Q.22 At 298 K, the bond dissociation energies of C–H, C–C and C=C are
415, 344 and 615 kJ mol−1, respectively.
The enthalpy of atomization of carbon is 717 kJ mol−1 and that of hydrogen is
218 kJ mol−1.
The heat of formation of naphthalene at 298 K is
________ kJ mol−1.
Given Data
- C–H bond energy = 415 kJ mol−1
- C–C bond energy = 344 kJ mol−1
- C=C bond energy = 615 kJ mol−1
- Atomization enthalpy of carbon = 717 kJ mol−1
- Atomization enthalpy of hydrogen = 218 kJ mol−1
Step 1: Molecular Formula of Naphthalene
Naphthalene has the molecular formula:
C10H8
Step 2: Number of Bonds in Naphthalene
| Bond Type | Number of Bonds |
|---|---|
| C–H | 8 |
| C–C (single) | 6 |
| C=C (double) | 5 |
Step 3: Total Bond Energy of Naphthalene
Energy of C–H bonds = 8 × 415 = 3320 kJ
Energy of C–C bonds = 6 × 344 = 2064 kJ
Energy of C=C bonds = 5 × 615 = 3075 kJ
Total bond energy released = 3320 + 2064 + 3075 = 8459 kJ
Step 4: Atomization Energy of Elements
Carbon atomization energy = 10 × 717 = 7170 kJ
Hydrogen atomization energy = 8 × 218 = 1744 kJ
Total atomization energy = 7170 + 1744 = 8914 kJ
Step 5: Heat of Formation Calculation
Heat of formation is calculated using:
ΔHf = (Energy required for atomization) − (Energy released during bond formation)
ΔHf = 8914 − 8459 = +455 kJ mol−1
Final Answer
Heat of formation of naphthalene at 298 K = +455 kJ mol−1
Conclusion
The heat of formation of naphthalene is positive because the total
atomization energy of carbon and hydrogen is greater than the energy
released during bond formation. This calculation is based on Hess’s
law and average bond energies.
