Question

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:

Answer 1

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

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

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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.