To find the entropy change when 6 moles of liquid acetone vaporizes, we can use the relationship between heat of vaporization and entropy change. The formula is:
ΔS = ΔH_{vap} / T
Where:
- ΔS is the change in entropy (in J/K),
- ΔH_{vap} is the heat of vaporization (in J/mol), and
- T is the absolute temperature (in Kelvin).
First, we need to convert the heat of vaporization from kJ/mol to J/mol:
ΔH_{vap} = 29.1 kJ/mol = 29100 J/mol
Next, convert the boiling point from Celsius to Kelvin:
T = 56 °C + 273.15 = 329.15 K
Now we can calculate the entropy change for 1 mole of acetone:
ΔS_{1 mol} = ΔH_{vap} / T = 29100 J/mol / 329.15 K ≈ 88.4 J/K
Since we are vaporizing 6 moles, we multiply this value by 6:
ΔS_{total} = 6 mol * 88.4 J/K = 530.4 J/K
Thus, the entropy change when 6 moles of liquid acetone vaporizes at 56 degrees Celsius is approximately 530.4 J/K.