Chemistry - Thermochemistry Question with Solution | TestHub
Estimate Δf H° [Pyridine(l)] (in kJ/mol) from the given data:
Data, Δf H° (kJ/mol): CH2=N-CH3(g) +44; CH3-NH-CH3(g) -18; cyclohexene(g) -37; cyclohexane(g) -156; piperidine(g) -50 (ring structures in the figure, in this order).
Magnitude of resonance energy of pyridine = 125 kJ/mol. ΔHvap, 298 K [Pyridine(l)] = 40 kJ/mol.
Answer:
Solution:
Hydrogenation of C=N: CH₂=N-CH₃ + H₂ → CH₃-NH-CH₃, ΔH = -18 - 44 = -62 kJ.
Hydrogenation of one C=C bond (e.g., cyclohexene → cyclohexane) has a ΔH = -156 - (-37) = -119 kJ.
For a hypothetical Kekulé pyridine (with two C=C bonds and one C=N bond, assuming no resonance) undergoing hydrogenation with 3H₂, forming piperidine, the enthalpy change would be:
ΔH = 2(-119) + (-62) = -300 kJ
Therefore, the enthalpy of formation for Kekulé pyridine in the gaseous state is Δf H(Kekulé, g) = -50 + 300 = +250 kJ/mol.
Actual pyridine (g) is more stable due to resonance energy. The resonance energy is 250 - 125 = 125 kJ/mol.
For pyridine in the liquid state, the value is 125 - 40 = 85 kJ/mol.
The final answer is 0085.
