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

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

Answer 2

`\Delta S_(vap)=0.096(kJ)/(mol*K) =96(J)/(mol*K)`

Step-by-step 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)=(\Delta H_(vap))/(T_b)`

Thus, for the given data we obtain:

`\Delta S_(vap)=(32.21kJ/mol)/((64+273.15)K) \\\\\Delta S_(vap)=0.096(kJ)/(mol*K) =96(J)/(mol*K)`

Best regards.