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: $(\ln 2 = 0.693, \ln 3 = 1.1)$.

The rate constant for a first order reaction is $60 \text{ s}^{-1}$. How much time will it take to reduce the concentration of the reactant to $1/20^{th}$ of its initial value (in $\text{ s}$)?

Consider the following two first order reactions occurring at $298 \ K$ with same initial concentration of $A$: $(1)$ $A \rightarrow B$; rate constant, $k=0.693 \ min^{-1}$ $(2)$ $A \rightarrow C$; half-life, $t_{1/2}=0.693 \ min$. Choose the correct option.

The integrated rate equation is $Rt = \log C_0 - \log C_t$. The straight line graph is obtained by plotting

For a first-order reaction,the concentration decreases from $800 \ mol/dm^3$ to $50 \ mol/dm^3$ in $2 \times 10^4 \ s$. What is the rate constant of the reaction?

Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of $3.33 \ h$ at $25^{\circ} C$. After $9 \ h$,the fraction of sucrose remaining is $f$. The value of $\log _{10} (\frac{1}{f})$ is ..... $\times 10^{-2}$. (Rounded off to the nearest integer) [Assume : $\ln 10 = 2.303, \ln 2 = 0.693$ ]