Examveda

Which of the following will give maximum gas conversion?

A. Fixed bed reactor

B. Fluidised bed reactor

C. Semi-fluidised bed reactor

D. Plug-flow catalytic reactor

Answer: Option C


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Comments (1)

  1. Atanu Chatterjee
    Atanu Chatterjee:
    4 months ago

    The correct answer is:

    D. Plug-flow catalytic reactor
    Explanation:
    To maximize gas conversion in a chemical reactor, the flow pattern and catalytic effectiveness are crucial.

    Why Plug-flow catalytic reactor gives maximum conversion:
    In a plug-flow reactor (PFR), the reactants move through the reactor without back-mixing.

    The concentration of reactants is highest at the inlet, which maximizes the reaction rate at the start.

    As the gas flows forward, the conversion gradually increases along the reactor length.

    When a catalyst is used, the effectiveness increases due to high reactant concentration at the catalyst surface in the inlet region.

    Comparisons:
    A. Fixed bed reactor
    → May behave like a PFR if ideal, but real systems often have non-idealities (channeling, heat/mass transfer issues).

    B. Fluidised bed reactor
    → Behaves closer to a CSTR due to excellent mixing. While good for heat transfer and solids handling, it gives lower conversion per pass than PFRs.

    C. Semi-fluidised bed reactor
    → Intermediate between fixed and fluidised beds; not optimal for maximum conversion.

    ✅ Final Answer:
    D. Plug-flow catalytic reactor

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