The ratio of the time taken to complete $60\%$ and $20\%$ of a first-order reaction $[A \rightarrow \text{product}]$ is : $(\log 2 = 0.3)$

Consider the two different first order reactions given below:
$A + B \rightarrow C$ (Reaction $1$)
$P \rightarrow Q$ (Reaction $2$)
The ratio of the half-life of Reaction $1$ : Reaction $2$ is $5 : 2$. If $t_1$ and $t_2$ represent the time taken to complete $2/3$ and $4/5$ of Reaction $1$ and Reaction $2$,respectively,then the value of the ratio $t_1 : t_2$ is $. . . . \times 10^{-1}$ (nearest integer).
[Given: $\log_{10}(3) = 0.477$ and $\log_{10}(5) = 0.699$]

For a first order reaction,show that the time required for $99 \%$ completion is twice the time required for the completion of $90 \%$ of the reaction.

The half-life of a substance in a certain enzyme-catalysed reaction is $138 \; s$. The time required for the concentration of the substance to fall from $1.28 \; mg \; L^{-1}$ to $0.04 \; mg \; L^{-1}$ is ....... $s$.

For a first order reaction $A \to B$,the reaction rate at reactant concentration of $0.01 \ M$ is found to be $2.0 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$. The half-life period of the reaction is .......... $sec$.

Calculate the half-life period of a first-order reaction if the rate constant of the reaction is $0.02 \ min^{-1}$. (in $min$)