Worked example: Determining a rate law using initial rates data
The rate law for a chemical reaction can be determined using the method of initial rates, which involves measuring the initial reaction rate at several different initial reactant concentrations. In this video, we'll use initial rates data to determine the rate law, overall order, and rate constant for the reaction between nitrogen dioxide and hydrogen gas. Created by Jay.
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- When we talk about initial rate of a reaction, is that a INSTANTANEOUS RATE of a product or sum of all the products or sum of all reactant ? I have an practice question in my AP Chemistry book by Pearson and they dont have answer key.
Given a reaction C2H5Br + OH- ---> C2H5OH + Br- , has rate law has rate= k[C2H5Br][OH] . When [C2H5Br}= 0.0477 and [OH-]=0.100 M , the rate of disappearance of ethyl bromide is 1.7 x 10^-7 M/s. What is the value of k, rate constant?
If it is a sum of all reactants, I got k= 7.12 x 10^-5(11 votes)
- An instantaneous rate is the slope of a tangent to the graph at that point.
An average rate is the slope of a line joining two points on a graph.
If the two points are very close together, then the instantaneous rate is almost the same as the average rate.
So the initial rate is the average rate during the very early stage of the reaction and is almost exactly the same as the instantaneous rate at t = 0.
If rate = k[C₂H₅Br][OH⁻], then
k = rate/([C₂H₅Br][OH⁻]) = 1.7 × 10⁻⁷mol·L⁻¹s⁻¹/(0.0477 mol·L⁻¹ × 0.100 mol·L⁻¹) =
3.6 x 10⁻⁵ L·mol⁻¹s⁻¹(17 votes)
- You've mentioned in every video, the unit of concentration of any reactant is (M) that is (Mol) and the unit of rate of reaction to be (M/s). But what we've been taught is that the unit of concentration of any reactant is (mol.dm^-3) and unit of rate of reaction is (mol.dm^-3.s^-1) . Can you please explain that?
Thank you.(4 votes)
- M is the symbol for molarity, not moles.
1 M = 1 mol / L = 1 mol / dm³(15 votes)
- how can you raise a concentration of a certain substance without changing the concentration of the other substances?(7 votes)
- One of the reagents concentrations is doubled while the other is kept constant in order to first determine the order of reaction for that particular reagent. This is done because in the equation for the rate law, the rate equals the concentrations of the reagents raised to a particular power. The order of reaction with respect to a particular reagent gives us the power it is raised to(3 votes)
- In our book, they want us to tell the order of reaction by just looking at the equation, without concentration given! How would you decide the order in that case?
- Late, but maybe someone will still find this useful.
Sometimes you can tell the reaction order from stoichiometry, but ONLY if the reaction is elementary - that means that it happens in one step, with no intermediates. I'm assuming that's what your textbook meant. In that case, the reaction order with respect to a reactant is equal to its stoichiometric coefficient, and the overall reaction order is the sum of those. For example:
A + B -> C
The stoichoimetric coefficient of A is 1, so the reaction is 1st order in A. It's also 1 for B, so the reaction is 1st order in B as well. Then you add their orders together and find that the reaction is 2nd order overall.
In practice, reaction order is only determined experimentally, and it's actually what helps us determine the reaction mechanism. For example, if reaction orders line up with stoichiometric coefficients in the equation, you know it's a one-step mechanism. If it doesn't, you can tell there are more steps, and you have to devise a mechanism that agrees with the experimentally determined rate. It only really works the other way around in, yeah, textbook problems ("assuming xyz is an elementary reaction, determine reaction order").(8 votes)
- I get k constant as 25 not 250 - could you check?(2 votes)
- Make sure the number of zeros are correct. I'm getting 250 every time. You should be doing 1.25x10^-5 / ((.005^2) x (.002))(7 votes)
- is it possible to find the reaction order ,if concentration of both reactant is changing .
is there anyway to cancel out the effect of other reactant ??(2 votes)
- Yes. To the first part, the changing concentrations have nothing to do with the order, and in fact, the way in which they change is in fact the order. So, if they were not changing, then we cannot determine the order.
As to the second part, yes. In fact, cancelling out the effect of all but one reactant is the standard method of finding the order. The method used is called flooding. In this, all but one are taken in very high concentrations (excess) such that even if the entire reaction were to take place and our reactant (in small amount) were to be entirely consumed, the net change in the concentrations of the remaining would be negligible, and hence can be taken as zero (constant concentration). Hence all their concentration along with the actual rate constant now form a new constant which can be computed while measuring the rate with respect to one reactant only. By doing this for every reactant, the original constant can be determined. To note, although we are taking the concentrations to be constant, it is just a mathematical step (the change being closed to zero) and not in reality, since their concentrations need to change for the reaction to occur (or at least, they need to be used up).(4 votes)
- at1:20so we have to use the experiment/trial that have a constant concentration when we want to determine the order of reaction?(2 votes)
- You need to run a series of experiments where you vary the concentration of one species each time and see how that changes the rate.(2 votes)
- What if i was solving for y (order) of a specific concentration and found that 2^y=1.41? I know that y has to be an integer so what would i round 1.41 to in order to find y?(2 votes)
- "y" doesn't need to be an integer - it could be anything, even a negative number.(2 votes)
- What if one of the reactants is a solid? How would you measure the concentration of the solid?(1 vote)
- You can't measure the concentration of a solid. But what would be important if one of the reactants was a solid is the surface area of the solid. The finer the solid is ground (and hence the larger the surface area), the faster the reaction will take place.(4 votes)
- What if the concentrations of [B] were not constant? But [A] has 2 experiments where it's conc. is constant, so you can find the order for [B] using this method.
But how would you find the order for [A]?(1 vote)
- You need data from experiments where [B] is constant and [A] is increased otherwise you cannot work out the order with respect to A.
You are always going to be given enough information to solve these problems.(3 votes)
- [Voiceover] Now that we understand how to write rate laws, let's apply this to a reaction. Here we have the reaction of nitric oxide, which is NO, and hydrogen to give us nitrogen and water at 1280 degrees C. In part A, our goals is to determine the rate law. From the last video, we know that the rate of the reaction is equal to K, which is the rate constant, times the concentration of nitric oxide. Nitric oxide is one of our reactants. So the rate of the reaction is proportional to the concentration of nitric oxide to some power X. We don't know what X is yet. We also know the rate of the reaction is proportional to the concentration of our other reactant, which is hydrogen, so we put hydrogen in here. But we don't know what the power is so we put a Y for now. Alright, we can figure out what X and Y are by looking at the data in our experiments. So let's say we wanted to first figure out what X is. To figure out what X is we need to know how the concentration of nitric oxide affects the rate of our reaction. We're going to look at experiments one and two here. The reason why we chose those two experiments is because the concentration of hydrogen is constant in those two experiments. The concentration of hydrogen is point zero zero two molar in both. If we look at what we did to the concentration of nitric oxide, we went from a concentration of point zero zero five to a concentration of point zero one zero. We increased the concentration of nitric oxide by a factor of two. We doubled the concentration. What happened to the initial rate of reaction? Well the rate went from one point two five times 10 to the negative five to five times 10 to the negative five. The rate increased by a factor of four. We increased the rate by a factor of four. If you have trouble doing that math in your head, you could just use a calculator and say five times 10 to the negative five and if you divide that by one point two five times 10 to the negative five, this would be four over one, or four. This rate is four times this rate up here. Now we know enough to figure out the order for nitric oxide. Remember from the previous video, what we did is we said two to the X is equal to four. Over here, two to the X is equal to four. Obviously X is equal to two, two squared is equal to four. So we can go ahead and put that in for our rate law. Now we know our rate is equal to K times the concentration of nitric oxide this would be to the second power. So the reaction is second order in nitric oxide. Next, let's figure out the order with respect to hydrogen. So this time we want to choose two experiments where the concentration of nitric oxide is constant. That would be experiment two and three where we can see the concentration of nitric oxide has not changed. It's point zero one molar for both of those experiments. But the concentration of hydrogen has changed. It goes from point zero zero two to point zero zero four. So we've increased the concentration of hydrogen by a factor of 2 and what happened to the rate of reaction? Well it went from five times 10 to the negative five to one times 10 to the negative four so we've doubled the rate. The rate has increased by a factor of two. Sometimes the exponents bother students. How is this doubling the rate? Well, once again, if you can't do that in your head, you could take out your calculator and take one times 10 to the negative four and divide that by five times 10 to the negative five and you'll see that's twice that so the rate goes up by a factor of two. Now we have two to what power is equal to two? So two to the Y is equal to two. Obviously Y is equal to one. Two to the first power is equal to two. So know we know that our reaction is first order in hydrogen. We can go ahead and put that in here. We can put in hydrogen and we know that it's first order in hydrogen. We've now determined our rate law. In part B they want us to find the overall order of the reaction and that's pretty easy to do because we've already determined the rate law in part A. We know that the reaction is second order in nitric oxide and first order in hydrogen. To find the overall order, all we have to do is add our exponents. Two plus one is equal to three so the overall order of the reaction is three. Let's compare our exponents to the coefficients in our balanced equation for a minute here. It's very tempting for students to say oh, we have a two here for our coefficient for nitric oxide, is that why we have a two down here for the exponent in the rate law? But if you look at hydrogen, hydrogen has a coefficient of two and we determined that the exponent was a one down here in the rate law. You can't just take your coefficients and your balanced chemical equation and put them in for your exponents in your rate law. You need to look at your experimental data to determine what your exponents are in your rate law. Later we'll get more into mechanisms and we'll talk about that a little bit more. Alright, let's move on to part C. In part C they want us to find, or calculate, the rate constant K. We could calculate the rate constant K by using the rate law that we determined in part A and by choosing one of the experiments and plugging in the numbers into the rate law so it doesn't matter which experiment you choose. You could choose one, two or three. I'm just going to choose one here, so experiment one. We're going to plug all of our information into the rate law that we just determined. For example, in our rate law we have the rate of reaction over here. Well, for experiment one, the initial rate of reaction was one point two five times 10 to the negative five. And it was molar per second so we're going to plug this in to our rate law. So let's go down here and plug that value in, one point two five times 10 to the negative five and this was molar per second. Alright, so that takes care of the rate of the reaction. Next, we have that equal to the rate constant K, so we're trying to solve for K, times the concentration of nitric oxide squared. Let's go back up here and find the concentration of nitric oxide in the first experiment. The concentration is point zero zero five molar. We're going to plug in point zero zero five molar in here. We have point zero zero five molar. Next, we're going to multiply that by the concentration of hydrogen to the first power. We go back up to experiment one and we find the concentration of hydrogen which is point zero zero two molar so we plug that in. We have zero point zero zero two molar. Next, all we have to do is solve for K. Let's go ahead and do that so let's get out the calculator here. We could say point zero zero five squared gives us two point five times 10 to the negative five, we need to multiply that by point zero zero two. On the right side we'd have five times 10 to the negative eight. So we have five times 10 to the negative eight. The thing about your units, this would be molar squared times molar over here and all of this times our rate constant K is equal to one point two five times 10 to the negative five molar per second. Let's go ahead and find the number first and then we'll worry about our units here. To find what K is, we just need to take one point two five times 10 to the negative five and if we divide that by five times 10 to the negative eight then we get that K is equal to 250. We can go ahead and put that in here. K is equal to 250, what would the units be? Well, we have molar on the left, we have molar on the right, so we could cancel one of those molars out. On the left we have one over seconds and on the right we have molar squared so we divide both sides by molar squared and we get, for our units for K, this would be one over molar squared times seconds. We've found the rate constant for our reaction. And notice this was for a specific temperature. Our reaction was at 1280 degrees C so this is the rate constant at 1280 degrees C. Finally, let's do part D. What is the rate of the reaction when the concentration of nitric oxide is point zero one two molar and the concentration of hydrogen is point zero zero six molar. Well, we can use our rate law. Our rate law is equal to what we found in A, our rate law is equal to K times the concentration of nitric oxide squared times the concentration of hydrogen to the first power. Our goal is to find the rate of the rate of reaction. We're solving for R here and we know what K is now. K is 250 one over molar squared times seconds. The concentration of nitric oxide is point zero one two, so we have point zero one two molar and then we square that. All I did was take this and plugged it into here and now we're going to take the concentration of hydrogen, which is point zero zero six molar and plug that into here. Let's go ahead and do that, so that would be times point zero zero six molar, let me go ahead and put in the molar there, so point zero zero six molar to the first power. And we solve for our rate. The rate is equal to, let's do the numbers first. We have point zero one two squared. We're going to multiply that, so times point zero zero six and then we also need to multiply that by our rate constant K so times 250. This gives us our answer of two point one six times 10 to the negative four. Let's round that to two point two so we have two point two times 10 to the negative four. In terms of our units, if we think about what happens to the units here, we would have molarity squared, right here molarity squared molarity squared so we end up with molar per seconds which we know is our units for the rate of reaction, so molar per seconds. We found the rate of our reaction.