JEE Advanced 2020 — Paper 2 — Question 24

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.

$\mathrm{P}=\mathrm{P}_{0} \times \mathrm{e}^{-\mathrm{kt}}$ or $\frac{\mathrm{P}}{\mathrm{P}_{0}}=\mathrm{e}^{-\mathrm{kt}}$

$\ln \left(\frac{\mathrm{P}}{\mathrm{P}_{0}}\right)=-\mathrm{kt}$

$y=m x$

So, the slope of the graph $\ln \left(\frac{\mathrm{P}}{\mathrm{P}_{0}}\right)$ versus time (t) should be negative.

Hence option D is incorrect.

Also $\frac{\mathrm{P}}{\mathrm{P}_{0}}=\mathrm{e}^{-\mathrm{kt}}$

$$\begin{aligned} & \left(1-\frac{\mathrm{P}}{\mathrm{P}_{0}}\right)=\left(1-\mathrm{e}^{-\mathrm{kt}}\right) \\& \frac{\mathrm{P}_{0}-\mathrm{P}}{\mathrm{P}_{0}}=\left(1-\mathrm{e}^{-\mathrm{kt}}\right) \end{aligned}$$

Or $\frac{[\mathrm{Q}]}{\mathrm{P}_{0}}=\left(1-\mathrm{e}^{-\mathrm{kt}}\right)$

So, graph between $\frac{[\mathrm{Q}]}{\mathrm{P}_{0}}$ Vs. t, should be as shown in figure (1), so, option (C)is incorrect

For a first order reaction

$\mathrm{r}=\mathrm{k}[\mathrm{P}]^{1}$

At $\mathrm{t}=0, \quad \mathrm{r}_{\mathrm{i}}=\mathrm{k}\left[\mathrm{P}_{0}\right]$

So, initial rate of reaction is directly proportional to initial conc. of P .

Hence option B is also incorrect.

For a first order reaction

$\mathrm{t}_{1 / 2}=\frac{\ln 2}{\mathrm{k}}\left[\mathrm{P}_{0}\right]^{0}$

So, $t_{1 / 2}$ is independent of initial conc. of reactant.

So, option (A) is correct.