Problem 48 - Chemistry and Chemical Reactivity

Question

What is the rate law for each of the following elementary reactions? (a) \(\mathrm{Cl}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \rightarrow \mathrm{I}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) (b) \(\mathrm{O}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{g}) \rightarrow 2 \mathrm{O}_{2}(\mathrm{g})\) (c) \(2 \mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\)

Step-by-Step Solution

Answer

(a) Rate = k [Cl][ICl]; (b) Rate = k [O][O3]; (c) Rate = k [NO2]^2.

Step 1: Analyze Reaction (a)

For the elementary reaction \( \mathrm{Cl} (\mathrm{g}) + \mathrm{ICl} (\mathrm{g}) \rightarrow \mathrm{I} (\mathrm{g}) + \mathrm{Cl}_{2} (\mathrm{g}) \), identify the reactants. The stoichiometry implies each molecule of \( \mathrm{Cl} \) and \( \mathrm{ICl} \) contributes to the reaction. Thus, both reactants have order 1.

Step 2: Formulate Rate Law for Reaction (a)

Since this is an elementary reaction, the rate law is directly based on the stoichiometric coefficients of the reactants. The rate law can be written as: \[ \text{Rate} = k [\mathrm{Cl}][\mathrm{ICl}] \] where \( k \) is the rate constant.

Step 3: Analyze Reaction (b)

Consider the reaction \( \mathrm{O} (\mathrm{g}) + \mathrm{O}_{3} (\mathrm{g}) \rightarrow 2 \mathrm{O}_{2} (\mathrm{g}) \). Here, the reactants are \( \mathrm{O} \) and \( \mathrm{O}_{3} \), each affecting the rate of reaction linearly, reflecting an order of 1 for both.

Step 4: Formulate Rate Law for Reaction (b)

The rate law for this elementary step, with both reactants having a stoichiometric coefficient of 1, is: \[ \text{Rate} = k [\mathrm{O}][\mathrm{O}_{3}] \] with \( k \) as the rate constant.

Step 5: Analyze Reaction (c)

This reaction is \( 2 \mathrm{NO}_{2} (\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4} (\mathrm{g}) \), an elementary reaction with \( 2 \mathrm{NO}_{2} \) as the only reactant impacting the rate. Since two molecules of \( \mathrm{NO}_{2} \) participate, this reflects a second-order with respect to \( \mathrm{NO}_{2} \).

Step 6: Formulate Rate Law for Reaction (c)

Given the stoichiometry of the reactant in the elementary reaction, the rate law is: \[ \text{Rate} = k [\mathrm{NO}_{2}]^2 \] where \( k \) represents the rate constant.

Key Concepts

Elementary Reactions

Elementary reactions are the simplest kind of reactions that occur in a single step. These reactions involve the direct interaction of reactant molecules to form products without any intermediates along the way. Because of this direct interaction, the rate law for an elementary reaction is derived directly from its balanced chemical equation.

For example, in the reaction \( \mathrm{Cl}(\mathrm{g}) + \mathrm{ICl}(\mathrm{g}) \rightarrow \mathrm{I}(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g}) \), we see that both \( \mathrm{Cl} \) and \( \mathrm{ICl} \) are reactants. In this elementary reaction, each particle reacts in a one-to-one manner, which means each has a reaction order of 1.

Elementary reactions make it easy to write the rate law. Simply look at the stoichiometric coefficients of the reactants, and write the rate law proportionally. This straightforward approach only works for elementary reactions, so make sure to distinguish them from more complex reactions in mechanism.

Reaction Order

Reaction order provides insight into how the concentration of each reactant influences the rate of a reaction. It is reflected in the exponents of concentration terms in the rate law equation. In elementary reactions, the reaction order corresponds directly to the stoichiometry of the reactants.

For instance, in the reaction \( \mathrm{O}(\mathrm{g}) + \mathrm{O}_3(\mathrm{g}) \rightarrow 2 \mathrm{O}_2(\mathrm{g}) \), each reactant appears with a stoichiometric coefficient of 1. Therefore, the rate law is \( \text{Rate} = k [\mathrm{O}][\mathrm{O}_3] \), indicating a first-order dependence on both \( \mathrm{O} \) and \( \mathrm{O}_3 \).

In contrast, a reaction like \( 2 \mathrm{NO}_2(\mathrm{g}) \rightarrow \mathrm{N}_2\mathrm{O}_4(\mathrm{g}) \) involves two molecules of \( \mathrm{NO}_2 \), leading to a second-order reaction with respect to \( \mathrm{NO}_2 \) written as \( \text{Rate} = k [\mathrm{NO}_2]^2 \).

This clear correspondence between stoichiometry and order makes analyzing elementary reactions straightforward.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. In the context of rate laws, it helps us understand the quantitative relationships guiding reaction kinetics. For elementary reactions, the stoichiometry directly informs the rate law.

Take the reaction \( \mathrm{Cl}(\mathrm{g}) + \mathrm{ICl}(\mathrm{g}) \rightarrow \mathrm{I}(\mathrm{g}) + \mathrm{Cl}_2(\mathrm{g}) \) as an example. The stoichiometry shows each reactant participating once, which directly leads to a rate law of \( \text{Rate} = k [\mathrm{Cl}][\mathrm{ICl}] \).

In more complex reactions, you may have intermediates and different pathways, meaning stoichiometry alone doesn't determine the rate law. However, for elementary reactions, the stoichiometric coefficients of reactants provide a simple yet powerful way to identify the reaction rate. This simplicity highlights the importance of recognizing when a reaction is elementary to make accurate kinetic predictions.