Calculate the entropy change (J/K) for the vaporization of 18 g of a hydrocarbon (114 g/mole]), at its boiling point of 58.2°C. The enthalpy of vaporization of this hydrocarbon is 27 kJ/mol.

The entropy change for the vaporization of 18 grams of a hydrocarbon at its boiling point is 12.87 J/K.

Explanation:

Mass of hydrocarbon = 18 g

Molar mass of hydrocarbon = 114 g/mol

Moles of hydrocarbons =

The enthalpy of vaporization of hydrocarbon =

Heat required to heat 0.1579 moles of hydrocarbon = Q

Q = 4.263 kJ = 4,263 J ( 1 kJ = 1000 J)

Boiling point of hydrocarbon = T = 58.2°C = 58.2 + 273 K = 331.2 K

Entropy change for the vaporization of 18 g of a hydrocarbon:

The entropy change for the vaporization of 18 grams of a hydrocarbon at its boiling point is 12.87 J/K.

answer; 4 electrons;