Chemistry - Thermochemistry Question with Solution | TestHub
The hydrogen bond between F- and CH3COOH is very strong and its strength may be analysed by setting up a Born-Haber cycle with the following data (in kJ/mol): lattice energy of KF·CH3COOH, 734; enthalpy of vaporization of CH3COOH, 20; enthalpy of solution of KF, 35; solvation energy of K+(g), -325; solvation energy of F-, -389; enthalpy of formation of KF·CH3COOH(s) from KF(s) and CH3COOH(l), -25. Find the energy of the hydrogen bond between F- and CH3COOH in the gas phase (in kJ/mol).
Answer:
Solution:
The lattice enthalpy of KF, Δ Hsol, is given by the equation:
Δ Hsol = U - (solvation of K+ + F-)
This gives the lattice energy UKF = 35 + 325 + 389 = 749 kJ/mol.
Consider two routes for the reaction:
Route 1:
KF(s) + CH₃COOH(l) tarrow KF · CH₃COOH(s) (enthalpy change: -25 kJ/mol)
Then, KF · CH₃COOH(s) tarrow K+(g) + F- ⋯ HOOCCH₃(g) (enthalpy change: +734 kJ/mol)
The net enthalpy change for Route 1 is -25 + 734 = +709 kJ/mol.
Route 2:
KF(s) tarrow K+(g) + F-(g) (enthalpy change: +749 kJ/mol)
CH₃COOH(l) tarrow CH₃COOH(g) (enthalpy change: +20 kJ/mol)
F-(g) + CH₃COOH(g) tarrow F- ⋯ HOOCCH₃(g) (enthalpy change: -x kJ/mol)
Equating the net enthalpy changes for both routes:
709 = 749 + 20 - x
709 = 769 - x
x = 769 - 709
x = 60 kJ/mol
The value of x is 60 kJ/mol. The answer is 0060.
