Problem 67 - Chemistry The Central Science In Si Units

Question

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CN}^{-}(a q) \longrightarrow \mathrm{HCN}(a q)\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}(\mathrm{solv})+\mathrm{OH}^{-}(\mathrm{solv}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{solv})+\mathrm{Cl}^{-}(\mathrm{solv})\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightarrow 2 \mathrm{NO}_{2}\)

Step-by-Step Solution

Answer

(a) Bimolecular, Rate = k[H2O][CN-]; (b) Bimolecular, Rate = k[CH3Cl][OH-]; (c) Unimolecular, Rate = k[N2O4].

Step 1: Identify the Reactants in Reaction (a)

For the reaction \( \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CN}^{-}(a q) \longrightarrow \mathrm{HCN}(a q) \), the reactants are \( \mathrm{H}_{2} \mathrm{O} \) and \( \mathrm{CN}^{-} \).

Step 2: Determine Molecularity of Reaction (a)

The molecularity is the number of reactant molecules involved in an elementary reaction. Reaction (a) involves two reactants: \( \mathrm{H}_{2} \mathrm{O} \) and \( \mathrm{CN}^{-} \). Therefore, the molecularity is bimolecular.

Step 3: Write Rate Law for Reaction (a)

For a bimolecular reaction, the rate law is determined by the concentration of each reactant. Thus, the rate law for this reaction is: \[ \text{Rate} = k[\mathrm{H}_{2} \mathrm{O}][\mathrm{CN}^{-}] \] where \( k \) is the rate constant.

Step 4: Identify the Reactants in Reaction (b)

For the reaction \( \mathrm{CH}_{3} \mathrm{Cl} (\text{solv}) + \mathrm{OH}^{-} (\text{solv}) \rightarrow \mathrm{CH}_{3} \mathrm{OH} (\text{solv}) + \mathrm{Cl}^{-} (\text{solv}) \), the reactants are \( \mathrm{CH}_{3} \mathrm{Cl} \) and \( \mathrm{OH}^{-} \).

Step 5: Determine Molecularity of Reaction (b)

This reaction also involves two reactant species: \( \mathrm{CH}_{3} \mathrm{Cl} \) and \( \mathrm{OH}^{-} \). Hence, it is a bimolecular reaction.

Step 6: Write Rate Law for Reaction (b)

As a bimolecular reaction, the rate law is based on the concentration of each reactant. The rate law for this reaction is: \[ \text{Rate} = k[\mathrm{CH}_{3} \mathrm{Cl}][\mathrm{OH}^{-}] \] with \( k \) being the rate constant.

Step 7: Identify the Reactants in Reaction (c)

For the reaction \( \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightarrow 2 \mathrm{NO}_{2} \), the reactant is \( \mathrm{N}_{2} \mathrm{O}_{4} \).

Step 8: Determine Molecularity of Reaction (c)

This reaction involves only one reactant molecule, \( \mathrm{N}_{2} \mathrm{O}_{4} \). Therefore, it is a unimolecular reaction.

Step 9: Write Rate Law for Reaction (c)

For a unimolecular reaction, the rate law depends only on the concentration of the single reactant. Thus, the rate law for this reaction is: \[ \text{Rate} = k[\mathrm{N}_{2} \mathrm{O}_{4}] \] where \( k \) represents the rate constant.

Key Concepts

Elementary Reactions

Elementary reactions are the simplest types of reactions in chemistry, involving the direct transformation of reactants into products in a single step. Unlike complex reactions, elementary reactions do not have intermediate steps. This means they reveal the true dynamics of how molecules interact at a molecular level.
An easy way to identify an elementary reaction is to see if the reaction can be described by a simple equation without any multiple sequential steps.

• Each step indicates a single collision or transformation.
• The stoichiometry directly gives insight into the reaction's molecularity.

Understanding elementary reactions is crucial as they form the building blocks for more complex mechanisms in chemical kinetics. This simplicity helps in determining both the molecularity and rate law of the reaction effectively.

Rate Law

Rate laws express the relationship between the concentration of reactants and the speed or rate of the reaction. For elementary reactions, the rate law can be directly inferred from the stoichiometry of the reaction, which is not typically the case with complex reactions.
Consider the reaction \[ ext{Rate} = k[ ext{A}][ ext{B}] \]Here, \( k \) is the rate constant, and \( [ ext{A}] \) and \( [ ext{B}] \) are the concentrations of reactants A and B respectively.

• The exponents show the order with respect to each reactant.
• For elementary reactions, these orders are usually the same as the stoichiometric coefficients.

Recognizing this connection is essential for understanding how changes in concentration could impact the speed of the reaction.

Bimolecular Reaction

Bimolecular reactions involve two reactant molecules colliding to form products. These reactions can be between two identical molecules or between two different molecules.
For instance, consider the reactions:- \( \ \ \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{CN}^{-}(aq) \rightarrow \mathrm{HCN}(aq) \)- \( \ \ \mathrm{CH}_{3} \mathrm{Cl}(solv) + \mathrm{OH}^{-}(solv) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(solv) + \mathrm{Cl}^{-}(solv) \)The characteristic feature of bimolecular reactions is their reliance on the concentration of both reactant molecules involved. The rate law for such reactions usually follows the form: \[ ext{Rate} = k[ ext{Reactant1}][ ext{Reactant2}] \]Bimolecular reactions are foundational in chemical kinetics, highlighting how the concentration of reactants can affect reaction velocities. Understanding these reactions help predict how the systems will behave under different concentration scenarios.

Unimolecular Reaction

Unimolecular reactions involve a single reactant molecule transforming into product(s) either spontaneously or through simple reorganization. A classic example from the exercise is: \( \ \ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightarrow 2 \mathrm{NO}_{2}(g) \)In this reaction, only one molecule of \( \mathrm{N}_{2} \mathrm{O}_{4} \) is involved to form two molecules of \( \mathrm{NO}_{2} \), making it unimolecular. The simplification of one molecule typically undergoing change results in a rate law expressed as: \[ ext{Rate} = k[ ext{Reactant}] \]Such reactions emphasize the role of molecular instability or specific environmental conditions that trigger the transformation. Studying unimolecular reactions provides insight into reaction rates that do not require intermolecular collisions, thus expanding our understanding of kinetic phenomena beyond simple collisions.