In this video, we're going to focus on reaction mechanisms, how to write the rate law for the overall reaction, and things like that. So a reaction mechanism is basically a step-by-step pathway by which a particular reaction occurs. And each individual step in a reaction mechanism are known as elementary reactions. So let me give you an example.
So let's say this is the first step of a certain reaction. A plus B becomes C plus D. So I'm going to put a 1 here to indicate that's step 1. Now step 2, this is going to be D plus E. And that turns into F plus B. So the series of these two steps is known as a reaction mechanism.
Each individual step is an elementary reaction. And we can write the rate log expression for each elementary reaction based on the coefficients that we see for the reactants of that reaction. So let's say if we want to write the rate log expression for the first step of this reaction mechanism.
Notice the coefficients of A and B. It's 1 to 1. And we're going to call the rate constant associated with the first step, K1. The rate constant for the second step, we'll call it K2. So this is going to be rate is equal to K1 times the concentration of A times the concentration of B. And this is going to meet the first power.
So that's the rate law expression for the first step of this reaction mechanism. Now for the second step, or the second elementary reaction, the rate law expression is going to be k2 times d times e. So that's how you can write the rate law expression for a certain step in a reaction mechanism.
By the way, in this reaction, can you identify the catalyst and the intermediate? Which species represents the catalyst and which one is the intermediate? If we were to add these two elementary reactions, the catalyst and the intermediate will be cancelled as we try to get the overall reaction.
To distinguish the catalyst from the intermediate, you need to know that the catalyst is consumed first in a reaction, and then it is produced later. The reverse is true for the intermediate. The intermediate is produced first in a reaction. and then it is consumed later. So notice that we can cancel species B and we can cancel D.
One of them is the intermediate and the other is the catalyst. So is B the catalyst or the intermediate? What would you say? Notice that B appears first on the left side of the reaction.
So it is consumed first. Then it reappears on the right side of another reaction later. So it's consumed first and then it's produced later.
That is the catalyst. The catalyst is present in the beginning of the reaction and at the end of the reaction. The intermediate is only present in the middle. It is produced first, and then it is consumed later.
So if you look at D, D is on the right side of step 1. It's on the right side of that reaction. So it is produced first. Then, it shows up later on the left side of the second reaction. So it's consumed later.
It is produced first, consumed later. So it is the intermediate. So remember, the catalyst is at the beginning and at the end of the reaction. It is consumed first and then produced later, but the intermediate is in the middle.
It is produced first and then consumed later. So that's a quick and simple way. that you could use to identify the intermediate and the catalyst in a reaction mechanism.
Now remember, a catalyst is a substance that can speed up a chemical reaction. And the way that it accomplishes this is by providing the reaction with an alternative pathway for the reactants to become products. And another benefit of a catalyst is that it lowers the activation energy.
And by those means, it can speed up a chemical reaction. So now let's write the overall reaction for this reaction mechanism. B and D has been cancelled.
On the left we have A and E. So this is going to be A plus E. And on the right we have C and F. So this is the overall equation. Now sometimes you need to be able to determine the molecularity of each individual step in a reaction mechanism.
So the first term you need to be familiar with is a unimolecular reaction. In a unimolecular reaction, if you think of the word uni, uni means one. So we have one molecule that is going to be the reactant. So a good example of a unimolecular elementary reaction is A going into B. The rate law expression for that is rate is equal to K times A raised to the first order.
The overall order of a unimolecular reaction is first order. Now the next thing we need to talk about is a bimolecular elementary reaction. So remember, an elementary reaction is simply one step in the overall reaction mechanism. A good example of a bimolecular reaction, and remember the prefix bi means two, so we're dealing with two reactants. One example is A plus A turns into B.
Another example is A plus B turns into C. So both of these elementary reactions are bimolecular reactions. In each case, we have two reactants. reacting with each other. So the rate law expression for the first one is going to be k times a times a, or we can just write a squared.
And for the second one, it's going to be rate is equal to K times A to the first power times B to the first power. In each case, the overall order of the reaction, in this case for each reaction, it's second order. So that's going to be the overall order for a bimolecular reaction.
Now the next type of reaction we need to consider is a term molecular reaction. When I think of the prefix ter, for some reason I associate it with tri, and tri means three. Now, term molecular reactions are quite rare because statistically speaking, it's difficult for three molecules to collide at the same time with the right amount of energy and with the right orientation for the reaction to occur. So these reactions are very rare, and they tend to be quite slow too. But here's an example of a term molecular reaction.
A plus B plus C become in D. Another example is 2A. We can write as A plus A, or just 2A plus B, turning into C.
Or we could say 3A becomes B. All of these examples are term molecular reactions. The rate law expression for the first one... it's going to be rate is equal to k times a times b times c. For the second one we have rate is equal to k times a squared because we have 2a and then times b.
For the last one, its rate is equal to k times a to the third power. So the order of each reactant is simply the coefficient of the reactant for an elementary reaction. You can only do that for elementary reactions and not the overall reaction. So as we can see, the overall order for each of these elementary reactions is 3. So a term molecular reaction, the overall order for that type of reaction is third order. It's third order overall.
Now consider this reaction where we have O3 reacting with NO2 to become NO3 plus O2. So that's the first step. And for the second step, we have NO3 plus NO2. Turn it into N2. 05 now let's say the first step is slow, but the second step is fast With this information, go ahead and write the overall reaction and determine the rate law for the overall reaction.
And identify any intermediate or catalyst, if applicable. So first, let's add up this reaction. Notice that NO3 cancels.
Is this a catalyst or intermediate? So NO3 doesn't appear in the beginning or at the end. It's only in the middle of the reaction.
It is produced first, and then it is consumed later. So because it only shows up in the middle of the reaction, this is going to be an intermediate. Now, let's write down what's left over. So on the left side we have one O3 molecule and we have two NO2 molecules. On the right side we have one N2O5 molecule and one O2 molecule.
So that's the overall reaction for this reaction mechanism. Now how can we write the rate law expression for the overall reaction? The rate of the overall reaction is going to be dependent on the rate determinant step.
Which step is the rate determinant step? Is it the first step? Or is it the second step?
The rate determinant step is going to be the first elementary reaction, because that's the slow step. The rate will always depend on the slow step. So the rate for the overall reaction will be equal to the rate of the slow step, which is... The rate of the slow step is K1 times O3 times NO2. So that's a bimolecular reaction.
So if they were to ask you, what is the molecularity of the first step, you would say it's bimolecular, because the overall order for that step is 2. We have two molecules reacting with each other. So this is the rate law expression for the slow step. To write the rate law for the overall reaction, instead of writing k1, we can simply write k.
So we can say it's k times O3 times NO2. So k would be just a generic term for... constant for the overall reaction, but this is the rate log expression that we're looking for So it's first order with respect to O3 first order with respect to NO2 and its second order overall Go ahead and try this example problem. So write the overall reaction for the mechanism shown below. Identify any catalysts and intermediates that may be in a reaction, and determine the molecularity for each elementary reaction.
And then write the rate law for the overall reaction as well. So let's begin. Let's identify any catalysts or intermediates that may be present first. Notice that we can cancel iodide and we can cancel I O minus.
So I O minus is the hypoiodide ion. Which one of these is the catalyst and which one is the intermediate? What would you say?
So looking at I-, it is present at the beginning of the reaction and at the end. It is consumed first, and then it is produced later. So this, this is the catalyst. I-O-is in the middle of the reaction. It is produced first, and then it is consumed later.
So this is going to be the intermediate. So that's how you can distinguish the two from each other, and that's how you can also identify them as well. So now let's go ahead and write the overall reaction. On the left side, we have two peroxide molecules reacting with each other. So that's hydrogen peroxide.
On the right, it's going to produce two water molecules. and we're going to get an oxygen molecule. So the decomposition of hydrogen peroxide into water and oxygen gas is accelerated by the presence of an iodide catalyst.
Now what is the molecularity for the first elementary reaction and what about for the second step? For the first step we have two reactant molecules so it's bimolecular for the first step And it's also bimolecular for the second step. So that's the molecularity for each step in this reaction mechanism.
Now to write the rate law expression, I need to tell you which one is slow and which one is fast. Let's say the first step is the slow step. And the second step is the fast step. So with that information, what is the rate law expression for the overall reaction? Feel free to pause the video and work on that.
The rate law expression for the overall reaction is going to be equal to the rate of the slow step, the rate determinant step. So let's write the rate law expression for the slow step. So this is K1 and this is K2.
So the rate for this low step is going to be K1 times the concentration of hydrogen peroxide. The coefficient for that is a 1, and then times the concentration of iodide, which is also to the first power. So that's the rate law expression for the first step.
But now to write the rate law expression for the overall reaction, which is this reaction here. Notice that iodide is not part of the overall reaction. The overall reaction contains only one reactant, H2O2, so therefore the rate law expression should only contain that reactant.
So instead of writing K1, we can simply write K for the rate constant for the overall reaction. Even though this K looks big, this should be lowercase k. Now this part is going to stay the same. The rate law for the overall reaction is going to equal the concentration.
It's proportional to the concentration of H2O2 to the first power. Now, since iodide is not in the overall rate law expression, we can get rid of it. So this will be the rate law expression for the overall reaction, and this is the rate law expression for the first step, with iodide included. So that's it for this problem.