For the reaction $2A + B \to \text{product}$,the rate law is given by $\frac{-d[A]}{dt} = k[A]$. At a time $t = \frac{t_{1/2}}{\ln 2}$,what is the concentration of the reactant $A$?

$A$ first order reaction has a rate constant $1.1 \times 10^{-3} \ s^{-1}$. How long will $6 \ g$ of this reactant take to reduce to $3 \ g$ (in $s$)?

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 $\log_{10}(5) = 0.699$]

$A$ molecule undergoes two independent first order reactions whose respective half-lives are $12 \ min$ and $3 \ min$. If both the reactions are occurring simultaneously,the time taken for the $50 \%$ consumption of the reactant is $.......... \ min$ (Nearest integer).

For a reaction $A \rightarrow$ product,the rate constant is $2 \times 10^{-2} \ s^{-1}$. The initial concentration of $A$ is $1.0 \ mol \ dm^{-3}$. What is the value of $\log \frac{1}{[A]_{t}}$ after $100 \ s$?

If the rate law is $r = K[A]$,then the concentration of the reactant remaining after time $t = 1/k$ is: ($[A]_0$ is the concentration of the reactant at $t = 0$)