Problem 68 - Chemistry: The Central Science

Question

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\) (b) \(\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(g) \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3}(g)\) (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)\)

Step-by-Step Solution

Answer

(a) The molecularity of the reaction is 2, and the rate law is \(Rate = k[\mathrm{NO}]^{2}\). (b) The molecularity of the reaction is 1, and the rate law is \(Rate = k[\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}]\). (c) The molecularity of the reaction is 1, and the rate law is \(Rate = k[\mathrm{SO}_{3}]\).

Step 1: Find the Molecularity

The molecularity is the number of reactant molecules involved in the reaction. Here, there are two molecules of NO involved in the reaction, so the molecularity is 2.

Step 2: Write the Rate Law

The rate law for this reaction is: \[Rate = k[\mathrm{NO}]^{2}\] (b) $\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(g) \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3}(g)$

Step 3: Find the Molecularity

In this case, there is only one reactant molecule involved in the reaction, the \(\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}(g)\). Therefore, the molecularity is 1.

Step 4: Write the Rate Law

The rate law for this reaction is: \[Rate = k[\mathrm{H}_{2} \mathrm{C}-\mathrm{CH}_{2}]\] (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)\)

Step 5: Find the Molecularity

In this case, there is only one reactant molecule involved in the reaction, the \(\mathrm{SO}_{3}(g)\). Therefore, the molecularity is 1.

Step 6: Write the Rate Law

The rate law for this reaction is: \[Rate = k[\mathrm{SO}_{3}]\]

Key Concepts

Molecularity

Molecularity refers to the number of reactant molecules that participate in an elementary reaction. It is a theoretical concept used in chemical kinetics to better understand the reaction mechanism.
Molecularity can be of three types:

• Unimolecular reactions: Involve a single reactant molecule. For instance, the decomposition of \(\mathrm{SO}_{3} \rightarrow \mathrm{SO}_{2} + \mathrm{O} \)is a unimolecular reaction with molecularity of 1.
• Bimolecular reactions: Involve two reactant molecules. For example, \(2 \mathrm{NO} \rightarrow \mathrm{N}_{2}\mathrm{O}_{2} \)is a bimolecular reaction with molecularity of 2.
• Termolecular reactions: Involve three reactant molecules, which are rare due to the high probability requirements of three molecules simultaneously colliding.

Understanding molecularity is crucial for predicting the mechanism of a chemical reaction and helps in writing the correct rate law for elementary reactions. Note, molecularity only applies directly to elementary reactions and is not used for complex reactions that occur in more than one step.

Elementary Reactions

Elementary reactions are single-step processes that describe the exact number of molecules (atoms, ions) and their arrangement that participate in a chemical reaction.
These reactions provide the simplest depiction of what happens at the molecular level during a reaction. Each elementary step has its own molecularity, distinguishing it from more complex, multi-step reactions.

• Elementary reactions are precise and describe a specific event, such as a collision of molecules leading to a transformation.
• The stoichiometric coefficients of an elementary reaction equal the exponents in the rate law for that reaction.
• They give insight into reaction mechanisms, allowing chemists to hypothesize which steps might occur as intermediate stages in a multi-step reaction pathway.

Due to their one-step nature, understanding elementary reactions is foundational for forming complex reaction mechanisms, making them essential for studying chemical kinetics.

Rate Law

The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. For elementary reactions, the rate law is directly derived from the molecularity.

• Unimolecular reactions: For reactions like \(\mathrm{H}_{2}\mathrm{C}-\mathrm{CH}_{2} \rightarrow \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{3} \)and \(\mathrm{SO}_{3} \rightarrow \mathrm{SO}_{2} + \mathrm{O} \), the rate law is expressed as:\[Rate = k [A]\]where \(A\)is the concentration of the reactant.
• Bimolecular reactions: For reactions like \(2 \mathrm{NO} \rightarrow \mathrm{N}_{2}\mathrm{O}_{2} \), the rate law becomes:\[Rate = k [\mathrm{NO}]^2\]indicating that the reaction rate is proportional to the square of the NO concentration.

In any rate law, \(k\)is the rate constant, which varies with temperature. Understanding the rate laws allows scientists to predict how the concentration of reactants affects reaction speed, aiding in the manipulation of conditions to achieve desired reaction rates in industrial or lab settings.