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A proposed mechanism for the thermal decomposition of acetaldehyde $$\mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \stackrel{k_{\text {obs }}}{\longrightarrow} \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}(\mathrm{g})$$ is $$\begin{array}{c} \mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \stackrel{k_{1}}{\longrightarrow} \mathrm{CH}_{3}(\mathrm{g})+\mathrm{CHO}(\mathrm{g}) \\ \mathrm{CH}_{3}(\mathrm{g})+\mathrm{CH}_{3} \mathrm{CHO}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CH}_{3} \mathrm{CO}(\mathrm{g}) \\ \mathrm{CH}_{3} \mathrm{CO}(\mathrm{g}) \stackrel{k_{3}}{\longrightarrow} \mathrm{CH}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \\ 2 \mathrm{CH}_{3}(\mathrm{g}) \stackrel{k_{4}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6} \end{array}$$ Is this reaction a chain reaction? If so, identify the initiation, propagation, inhibition, and termination step(s). Determine the rate laws for \(\mathrm{CH}_{4}(\mathrm{g}), \mathrm{CH}_{3}(\mathrm{g}),\) and \(\mathrm{CH}_{3} \mathrm{CO}(\mathrm{g}) .\) Show that if you assume the steady-state approximation for the intermediate species, \(\mathrm{CH}_{3}(\mathrm{g})\) and \(\mathrm{CH}_{3} \mathrm{CO}(\mathrm{g}),\) the rate law for methane formation is given by $$\frac{d\left[\mathrm{CH}_{4}\right]}{d t}=\left(\frac{k_{1}}{k_{4}}\right)^{1 / 2} k_{2}\left[\mathrm{CH}_{3} \mathrm{CHO}\right]^{3 / 2}$$

Step by step solution

01

Understanding Mechanism as Chain Reaction

A chain reaction involves an initiation step creating reactive intermediates, followed by propagation steps involving these intermediates, and possibly termination steps leading to stable products that do not further react. The decomposition of acetaldehyde involves: an initiation step [1] creating CH3 and CHO radicals, a propagation step [2] where CH3 radicals react with another CH3CHO to form CH4 and CH3CO, and a regeneration of CH3 from CH3CO in step [3]. Step [4] is a termination step where CH3 radicals combine to form C2H6.

02

Identifying Steps

- Initiation: Step 1, where CH3CHO decomposes into CH3 and CHO. - Propagation: Steps 2 and 3, where CH3 radicals react to produce CH4 and regenerate CH3 from CH3CO respectively. - Inhibition: Absent explicitly but presence can be inferred with radical recombination. - Termination: Step 4, where two CH3 radicals form C2H6.

03

Rate Laws for Species

The rate of CH4 formation mainly depends on step 2. It can be represented as \[ \frac{d\left[\mathrm{CH}_4\right]}{dt} = k_2 [\mathrm{CH}_3][\mathrm{CH}_3\mathrm{CHO}] \] For CH3: \[ \frac{d[\mathrm{CH}_3]}{dt} = k_1 [\mathrm{CH}_3\mathrm{CHO}] - k_2 [\mathrm{CH}_3][\mathrm{CH}_3\mathrm{CHO}] - 2k_4 [\mathrm{CH}_3]^2 \] For CH3CO: \[ \frac{d[\mathrm{CH}_3\mathrm{CO}]}{dt} = k_2 [\mathrm{CH}_3][\mathrm{CH}_3\mathrm{CHO}] - k_3 [\mathrm{CH}_3\mathrm{CO}] \]

04

Applying Steady-State Approximation

Assume that \( \frac{d[\mathrm{CH}_3]}{dt} = 0 \) and \( \frac{d[\mathrm{CH}_3\mathrm{CO}]}{dt} = 0 \). Thus, For CH3, the rate equation simplifies to: \[ k_1 [\mathrm{CH}_3\mathrm{CHO}] = k_2 [\mathrm{CH}_3][\mathrm{CH}_3\mathrm{CHO}] + 2k_4 [\mathrm{CH}_3]^2 \] Solving for [CH3], we find \[ [\mathrm{CH}_3] = \left(\frac{k_1 [\mathrm{CH}_3\mathrm{CHO}]}{2k_4}\right)^{1/2} \]. Similarly, for CH3CO: \[ [\mathrm{CH}_3\mathrm{CO}] = \frac{k_2 [\mathrm{CH}_3][\mathrm{CH}_3\mathrm{CHO}]}{k_3} \]

05

Deriving Final Rate Law for Methane

Substitute [CH3] from the steady-state condition into the rate law for CH4:\[ \frac{d\left[\mathrm{CH}_4\right]}{dt} = k_2 \left(\frac{k_1 [\mathrm{CH}_3\mathrm{CHO}]}{2k_4}\right)^{1/2} [\mathrm{CH}_3\mathrm{CHO}] \] This results in: \[ \frac{d\left[\mathrm{CH}_4\right]}{dt} = \left(\frac{k_1}{k_4}\right)^{1/2} k_2 [\mathrm{CH}_3\mathrm{CHO}]^{3/2} \] which is the given rate law for methane formation.

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Key Concepts

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

Thermal Decomposition

Thermal decomposition is a type of chemical reaction where a single compound breaks down into two or more simpler substances when heated. In the context of acetaldehyde (\( \mathrm{CH}_{3} \mathrm{CHO} \)), this reaction proceeds through a chain mechanism involving radicals.

• During the thermal decomposition, acetaldehyde breaks down into methane (\( \mathrm{CH}_{4} \)) and carbon monoxide (\( \mathrm{CO} \)).
• The process starts at high temperatures which supply the energy necessary to break the bonds within acetaldehyde.

This reaction is quite interesting because it creates highly reactive radicals like methyl (\( \mathrm{CH}_{3} \)) and formyl (\( \mathrm{CHO} \)). These radicals are crucial for subsequent steps in the reaction mechanism.

Rate Laws

Rate laws describe how the concentration of reactants affects the rate of a chemical reaction. They are mathematical representations that quantify the speed or velocity of a reaction.

• The rate law for the formation of methane (\( \mathrm{CH}_4 \)) is derived from the reaction mechanism.
• It is expressed as \( \frac{d[\mathrm{CH}_4]}{dt} = k_2 [\mathrm{CH}_3][\mathrm{CH}_3\mathrm{CHO}] \), indicating the dependence on both the methyl radical and acetaldehyde concentrations.

To determine the rate laws for the intermediates, such as methyl (\( \mathrm{CH}_3 \)) and acetyl (\( \mathrm{CH}_3\mathrm{CO} \)), same principles apply but consider their transient nature, typically using approximations like the steady-state.

Steady-State Approximation

The steady-state approximation is a useful tool in chemical kinetics, especially when dealing with complex mechanisms involving intermediates. It assumes that the concentration of certain intermediates remains constant throughout the reaction.

• This simplification arises because the rate of formation and the rate of consumption of the intermediate are equal.
• For the thermal decomposition of acetaldehyde, \( \mathrm{CH}_3 \) and \( \mathrm{CH}_3\mathrm{CO} \) radicals are treated under this assumption.

By applying the steady-state approximation, we can simplify the calculation of rate laws. For example, the concentration of \( \mathrm{CH}_3 \) is solved as \( [\mathrm{CH}_3] = \left(\frac{k_1 [\mathrm{CH}_3\mathrm{CHO}]}{2k_4}\right)^{1/2} \), which simplifies the rate expression for methane production.

Acetaldehyde

Acetaldehyde is a simple aldehyde with the chemical formula \( \mathrm{CH}_{3} \mathrm{CHO} \). It is an important reactant in the discussed decomposition reaction.

• Being volatile and flammable, acetaldehyde undergoes thermal decomposition at elevated temperatures.
• It serves as both a reactant and a source for intermediate radicals in the reaction mechanism.

Understanding the role of acetaldehyde in this mechanism aids in grasping how initial steps lead to radical formation. These radicals are critical in driving subsequent steps to ultimately produce stable products such as methane and carbon monoxide.