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
Join The Discussion
Comments (1)
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. $$\frac{1}{{{{\text{k}}_{\text{e}}}{\text{ff}}}} = \frac{1}{{\text{k}}} + \frac{1}{{{{\text{k}}_{\text{g}}}}}$$
The half life period of a first order reaction is given by (where, K = rate constant. )
A. 1.5 K
B. 2.5 K
C. $$\frac{{0.693}}{{\text{K}}}$$
D. 6.93 K
Catalyst is a substance, which __________ chemical reaction.
A. Increases the speed of a
B. Decreases the speed of a
C. Can either increase or decrease the speed of a
D. Alters the value of equilibrium constant in a reversible
A. $$ \propto {\text{CA}}$$
B. $$ \propto \frac{1}{{{\text{CA}}}}$$
C. Independent of temperature
D. None of these

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