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

The decomposition of substance $A$ follows first-order kinetics. Flask $I$ contains $1 \ L$ of $1 \ M$ solution of $A$,and flask $II$ contains $100 \ mL$ of $0.6 \ M$ solution of $A$. After $8 \ hours$,the concentration of $A$ in flask $I$ becomes $0.25 \ M$. How many hours will it take for the concentration of $A$ in flask $II$ to become $0.3 \ M$?

The rate of a first-order reaction is $1.5 \times 10^{-2} \, mol \, L^{-1} \, min^{-1}$ when the concentration of the reactant is $0.5 \, M$. Find the half-life $(t_{1/2})$ of the reaction in $min$.

The integrated rate equation for a first-order reaction is:

In a first order reaction,the concentration of the reactant decreases from $0.6 \ M$ to $0.3 \ M$ in $15 \ min$. The time taken for the concentration to change from $0.1 \ M$ to $0.025 \ M$ in minutes is

$A$ first order reaction with respect to the reactant $A$ has a rate constant of $6 \; sec^{-1}$. If we start with $[A] = 0.5 \; mol/L$,then in what time the concentration of $A$ becomes $0.05 \; mol/L$ ........... $sec$