Aparticular 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)

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.

plasma - state of matter made up of electrically charged particles