Question

Using the thermo. values listed in Appendix B (pg. A 5-7) of your textbook

CO(g): deltaH:-110.5(kJ/mol) deltaG:-137.2(kJ/mol) deltaS:197.5(J/mol*K)

CO2(g): deltaH: -393.5(kJ/mol). deltaG: -394.4(kJ/mol) deltaS: 213.7(J/mol*K)

CO2(aq) deltaH:-412.9(kJ/mol) deltaG: -386.2(kJ/mol) deltaS:121

H2(g): deltaH:0. delta G: 0. delta S:130.6   

(a) Calculate deltaS°, deltaH° & deltaG° for the vaporization of methanol, CH3OH:

(b) Is this phase transition spontaneous at room temperature (show your work) ?

(c) If the phase transition is not spontaneous at room temperature, at what higher temperature will it become spontaneous (show your work) ?

(d) Exactly what is the temperature that you calculated in part (c) called ?

Answer 1

Data from appendix B pg A 5 to A 7 (The values of enthalpy entropy and gibbs free energy given in the queston are not relevent to the solution)

Note: The "?" appearing denote delta (looks like a triangle)

H3COH (l) H3COH(g)

a)

i) ?H°rxn

?H°rxn = ? ?H°f (products) - ??H°f (reactants)

For given vaporization reaction

?H°vap = ??H°f H3COH(g) - ?H°f H3COH(l)

Using above thermodynamic table,

?H°vap = ?H°f H3COH(g) - ?H°f H3COH(l)

?H°vap = (-201.2)- (-238.6)

?H°vap = +37.4 kJ

?H°vap = + 37400 J

i) ?S°vap

?S°vap = ? S°f (products) - ?S°f (reactants)

For given vaporization reaction

?S°vap = ?S°f H3COH(g) - S°f H3COH(l)

Using above thermodynamic table,

?S°vap = 238 - 127

?S°vap = 111J/K.mol

iii) ?G°vap = ?

i) ?G°rxn

?G°rxn = ? ?G°f (products) - ??G°f (reactants)

For given vaporization reaction

?G°vap = ?G°f H3COH(g) - ?G°f H3COH(l)

Using above thermodynamic table,

?G°vap = (-161.9)- (-166.2)

?G°vap = +4.3 kJ

?G°vap = + 4300 J

b)

?G°vap = ? at 298 K (Room temperature)

Gibbs equation gives,

?G°vap = ?H°vap -T?S°vap

Assuming ?H°vap -T?S°vap do not change over a rang of temperature we have,

?G°vap = 37400 - 298*111.

?G°vap = +4300 J

?G°vap = +ve value

Hence at room temperature, vaporization of ethanol is a Non-spntaneous process.

c)

Let us find T = ? for which ?G°vap = 0 because above that temperature ?G°vap < 0 i.e. -ve value and process will be spontaneous.

Using ?G°vap = 0, ?H°vap =37400 J, ?S°vap = 111.0 J/K.mol T = ?

Using Gibbs equation,

0 = 37400 - T*111

T = 37400 / 111

T = 336.9 K

It meant at 337 K i.e. at 64.0 oC Vaporization of methanol will be a spontaneous process.

d)

This is nothing but the Boiling Point of Methanol.