# Pchem kinetics ques

1. question
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On a catalyst surface, the formation reaction of CO2 by the oxidation of CO with O2 proceeds in the following mechanism.

(1) O2(g) + 2V <---> 2O(ads) (equilibrium constant KO2)
(2) CO(g) + V <--> CO(ads) (equilibrium constant (KCO)

(g) denotes gas phase
O(ads) denotes Oxygen atom on absorption site
CO(ads) denotes CO on absorption site
V denotes vacant absorption site

Partial pressures of O2(g) and CO(g) are defined as PO2 and PCO respectively. Densities of O(ads), CO(ads), and V are [O(ads)], [CO(ads)], and [v] respectively. Fractional coverage of O(ads) and CO(ads) are defined are theta(Oxygen) and theta(CO) respectively.
Only reaction 1- 3 are considered and reaction 3 is the rate limiting step. More assumptions are that all adsorption sites are equivalent and a single adsorbate occupies an adsorption site at once.

Here are the questions[b/]

1. Show the adorption rate, ra, and desorption rate, rd, using all or parts of ka, kd, PO2, [V], and [O(ads)], where ka and kd denote the rate constants of adsorption and desorption in eqn (1).

2. Show KO2 and KCO using all or parts of the densities and partial pressures described above.

3. Show theta(Oxygen) and theta (CO) using all or parts of PO2, PCO, KO2, and KCO.

4. Derive the rxn rate rCO2 of eqn 3 using all or parts of k, PO2, PCO, K)2, KCO, and n.

5. Here, PCO dependence of rCO2 is considered under the condition of PCO >> PO2. Express rCO2 as a function of PCO in the two cases of a sufficiently low PCO (KCO x PCO << 1) and a sufficiently high PCO (KCO x PCO >> 1). In addition draw a rough sketch of the relation between rCO2 and PCO.

## Homework Equations

The only equations that I could think of are:
general rate laws

rate(r) = k x [reactant 1] x {reactant 2] x ....
where k is the rate constant

* if at equilibrium, the concentration of the reactants must equal the concentration of products

Raoult's partial pressure which relation partial pressure of pure solvents to solutions and concentrations

P (solution) = X x P (pure solvent)
* X = mol fraction of solvent = (mol solvent)/ (mol solvent + mol solute)

Catalyst equations

theta = (# adsorption sites occupied)/ (# sites available)

rate of change of surface coverage (d theta/ d t) = K(ads) x P x N x ( 1- theta)
rate of change of desorption (d theta/ d t) = - K(desorption) x N x theta
* N = total number of sites

## The Attempt at a Solution

So my attempts are weak

for 1

ra = Ka x [V]
* here I think [O2] is constant so it is not taken into account of the rate of adsorption

For 2, 3, 4, and 5 I am totally lost.

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AGNuke
Gold Member
* if at equilibrium, the concentration of the reactants must equal the concentration of products
Is it so? Try writing its its Equilibrium constant formula, you'll know

Is it so? Try writing its its Equilibrium constant formula, you'll know
But that's just a standard definition of equilibrium...why do you say that it's not?

But that's just a standard definition of equilibrium...why do you say that it's not?
That's not the definition of equilibrium ever. It's about rates.

AGNuke
Gold Member
why do you say that it's not?
$$K_{O_2}=\frac{[O]^2}{[O_2]}$$

See, its square of product concentration by Concentration of reactant.

Hmmm I guess that i was mixing up determing reaction order and writing the equilibrium constant. Eitherwy, I need massive help with understanding where the problem is going. I've read up on reaction rates and kinetics from Atkins physical chemistry and from my gen chem book, but I still can't decipher the question. Old you guys help e with that and offer some steps t o follow to understand and then solve the problem?

epenguin
Homework Helper
Gold Member
That's not the definition of equilibrium ever. It's about rates.
That may be true but since the student is nowhere it might be well he start assuming the adsorbed O and adsorbed CO are in equilibrium with the gas O2 and gas CO, and that the catalysis is slow enough not to change this equilibrium significantly - if he can solve that it is a more than good start. Refine later.

Start at the end Hoshi, think only of what is already adsorbed onto the catalyst. What is the equation for reaction rate in terms of that? As a hint your first equation
ra = Ka x [V]
has got to be wrong. It is saying that the reaction rate is proportional to the amount of catalyst surface where there is nothing there that could react!

Your physical chemistry textbooks should show how to treat such situations. It is practically formally identical to enzyme kinetics if you have done any of that.

It might help to start with solving under the simplifying conditions of 5.
The question asks you to start with full generality and then look at simpler special cases, but the opposite might be easier.

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epenguin
Homework Helper
Gold Member
On second thoughts, from the way the question is affronted I think Hoshi should first try to solve, orstudy how it is done, a simpler mechanisms such as a surface catalysis of a simple reaction A → B via adsorption:
Agas⇔ Aadsorbed →Badsorbed → Bgas. The question is just a .more complicated example involving the same approach and principles.