In this screencast, I'm going to discuss how the apparent activation energy, which means
if we measure the log of an experimental rate constant versus inverse temperature, how that
apparent activation energy depends on the activation energy of the reaction rate constant,
and on the heats of adsorption for the reactants and/or products, or even other components
that are present.
And also, as a result, it means this does not have to be a straight line.
An apparent activation energy can change with temperature.
So let's look at a simple example in which A plus B is reacting to products.
And we're going to assume that the product concentration, so this, we're going to assume
is a low concentration just to simplify a bit the calculations and make it a little
easier to see the behavior that we're interested in.
So if we had the type of reaction rate expression that we see for Langmuir-Hinshelwood mechanisms,
we'd have a rate that's proportional to a rate constant, and then the adsorption equilibrium
constants for A and B, the pressure of A and the pressure of B, and in the denominator
we have a term that has adsorption equilibrium constants for A and B and the pressures for
A and B squared.
So depending on which regime of coverage on the surface, because this results from saying
we have a rate constant times the coverage of A, the fractional coverage, and the fractional
coverage of B. So let's look at some limiting cases.
So we're going to assume that A is weakly adsorbed relative to B, meaning coverage of
A on the surface is low, and B is strongly adsorbed, so it's coverage on the surface
is high.
So this means that we're assuming K_B times the pressure of B is significantly greater
than K_A times the pressure of A. K_B times the pressure of B is also greater than 1,
so we're going to simplify the rate expression.
What I've done is say, this term and this term is small in the denominator, so the denominator
just becomes k_B squared, pressure B squared, which simplifies then to this form that we
see here.
Now, each of these terms have a temperature dependence, and therefore they're going to
affect the apparent activation energy if we look at this rate as being k experimental,
pressure of A, the pressure of B, to the minus 1 over some temperature regime where the coverage
of B is much higher than A, and high enough that we can drop the term here in the rate
expression.
So the rate constant of the reaction is going to have a temperature dependence, where it
exponentially depends on the activation energy for the reaction, so this is the activation
energy for the surface reaction, and then the adsorption equilibrium constants, both
K_A and K_B, are going to have a temperature dependence, and that is going to depend on
the heat of adsorption, so I'm going to use the notation here to make it easier to follow,
that lambda is a positive number.
In other words, lambda A is minus delta H of A for adsorption, so that lambda of A is
greater than 0, because almost all adsorption is exothermic, meaning this term is negative.
So if we substitute these three expressions into our rate expression, and this rate we
can look at as the apparent rate constant, so I'll call this k_0, apparent, e to the
minus E apparent over RT, again pressure of A and pressure of B. So we can look at the
temperature dependence terms, which means we can say this term is equal to the product
of these three terms, and therefore relate the apparent activation energy, so minus E
apparent, is going to be equal to the activation energy for the surface reaction, plus the
heat of adsorption for A, minus the heat of adsorption for B. So then we can see for our
expression for apparent activation energy, it can be less than the surface reaction activation
energy, E, or it even could be greater.
It depends on the relative values of these two.
Now, we said that B is strongly adsorbed relative to A, so that would mean the heat of adsorption
for B is greater than the heat of adsorption for A, so what we then would expect is that
E apparent is going to be greater than the surface reaction activation energy.
Now, we can certainly imagine other possibilities.
We could have both A and B being weakly adsorbed, so that the denominator, if we look back at
our original rate expression, so that 1 in the denominator, here is large relative to
the other two terms, and then our rate expression.
So I've just done the same type of substitution to determine the apparent activation energy
and how it relates to heats of adsorption in the surface reaction activation energy,
and now we can see that in this case, E apparent is going to be less than E. Remember, we see
that the adsorption values are both positive numbers, so the net effect is just subtracting
a value from the surface reaction activation energy.
So what this means physically is that, as we raise the temperature, so as temperature
increases, the rate constant for the reaction increases, but the coverages of A and of B
can decrease, do decrease, and depending on the relative values, it determines how the
reaction rate depends on temperature for different possibilities, of which we've shown a couple
in this screencast.
But there's a competition where coverage decreasing causes the rate to decrease, rate constant
increasing causes the rate to increase.
Indeed, it's possible in some cases, as we raise the temperature, the rate would go through
a maximum because that coverage decrease could dominate, namely these two terms could be
large enough to make the apparent activation energy negative.