If the rate constant is $1.155 \times 10^{-3} \ s^{-1}$,after how many seconds will the concentration of the reactant be reduced to half in a first-order reaction?

For the gaseous reaction $A(g) \longrightarrow B(g) + C(g) + D(g)$,starting with an initial pressure of $400 \ atm$. If the total pressure after $2 \ hr$ is $800 \ atm$,the value of the rate constant is: (consider the reaction to be $1^{st}$ order): (in $hr^{-1}$)

For a first order reaction $A \rightarrow \text{Products}$,initial concentration of $A$ is $0.1 \, M$,which becomes $0.001 \, M$ after $5 \, \min$. Rate constant for the reaction in $\min^{-1}$ is .... .

Show that in a first order reaction,the time required for completion of $99.9 \%$ is $10$ times the half-life $(t_{1/2})$ of the reaction.

For a first-order reaction at $100\,^oC$,the rate constant is $K = 1.5 \times 10^{-2} \, s^{-1}$. If the initial concentration of the reactant is $100 \, mol \, L^{-1}$,what will be the concentration of the reactant after $10 \, min$?

The half-life period of a first-order reaction is $1386 \text{ s}$. The rate constant of the reaction is: