Equilibrium Constant

Question

BaCl2⋅H2O(s) loses water when it is heated in an oven:

BaCl2⋅H2O(s)↽−−⇀BaCl2(s) + H2O(g)

Δ𝐻∘=63.11 kJ/mol at 25 ∘C

Δ𝑆∘=+148 J/(K⋅mol) at 25 ∘C

Write the equilibrium constant for this reaction. Find the vapor pressure of H2O (𝑃H2O) above BaCl2⋅H2O(s) at 298 K.

And...

If Δ𝐻∘ and Δ𝑆∘ are not temperature dependent (a poor assumption), estimate the temperature at which 𝑃H2O above BaCl2⋅H2O(s) will be 1 bar.

Answer 1

∆Gº = ∆H -T∆S

∆Gº = 63.11 kJ/mol - (298K)(0.148 kJ/Kmol) = 63.11 - 44.10

∆Gº = 19.01 kJ/mol

∆Gº = -RT ln K

19.01 = -(0.008314)(298) ln K

ln K = 19.01 / (-2.48)

ln K = -7.67

K = 4.67x10^-4

K = 4.67x10^-4 = (H₂O) = 4.67x10^-4