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.