For a first order reaction,the concentration of reactant was reduced from $0.03 \ mol \ L^{-1}$ to $0.02 \ mol \ L^{-1}$ in $25 \ min$. What is its rate (in $mol \ L^{-1} \ s^{-1}$)?

$t_{1/4}$ can be taken as the time taken for the concentration of a reactant to drop to $3/4$ of its initial value. If the rate constant for a first order reaction is $K$,the $t_{1/4}$ can be written as (in $/K$)

$T_{50}$ of a first-order reaction is $10 \ min$. Starting with $10 \ mol \ L^{-1}$,the rate after $20 \ min$ is:

$75\%$ of the first order reaction was completed in $32$ minutes. When was $50\%$ of the reaction completed? (in $\text{minutes}$)

Which of the following plots is(are) correct for the given reaction? $( [P]_0$ is the initial concentration of $P$ $)$

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 $\log_{10}(5) = 0.699$]