Answer :
Final answer:
To find the entropy change for the vaporization of 20.0 g of ethanol, convert the mass to moles, use the enthalpy of vaporization for the total energy, and divide by the temperature in Kelvin to get Delta S sys ≈ 47.5 J/K.
Explanation:
To calculate the entropy change (Delta S sys) for the vaporization of 20.0 g of ethanol at 79.4 °C, we must first find the molar quantity of ethanol vaporized and then use the given enthalpy of vaporization.
First, convert the mass of ethanol to moles using its molecular weight (MW):
- Number of moles = Mass / MW
- Number of moles = 20.0 g / 46.07 g/mol
- Number of moles ≈ 0.434 mol
Since the enthalpy (Delta H) required to vaporize 1 mole of ethanol is 38.6 kJ, we can find the total enthalpy for 0.434 mol.
- Total Delta H = 38.6 kJ/mol * 0.434 mol
- Total Delta H ≈ 16.749 kJ
Then, we use the formula Delta S = Delta H / T, where T must be in Kelvin:
- T = 79.4 °C + 273.15 = 352.55 K
- Delta S sys = Total Delta H / T
- Delta S sys ≈ 16.749 kJ / 352.55 K
- Delta S sys ≈ 0.0475 kJ/K or 47.5 J/K
Therefore, the Delta S sys for the vaporization of 20.0 g of ethanol at 79.4 °C is approximately 47.5 J/K.
