Problem 67 - Chemistry The Central Science

Question

What is the molecularity of each of the following elementary reactions? Write the rate law for each. \(\begin{array}{l}{\text { (a) } \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Cl}(g)} \\ {\text { (b) } \mathrm{OCl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HOCl}(a q)+\mathrm{OH}^{-}(a q)} \\\ {\text { (c) } \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NOCl}_{2}(g)}\end{array}\)

Step-by-Step Solution

Answer

The molecularity and rate law for each elementary reaction are: (a) Molecularity: 1 (unimolecular), Rate Law: \(Rate = k[\mathrm{Cl}_{2}]\) (b) Molecularity: 2 (bimolecular), Rate Law: \(Rate = k[\mathrm{OCl}^-][\mathrm{H}_{2}\mathrm{O}]\) (c) Molecularity: 2 (bimolecular), Rate Law: \(Rate = k[\mathrm{NO}][\mathrm{Cl}_{2}]\)

Step 1: Reaction (a) - Molecularity and Rate Law

For the first reaction: \(\mathrm{Cl}_{2}(g) \longrightarrow 2\mathrm{Cl}(g)\) This is an elementary reaction, which means, we can directly obtain the molecularity by counting the number of molecules involved. In this case, there is only one molecule of Cl2 involved. Molecularity: 1 (unimolecular) Now, we can write the rate law for this reaction. For an elementary reaction, the rate law is proportional to the concentration of the reactants. So, Rate = k[Cl2], where k is the rate constant.

Step 2: Reaction (b) - Molecularity and Rate Law

For the second reaction: \(\mathrm{OCl}^{-}(a q)+\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{HOCl}(a q)+\mathrm{OH}^{-}(a q)\) This is also an elementary reaction. Count the number of molecules involved in the reaction. In this case, we have one molecule of OCl^- and one molecule of H2O involved. Molecularity: 2 (bimolecular) Now, let's write the rate law for this reaction. The rate law is proportional to the concentration of the reactants: Rate = k[OCl^-][H2O], where k is the rate constant.

Step 3: Reaction (c) - Molecularity and Rate Law

For the third reaction: \(\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NOCl}_{2}(g)\) Again, this is an elementary reaction. Counting the number of molecules involved, we get one molecule of NO and one molecule of Cl2. Molecularity: 2 (bimolecular) Now, let's write the rate law for this reaction, which will be proportional to the concentration of the reactants: Rate = k[NO][Cl2], where k is the rate constant. The molecularity and rate law for each elementary reaction are as follows: (a) Molecularity: 1, Rate Law: Rate = k[Cl2] (b) Molecularity: 2, Rate Law: Rate = k[OCl^-][H2O] (c) Molecularity: 2, Rate Law: Rate = k[NO][Cl2]

Key Concepts

Elementary Reactions

An elementary reaction is a single step reaction with no intermediate stages. It's the most basic form of a reaction and occurs exactly as it's written in the chemical equation. These reactions involve reactant molecules such as atoms, ions, or radicals directly interacting to form products. Because they are single steps, the molecularity can be determined simply by counting the reactants involved at this initial stage.

• Molecularity: This refers to the number of molecules coming together to react in an elementary reaction. It's always a small whole number.
• Types of Elementary Reactions: Based on molecularity, these reactions can be unimolecular, involving one species, or bimolecular, involving two.

These are crucial because they provide the basis for understanding the mechanism of complex reactions by breaking them down into simpler steps.

Rate Law

The rate law expresses the speed of a chemical reaction, specifically how the rate depends on the concentration of the reactants. For elementary reactions, it's straightforward because it directly corresponds with the stoichiometry of the reaction.

• Proportionality to Reactant Concentrations: The rate of reaction for an elementary step is directly proportional to the concentration of the reactants raised to the power of their stoichiometric coefficients in the balanced equation.
• Rate Constant: This is a specific constant for each reaction at a given temperature, denoted by the symbol \(k\).

Understanding the rate law helps in predicting how changes in concentrations affect the reaction speed, which is essential for controlling reactions in real-world applications.

Unimolecular and Bimolecular Reactions

Reactions are classified based on how many molecules participate in a single step, thinking about molecularity aids in understanding basic reaction dynamics.

• Unimolecular Reactions: These involve a single reactant molecule splitting into products. An example is the decomposition of \( ext{Cl}_2\) into two chlorine radicals. The rate law for such a reaction is simple: rate = \(k[ ext{Cl}_2]\), directly tied to the concentration of the chlorine gas.
• Bimolecular Reactions: These involve the collision between two reactant molecules to produce products. For instance, the interaction of \( ext{NO}\) and \( ext{Cl}_2\) to form \( ext{NOCl}_2\). The rate for this scenario is rate = \(k[ ext{NO}][ ext{Cl}_2]\), emphasizing the dependence on both reactants meeting.

This molecular approach lays the foundation for exploring more complex chemical processes, demonstrating how essential it is to understand the primary interactions that drive chemical change.