Question

At 25°C and 1 atm pressure, the enthalpies of combustion are as given below:

Substance	H2	C (graphite)	C2H6(g)
`(Δ_"c""H"°)/"kJ mol"^-1`	−286	−394	−1560

The enthalpy of the formation of ethane is:

पर्याय

• + 54 kJ mol⁻¹

• − 68 kJ mol⁻¹

• − 86 kJ mol⁻¹

• + 97 kJ mol⁻¹

Answer 1

− 86 kJ mol⁻¹

Explanation:

\[\ce{C2H6(g) + 7/2O2(g) -> 2CO2(g) + 3H2O(l)}\]

Heat of combustion = `∑Δ_"f""H"_"(products)" − ∑Δ_"f""H"_"(reactants)"`

Δ_cH(C₂H₆, g) = 2Δ_cH(C, graphite) + 3Δ_cH(H₂, g) − Δ_fH(O₂, g) − Δ_fH(C₂H₆, g)

⇒ −1560 = 2 (−394) + 3 (−286) − 0 − Δ_fH(C₂H₆, g)

⇒ Δ_fH(C₂H₆, g) = − 86 kJ mol⁻¹