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

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

### Reaction Rates and Rate Laws

- Customarily, all rates are positive.
- Reaction Rates are most often expressed with respect to a Reactant.
- An Instantaneous Reaction Rate is determined from the Concentration (y) vs. Time (x) Graph by choosing a point tangent to the graph, drawing the tangent line, then choosing 2 points on the line from which to calculate the slope. This slope is the Instantaneous Rate at that moment.
- Rate Law and Differential Rate Law are synonymous. A Rate Law is an expression that shows how rate is a function of reactant concentration.
- The overall order of a reaction is the sum of the orders for each species in the Rate Law.
- Order and Rate Constant must be determined experimentally.

### Reaction Rates and Rate Laws

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 2:18
- Rate: 2 NO₂ (g) → 2NO (g) + O₂ (g)
- Reaction Rates Graph
- Time Interval & Average Rate
- Instantaneous Rate
- Rate of Reaction is Proportional to Some Power of the Reactant Concentrations
- Example 1

### AP Chemistry Online Prep Course

### Transcription: 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 NO _{2}; and again, square brackets always means concentration in chemistry, in moles per liter.*0265

*NO _{2}, concentration of NO, concentration of O_{2}...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 NO _{2}.*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 NO _{2}.*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 NO _{2}, 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 O _{2}.*0587

*Notice, it is half of the NO, which works with the stoichiometry; for every 1 mole of O _{2}, 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 ΔNO _{2} 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 NO _{2}--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 -ΔNO _{2}/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: -ΔNO _{2} 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 -ΔNO _{2}/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 x _{2}; that is x_{1}; y_{2}, y_{1}); you just take y_{2}-y_{1}/x_{2}-x_{1} 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 NO _{2} going to NO + O_{2}; 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 NO _{2} consumption: well, I know that that equals the rate of NO production; it is also equal to twice the rate of O_{2} 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 -ΔNO _{2}/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 ΔO _{2}/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 NO _{2} → NO + O_{2}, 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 NO _{2}, 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 A ^{n}, 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 N _{2}O_{5} → 4 NO_{2} + O_{2}.*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 N _{2}O_{5} → 4 NO_{2} + O_{2}, in CCl_{4}.*1679

*OK, so now we have some time measurements, and we have some concentration measurements: N _{2}O_{5}...*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 N _{2}O_{5}, 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 N _{2}O_{5} 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 (N _{2}O_{5}), 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 N _{2}O_{5}.*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

1 answer

Last reply by: Professor Hovasapian

Thu Feb 23, 2017 1:10 AM

Post by Kaye Lim on February 21 at 12:00:51 AM

Hello Raffi,

For example 1, I thought that rate of depletion of N2O5 is different than reaction rate (you mentioned this earlier in the lecture), why would the slope of the tangent line for 0.45 and 0.90M concentration be reaction rate. Is the slope of tangent line for a certain y value in [A] vs. t graph be instantaneous reaction rate or is it instantaneous rate of depletion of N2O5?

-for the slope of small time interval, is it average rate of depletion of the reactant in that time interval? or is it average reaction rate of the reactant in that time interval?

4 answers

Last reply by: Christian Fischer

Tue Jul 15, 2014 4:21 AM

Post by Christian Fischer on July 9, 2014

Hi Raffi, can you help me understand what exactly is meant by "rate of a reaction": According to the rate law for the reaction A--> B + C

I would guess "the rate of the reaction" is the same as the rate which reactant A is turned into product. But for the reaction A + B --> C+D

It looks as if the "Rate of the reaction" only means "the rate of one of the reactants in the reaction", so I get confused because to me it seems like the rate of the reaction always means "the rate at which one of multiple reactants turn intp product" so it has nothing to do with the reaction as a whole, but only a single reactant.

1 answer

Last reply by: Professor Hovasapian

Fri Feb 7, 2014 6:56 PM

Post by Tim Zhang on February 7, 2014

could you tell me am I correct on this statement? the graph of the rate function appears be a curve is because the reverse formation of the product in a chemical reaction?

4 answers

Last reply by: Professor Hovasapian

Tue Apr 23, 2013 4:22 PM

Post by Antie Chen on April 22, 2013

Hello, Raffi.

I have some question about the equation rate=-(delta)[A]/(delta)t=K[A]_a*[B]_b

K=[C]_c*[D]_d/[A]_a*[B]_b, so the rate=[C]_c*{D]_d?

and I also confused about the unit of the equation...

In addition, your said that n is called the order, but I can't got it, can you explain me?

2 answers

Last reply by: Gayatri Arumugam

Mon Nov 26, 2012 9:35 PM

Post by Gayatri Arumugam on November 8, 2012

At 14 min, wouldn't the 0--> 50 seconds be 4.2x10 -5?