Now let's talk about the reaction mechanisms. So when you're writing a reaction, were usually writing the overall reaction. For example, Reaction H two plus to I C l goes to to eight cell Plus I to now when we write this this what we mean is this is the overall reaction. But what really happens is distraction is a two step reaction. So at step one, we have h two plus I c l and this will give us a chai plus a sale. And in step two, you will have h I plus in the i c l And this will be HCL plus I t. Now, how do we get to the overall reaction using thes two steps So you fight am to add these two reactions together. What will happen is this h I is going to cancel this h i and we will have h two plus one one total to Isil And this will give us 11 total to HCL plus this idea over here. So what we wanna do is at the left hand side, on the left hand side, in the overall and at the right hand side to the right hand side on the overall. And as you can see, there was this h I. But it got canceled with E H I on the left hand side. So basically what happened is it's step one in the products we got an H I. And in step two, we use that h I regard from the products off the step one. So if you see a molecule like that first is in the reaction, uh, in the products off the first step or second step and in the consecutive step, if you're using that, then we call them the intermediate. And these each step, each individual step is called elementary Steps. Now we're going toe Investigate mawr about these in the later videos. Now, before we move on to direct determining steps, let's talk about the Miller molecular ity. So let's say these are our elementary steps and we have molecule Larry T. And we have the rates law. Let's say in our elementary step behalf and a as in the wreck tints and this goes to products. The molecular iti is going to be one and we call this uni molecular and the red law is going to be rate equal. Sasuke. Hey, if we have something like a plus a going to products molecular, it is going to be too. We call this bye molecular and the red law is going to be K a square. So this is the first order and there's a second order. Let's say I have a plus B this time going to products again. The molecular T is going to be, too, because I have two molecules that are having a coalition and we call this by molecular Great law is going to be, say, a M B. Now This is also overall a second order because one plus one with Give us to Let's save your house a plus a plus a going to products molecular It is going to be a tree this time and we call this term molecular and read Law is going to be K. Thank you. Let's say we have a plus a plus c going to the products again. This is going to be a term molecular and read law is going to be K a squirt. See again. Overall, two plus one is going to be equal to three. So the molecular T is going to be tree. Maybe the most important part in reaction mechanisms is the rate determining steps. So think about it like this. Say you're this little guy and your walking with your grandma. Even though you're fast, you're going to walk with your grandma so your speed is going to be the speed off your grandma know your speed. So you these groups, overall speed is going to be your grandma's speed. And this pretty much applies to chemistry to. So let's say we have this overall reaction mechanism. No, to plus CEO, use this, you know, plus co two. Now, you would expect this reaction to have a raid law like K NL too, and CEO, But this is quite not the case. Instead, what we see is K No. Two square. So what actually happens here is this reaction is two steps. It's step one we have, you know, to plus and no two goes to in a tree. Plus, you know, and this is going to be our slow step and in the step to we'll have and all Tree plus CEO go Sue, you know, to plus co two. And this is our faster If you wanna look at the overall? What happens is this N o tree is going to cancel with Zeno Tree and this n 02 is going to cancel with this n 02 So on the left hand sides, we'll have no to plus CEO on the right hand side. We'll have, you know, plus co two and this is our overall, so we need to look at the slow step. The slow step is basically our grandma in this reaction. So if you want to write the slow steps read Law, what we're going to write is K N 02 No, to these are our reactor's right, And this is going to be equal to K, you know, to squared. So this is basically second order with respect to N o. T. If you actually want to draw an energy diagram for what we just saw this two step mechanism and this is going to be energy versus reaction progress, we'll see this. So what does it mean? We have our reacting here, and these reactions have to pass this hill what we call the energy of activation and then they're going to be in the second reaction over here. Reactors off the second reaction, and then they're going to pass this energy off activation, and then it's going to give us the products. So what we call is this is the e A one and this is going to be R E A to. And if you remember, in our previous reaction, what we had no two plus n 02 gives us, you know, three plus, you know, and this was the slow. And then after that, we will have an l Tree Plus CEO would give us n 02 plus Co. Two. And this is fast. So the very first step is a slow step. It means it's energy off. Activation is higher, then energy off actuacion to here. So since the entire reaction is dependent on these molecules passing true the energy off activations. But the first one is going to take much larger energy than this second one over here. That's why we can show it as a slow step, having ah higher energy off actuacion. Whereas the second step, the Fast Step, has a lower energy evacuation. Now let's talk about the mechanisms with a fast initial step. So this is the place where the students are getting confused the mosque, but actually is going to be very easy. And I'm going to show you how so we have react INTs and I'm going to use different colors, so it will be easier to see for you guys. The action is going to products, right? The use a double arrow and what this double arrow means is equilibrium. So what does that include? Dream. So we have a forward reaction from the left to the right. Reaction is becoming products. If you also have reverse reaction from the products on the right hand side, back to reactions, and it means that they could be, um, rates off the reactions getting products and products getting reactions are the same. So how we can write this is let's right to rate are red law for the reactions. Okay, are right because we only have the molecule are that we use for directions and some K and the same rate is going to be equal to the reverse. So let's say this is K and let's say this is K minus one. So diverse is going to be K minus one p so As you can see, the forward direction is now equal to the reverse reaction. Hence we can write. This equilibrium is an example. We can give this reaction to H two plus to and oh is going to to water molecules and into. And this is our overall reaction when we actually investigate this reaction. What we have is three steps. Step one is going to be two and no in equilibrium. It's into 02 and this is a fast step. And then we will have h two plus in 202 This guy is going to H 20 plus and to and this is our slow step. And if you remember, what we call this is the rate limiting and lastly, we will have into plus H two goes to into plus H two. And this is also fast. And if he ad left hand side is the left hand side and right hand sides the right hand side. What we're going to see is this until two is canceling it This until two. And this until is getting canceled with this end to and we're going to get this Orel reaction. So a quick recap what we call this until two that air we see in the products after the first step and then in the reactions after the second step. And over here in the second step we have in the products, and in the third step we had in the reactions we call them intermediates. So if you want to write the rate law, what you're supposed to ride is the slow. The rate limiting steps reactions, right? So let's say this is K one. Let's say this is K minus one. This is K two this Katri. So what we're supposed to write for over all right is going to be K two h two into, 02 But the problem is, we do not have this into 02 until two is an intermediate, so we cannot define the rates with until two. That's why we need to change this until two. In such way, that became Write it in some other terms. So let's try to do that. I am going to erase this first part, our overall reaction. Well, actually, let me erase this top part, so it's going to be easier for us. So we want this guy various. This guy in the entire reaction. We can see it over here in the very first step. So let's strike the red law for the first step. We have K one the forward Reaction times and, uh oh, squirt. This too comes up as a square, and this is going to be equal to the reverse K K minus one times and to or two. Now, if I sold for until to what I'm going to get is K one over K. Neither one and oh to the power of to this is our until two. Now, over here, instead of writing this nto to became right this that we found from the first step. So combining this equation 12 does equation to let's say this is one and this is to we can just right que two times k one over k negative one h two and you know, square. And this is going to be our rate law for overall reaction again. What we want to do is you want to get rid off this intermediate in the slow step and write it with something we already have in the reaction. Like over here, we have no. Over here, we have no You want to define our red law with whatever we have in the overall reaction, we have h two n n Oh, so that's how we could define it. Going back to the fast step and getting the solution for the intermediate right again in writing the solution inside and getting the result.