Question

The reaction of 1-bromopropane and sodium hydroxide in ethanol occurs by an \(\mathrm{S}_{\mathrm{N}} 2\) mechanism. What happens to the rate of this reaction under the following conditions? (a) The concentration of \(\mathrm{NaOH}\) is doubled. (b) The concentrations of both \(\mathrm{NaOH}\) and 1-bromopropane are doubled. (c) The volume of the solution in which the reaction is carried out is doubled.

Short answer

Answer: (a) The reaction rate will double, (b) the reaction rate will quadruple, and (c) the reaction rate will be reduced to one-fourth of the original rate.

Step-by-step solution

Rate Law for \(\mathrm{S}_{\mathrm{N}} 2\) Reactions

For an \(\mathrm{S}_{\mathrm{N}} 2\) reaction, the overall rate depends on the concentrations of both the nucleophile (sodium hydroxide, \(\mathrm{NaOH}\)) and the substrate (1-bromopropane). The rate law for this type of reaction can be written as: Rate \(= k[\mathrm{NaOH}][1\text{-bromopropane}]\) Where k is the rate constant, and the concentrations of NaOH and 1-bromopropane are raised to the power of 1 since it's a bimolecular reaction.

Scenario (a): Concentration of \(\mathrm{NaOH}\) is doubled

If the concentration of \(\mathrm{NaOH}\) is doubled, then the rate equation becomes: New Rate \(= k[2 \times \mathrm{NaOH}][1\text{-bromopropane}]\) Since the reaction is first-order with respect to \(\mathrm{NaOH}\), the new rate would be two times the original rate: New Rate \(= 2 \times k[\mathrm{NaOH}][1\text{-bromopropane}]\) So, the reaction rate will double in this scenario.

Scenario (b): Concentrations of both \(\mathrm{NaOH}\) and 1-bromopropane are doubled

In this case, both the concentration of \(\mathrm{NaOH}\) and 1-bromopropane are doubled, so the rate equation becomes: New Rate \(= k[2 \times \mathrm{NaOH}][2 \times 1\text{-bromopropane}]\) Now, the rate would be four times the original rate due to both the concentration of \(\mathrm{NaOH}\) and 1-bromopropane being doubled: New Rate \(= 4 \times k[\mathrm{NaOH}][1\text{-bromopropane}]\) So, the reaction rate will quadruple in this scenario.

Scenario (c): The volume of the solution is doubled

Doubling the volume of the solution effectively means the concentrations of all reactants will be reduced by half. Since the reaction rate is dependent on the product of the concentrations, we can write: New Rate \(= k \left[\frac{1}{2} \times \mathrm{NaOH}\right] \left[\frac{1}{2} \times 1\text{-bromopropane}\right]\) Multiplying the fractions, we get: New Rate \(= \frac{1}{4} \times k[\mathrm{NaOH}][1\text{-bromopropane}]\) So, the reaction rate will be reduced to one-fourth of the original rate in this scenario.

Key concepts

Chemical Kinetics and Reaction Rates

Chemical kinetics is the study of the speed or rate at which chemical reactions occur. It's a vital aspect of chemistry because it helps us understand how changes in conditions can affect the progress of a reaction. In an educational context, understanding kinetics allows students to predict and control the speed of chemical reactions, which is crucial in experiments and industrial processes.

To quantify the rate of a chemical reaction, we use the rate law. For instance, in an SN2 reaction—a type of bimolecular nucleophilic substitution—the rate law shows a direct relationship between the concentration of the reactants and the speed of the reaction. The rate law for SN2 reactions can be generally represented as:
Rate = k[A][B]

Where 'k' is the rate constant, and [A] and [B] represent the molar concentrations of the reactants. In the SN2 mechanism, the reaction is second-order overall: first-order with respect to each reactant. This means that the rate of reaction is directly proportional to the concentration of each reactant. Therefore, if you change the concentration of one or both reactants, you significantly affect the reaction rate.

Bimolecular Reactions

A bimolecular reaction involves two reactant molecules coming together to form products. The SN2 reaction is a classic example of this type of reaction, in which two species are involved in the rate-determining step. One reactant is typically a nucleophile, like sodium hydroxide (NaOH), and the other is an electrophile, such as 1-bromopropane.

In terms of kinetics, bimolecular reactions such as SN2 have a unique rate law, where the rate is proportional to the product of the concentrations of the two reactants. The rate law reflects the mechanism of these reactions, which requires a collision between two molecules. It's important to note that not only must these molecules collide, but they must also do so with sufficient energy and in the correct orientation to react effectively. This information is invaluable when considering how changes in concentration, volume, or other factors might impact the rate of the reaction.

Concentration Effect on Reaction Rate

In chemical reactions, the concentration of reactants is a crucial factor that influences the rate at which a reaction proceeds. For a bimolecular reaction such as the SN2 mechanism, the rate is directly dependent on the concentration of both reactants. When you increase the concentration of one reactant, like in the case of doubling the NaOH, the reaction rate will also double, because more reactant molecules are available to collide and react.

If the concentrations of both reactants are simultaneously increased, the effect on the rate is more dramatic. Doubling both the NaOH and 1-bromopropane results in the reaction rate quadrupling, because the likelihood of effective collisions rises significantly. Conversely, if we decrease the concentration of reactants, by doubling the volume of the reaction for instance, we observe a significant decrease in the reaction rate. Understanding these relationships is essential for controlling the speed of chemical reactions and is a fundamental concept in chemical kinetics. Thus, it’s critical for students to grasp how changes in concentration can manipulate reaction rates in a predictable manner.