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

Answer:

Approximately 100 °C.

Explanation:

Hello,

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

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

We can solve for the temperature as follows:

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

Thus, with the proper units, we obtain:

$T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C$

Hence, answer is approximately 100 °C.

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