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Chapter 13: Problem 51

Determine the molecularity and write the rate law for each of the following elementary steps: (a) \(\mathrm{X} \longrightarrow\) products (b) \(\mathrm{X}+\mathrm{Y} \longrightarrow\) products (c) \(\mathrm{X}+\mathrm{Y}+\mathrm{Z} \longrightarrow\) products (d) \(\mathrm{X}+\mathrm{X} \longrightarrow\) products (e) \(\mathrm{X}+2 \mathrm{Y} \longrightarrow\) products

Short Answer

Expert verified
Molecularity and rate laws for given reactions are: (a) Molecularity: Unimolecular, Rate Law: Rate = \(k[\mathrm{X}]\). (b) Molecularity: Bimolecular, Rate Law: Rate = \(k[\mathrm{X}][\mathrm{Y}]\). (c) Molecularity: Termolecular, Rate Law: Rate = \(k[\mathrm{X}][\mathrm{Y}][\mathrm{Z}]\). (d) Molecularity: Bimolecular, Rate Law: Rate = \(k[\mathrm{X}]^2\). (e) Molecularity: Termolecular, Rate Law: Rate = \(k[\mathrm{X}][\mathrm{Y}]^2\).

Step by step solution

01

Identify Molecularity and Write Rate Law for \(\mathrm{X} \longrightarrow\) products

Since there is only one reactant, the molecularity of the reaction is uni-molecular. The rate law is Rate = \(k[\mathrm{X}]\), where \(k\) is the rate constant and \([\mathrm{X}]\) is the concentration of X.
02

Identify Molecularity and Write Rate Law for \(\mathrm{X}+\mathrm{Y} \longrightarrow\) products

This reaction has two reactants, hence it is bi-molecular. The rate law is Rate = \(k[\mathrm{X}][\mathrm{Y}]\).
03

Identify Molecularity and Write Rate Law for \(\mathrm{X}+\mathrm{Y}+\mathrm{Z} \longrightarrow\) products

This elementary step includes three reactants, so it has a molecularity of three, which makes it a termolecular reaction. Its rate law is Rate = \(k[\mathrm{X}][\mathrm{Y}][\mathrm{Z}]\).
04

Identify Molecularity and Write Rate Law for \(\mathrm{X}+\mathrm{X} \longrightarrow\) products

This has two molecules of the same reactant, X, hence it is bi-molecular. The rate law is given by Rate = \(k[\mathrm{X}]^2\).
05

Identify Molecularity and Write Rate Law for \(\mathrm{X}+2 \mathrm{Y} \longrightarrow\) products

There are three reactant pieces, one X and two Y, which makes it a termolecular reaction. However, note that it involves two molecules of Y. The rate law would be given by Rate = \(k[\mathrm{X}][\mathrm{Y}]^2\).

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Most popular questions from this chapter

The rate law for the decomposition of ozone to molecular oxygen \(2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)\) is rate \(=k \frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]}\) The mechanism proposed for this process is \(\mathrm{O}_{3} \stackrel{k_{1}}{k_{-1}} \mathrm{O}+\mathrm{O}_{2}\) \(\mathrm{O}+\mathrm{O}_{3} \stackrel{k_{2}}{\longrightarrow} 2 \mathrm{O}_{2}\) Derive the rate law from these elementary steps. Clearly state the assumptions you use in the derivation. Explain why the rate decreases with increasing \(\mathrm{O}_{2}\) concentration.

What do we mean by the mechanism of a reaction? What is an elementary step? What is the molecularity of a reaction?

The rate constant for the second-order reaction $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ is \(0.54 / M \cdot \mathrm{s}\) at \(300^{\circ} \mathrm{C}\). How long (in seconds) would it take for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(0.62 M\) to \(0.28 M ?\)

The activation energy \(\left(E_{\mathrm{a}}\right)\) for the reaction $$ 2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \quad \Delta H^{\circ}=-164 \mathrm{~kJ} / \mathrm{mol} $$ is \(240 \mathrm{~kJ} / \mathrm{mol} .\) What is \(E_{\mathrm{a}}\) for the reverse reaction?

What are the units of the rate constant for a thirdorder reaction?
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