Consider the evaporation of methanol at 25.0 ?C : CH3OH(l)?CH3OH(g) . Why methan
ID: 868632 • Letter: C
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
Explanation / Answer
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