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) 0.104 J/(mol middot K)
B) 104 J/(mol middot K)
C) 545 J/(mol middot K)
D) -35.2 J(mol middot K)
E) 0.545 J(mol middot K)

B) 104 J/mol.K

Explanation:

Step 1: Given data

Step 2: Convert "Tb" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 64.6°C + 273.15

K = 337.8 K

Step 3: Calculate the standard change in entropy for the vaporization of methanol (ΔS°vap)

The vaporization is the phase transition from the liquid phase to vapor. We can calculate the standard change in entropy for the vaporization using the following expression.

ΔS°vap = ΔH°vap / Tb

ΔS°vap = (35.2 × 10³ J/mol) / 337.8 K

ΔS°vap = 104 J/mol.K

wher is the rest of the question