Heterogeneous Reaction Kinetics Questions and Answers

Answer: A

The assumptions are strong but they give a tractable model that describes many systems well enough. The BET isotherm relaxes the monolayer assumption and is used for surface area measurement. Simplicity is what makes it useful.

Answer: B

Both are true and the reason gives the mechanism: once every site is occupied, further pressure can add nothing, which is exactly the plateau the isotherm shows. Coverage cannot exceed one monolayer. Saturation is the defining feature of the isotherm.

Answer: C

Correct answer: A-2, B-4, C-1, D-3

Four models, three of mechanism and one of measurement. Langmuir–Hinshelwood has both reactants adsorbed before they meet on the surface. Eley–Rideal has one adsorbed species struck by one arriving directly from the gas. Mars–van Krevelen has the catalyst’s own lattice oxygen consumed and then replenished, which is how selective oxidation catalysts work. BET is not a mechanism at all but the multilayer adsorption model used to measure surface area.

Answer: D

The denominator accounts for the competition of species for the surface, which is why the observed order can change with pressure. A power law is an empirical fit that hides this behaviour. Competition for sites is the cause.

Answer: A

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the mechanism. A Langmuir–Hinshelwood rate has adsorption terms in its denominator, and their relative size shifts with pressure. At low coverage the denominator is near unity and the rate looks first order; at high coverage the adsorption term dominates and the rate becomes nearly independent of pressure, so the apparent order falls towards zero. Nothing about the chemistry has changed - only the coverage.

Answer: B

Correct answer: 1 and 2 only

Statements 1 and 2 are correct: with both reactants competing for sites the rate can rise, peak and fall as one of them is raised, and over a narrow range the expression is well approximated by a power law. Statement 3 claims too much - fitted adsorption constants frequently come out negative or physically implausible, and are then merely parameters that happen to fit.

Answer: C

Correct answer: strong adsorption blocking the surface

A maximum arises when one reactant adsorbs so strongly that raising its pressure further displaces the other from the surface, so the two can no longer meet. Beyond the peak the rate falls even though more reactant is present. It is a signature of competitive adsorption and cannot be produced by any simple power law, which is one of the reasons the Langmuir–Hinshelwood form is used.

Answer: D

Correct answer: the intrinsic rate with pore diffusion

\[ \phi = L\sqrt{\frac{k}{D_{eff}}}, \] a ratio of the rate at which reactant would be consumed to the rate at which it can diffuse into the pellet. A small value means diffusion keeps up and the whole pellet is used; a large value means the reaction is finished near the outside.

Answer: A

Correct answer: Both A and R are true, and R is the correct explanation of A

Both are true and the reason gives the mechanism. A large modulus means the reaction is fast relative to diffusion, so reactant is consumed within a thin outer shell and the interior of the pellet sees almost none. The effectiveness factor - the actual rate divided by the rate if the whole pellet were at surface conditions - is correspondingly small, and most of the expensive catalyst inside is doing nothing.

Answer: B

Correct answer: about 0.1

For a first order reaction in a slab, \[ \eta = \frac{\tanh\phi}{\phi}, \] and at \( \phi=10 \) the hyperbolic tangent is essentially 1, so \( \eta \approx 1/10 = 0.1 \). The useful rule is that for any large modulus the effectiveness factor is close to \( 1/\phi \) whatever the pellet shape, so only about a tenth of this catalyst is being used.