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

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Correct answer: C. - 86 . 0kJ mol ^ - 1

Given (i) C _ 2 H _ 6 (g) + (7)/(2) O _ 2 (g) ⟶ 2 CO _ 2 (g) + 3 H _ 2 O(l) Δ H _ comb ^ ° = - 1560 . 0kJ / mole (ii) C(s) + O _ 2 (g) ⟶ CO _ 2 (g) Δ H _ comb ^ ° = - 394 . 0kJ / mole (ii) H _ 2 (g) + (1)/(2) O _ 2 (g) ⟶ H _ 2 O(g) Δ H _ comb ^ ° = - 286 . 0kJ / mole Target 2C(s) + 3 H _ 2 (g) ⟶ C _ 2 H _ 6 (g) Δ H _ rx ^ ° = Δ H _ f ^ ° ( C _ 2 H _ 6 , g) Δ H _ r ^ ° = Δ H _ c ^ ° (reactant) - Δ H _ c ^ ° (Product) = 2 × (- 394) + 3(- 286) - (- 1560) = - 788 - 858 + 1560 = - 86 . 0KJ / mole

Substance	${H}_{2}$	$C$ (graphite)	${C}_{2}{H}_{6}\left(g\right)$
$\dfrac{{\Delta}_{c}{H}^{Θ}}{{kJmol}^{- 1}}$	$- 286 . 0$	$- 394 . 0$	$- 1560 . 0$

At 25 ° C and 1 atm pressure, the enthalpies of combustion are as...

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