Texas Tech University
Reaction Mechanisms - Example 4


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one more questions. So they're telling us there is his reaction. X plus why goes to X Y and this is overall. And what they're saying is this is experimentally second order in X and first order in. Why? So what? Me? What that means is que X is going to be second order and why is going to be the first order? This is going to be our rates law, right? So And the question they're saying, Is this going to occur in one step or is this the overall? So it's definitely not going to occur in one step because we have extra part of two. That means we need an extra X in the elementary, step off this, uh, great determining step. So we ruled that out and they're proposing us and mechanism vit says two x k one k minus one is going to be X two first, and this is going to be the fast step and then x two plus, why is going to be X Y plus X, and this is going to be slow step. And they're asking us if this is a good way to show this overall, um, reaction So Let's first add the two sides together and see if this really gives us overall reaction. So we have this extremely intermediate. So we also have this X over here, and it's going to cancel with one off them. So what we will have is X plus Why equals two X y So, yes, this is the same with the overall, but it doesn't mean that it is going to give us the same rate with the red law. We're going to check for that. First. I just need to mention this X two is Theo Intermediate? We see. But X is not the intermediate because X is in the reactant and then it gets in the products. It does not get consumed intermittent. It means we don't have it in the reactions at the starting off the experiment and we have it in the products and then it will get consumed and reactions. So if you want to write the slow steps straight law, which is going to be our red law? If it does, I'm going to write K two here. So it will be k two x two. Why now? The problem is X two is intermediate so we need to write it in terms off something else. If I look at the first step, I have forward reaction. Okay. Once X squared equals diverse traction came on swan X two. So x two is going to be K one over K minus one x to the power of to. And now, if I write this part instead off this x two, what I'm going to get is K one que two or K minus one x square. And why now? This is consistent with the rate law that they could observe. And we can conclude that this two step mechanism is a consistent mechanism with our rate law.

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