Question

What is the entropy change of the surroundings
when 10.8 g of acetone (CH3COCH3)
condenses at its normal boiling of 56.2◦C?

Answer 1

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,

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

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