Answer:
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 = [tex]\frac{18 g}{114 g/mol}=0.1579 mol[/tex]
The enthalpy of vaporization of hydrocarbon = [tex]\Delta H_{vap}=27 kJ/mol[/tex]
Heat required to heat 0.1579 moles of hydrocarbon = Q
[tex]Q=\Delta H_{vap}\times 0.1579 mol=27 kJ/mol\times 0.1579 mol=4.263 kJ[/tex]
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:
[tex]\frac{\Delta H_{vap}}{T}=\frac{4,263 J}{331.2 K}=12.87 J/K[/tex]
The entropy change for the vaporization of 18 grams of a hydrocarbon at its boiling point is 12.87 J/K.
