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 $hr^{-1}$)

Half-life period of a first-order reaction is $10 \ min$. Starting with an initial concentration of $12 \ M$,the rate after $20 \ min$ is:

The rate constant for the oxidation of hydrogen peroxide by $KMnO_4$ is $6.93 \times 10^{-5} \ s^{-1}$. How much time will it take for the volume of a standard $KMnO_4$ solution to decrease from $20 \ mL$ to $8 \ mL$?

Consider the two different first order reactions given below:
$A + B \rightarrow C$ (Reaction $1$)
$P \rightarrow Q$ (Reaction $2$)
The ratio of the half-life of Reaction $1$ : Reaction $2$ is $5 : 2$. If $t_1$ and $t_2$ represent the time taken to complete $2/3$ and $4/5$ of Reaction $1$ and Reaction $2$,respectively,then the value of the ratio $t_1 : t_2$ is $. . . . \times 10^{-1}$ (nearest integer).
[Given: $\log_{10}(3) = 0.477$ and $\log_{10}(5) = 0.699$]

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: $\log_{10}(5) = 0.6989$)

$A \rightarrow P$ is a first order reaction. At $300 \ K$,this reaction was started with $[A] = 0.5 \ mol \ L^{-1}$. The rate constant of the reaction was $0.125 \ min^{-1}$. The same reaction was started separately with $[A] = 1 \ mol \ L^{-1}$ at $300 \ K$. The rate constant (in $min^{-1}$) now is: