Question

The enthalpy of vaporization (ΔH°vap) of benzene is 30.7 kJ/mol at its normal boiling point of 353.3 K. What is ΔS°vap at this temperature? a. 383 J/(mol·K) b. 0.0115 J/(mol·K) c. 86.9 J/(mol·K) d. 0.087 J/(mol·K) e. 11.5 J/(mol·K)

Answer 1

Answer: The correct answer is Option c.

Explanation:

Vaporization is defined as the physical process in which liquid particles get converted to gaseous particles.

$Liquid\rightleftharpoons Gas$

The value of standard Gibbs free energy is 0 for equilibrium reactions.

To calculate $\Delta S^o_{vap}$ for the reaction, we use the equation:

$\Delta S^o_{vap}=\frac{\Delta H^o_{vap}}{T}$

where,

$\Delta S^o_{vap}$ = standard entropy change of vaporization

$\Delta H^o_{vap}$ = standard enthalpy change of vaporization = 30.7 kJ/mol = 30700 J/mol    (Conversion factor: 1 kJ = 1000 J)

T = temperature of the reaction = 353.3 K

Putting values in above equation, we get:

$\Delta S^o_{vap}=\frac{30700J/mol}{353.3K}=86.9J/(mol.K)$

Hence, the correct answer is Option c.