Respuesta :
Answer : The correct option is, (b) +115 J/mol.K
Explanation :
Formula used :
[tex]\Delta S=\frac{\Delta H_{vap}}{T_b}[/tex]
where,
[tex]\Delta S[/tex] = change in entropy
[tex]\Delta H_{vap}[/tex] = change in enthalpy of vaporization = 40.5 kJ/mol
[tex]T_b[/tex] = boiling point temperature = 352 K
Now put all the given values in the above formula, we get:
[tex]\Delta S=\frac{\Delta H_{vap}}{T_b}[/tex]
[tex]\Delta S=\frac{40.5kJ/mol}{352K}[/tex]
[tex]\Delta S=115J/mol.K[/tex]
Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K
The entropy of vaporization is the ratio of the heat of vaporization to that of the boiling point. The standard entropy of vaporization of ethanol will be +115 J/mol.K.
What is entropy?
Entropy is the amount of unavailable energy that is equal to the thermal energy of the system per unit of temperature.
The change in entropy can be calculated as:
[tex]\rm \Delta S = \dfrac{\Delta H_{vap}}{T_{b}}[/tex]
Here, change in enthalpy of vaporization = 40.5 kJ/mol and boiling point temperature = 352 K
Substituting values in the above equation:
[tex]\begin{aligned} \rm \Delta S &= \rm \dfrac{\Delta H_{vap}}{T_{b}}\\\\&= \dfrac{40.5}{352}\\\\&= 115\;\rm J/mol.K \end{aligned}[/tex]
Therefore, option b. +115 J/mol.K is the standard entropy of vaporization.
Learn more about the entropy of vaporization here:
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