Calculate the change in molar entropy and change in Gibbs' free energy when 1 mol of liquid butanol vaporizes at 25ºC to a gas that is at 0.001 atm?

The molar enthalpy for the evaporation of k-butanol at #25^@ "C"# is #"10.21 kcal/mol"#. The pressure of the saturated steam as a function of temperature is given by the following empirical expression:

#log(p) = -1971.7/(theta + 230) + 8.5856#,

where #p# is the pressure in #"mmHg"# and #theta# is the temperature in #""^@ "C"#.

Original question given here:
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1 Answer
Apr 23, 2018

#DeltabarS_"vap" = "18.62 J/mol"cdot"K"#

#DeltabarG_"vap" = "37.17 kJ/mol"cdot"K"#


For this, we are changing pressure at constant temperature. For the total derivative of the molar entropy as a function of temperature and pressure,

#dbarS(T,P) = ((delbarS)/(delP))_TdP + cancel(((delbarS)/(delT))_PdT)^(0)#

at constant temperature. From the Gibbs' free energy Maxwell relation,

#dbarG = -barSdT + barVdP#,

we know that #((delbarS)/(delP))_T = -((delbarV)/(delT))_P#, so that

#dbarS = -((delbarV)/(delT))_PdP#.

Assuming k-butanol is an ideal gas,

#PbarV = RT -> barV = (RT)/P#,

so #((delbarV)/(delT))_P = R/P# and:

#intdbarS = -R int_(P_1)^(P_2)1/PdP#

As a result, #DeltabarS# at constant temperature is given by:

#DeltabarS = -Rln(P_2/P_1)#

Now, the vapor pressure above the liquid is given as:

#log p = -(1971.7)/(theta + 230) + 8.5856#

where #p# is in #"mm Hg"# and #theta# is in #""^@ "C"#.

At #25^@ "C"#, this gives #log p = 0.8534#, so

#P_1 = 10^(0.8534)/("760 torr/atm") = "0.009389 atm"#.

The second pressure is given for the evolved gas upon vaporization as #"0.001 atm"#.

So, the change in entropy for vaporization of liquid k-butanol at #"298.15 K"# is:

#color(blue)(DeltabarS_"vap") = -"8.314472 J/mol"cdot"K"ln("0.001 atm"/"0.009389 atm")#

#= color(blue)("18.62 J/mol"cdot"K")#

At constant temperature,

#DeltaG = DeltaH - TDeltaS#,

so the change in Gibbs' free energy of vaporization is:

#color(blue)(DeltabarG_"vap") = "10.21 kcal/mol" xx "4.184 J"/"cal" - "298.15 K" cdot 18.62 xx 10^(-3) "kJ/mol"cdot"K"#

#= color(blue)("37.17 kJ/mol"cdot"K")#

This should make sense because the boiling point should be much higher than #25^@ "C"#, and thus this evaporation is nonspontaneous.