NTA Abhyas JEE Main2020ChemistryThermodynamics (C)Practice
If Bond energies of C - C , C = C , C - H bonds are 83 Kcal, 140 Kcal & 99 Kcal resp. The calculate the heat of formation of benzene. H e a t o f a t o m i s t i o n o f C = 170.9 K c a l H e a t o f a t o m i s a t i o n o f H = 52.1 K c a l
Options
- A- 65 Kcal
- B- 70 Kcal
- C- 75 Kcal
- D- 80 Kcal
Correct answer
C. - 75 Kcal
Step-by-step solution
Aim: 6 C s + 3 H 2 g → C 6 H 6 g ΔH = ? For reactant Heat of atomization of 6 moles of C = 6 × 170.9 Hear of atomization of 6 moles of H = 6 × 52.1 For product Heat of formation of 6 moles of C - H bonds = - 6 × 99 Heat of formation of 3 moles of C - C bonds = - 3 × 83 Heat of formation of 3 moles of C = C bonds = - 3 × 140 On adding : ΔH = 6 × 170.9 + 6 × 52.1 + 6 × − 99 + 3 × − 83 + 3 × − 140 = − 75 Kcal .