Answer to Question #345976 in General Chemistry for Kakarot

Question

The heat of vaporization of methanol, CH3OH, is 35.20 kJ/mol. Its boiling

point is 64.6°C. What is the change in entropy for the vaporization of

methanol?

Answer 1

Step 1 : given data

– standard enthalpy of vaporization (∆H°_vap) = 35,2 kJ/mol;

– normal boling point (T_b) = 64,6°C

Step 2 : Convert T_b to Kelvin

We will use the following expression.

K = °C + 273

K = 64,6 + 273 = 337,6 K

Step 3 : Calculate the standard change entropy for the vaporization of methanol (∆S°_vap)

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

∆S°_vap = ∆H°_vap / T_b

∆S°_vap = (35,2 * 10³ J/mol) / 337,6 K

∆S°_vap = 104,2 J/mol•K ≈ 104 J/mol•K

ANSWER: 104 J/mol•K