$\Delta$H_vap for ethanol is 38.7 kJ/mol at its boiling point (78^$\circ$C). What is $\Delta$S_surr when 1.00 mole of ethanol is boiled A) -310 J/K·mol B) -210 J/K·mol C) -110 J/K·mol D) -100 J/K·mol E) None of the above

Question

Quizplus · Accepted Answer

We know that ΔHvap = +38.7 kJ/mol (endothermic process as it requires energy to overcome intermolecular forces to go from liquid to gas). 
At constant pressure, ΔH = q 
q = ΔHvap 
q = +38.7 kJ/mol 
Since 1.00 mole of ethanol is boiled, the process is occurring at constant pressure, so q = qsurroundings = -qsystem. 
Therefore, qsurroundings = -38.7 kJ/mol = -38,700 J/mol 
Now, at constant temperature, ΔS = -qsurroundings/T 
ΔS = -(-38,700 J/mol)/(351.15 K) = +110.33 J/K·mol 
ΔS will be negative for the surroundings (exothermic process, heat flows from system to surroundings), so we must change the sign: 
ΔSsurroundings = -110.33 J/K·mol = -110 J/K·mol (rounded to nearest integer) 
Therefore, ΔSsurr = -110 J/K·mol. 
The correct answer is (C).

Δ\DeltaΔ Hvap for Ethanol Is 38 ∘\circ∘ C) What Is Δ\DeltaΔ Ssurr When 1

$\Delta$ H_vap for Ethanol Is 38 $\circ$ C) What Is $\Delta$ S_surr When 1