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

Substance	$$H_2$$	C (graphite)	$$C_2H_6(g)$$
$$\frac{\Delta_c H^\ominus}{kJ mol^{-1}}$$	$$-286.0$$	$$-394.0$$	$$-1560.0$$

The enthalpy of formation of ethane is

We need to find the enthalpy of formation of ethane: $$2C(\text{graphite}) + 3H_2(g) \to C_2H_6(g)$$

Using Hess's Law, the enthalpy of formation can be calculated from combustion data:

$$\Delta_f H^\circ = \sum \Delta_c H^\circ (\text{reactants}) - \Delta_c H^\circ (\text{product})$$

$$\Delta_f H^\circ (C_2H_6) = 2 \times \Delta_c H^\circ (C) + 3 \times \Delta_c H^\circ (H_2) - \Delta_c H^\circ (C_2H_6)$$

Substituting the given values:

$$\Delta_f H^\circ = 2 \times (-394.0) + 3 \times (-286.0) - (-1560.0)$$

$$= -788.0 - 858.0 + 1560.0$$

$$= -1646.0 + 1560.0$$

$$= -86.0 \text{ kJ mol}^{-1}$$

The correct answer is Option C: $$-86.0$$ kJ mol$$^{-1}$$.