Question

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

Short answer

The molecularity of each reaction is as follows: (a) \(2\mathrm{NO(g)} \longrightarrow \mathrm{N}_{2}\mathrm{O}_{2}(g)\) is a bimolecular reaction with a rate law of Rate = k[\(\mathrm{NO}\)]^2 (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g) + \mathrm{O}(g)\) is a unimolecular reaction with a rate law of Rate = k[\(\mathrm{SO}_{3}\)]

Step-by-step solution

a) Determine the molecularity of the reaction

To find the molecularity of the reaction, we simply need to count the number of molecules participating in the reaction on the reactant side: \(2\mathrm{NO(g)} \longrightarrow \mathrm{N}_{2}\mathrm{O}_{2}(g)\) In this reaction, there are 2 molecules of NO. So, the molecularity of this reaction is 2 (bimolecular).

a) Write the rate law for the reaction

For a bimolecular reaction, the rate law is given by: Rate = k[reactant 1]^m * [reactant 2]^n In this case, there's only one reactant, NO. Thus, the rate law for this reaction is: Rate = k[\(\mathrm{NO}\)]^2

c) Determine the molecularity of the reaction

Again, we count the number of molecules participating in the reaction on the reactant side: \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g) + \mathrm{O}(g)\) In this reaction, there's only one molecule of SO3. So, the molecularity of this reaction is 1 (unimolecular).

c) Write the rate law for the reaction

For a unimolecular reaction, the rate law is given by: Rate = k[reactant]^m In this case, the only reactant is SO3. Thus, the rate law for this reaction is: Rate = k[\(\mathrm{SO}_{3}\)]

Key concepts

Understanding Rate Laws

Rate laws are mathematical equations that describe how the speed of chemical reactions depend on the concentration of reactants. These equations play a crucial role in predicting how a reaction progresses over time under various conditions.

For an elementary reaction, the rate law can be directly derived from the reaction's molecularity, which is the number of reactant molecules involved in the rate-determining step. The general form of a rate law states that the rate of reaction is proportional to the product of the concentrations of the reactants raised to some power:
Rate = k[Reactant 1]^{m}[Reactant 2]^{n}...
where k is the rate constant and m, n are the reaction orders with respect to each reactant. It's important to note that the values of m and n are determined experimentally and are not necessarily the stoichiometric coefficients found in the chemical equation.

In the exercise provided, for reaction a), where the formation of \( \mathrm{N}_{2}\mathrm{O}_{2}(g) \) from \( 2 \mathrm{NO}(g) \) occurs, the rate law is found to be proportional to the square of the concentration of NO.

Demystifying Bimolecular Reactions

Bimolecular reactions refer to reactions where two reactant molecules collide and react with each other. The molecularity of such a reaction is two, indicating this two-participant interaction.

A classic example of a bimolecular reaction can be seen in the equation given in the exercise for reaction a):-
\(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}\mathrm{O}_{2}(g)\)
Here, two NO molecules collide to form N2O2. Since the molecularity is two, and the reaction involves only one type of molecule as a reactant, the rate law, as mentioned in the solution, is expressed as:
Rate = k[NO]^2

This quadratic dependence on a single reactant's concentration is characteristic of bimolecular reactions that involve identical reactants.

Unraveling Unimolecular Reactions

Unimolecular reactions are reactions that involve only a single reactant molecule which makes a transition to the products without the need for a collision with another reactant molecule. Essentially, the reaction's molecularity is one.

An example of a unimolecular reaction is provided in the exercise for reaction c):-
\(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g) + \mathrm{O}(g)\)
In this scenario, a single \(\mathrm{SO}_{3}\) molecule decomposes into \(\mathrm{SO}_{2}\) and O. The rate law for a unimolecular reaction is simply first-order in the reactant, reflecting that the reaction rate is directly proportional to the concentration of the single reactant. This is confirmed in the exercise where the rate law is given by:
Rate = k[SO3]

Unimolecular reactions are often observed in processes such as radioactive decay or isomerization, where the internal energy of the molecule is sufficient to cause transformation.