Question

The boiling point of ethanol is 78.4 °C, and the enthalpy change for the conversion of liquid to vapor is ΔHvap = 38.56 kJ/mol. What is the entropy change for vaporization, ΔSvap, in J/(K ⋅ mol)?

Answer 1

Explanation:

The given data is as follows.

    Enthalpy of vaporization, $\Delta H_{vap}$ = 38.56 kJ/mol

    Temperature (T) = $78.4^{o}C$

                                = (78.4 + 273) K

                                = 351.4 K

Now, we will calculate the change in entropy using the formula as follows.

           $\Delta H_{vap} = T \Delta S_{vap}$

           $\Delta S_{vap} = \frac{\Delta H_{vap}}{T}$

                       = $\frac{38.56 kJ/mol}{351.4 K}$

                       = 0.109 kJ/mol K

Thus, w ecan conclude that the entropy change for vaporization is 0.109 kJ/mol K.