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)

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.

shrutiagrawal1798 shrutiagrawal1798 · Answer 2 · 2022-02-11T07:48:33+01:00

The experiment to be run at 75 degree Celsius did not consider the particular solvent suitable.

Entropy is defined as the disorder of the molecules in the reaction. The reaction is processed with the value of positive entropy for more disorder and negative entropy for less disorder.

What will happen to the solvent with positive entropy change?

The given solvent has the value of +112.9 J/K.mol for the entropy change. The value indicates the solution to have higher disorder. The increased disorder for the reaction is the representation for the boiling of the solvent.

Thus, the solvent is not suitable for the experiment as it boils at 75 degree Celsius.

Learn more about change in entropy, here:

https://brainly.com/question/1301642