Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The Rice-Herzfeld mechanism for the thermal decomposition of acetaldehyde \(\left(\mathrm{CH}_{3} \mathrm{CO}(g)\right)\) is \\[ \begin{array}{l} \mathrm{CH}_{3} \mathrm{CHO}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{CH}_{3} \cdot(g)+\mathrm{CHO} \cdot(g) \\ \mathrm{CH}_{3} \cdot(g)+\mathrm{CH}_{3} \mathrm{CHO}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CH}_{4}(g)+\mathrm{CH}_{2} \mathrm{CHO} \cdot(g) \\ \mathrm{CH}_{2} \mathrm{CHO} \cdot(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{CO}(g)+\mathrm{CH}_{3} \cdot(g) \\ \mathrm{CH}_{3} \cdot(g)+\mathrm{CH}_{3} \cdot(g) \stackrel{k_{4}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g) \end{array} \\] Using the steady-state approximation, determine the rate of methane \(\left(\mathrm{CH}_{4}(g)\right)\) formation.

Short Answer

Expert verified
The rate of methane formation in the Rice-Herzfeld mechanism for the thermal decomposition of acetaldehyde can be found using the steady-state approximation and is given by: \(r_{\mathrm{CH}_{4}} = k_{2}\sqrt{\frac{k_{1}}{2k_{4}}[\mathrm{CH}_{3} \mathrm{CHO}]}[\mathrm{CH}_{3} \mathrm{CHO}]\)

Step by step solution

01

List the given reactions and constants

The Rice-Herzfeld mechanism is given as: \[ \begin{array}{l} \mathrm{CH}_{3} \mathrm{CHO}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{CH}_{3} \cdot(g)+\mathrm{CHO} \cdot(g) \\ \mathrm{CH}_{3} \cdot(g)+\mathrm{CH}_{3} \mathrm{CHO}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CH}_{4}(g)+\mathrm{CH}_{2} \mathrm{CHO} \cdot(g) \\ \mathrm{CH}_{2} \mathrm{CHO} \cdot(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{CO}(g)+\mathrm{CH}_{3} \cdot(g) \\ \mathrm{CH}_{3} \cdot(g)+\mathrm{CH}_{3} \cdot(g) \stackrel{k_{4}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g) \end{array} \]
02

Apply the steady-state approximation to the reactive intermediates

Using the steady-state approximation, the sum of all rates of formation and consumption of the reactive intermediates (CH₃·(g) and CH₂CHO·(g)) equals zero: For CH₃·(g): \(\frac{d[\mathrm{CH}_{3} \cdot]}{dt} \approx 0 = k_{1}[\mathrm{CH}_{3} \mathrm{CHO}] - k_{2}[\mathrm{CH}_{3} \cdot][\mathrm{CH}_{3} \mathrm{CHO}] + k_{3}[\mathrm{CH}_{2} \mathrm{CHO} \cdot] - 2k_{4}[\mathrm{CH}_{3} \cdot]^{2}\) For CH₂CHO·(g): \(\frac{d[\mathrm{CH}_{2} \mathrm{CHO} \cdot]}{dt} \approx 0 = k_{2}[\mathrm{CH}_{3} \cdot][\mathrm{CH}_{3} \mathrm{CHO}] - k_{3}[\mathrm{CH}_{2} \mathrm{CHO} \cdot]\)
03

Solve the simultaneous equations for the reactive intermediates

Now solve the two simultaneous equations for [\(\mathrm{CH}_{3} \cdot\)] and [\(\mathrm{CH}_{2} \mathrm{CHO} \cdot\)]: From the second equation: \([\mathrm{CH}_{2} \mathrm{CHO} \cdot] = \frac{k_{2}}{k_{3}}[\mathrm{CH}_{3} \cdot][\mathrm{CH}_{3} \mathrm{CHO}]\) Insert this expression into the first equation and get: \(k_{1}[\mathrm{CH}_{3} \mathrm{CHO}] - k_{2}[\mathrm{CH}_{3} \cdot][\mathrm{CH}_{3} \mathrm{CHO}] + k_{3} \cdot \frac{k_{2}}{k_{3}}[\mathrm{CH}_{3} \cdot][\mathrm{CH}_{3} \mathrm{CHO}] - 2k_{4}[\mathrm{CH}_{3} \cdot]^{2} = 0\) Simplify the equation and get: \(k_{1}[\mathrm{CH}_{3} \mathrm{CHO}] - 2k_{4}[\mathrm{CH}_{3} \cdot]^{2} = 0\) Now solve for [\(\mathrm{CH}_{3} \cdot\)]: \([\mathrm{CH}_{3} \cdot] = \sqrt{\frac{k_{1}}{2k_{4}}[\mathrm{CH}_{3} \mathrm{CHO}]}\)
04

Find the rate of methane formation

Since the rate of methane formation is determined by the second reaction, we can write: \(r_{\mathrm{CH}_{4}} = k_{2}[\mathrm{CH}_{3} \cdot][\mathrm{CH}_{3} \mathrm{CHO}]\) Substitute the expression for [\(\mathrm{CH}_{3} \cdot\)]: \(r_{\mathrm{CH}_{4}} = k_{2}\sqrt{\frac{k_{1}}{2k_{4}}[\mathrm{CH}_{3} \mathrm{CHO}]}[\mathrm{CH}_{3} \mathrm{CHO}]\) Hence, the rate of methane formation is given by: \(r_{\mathrm{CH}_{4}} = k_{2}\sqrt{\frac{k_{1}}{2k_{4}}[\mathrm{CH}_{3} \mathrm{CHO}]}[\mathrm{CH}_{3} \mathrm{CHO}]\)

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 Vaia!

Key Concepts

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

Rice-Herzfeld Mechanism
The Rice-Herzfeld mechanism is a series of chemical reactions that describe the thermal decomposition of certain organic compounds. It's a fundamental concept for students learning about reaction mechanisms in chemical kinetics. This mechanism, particularly for the decomposition of acetaldehyde into methane and other products, is essential when understanding how complex reactions can be broken down into simpler steps.
In the realm of organic chemistry, this mechanism serves as a classic example of radical reactions where transient species, such as free radicals, play a pivotal role. Studying such mechanisms helps in understanding the dynamics of chemical processes and allows chemists to manipulate these reactions for various applications, such as synthesis of new compounds or materials.
Key Steps in the Rice-Herzfeld Mechanism:
  • Initiation of the radical process by homolytic bond dissociation.
  • Propagation where radicals react with stable molecules to form new radicals.
  • Termination when two radicals combine to form a stable product.
For students, grasping the Rice-Herzfeld mechanism's steps and recognizing how radical species influence reaction rates is crucial for mastering topics in chemical kinetics and organic chemistry.
Thermal Decomposition of Acetaldehyde
The thermal decomposition of acetaldehyde (\( \text{CH}_3\text{CHO} \text{(g)} \) is a chemical reaction where acetaldehyde is broken down into simpler molecules upon heating. This process is fundamental to both industrial chemistry and laboratory studies, where understanding the breakdown of compounds at high temperatures is necessary for various applications such as combustion, pyrolysis, and material synthesis.
Understanding the kinetics and mechanisms, such as the Rice-Herzfeld mechanism, behind the thermal decomposition of acetaldehyde helps us control the reaction conditions to favor the desired products. For instance, predicting the formation of methane (\( \text{CH}_4 \text{(g)} \) or ethane (\( \text{C}_2\text{H}_6 \text{(g)} \) during the decomposition process has significant implications in energy production and environmental impact assessment.
The breakdown of acetaldehyde involves radical intermediates, and the reaction is typically carried out in a controlled environment where temperature can affect the rate and products of decomposition. Students looking to specialize in physical chemistry or chemical engineering would benefit from a deep understanding of such reactions.
Rate of Methane Formation
The rate of methane formation from the decomposition of acetaldehyde is an essential concept for students exploring reaction kinetics within the context of energy-related processes. Methane, being a fundamental hydrocarbon, is widely used as a fuel and as a starting material for various chemical syntheses.
In the provided exercise, the steady-state approximation assists in simplifying the complex set of reactions to make it possible to calculate the rate of methane (\( \text{CH}_4 \text{(g)} \) formation mathematically. This approximation assumes that the concentration of intermediate radicals stays constant throughout the reaction. By applying this principle, it's possible to derive a mathematical expression that links the rate of methane formation to the concentrations of reactants and the rate constants of the intermediate steps.
This understanding is not only fundamental in academic settings but also critical in industries that rely on the production of methane and other hydrocarbons, where maximizing yield and efficiency are of utmost importance. Deep diving into this topic, students can draw connections between theoretical calculations and practical applications in real-world scenarios.
Chemical Kinetics
Chemical kinetics is the study of the rates of chemical processes and the factors affecting these rates. It's a cornerstone of both academic research and practical applications in chemical and pharmaceutical industries, environmental science, and materials engineering.
The Rice-Herzfeld mechanism and the study of methane formation from acetaldehyde decomposition serve as a model for teaching the principles of kinetics. Through this example, students encounter core concepts such as reaction mechanisms, rate laws, and the steady-state approximation. These principles enable the prediction and control of the speed at which reactions occur, which is vital for designing chemical reactors and optimizing production processes.
By understanding chemical kinetics, students gain a valuable toolset to analyze reactions critically, derive rate equations, and apply these concepts to novel situations. Notably, the ability to manipulate rates of reaction is instrumental in creating more efficient chemical processes, reducing waste, and improving safety in industrial settings.

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

Determine the predicted rate law expression for the following radical-chain reaction: \\[ \begin{array}{l} \mathrm{A}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{A} \\ \mathrm{A} \cdot \stackrel{k_{2}}{\longrightarrow} \mathrm{B} \cdot+\mathrm{C} \end{array} \\] $$\begin{array}{l} \mathrm{A} \cdot+\mathrm{B} \cdot \stackrel{k_{3}}{\longrightarrow} \mathrm{P} \\\ \mathrm{A} \cdot+\mathrm{P} \stackrel{k_{4}}{\rightarrow} \mathrm{B} \end{array}$$

The reaction of nitric oxide \((\mathrm{NO}(g))\) with molecular hydrogen \(\left(\mathrm{H}_{2}(g)\right)\) results in the production of molecular nitrogen and water as follows: \\[ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \\] The experimentally-determined rate-law expression for this reaction is first order in \(\mathrm{H}_{2}(g)\) and second order in \(\mathrm{NO}(g)\) a. Is the reaction as written consistent with the experimental order dependence for this reaction? b. One potential mechanism for this reaction is as follows: \\[ \begin{array}{l} \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} \\] Is this mechanism consistent with the experimental rate law? c. An alternative mechanism for the reaction is \\[ \begin{array}{l} 2 \mathrm{NO}(g) \frac{k_{1}}{\sum_{k-1}} \mathrm{N}_{2} \mathrm{O}_{2}(g) \text { (fast) } \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}_{2}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{H}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} \\] Show that this mechanism is consistent with the experimental rate law.

a. For the hydrogen-bromine reaction presented in Problem P36.7 imagine initiating the reaction with only \(\mathrm{Br}_{2}\) and \(\mathrm{H}_{2}\) present. Demonstrate that the rate law expression at \(t=0\) reduces to \\[ \left(\frac{d[\mathrm{HBr}]}{d t}\right)_{t=0}=2 k_{2}\left(\frac{k_{1}}{k_{-1}}\right)^{1 / 2}\left[\mathrm{H}_{2}\right]_{0}\left[\mathrm{Br}_{2}\right]^{1 / 2} \\] b. The activation energies for the rate constants are as follows: $$\begin{array}{cc} \text { Rate Constant } & \Delta \boldsymbol{E}_{a}(\mathbf{k J} / \mathbf{m o l}) \\ \hline k_{1} & 192 \\ k_{2} & 0 \\ k_{-1} & 74 \end{array}$$ c. How much will the rate of the reaction change if the temperature is increased to \(400 .\) K from \(298 \mathrm{K} ?\)

In a FRET experiment designed to monitor conformational changes in T4 lysozyme, the fluorescence intensity fluctuates between 5000 and 10,000 counts per second. Assuming that 7500 counts represents a FRET efficiency of \(0.5,\) what is the change in FRET pair separation distance during the reaction? For the tetramethylrhodamine/texas red FRET pair employed \(r_{0}=50 . \AA\)

Consider the collision-induced dissociation of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) via the following mechanism: \\[ \begin{array}{l} \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{N}_{2} \mathrm{O}_{5}(g) \frac{k_{1}}{\overline{k_{-1}}} \mathrm{N}_{2} \mathrm{O}_{5}(g)^{*}+\mathrm{N}_{2} \mathrm{O}_{5}(g) \\ \mathrm{N}_{2} \mathrm{O}_{5}(g) * \stackrel{k_{2}}{\longrightarrow} \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g) \end{array} \\] The asterisk in the first reaction indicates that the reactan is activated through collision. Experimentally it is found that the reaction can be either first or second order in \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) depending on the concentration of this species. Derive a rate law expression for this reaction consistent with this observation.

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.

Sign-up for free