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Artist 52 [7]
3 years ago
5

What is the entropy change of the surroundings

Chemistry
1 answer:
KiRa [710]3 years ago
5 0

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:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of acetone = 58 g/mol

Number of moles = \frac{10.8}{58}=0.1862moles

If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then

0.1862 moles will have = \frac{32.3}{1}\times 0.1862=5.828kJ/mol

To calculate the entropy change for the system, we use the formula:

\Delta S_{sys}=\frac{\Delta H_{vap}}{T(\text{ in K)}}

Temperature = 56.2°C = (273 + 56.2)K = 329.2K

Putting values in above equation, we get

\Delta S_{sys}=\frac{5.828}{329.2}=0.0177kJ/Kmol=17.7J/Kmol   (Conversion Factor: 1 kJ = 1000J)

At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,

\Delta S_{system}+\Delta S_{surrounding}=0

\Delta S_{surrouding}=-\Delta S_{system}=-17.7J/Kmol

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