Texas Tech University
The Rate Law - Overview


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Now let's talk about the effect of concentration in the red law This safe, they have a molecule A and this goes to products. So how can we determine the effect off the concentration A into the red law? What we How we can write this rates equals two que a to the power of end. Now, this is a general way to show the rate law que is going to be our rates constant and end is going to be our reaction order. Now what kind of orders we can have what we're going to see in, um, general chemistry, too. We're going to see on Lee three types off read lost. So what are these three reaction orders? That's right. Our right long again rates equals two K aid to the poor off end. And if our n equals 20 this is zeroth order. And what would be the red law que a to the power off zero, which is just came. So this is independent off the concentration off a at N equals to one. This is first order. And let's strike this guy with blue. So there's going to be que 82 d poorer front and if and equals two, this is going to be second order on this show of this guy. It's red. So this is going to be K Concentrations Square now. I want to draw two graphs to illustrate this. Furthermore, what we're going to have in the first graph is reacting concentration versus time, whereas in the second graph, what we will have is the rate versus reacted concentration. So let's just show this as concentration off a versus time. And this is going to be rates concentration over seconds. Where is this time? So lists. Look at the zeroth order now. Zero order is going to be independent off the concentration. So let's say you're going to have a lion like this. This is going to be a straight line because this is independent. Off this concentration we have Let's say one 0.80 point 60.40 point two and finally zero. As Thekla concentration decreases, the rate does not change. If that was the first order, they suffering the same place. But after a while this would change. The concentration would not decrease as fast. The reason why is now this this is the first order and we multiply K with a And now this A is getting smaller and smaller and you're multiplying k with Cem number that zero point something so the rate is decreasing. That's why the consumption off A is also decreasing. And if you're looking at the second order, it's going to be even slower after some point, because now you're taking whatever number this concentration is and taking its square. So making it even a smaller number and multiplying by K the rate is going to be even slower. So in the same fashion, let's look at the rate versus the concentration. So what would you expect for the zeroth order? Straight line? Baha, This is just not going to change. Let's write thes zero and this is going to be our zeroth order. The rate does not change because rate is already K. It's independent off the concentration A and I am sorry for this thi this guy was supposed to be concentration, faith, So this is not going to change it all. If you look at the first order now, what we're going to see is the rates increasing with the increasing a again on the right. The A is increasing. If a was zero, our first order over here is going to be zero times k. So this is going to be zero. It will actually be better. You find drawing with blue because I use blue for first order. But as we increase a, then the rate starts to increase. What about the second order? So over here the second order is also going thio increase. But it's going to increase exponentially because we have a two over here. So as the concentration off a increases, this red guy is going to increase even faster after it is at a certain number off the concentration off a. Now let's go through all this once again at the very first one, we have the concentration off a versus time and the concentration off A as it drops with the time it drops in a linear fashion without changing after every interval, it changed the same for our first order. What happens is as the a decreases with the time passes, the decrease off a is getting slower because now we're multiplying K with a number that has decimal point. And over here the second order. It's getting even slower because now it's the square off this number that has a decimal point. And when we look at the rates, what we, however, here ISS at zeros ordered the rate doesn't change with the changing concentration for this guy. The first order great increases with increasing A in a linear fashion because it's dependent on the power off one just concentration off A. Whereas it is the second order, this rate increases in exponential fashion, since this has a A to the power of to term. And lastly, let's write a red law for this general reaction that we had before. A A plus d B goes to C C plus de. The to rate law has three form off rates equals two que are constant A to the poor off em be to the power of end. So these powers, we usually determined it from the data that we have that we collected in the lab and this are the concentration off the reactions A and B. This could as well be something like if the reaction is like a plus, B plus C goes to d plus E than our rates will be equal to K A to the poor off em B to the power off and and see to the power of X. We're on Lee writing here what we have in the reactions. And if you're talking about, let's say what is this reaction in the order off with respect to a So with respect toe a the order off. This reaction is going to be M. What's the order with respect to be is going to be end. And was the order with respect to see this is going to be X. If they ask us what is the overall order, this is going to be equal to all the powers added to each other. Mm plus n plus X. Again, The overall order is all the powers, whereas the respectively ABC orders are M N N X. Now let's talk about the units off K constant. So one common mistake these things do is they want to just memorize thes K Constance units for zero first and second order. But once they're in an exam in a test or in the homework, if they see a different, um, order like a turd or fourth order, they cannot come up with the answer or even verse. They forget sometimes what they memorized and they just get confused. They give the wrong answer, but I don't want to guys to do that. I actually want to just teach you how to find decay constant using some simple math. So let's start with zero soldier. What we have, the rate equals two K times eight to the part of Syria. Hence there's no eight. So what is the units for rate? This was concentration over time. So let's just right for this simplicity, concentration as a majority and time in seconds. And this is going to be our k constancy in its That's just simply this. If you go for first order, what we have is rate equals two. Okay. Hey, So then again, we have M O R s on the left hand side and on the right hand side, what we have is K and concentration again. So the UNICEF or K, is just going to be one of our seconds. This AM's are going to Catholic China. So let's write it for second order. What we will have again two rates equals two. Say a to the power of to so morality over seconds equals two K or two square. So what we're going to have is going to be one over more lt second. So if you can recall instead off writing one over M s or year one of our s or hear em over is what you could do is MM. As the poor of negative one. If it is in the bottom side, you write it as a native right over here. This is going to be s native one. And this is going to be mm. Native one s negative one. So that's right. Mhm. Third order regard. Writing all these is you can see the fashion. What happens is M was one and then zero the minus one and here is going to be a minus two. So it's going to be m squared s. And it's going to be aim to the part of Negative, too is to part of native one. So let's try to write something a little bit more cultivated. Let's see, I have a plus. B is giving us see, and what we have is okay. A to the power off one. Mhm. Be to the poor if 1/2. Now, what do we do? Well, we're going to do the very same thing. So this is going to be equal to the rate. And what is our rates? Mm. Over s equals two for a We have m to the power of one and four B. We have mm to the poor. Off 1/2. Now, these are in multiplication, so if you're going toe added the powers. So what we're going to have is one plus 1/2. She over to equals two. AM Over s You are okay over here. So we will have one over s. And if you write the end terms together MM. Divided by I am she over to? That is a very ugly princess sees. Let me do that again. All right, so there's going to be equal to our okay. So continuing. What we do over here is we have em to the poor off one, and this is on the day numerator. So what we're going to do is really we're going to substructure these values one minus tree over to and we have one over s equals two K. And once we do this Math. This is going to be one over. AM to the part off 1/2. Yes, he calls to K, And if I want to write it in of in a way, what we're going to do is is going to be m to the power off. Negative 1/2. As to the poor, if negative one is going to be the units for R K for this reaction. So, as you can see, you don't need to memorize anything. And once you cease, um, mawr complicated question like this, you can easily vert through the math and come up with the right answer.

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