A particular solvent with ΔS∘vap=112.9J/(K⋅mol)ΔSvap∘=112.9J/(K⋅mol) and ΔH∘vap=38.0kJ/molΔHvap∘=38.0kJ/mol is being considered

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3 years ago

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A particular solvent with ΔS∘vap=112.9J/(K⋅mol)ΔSvap∘=112.9J/(K⋅mol) and ΔH∘vap=38.0kJ/molΔHvap∘=38.0kJ/mol is being considered

for an experiment. In the experiment, which is to be run at 75 ∘C∘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?

Physics

1 answer:

Pachacha [2.7K] · Answer 1 · 2021-12-20T06:46:18+03:00

Answer:

Explanation:

The concept of gibb's free energy is applied in the problem. Mathematically from Gibb's free energy ; ΔG = ΔH - TΔS

Entropy = ΔS = is the degree of disorderliness of the system.

at equilibrium, ΔG = 0

as such. ΔH = TΔS

Since the enthalpy of vaporization is given ΔHvap = 38.0kJ/mol = 38000J/mol

Also ΔSvap = 112.9J/mol.K

Temperature of the experiment given = 75 degree celsius

calculate the boiling point of the solvent ; T = ΔHvap/ΔSvap

T = 38000J/mol / 112.9J/mol.K

= 335.57K , in celsius ; 335.57K - 273

= 62.57 degree celsius

the temperature calculated indicates that the solvent is not suitable for the experiment because its boiling point is lesser than the temperature at which the experiment want to be carried out.