Examveda

For an isothermal second order aqueous phase reaction, A → B, the ratio of the time required for 90% conversion to the time required for 45% conversion is

A. 2

B. 4

C. 11

D. 22

Answer: Option C


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Comments (3)

  1. Chintan Bhalerao
    Chintan Bhalerao:
    4 years ago

    Here, for 2nd order reaction we have a equation:
    (-rA) = -dCA/dt = KCA^2
    Solving the equation we will get (1/CA - 1/CA0) = Kt
    Now we have to solve for 90% conversion so we have CA = 0.1 (as 90% is consumed) and assume CA0 = 1
    Similarly for 45% conversion we have CA = 0.55 and CA0 = 1
    for 90% conversion we will get Kt1 = 9 and for 45% conversion we will get Kt2 = 0.82
    Now just divide kt1/kt2, so we will get t1/t2 = 11
    I hope you got the solution

  2. Jayashree Mondal
    Jayashree Mondal:
    6 years ago

    Can I have the calculation please?

  3. THARANI K
    THARANI K:
    6 years ago

    can I know how to calculate this please

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