KVPY2018ChemistryThermodynamics (C)
Nitroglycerine MW = 227 . 1 detonates according to the following equations, 2 C 3 H 5 NO 3 3 l ⟶ 3   N 2 g + 1 2 O 2 g + 6 CO 2 g + 5 H 2 O g The standard molar enthalpies of formation, ΔH f ° for all the compounds are given below. ΔH f ° C 3 H 5 NO 3 3 = - 364   kJ / mol ΔH f ° CO 2 g = - 393 . 5   kJ / mol ΔH f ° H 2 O g = - 241 . 8   kJ / mol Δ
Options
- A- 100 . 5   kJ / mol
- B- 62 . 5   kJ / mol
- C- 80 . 3   kJ / mol
- D- 74 . 9   kJ / mol
Correct answer
B. - 62 . 5   kJ / mol
Step-by-step solution
ΣH Reaction ° = ΣH f P ° - ΣH f ( R ) ° ΔH Reaction ° = 3 × 0 + 1 2 × 0 + 6 × - 393 . 5 + 5 × - 241 . 8 - 2 × - 364 = - 2361 - 1209 + 728 = - 2842 kJ / mol For 2 moles of nitroglycerine enthalpy - 2842 kJ / mol ∴ For 1 mole of nitroglycerine enthalpy = - 2842 2 = - 1421 kJ / mol 227 . 1 g of C 3 H 5 NO 3 3 has enthalpy - 1421 kJ / mol . ∴ 10 g of C 3 H 5 NO 3 3 has enthalpy = - 1421 227 . 1 × 10 = - 62 . 5 kJ / mol