Now let's make an example. Ozone is naturally composed the oxygen by the reaction to all three tree all to now. This is our overall reaction, but experimentally observed rate law equals suit. Okay, hold on the square and oxygen to the power of NATO worth. So there's a proposed mechanism for, uh showing this rate. And that proposed mechanism is all three. First, isn't equilibrium vits go to plus oxygen K one K minus one. And the second step is old tree Plus oh, just goes to 202 and this is going to be the fast And there's going to be the slow steps. And what they're asking is, is it consistent with this red law? So let's solve it normally, what we would write for the slow steps rate law would be I am going to quickly name this K two and que two all tree and oath, right? But the problem is, we do not have oh in the overall reaction, because what happens is this oxygen is an intermediate, so they cancel. So instead, off this oxygen, we need to write something else. Let's try to see what we can write for him. If I go back to the first fast step. What we have is K one times Otri will be equal to This is the forward reaction is going to be equal to the reverse direction K minus one times 02 times oxygen. Now I want to write soul for this oxygen because I want to write this oxygen in terms off anything that we actually have to see in the overall reaction. So what we can do for that ISS K one divide by Okay, native month or minus one. We have old tree and we divide this bye o t and this is going to be equal to Oh, So now basically, I can take this part over here and write it in here in this equation. So what? I'm going to get ISS. Let's use the same color against We already had K two, all three, and now instead off oxygen. What we write is K one over K negative. Once all three do I buy or two. And if I rearrange this, what I'm going to get is que two okay, one over K minus one instead. Off here you may see the question a capital K or even a small K. It doesn't matter. That means just a constant all tree square. Do I buy? 02 And if you look at it, it is the same rate that we see here. This k is the same. It this portion off R K. So is it consistent? Yes, it is a constant stint. Uh, vit the rate and the overall reaction.