The standard enthalpy of vaporization of methanol is 35.2 kJ/mol at its normal boiling point,64.6°C.What is the standard change in entropy for the vaporization of methanol at its normal boiling point?
A)104 J/(mol ∙ K)
B)35.2 J/(mol ∙ K)
C)0.544 J/(mol ∙ K)
D)0.104 J/(mol ∙ K)
E)544 J/(mol ∙ K)

Question

Quizplus · Accepted Answer

The standard formula for calculating the standard change in entropy for vaporization is:
ΔSvap = ΔHvap/T
where ΔHvap is the standard enthalpy of vaporization and T is the boiling point temperature in Kelvin.
Converting 64.6°C to Kelvin, we get:
T = 337.8 K
Substituting the given values, we get:
ΔSvap = 35.2 kJ/mol / 337.8 K
ΔSvap = 104 J/(mol*K)
Therefore, the best choice is A.

The Standard Enthalpy of Vaporization of Methanol Is 35