Question

Calculate the standard entropy of vaporization of ethanol at its boiling point, 352 K. The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ/mol.
a. +40.5 J/mol K
b. +115 J/mol K
c. -40.5 J/mol K
d. -115 J/mol K

Answer 1

Answer : The correct option is, (b) +115 J/mol.K

Explanation :

Formula used :

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

where,

$\Delta S$ = change in entropy

$\Delta H_{vap}$ = change in enthalpy of vaporization = 40.5 kJ/mol

$T_b$ = boiling point temperature = 352 K

Now put all the given values in the above formula, we get:

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

$\Delta S=\frac{40.5kJ/mol}{352K}$

$\Delta S=115J/mol.K$

Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K