A particular solvent with ΔS∘vap=112.9J/(K⋅mol) and ΔH∘vap=38.0kJ/mol is being considered for an experiment. In the experiment,

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A particular solvent with ΔS∘vap=112.9J/(K⋅mol) and ΔH∘vap=38.0kJ/mol is being considered for an experiment. In the experiment,

which is to be run at 75 ∘C, the solvent must not boil. Based on the overall entropy change associated with the vaporization reaction, would this solvent be suitable and why or why not? View Available Hint(s)

Chemistry

1 answer:

Dovator [93] · Answer 1 · 2022-08-15T08:30:15+03:00

Answer:

The solvent will be not be suitable because ΔS > 0

Explanation:

The entropy measures the disorder of the system, and by the second law of the thermodynamics, it's always increasing. So, a glass that broked (increased in disorder) does not intend to be fixed naturally.

So, when ΔS > 0, it means that the disorder in the system increased, the reaction occurred spontaneously. ΔSºvap is positive for that solvent, so it means that it will boil.