Question

A secondary alkyl halide (A) hydrolyzed with alkali(B) in aqueous medium simultaneously vis SN1 and SN2 pathways with rate constants k1 and k2 respectively. From kinetic data, it was found that a plot of \[{\text{ = }}\dfrac{{ - 1d\left[ A \right]}}{{\left[ A \right]\,dt}}\] vs \[\left[ B \right]\]isa straight line with a slope equal to \[2.7 \times {10^{ - 4}}\,{\text{L}}\,{\text{mo}}{{\text{l}}^{ - 1}}\,{\text{mi}}{{\text{n}}^{ - 1}}\] and intercept equal to \[1.02 \times {10^{ - 3}}\]. Minimum initial concentration of \[\left[ A \right] = 0.2\,{\text{M}}\] and \[\left[ B \right]\], i.e., \[\left[ {O{H^ - }} \right] = 0.5\,{\text{M}}\]. The value of the overall rate constant of the hydrolysis of A(in \[{\text{L}}\,{\text{mo}}{{\text{l}}^{ - 1}}\,{\text{mi}}{{\text{n}}^{ - 1}}\]) is:
A.\[2.7 \times {10^{ - 4}}\]
B.\[1.02 \times {10^{ - 3}}\]
C.\[1.29 \times {10^{ - 3}}\]
D.None of the above

Answer 1

Hint:The rate of reaction is determining the speed of the reaction. The branch of physical chemistry which studies the rate of the reaction is known as Chemical kinetics.
The rate of the reaction is given in the reactants as well as the products. The change in concentration of species either reactant or product with time gives the rate of the reaction.
The unit of the rate is nothing but the concentration per unit time.

Complete step-by-step answer:The hydrolysis reaction of the alkyl halide with alkali is as follows:
\[{\text{RX}}\, + \,{\text{O}}{{\text{H}}^ - } \to \,{\text{Products}}\]
The rate of reaction as per the SN1 mechanism depends on alkyl halide only.
\[{\text{rate}}\,{\text{ = }}\dfrac{{ - \left[ {{\text{dA}}} \right]}}{{{\text{dt}}}}{\text{ = }}{{\text{k}}_{\text{1}}}\left[ {{\text{RX}}} \right]{\text{ = }}{{\text{k}}_{\text{1}}}\left[ {\text{A}} \right]\] (i) The rate of reaction as per the SN2 mechanism depends on both alkyl halide and alkali concentration.
\[{\text{rate}}\,{\text{ = }}\dfrac{{ - \left[ {{\text{dA}}} \right]}}{{{\text{dt}}}}{\text{ = }}{{\text{k}}_2}\left[ {{\text{RX}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]{\text{ = }}{{\text{k}}_2}\left[ {\text{A}} \right]\left[ {\text{B}} \right]\] (ii)
The combining equation (i) and (ii).
\[{\text{rate}}\,{\text{ = }}\dfrac{{ - \left[ {{\text{dA}}} \right]}}{{{\text{dt}}}}{\text{ = }}{{\text{k}}_{\text{1}}}\left[ {{\text{RX}}} \right]{\text{ + }}{{\text{k}}_2}\left[ {{\text{RX}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]\]
\[{\text{rate}}\,{\text{ = }}\dfrac{{ - \left[ {{\text{dA}}} \right]}}{{{\text{dt}}}}{\text{ = }}{{\text{k}}_{\text{1}}}\left[ {\text{A}} \right]{\text{ + }}{{\text{k}}_2}\left[ {\text{A}} \right]\left[ {\text{B}} \right]\]
\[\dfrac{{ - \left[ {{\text{dA}}} \right]}}{{{\text{dt}}}}{\text{ = }}\left[ {\text{A}} \right]\left( {{{\text{k}}_{\text{1}}}{\text{ + }}{{\text{k}}_2}\left[ {\text{B}} \right]} \right)\]
\[\dfrac{1}{{\left[ {\text{A}} \right]}}\left( {\dfrac{{ - \left[ {{\text{dA}}} \right]}}{{{\text{dt}}}}} \right){\text{ = }}\left( {{{\text{k}}_{\text{1}}}{\text{ + }}{{\text{k}}_2}\left[ {\text{B}} \right]} \right)\]……(iii)
The equation of the straight line is
\[{\text{y = mx + c}}\]……(iv)
Here, m is the slope and c is the intercept.
Thus, equation (iii) is in the form of the equation of the straight line.
If the graph is plotted between the \[{\text{ = }}\dfrac{{ - 1d\left[ A \right]}}{{\left[ A \right]\,dt}}\] vs \[\left[ B \right]\] slope obtained is \[{{\text{k}}_2}\] and intercept is \[{{\text{k}}_{\text{1}}}\].
Therefore, \[{{\text{k}}_2}\] is\[2.7 \times {10^{ - 4}}\,{\text{L}}\,{\text{mo}}{{\text{l}}^{ - 1}}\,{\text{mi}}{{\text{n}}^{ - 1}}\], and \[{{\text{k}}_{\text{1}}}\] is \[1.02 \times {10^{ - 3}}\].
The overall rate constant is obtained by adding two rate constants.
\[{\text{overall rate constant}} = {{\text{k}}_1} + {{\text{k}}_2}\]
Substitute \[2.7 \times {10^{ - 4}}\,{\text{L}}\,{\text{mo}}{{\text{l}}^{ - 1}}\,{\text{mi}}{{\text{n}}^{ - 1}}\] for \[{{\text{k}}_2}\], and \[1.02 \times {10^{ - 3}}\] for \[{{\text{k}}_{\text{1}}}\].
\[{\text{overall rate constant}} = \left( {2.7 \times {{10}^{ - 4}}\, + 1.02 \times {{10}^{ - 3}}} \right){\text{L}}\,{\text{mo}}{{\text{l}}^{ - 1}}\,{\text{mi}}{{\text{n}}^{ - 1}}\]
\[{\text{overall rate constant}} = 1.29 \times {10^{ - 3}}\,{\text{L}}\,{\text{mo}}{{\text{l}}^{ - 1}}\,{\text{mi}}{{\text{n}}^{ - 1}}\]
Thus, the overall rate constant of the reaction is \[1.29 \times {10^{ - 3}}{\text{L}}\,{\text{mo}}{{\text{l}}^{ - 1}}\,{\text{mi}}{{\text{n}}^{ - 1}}\].
Here, option(A)\[2.7 \times {10^{ - 4}}\]is incorrect.
Now, option(B)\[1.02 \times {10^{ - 3}}\] is also incorrect.
Option(C) \[1.29 \times {10^{ - 3}}\] is the correct answer to the question.

Here, option(D) None of the above is incorrect.

Note:The reactions in which one functional group of the one reactant is replaced by another functional group is known as substitution reaction.
There are two types of substitution reactions, that is electrophilic and nucleophilic substitution reactions.
In the case of the electrophilic substitution reactions, electrophile substitutes the functional group while in the case of nucleophilic substitution nucleophiles substitute the functional group of the molecule.
In the case of the nucleophilic substitution reactions, there are two types SN1 and SN2. SN1is the unimolecular nucleophilic substitution reaction while SN2is the bimolecular nucleophilic substitution reaction.