Calculate the time in seconds required to reduce the concentration of reactant to half of initial concentration for a first order reaction if the rate constant is $1.386 \times 10^{-3} \ s^{-1}$. (in $s$)

$A$ first order reaction is $10 \%$ completed in $20 \, \text{minutes}$. What will be the time for $19 \%$ completion of the reaction? (in $\text{min}$)

What is the half-life time of a first-order reaction if the initial concentration of the reactant is $0.01 \ mol \ L^{-1}$ and the rate of reaction is $0.00352 \ mol \ L^{-1} \ minute^{-1}$?

For a first-order reaction,the time required for $93.75\%$ completion is $4$ times the half-life of the reaction. What is the relationship between the time required for $93.75\%$ completion and the half-life $(t_{0.5})$?

If the rate constant of a first order reaction is $4.606 \times 10^{-3} \ s^{-1}$,then find the time required for $400 \ g$ of the reactant to reduce to $50 \ g$. (in $min$)

$A$ 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 : $\ln 2 = 0.693$ )