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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?

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IthaloAbreu

Answer:

This solvent is not suitable because ΔS°vap > 0.

Explanation:

Let's consider a system at a higher temperature T1, and its surroundings at a lower temperature T2. Let's call q the heat that goes irreversible from the system to the surroundings:

ΔSsystem = -q/T1 (it's losing heat, so q must be negative)

ΔSsurroundings= q/T2

ΔSprocess = ΔSsystem + ΔSsurroundings

ΔSprocess = -q/T1 + q/T2

ΔSprocess = q*(1/T2 - 1/T1)

ΔSprocess = q*[(T1 - T2)/(T1*T2)]

As pointed above, T1> T2, so ΔSprocess > 0 and because of that, the reaction is spontaneous. It means that if ΔS°vap > 0, the solvent will vaporize. So, as we can notice, the solvent given is not suitable.