JEE Main202224 Jun 2022Evening ShiftChemistryThermodynamics (C)Actual
At 25 ° C and 1 atm pressure, the enthalpies of combustion are as given below: Substance H 2 C (graphite) C 2 H 6 g Δ c H Θ kJmol - 1 - 286 . 0 - 394 . 0 - 1560 . 0 The enthalpy of formation of ethane is
Options
- A+ 54 . 0   kJ   mol - 1
- B- 68 . 0   kJ   mol - 1
- C- 86 . 0   kJ   mol - 1
- D+ 97 . 0   kJ   mol - 1
Correct answer
C. - 86 . 0   kJ   mol - 1
Step-by-step solution
Given (i) C 2 H 6 g + 7 2 O 2 g ⟶ 2 CO 2 g + 3 H 2 O l ΔH comb ° = - 1560 . 0   kJ / mole (ii) C s + O 2 g ⟶ CO 2 g ΔH comb ° = - 394 . 0   kJ / mole (ii) H 2 g + 1 2 O 2 g ⟶ H 2 O g ΔH comb ° = - 286 . 0   kJ / mole Target 2 C s + 3 H 2 g ⟶ C 2 H 6 g ΔH rx ° = ΔH f ° C 2 H 6 ,   g ΔH r ° = ΔH c ° reactant - ΔH c ° Product = 2 × - 394 + 3 - 286 - - 1560 = - 788 - 858 + 1560 = - 86 . 0