Question 1

Why is radical fluorination of methane thermodynamically favorable but impractical?

Accepted Answer

Radical fluorination of methane has a large negative enthalpy change (ΔH = -431 kJ/mol), making it thermodynamically favorable. However, the reaction is highly exothermic and explosive, releasing extreme energy that makes it dangerous and unsuitable for laboratory or synthetic applications.

Question 2

What determines whether radical halogenation of alkanes is thermodynamically favorable?

Accepted Answer

For alkane halogenation, entropy change is negligible because reactant and product molecule counts are equal, so Gibbs free energy (ΔG) depends solely on enthalpy (ΔH). The strength of bonds broken versus bonds formed determines ΔH. A negative ΔH indicates a thermodynamically favorable reaction.

Question 3

Why is radical iodination of alkanes thermodynamically impossible?

Accepted Answer

Radical iodination has a positive enthalpy change (ΔH = +55 kJ/mol), resulting in a positive Gibbs free energy value. This positive ΔG makes the reaction thermodynamically unfavorable and prevents it from occurring under normal conditions, making iodination impossible to achieve.

Question 4

How do chlorination and bromination compare in terms of reaction rate?

Accepted Answer

Bromination is slower than chlorination. The first propagation step (hydrogen abstraction) is exothermic for chlorination with low activation energy, but endothermic for bromination with high activation energy. Despite bromination's negative overall ΔH (-33 kJ/mol), the endothermic first step limits its rate.

Question 5

What role does bond dissociation energy play in radical halogenation thermodynamics?

Accepted Answer

Bond dissociation energy determines the enthalpy change by comparing energy required to break bonds versus energy released when forming new bonds. Lower bond dissociation energies for bonds broken and higher energies for bonds formed produce more negative ΔH values, favoring thermodynamically favorable halogenation reactions.

Question 6

Why are only chlorination and bromination practically feasible among halogenation reactions?

Accepted Answer

Chlorination (ΔH = -104 kJ/mol) and bromination (ΔH = -33 kJ/mol) both have negative enthalpy changes, making them thermodynamically favorable. Fluorination is too explosive, and iodination is thermodynamically unfavorable. This makes chlorination and bromination the only practical radical halogenation methods for alkanes.

Question 7

How does the rate-determining step affect bromination versus chlorination?

Accepted Answer

The rate-determining step is hydrogen abstraction, the first propagation step. For chlorination, this step is exothermic with small activation energy, enabling fast reaction. For bromination, this step is endothermic with large activation energy, creating a kinetic barrier that slows the overall reaction despite favorable thermodynamics.

Halogenação Radical: Termodinâmica

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