Problem: Consider the evaporation of methanol at 25.0 ˚C: CH3OH(l)  →  CH3OH(g).Find ΔGf at 25.0 ˚C under the given nonstandard conditions.i. PCH3OH = 150.0 mmHgii. PCH3OH = 100.0 mmHgiii. PCH3OH = 10.0 mmHg

FREE Expert Solution

G=G°+RTlnQ


Step 1:

CH3OH(l)  →  CH3OH(g)

ΔG°=ΔG°f, prod-ΔG°f, react

ΔG°=(1 mol CH3OH)-162.3 kJ1 mol CH3OH-(1 mol CH3OH)-166.6 kJ1 mol CH3OH

ΔG° = 4.3 kJ/mol

ΔG°=4.3 kJmol×103 J1 kJ

ΔG° = 4300 J/mol


Step 2:

G=G°+RTlnQG=G°+RTln(PproductPreactant)G=G°+RTln (PCH3OH(g))


i. PCH3OH = 150.0 mmHg

PCH3OH=150.0 mmHg×1 atm760 mmHg

PCH3OH = 0.1974 atm


G=G°+RTln (PCH3OH(g))G=4300 Jmol+(8.314 Jmol·K)(298.15 K)ln (0.1974)

ΔG = 277.66 J/mol


ii. PCH3OH = 100.0 mmHg

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Problem Details

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

Find ΔGf at 25.0 ˚C under the given nonstandard conditions.
i. PCH3OH = 150.0 mmHg
ii. PCH3OH = 100.0 mmHg
iii. PCH3OH = 10.0 mmHg

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