Question

Under the same reaction condition, the initial concentration of $1.386{\text{mol dm}}^{-3}$ of a substance becomes half in $40$ seconds and $20$ seconds through first order and zero order kinetics, respectively. Ratio $\left(k_1/k_{\circ }\right)$ of the rate constants for the first order $\left(k_1\right)$ and zero order $\left(k_{\circ }\right)$ of the reaction is

• A. $0.5{\text{mol dm}}^{-3}$
• B. $1.0{\text{mol dm}}^{-3}$
• C. $1.5{\text{mol dm}}^{-3}$
• D. $2.0{\text{mol dm}}^{-3}$

Answer 1

The correct option is A $0.5{\text{mol dm}}^{-3}$

The rate expressions are:

First order $ln\left\{\right[A{]}_t/\left[A{]}_{\circ }\right\}=-k_{1}t$ and Zero order $[A{]}_t-[A{]}_{\circ }=-k_{\circ }t$

Hence, for first order reaction $t_{1/2}=\frac{1}{k_1}ln\frac{1}{2}=\frac{0.693}{k_1}=40s$

For zero order reaction
$t_{1/2}=\frac{[A{]}_{\circ }}{2k_{\circ }}=20s\text{Thus}\frac{k_1}{k_{\circ }}=\frac{0.693/\left(40s\right)}{\left(1.386mold{m}^{-3}\right)/(2\times 20s)}=0.5mo{l}^{-1}d{m}^3$