Chemistry - Thermochemistry Question with Solution | TestHub
Calculate ΔHvap [CH3COOH(l)] in kJ/mol from the following data: ΔHSolution [KF·CH3COOH(s)] in glacial acetic acid = -3 kJ/mol; ΔHSolution [KF(s)] in glacial acetic acid = +35 kJ/mol; the strength of H-bond between F-(g) and CH3COOH(g) = +46 kJ/mol; lattice enthalpy of KF·CH3COOH(s) = +734 kJ/mol; lattice enthalpy of KF(s) = +797 kJ/mol.
Answer:
Solution:
The reaction is:
KF(s) + CH3COOH(l) tarrow KF · CH3COOH(s)
with enthalpy change Δ H1. Both KF(s) and KF\u00b7CH3COOH(s) dissolve to the same solution, so:
Δ H1 + (-3) = 35 ⇒ Δ H1 = +38 kJ/mol
Now consider the thermochemical cycle:
Δ H1 + 734 = 797 + Δ Hvap - 46
where KF(s) tarrow K+(g) + F-(g), the acid is vaporized, and hydrogen-bond formation releases 46 kJ/mol.
Substituting:
38 + 734 = 797 + Δ Hvap - 46
772 = 751 + Δ Hvap
Δ Hvap = 21 kJ/mol
The answer is 0021.
