Consider the two hypothetical reactions given below:
I $a A ightarrow$ Products, $k=x \mathrm{~mol}^{-1} \mathrm{~L} \min ^{-1}$
II $b B ightarrow$ Products, $k=y \min ^{-1}$
The half-lives of both the reactions are the same, equal to $1 \mathrm{hr}$ when molar concentration of the reactant is $1.0 \mathrm{M}$ in each case. If these reactions are started at the same time taking $1 \mathrm{M}$ of the reactant in each case, the ratio $[A] /[B]$ after $3 \mathrm{hr}$ will be :

Question

Aicharya · Accepted Answer

Units of $k$ indicate that reaction $I$ is of second order and reaction $I I$ is first order. For $I$ reaction, $t_{1 / 2} \propto 1 / a$, first $t_{1 / 2=} 1 \mathrm{hr}$, second $t_{1 / 2}=2 \mathrm{hr}$
$[A]=1 \mathrm{M} \stackrel{1 \mathrm{hr}}{\longrightarrow} 0.5 \mathrm{M} \stackrel{2 \mathrm{hr}}{\longrightarrow} 0.25 \mathrm{M}$
$[B]=1 \mathrm{M} \stackrel{1 \mathrm{hr}}{\longrightarrow} 0.5 \mathrm{M} \stackrel{1 \mathrm{hr}}{\longrightarrow}$
$0.25 \mathrm{M} \stackrel{1 \mathrm{hr}}{\longrightarrow} 0.125 \mathrm{M}$
$\frac{[A]}{[B]}=\frac{0.25 \mathrm{M}}{0.125 \mathrm{M}}=2$ *

Consider the two hypothetical reactions given below: I $a A ightarrow$ Products, $k=x \mathrm{~mol}^{-1}…

Solution