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It takes 38.6 kJ of energy to vaporize 1.00 mol of ethanol (molecular weight: 46.07 g/mol). What will be the change in entropy of the system (\(\Delta S_{\text{sys}}\)) for the vaporization of 18.5 g of ethanol at 80.0 °C?

Answer :

Final answer:

The change in entropy (∆Ssys) for the vaporization of 18.5 g of ethanol at 80.0 °C is 43.84 J/K, calculated using the energy required to vaporize 1.00 mol of ethanol (38.6 kJ) and the temperature in Kelvin.

Explanation:

The change in entropy (∆Ssys) for the vaporization of 18.5 g of ethanol at 80.0 °C, we must use the given energy required to vaporize 1.00 mol of ethanol, which is 38.6 kJ. First, we need to convert the mass of ethanol to moles using its molar mass (46.07 g/mol). Next, the energy required for vaporizing this amount of ethanol is calculated. Finally, we'll use the formula ∆S = qrev/T to find the entropy change, where qrev is the heat involved in the reversible process (in J, not kJ) and T is the temperature in Kelvin.

Step-by-step calculation:

Calculate moles: 18.5 g ethanol × (1 mol/46.07 g) = 0.401 mol ethanol.

  1. Calculate total energy required: 0.401 mol × (38.6 kJ/mol) = 15.48 kJ.
  2. Convert energy to joules: 15.48 kJ × 1000 J/kJ = 15480 J.
  3. Convert temperature to Kelvin: 80.0 °C + 273.15 = 353.15 K.
  4. Calculate entropy change: ∆Ssys = 15480 J / 353.15 K = 43.84 J/K.

Therefore, the change in entropy of the system (∆Ssys) for the vaporization of 18.5 g of ethanol at 80.0 °C is 43.84 J/K.

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