For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

### 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

### AP Chemistry Online Prep Course

### 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 NH _{4}^{+} concentration; NO_{2}^{-} 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 NO _{2}^{-} 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 NH _{4}^{+} raised to some power (m--actually, let me use n, because I want to...); and the other is NO_{2}^{-}, 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 NH _{4}^{+}, which in Experiment 1 is 0.100 to the m power, and the concentration of NO_{2}^{-}, 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 NH_{4}^{+}, times the concentration of NO_{2}^{-}.*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

*That is our task, OK?*1219

*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: BrO _{3}^{-}, 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) -ΔBrO _{3}^{-}, over Δt, just a symbol, is equal to some constant, K.*1625

*BrO _{3}^{-} 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

1 answer

Last reply by: Professor Hovasapian

Sun Jul 3, 2016 7:31 PM

Post by Jeffrey McNeary on July 1, 2016

I think it's really cool how there is a concrete, quantitative relationship between concentration of reactants and the rate at which those reactants are used up, the quantitative relationship being: rate = k [ ] ^n. How crazy is it that we can sometimes capture the physical world in a mathematical equation...

However, why do we need to know how to use the method of initial rates which requires slow, tedious calculations by hand, when I'm sure a calculator could immediately find the values of k and n?

Is it because the AP curriculum simply requires it, or is there some significance that I am missing? I don't mean to be rude in any way. Something isn't clicking in my head...

1 answer

Last reply by: Professor Hovasapian

Thu Jan 7, 2016 10:28 PM

Post by Gaurav Kumar on December 30, 2015

Hi Professor,

I just want to make sure I got the concept down correctly. When you write the experimental values, those are not solved for correct? Those values are "given" values? If not, I am a little confused on how you get the values.

Thank you

1 answer

Last reply by: Professor Hovasapian

Fri Mar 14, 2014 5:22 PM

Post by Daniel Nguyen on February 17, 2014

For example 2, in experiment 3, shouldn't the BrO3- ion concentration be 0.10?

2 answers

Last reply by: Professor Hovasapian

Wed Jan 8, 2014 5:07 AM

Post by Peter Tin on January 7, 2014

For example 2. Is the overall order of this reaction = 3?

1 answer

Last reply by: Professor Hovasapian

Fri May 17, 2013 6:18 PM

Post by Nawaphan Jedjomnongkit on May 17, 2013

Thank you for the lecture, from example 1 equation , how to balance the equation? From what you give it's non balance and I don't know how to balance it.

2 answers

Last reply by: Sally Acebo

Tue Feb 3, 2015 4:25 PM

Post by Rajendran Rajaram on May 6, 2013

aaaaaaaaaaaaaah, The concentration of the nitrite ion in experiment 1 and 3 should be .05 not .005!

2 answers

Last reply by:

Mon Apr 8, 2013 7:49 AM

Post by success10 on April 6, 2013

Professor Raffi Hovasapian: I hope this finds you well. Thank you for your lectures, I am enjoying them immensely, both in sciences and math.

I have a doubt related to this and the previous lectures. Given for example the reaction in example 2 (Bro2- + 5Br- + 6H+ -->3Br2 + 3H2O), suppose I start with a concentration 10 M of each reactant in 1 litter of solution, so that I have 10 mols of each reactant at time 0. Suppose that at time 1 sec I mol of BrO2- has been consumed, then the new [BrO2-] is 9M and the Delta[BrO2-] is 1 M. Using the stechiometry of the reaction 6 mols of H+ should have been consumed, then the new [H+]= 4M and the Delta[H+] = 6M. Then if I define the rate by the BrO2 it is 1M/s, but if I define it by the H+ it is 6M/s. However, from the lectures I infer the rate is the same no matter which reactant is used, and equal to k[BrO2-][Br-][H+]^2 (The same product no matter which reactant was used to define it). I am missing something I cannot find it. Could you please help me with this point? Thank you. Silvia

1 answer

Last reply by: Professor Hovasapian

Mon Oct 29, 2012 6:18 PM

Post by Dana Meredith on October 28, 2012

Correction, only experiment #1, the concentration of the nitrite ion should be .05

0 answers

Post by Dana Meredith on October 28, 2012

The concentration of the nitrite ion in experiment 1 and 3 should be .05 not .005

1 answer

Last reply by: Justin Jones Jones

Wed Jun 20, 2012 7:12 AM

Post by Xinyue Shen on May 5, 2012

Example 1, I believe the concentration of NO2 should 0.05 instead of 0.005!

If I misunderstood something, please point that out.