Under the same reaction conditions, initial concentration of $1.386 \mathrm{~mol} \mathrm{dm}^{-3}$ of a…
Under the same reaction conditions, initial concentration of $1.386 \mathrm{~mol} \mathrm{dm}^{-3}$ of a substance becomes 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 $\left(k_1\right)$ and zero $\operatorname{order}\left(k_0\right)$ of the reaction is
$0.5 \mathrm{~mol}^{-1} \mathrm{dm}^3$
$1.0 \mathrm{~mol} \mathrm{dm}^3$
$1.5 \mathrm{~mol} \mathrm{dm}^3$
$2.0 \mathrm{~mol}^{-1} \mathrm{dm}^3$
Solution
First order kinetics, $k_1=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{40} \mathrm{~s}^{-1}$
Zero order kinetics, $k_1=\frac{C_0}{2 t_{1 / 2}}=\frac{1.386}{2 \times 20}$
Hence, $\frac{k_1}{k_0}=\frac{0.693}{1.386}=0.5$