Question

The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate boiling point of 1-pentanol? 100 oC 375 oC 0 oC 25 oC

Answer 1

Approximately 100 °C.

Step-by-step explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

`\Delta S_(vap)=(\Delta H_(vap))/(T)`

We can solve for the temperature as follows:

`T=(\Delta H_(vap))/(\Delta S_(vap))`

Thus, with the proper units, we obtain:

`T=(55500J/mol)/(148J/(mol*K)) =375K\\\\T=102 \°C`

Hence, answer is approximately 100 °C.

Best regards.