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 reaction,$N_2O_5 \longrightarrow 2NO_2 + \frac{1}{2} O_2$ is of first order for $N_2O_5$ with rate constant $6.2 \times 10^{-4} \ s^{-1}$. What is the value of rate of reaction when $[N_2O_5] = 1.25 \ mol \ L^{-1}$?

$A$ flask is filled with equal moles of $A$ and $B$. The half-lives of $A$ and $B$ are $100 \, s$ and $50 \, s$ respectively and are independent of the initial concentration. The time required for the concentration of $A$ to be four times that of $B$ is $.... \, s.$
(Given : $\ln 2 = 0.693$ )

$[A] \rightarrow [B]$. If the formation of compound $[B]$ follows first-order kinetics and after $70 \ min$ the concentration of $[A]$ was found to be half of its initial concentration,then the rate constant of the reaction is $x \times 10^{-6} \ s^{-1}$. The value of $x$ is $......$ (Nearest Integer).

$A$ first order reaction with respect to the reactant $A$ has a rate constant of $6 \; sec^{-1}$. If we start with $[A] = 0.5 \; mol/L$,then in what time the concentration of $A$ becomes $0.05 \; mol/L$ ........... $sec$

An organic compound undergoes first order decomposition. The time taken for decomposition to $\left(\frac{1}{8}\right)^{\text{th}}$ and $\left(\frac{1}{10}\right)^{\text{th}}$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$? (Given: $\log 2 = 0.3$)