Question

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.

Answer 1

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

Step-by-step explanation:

Mass of hydrocarbon = 18 g

Molar mass of hydrocarbon = 114 g/mol

Moles of hydrocarbons =
`(18 g)/(114 g/mol)=0.1579 mol`

The enthalpy of vaporization of hydrocarbon =
`\Delta H_(vap)=27 kJ/mol`

Heat required to heat 0.1579 moles of hydrocarbon = Q

`Q=\Delta H_(vap)* 0.1579 mol=27 kJ/mol* 0.1579 mol=4.263 kJ`

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:

`(\Delta H_(vap))/(T)=(4,263 J)/(331.2 K)=12.87 J/K`

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