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Chapter 27: Problem 48

A simplified mechanism for this reaction is $$\begin{array}{c} \text { electric spark }+\mathrm{H}_{2}(\mathrm{g}) \Longrightarrow 2 \mathrm{H}(\mathrm{g}) \\ \mathrm{H}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \stackrel{k_{1}}{\longrightarrow} \mathrm{OH}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \\ \mathrm{O}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{OH}(\mathrm{g})+\mathrm{H}(\mathrm{g}) \\ \mathrm{H}_{2}(\mathrm{g})+\mathrm{OH}(\mathrm{g}) \stackrel{k_{3}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}(\mathrm{g}) \\ \mathrm{H}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \stackrel{k_{4}}{\longrightarrow} \mathrm{HO}_{2}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \end{array}$$ A reaction that produces more molecules that can participate in chain- propagation steps than it consumes is called a branching chain reaction. Label the branching chain reaction(s), inititation reaction(s), propagation reaction(s), and termination reaction(s) for this mechanism. Use the following bond dissociation energies to evaluate the energy change for steps (2) and (3) $$\begin{array}{cc} \text { Molecule } & D_{0} / \mathrm{kJ} \cdot \mathrm{mol}^{-1} \\ \hline \mathrm{H}_{2} & 432 \\ \mathrm{O}_{2} & 493 \\ \mathrm{OH} & 424 \end{array}$$

Short Answer

Expert verified
Initiation: Reaction 1; Propagation: Reactions 2, 3, 4; Branching: Reaction 2; Termination: Reaction 5.

Step by step solution

01

Identify the Initiation Reaction

An initiation reaction is one that starts the reaction mechanism. In this case, the first reaction is initiated by an electric spark, which breaks down \( \mathrm{H}_2 \) into two hydrogen atoms (\( 2\mathrm{H}(\mathrm{g}) \)). This is the initiation step.
02

Identify the Propagation Reactions

Propagation reactions involve the formation and consumption of reactive intermediates (radicals). Reactions 2, 3, and 4 are propagation steps as they regenerate reactive species: \( \mathrm{H}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{OH}(\mathrm{g}) + \mathrm{O}(\mathrm{g}) \), \( \mathrm{O}(\mathrm{g}) + \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{OH}(\mathrm{g}) + \mathrm{H}(\mathrm{g}) \), and \( \mathrm{H}_{2}(\mathrm{g}) + \mathrm{OH}(\mathrm{g}) \rightarrow \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) + \mathrm{H}(\mathrm{g}) \).
03

Evaluate Energy Changes for Propagation Reactions

To determine if a reaction is a branching chain step, calculate the energy changes for reactions 2 and 3 using bond dissociation energies. For reaction 2: \( \mathrm{H} + \mathrm{O}_{2} \rightarrow \mathrm{OH} + \mathrm{O} \), the approximation is \( 493 - 424 = 69 \, \mathrm{kJ/mol} \) absorbed. For reaction 3: \( \mathrm{O} + \mathrm{H}_{2} \rightarrow \mathrm{OH} + \mathrm{H} \), the approximation is \( 432 - 424 = 8 \, \mathrm{kJ/mol} \) released. Both steps either absorb or release a small amount of energy, maintaining the reaction network's activity.
04

Identify the Branching Chain Reaction

A branching chain reaction produces more radicals than it consumes, thus accelerating the overall reaction. In the given mechanism, Reaction 2 is a branching chain step because it generates two radicals (one \( \mathrm{OH}(\mathrm{g}) \) and one \( \mathrm{O}(\mathrm{g}) \)) from one radical (one \( \mathrm{H}(\mathrm{g}) \)).
05

Identify the Termination Reaction

A termination reaction occurs when radicals are effectively removed from the system, thus decreasing the reaction's capability to propagate. Reaction 5 involves \( \mathrm{H} \left(\mathrm{g}\right) + \mathrm{O}_{2} \left(\mathrm{g}\right) + \mathrm{M} \left(\mathrm{g}\right) \rightarrow \mathrm{HO}_{2} \left(\mathrm{g}\right) + \mathrm{M} \left(\mathrm{g}\right) \), as it neutralizes an \( \mathrm{H} \) radical.

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Key Concepts

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

Initiation reaction
In a chemical mechanism, the initiation reaction serves as the starting point. It's the spark that lights the fuse, creating the first reactive species or radicals to kick off the chain mechanism. In the example of the given reaction mechanism, this is initiated by an electric spark. The spark provides the energy required to break the molecular hydrogen (\(\mathrm{H}_2\) into two hydrogen atoms (\(2\mathrm{H}(\mathrm{g})\)).
These hydrogen atoms are crucial as they are reactive intermediates that can participate in further reactions. Initiation reactions are vital because without them, a chain reaction cannot begin.
Propagation reaction
Propagation reactions are the ongoing "heartbeat" of a chain reaction. These reactions involve the consumption and regeneration of radicals, allowing the chain to continue. In the example mechanism, reactions 2, 3, and 4 fall under the propagation category.
  • Reaction 2: The reaction between hydrogen radical and oxygen molecule forms hydroxyl radical (\(\mathrm{OH}(\mathrm{g})\)) and a separate oxygen atom (\(\mathrm{O}(\mathrm{g})\)).
  • Reaction 3: Involves forming additional hydroxyl radicals with the continuous regeneration of hydrogen radicals.
  • Reaction 4: Further continues the cycle by more consumption, continuing to fuel subsequent reactions.

Through propagation, the cycle of reactions keeps the chemical process going efficiently, using a small number of radicals to transform larger amounts of reactive elements.
Branching chain reaction
A branching chain reaction is unique because it not only propagates the chain but also multiplies the number of active radicals involved. This acceleration in the reaction can exponentially increase the overall rate.
In the mechanism provided, Reaction 2 acts as a branching chain reaction. This particular reaction starts with one hydrogen radical but yields two different radicals: \(\mathrm{OH}(\mathrm{g})\) and \(\mathrm{O}(\mathrm{g})\). With more radicals produced than consumed, the reaction rapidly builds momentum, potentially leading to explosive outcomes if not controlled or limited.
Termination reaction
Termination reactions play the important role of halting the chain reactions by effectively removing reactive intermediates from the system. These reactions prevent the mechanism from continuing endlessly.
In the provided mechanism, reaction 5 exemplifies a termination step: \(\mathrm{H}(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) + \mathrm{M}(\mathrm{g}) \rightarrow \mathrm{HO}_2(\mathrm{g}) + \mathrm{M}(\mathrm{g})\). In this reaction, hydrogen radicals are neutralized by forming a less reactive molecule while the inert gas \(\mathrm{M}\) remains unchanged. The radicals are thus effectively "trapped" as non-reactive species, bringing the chain mechanism closer to its conclusion.
Bond dissociation energy
Bond dissociation energy is a key concept in understanding energy changes within a reaction mechanism. It refers to the energy needed to break a bond in a molecule, forming two separate radicals or atoms.
In the context of the reactions provided, calculating the energy changes in steps 2 and 3 can provide insights into the reaction feasibility and stability.
  • Reaction 2: Involves breaking the \(\mathrm{O}_2\) bond, requiring approximately 493 kJ/mol, while forming \(\mathrm{OH}\) and \(\mathrm{O}\) consumes around 424 kJ/mol. This means an additional 69 kJ/mol is absorbed, making it energy-intensive.
  • Reaction 3: Deals with breaking and forming of bonds that result in a small release of energy (8 kJ/mol). Here, bond energies confirm the small energy alteration aligns with the continuous nature of the propagation step.

Overall, understanding bond dissociation energies helps predict how energy is distributed or adjusted in reactions, influencing both speed and mechanism direction.

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Most popular questions from this chapter

The thermal decomposition of ethylene oxide occurs by the mechanism $$\begin{array}{l} \mathrm{H}_{2} \mathrm{COCH}_{2}(\mathrm{g}) \stackrel{k_{1}}{\longrightarrow} \mathrm{H}_{2} \mathrm{COCH}(\mathrm{g})+\mathrm{H}(\mathrm{g}) \\ \mathrm{H}_{2} \mathrm{COCH}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{CH}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \\ \mathrm{CH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{COCH}_{2}(\mathrm{g}) \stackrel{k_{3}}{\longrightarrow} \mathrm{H}_{2} \mathrm{COCH}(\mathrm{g})+\mathrm{CH}_{4}(\mathrm{g}) \\ \mathrm{CH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{COCH}(\mathrm{g}) \stackrel{k_{4}}{\longrightarrow} \text { products } \end{array}$$ Which of these reaction(s) are the initiation, propagation, and termination step(s) of the reaction mechanism? Show that if the intermediates \(\mathrm{CH}_{3}\) and \(\mathrm{H}_{2} \mathrm{COCH}\) are treated by the steady-state approximation, the rate law, \(d[\text { products }] / d t,\) is first order in ethylene oxide concentration.

Consider the following mechanism for the recombination of bromine atoms to form molecular bromine $$\begin{array}{c} 2 \operatorname{Br}(\mathrm{g}) \stackrel{k_{1}}{\rightleftharpoons} \operatorname{Br}_{2}^{*}(\mathrm{g}) \\ \mathrm{Br}_{2}^{*}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{Br}_{2}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \end{array}$$ The first step results in formation of an energized bromine molecule. This excess energy is then removed by a collision with a molecule \(\mathrm{M}\) in the sample. Show that if the steady-state approximation is applied to \(\mathrm{Br}_{2}^{*}(\mathrm{g}),\) then $$\frac{d[\mathrm{Br}]}{d t}=-\frac{2 k_{1} k_{2}[\mathrm{Br}]^{2}[\mathrm{M}]}{k_{-1}+k_{2}[\mathrm{M}]}$$ Determine the limiting expression for \(d[\mathrm{Br}] / d t\) when \(v_{2} \gg v_{-1} .\) Determine the limiting expression for \(d[\operatorname{Br}] / d t\) when \(v_{2} \ll v_{-1}\).

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 $$\frac{d\left[\mathrm{CH}_{4}\right]}{d t}=\left(\frac{k_{1}}{k_{4}}\right)^{1 / 2} k_{2}\left[\mathrm{CH}_{3} \mathrm{CHO}\right]^{3 / 2}$$

The rate law for the hydrolysis of ethyl acetate by aqueous sodium hydroxide at \(298 \mathrm{K}\) $$\mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \stackrel{k_{\mathrm{obs}}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{aq})$$ is $$\frac{d\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right]}{d t}=k_{\mathrm{obs}}\left[\mathrm{OH}^{-}\right]\left[\mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}\right]$$ Despite the form of this rate law, this reaction is not an elementary reaction but is believed to occur by the following mechanism $$\begin{array}{l} \mathrm{CH}_{3} \mathrm{CO}^{-}(\mathrm{OH}) \mathrm{OCH}_{2} \mathrm{CH}_{3}(\mathrm{aq}) \stackrel{k_{2}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}(\mathrm{aq}) \\ \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}(\mathrm{aq}) \stackrel{k_{3}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{aq}) \end{array}$$ Under what conditions does this mechanism give the observed rate law? For those conditions, express \(k_{\text {obs }}\) in terms of the rate constants for the individual steps of the reaction mechanism.

A mechanism for ozone creation and destruction in the stratosphere is $$\begin{array}{c} \mathrm{O}_{2}(\mathrm{g})+h v \stackrel{j_{1}}{\longrightarrow} \mathrm{O}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \\ \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \stackrel{k_{2}}{\longrightarrow} \mathrm{O}_{3}(\mathrm{g})+\mathrm{M}(\mathrm{g}) \\ \mathrm{O}_{3}(\mathrm{g})+h v \stackrel{j_{3}}{\longrightarrow} \mathrm{O}_{2}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \\ \mathrm{O}(\mathrm{g})+\mathrm{O}_{3}(\mathrm{g}) \stackrel{k_{4}}{\longrightarrow} \mathrm{O}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \end{array}$$ where we have used the symbol \(j\) to indicate that the rate constant is for a photochemical reaction. Determine the rate expressions for \(d[\mathrm{O}] / d t\) and \(d\left[\mathrm{O}_{3}\right] / d t .\) Assume that both intermediate species, \(\mathrm{O}(\mathrm{g})\) and \(\mathrm{O}_{3}(\mathrm{g}),\) can be treated by the steady-state approximation and thereby show that $$[\mathrm{O}]=\frac{2 j_{1}\left[\mathrm{O}_{2}\right]+j_{3}\left[\mathrm{O}_{3}\right]}{k_{2}\left[\mathrm{O}_{2}\right][\mathrm{M}]+k_{4}\left[\mathrm{O}_{3}\right]}$$ and $$\left[\mathrm{O}_{3}\right]=\frac{k_{2}[\mathrm{O}]\left[\mathrm{O}_{2}\right][\mathrm{M}]}{j_{3}+k_{4}[\mathrm{O}]}$$ Now substitute Equation 1 into Equation 2 and solve the resulting quadratic formula for \(\left[\mathrm{O}_{3}\right]\) to obtain $$\left[\mathrm{O}_{3}\right]=\left[\mathrm{O}_{2}\right] \frac{j_{1}}{2 j_{3}}\left\\{\left(1+4 \frac{j_{3}}{j_{1}} \frac{k_{2}}{k_{4}}[\mathrm{M}]\right)^{1 / 2}-1\right\\}$$ Typical values for these parameters at an altitude of \(30 \mathrm{km}\) are \(j_{1}=2.51 \times 10^{-12} \mathrm{s}^{-1}\) \(j_{3}=3.16 \times 10^{-4} \mathrm{s}^{-1}, k_{2}=1.99 \times 10^{-33} \mathrm{cm}^{6} \cdot\) molecule \(^{-2} \cdot \mathrm{s}^{-1}, k_{4}=1.26 \times 10^{-15} \mathrm{cm}^{3}\) molecule \(^{-1} \cdot \mathrm{s}^{-1}, \quad\left[\mathrm{O}_{2}\right]=3.16 \times 10^{17} \quad\) molecule \(\cdot \mathrm{cm}^{-3}, \quad\) and \(\quad[\mathrm{M}]=3.98 \times 10^{17}\) molecule \(\cdot \mathrm{cm}^{-3} .\) Find \(\left[\mathrm{O}_{3}\right]\) and \([\mathrm{O}]\) at an altitude of \(30 \mathrm{km}\) using Equations 1 and 2 Was the use of the steady-state assumption justified?
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