Answer :
Final answer:
At 55°C, the change in entropy of the surroundings (ΔSsurr) for the vaporization of ethanol is -117.7 J/K·mol, and the change in entropy of the universe (ΔSuniv) is -7.7 J/K·mol.
Explanation:
The student's question asks about the change in entropy of the surroundings (ΔSsurr) and the change in entropy of the universe (ΔSuniv) at 55°C when ethanol vaporizes.
To calculate these, we need to use the provided enthalpy of vaporization for ethanol (38.6 kJ/mol) and the entropy of vaporization for ethanol (110 J/K·mol).
First, remember that the entropy change of the surroundings is defined by the formula ΔSsurr = -ΔHsys / T, where ΔHsys is the enthalpy change of the system (ethylene vaporization in this case), and T is the temperature in Kelvin.
To convert the temperature from Celsius to Kelvin, add 273.15 to the Celsius temperature (55°C + 273.15 = 328.15 K). Now, plug in the values: ΔSsurr = -38,600 J/mol / 328.15 K = -117.7 J/K·mol.
The change in entropy of the universe is the sum of the entropy change of the system and surroundings:
ΔSuniv = ΔSsys + ΔSsurr. Given ΔSsys = 110 J/K·mol, we find ΔSuniv = 110 J/K·mol + (-117.7 J/K·mol) = -7.7 J/K·mol.
Therefore, the change in entropy of the surroundings is -117.7 J/K·mol and the change in entropy of the universe is -7.7 J/K·mol at 55°C when ethanol vaporizes.
