Ethanol has a heat of vaporization of 38.56kJ/mol and a normal boiling point of

Question

Ethanol has a heat of vaporization of 38.56kJ/mol and a normal boiling point of 78.4 C . Remember that the normal boiling point is the temperature at which the vapor pressure equals the atmospheric pressure where the word "normal" means assume the atmospheric pressure is 760 torr. So a normal boiling point of 78.4 C is telling you that the vapor pressure is 760 torr at 78.4 C .

What is the vapor pressure of ethanol at 19 C ? You'll be using the two point form of the van't Hoff equation for the vaporization of ethanol; this equation is also known as the Clausius-Clapeyron equation when it is specifically applied to vaporization reactions, for example with ethanol:

CH3CH2OH(l)CH3CH2OH(g)

Don't forget to convert all temperatures to Kelvin!

Explanation / Answer

The Clausius-Clapeyron equation is

where P1 = vapor pressure at absolute temperature T1 and P2 = vapor pressure at absolute temperature T2. Hvap is the enthalpy of vaporization and is assumed to remain constant over the temperature range T1 to T2.

Given P1 = 760 torr at T1 = 78.4C = (78.4 + 273) K = 351.4 K; Hvap = 38.56 kJ/mol and T2 = 19C = (19 + 273) K = 292 K, plug in values and write

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The vapor pressure of ethanol at 19C is 51.1 torr (ans).

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