Assertion : For a first order reaction,$t_{1/2}$ is independent of initial concentration.
Reason : For a first order reaction,rate constant $k \propto [R]$.

Consider the following reaction,the rate expression of which is given below:
$A + B \rightarrow C$
$\text{rate} = k[A]^{1/2}[B]^{1/2}$
The reaction is initiated by taking $1 \ M$ concentration of $A$ and $B$ each. If the rate constant $(k)$ is $4.6 \times 10^{-2} \ s^{-1}$,then the time taken for $A$ to become $0.1 \ M$ is . . . . . . . . . . $sec$. (nearest integer)

Following data was obtained for the first order decomposition of $SO_2Cl_{2(g)}$ at constant volume:
$SO_2Cl_{2(g)} \to SO_{2(g)} + Cl_{2(g)}$

$S. No.$	$Time$ $(s)$	$Total$ $pressure$ $(atm)$
$1$	$0$	$0.5$
$2$	$100$	$0.6$

Calculate the rate constant.

$A$ first order reaction has a rate constant $0.00813 \ min^{-1}$. How long will it take for $60 \%$ completion (in $min$)?

The following data were obtained during the first-order decomposition of $2 A_{(g)} \rightarrow B_{(g)} + C_{(s)}$ at a constant volume and at a particular temperature. The rate constant in $min^{-1}$ is:

$S$.no.	Time	Total pressure in Pascal
$1.$	At the end of $10 \ min$	$300$
$2.$	After completion	$200$

$N_{2}O_{5(g)} \rightarrow 2NO_{2(g)} + \frac{1}{2}O_{2(g)}$
In the above first order reaction,the initial concentration of $N_{2}O_{5}$ is $2.40 \times 10^{-2} \ mol \ L^{-1}$ at $318 \ K$. The concentration of $N_{2}O_{5}$ after $1 \ hour$ was $1.60 \times 10^{-2} \ mol \ L^{-1}$. The rate constant of the reaction at $318 \ K$ is $..... \times 10^{-3} \ min^{-1}$. (Nearest integer)
[Given: $\log 3 = 0.477, \log 5 = 0.699$]