In this video, we're going to apply the steady state approximation to a reaction sequence
that proceeds through a chain mechanism. So the overall gas phase reaction is ethylene
reacting with hydrogen to make ethane. And the way this reaction proceeds is, there's
an initiation step to form these active intermediates. So these are much more reactive, they're going
to be present at low concentrations. One of the hydrogen atoms can then collide with an
ethylene to make, again, one of these reactive intermediates - an ethyl radical. That
can collide with a hydrogen to make our final product, recreate a hydrogen atom that
can go back in this step and react again. And so we're going to run through this sequence
many times, once when we've initiated the reaction, once we've created the hydrogen
radical, many times in the chain mechanism - we go through the chain - and then there's
some probability that a hydrogen and ethyl radical will recombine in a termination reaction
to make product. The amount of product made by this step is small because, remember, both
of these are present in low concentrations because they're very reactive. So hydrogen
atom is most likely to collide with either hydrogen molecule or an ethylene molecule.
And so they have a much lower probability of colliding to go through this termination
step. So what we want to do is, using this mechanism and the idea of the steady state
approximation, determine how the rate of reaction is going to depend on the concentration of
hydrogen and ethylene. And so the steady state approximation says that very quickly after
we initiate, we build up to a steady state concentration of hydrogen. So we're going
to assume that concentration doesn't change with time. And then also that means that the
concentration of this doesn't change with time, and it's zero. And so then to apply
this, we're assuming also that these elementary steps. And what that means is, we can write
down the rate expressions directly - the rate of this first reaction would be first order
in hydrogen and first order in ethylene. So now we look at all the steps that involve
hydrogen atoms, and I'll write down the expression and then describe what each of the terms
is. OK, so let's look at the change in the concentration of the hydrogen atoms with respect
to time. In the first step, making hydrogen atoms, first order in ethylene, first order
in hydrogen. So this is a plus sign here. In the second step we're using up the hydrogen
atoms, so a minus sign. And again first order in the hydrogen atom concentration, first
order in ethylene. In the third step, we're making hydrogen atoms, so that's a plus
sign. And the fourth step, we're using them up, so again a minus sign, the termination
rate constant, the concentration of the ethyl radical, the concentration of hydrogen atoms.
And then we do the same thing for C2H5 dot, the ethyl radical. We make it initiation step,
we make it in the second step, and we use it up third step and in the termination reaction.
And so now we could add these two equations together, and if we add them together, of
course we notice - because of the sign changes between these, this K3 term is going to cancel,
this K2 term is going to cancel. And so when I add them, left side is zero, I have the
initiation, C2H4 times hydrogen minus Kt, or Ki; the rate of initiation is equal to
the rate of termination, which is indeed what we would expect. If everything is at steady
state, then the rate that we're going to be making hydrogen in the initiation step is
the rate that we're using it up in the termination step. So adding the two equations shows us
that rate of initiation equals rate of termination. If instead I subtract this equation from this
equation, notice the first terms - the Ki terms - will cancel. K2 terms will cancel.
These then will add. I'll have two of these, a minus sign, and two of the K3 terms with
a plus sign. And so here's what we end up with when we subtract the two equations. And
just rearranging and cancelling the 2s, it says the rate of the second step equals the
rate of the third step. So it says the rate of this step equals the rate of this step.
And this makes sense from out argument that we're at steady state - in this step, we're
making C2H5, if we don't use it up at the same rate then we would be building up the
concentration of C2H5 and we would not be at steady state. So applying the steady state
approximation, now we have a relation between H radicals and C2H5, the ethyl radicals. So
this says that the rate of propagation, which is what the second and third steps are, so the
rates of the propagation steps are the same in this chain mechanism. The overall rate
is how fast we're making product. So we're making product in the third step. Indeed,
we're also making product in the fourth step, but remember this happens a few times compared
to many times as this goes through this chain mechanism. So the overall rate is going to
be K3 times the dihydrogen gas phase concentration and the ethyl radical concentration. So the
overall rate, of course, is the rate that C2H6 forms. So we'll work with this rate,
and we want to be able to write the concentration of hydrogen in terms of measurable concentrations.
And so what we're going to do is use the rate of initiation is equal to the rate of termination.
And if we rearrange that, the concentration of hydrogen atoms, initiation rate constant,
termination rate constant, H2, C2H4, and then the C2H5 radical. And then we're going to
use the rate of propagation, the two steps - use this equation - to write that C2H5 concentration
is equal to - just rearranging - this is now in terms of concentration of hydrogen atoms.
So we can substitute this here, and we'll end up something where we can solve for the
concentration of hydrogen atoms. So if we just combine those two equations and simplify,
we'll get the hydrogen atom concentration in terms of these rate constants. Now, since
our overall rate is the hydrogen atom concentration times C2H4, then our overall rate is K2 times
the hydrogen atom concentration times ethylene. And so our final rate expression would be
K2, K3, Ki, over K termination to the one half because we have K2 to the first power
in the numerator and K2 to the one half power in the denominator. And then first order in
hydrogen, first order in ethylene. And so note that this says that if the rate constant
for the termination reaction is larger, the rate's going to be
slower, but if the initiation rate constant was larger, this rate would be faster. Right,
so this is the overall rate that forms C2H6 ethane by hydrogenating ethylene in the gas
phase by a chain reaction.