$A$ first-order reaction is $50\%$ complete in $45 \text{ minutes}$. How many hours will it take for the reaction to be $99.9\%$ complete?

$A$ flask is filled with equal moles of $A$ and $B$. The half-lives of $A$ and $B$ are $100 \, s$ and $50 \, s$ respectively and are independent of the initial concentration. The time required for the concentration of $A$ to be four times that of $B$ is $.... \, s.$
(Given : $\ln 2 = 0.693$ )

The time required for $90\%$ completion of a certain first order reaction is $1 \text{ hour}$. Calculate the time required for $99.9\%$ completion of the same reaction.

For the reaction $2A + B \to \text{Product}$,the rate law is given as $\frac{-d[A]}{dt} = K[A]$. At a time when $t = \frac{1}{K}$,the concentration of the reactant $A$ is ($Co =$ initial concentration).

For the reaction shown in the image,the half-life does not depend on the concentration of the reactant. After $10 \, \text{min}$,the volume of $N_2$ gas evolved is $20 \, \text{L}$ and after the completion of the reaction,it is $100 \, \text{L}$. Hence,the rate constant is:

The reaction $N_{2}O_{5} \longrightarrow 2NO_{2} + \frac{1}{2}O_{2}$ is first order in $N_{2}O_{5}$ having rate constant $6.2 \times 10^{-4} \ s^{-1}$. What is the value of rate of reaction when concentration of $N_{2}O_{5}$ is $1.25 \ mol \ L^{-1}$?