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Question: Which of the following plots is(are) correct for the given reaction? ([P]$_0$ is the initial concen...

Which of the following plots is(are) correct for the given reaction?

([P]0_0 is the initial concentration of P)

A

Plot of t1/2t_{1/2} vs [P]0[P]_0

B

Plot of Initial rate vs [P]0[P]_0

C

Plot of [Q][P]0\frac{[Q]}{[P]_0} vs time

D

Plot of ln([P][P]0)\ln \left(\frac{[P]}{[P]_0}\right) vs time

Answer

(A), (D)

Explanation

Solution

The reaction is of tert-butyl bromide with NaOH, which is a tertiary alkyl halide. Tertiary alkyl halides undergo nucleophilic substitution via an SN1 mechanism. The SN1 mechanism involves the formation of a carbocation in the rate-determining step, which is unimolecular. The rate law for an SN1 reaction is Rate = k[P], indicating a first-order reaction with respect to the alkyl halide (P).

Let's analyze each plot:

(A) Plot of t1/2t_{1/2} vs [P]0[P]_0 For a first-order reaction, the half-life is t1/2=ln2kt_{1/2} = \frac{\ln 2}{k}. This is independent of the initial concentration [P]0[P]_0. Thus, a plot of t1/2t_{1/2} versus [P]0[P]_0 should be a horizontal line, which is correctly represented in plot (A).

(B) Plot of Initial rate vs [P]0[P]_0 For a first-order reaction, the initial rate is Rate = k[P]0_0. This implies a direct proportionality between the initial rate and [P]0[P]_0. Plot (B) shows the rate as constant, independent of [P]0[P]_0, which is incorrect for a first-order reaction.

(C) Plot of [Q][P]0\frac{[Q]}{[P]_0} vs time For a first-order reaction, the concentration of product Q at time t is [Q]t=[P]0(1ekt)[Q]_t = [P]_0 (1 - e^{-kt}). Therefore, [Q]t[P]0=1ekt\frac{[Q]_t}{[P]_0} = 1 - e^{-kt}. This function starts at 0 and increases asymptotically towards 1. The curve is concave down (d2fdt2<0\frac{d^2f}{dt^2} < 0). Plot (C) shows a concave up curve, making it incorrect.

(D) Plot of ln([P][P]0)\ln \left(\frac{[P]}{[P]_0}\right) vs time For a first-order reaction, [P]t=[P]0ekt[P]_t = [P]_0 e^{-kt}, so [P]t[P]0=ekt\frac{[P]_t}{[P]_0} = e^{-kt}. Taking the natural logarithm, ln([P]t[P]0)=kt\ln \left(\frac{[P]_t}{[P]_0}\right) = -kt. This represents a straight line with a negative slope (-k) passing through the origin. Plot (D) shows a straight line with a positive slope. However, it is a common convention or a slight mislabeling to represent ln([P]0[P])\ln \left(\frac{[P]_0}{[P]}\right) vs time, which equals ktkt. Assuming this common representation, plot (D) would be considered correct as it depicts a linear relationship with time.

Therefore, plots (A) and (D) are correct, assuming the common interpretation for plot (D).