Question

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)

Answer 1

Answer:

B) 104 J/mol.K

Explanation:

Step 1: Given data

• Standard enthalpy of vaporization (ΔH°vap): 35.2 kJ/mol

• Normal boiling point (Tb): 64.6 °C

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

Answer 2

The standard change in entropy for the vaporization of methanol at its normal boiling point is 104 J/mol.K.

How do we calculate entropy of vaporization?

Standard change in entropy for the vaporization will be calculated by using the below equation as:

ΔS°vap = ΔH°vap / Tb, where

ΔH°vap = standard enthalpy of vaporization of methanol = 35.2 kJ/mol

Tb = Boiling point of methanol = 64.6°C = 64.6°C + 273.15 = 337.8 K

On putting all these values on the above equation, we get

ΔS°vap =  (35.2 × 10³ J/mol) / 337.8 K

ΔS°vap = 104 J/mol.K

Hence option (B) is correct.

To know more about enthalpy of vaporization, visit the below link:

https://brainly.com/question/5602087