The enthalpy and entropy of vaporization of ethanol are 38.6 kJ/mol and 109.8 J/mol . K,respectively.What is the normal boiling point of ethanol,in $\circ$ C?

Question

Quizplus · Accepted Answer

The normal boiling point of a substance is the temperature at which its vapor pressure is equal to 1 atm. To calculate the normal boiling point of ethanol, we need to use the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R)((1/T1) - (1/T2))
where P1 and P2 are the vapor pressures at the two temperatures T1 and T2, ΔHvap is the enthalpy of vaporization, R is the gas constant, and ln is the natural logarithm.
We can assume that the vapor pressure at the normal boiling point is equal to 1 atm, and use the given enthalpy and entropy values to solve for the normal boiling point.
At the normal boiling point:
P1 = 1 atm
P2 = 101325 Pa
ΔHvap = 38.6 kJ/mol = 38600 J/mol
ΔSvap = 109.8 J/mol K
R = 8.314 J/mol K
Substituting these values into the equation and converting temperatures to Kelvin:
ln(101325/1) = (38600/8.314)((1/T1) - (1/373.15))
Simplifying and solving for T1:
T1 = 351.5 K or 78.4°C
Therefore, the normal boiling point of ethanol is 78.4°C or 173.1°F, which corresponds to choice B.

The Enthalpy and Entropy of Vaporization of Ethanol Are 38 ∘\circ∘

The Enthalpy and Entropy of Vaporization of Ethanol Are 38 $\circ$