Q. Under the same reaction condition, initial concentration of $1.386\ mol\ dm^{- 3}$ of a substance become half in 40 s and 20 s through first order and zero order kinetics respectively. Ratio $\left( \frac{K_{1}}{K_{0}} \right)$ of the rate constants for first order ($k_{1})$ and zero order $(k_{0})$ of the reaction is
A) $0.5\ mol^{- 1}dm^{- 3}$
B) $1.0\ mol^{- 1}dm^{- 3}$
C) $1.5\ mol^{- 1}dm^{- 3}$
D) $2\ mol^{- 1}dm^{- 3}$
Explanation
(a)
First order kinetics,$k_{1} = \frac{0.693}{t_{1/2}} = \frac{0.693}{40}s^{- 1}$
Zero order kinetics,$\ k_{0} = \frac{C_{0}}{2t_{1/2}} = \frac{1.386}{2 \times 20}$
Hence,$\frac{k_{1}}{k_{0}} = \frac{0.693}{1.386} = 0.5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.