1. Write the formation reaction of ethane

The standard heat of formation of ethane corresponds to:

$2C(\text{graphite}) + 3H_2(g) \rightarrow C_2H_6(g)$

We need to find:

$\Delta_f H^\circ_{298}(C_2H_6)$

2. Use Hess's law with combustion data

Given heats of combustion:

• Ethane: $C_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O \qquad \Delta H = -1560\,\text{kJ/mol}$

• Graphite: $C + O_2 \rightarrow CO_2 \qquad \Delta H = -393.5\,\text{kJ/mol}$

• Hydrogen: $H_2 + \frac{1}{2}O_2 \rightarrow H_2O \qquad \Delta H = -286\,\text{kJ/mol}$

3. Find enthalpy change for converting elements to combustion products

From elements needed to form ethane:

$2C + 3H_2 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O$

Its enthalpy is:

$\Delta H = 2(-393.5) + 3(-286)$

$= -787 - 858 = -1645\,\text{kJ/mol}$

4. Relate this to formation of ethane and then combustion of ethane

This same overall reaction can be written in two steps:

• Formation of ethane: $2C + 3H_2 \rightarrow C_2H_6 \qquad \Delta H = \Delta_f H^\circ$

• Combustion of ethane: $C_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O \qquad \Delta H = -1560$

So,

$\Delta_f H^\circ + (-1560) = -1645$

Therefore,

$\Delta_f H^\circ = -1645 + 1560 = -85\,\text{kJ/mol}$

5. Final answer

$\boxed{-85}$

So the standard heat of formation of ethane is $-85\,\text{kJ/mol}$.