Problem 68 - Chemistry The Molecular Science

Question

Assuming that each reaction is elementary, predict the rate law. (a) \(\mathrm{Br}(\mathrm{g})+\mathrm{IBr}(\mathrm{g}) \longrightarrow \mathrm{I}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{~g})\) (b) \(\mathrm{Cl}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HCl}(\mathrm{g})+\mathrm{H}(\mathrm{g})\) (c) \(2 \mathrm{NO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) (d) Cyclopropane(g) \(\longrightarrow\) propene(g)

Step-by-Step Solution

Answer

(a) \(rate = k [\text{Br}][\text{IBr}]\); (b) \(rate = k [\text{Cl}][\text{H}_2]\); (c) \(rate = k [\text{NO}_2]^2\); (d) \(rate = k [\text{cyclopropane}]\).

Step 1: Understanding Elementary Reactions

An elementary reaction is a single step reaction with a simple stoichiometric equation. Each molecule that participates in the reaction will directly affect the reaction rate.

Step 2: Rate Law for Reaction (a)

For the reaction \( ext{Br(g)} + ext{IBr(g)} \rightarrow \text{I(g)} + \text{Br}_2\text{(g)}\), the rate law is based on the stoichiometry of the reaction and is \(rate = k [\text{Br}][\text{IBr}]\) because both Br and IBr are reactants in the rate-determining step.

Step 3: Rate Law for Reaction (b)

For the reaction \(\text{Cl(g)} + \text{H}_2\text{(g)} \rightarrow \text{HCl(g)} + \text{H(g)}\), the rate law is determined by the stoichiometric coefficients: \(rate = k [\text{Cl}][\text{H}_2]\).

Step 4: Rate Law for Reaction (c)

For \(2\text{NO}_2(\text{g}) \rightarrow \text{N}_2\text{O}_4(\text{g})\), the reaction involves two NO2 molecules colliding, hence the rate law is \(rate = k [\text{NO}_2]^2\).

Step 5: Rate Law for Reaction (d)

Cyclopropane converting to propene is a unimolecular reaction, thus its rate law is \(rate = k [\text{cyclopropane}]\).

Key Concepts

Elementary Reactions

Elementary reactions are the building blocks of chemical kinetics. They represent a single step process in which reactants are directly transformed into products. Each participating molecule in an elementary reaction directly influences the rate of reaction. This means:

• The rate law, which describes how the concentration of reactants affects the rate, can be written directly from the balanced chemical equation of the reaction.
• No intermediate steps or reactions take place, unlike complex reactions that involve multiple steps.

For example, in the reaction \(\mathrm{Br} + \mathrm{IBr} \rightarrow \mathrm{I} + \mathrm{Br}_2\), since it is a simple one-step reaction, each molecule of \(\mathrm{Br}\) and \(\mathrm{IBr}\) has a direct effect on the reaction rate. The rate law is therefore expressed as \(\text{rate} = k [\text{Br}][\text{IBr}]\). Recognizing that a reaction is elementary is key to predicting its behavior accurately.

Reaction Kinetics

Understanding how fast a reaction takes place, known as its kinetics, is crucial for predicting and controlling chemical processes. The rate law provides this information by linking the concentration of reactants to the rate of reaction:

• The rate constant \(k\) is a unique factor for each reaction, influenced by temperature and other environmental conditions.
• For bimolecular reactions, where two molecules collide and react, like \(\mathrm{Cl}(\mathrm{g}) + \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{HCl}(\mathrm{g}) + \mathrm{H}(\mathrm{g})\), the rate law is \(\text{rate} = k [\text{Cl}][\text{H}_2]\).

This straightforward relationship from the stoichiometry makes elementary reactions simple to study and predict. Further, in reactions like \(2\text{NO}_2(\text{g}) \rightarrow \text{N}_2\text{O}_4(\text{g})\), the involvement of two \(\text{NO}_2\) molecules means a second-order dependency, \(\text{rate} = k [\text{NO}_2]^2\). Reaction kinetics help in understanding not just how fast a reaction proceeds, but also how conditions might alter this speed.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is essential in determining the rate laws for elementary reactions because:

• The coefficients in a balanced equation reveal how molecules are consumed and formed, thus helping predict the rate law.
• In \(\text{2NO}_2(\text{g}) \rightarrow \text{N}_2\text{O}_4(\text{g})\), stoichiometry tells us two molecules of \(\text{NO}_2\) react to form one molecule of \(\text{N}_2\text{O}_4\), influencing the rate as \(\text{rate} = k [\text{NO}_2]^2\).

For a unimolecular process like cyclopropane converting to propene, we see a one-to-one correspondence (\(\text{cyclopropane} \rightarrow \text{propene}\)), leading to a rate law of \(\text{rate} = k [\text{cyclopropane}]\). By understanding stoichiometry, students can better grasp how reactions proceed and how concentrations relate to changes over time.