$A$ first order reaction is $50 \%$ complete in $16 \ min$. How much time is required to complete $87.5 \%$ of the reaction (in $min$)?

The units of rate constant for a first-order reaction are:

$A$ flask contains a mixture of compounds $A$ and $B.$ Both compounds decompose by first-order kinetics. The half-lives for $A$ and $B$ are $300 \ s$ and $180 \ s,$ respectively. If the concentrations of $A$ and $B$ are equal initially,the time required for the concentration of $A$ to be four times that of $B$ (in $s$): (Use $\ln 2 = 0.693$)

If a $75\%$ first-order reaction is completed in $32 \text{ minutes}$,how many minutes will it take for the same reaction to be $50\%$ completed?

Derive the equation showing the relation between the concentration $[R]_1$ and $[R]_2$ at time $t_1$ and $t_2$ for a first-order reaction.

$A \rightarrow P$ is a first order reaction. At $300 \ K$,this reaction was started with $[A] = 0.5 \ mol \ L^{-1}$. The rate constant of the reaction was $0.125 \ min^{-1}$. The same reaction was started separately with $[A] = 1 \ mol \ L^{-1}$ at $300 \ K$. The rate constant (in $min^{-1}$) now is: