Understanding Rate Law Determination

Hello and welcome to The Chemistry Solution. This tutorial is on how to determine the rate law from experimental data using initial reaction rates. The first thing that we need to remember is that rate laws must be experimentally determined. So you can't look at an overall chemical reaction and determine the rate law from the stoichiometry of that reaction. And so in order to determine a rate law, you're going to need some experimental data. So this data can be collected in the lab, typically in a lecture course that experimental data will be provided. But if you were going to do this in the lab, what you would do is you would want to run your chemical reaction for a short time and alter the concentration of just one reactant at a time. When you do this, you want to determine the initial instantaneous rate for each set of conditions. and then that will allow you to compare how the reaction rate changes as the concentration of that one reactant changed. And then you know that any change in the reaction rate was due to that change in the concentration of your reactant. You want to look at each reactant individually and determine whether each reactant is 0 or 1st or 2nd or maybe even 3rd order, and that will allow you to determine the overall rate law. Then once you have the rate law, you can use your reaction rate data to determine the value of the rate constant. So we're going to go through a couple examples of how you would do that. Okay, so let's look at this data here. We have the following experimental data that was obtained for this reaction A plus B going to C. The first thing that I want to do is determine the rate law. We know that the rate law is going to be equal to sum rate constant K, which we haven't calculated yet, but we will be able to do once we determine the reaction order, times the concentration of reactant A raised to some power times the concentration of reactant B raised to some power. And so our job right now is to determine what these powers are that I've listed as X and Y in this rate law. So we're going to start and we're going to look at each of these reactants individually and determine the order for each reactant individually. So let's start by determining the reaction order for reactant A and I'm choosing to start with A here. You could just as easily start by determining the reaction order for reactant B. Whatever order you want to go in is fine. So in order to do that, I want to compare two different experiments where the concentration of A is changed but the concentration of B. remains constant. Okay, so if you look at experiment one and two, you can see that the concentration of A has changed, but we're holding the concentration of B constant. This is really important because we want to know that this change in reaction rate is just due to the change in the concentration of reactant A. And so let's look at that. How does the change in the concentration of A affect the reaction rate? So you can approach these problems from kind of a logical sense and I'll show you how to do that and then I'll also show you how to approach this from a more mathematical technique. So if you look at A here you can see that the concentration of A has doubled from 0.1 to 0.2 molar and you can see that the reaction rate doubled. So if the effect on the reaction rate is the same as how the concentration of your reactant changed So for example, if the rate doubles when the concentration doubles, then the rate law is first order with respect to that reactant. And so you can see that if you were to double the concentration of A here and that led to a doubling of the rate, that would mean that this reaction is first order with respect to reactant A. And it doesn't need to necessarily just double. If you tripled the concentration of A and the rate tripled, or you quadrupled the concentration of A and the rate quadrupled, when you see the same sort of change in your rate as you do in your concentration of a reactant, that means that your reaction is first order with respect to that reactant. But let me show you how we could do this mathematically. So we can look at the relationship between the rates and the concentrations we have of our reactants using proportionalities. And so what I've done here is I've set this up where we're comparing the rate of experiment two, and then here's our rate law to the rate of experiment one, and here's our rate law. So what I'm going to do here is I'm going to put this rate for experiment two in here along with these concentrations that we have for our reactants and compare that to the rate and the conditions for experiment one which I'll put down here on the bottom. So let's go ahead and fill in those numbers there. What this is going to allow us to do is it's going to allow us to solve for this exponent x. So you can see that a lot of terms in this proportionality are going to cancel out. Your rate constant K is going to cancel out. We don't even know what that is yet, but we know that it's a constant, and you see it on the top and bottom, so that's going to cancel out. We have our concentration of B that is the same in both of these experiments here, so even though we don't know why, 0.2 to the y will cancel out with 0.2 to the y. And so this is going to allow us to then solve for this exponent x. And so you can see that if you do the math here and take 7 times 10 to the negative 4th divided by 3.5 times 10 to the negative 4th, that is going to be equal to 2. And then on the right side, you'll get 0.2 divided by 0.1 raised to the x power. Remember that when you're working with exponents, if you have 0.2, let's say, to the x, divided by 0.1 to the x, that is the same as saying 0.2 divided by 0.1 entirely to the x. And so this is going to allow us to solve for that exponent here. You can see that we get the 2 is equal to 2 to the x power. So what is x going to be? Well, in this case, x is going to be equal to 1. And that's what we predicted from looking at this experimental data, that this reaction would be first order with respect to A. So let's go ahead and put that in here. And then now what we're going to be able to do is do this exact same process for reactant B. So we're going to do the same thing. First we need to compare experiments where the concentration of B is changed and the concentration of A remains constant. So in this data that we have here, that's going to be using experiments 1, 3. Again what you're going to be able to see here is that where the concentration of A is held constant, any change in the reaction rate has to be due to the concentration change of our reactant B. So how does the change in the concentration of B affect our reaction rate? So notice in this case the concentration of B was doubled, but look what happened to the change in the reaction rate. the change in the reaction rate didn't just double, the change in the reaction rate quadrupled. And so if you see an effect on the rate that is equal to the effect on the change in concentration squared, so like in this example here we saw that our rate quadrupled when the concentration was doubled, that means that our rate law is second order with respect to this reactant. And again, let me show you how to do this mathematically. So let's compare these two reactions again. Notice when I set up these proportionalities, I put the experiment with the higher rate on the top and the experiment with the lower rate on the bottom. You don't necessarily have to do it that way. You could flip these around and you're going to come up with the same answer, but the math is just a little bit easier to do in your head if you have the larger reaction rate on top. and the conditions that gave you that larger reaction rate on top. So again, we'll just use this proportionality here, and you can see that our rate for experiment 3 is equal to the rate law, which is going to be some rate constant K, we haven't determined that yet, times the concentration of A. Now we know that this reaction depends on the concentration of A to the first power. That doesn't really matter to us at this point because those terms are going to cancel out anyway, but you can still put that in times the concentration of B raised to some power Y. And this is what we're going to try to solve for here is this value of Y. And then put in the same information from experiment one down here. Okay, when you do that, your numbers are going to look like this. And again, you're going to see that there are terms that are going to cancel out. So you can see the rate constant again will cancel out and you'll see that the concentration of A raised to the first power will also cancel out and so now this will allow you to solve for Y. What you're going to get is when you take 1.4 times 10 to the negative third divided by 3.5 times 10 to the negative fourth, you're going to get that that is equal to four. And then over here on this right side, 0.4 divided by 0.2 raised to the y power is two to the y power. Well, what power do you need to raise to two to give you four? Well, two squared will give you 4, and so the value of y here is 2. So we've just confirmed that this reaction is second order with respect to reactant B. Okay, so now that we know this reaction order with respect to each of our reactants, what we can do is we can solve for this value k. Remember we have rates for different concentrations of reactants. Now we know what these exponents are. and so that's going to allow us to solve for our rate constant the nice thing about this is remember that your rate constant k is a constant so you can use the reaction rate data from any one of these experiments that you like and you should always come up with the same value of k so i'm going to pick one here i'm going to use the data from experiment one but you can try this on your own. Double check that the data from experiment two or experiment three gives you the same value of the rate constant as I get here. Okay, so here's my rate law. I'm going to use again the data from experiment one, and so I'm going to put these numbers here. Here's my rate when the concentration of A was 0.1 molar and the concentration of B was 0.2 molar. Don't forget to include these exponents here that we just determined for our reaction order. So now we can divide each side by 0.1 and by 0.2 squared. Notice on the right side that will cancel out leaving us with just k and then the other thing that we want to watch here is units. Notice we have molar that will cancel out molar down here. We're still going to have inverse seconds up here on the top and we have molar squared on the bottom still. So inverse molar squared inverse seconds are going to be our units for this rate constant. And when you go ahead and do the math here, you will get that your rate constant has a value of 0.0875. And then the units, like we just said, are inverse molar squared inverse seconds. So we were able to use experimental data to determine not only our reaction order, the order with respect to each reactant, but also the value of our rate constant. And so now we know the entire rate law for this hypothetical reaction, A plus B going to C. Okay, let's try this one more time. And now might be a good time to pause the video and try this one on your own. So I'm giving you kind of a similar reaction here, A plus B going to C. And here's some experimental data, so let's determine the rate law. So we know the rate law is going to be equal to some rate constant k times the concentration of A raised to some power times the concentration of B raised to some power. I'm going to approach this just like I did the last problem, and I'm going to start by determining the reaction order for reactant A. Again, if you wanted to start with reactant B, that is just fine. So I want to compare experiments where the concentration of A is changed and the concentration of B remains constant. Okay, that's going to be experiments one and three in this data table here. Your concentration of A is changed and your concentration of B is held constant. So how does the change in the concentration of A affect the reaction rate? Notice that the concentration of A here was tripled from 0.2 to 0.6 and notice that my reaction rate increased by a factor of 9. So again, if you change the concentration by some amount and the effect that you see on the reaction rate is equal to this amount squared, then we know that our reaction is second order. with respect to that reactant. But let's just prove this to ourselves mathematically. And so let's look at the relationship between the rates and the reactant concentrations in experiments one and three. Again, when I like to set up these proportionalities, I like to have the data for the experiment that gives me the largest reaction rate on top. So you can fill in all these numbers here and same thing for the data from experiment one. And notice again you're going to have a lot of things cancel out. The rate constant will cancel out and the concentration of B raised to the y power will also cancel out. That's going to allow us to solve for this term x. And so when you do that, you get 1.35 times 10 to the negative third divided by 1.5 times 10 to the negative fourth, which is 9, is going to be equal to 0.6 to the power of x divided by 0.2 to the power of x, which is 3 to the power of x. And so when you solve for x, you get that x is 2. And that's what we predicted looking at the data. And so anytime you increase the concentration of some reactant by a certain factor and the reaction rate increases by the amount of that factor squared, that indicates that your reaction is second order with respect to that reactant. Let's do the same thing for B. So let's compare experiments where the concentration of B is changed and the concentration of A remains constant. So that's going to be experiments one and two. Well, look at what's happening here. So here, we've changed the concentration of B. We've doubled it. Notice we see no change in the reaction rate. So what does that mean? What that means is that the rate law is going to be zero order with respect to this reactant, meaning that the concentration of B doesn't affect the rate. Again, let's just prove this to ourselves mathematically. Let's compare the data for experiments 1 and 2. You can fill in the rates and the respective concentrations. You'll see again k will cancel out, and the concentration of a squared will cancel out, and you will get 1.5 times 10 to the negative 4th divided by 1.5 times 10 to the negative 4th, which is 1, is going to be equal to 0.4 raised to the power of y divided by 0.2 raised to the power of y, which is 2. Well, 2 raised to what power will give you 1? Remember how these exponents work, that anything raised to the 0 power is equal to 1. And so in this case, y has to be 0. So you could write your rate law like this. Or if you have a reactant concentration that's raised to the 0 power, a lot of times that's just omitted from the rate law. And we would just write our rate law as our rate is equal to the rate constant. times the concentration of A squared. It's independent of the concentration of B as long as there's some B present. Okay, let's just finish up this rate law here and let's use some reaction rate data and our newly determined rate law to determine the value of the rate constant. Again, you can pick any experiment number. I picked experiment number one, and so I'm just going to fill in my rate, the concentration of A, that gave me that rate. I know that this reaction is second order with respect to A, and now we can solve for our rate constant. So we'll divide each side by 0.2 molar squared. Notice again, we want to pay attention to the units of our rate constant. And so here we're going to have molar, which will cancel out one unit of molar here, but we have molar squared on bottom. So we'll still have inverse seconds. And then inverse molar will be the units for our rate constant. Okay, so let's cancel out the things that are the same on both sides. And then when you do the math, you will get a value for your rate constant of 3.75 times 10 to the negative third. Inverse molar, inverse seconds. A couple of last things that I want to mention. Typically in these types of problems, you are going to be solving for exponents and reaction orders that are integer numbers. Reaction orders aren't necessarily always integers, but that indicates a more complex reaction with regard to chemical kinetics. And so if you're taking just a general chemistry course, you are typically going to be looking for reaction orders that are integers. The other scenario that I didn't cover in here, but maybe we should just talk about quickly, is what if you had a change in concentration, say, say this change in concentration of A went from 0.2 molar to 0.6 molar, say it increased by a factor of 3, and let's say the rate... increased by a factor of 27. What would that mean? Well, think about what power would you have to raise 3 to to give you a corresponding increase in reaction rate of 27. That would mean that your reaction would have to be third order with respect to that reactant. So those are typically the three you're going to see. So if you see the same increase in reaction rate as you do an increase in concentration of reactant, concentration of reactant doubles, reaction rate doubles, or concentration of reactant triples, reaction rate triples, that always indicates first order. But if you see a change in concentration, say, that is doubled and the reaction rate increases by a factor of four, or your concentration of a reactant triples and your reaction rate increases by a factor of nine, that's going to indicate a second order reaction. And then I talked about what you would see if a reaction was third order with respect to a particular reactant. Okay, so that wraps things up for this tutorial. I hope that you found this helpful.

And so in order to determine a rate law, you're going to need some experimental data. So this data can be collected in the lab, typically in a lecture course that experimental data will be provided. But if you were going to do this in the lab, what you would do is you would want to run your chemical reaction for a short time and alter the concentration of just one reactant at a time.

When you do this, you want to determine the initial instantaneous rate for each set of conditions. and then that will allow you to compare how the reaction rate changes as the concentration of that one reactant changed. And then you know that any change in the reaction rate was due to that change in the concentration of your reactant.

You want to look at each reactant individually and determine whether each reactant is 0 or 1st or 2nd or maybe even 3rd order, and that will allow you to determine the overall rate law. Then once you have the rate law, you can use your reaction rate data to determine the value of the rate constant. So we're going to go through a couple examples of how you would do that. Okay, so let's look at this data here. We have the following experimental data that was obtained for this reaction A plus B going to C.

The first thing that I want to do is determine the rate law. We know that the rate law is going to be equal to sum rate constant K, which we haven't calculated yet, but we will be able to do once we determine the reaction order, times the concentration of reactant A raised to some power times the concentration of reactant B raised to some power. And so our job right now is to determine what these powers are that I've listed as X and Y in this rate law.

So we're going to start and we're going to look at each of these reactants individually and determine the order for each reactant individually. So let's start by determining the reaction order for reactant A and I'm choosing to start with A here. You could just as easily start by determining the reaction order for reactant B. Whatever order you want to go in is fine.

So in order to do that, I want to compare two different experiments where the concentration of A is changed but the concentration of B. remains constant. Okay, so if you look at experiment one and two, you can see that the concentration of A has changed, but we're holding the concentration of B constant. This is really important because we want to know that this change in reaction rate is just due to the change in the concentration of reactant A. And so let's look at that.

How does the change in the concentration of A affect the reaction rate? So you can approach these problems from kind of a logical sense and I'll show you how to do that and then I'll also show you how to approach this from a more mathematical technique. So if you look at A here you can see that the concentration of A has doubled from 0.1 to 0.2 molar and you can see that the reaction rate doubled. So if the effect on the reaction rate is the same as how the concentration of your reactant changed So for example, if the rate doubles when the concentration doubles, then the rate law is first order with respect to that reactant. And so you can see that if you were to double the concentration of A here and that led to a doubling of the rate, that would mean that this reaction is first order with respect to reactant A.

And it doesn't need to necessarily just double. If you tripled the concentration of A and the rate tripled, or you quadrupled the concentration of A and the rate quadrupled, when you see the same sort of change in your rate as you do in your concentration of a reactant, that means that your reaction is first order with respect to that reactant. But let me show you how we could do this mathematically.

So we can look at the relationship between the rates and the concentrations we have of our reactants using proportionalities. And so what I've done here is I've set this up where we're comparing the rate of experiment two, and then here's our rate law to the rate of experiment one, and here's our rate law. So what I'm going to do here is I'm going to put this rate for experiment two in here along with these concentrations that we have for our reactants and compare that to the rate and the conditions for experiment one which I'll put down here on the bottom.

So let's go ahead and fill in those numbers there. What this is going to allow us to do is it's going to allow us to solve for this exponent x. So you can see that a lot of terms in this proportionality are going to cancel out.

Your rate constant K is going to cancel out. We don't even know what that is yet, but we know that it's a constant, and you see it on the top and bottom, so that's going to cancel out. We have our concentration of B that is the same in both of these experiments here, so even though we don't know why, 0.2 to the y will cancel out with 0.2 to the y.

And so this is going to allow us to then solve for this exponent x. And so you can see that if you do the math here and take 7 times 10 to the negative 4th divided by 3.5 times 10 to the negative 4th, that is going to be equal to 2. And then on the right side, you'll get 0.2 divided by 0.1 raised to the x power. Remember that when you're working with exponents, if you have 0.2, let's say, to the x, divided by 0.1 to the x, that is the same as saying 0.2 divided by 0.1 entirely to the x.

And so this is going to allow us to solve for that exponent here. You can see that we get the 2 is equal to 2 to the x power. So what is x going to be? Well, in this case, x is going to be equal to 1. And that's what we predicted from looking at this experimental data, that this reaction would be first order with respect to A. So let's go ahead and put that in here.

And then now what we're going to be able to do is do this exact same process for reactant B. So we're going to do the same thing. First we need to compare experiments where the concentration of B is changed and the concentration of A remains constant.

So in this data that we have here, that's going to be using experiments 1, 3. Again what you're going to be able to see here is that where the concentration of A is held constant, any change in the reaction rate has to be due to the concentration change of our reactant B. So how does the change in the concentration of B affect our reaction rate? So notice in this case the concentration of B was doubled, but look what happened to the change in the reaction rate.

the change in the reaction rate didn't just double, the change in the reaction rate quadrupled. And so if you see an effect on the rate that is equal to the effect on the change in concentration squared, so like in this example here we saw that our rate quadrupled when the concentration was doubled, that means that our rate law is second order with respect to this reactant. And again, let me show you how to do this mathematically. So let's compare these two reactions again.

Notice when I set up these proportionalities, I put the experiment with the higher rate on the top and the experiment with the lower rate on the bottom. You don't necessarily have to do it that way. You could flip these around and you're going to come up with the same answer, but the math is just a little bit easier to do in your head if you have the larger reaction rate on top. and the conditions that gave you that larger reaction rate on top. So again, we'll just use this proportionality here, and you can see that our rate for experiment 3 is equal to the rate law, which is going to be some rate constant K, we haven't determined that yet, times the concentration of A.

Now we know that this reaction depends on the concentration of A to the first power. That doesn't really matter to us at this point because those terms are going to cancel out anyway, but you can still put that in times the concentration of B raised to some power Y. And this is what we're going to try to solve for here is this value of Y. And then put in the same information from experiment one down here.

Okay, when you do that, your numbers are going to look like this. And again, you're going to see that there are terms that are going to cancel out. So you can see the rate constant again will cancel out and you'll see that the concentration of A raised to the first power will also cancel out and so now this will allow you to solve for Y. What you're going to get is when you take 1.4 times 10 to the negative third divided by 3.5 times 10 to the negative fourth, you're going to get that that is equal to four.

And then over here on this right side, 0.4 divided by 0.2 raised to the y power is two to the y power. Well, what power do you need to raise to two to give you four? Well, two squared will give you 4, and so the value of y here is 2. So we've just confirmed that this reaction is second order with respect to reactant B.

Okay, so now that we know this reaction order with respect to each of our reactants, what we can do is we can solve for this value k. Remember we have rates for different concentrations of reactants. Now we know what these exponents are.

and so that's going to allow us to solve for our rate constant the nice thing about this is remember that your rate constant k is a constant so you can use the reaction rate data from any one of these experiments that you like and you should always come up with the same value of k so i'm going to pick one here i'm going to use the data from experiment one but you can try this on your own. Double check that the data from experiment two or experiment three gives you the same value of the rate constant as I get here. Okay, so here's my rate law.

I'm going to use again the data from experiment one, and so I'm going to put these numbers here. Here's my rate when the concentration of A was 0.1 molar and the concentration of B was 0.2 molar. Don't forget to include these exponents here that we just determined for our reaction order.

So now we can divide each side by 0.1 and by 0.2 squared. Notice on the right side that will cancel out leaving us with just k and then the other thing that we want to watch here is units. Notice we have molar that will cancel out molar down here. We're still going to have inverse seconds up here on the top and we have molar squared on the bottom still.

So inverse molar squared inverse seconds are going to be our units for this rate constant. And when you go ahead and do the math here, you will get that your rate constant has a value of 0.0875. And then the units, like we just said, are inverse molar squared inverse seconds.

So we were able to use experimental data to determine not only our reaction order, the order with respect to each reactant, but also the value of our rate constant. And so now we know the entire rate law for this hypothetical reaction, A plus B going to C. Okay, let's try this one more time. And now might be a good time to pause the video and try this one on your own.

So I'm giving you kind of a similar reaction here, A plus B going to C. And here's some experimental data, so let's determine the rate law. So we know the rate law is going to be equal to some rate constant k times the concentration of A raised to some power times the concentration of B raised to some power.

I'm going to approach this just like I did the last problem, and I'm going to start by determining the reaction order for reactant A. Again, if you wanted to start with reactant B, that is just fine. So I want to compare experiments where the concentration of A is changed and the concentration of B remains constant.

Okay, that's going to be experiments one and three in this data table here. Your concentration of A is changed and your concentration of B is held constant. So how does the change in the concentration of A affect the reaction rate? Notice that the concentration of A here was tripled from 0.2 to 0.6 and notice that my reaction rate increased by a factor of 9. So again, if you change the concentration by some amount and the effect that you see on the reaction rate is equal to this amount squared, then we know that our reaction is second order.

with respect to that reactant. But let's just prove this to ourselves mathematically. And so let's look at the relationship between the rates and the reactant concentrations in experiments one and three.

Again, when I like to set up these proportionalities, I like to have the data for the experiment that gives me the largest reaction rate on top. So you can fill in all these numbers here and same thing for the data from experiment one. And notice again you're going to have a lot of things cancel out. The rate constant will cancel out and the concentration of B raised to the y power will also cancel out. That's going to allow us to solve for this term x.

And so when you do that, you get 1.35 times 10 to the negative third divided by 1.5 times 10 to the negative fourth, which is 9, is going to be equal to 0.6 to the power of x divided by 0.2 to the power of x, which is 3 to the power of x. And so when you solve for x, you get that x is 2. And that's what we predicted looking at the data. And so anytime you increase the concentration of some reactant by a certain factor and the reaction rate increases by the amount of that factor squared, that indicates that your reaction is second order with respect to that reactant. Let's do the same thing for B.

So let's compare experiments where the concentration of B is changed and the concentration of A remains constant. So that's going to be experiments one and two. Well, look at what's happening here.

So here, we've changed the concentration of B. We've doubled it. Notice we see no change in the reaction rate. So what does that mean? What that means is that the rate law is going to be zero order with respect to this reactant, meaning that the concentration of B doesn't affect the rate.

Again, let's just prove this to ourselves mathematically. Let's compare the data for experiments 1 and 2. You can fill in the rates and the respective concentrations. You'll see again k will cancel out, and the concentration of a squared will cancel out, and you will get 1.5 times 10 to the negative 4th divided by 1.5 times 10 to the negative 4th, which is 1, is going to be equal to 0.4 raised to the power of y divided by 0.2 raised to the power of y, which is 2. Well, 2 raised to what power will give you 1? Remember how these exponents work, that anything raised to the 0 power is equal to 1. And so in this case, y has to be 0. So you could write your rate law like this. Or if you have a reactant concentration that's raised to the 0 power, a lot of times that's just omitted from the rate law.

And we would just write our rate law as our rate is equal to the rate constant. times the concentration of A squared. It's independent of the concentration of B as long as there's some B present. Okay, let's just finish up this rate law here and let's use some reaction rate data and our newly determined rate law to determine the value of the rate constant.

Again, you can pick any experiment number. I picked experiment number one, and so I'm just going to fill in my rate, the concentration of A, that gave me that rate. I know that this reaction is second order with respect to A, and now we can solve for our rate constant. So we'll divide each side by 0.2 molar squared. Notice again, we want to pay attention to the units of our rate constant.

And so here we're going to have molar, which will cancel out one unit of molar here, but we have molar squared on bottom. So we'll still have inverse seconds. And then inverse molar will be the units for our rate constant.

Okay, so let's cancel out the things that are the same on both sides. And then when you do the math, you will get a value for your rate constant of 3.75 times 10 to the negative third. Inverse molar, inverse seconds.

A couple of last things that I want to mention. Typically in these types of problems, you are going to be solving for exponents and reaction orders that are integer numbers. Reaction orders aren't necessarily always integers, but that indicates a more complex reaction with regard to chemical kinetics. And so if you're taking just a general chemistry course, you are typically going to be looking for reaction orders that are integers.

The other scenario that I didn't cover in here, but maybe we should just talk about quickly, is what if you had a change in concentration, say, say this change in concentration of A went from 0.2 molar to 0.6 molar, say it increased by a factor of 3, and let's say the rate... increased by a factor of 27. What would that mean? Well, think about what power would you have to raise 3 to to give you a corresponding increase in reaction rate of 27. That would mean that your reaction would have to be third order with respect to that reactant.

So those are typically the three you're going to see. So if you see the same increase in reaction rate as you do an increase in concentration of reactant, concentration of reactant doubles, reaction rate doubles, or concentration of reactant triples, reaction rate triples, that always indicates first order. But if you see a change in concentration, say, that is doubled and the reaction rate increases by a factor of four, or your concentration of a reactant triples and your reaction rate increases by a factor of nine, that's going to indicate a second order reaction. And then I talked about what you would see if a reaction was third order with respect to a particular reactant.

Okay, so that wraps things up for this tutorial. I hope that you found this helpful.

Transcript for:Understanding Rate Law Determination

Transcript for:
Understanding Rate Law Determination