Question

The boiling point of a liquid is 64 °c and the enthalpy change for the conversion of this liquid to the gas is 32.21 kj/mole. what is the entropy change for vaporization, δsvap?

Answer 1

Answer:

$\Delta S_{vap}=0.096\frac{kJ}{mol*K} =96\frac{J}{mol*K}$

Explanation:

Hello,

In this case, the entropy of vaporization (conversion from liquid to gas) is mathematically defined in terms of enthalpy and the boiling temperature in K as shown below:

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

Thus, for the given data we obtain:

$\Delta S_{vap}=\frac{32.21kJ/mol}{(64+273.15)K} \\\\\Delta S_{vap}=0.096\frac{kJ}{mol*K} =96\frac{J}{mol*K}$

Best regards.

Answer 2

Entropy change of vaporization is simply the ratio of enthalpy change and the temperature in Kelvin.

Temperature = 64 + 273.15 = 337.15 K

 

Hence,

δsvap = (32.21 kJ / mole) / 337.15 K

δsvap = 0.0955 kJ / mole K = 95.5 J / mole K