In a first order reaction,$60 \%$ of the reactant decomposes in $4.606 \ min$. What is the half-life of the reaction (in $min$)? (Given: $k = 0.1989 \ min^{-1}$)

For a first order reaction,to obtain a positive slope,we need to plot $...$ where $[A]$ is the concentration of reactant $A$.

For a first order reaction,the intercept of the graph between $\log \left(\frac{[A]_0}{[A]_t}\right)$ ($Y$-axis) and time ($X$-axis) is equal to

For the thermal decomposition of $N_2O_{5(g)}$ at constant volume,the following table can be formed for the reaction mentioned below:
$2N_2O_{5(g)} \rightarrow 2N_2O_{4(g)} + O_{2(g)}$

$S.NO$	$Time/s$	Total pressure $(atm)$
$1.$	$0$	$0.6$
$2.$	$100$	$X$

$X = . . . . . . \times 10^{-3} \ atm$ [nearest integer]
Given: Rate constant for the reaction is $4.606 \times 10^{-2} \ s^{-1}$.

Given below are two statements:
Statement $I$: The graph of $t_{1/2}$ versus initial concentration $[R]_0$ for a first-order reaction is a horizontal line.
Statement $II$: The graph of $\log \frac{[R]_0}{[R]}$ versus time $t$ for a first-order reaction is a straight line passing through the origin with a slope equal to $\frac{k}{2.303}$.
In the light of the above statements,choose the correct answer from the options given below:

The half-life of a first-order reaction is $20 \text{ minutes}$. What is the rate constant (in $\text{min}^{-1}$) for this reaction?