The order of the above reaction is one $\left( S _{ N } 1\right)$ because substrate is a $3^{0}$-halide
So, for a first order reaction.
$P=P_{0} \times e^{-k t}$ or $\frac{P}{P_{0}}=e^{-k t}$
$\ln \left(\frac{P}{P_{0}}\right)=-k t$
$y = mx$
So, the slope of the graph $\ln \left(\frac{ P }{ P _{0}}\right)$ versus time ( $t$ ) should be negative.
Hence option D is incorrect.
Also $\frac{ P }{ P _{0}}= e ^{- kt } $
$\left(1-\frac{ P }{ P _{0}}\right)=\left(1- e ^{- kt }\right)$
$\frac{ P _{0}- P }{ P _{0}}=\left(1- e ^{- kt }\right)$
Or $ \frac{[Q]}{P_{0}}=\left(1-e^{-k t}\right)$
So, graph between $\frac{[ Q ]}{ P _{0}}$ Vs. $t$, should be as shown in figure (1), so, option (C)is incorrect
For a first order reaction
$ r = k [ P ]^{1} $
At $ t =0, r _{ i }= k \left[ P _{0}\right]$
So, initial rate of reaction is directly proportional to initial conc. of $P$.
Hence option B is also incorrect.
For a first order reaction
$t _{1 / 2}=\frac{\ln 2}{ k }\left[ P _{0}\right]^{0}$
So, $t _{1 / 2}$ is independent of initial conc. of reactant.
So, option (A) is correct.
