Respuesta :
Answer: The entropy change of the surroundings will be -17.7 J/K mol.
Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol
Amount of Acetone given = 10.8 g
Number of moles is calculated by using the formula:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of acetone = 58 g/mol
Number of moles = [tex]\frac{10.8}{58}=0.1862moles[/tex]
If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then
0.1862 moles will have = [tex]\frac{32.3}{1}\times 0.1862=5.828kJ/mol[/tex]
To calculate the entropy change for the system, we use the formula:
[tex]\Delta S_{sys}=\frac{\Delta H_{vap}}{T(\text{ in K)}}[/tex]
Temperature = 56.2°C = (273 + 56.2)K = 329.2K
Putting values in above equation, we get
[tex]\Delta S_{sys}=\frac{5.828}{329.2}=0.0177kJ/Kmol=17.7J/Kmol[/tex] (Conversion Factor: 1 kJ = 1000J)
At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,
[tex]\Delta S_{system}+\Delta S_{surrounding}=0[/tex]
[tex]\Delta S_{surrouding}=-\Delta S_{system}=-17.7J/Kmol[/tex]
