The normal boiling point of ethanol is 78.85 °C and the molar enthalpy of vaporization, AHvap, is 71.8 kJ/mol. What is the value of AS in J/K when 3.40 mol sys of ethanol (1) vaporizes at 78.85 °C? 5.21 x 107 J/K 694 J/K 0.204 J/K 0.694 J/K 204 J/K
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![**Educational Content on the Thermodynamic Properties of Ethanol**
The normal boiling point of ethanol is 78.85 °C, and the molar enthalpy of vaporization (ΔH_vap) is 71.8 kJ/mol. The problem presented is to determine the value of ΔS_sys, in J/K, for the vaporization of 3.40 mol of ethanol (l) at 78.85 °C.
Options:
- \(5.21 \times 10^7 \, \text{J/K}\)
- \(694 \, \text{J/K}\)
- \(0.204 \, \text{J/K}\)
- \(\textbf{0.694 \, \text{J/K}}\) (selected)
- \(204 \, \text{J/K}\)
**Detailed Explanation:**
The question involves calculating the change in entropy (ΔS_sys) for a given number of moles of ethanol upon vaporization at its boiling point. The entropy change can be determined using the relation:
\[
\Delta S_{sys} = \frac{\Delta H_{vap}}{T}
\]
Where:
- \(\Delta H_{vap}\) is the molar enthalpy of vaporization.
- \(T\) is the temperature in Kelvin.
To solve the problem, convert the temperature from Celsius to Kelvin:
\[ T = 78.85 + 273.15 = 352.00 \, \text{K} \]
Calculate the total enthalpy change for 3.40 mol:
\[ \Delta H_{total} = 3.40 \times 71.8 \times 10^3 \, \text{J/mol} \]
Then, use the equation to find \( \Delta S_{sys} \):
\[ \Delta S_{sys} = \frac{\Delta H_{total}}{T} \]
The answer should reflect one of the options provided, with \( \textbf{0.694 \, \text{J/K}} \) being the correct choice according to the selection.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F83e4d0e0-41e1-45c9-8283-0ae60fe430d6%2Fc4842c1b-706f-42fc-8976-d6baf2df7652%2Fk1fdq4s_processed.jpeg&w=3840&q=75)
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