Consider the following cases of standard enthalpy of reaction ($$\Delta H_{r}^{\circ}$$ in kJ $$mol^{-1}$$)
$$C_{2}H_{6}(g)+\frac{7}{2}O_{2}(g) \rightarrow 2CO_{2}(g)+3H_{2}O(1)\Delta H_{1}^{\circ}=-1550$$
$$C(graphite)+O_{2}(g)\rightarrow CO_{2}(g)$$ $$\Delta H_{2}^{\circ}=-393.5$$ The magnitude of $$\Delta H_{fC_{2}H_{6}(g)}^{\circ}$$ is_______
$$H_{2}(g)+\frac{1}{2}O_{2}(g)\rightarrow H_{2}O(1)$$ $$\Delta H_{3}^{\circ}=-286$$
$$kJ mol^{-1}$$ (Nearest integer).

We need to find the magnitude of the standard enthalpy of formation of ethane ($$C_2H_6$$) using Hess's Law.

The formation reaction of ethane from its elements in their standard states is:

$$2C(graphite) + 3H_2(g) \rightarrow C_2H_6(g) \quad \Delta H_f^{\circ} = ?$$

Reaction 1 (combustion of ethane): $$C_2H_6(g) + \frac{7}{2}O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$$, $$\Delta H_1^{\circ} = -1550$$ kJ/mol

Reaction 2 (combustion of carbon): $$C(graphite) + O_2(g) \rightarrow CO_2(g)$$, $$\Delta H_2^{\circ} = -393.5$$ kJ/mol

Reaction 3 (combustion of hydrogen): $$H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)$$, $$\Delta H_3^{\circ} = -286$$ kJ/mol

We can construct the formation reaction by combining the given reactions:

Target = 2 × (Reaction 2) + 3 × (Reaction 3) - (Reaction 1)

This works because:

- 2 × Reaction 2 gives: $$2C(graphite) + 2O_2 \rightarrow 2CO_2$$

- 3 × Reaction 3 gives: $$3H_2 + \frac{3}{2}O_2 \rightarrow 3H_2O$$

- Reverse of Reaction 1 gives: $$2CO_2 + 3H_2O \rightarrow C_2H_6 + \frac{7}{2}O_2$$

Adding these: $$2C + 3H_2 \rightarrow C_2H_6$$ (the $$O_2$$, $$CO_2$$, and $$H_2O$$ cancel out)

$$\Delta H_f^{\circ} = 2\Delta H_2^{\circ} + 3\Delta H_3^{\circ} - \Delta H_1^{\circ}$$

$$= 2(-393.5) + 3(-286) - (-1550)$$

$$= -787 - 858 + 1550$$

$$= -1645 + 1550 = -95 \text{ kJ/mol}$$

The magnitude is $$|\Delta H_f^{\circ}| = 95$$ kJ/mol.

The answer is 95.