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Chapter 11: Problem 74

One reason suggested for the instability of long chains of silicon atoms is that the decomposition involves the transition state shown below: The activation energy for such a process is \(210 \space\mathrm{kJ} / \mathrm{mol}\), which is less than either the \(\mathrm{Si}-\mathrm{Si}\) or the \(\mathrm{Si}-\mathrm{H}\) bond energy. Why would a similar mechanism not be expected to play a very important role in the decomposition of long chains of carbon atoms as seen in organic compounds?

Short Answer

Expert verified
The main reason why a similar decomposition mechanism to that of silicon chains is not expected to play a significant role in the decomposition of long chains of carbon atoms in organic compounds is because carbon chains have higher bond energies and a higher activation energy for decomposition. This results in carbon chains being more stable and resistant to decomposition compared to silicon chains.

Step by step solution

01

Identify the decomposition process

The decomposition process of silicon chains involves a transition state with an activation energy of 210 kJ/mol. This activation energy is less than the Si-Si or Si-H bond energies. This fact implies that the decomposition process is easier/more feasible due to having less energy needed to break the bonds.
02

Discuss the stability of carbon chains

When comparing silicon chains to carbon chains in organic compounds, carbon chains are more stable due to having stronger C-C and C-H bonds and a lower activation energy for decomposition. Carbon chains have higher bond energies, which means that the energy required to break the interactions between the atoms is higher, leading to more stable structures.
03

Examine the decomposition mechanism of carbon chains

Similar to silicon chains, carbon chains can also undergo chemical reactions and decompositions. However, these processes usually involve different mechanisms than those in silicon chains, and often they require a higher activation energy. This higher activation energy makes the carbon chains more stable and resistant to decomposition.
04

Conclusion

A similar decomposition mechanism to that of silicon chains is not expected to play a very important role in the decomposition of long chains of carbon atoms in organic compounds. The reason is that carbon chains have higher bond energies and a higher activation energy for decomposition, making them more stable and resistant to decomposition compared to silicon chains.

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Most popular questions from this chapter

A reaction of the form $$aA \longrightarrow Products$$gives a plot of \(\ln [\mathrm{A}]\) versus time (in seconds), which is a straight line with a slope of \(-7.35 \times 10^{-3} .\) Assuming \([\mathrm{A}]_{0}=\) \(0.0100 M,\) calculate the time (in seconds) required for the reaction to reach \(22.9 \%\) completion.

A first-order reaction has rate constants of \(4.6 \times 10^{-2} \mathrm{s}^{-1}\) and \(8.1 \times 10^{-2} \mathrm{s}^{-1}\) at \(0^{\circ} \mathrm{C}\) and \(20 .^{\circ} \mathrm{C},\) respectively. What is the value of the activation energy?

How does temperature affect \(k,\) the rate constant? Explain.

You and a coworker have developed a molecule that has shown potential as cobra antivenin (AV). This antivenin works by binding to the venom (V), thereby rendering it nontoxic. This reaction can be described by the rate law $$\text { Rate }=k[\mathrm{AV}]^{1}[\mathrm{V}]^{1}$$ You have been given the following data from your coworker: $$[\mathrm{V}]_{0}=0.20 \space\mathrm{M}$$ $$[\mathrm{AV}]_{0}=1.0 \times 10^{-4} \space\mathrm{M}$$A plot of \(\ln [\mathrm{AV}]\) versus \(t\) (s) gives a straight line with a slope of \(-0.32 \mathrm{s}^{-1} .\) What is the value of the rate constant \((k)\) for this reaction?

The activation energy for a reaction is changed from \(184 \space\mathrm{kJ} /\) mol to \(59.0 \space\mathrm{kJ} / \mathrm{mol}\) at \(600 .\) K by the introduction of a catalyst. If the uncatalyzed reaction takes about 2400 years to occur, about how long will the catalyzed reaction take? Assume the frequency factor \(A\) is constant, and assume the initial concentrations are the same.
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