Examveda

Pure A in gas phase enters a reactor 50% of this A is converted to B through the reaction, A → 3B. Mole fraction of A in the exit stream is

A. $$\frac{1}{2}$$

B. $$\frac{1}{3}$$

C. $$\frac{1}{4}$$

D. $$\frac{1}{5}$$

Answer: Option C


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Comments (2)

  1. SATYAM SINGH
    SATYAM SINGH:
    5 years ago

    Let 1 mole of A taken initially
    After reaction the mixture has 1.5 mole of B and 0.5 mole of A
    So mole fraction of A=0.5/2=1/4

  2. Parth Nayi
    Parth Nayi:
    5 years ago

    How it is possible

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