In a first order reaction, $20 \text{ mmol}$ of reactant is reduced to $10 \text{ mmol}$ in $0.3010 \text{ min}$. Find the rate constant of the reaction. (in $\text{ min}^{-1}$)

Mathematical representation for $t_{1/4}$ life for a first-order reaction is given by:

In a reaction $A + B \rightarrow C$,initial concentrations of $A$ and $B$ are related as $[A]_0 = 8[B]_0$. The half-lives of $A$ and $B$ are $10 \ min$ and $40 \ min$,respectively. If they start to disappear at the same time,both following first-order kinetics,after how much time will the concentration of both the reactants be the same (in $min$)?

For a first order reaction,to obtain a positive slope,we need to plot $...$ where $[A]$ is the concentration of reactant $A$.

The rate constant for a first order reaction is $60 \text{ s}^{-1}$. How much time (in seconds) will it take to reduce the initial concentration of the reactant to its $1/16^{th}$ value?

In a first order reaction,the concentration of the reactant decreases from $20 \ mmol \ dm^{-3}$ to $8 \ mmol \ dm^{-3}$ in $40 \ minute$. Find the rate constant of the reaction.