Problem 63 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

The molecularity and rate laws for the given elementary reactions are as follows: (a) Molecularity: 1; Rate law: \(Rate = k[Cl_2]\) (b) Molecularity: 2; Rate law: \(Rate = k[OCl^-][H_2O]\) (c) Molecularity: 2; Rate law: \(Rate = k[NO][Cl_2]\)

Step 1: (a) Identify molecularity and write the rate law for the reaction: \(\mathrm{Cl}_{2}(g) \stackrel{-\longrightarrow}{\longrightarrow} 2 \mathrm{Cl}(g)\)

In this reaction, we have one molecule of \(\mathrm{Cl}_{2}(g)\) dissociating into two molecules of \(\mathrm{Cl}(g)\). Since we have only one reactant molecule involved in this elementary reaction, the molecularity is 1. To write the rate law, we use the concentration of the reactant (\([Cl_2]\)) and its order (which is the stoichiometric coefficient of the reactant, in this case, 1). The rate law for this reaction is: \(Rate = k[Cl_2]\)

Step 2: (b) Identify molecularity and write the rate law for the reaction: \(\mathrm{OCl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{HOCl}(a q)+\mathrm{OH}^{-}(a q)\)

In this reaction, we have one molecule of \(\mathrm{OCl}^{-}(a q)\) and one molecule of \(\mathrm{H}_{2} \mathrm{O}(l)\) involved in this elementary reaction. The sum of the reactant molecules is 1 + 1, so the molecularity of this reaction is 2. To write the rate law, we use the concentration of both reactants (\([OCl^-]\) and \([H_2O]\)) and their respective orders (which are the stoichiometric coefficients of the reactants, in this case, both are 1). The rate law for this reaction is: \(Rate = k[OCl^-][H_2O]\)

Step 3: (c) Identify molecularity and write the rate law for the reaction: \(\mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \operatorname{NOCl}_{2}(g)\)

In this reaction, we have one molecule of \(\mathrm{NO}(g)\) and one molecule of \(\mathrm{Cl}_{2}(g)\) involved in this elementary reaction. The sum of the reactant molecules is 1+1, so the molecularity of this reaction is 2. To write the rate law, we use the concentration of both reactants (\([NO]\) and \([Cl_2]\)) and their respective orders (which are the stoichiometric coefficients of the reactants, in this case, both are 1). The rate law for this reaction is: \(Rate = k[NO][Cl_2]\)

Key Concepts

Rate Law

The rate law of a chemical reaction expresses the relationship between the reaction rate and the concentration of its reactants. It is typically written as: - \[ \text{Rate} = k[A]^m[B]^n \] where:

• \(k\) is the rate constant, which is specific to the reaction at a given temperature
• \([A]\) and \([B]\) represent the concentrations of reactants
• \(m\) and \(n\) are the reaction orders with respect to each reactant, often determined experimentally

Reactants in an elementary reaction have reaction orders that are identical to their stoichiometric coefficients. This method simplifies determining the effect of each reactant's concentration on the reaction rate. For example, in the equation \(Rate = k[Cl_2]\), the reaction order with respect to \(Cl_2\) is 1, indicating that the rate is directly proportional to \([Cl_2]\). Understanding the rate law helps in predicting how changing reactant concentrations affect the speed of a reaction, offering valuable insights into controlling chemical processes.

Elementary Reaction

An elementary reaction is a single-step process in a chemical reaction mechanism. Unlike complex reactions which occur in multiple steps, an elementary reaction involves direct interactions of reactants to form products. There are key characteristics of elementary reactions to keep in mind: -

• They consist of a single transition state without intermediates.
• The rate law can be directly written from the stoichiometry of the reactants.
• Every molecule involved participates directly in the transition state.

Understanding elementary reactions can help in discerning the exact molecular interactions in reactions. For instance, in the reaction \(NO + Cl_2 \rightarrow NOCl_2\), each molecule of NO and Cl_2 partakes simultaneously in forming the product. Elementary reactions are fundamental components in understanding broader mechanisms and must be mastered to unravel complex pathways.

Stoichiometric Coefficient

Stoichiometric coefficients are numbers that appear before molecules in balanced chemical equations. They reflect the ratio of molecules or moles of reactants and products involved in the chemical reaction. They are crucial for both balancing equations and writing rate laws: -

• They ensure mass conservation by indicating how molecules combine and convert into products.
• They direct the formation of the rate law in elementary reactions by giving the reaction orders directly.

For instance, in the dissociation \(Cl_2 \rightarrow 2Cl\), the stoichiometric coefficient of \(Cl_2\) is 1, and for \(Cl\), it is 2. While stoichiometric coefficients aid significantly in rate law formulation, it's important to distinguish between these and reaction orders, particularly in non-elementary reactions. The coefficients ensure accuracy and consistency in calculations, reflecting the exact proportions of each substance involved in the reaction.