Question

1)

Consider the equilibrium of methanol vapor and the liquid.

CH3OH(l)↽−−⇀CH3OH(g)

Thermodynamic Table at 25 ∘C

Substance	ΔHf∘ (kJ/mol)	S∘ (J/mol‑K)	ΔGf∘ (kJ/mol)
CH3OH(l)	−239.2	126.8	−166.6
CH3OH(g)	−201.0	239.9	−162.3

What is the vapor pressure of the methanol at −30 ∘C?

Pvap=

atm

What is the vapor pressure of the methanol at 40 ∘C?

Pvap=

2)

Substance	ΔG°f(kJ/mol)
M2O(s)	−8.70
M(s)	0
O2(g)	0

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

M2O(s)↽−−⇀ 2M(s)+12O2(g)

What is the standard change in Gibbs energy for the reaction, as written, in the forward direction?

ΔGrxn°=

kJ/mol

What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?

K=

What is the equilibrium pressure of O₂(g) over M(s) at 298 K?

PO2=

Answer 1

Question 1.

Part i.

Boiling point of methanol (T1) = 64.7 ^oC = 64.7 + 273.15 = 337.85 K at the pressure (P1) = 1 atm

T2 = -30 + 273.15 = 243.15 K

The enthalpy of vaporization of methanol ($\Delta$H_vap) = -201 - (239.2) = 38.2 kJ/mol = 38200 J/mol

According to the Clausius-Clapeyron equation: ln(P2/P1) = $\Delta$H_vap/R (1/T1 - 1/T2)

i.e. ln(P2/1) = 38200/8.314 * (1/337.85 - 1/T2)

i.e. ln(P2) = -5.297

i.e. P2 = e^-5.297 = 0.00501 atm

Part ii.

T2 = 40 + 273.15 = 313.15 K

i.e. ln(P2/1) = 38200/8.314 * (1/337.85 - 1/313.15)

i.e. ln(P2) = -1.073

i.e. P2 = e^-1.073 = 0.34209 atm