/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 102 Write the rate laws for the foll... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write the rate laws for the following elementary steps and identify them as uni-, bi-, or termolecular steps: a. \(\mathrm{Cl}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g)\) b. \(2 \mathrm{NO}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\) \(^{\bullet} \mathrm{c} .^{14} \mathrm{C} \rightarrow_{7}^{14} \mathrm{N}+_{-1}^{0} \beta\)

Short Answer

Expert verified
Question: Identify the type of each elementary step and write the corresponding rate law. a. Cl(g) + O3(g) -> ClO(g) + O2(g) b. 2 NO2(g) -> N2O4(g) c. \(^{14}_7 C \rightarrow^{14}_7 N +^{-1}_0 \beta\) Answer: a. Bimolecular step, Rate = k[Cl][O3] b. Bimolecular step, Rate = k[NO2]^2 c. Unimolecular step, Rate = k[\(^{14}_7 C\)]

Step by step solution

01

a. Cl(g) + O3(g) -> ClO(g) + O2(g)

In this elementary step, one molecule of Cl(g) and one molecule of O3(g) react with each other to produce one molecule of ClO(g) and one molecule of O2(g). So, the overall rate law will be represented by: Rate = k[Cl][O3] Since there are two reactant molecules involved in this step, we classify it as a bimolecular step.
02

b. 2 NO2(g) -> N2O4(g)

In this elementary step, two molecules of NO2(g) react with each other to produce one molecule of N2O4(g). So, the overall rate law will be represented by: Rate = k[NO2]^2 Since there are two reactant molecules of the same species involved in this step, we classify it as a bimolecular step.
03

c. \(^{14}_7 C \rightarrow^{14}_7 N +^{-1}_0 \beta\)

In this elementary step, one atom of \(^{14}_7 C\) decays, producing one atom of \(^{14}_7 N\) and one beta particle (\(^{-1}_0 \beta\)). Since only one reactant particle is involved in this step, the overall rate law will be represented by: Rate = k[\(^{14}_7 C\)] As there is only one reactant particle involved in this step, we classify it as a unimolecular step.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Elementary Steps
Elementary steps are crucial components in chemical reaction mechanisms. Each elementary step represents a simple reaction, where reactants directly form products without any intermediates or complex stages. They're like building blocks of a reaction, depicting the actual collision events happening in the reaction at a molecular level.
In the context of rate laws, each elementary step has a clear relationship between the concentration of the reactants and the speed at which the reaction occurs. This relationship is expressed as the rate law, which is generally determined directly by the stoichiometry of the reactants in the step.
For instance, consider the elementary step \[Cl(g) + O_3(g) \rightarrow ClO(g) + O_2(g)\]Here, the rate law given by the stoichiometry is \[Rate = k[Cl][O_3]\]This expression implies that the rate of reaction is directly proportional to the concentrations of Cl and O_3. Thus, elementary steps provide essential insights into the detailed dynamics of how reactions progress.
Molecularity
Molecularity refers to the number of reactant particles involved in an elementary step. It's an important concept in classifying elementary reactions.
There are three main types of molecularity:
  • **Unimolecular**: Involves a single reactant particle breaking down to form products. An example is \[^{14}_7 C \rightarrow^{14}_7 N +^{-1}_0 \beta\]This step has a molecularity of one, making it unimolecular.
  • **Bimolecular**: Involves two reactant particles. These could be two of the same kind or different. For example, in \[2NO_2(g) \rightarrow N_2O_4(g)\]two NO2 molecules participate, classifying the step as bimolecular.
  • **Termolecular**: Involves three reactant particles. This is less common due to the low probability of three particles colliding simultaneously.
Molecularity is integral to understanding not just the classification of reactions, but also their possible mechanisms and complexities.
Chemical Kinetics
Chemical kinetics is a branch of chemistry concerned with the rates of chemical reactions and the factors affecting these rates. Understanding kinetics is vital to comprehending how and why reactions occur at the speed they do.
This field covers several key areas:
  • **Rate Laws**: These are mathematical expressions derived from experimental data that indicate the relationship between the concentration of reactants and the reaction rate. For elementary reactions, these laws can often be deduced from the stoichiometry of the reactants involved.
  • **Reaction Mechanisms**: These are detailed descriptions of the steps through which reactants transform into products. Each step in a mechanism is represented as an elementary step.
  • **Factors Affecting Reaction Rates**: Temperature, concentration, and catalysts are primary influences that can accelerate or decelerate a reaction.
In chemical kinetics, what happens at the molecular level is scrutinized to predict and control the outcomes of chemical processes. This helps in fields ranging from material synthesis to pharmaceuticals, making it a cornerstone of applied chemistry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Catalytic Converters and Combustion Catalytic converters in automobiles combat air pollution by converting \(\mathrm{NO}\) and \(\mathrm{CO}\) into \(\mathrm{N}_{2}\) and \(\mathrm{CO}_{2}\) : $$2 \mathrm{CO}(g)+2 \mathrm{NO}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{CO}_{2}(g)$$ a. How is the rate of formation of \(\mathrm{N}_{2}\) related to the rate of consumption of CO? b. How is the rate of formation of \(\mathrm{CO}_{2}\) related to the rate of consumption of NO? c. How is the rate of consumption of CO related to the rate of consumption of NO?

\(\mathrm{N}_{2} \mathrm{O}_{5},\) when dissolved in water, decomposes to produce \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\) $$2 \mathrm{N}_{2} \mathrm{O}_{5}(a q) \rightarrow 4 \mathrm{NO}_{2}(a q)+\mathrm{O}_{2}(a q)$$ The rate of formation of \(\mathrm{O}_{2}\) is \(4.0 \times 10^{-3} M \cdot \mathrm{min}^{-1}\) a. What is the rate of formation of \(\mathrm{NO}_{2}(a q) ?\) b. What is the rate of disappearance of \(\mathrm{N}_{2} \mathrm{O}_{5}(a q) ?\)

The metabolism of table sugar (sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) ) begins with the hydrolysis of the disaccharide to glucose and fructose (both \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) ): $$\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)$$ The kinetics of the reaction were studied at \(24^{\circ} \mathrm{C}\) in a reaction system with a large excess of water, so the reaction was pseudo first order in sucrose. Determine the rate law and the pseudo-first-order rate constant for the reaction from the following data: $$\begin{array}{cc} \text { Time (s) } & {\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right](\mathrm{M})} \\ 0 & 0.562 \\ \hline 612 & 0.541 \\ \hline 1600 & 0.509 \\ \hline 2420 & 0.484 \\ \hline 3160 & 0.462 \\ \hline 4800 & 0.442 \\ \hline \end{array}$$

If the rate of the reverse reaction is much slower than the rate of the forward reaction, does the method used to determine a rate law from initial concentrations and initial rates also work at some other time \(t ?\)

Does a catalyst affect both the rate and the rate constant of a reaction?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.