Examveda

At a given value of E/R (ratio of activation energy and gas constant), the ratio of the rate constants at 500°K and 400°K is 2, if Arrhenious law is used. What will be this ratio, if transition state theory is used with the same value of E/R?

A. 1.6

B. 2

C. 2.24

D. 2.5

Answer: Option D


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Comments (1)

  1. Chintan Bhalerao
    Chintan Bhalerao:
    4 years ago

    Firstly calculate E/R by using Arrhenious Theory. Then we have the expression for transition state theory i.e. K = Te^E/RT solve the expression you will get:
    ln(K2/K1) = ln(T2/T1) + E/R(1/T1 + 1/T2)
    we have all the values so put the values in the equation and you will get K2/K1 = 2.5

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