Problem 59
Question
Write the rate law for each of these elementary reactions. (a) \(\mathrm{Cl}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \longrightarrow \mathrm{I}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{CH}_{3} \mathrm{~N}=\mathrm{NCH}_{3}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})\) (c) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})\)
Step-by-Step Solution
Verified Answer
(a) \(\text{Rate} = k[\mathrm{Cl}][\mathrm{ICl}]\); (b) \(\text{Rate} = k[\mathrm{CH}_{3}\mathrm{~N}=\mathrm{NCH}_{3}]\); (c) \(\text{Rate} = k[\mathrm{N}_{2}\mathrm{O}_{4}]\).
1Step 1: Understand the Concept of Elementary Reactions
In an elementary reaction, the rate law can be written directly from the stoichiometry of the reaction. This is because, in elementary reactions, the reaction occurs in a single step, and the rate law corresponds exactly to the molecularity (number of molecules colliding).
2Step 2: Determine the Rate Law for Reaction (a)
For the reaction \( \mathrm{Cl}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \longrightarrow \mathrm{I}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \), the reaction is bimolecular as it involves two reactant species colliding. The rate law is: \[ \text{Rate} = k[\mathrm{Cl}][\mathrm{ICl}] \]
3Step 3: Determine the Rate Law for Reaction (b)
For the reaction \( \mathrm{CH}_{3} \mathrm{~N}=\mathrm{NCH}_{3}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g}) \), it is a unimolecular decomposition reaction. The rate law is: \[ \text{Rate} = k[\mathrm{CH}_{3}\mathrm{~N}=\mathrm{NCH}_{3}] \]
4Step 4: Determine the Rate Law for Reaction (c)
For the reaction \( \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{~g}) \), this is also a unimolecular reaction. The rate law is based on a single molecule decomposing: \[ \text{Rate} = k[\mathrm{N}_{2}\mathrm{O}_{4}] \]
Key Concepts
Rate LawMolecularityUnimolecular Reactions
Rate Law
In chemistry, a rate law is an equation that describes the rate of a reaction. It tells us how the concentration of reactants affects the rate at which a reaction proceeds. With elementary reactions, we can define the rate law directly based on the stoichiometry of the reactants. This is because the reaction occurs in a single step, involving direct collisions between molecules.
- For a reaction involving two reactant particles, it is termed a bimolecular reaction and the rate law typically involves the product of both concentrations.
- For a reaction involving one molecule, termed a unimolecular reaction, the rate law includes only the concentration of that single species.
Molecularity
Molecularity refers to the number of reacting species (atoms, ions, or molecules) involved in an elementary reaction. Unlike the overall order of reaction which can be non-integral, molecularity is always a positive integer. It provides insight into the mechanism of the reaction.
- Unimolecular: Involves a single molecule undergoing change. For example, the decomposition of a compound. These reactions have a molecularity of one.
- Bimolecular: Involves two reacting species colliding and reacting together, such as in reaction (a) of the exercise.
Unimolecular Reactions
Unimolecular reactions are a class of reactions where a single molecule undergoes a transformation into different products. This typically involves processes like rearrangement, dissociation, or isomerization.
- An example is when a molecule loses atoms or groups of atoms to form simpler products.
- In the exercise, reactions (b) and (c) are examples of unimolecular reactions. They follow the general rate law: \( \text{Rate} = k[A] \), where \([A]\) is the concentration of the reactant.
Other exercises in this chapter
Problem 57
For the gas-phase reaction $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}(\mathrm{g}) \longrightarrow \mathrm{CH}_{2} \mathrm{CH}_{2}(\mathrm{~g})+\mathrm{HI}(\m
View solution Problem 58
For the gas-phase reaction cis-CHClCHCl(g)\longrightarrow trans-CHClCHCl(g) the activation energy \(E_{\mathrm{a}}\) is \(234 \mathrm{~kJ} / \mathrm{mol}\) and
View solution Problem 61
Draw an energy versus reaction progress diagram (similar to the one in Question 60 ) for each of the reactions whose activation energy and enthalpy change are g
View solution Problem 62
Draw an energy versus reaction progress diagram (similar to the one in Question 60 ) for each of the reactions whose activation energy and enthalpy change are g
View solution