Fluid Solid Reactions Questions and Answers

Answer: A

Correct answer: the solid is consumed by the reaction

A catalyst is unchanged by the reaction it promotes and can in principle serve indefinitely; in a non-catalytic fluid–solid reaction the solid is a reactant and is used up. Ore roasting, coal combustion and the calcination of limestone are all of this kind, and the consequence is that the particle changes with time - which is what the shrinking core model exists to describe.

Answer: B

An unreacted core shrinks while a layer of product ash grows around it. The progressive conversion model is the alternative, where the fluid penetrates everywhere and reacts throughout. A front moves inward in the core model.

Answer: C

Both are true and the reason gives the criterion: a highly porous solid admits the fluid everywhere and the progressive conversion picture then fits better. Porosity is the deciding property. Dense solids give a sharp reaction front.

Answer: D

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

Each controlling resistance leaves its own signature in the conversion-time curve. Under gas film control the time is directly proportional to conversion. Under ash layer diffusion it follows \( 1-3(1-X)^{2/3}+2(1-X) \), a cube root expression. Under chemical reaction control it is \( 1-(1-X)^{1/3} \). And when two resistances are comparable there is no simple closed form. Fitting data to the three forms is how the controlling step is identified.

Answer: A

Correct answer: the radius to the first power only

With the mass transfer coefficient taken as constant, \[ \tau = \frac{\rho_B R}{3bk_gC_{Ag}}, \] so the time varies directly with the radius - the weakest dependence of the three regimes, since ash layer control gives \( R^{2} \) and reaction control gives \( R \) as well but with a different constant. In practice the coefficient itself depends on particle size, which steepens the observed dependence somewhat.

Answer: B

The first two are correct. The third is wrong, since film control gives a linear dependence for small particles in the Stokes regime, and no case is entirely independent of size. Each resistance gives a different power.

Answer: C

Both are true and the reason gives the diagnostic: a linear dependence points at film or reaction control and a square dependence at ash layer diffusion. Two sizes are enough to distinguish them. The exponent on radius is the clue.

Answer: D

Ash diffusion control gives a time proportional to the square of the radius, so doubling the radius quadruples the time. Reaction control would only have doubled it. The square dependence is the strongest of the three.

Answer: A

Reaction then proceeds everywhere at once and the particle keeps its size while its composition changes. The two models are the extremes of a range and real particles often lie between them. The particle keeps its size in this model.

Answer: B

Correct answer: diffusion within the unreacted core

The model treats the unreacted core as non-porous, so nothing diffuses into it - reaction happens only at its surface, which retreats inward as conversion proceeds. The three genuine resistances are the gas film outside the particle, the porous ash layer between the surface and the core, and the chemical step at the core surface itself.