Langmuir-Hinshelwood Kinetics: Derivation and Model Verification

[00:00:06]So this is a problem that involves coming up with a Langmuir-Hinshelwood mechanism and then evaluating whether that mechanism can explain some kinetic data in the second part of the problem.

[00:00:16]In this case, we're dealing with the hydrogenation of ethylene to ethane conducted over a nickel catalyst, where the mechanism is proposed to exist as just a few simple steps shown here, which involves the adsorption of ethylene onto a surface site in reaction 1 to for adsorbed ethylene, a reversible step, the dissociative adsorption of hydrogen onto two surface sites to form two hydrogen atoms, and then adsorbed ethylene reacting with two adsorbed hydrogen atoms to produce ethane directly in the gas phase, which is pretty typical because ethane is very weakly adsorbed, which frees up those three surface sites to participate in another reaction in the cycle.

[00:00:56]We're given in the problem statement in this case that the rate-determining step is step three, so the other steps will be assumed to be equilibrated, and we'll describe those in terms of adsorption equilibrium constants.

[00:01:12]So let's look at how to set this up in a mathematical Langmuir-Hinshelwood format, which is what A asks us to do, and it asks us to do that in terms of equilibrium constants for the reactions, and the forward reaction rate constant, as well as the total concentration of surface sites, CT, available for reaction.

[00:01:32]So if we want to set this up, we need to start writing our equilibrium expressions, so the first one for the adsorption of C2H4 basically says that the equilibrium constant for process 1, for adsorption, is equal to the concentration of the products of that reaction, which is adsorbed ethylene, over the concentration of the reactants, which are vacant surface sites and the concentration, normally we express this in terms of the partial pressure, of C2H4.

[00:02:04]So we can define that equilibrium relationship because that adsorption step is equilibrated, and it's going to be more convenient to express that this way, that C2H4 adsorbed is equal to K for step 1 times P_C2H4 times the concentration of vacant surface sites.

[00:02:28]Alright, very similarly, if we look at step 2, the dissociative adsorption of hydrogen, we can write that in terms, I'll just go ahead and skip the first part of this, so we can write that in terms of the concentration of sites on the surface, the sites containing hydrogen on the surface squared, is equal to K2 P_H2 star squared.

[00:02:55]And in this case, we carry along the squares because the stoichiometry dictates that we have two surface sites participating the reaction and two hydrogen atoms participating in the reaction.

[00:03:08]And I initially skipped this, but maybe I should go ahead and show it, that that means that products over reactants will be H* squared, the concentration of hydrogen squared in our product, over the concentration of star squared, one of our reactants, and then it's first order in hydrogen concentration or pressure, so we don't square that term.

[00:03:28]So these are two equations that are going to be useful to us, and another way of writing this equation, I'm going to call that 2A, is taking the square root of both sides, we're going to find that to be useful, so that will give us H* is equal to the square root of K2 P_H2 times the concentration of vacant surface sites.

[00:03:50]So these are equilibrated steps, and then the other element of our Langmuir-Hinshelwood expression is a site balance, where the total number of sites, the concentration of total sites, is equal to the sum of all the different types of sites you have on your surface.

[00:04:07]So one of those types is vacant sites.

[00:04:09]We also have sites that contain ethylene, and we also have sites that contain hydrogen.

[00:04:16]And all of those add up to some total concentration of sites that must be conserved during the reaction.

[00:04:22]For a catalytic reaction, we don't generate or remove sites in the simplest approximation.

[00:04:27]Alright, we can substitute in for this, and I'm going to go ahead and factor out the concentration of sites, so we have 1 if we factor that out here, C2H4* from equation 1 is equal to K1 times P_C2H4 times this concentration of vacant sites, which we've factored out through this term, so that's equal to 1 plus K1 P_C2H4 plus H*, we see is equal to the square root of K2 P_H2 times the concentration of vacant sites, which we've again factored out, so we just get the square root of K2 P_H2 here.

[00:05:11]So that's the total concentration of sites in terms of the concentration of vacant sites, so we've eliminated some unknowns from this expression, which will be useful for us.

[00:05:22]And then finally, we can write this rate of reaction.

[00:05:28]We need to express this as a rate, which is what the problem asked us to do.

[00:05:31]That's equal to the rate of reaction 3, which is shown here, and we're given that this is a reversible reaction so we have to include both the rate of the forward and reverse process, so let's go ahead and do that below.

[00:05:46]So it's equal to k3, the rate constant for the forward process, times the concentration of ethylene adsorbed on the surface times the concentration of hydrogen adsorbed on the surface squared, because two hydrogen atoms participate in that reaction, and we assume an elementary reaction in this Langmuir-Hinshelwood mechanism, minus the partial pressure of ethylene, since it's the concentration of the product that we need to take into account for the reverse reaction, times the concentration of vacant sites cubed, and we divide that by K3, again because there are three vacant sites that participate in the reverse reaction, three vacant sites plus ethane.

[00:06:38]Once again, what we would like to do is be able to express all of this in terms of rate constants, equilibrium constants, and the concentration of total sites, which is a quantity that's conserved, because it's very difficult to actually measure the concentration of vacant sites, hydrogen containing sites, and ethylene containing sites.

[00:07:01]So in order to do that, the strategy that we'll use is to substitute everything in so that we get only the concentration of vacant sites in here, we have enough relationships to do that.

[00:07:13]So let's see how this works.

[00:07:15]So this rate is then equal to k3.

[00:07:21]For C2H4*, we have a relation up here from equation 1 that relates that to star, to the vacant sites.

[00:07:28]We have an expression for H* squared.

[00:07:32]We have this part of our expression, and I want to make sure that this is a capital K3.

[00:07:37]So if we collect terms, this becomes a vacant sites cubed, both in this part of the expression and here, and we can factor that out, a vacant sites cubed out here, and we have in expression I'll call it 3 here, the concentration of vacant sites, we can express that in terms of only constants and pressures, things that we can measure and fit, and the concentration of total sites which is conserved.

[00:08:07]So we will combine equation 3 with equation 4, and I'll re-express that.

[00:08:14]We get that the rate is equal to the total concentration of sites cubed times k2, we collect terms, we get these K1, K2, pressure of C2H4, pressure of hydrogen, minus the pressure of C2H6 over K3, and then we get the typical Langmuir-Hinshelwood denominator term, which is 1, which stands in for the concentration, the relative proportion of vacant sites, the more important the 1 is compared to the other terms, plus K1 times the pressure of C2H4, which relative to the other terms is proportional to the number of sites covered with ethylene, plus square root of K2, pressure of hydrogen, and then this entire term is taken to the third power, which originates from the third power right here.

[00:09:11]So we see now we've expressed, we've done what the problem statement asked, and we've expressed everything in terms of equilibrium adsorption constants which we can fit to a model, as well as a rate constant which we can also just fit if we vary the pressures of hydrogen and ethylene and measure the rate, and the total concentration of sites which is often lumped in with k3, because it should be fixed during the reaction as long as you don't have substantial catalytic deactivation.

[00:09:38]Now, if we turn to part D of this problem, it shows you some experimental data, in which the partial pressure of hydrogen has been varied at a fixed partial pressure of the other reactant, ethylene, and the outlet flow rate of ethane has been measured in a differential type of reactor.

[00:10:01]So what we see, you can go through a more rigorous process here but we'll just look at this very simply because the trends are pretty simple.

[00:10:07]If we quadruple the inlet flowrate, or the inlet concentration of hydrogen from 10 to 40 and then from 40 to 160, so each time we're quadrupling that, then we only double the outlet, we only double the productivities, so I only double the outlet flowrate of ethane, of our product of this reaction.

[00:10:28]That means that here, the rate is proportional to the pressure of hydrogen to the 1/2 power, because when you quadrupled the amount of hydrogen, you only doubled the rate, so that's a square root dependence.

[00:10:41]Let's see how this compares to our expression.

[00:10:45]We furthermore note that it is known that the hydrogen coverage on the surface is very low under these conditions, and that happens to be true for this kind of catalyst.

[00:10:55]We know that the coverage of hydrogen is very low, so that means that this term, which describes the coverage of hydrogen, is very low compared to these two terms, the 1 plus K1 P_C2H4.

[00:11:08]So we can assume that this is negligible and goes to zero.

[00:11:13]We furthermore can assume, although we don't have to for this problem, but it makes it simpler, that because we're not feeding any ethane and this is a differential reactor, so that the conversion is low, that this term goes to zero.

[00:11:26]So there's very little ethane in our reactor, and we can assume that this goes to zero, which happens to also be a very good assumption for this particular reaction.

[00:11:36]And so therefore, what our model predicts is that the reaction should be first order in hydrogen, notice that we get this first order dependence in our model.

[00:11:44]What the experimental data told us, remember, is that it's actually half order.

[00:11:49]So this model that we've proposed is not consistent with the experimental data, and we have to go and propose another model that could be consistent with the experimental data.

[00:12:01]And we won't include this in this screencast, but one productive thing that we might think about doing, and which will actually lead to a model that's consistent with the data, is to, instead of assuming that both hydrogens add at once, we could assume a sequential process where first, one hydrogen added in order to make ethyl in the rate-determining step, and then we had a second process where the next hydrogen was added in a fast step that we assume does not contribute to the rate.

[00:12:34]And if we use this, we'll see that we can get an expression that is consistent with the experimental data here, and makes sense anyway since as we've talked about previously in class, these kind of three-body collisions where 2 hydrogen atoms simultaneously collide with an ethylene molecule are unlikely, and that's true both on surfaces and in three-dimensional space.