The enthalpy of vaporization for toluene (C₇H₈) is 380.00 kJ/kg, and its boiling point is 110.6 $\circ$ C. What is the entropy change when 0.75 mole of toluene vaporizes at 110.6 $\circ$ C? A)743 J/K B)139 J/K C)68.5 J/K D)36.3 J/K E)253 kJ/K

Question

Quizplus · Accepted Answer

From the given information:

Enthalpy of vaporization (ΔHvap) = 380.00 kJ/kg

Boiling point (Tb) = 110.6°C = 383.75 K

Number of moles of toluene vaporized (n) = 0.75 mol

To calculate the entropy change (ΔS) during the vaporization of toluene, we can use the following formula:

ΔS = ΔHvap / Tb * n

Substituting the given values, we get:

ΔS = 380.00 kJ/kg / 383.75 K * 0.75 mol

ΔS = 68.5 J/K

Therefore, the correct answer is (C).

The Enthalpy of Vaporization for Toluene (C7H8) Is 380 ∘\circ∘ C What Is the Entropy Change When 0

The Enthalpy of Vaporization for Toluene (C₇H₈) Is 380 $\circ$ C What Is the Entropy Change When 0