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

Short Answer

Expert verified
(a) Rate = \( k[\mathrm{Cl}][\mathrm{ICl}] \). (b) Rate = \( k[\mathrm{O}][\mathrm{O}_{3}] \). (c) Rate = \( k[\mathrm{NO}_{2}]^{2} \).

Step by step solution

01

Understand Rate Laws

For an elementary reaction, the rate law can be written directly from its stoichiometry. Each reactant's concentration is raised to a power corresponding to its coefficient in the balanced equation.
02

Analyze Reaction (a)

The reaction is \( \mathrm{Cl}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \longrightarrow \mathrm{I}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \). This is an elementary reaction with a 1:1 stoichiometric ratio for each reactant, so the rate law is \( \text{Rate} = k[\mathrm{Cl}][\mathrm{ICl}] \), where \( k \) is the rate constant.
03

Analyze Reaction (b)

The reaction is \( \mathbf{O}(\mathrm{g})+\mathbf{O}_{3}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{g}) \). Both reactants appear with a stoichiometric coefficient of 1, so the rate law is \( \text{Rate} = k[\mathrm{O}][\mathrm{O}_{3}] \).
04

Analyze Reaction (c)

The reaction is \( 2 \mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \). The stoichiometry of the reactant \( \mathrm{NO}_{2} \) is 2, so the rate law is \( \text{Rate} = k[\mathrm{NO}_{2}]^{2} \).

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Key Concepts

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

Elementary Reactions
Elementary reactions are the simplest type of chemical reactions that occur in a single step. Each step in an elementary reaction progresses without any detectable intermediates coming into play. Understanding these reactions is crucial as they form the building blocks for more complex reactions.

When an elementary reaction occurs, it proceeds exactly as written in the chemical equation. For example, in the reaction \( \mathrm{Cl}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \longrightarrow \mathrm{I}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \), the reaction takes place in one go without formation of any intermediate products. As such, the rate law can be directly derived from the chemical equation.

In general, elementary reactions follow the molecularity concept, which directly relates to the number of reacting species involved in the reaction. This aspect makes it easier to determine the rate law directly, as each reactant contributes to the overall rate according to its stoichiometric coefficient.
Stoichiometry
Stoichiometry refers to the quantitative relationships between reactants and products in chemical reactions. It is crucial for writing accurate rate laws in the context of elementary reactions.

In stoichiometry, each reactant in a balanced chemical equation provides a coefficient, which indicates the proportion in which reactants are consumed or products are produced. This is particularly useful in determining the rate laws, especially for elementary reactions.

- In reaction \( \mathrm{Cl}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \), both reactants have a stoichiometric coefficient of 1, leading to a rate law of \( \text{Rate} = k[\mathrm{Cl}][\mathrm{ICl}] \).- For \( 2 \mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \), the stoichiometry shows two \( \mathrm{NO}_{2} \) molecules are involved, hence the rate law \( \text{Rate} = k[\mathrm{NO}_{2}]^{2} \).

It is important to match stoichiometric coefficients with the reactants in the rate law equation to ensure the accuracy and predict the correct reaction kinetics.
Reaction Kinetics
Reaction kinetics explores the speed at which chemical reactions occur and the factors that influence this speed. A key component is the formation of rate laws which describe how the reaction rate is affected by the concentration of reactants.

For elementary reactions, the rate law can be straightforwardly written with the concentrations of reactants in a direct correlation to their stoichiometric coefficients. This provides a clear pathway to predict how altering concentration impacts the rate.

- For instance, if you double the concentration of \( \mathrm{Cl} \) in the reaction \( \mathrm{Cl}(\mathrm{g})+\mathrm{ICl}(\mathrm{g}) \rightarrow \mathrm{I}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \), the reaction rate will double, provided \( \mathrm{ICl} \) stays constant. - Similarly, for \( 2 \mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \), doubling \( \mathrm{NO}_{2} \) concentration increases the reaction rate by a factor of 4, due to the squared term in the rate law \( \text{Rate} = k[\mathrm{NO}_{2}]^{2} \).

Understanding reaction kinetics not only aids in theoretical prediction but also guides practical applications such as controlling industrial chemical processes and developing new products.

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Most popular questions from this chapter

The decomposition of gaseous dimethyl ether at ordinary pressures is first order. Its half-life is 25.0 min at \(500^{\circ} \mathrm{C}\) $$\mathrm{CH}_{3} \mathrm{OCH}_{3}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})$$ (a) Starting with \(8.00 \mathrm{g}\) of dimethyl ether, what mass remains (in grams) after 125 min and after 145 min? (b) Calculate the time in minutes required to decrease \(7.60 \mathrm{ng}\) (nanograms) to 2.25 ng. (c) What fraction of the original dimethyl ether remains after 150 min?

Gaseous azomethane, \(\mathrm{CH}_{3} \mathrm{N}=\mathrm{NCH}_{3},\) decomposes in a first-order reaction when heated: $$\mathrm{CH}_{3} \mathrm{N}=\mathrm{NCH}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})$$ The rate constant for this reaction at \(425^{\circ} \mathrm{C}\) is \(40.8 \mathrm{min}^{-1}\) If the initial quantity of azomethane in the flask is \(2.00 \mathrm{g}\) how much remains after 0.0500 min? What quantity of \(\mathrm{N}_{2}\) is formed in this time?

Gaseous ammonia is made by the reaction $$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})$$ Use the information on the formation of \(\mathrm{NH}_{3}\) given in the table to answer the questions that follow. $$\begin{array}{lll}\hline\left[\mathrm{N}_{2}\right](\mathrm{M}) & {\left[\mathrm{H}_{2}\right](\mathrm{M})} & \text { Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \\\\\hline 0.030 & 0.010 & 4.21 \times 10^{-5} \\\0.060 & 0.010 & 1.68 \times 10^{-4} \\\0.030 & 0.020 & 3.37 \times 10^{-4} \\\\\hline\end{array}$$ (a) Determine \(n\) and \(m\) in the rate equation: Rate \(=\) \(k\left[\mathrm{N}_{2}\right]^{n}\left[\mathrm{H}_{2}\right]^{m}\) (b) Calculate the value of the rate constant. (c) What is the order of the reaction with respect to \(\left[\mathrm{H}_{2}\right] ?\) (d) What is the overall order of the reaction?

The decomposition of ammonia is first order with respect to \(\mathrm{NH}_{3}\). (Compare with Study Question 58.) $$2 \mathrm{NH}_{3}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g})$$ (a) What is the rate equation for this reaction? (b) Calculate the rate constant, \(k\), given the following data: $$\begin{array}{lc}\hline\left[\mathrm{NH}_{3}\right](\mathrm{mol} / \mathrm{L}) & \text { Time }(\mathrm{s}) \\\\\hline 0.67 & 0 \\\0.26 & 19 \\\\\hline\end{array}$$ (c) Determine the half-life of \(\mathrm{NH}_{3}\)

In the reaction \(2 \mathrm{O}_{3}(\mathrm{g}) \longrightarrow 3 \mathrm{O}_{2}(\mathrm{g}),\) the rate of formation of \(\mathrm{O}_{2}\) is \(1.5 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot\) s. What is the rate of decomposition of \(\mathrm{O}_{3} ?\)

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