Question

Consider the evaporation of methanol at 25.0 ?C :
CH3OH(l)?CH3OH(g) .

Why methanol spontaneously evaporates in open air at 25.0 ?C : Methanol evaporates at room temperature because there is an equilibrium between the liquid and the gas phases. The vapor pressure is moderate (143 mmHg at 25.0 degrees centigrade), so a moderate amount of methanol can remain in the gas phase, which is consistent with the free energy values.

A.) Find ?G? at 25.0 ?C .

B.) Find ?G at 25.0 ?C under the following nonstandard conditions:
(i)PCH3OH= 158.0mmHg

C). Find ?G at 25.0 ?C under the following nonstandard conditions:
(ii) PCH3OH= 110.0mmHg

D.) Find ?G at 25.0 ?C under the following nonstandard conditions:
(iii) PCH3OH= 11.0mmHg

Answer 1

the given reaction is

Ch3OH(l) ---> Ch3OH (g)

A)

we know that

dGo = dGo products - dGo reactants

so

dGo = dGo Ch3OH(g) - dGo CH3OH(l)

dGo = -162.3 + 166.6

dGo = 4.3 kJ/mol

B)

now consider the reaction

Ch3Oh(l) ----> Ch3OH (g)

Kp = pCH3OH

we know that

dG = dGo + RT lnKp

Given

pCh3OH = 158 mm Hg = ( 158/760 ) atm

pCh3OH = 0.20789 atm

so

Kp = pCH3OH = 0.20789 atm

now

dG = dGo + RT lnKp

dG = ( 4.3 x 1000 ) + ( 8.314 x 298 x ln0.20789)

dG = 0.408 kJ/mol

so dGo under given conditions is 0.408 kJ/mol

C)

given

pCH3Oh = 110 mm Hg = ( 110/760) atm

pCH3OH =0.1447 atm

Kp = PcH3Oh = 0.1447

so

dG = dGo + RT lnKp

dG = ( 4.3 x 1000) + ( 8.314 x 298 x ln0.1447)

dG = -0.4887 kJ/mol

so

the dG value under given conditions is -0.4887 kJ/mol

D)

pCH3Oh = 11/ 760 atm

pCH3OH = 0.0014

dG = ( 4.3 x 1000 ) + ( 8.314 x 298 x ln 0.0014)

dG = -6.19 kJ/mol

so

the value of dG under given conditions is -6.19 kJ/mol