Question

Which of the following plots is (are) correct for the given reaction? ([P] ${}_0$ is the initial concentration of $P$ )

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

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.
$P=P_0\times e^{-kt}$ or $\frac{P}{P_0}=e^{-kt}$
$\ln \left(\frac{P}{P_0}\right)=-kt$
$y=mx$
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 $\frac{[Q]}{P_0}=\left(1-e^{-kt}\right)$
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.