Question

The entropy of sublimation of a certain compound is 20.7 J K–1 mol–1 at its normal sublimation point of 101 °C. Calculate the vapour pressure (in torr) of the compound at 9 °C. Assume that the enthalpy of sublimation is constant.

Answer 1

A sublimation reaction is of the form:

X(solid) -> X(gas)

where X is some compound.

At some arbitrary temperature, the Gibbs free energy change for this reaction would be given by:

ΔG_sub = ΔH_sub - T*ΔS_sub

Furthermore, at the equilibrium sublimation temperature, To, the solid and gas are in equilibrium, so

ΔG°_sub = 0 = ΔH_sub - To*ΔS_sub

The equilibrium constant for this reaction is given by:

Keq = Px

where Px is the partial pressure (vapor pressure) of X.

The equilibrium constant is related to the Gibbs free energy change of the reaction by:

ΔG = -R*T*ln(Keq)

So, in this case,

ΔG_sub = -R*T*ln(Px) = ΔH_sub - T*ΔS_sub

ln(Px) = -ΔH_sub/(R*T) - ΔS_sub/R

I assume that the "normal sublimation point" refers to standard pressure conditions, i.e., 1 atm (760 torr). At the normal sublimation point, Px° = 1 atm.

ln(Px) - ln(Px°) = (-ΔH_sub/(R*T) - ΔS_sub/R) - (-ΔH_sub/(R*T°) - ΔS_sub/R)

Assuming ΔH_sub and ΔS_sub are independent of temperature:

ln(Px/1atm) = (-ΔH_sub/R)*(1/T - 1/T°)

Px(T) = (1 atm)*exp[(ΔH_sub/R)*(1/T° - 1/T)]

This is the version of the Claussius-Clapeyron relevant to this problem. We know T and T°, so we need to figure out what ΔH_sub is from the information given.

We are told that ΔS_sub = 20.7 J/(mol*K) at 101°C = 376K, and we know that under these conditions, ΔG_sub = 0, so:

0 = ΔH_sub - T°*ΔS_sub

ΔH_sub = (376 K)*(20.7 J/(mol*K)) = 7783.2 J/mol

The vapor pressure at 9°C = 284 K, is then given by:

Px(284K) = (1 atm)*exp[(7783.2 J/mol)/(8.314 J/(mol*K))*(1/376K - 1/284K)]

Px(284K) = 0.446 atm = 338.96 torr

Thank you