Correct Reaction Plots for Nucleophilic Substitution Kinetics

The given chemical reaction is the hydrolysis of a tertiary alkyl halide (tert-butyl bromide):

$\text{(CH}_3\text{)}_3\text{C-Br} + \text{NaOH} \longrightarrow \text{(CH}_3\text{)}_3\text{C-OH} + \text{NaBr}$

Here, reactant $\mathbf{P}$ is tert-butyl bromide, and product $\mathbf{Q}$ is tert-butyl alcohol.

1. Kinetic Analysis of the Reaction

Tertiary alkyl halides undergo nucleophilic substitution predominantly via the $\text{S}_\text{N}1$ mechanism (unimolecular nucleophilic substitution) due to the formation of a stable tertiary carbocation intermediate $\text{(CH}_3\text{)}_3\text{C}^+$.

The rate-determining step is the unimolecular ionization of $\mathbf{P}$: $\text{(CH}_3\text{)}_3\text{C-Br} \xrightarrow{\text{slow}} \text{(CH}_3\text{)}_3\text{C}^+ + \text{Br}^-$

Thus, the reaction follows first-order kinetics with respect to $\mathbf{P}$: $\text{Rate} = -\frac{d[\mathbf{P}]}{dt} = k [\mathbf{P}]$ where $k$ is the first-order rate constant.

The integrated rate equation for a first-order reaction is: $[\mathbf{P}] = [\mathbf{P}]_0 e^{-kt}$

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2. Evaluation of the Options

• Option (A): Plot of $t_{1/2}$ vs $[\mathbf{P}]_0$ The half-life ($t_{1/2}$) of a first-order reaction is calculated as: $[\mathbf{P}] = \frac{[\mathbf{P}]_0}{2} \implies \frac{[\mathbf{P}]_0}{2} = [\mathbf{P}]_0 e^{-k t_{1/2}}$ $t_{1/2} = \frac{\ln 2}{k}$ Since $t_{1/2}$ is independent of the initial concentration $[\mathbf{P}]_0$, the plot of $t_{1/2}$ versus $[\mathbf{P}]_0$ is a horizontal straight line parallel to the $[\mathbf{P}]_0$ axis. Hence, Option (A) is CORRECT.

• Option (B): Plot of Initial rate vs $[\mathbf{P}]_0$ The initial rate of a first-order reaction is given by: $\text{Initial rate} = k [\mathbf{P}]_0$ This indicates a linear relationship passing through the origin with a positive slope $k$. Plot (B) depicts a horizontal line (indicative of zero-order kinetics), which is incorrect for an $\text{S}_\text{N}1$ reaction. Hence, Option (B) is INCORRECT.

• Option (C): Plot of $\frac{[\mathbf{Q}]}{[\mathbf{P}]_0}$ vs time By stoichiometry, the concentration of product $\mathbf{Q}$ formed at time $t$ is: $[\mathbf{Q}] = [\mathbf{P}]_0 - [\mathbf{P}] = [\mathbf{P}]_0 \left(1 - e^{-kt}\right)$ Rearranging gives: $\frac{[\mathbf{Q}]}{[\mathbf{P}]_0} = 1 - e^{-kt}$ Taking the first and second derivatives with respect to time $t$: $\frac{d}{dt}\left(\frac{[\mathbf{Q}]}{[\mathbf{P}]_0}\right) = k e^{-kt} > 0$ $\frac{d^2}{dt^2}\left(\frac{[\mathbf{Q}]}{[\mathbf{P}]_0}\right) = -k^2 e^{-kt} < 0$ Since the second derivative is negative for all $t > 0$, the graph must be concave downward (slope decreases over time). Plot (C) depicts a concave upward curve. Hence, Option (C) is INCORRECT.

• Option (D): Plot of $\ln\left(\frac{[\mathbf{P}]}{[\mathbf{P}]_0}\right)$ vs time Taking the natural logarithm of $[\mathbf{P}] = [\mathbf{P}]_0 e^{-kt}$: $\ln\left(\frac{[\mathbf{P}]}{[\mathbf{P}]_0}\right) = -kt$ This represents a straight line passing through the origin with a negative slope ($-k$). Plot (D) shows a straight line with a positive slope. Hence, Option (D) is INCORRECT.

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Conclusion

The only correct graph is given in option (A).