Consider the thermodynamic data given below: The autoionization of water can be described according — Thermodynamics / Acid-Base Equilibrium Chemistry Question
Problem Context
Consider the thermodynamic data given below:
The autoionization of water can be described according to the equation below. Its equilibrium constant, Kw, is 1.0 × 10–14 at 25 ºC.
H2O(l) ⇌ H+(aq) + OH–(aq)
Calculate ∆Hº for the autoionization of water.
Model Answer
∆Hº = (–229.9 kJ/mol) – (–285.8 kJ/mol) = +55.9 kJ/mol
Calculate ∆Gº (at 298 K) for the autoionization of water.
Model Answer
∆Gº = –RTlnKeq = –(8.314 J/mol•K)(298 K)ln(1.0 × 10-14) = +79.9 kJ/mol
Calculate ∆Sº for the autoionization of water and rationalize its sign.
Model Answer
∆Gº = ∆Hº – T∆Sº
79.9 kJ/mol = 55.9 kJ/mol – (298 K)(∆Sº)
∆Sº = –80.5 J/mol•K
The entropy change is negative because the ions strongly order the solvent molecules around them (much more so than the neutral water).
Calculate Sº for OH–(aq).
Model Answer
–80.5 J/mol•K = (Sº of OH– (aq)) – (69.95 J/mol•K)
Sº of OH– (aq) = –10.6 J/mol•K
Calculate Kw at 50 ºC.
Model Answer
At 50 ºC, ∆Gº = 55.9 kJ/mol – (323 K)(–80.5 J/mol•K) = 81.9 kJ/mol
ln(Kw) = –∆Gº/RT = (–81900 J/mol)/(8.314 J/mol•K)(323 K) = –30.5
Kw = 5.7 × 10-14