Answer :
Final answer:
The change in entropy for the vaporization of ethanol is approximately 0.110 kJ mol⁻¹ K⁻¹.
Explanation:
To calculate the change in entropy for the vaporization of ethanol, we can use the equation:
ΔS = ΔH / T
Where:
- ΔS is the change in entropy
- ΔH is the enthalpy change
- T is the temperature in Kelvin
Given that the enthalpy change associated with the vaporization of ethanol is 38.6 kJ mol⁻¹ and the boiling point temperature is 78 °C, we need to convert the temperature to Kelvin:
T(K) = T(°C) + 273.15
So, T(K) = 78 + 273.15 = 351.15 K
Now, we can substitute the values into the equation:
ΔS = 38.6 kJ mol⁻¹ / 351.15 K
Calculating this, we get:
ΔS ≈ 0.110 kJ mol⁻¹ K⁻¹
Therefore, the change in entropy for the vaporization of ethanol is approximately 0.110 kJ mol⁻¹ K⁻¹.
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