Examveda

If pore diffusion is the controlling step in a solid catalysed reaction, the catalyst

A. Porosity is very important

B. Porosity is of less importance

C. Internal surface area is utilised efficiently

D. None of these

Answer: Option A

Solution (By Examveda Team)

Concept: In a solid-catalyzed reaction, reactant molecules must diffuse through the pores of the catalyst to reach the active sites where the reaction takes place.

Pore diffusion control: When pore diffusion is the controlling step, the overall reaction rate is limited by the movement of reactants within the pores rather than by the surface reaction itself.

Importance of porosity: Higher porosity provides more void space for reactant molecules to diffuse easily, reducing internal mass transfer resistance and improving access to active sites.

Reason: If the porosity is low, diffusion is hindered, and a large portion of the catalyst's active surface remains underutilized, decreasing the overall efficiency of the catalyst.

Conclusion: When pore diffusion controls the reaction, porosity is very important for effective catalyst performance.

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Comments (1)

  1. Atanu Chatterjee
    Atanu Chatterjee:
    3 months ago

    A. Porosity is very important

    When pore diffusion limits the overall rate, the ease with which reactants can travel through the pore network to reach active sites becomes critical. High porosity (larger or more interconnected pores) reduces diffusion path‐lengths and resistance, helping to alleviate the internal mass‐transfer limitation. Conversely, if porosity is poor, reactants can’t efficiently penetrate to the catalyst’s internal surface, underutilizing much of its active area.

Related Questions on Chemical Reaction Engineering

A first order gaseous phase reaction is catalysed by a non-porous solid. The kinetic rate constant and the external mass transfer co-efficients are k and $${{\text{k}}_{\text{g}}}$$ respectively. The effective rate constant (keff) is given by

A. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\text{k}} + {{\text{k}}_{\text{g}}}$$

B. $${{\text{k}}_{\text{e}}}{\text{ff}} = \frac{{{\text{k}} + {{\text{k}}_{\text{g}}}}}{2}$$

C. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\left( {{\text{k}}{{\text{k}}_{\text{g}}}} \right)^{\frac{1}{2}}}$$

D. $$\frac{1}{{{{\text{k}}_{\text{e}}}{\text{ff}}}} = \frac{1}{{\text{k}}} + \frac{1}{{{{\text{k}}_{\text{g}}}}}$$