to see how a change in Concentration influences the Rate.
• Once we find the orders for the respective reactants involved, we can substitute the values of any of the Experiments to recover the Rate Constant.

### Method of Initial Rates

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 0:33
• Rate
• Idea
• Example 1: NH₄⁺ + NO₂⁻ → NO₂ (g) + 2 H₂O
• Example 2: BrO₃⁻ + 5 Br⁻ + 6 H⁺ → 3 Br₂ + 3 H₂O

### Transcription: Method of Initial Rates

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

Today, we are going to continue our discussion of reaction kinetics, chemical kinetics.0004

In the last lesson, we introduced rate laws--in particular, the differential rate law.0010

Let's just recap what that is, and we'll get into the actual method (or one of the methods) of determining that differential rate law for a particular reaction, given a particular set of data.0015

The method is called the method of initial rates; let's go ahead and get started.0029

We said that (you know, I think I'm going to use black ink now) the rate of a reaction, which is how fast it is going, symbolized by -Δ some species (reactant species), where this negative comes from...Δt is equal to some constant, times the product of the reactants (however many there are), raised to (A, B...let's say there are three reactants) certain powers.0035

These powers are called the order of that reactant.0079

n, m, s...the order of the reactant; this m, s, and n, and K, are experimentally determined.0085

The idea is this (we talked a little bit about it last time, but let's be more specific about it)--the idea is: we want to start a reaction, and we want to measure the rate of that reaction, the rate of depletion of this particular thing (the reactant) as quickly as possible, before any of the products have had a chance to build up and start working in reverse, because as product builds up, the reaction goes backwards.0103

That is the thing; chemistry, as it is going one way, is also going the other way; so, before it becomes too complicated, we want to see if we can measure the rate of the reaction--before anything else gets in the way.0131

Let's just write this down--so, the idea is this: we run several experiments with different initial rates...not initial rates; I apologize--initial concentrations.0144

We pick specific concentrations to start with--reactants and products--and we measure it immediately.0171

We start with different initial concentrations and measure the initial rates (that is what we are measuring--initial rates) before any products have an opportunity to build up and complicate matters.0179

OK, we then compare the rates of the different experiments...we then compare the rates among experiments...to see how the rate depends on the concentration (or specifically, the concentration change from one experiment to the other).0228

It depends on the concentration of a given reactant.0272

This will make a lot more sense when we actually do our first example, which we are going to do right now.0277

We are going to run a series of experiments--2, 3, 4, 5 of them--and we are going to modify concentrations (initial concentrations).0283

So, if I have an Experiment 1 and Experiment 2, and initial concentration 1 and initial concentration 2, I'm going to measure the rate; and depending on the concentration changes, I am going to compare the rates, and I'm going to see what the relationship is (if it doubles; does it go up by 4? does it go up by 6? does it go up by 19? 14.3?), and that is how I get those exponents.0295

That is how I get these numbers first.0318

Then, when I am done with that--once I get the orders of the reactions--then I can take any one of the data in the experiment and put it in here to find K.0320

Let's go ahead and do that; it will make a lot more sense.0329

OK, so let's take (you know what, I think I'm going to start with a fresh page here)...let's do the following reaction.0332

Let's take ammonium ion, plus the nitrite ion; when I mix those together, I am going to end up with nitrogen dioxide gas, plus 2 moles of liquid water.0341

Ammonium and nitrite ion forms nitrogen dioxide gas, which bubbles off; and I am left with liquid water.0361

Notice, I have two reactants--two reactants; I did three experiments on this; here is my data.0372

Experiment #1, Experiment #2, Experiment #3 (you know what, I probably don't need to make them this big--leave myself some room here--Experiment #2, Experiment #3)...0390

Now, I have NH4+ concentration; NO2- concentration; and I have initial rate, which is what I am measuring.0403

Now, my first experiment: I started (remember, we are dealing with reactants here; we are measuring the rates before any products have a chance to build up)...0.100 moles per liter, and here I started with an NO2- concentration of 0.0050 moles per liter.0418

When I ran this experiment, right as it started, I measured the rate to be 1.35x10-7.0441

Now, I ran a second experiment: 0.100; I left this concentration the same, but I doubled this one: 0.00...no, I doubled this concentration: 0.0100.0450

When I make a change from experiment to experiment, I change one species--I don't change everything.0467

The reason is because I need two experiments to compare; later, what I'm going to do is...notice, I have left this the same, and I have left this the same, but I have changed this; I'm going to use experiments 1 and 2 to compare, to find out how it's related to nitrite ion concentration--how the rate depends on this concentration.0473

I change one species at a time; I don't change all of them.0493

I left this the same and changed this one; when I do that, I ended up with a rate of 2.70x10-7.0496

Now, Experiment 3: 0.200; now, I doubled the ammonium concentration, and I left this concentration the same.0505

When I do that, I ended up with a rate of 5.4x10-7.0517

So, my first experiment; second experiment; third experiment: .1, .005, .1 (wait, I'm sorry, this is .1, not .01--I knew I had too many...I was going to say it looks a little odd); .100, and this is .100; that is right, .2.0523

OK, so .1, .1 initial; initial rate; and then, I do another experiment, and I start with .2 mol of this, .1 mol of that, and I measure an initial rate, 5.4.0548

Now, I can do my comparison.0561

OK, so here is where the good stuff starts.0563

Now, we said that the rate, which we will symbolize as -Δ (and I can pick any one--I am just going to go ahead and pick this one, so this is just a symbol); it's equal to some constant (K), times the NH4+ raised to some power (m--actually, let me use n, because I want to...); and the other is NO2-, raised to the power of m.0566

So this is it; this was our assumption: under these conditions of initial rate, as it turns out, the rate is dependent on the product of the concentrations of the reactants, raised to different powers.0601

I write this, and now I am going to use this experiment to determine m, to determine n and determine K.0615

OK, now, Experiment 1 says that the rate is equal to 1.35x10-7.0622

Well, the rate is equal to K, times the concentration of NH4+, which in Experiment 1 is 0.100 to the m power, and the concentration of NO2-, which is 0.0050 to the n power.0641

I just literally plug in these numbers into what I wrote down.0662

Experiment 2 says (I'll say this is rate 1, and this is rate 2, of course): rate 2 is 2.70x10-7, and that is equal to K times...now I use the values for Experiment 2: .100 to the n power, times 0.100 to the m power.0669

OK, now I compare the two rates.0698

Rate 1, divided by rate 2, is equal to 1.35x10-7, divided by 2.70x10-7, equals--this over that equals--K times 0.100 to the n, times 0.0050 to the m (there is a lot of writing here, but it is important to see everything), over K times 0.100 to the n, over 0.100 to the m.0704

Now, look what happens: 1.35x10-7, over 2.7x10-7 (let me make this a little more clear here--this looks a little confusing...10-7)...this is equal to one-half.0742

K cancels K; .100 to the n, .100 to the n--those cancel; what I end up with is .0050 to the m, over .100 to the m.0758

This is just .0050, divided by .100, to the m power.0772

Well (oh, I don't know where these lines are coming from!), it equals...this is just one-half to the m power.0780

Well, 1/2 equals 1/2 to the m power; that implies that m is equal to 1.0796

So, because m is equal to 1, that means the nitrite ion concentration, m, is 1; that means it is an order 1 in nitrite.0807

So now, I will do the same to find n; so I have already found m--now I am going to do the same thing for n.0829

This time, I am going to compare Experiment 2 and Experiment 3.0836

Let's go back to black.0841

OK, Experiment 2: the rate is equal to 2.7x10-7; it is equal to K times 0.100 to the n, times 0.100 to the m, over (oh, actually, no, not yet).0845

OK, and then we will do Experiment 3: the rate is equal to 5.40x10-7, is equal to K times 0.200 to the n, times 0.100 to the m.0880

Well, rate 2 divided by rate 3 (and you can do it in either order--you can do rate 3 over rate 2; it doesn't really matter) equals 2.70x10-7, over 5.40x10-7, equals K times 0.100 to the n, 0.100 to the m, divided by K times 0.200 to n, 0.100 to the m.0900

K cancels K; that cancels that; that is equal to 1/2; that is equal to 1/2, this time to the n power.0939

Well, 1/2 to what power is equal to 1/2?--that implies that n equals 1.0949

I found my differential rate law.0956

OK, now my differential rate law is this: my rate is equal to some constant (K), times the concentration of ammonium ion raised to the first power, times the concentration of nitrite raised to the first power.0960

The rate is first-order in each; the total order of the reaction is 2--you add up the orders.0978

So, m is equal to 1; n is equal to 1; their sum is equal to 2, so the overall order of this reaction is 2.0986

That is what I have done--I have compared rates to find this--but I am not done.1001

Now, I want to find K.1004

Now that we have our rate law, that the rate is proportional to the concentration of ammonium raised to the first power, times the concentration of nitrite raised to the first power, I can take any one of those experiments, 1, 2, or 3, plug in the values (I have a rate; I have this concentration; I have that concentration), and I can solve for K, and that is exactly what I am going to do.1005

I'm just going to go ahead and choose Experiment 2.1030

Experiment 2 says: the rate is 2.70x10-7, equals K times (oops, I don't have to actually do the brackets when I'm putting numbers) 0.100 raised to the first power, and the 0.100 raised to the first power.1034

And now, I just go ahead and do my division.1070

OK, let me see here; .1; .01; minus and minus; and I end up with K is equal to 2.70 (please check my arithmetic; my arithmetic is notorious for being wrong) times 10 to the -5.1079

There we go; and the units, in this particular case, is liters per mole-second.1098

The reason is because we have a rate, which is moles per liter per second, and then you are dividing by moles per liter, dividing by moles per liter, so this K ends up with liters per mole per second.1106

K--this is not the unit for all K's; this is just the unit for K for this particular reaction, which is second-order--first-order in both of these.1125

OK, method of initial rates: we found the rate law, and then we used one of the experiments to find K, and then we plug it back in, so our final answer is: the rate equals 2.70x10-5 times the concentration of NH4+, times the concentration of NO2-.1137

That is our differential rate law for this reaction.1164

We are done.1167

OK, let's do another example.1168

This time, we are going to use three reactants.1172

This time, we will do bromate ion, plus 5 bromide ion, plus 6 hydrogen ion, forms 3 molecules of bromine plus 3 molecules of water.1182

3 reactants: our rate law is going to have 3 things in there.1202

OK, you want to determine the order of each reactant; you want to determine the overall order; and we want to determine the rate constant.1207

Let's write it down: determine the order of each reactant, overall order, and rate constant.1221

OK, well, here is our experimental data...well, let's go ahead and write the rate law first--the general one.1242

The rate is equal to some constant (K); it is going to be that raised m power, Br- raised to n power, H+ raised to s power.1249

This is the general form; we are going to find m, n, and s, and we are going to find K, based on the following data.1268

So, we have four experiments; let's see...1274

Experiment: BrO3-, Br-, H+, and initial rate...1284

All right, Experiment 1; 2; 3; 4; 0.10, 0.10, 0.10; initial rate is 8x10-4.1296

0.20, 0.10, 0.10; notice, I have only changed one thing in this case--the concentration of the bromate: bromate, bromide, hydrogen ion.1314

We get 1.6x10-3--that is interesting; OK.1328

0.20, 0.20, 0.10; this time, from here to here, we have changed the bromide.1336

We end up with 3.2x10-3.1344

The fourth one: 0.10, 0.10, 0.20; we have gone back--that is the same; that is the same; that is what is different.1349

We end up with 3.2x10-3.1362

OK, so in order to find m, which goes with the bromate ion, I'm going to use Experiments 1 and 2, because bromate changes from Experiment 1 to 2.1369

Let's go ahead and do that.1384

I'm just going to go ahead and write it out without writing absolutely everything.1387

So, rate 1, divided by rate 2, equals 8x10-4, divided by 1.6x10-3, equals...well, the rate is equal to that whole expression--is equal to K times (and now, I'm going to stop using brackets; I'm just going to use parentheses; but I do mean concentrations) concentration 0.1 to the m; 0.1 to the n; 0.1 to the s, over K times 0.2 to the m, 0.1 to the n, 0.1 to the s.1390

We have a whole bunch of cancellations; the only thing that doesn't cancel is this.1444

So, this number is equal to 1/2, equals 1/2 to the m power; that implies that m equals 1.1450

Therefore, the bromate power is 1; we are done with that.1465

Now, let's compare rate 2 and rate 3; in other words, Experiment 2 and Experiment 3.1473

Rate 2, divided by rate 3, equals 1.6x10-3, over 3.2x10-3, equals K times .2 to the m, .1 to the n, .1 to the s, over K times .2 to the m, .2 to the n, .1 to the s.1480

Cancel, cancel, cancel, cancel, cancel, cancel; we end up with 1/2 equals 1/2, this time, to the n power, which implies that n equals 1.1517

Well, n equals 1; that is the bromide ion concentration.1530

So, bromide is also order 1.1533

OK, now we will do rate 1 compared to rate 4.1537

Rate 1 is 8x10-7 (is that right?); no, it's 8x10-4, not -7.1545

8x10-4; I was going to say: that is a little bit too much.1559

8x10-4, divided by 3.2x10-3, equals K times .10 to the m, .10 to the n, .10 to the s, over K times .10 to the m, .10 to the n (not s yet--.10 to the n--wow, symbols everywhere), .2 to the s.1565

Cancel, cancel, cancel, cancel, cancel, cancel; we end up with 1/4 equals 1/2 to the s power.1602

Well, 1/2 raised to what power gives me 1/4?--that implies that s equals 2, so the hydrogen ion concentration has an order of 2.1613

There we have it: we have our rate, which is equal to (I'm going to go ahead and use one of them as a symbol) -ΔBrO3-, over Δt, just a symbol, is equal to some constant, K.1625

BrO3- to the first power, Br- to the first power, H+ to the second power: this is our differential rate law.1648

You can just...that; this is just a symbol that says "the rate is."1659

The rate of depletion of bromate is equal to some constant, times this, this, this, squared.1666

That is what is going on.1673

Now, I take any one of the experiments; I put the values of the experiments in--that one, that one, that one--and I have the rate, because I calculated the rate--it's part of the data--and I solve for K.1674

So, I'm going to go ahead and take...oh, let's just take Experiment 1; it's not a problem.1686

OK, so I get 8.0x10-4 equals K times 0.10, times (that is the bromate concentration) the bromide concentration--is 0.10, and this other one is 0.10 squared.1692

When I solve for K, I get 8; in this particular case, it is liters cubed, over moles cubed, second.1711

And again, the unit doesn't really tell you all that much; it just tells you what is going on up here.1722

But, the real thing--this number is what is important.1727

The order of bromate is 1; the order of bromide is 1; the order of hydrogen ion is 2.1731

The overall order is 1 plus 1 plus 2; the overall order is 4.1739

The rate constant is 8.1743

OK, that is the method of initial rates: you write down "equals constant K, times the particular species, raised to..." and so on; it's usually not going to be more than 2 or 3; I think 3 is about the most that you are really going to get, as far as reactants are concerned.1749

Then, you take the data that has been collected: Experiment 1 has initial concentrations for each of the reactants, and a measured rate.1768

You compare the rate of one with the rate of the other by putting in values based on this equation, just like we did to derive those numbers.1778

Once you have the orders, you use them, plus one of the experimental values, concentration, concentration rate, to find K.1787

And now, you have your final differential rate law.1798

So, in this particular case, we get that the rate is equal to 8 times that, that, that: the rate of the reaction.1802

Now, I can plug in any concentration I want, randomly: OK, .26, 19, whatever; and based on this that I derived from experiment, I can tell you what the rate of the reaction is going to be--how fast it is going to be.1818

OK, so that was the method of initial rates; we did a couple of examples; next lesson, we are going to talk about the integrated rate law.1836

This was the differential rate law.1843

Thank you for joining us here at Educator.com; we will see you next time; goodbye.1845