Examveda

In solid catalysed reactions the diffusional effects are more likely to affect the overall rate of reaction for

A. Fast reactions in catalyst of small pore diameter

B. Fast reaction in catalyst of large pore diameter

C. Slow reactions in catalyst of small pore diameter

D. Slow reactions in catalyst of large pore diameter

Answer: Option C


Join The Discussion

Comments (3)

  1. Atanu Chatterjee
    Atanu Chatterjee:
    4 months ago

    The correct answer is:

    A. Fast reactions in catalyst of small pore diameter

    ✅ Explanation:
    In solid-catalyzed reactions, the overall rate can be affected by diffusional resistance, especially internal diffusion within catalyst pores.

    🔍 Diffusional effects become significant when:
    The reaction is fast:

    Reactants are consumed rapidly at active sites.

    To sustain the reaction, reactants must diffuse quickly to the internal surface.

    The pore diameter is small:

    This increases resistance to internal diffusion (pore diffusion).

    Reactants and products move more slowly, creating a concentration gradient and lowering the effective rate.

    📌 Therefore:
    If the reaction is fast and pore diameter is small, the system becomes diffusion-limited, i.e., the rate of mass transfer can't keep up with the rate of reaction.

    This leads to diffusional effects controlling the overall rate.

    ❌ Why other options are incorrect:
    B. Large pores reduce diffusion resistance.

    C. Slow reactions don't demand fast diffusion, so diffusion isn't rate-limiting.

    D. Both slow reaction and large pores mean minimal diffusion impact.

    ✅ Final Answer: A. Fast reactions in catalyst of small pore diameter

  2. ABRAR ALAM
    ABRAR ALAM:
    2 years ago

    C

  3. Pradeep Kumar
    Pradeep Kumar:
    6 years ago

    answer is B

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