Answer :
The enthalpy of vaporization for ethanol is 38.6 kJ/mol, and the entropy of vaporization for ethanol is 110 J/(K*mol). What is ΔSsurr and ΔSuniv at 55°C, the correct option is c) ΔSsurr = -110 J/(K*mol), ΔSuniv = 38.6 kJ/mol.
To determine ΔSsurr (change in entropy of the surroundings) and ΔSuniv (change in total entropy) at 55°C for the vaporization of ethanol, we can use the following equations:
[tex]\[ \Delta S_{\text{surr}} = -\frac{\Delta H_{\text{vap}}}{T} \]\[ \Delta S_{\text{univ}} = \Delta S_{\text{surr}} + \frac{\Delta H_{\text{vap}}}{T} \][/tex]
Given:
[tex]\[ \Delta H_{\text{vap}} = 38.6 \, \text{kJ/mol} \]\[ \Delta S_{\text{vap}} = 110 \, \text{J/(K\cdot mol)} \]\[ T = 55 \, \text{°C} = 55 + 273.15 \, \text{K} \][/tex]
[tex]( \Delta S_{\text{surr}} \):\[ \Delta S_{\text{surr}} = -\frac{38.6 \, \text{kJ/mol}}{55 + 273.15 \, \text{K}} \]\[ \Delta S_{\text{surr}} \approx -0.116 \, \text{kJ/(K\cdot mol)} \][/tex]
So, the correct answer is:
c) [tex]\( \Delta S_{\text{surr}} = -0.116 \, \text{kJ/(K\cdot mol)}, \, \Delta S_{\text{univ}} \approx -0.002 \, \text{kJ/(K\cdot mol)} \)[/tex]
I apologize for the oversight in my previous response. The correct values for[tex]\( \Delta S_{\text{surr}} \) and \( \Delta S_{\text{univ}} \)[/tex] are as provided above.
The enthalpy of vaporization for ethanol is 38.6 kJ/mol, and the entropy of vaporization for ethanol is 110 J/(K*mol). What is ΔSsurr and ΔSuniv at 55°C?
a) ΔSsurr = -110 J/(K*mol), ΔSuniv = -38.6 kJ/mol
b) ΔSsurr = 110 J/(K*mol), ΔSuniv = 38.6 kJ/mol
c) ΔSsurr = -110 J/(K*mol), ΔSuniv = 38.6 kJ/mol
d) ΔSsurr = 110 J/(K*mol), ΔSuniv = -38.6 kJ/mol
