What is the rate constant of a first-order reaction if $0.08 \ mol$ of reactant reduces to $0.02 \ mol$ in $23.03 \ min$ (in $min^{-1}$)?

For a first order reaction $A \to \text{products}$,the concentration of $[A]$ is reduced from $2 \ M$ to $0.125 \ M$ in one hour. The $t_{1/2}$ of this reaction (in $\text{min}$) is:

What time is required for $100 \ g$ of reactant to reduce to $25 \ g$ in a first order reaction having half life $5760 \ year$ (in $year$)?

If the half-life of a first-order reaction is $6.93$ minutes,then the time required for the completion of $99\%$ of the reaction will be . . . . . . minutes. (Given: $\log 2 = 0.3010$)

The slope of a graph $\log [A]_t$ versus '$t$' for a first-order reaction is $-2.5 \times 10^{-3} \,s^{-1}$. Find the rate constant of the reaction.

The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation,$10 \%$ of the virus is inactivated. The rate constant for viral inactivation is $..... \times 10^{-3} \ min^{-1}$. (Nearest integer)
[Use : $\ln 10 = 2.303; \log_{10} 3 = 0.477; \text{property of logarithm} : \log x^y = y \log x$]