Hi. It's Mr. Andersen and this is chemistry essentials video 42. It's on the rate limiting
step. And I'm going to start with an analogy. Imagine if we have a road and cars are moving
from point A to point B to point C. But nothing is slowing them down. Well this analogy I'm
later going to talk about chemical reactions when we're going from reactants to intermediates
and then finally to products. And there's nothing that's slowing down this reaction.
But let's say I want to collect money from these cars as they drive down the road. And
so I'm going to add an E-ZPass. And if you don't know what that is, basically cars have
a microchip in their car. As they drive underneath this it's going to count them. So they can
charge them for using that road. And so if we start the cars again, we're going to find
that this E-ZPass makes it easy. In other words the cars can ride right underneath it
and it's not going to slow them down. In other words the movement from B to C is fast. But
let's say I now want to install a toll booth. And this one works by having cars slow down
and then throwing their money in. Now the cars have to stop at toll 1. And so that's
going to slow down their overall speed down this road. And so the toll booth then becomes
the rate-limiting step. And so in an overall reaction remember it's not just one step,
it's step after step after step. And overall that's called the reaction mechanism. And
we can't just look at an overall equation and figure out the rate law of that reaction.
In other words we have to look at the speed of each of those elementary steps. Some are
really fast. Some are incredibly slow. And so we could look at those and figure out what's
the rate-limiting step. In other words what's the toll booth. What's slowing down this whole
reaction? If we know what that is then we can figure out the rate law. And so going
back to that analogy of the road, we really have two steps. Cars driving from point A
to point B. And then cars driving from point B to point C. And so if we look at what's
slowing them down or what's determining their rate, it's either the toll booth or the E-ZPass.
And so if we look at that overall reaction, driving from point A to point C, since that
first step is the time-limiting step, it's slowing it down, that's going to be the limiting
step of this whole reaction. So if we were to write the rate law for that it's going
to be based on how long does it take them to get through the toll times the concentration
of the cars. Now let's get to chemistry. Let's say we're looking at this reaction. Right
here we have nitrogen dioxide and carbon monoxide. And that's becoming nitrogen monoxide and
carbon dioxide. Now this progresses along two steps. And so if we lay those out, here's
step 1 and step 2. And we can see that they each have a different rate constant. But that
first step is going to be the slow part of the reaction. And the second is going to be
the fast. And so if we look at that overall reaction, again these are going to be the
reactants. So we've got nitrogen dioxide and then the carbon monoxide. And then these are
going to be the products, the nitrogen monoxide and the carbon dioxide. And so if we look
at that, what's slowing us down is that step 1. And so what's slowing down that whole equation?
It's going to be step 1. And so it's going to be this k1 or this slow rate of reaction
that's slowing down the overall reaction. So if we were to write the rate of equation
then, how are we going to do that? Well our rate equation is equal to k, that's going
to be our slow rate constant. And then what's going to be our reactants, it's simply going
to be the nitrogen dioxide reacting with itself. And so that's going to be our overall rate
reaction. Now one thing that you want to make sure that you're rate reaction is correct
is that we can have no intermediates here. Now what is an intermediate? If you look right
here we have nitrogen trioxide. That's not either a reactant or a product. In other words
that is an intermediate. And we can't include that in a rate reaction. And that's going
to be important as we look at this next reaction. So in the next one, same kind of a thing.
We've got two steps in this overall reaction. So here is step 1. And here is step 2. Now
if you look at step 1 we're finding something interesting. That this is a reversible reaction.
Not only are were going in this direction, but we could also go in the other direction.
Now we can see that this is the fast part of that reaction. And then we have the slow
part. So now we have essentially the E-ZPass first and then we have the toll booth after
that. So if we look at the overall equation, which of these is going to be the rate limiting
or the time limiting step? It's going to be step 2 now. And so what's going to be the
overall slowing of it down? It's going to be k2. Or it's going to be that second rate
constant. Now if we write the rate reaction we might just write it like this. So we've
got k2 right here. Now we've got step 2 which is di-nitrogen dioxide with oxygen gas. And
you might say I'm done. But remember, we can't include intermediates in that rate law. And
so we're not quite there yet. In other words that couldn't be the rate law. And that's
not going to help us when we're looking at a reaction. Because we don't see the intermediates.
We only see the reactants and we see the products. And so how are we going to get around this
problem? Well let's put this right here in red. This in other words is the thing that
we want to get rid of in this equation. And so the first thing we're going to do is write
this out. And so where did I get this equation? Well this is going to be the rate constant
of that first reaction. Again from the nitrogen monoxide to the di-nitrogen dioxide. And this
is going to be the reversible. In other words this is going to be going back. And so we
assume that the rate forward is going to be equal to the rate backwards. And so this is
a reversible reaction. How could you think about that with our road again? Think of it
as cars that can't get through this second toll and so they're just driving back and
forth between point A and point B. And so once we've got that then we can solve for
what is in here. In other words we can solve for what's in here. How do we do that? We're
going to divide both side by k to the negative 1. And now we can isolate that. So we've isolated
this on one side. That's the thing that we want to get rid of. And again we've divided
this side by k sub negative 1. And I can just move that up into my general equation. I can
substitute it. And so here's going to be the overall rate law of this equation. And so
I'm good now because I've got right here I've got one of my reactants. Right here I've got
one of my other reactants. And this is going to be the overall rate of this reaction. So
it seems a little bit confusing. But again, all we're doing is just doing some simplification
and then we're substituting that in so we can have a rate law that we can actually use.
So did you learn that the rate of the reaction is set by the slowest elementary reaction?
And we can use that to figure out our rate law. Well if you did, you learned what I wanted
you to and I hope that was helpful.