For a first-order reaction, $20 \%$ of the initial concentration remains after $10 \text{ min}$. What is the rate constant of the reaction (in $\text{ min}^{-1}$)? (Given: $\log_{10}(5) = 0.6989$)

At $298 \ K$,for a first order reaction $(A \rightarrow P)$ the following graph is obtained. The rate constant (in $s^{-1}$) and initial concentration (in $mol \ L^{-1}$) of '$A$' are respectively ($y$-axis $= \ln(a-x)$; $x$-axis $=$ time in sec).

If the rate constant of a reaction is $0.03 \ s^{-1}$,how much time does it take for $7.2 \ mol \ L^{-1}$ concentration of the reactant to get reduced to $0.9 \ mol \ L^{-1}$ (in $s$)? (Given: $\log 2 = 0.301$)

For a first order reaction,the concentration of reactant was reduced from $0.03 \ mol \ L^{-1}$ to $0.02 \ mol \ L^{-1}$ in $25 \ min$. What is its rate (in $mol \ L^{-1} \ s^{-1}$)?

For a first order reaction,$A \to P$,$t_{1/2}$ (half-life) is $10 \ days$. The time required for $\frac{1}{4}$ conversion of $A$ (in days) is: $(\ln 2 = 0.693, \ln 3 = 1.1)$.

Calculate the half-life period of a first-order reaction if the rate constant of the reaction is $0.02 \ min^{-1}$. (in $min$)