$A$ first order reaction takes $23.03 \ minutes$ for $20 \%$ decomposition. Calculate its rate constant.

If the half-life $(t_{1/2})$ for a first order reaction is $1 \text{ minute}$,then the time required for $99.9\%$ completion of the reaction is closest to

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 $t = 0$)

For a first order reaction $A_{(g)} \rightarrow 2B_{(g)} + C_{(g)}$ at constant volume and $300 \ K$,the total pressure at the beginning $(t=0)$ and at time $t$ are $P_0$ and $P_t$,respectively. Initially,only $A$ is present with concentration $[A]_0$,and $t_{1/3}$ is the time required for the partial pressure of $A$ to reach $1/3^{rd}$ of its initial value. The correct option$(s)$ is (are) (Assume that all these gases behave as ideal gases)

$A$ first order reaction has the rate constant,$k = 4.6 \times 10^{-3} \ s^{-1}$. The number of correct statement/s from the following is/are
Given : $\log 3 = 0.48$
$A.$ Reaction completes in $1000 \ s$.
$B.$ The reaction has a half-life of $500 \ s$.
$C.$ The time required for $10 \ \%$ completion is $25$ times the time required for $90 \ \%$ completion.
$D.$ The degree of dissociation is equal to $(1 - e^{-kt})$.
$E.$ The rate and the rate constant have the same unit.

If the concentration unit decreases by a factor of $n$,what will be the value of the rate constant for a first-order reaction?