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Chapter 14: Problem 67

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

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

(a) Consider the combustion of ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\) \(3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) .\) If the concentration of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is decreasing at the rate of \(0.036 \mathrm{M} / \mathrm{s},\) what are the rates of change in the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?(\mathbf{b})\) The rate of decrease in \(\mathrm{N}_{2} \mathrm{H}_{4}\) partial pressure in a closed reaction vessel from the reaction \(\mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) is 74 torr per hour. What are the rates of change of \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel?

(a) If you were going to build a system to check the effectiveness of automobile catalytic converters on cars, what substances would you want to look for in the car exhaust? (b) Automobile catalytic converters have to work at high temperatures, as hot exhaust gases stream through them. In what ways could this be an advantage? In what ways a disadvantage? (c) Why is the rate of flow of exhaust gases over a catalytic converter important?

You have studied the gas-phase oxidation of HBr by \(\mathrm{O}_{2}\) : $$4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g)$$ You find the reaction to be first order with respect to HBr and first order with respect to \(\mathrm{O}_{2}\) . You propose the following mechanism: $$ \begin{array}{c}{\operatorname{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow \operatorname{HOOBr}(g)} \\ {\operatorname{HOOBr}(g)+\operatorname{HBr}(g) \longrightarrow 2 \mathrm{HOBr}(g)} \\\ {\operatorname{HOBr}(g)+\operatorname{HBr}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g)}\end{array}$$ (a) Confirm that the elementary reactions add to give the overall reaction. (b) Based on the experimentally determined rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at 330 \(\mathrm{K}\) , \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(a l c)\) is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is 0.0477 \(\mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 \mathrm{M},\) the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\) (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

Many primary amines, RNH \(_{2},\) where \(R\) is a carbon-containing fragment such as \(C H_{3}, C H_{3} C H_{2},\) and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a \(C\) atom for \(^{4} \mathrm{R}^{\prime \prime}\) . (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?
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