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:
The molar enthalpy for the evaporation of k-butanol at
#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:
1 Answer
#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
#dbarS = -((delbarV)/(delT))_PdP# .
Assuming k-butanol is an ideal gas,
#PbarV = RT -> barV = (RT)/P# ,
so
#intdbarS = -R int_(P_1)^(P_2)1/PdP#
As a result,
#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
#P_1 = 10^(0.8534)/("760 torr/atm") = "0.009389 atm"# .
The second pressure is given for the evolved gas upon vaporization as
So, the change in entropy for vaporization of liquid k-butanol at
#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