If the rate constant of a reaction is $x \, sec^{-1}$,by what factor does the rate increase if the initial concentration of $A$ is tripled?

For a first-order reaction,the time taken for the initial concentration to decrease to $1/4$ of its value is $20 \ min$. The time taken for the initial concentration to decrease to $1/16$ of its value will be $......... \ min$.

The reaction $L \rightarrow M$ starts with $10 \, gL^{-1}$. After $30$ and $90$ minutes,$5 \, gL^{-1}$ and $1.25 \, gL^{-1}$ remain,respectively. The order of the reaction is:

In a first-order reaction,the concentration of the reactant decreases from $0.80 \, mol \, L^{-1}$ to $0.06 \, mol \, L^{-1}$ in $45 \, min$. Calculate the half-life $(t_{1/2})$. (in $, min$)

Initial concentration of reactant in a first order reaction is $0.08 \text{ mol dm}^{-3}$. What concentration would remain after $40 \text{ minutes}$? (Given $\frac{[A]_0}{[A]_t} = 5.00$)

For a first order reaction $A \rightarrow P$,$t_{1/2}$ (half-life) is $10 \text{ days}$. The time required for $1/4^{th}$ conversion of $A$ (in days) is: