Question

The normal boiling point of ethanol is 78.4 oC. Its enthalpy of vaporization is 38.6 kJ/mol. Estimate the vapor pressure of ethanol at 26.3 oC.

Answer 1

Answer: The vapor pressure of ethanol at
`26.3^(o)C` is 238.3 torr.

Step-by-step explanation:

Given:
`\Delta H_(vap)` = 38.6 kJ/mol

`T_(1) = 26.3^(o)C = (26.3 + 273) K = 299.3 K`

`T_(2) = 78.4^(o)C = (78.4 + 273) K = 351.4 K`

Formula used to calculate the vapor pressure of ethanol is as follows.

`ln(P_(2))/(P_(1)) = (\Delta H_(vap))/(R) [(1)/(T_(1)) - (1)/(T_(2))]\\`

Substitute the values into above formula as follows.

`ln(P_(2))/(P_(1)) = (\Delta H_(vap))/(R) [(1)/(T_(1)) - (1)/(T_(2))]\\ \\ln (760 torr)/(P_(1)) = (38600 J)/(8.314 J/mol K)[(1)/(299.3) - (1)/(351.4)]\\(760)/(P_(1)) = 3.18\\P_(1) = 238.3 torr`

Thus, we can conclude that the vapor pressure of ethanol at
`26.3^(o)C` is 238.3 torr.