Given :
$$\Delta H^{\ominus}_{\text{sub}}[C(\text{graphite})] = 710$$ kJ mol$$^{-1}$$
$$\Delta_{C-H} H^{\ominus} = 414$$ kJ mol$$^{-1}$$
$$\Delta_{H-H} H^{\ominus} = 436$$ kJ mol$$^{-1}$$
$$\Delta_{C=C} H^{\ominus} = 611$$ kJ mol$$^{-1}$$

The $$\Delta H_f^{\ominus}$$ for $$CH_2=CH_2$$ is _________ kJ mol$$^{-1}$$
(nearest integer value)

The standard enthalpy of formation $$\Delta H_f^{\ominus}$$ is defined for the reaction (all substances in their standard states):

$$2\,C(\text{graphite}) + 2\,H_2(g) \rightarrow CH_2=CH_2(g)$$ $$-(1)$$

Step 1 - Atomise the reactants.
To write the reaction entirely in terms of gaseous atoms, first convert graphite to gaseous carbon atoms and $$H_2(g)$$ to gaseous hydrogen atoms.

$$2\,C(\text{graphite}) \rightarrow 2\,C(g) \quad \Delta H = 2 \times 710 = 1420\ \text{kJ}$$

$$2\,H_2(g) \rightarrow 4\,H(g) \quad \Delta H = 2 \times 436 = 872\ \text{kJ}$$

Total energy required for atomisation of reactants:

$$\Delta H_{\text{atom,reactants}} = 1420 + 872 = 2292\ \text{kJ}$$ $$-(2)$$

Step 2 - Form the product molecule from gaseous atoms.
Ethene, $$CH_2=CH_2$$, contains one $$C=C$$ double bond and four $$C-H$$ single bonds.

Energy released on forming these bonds (negative sign because bond formation releases energy):

$$\Delta H_{\text{form,bonds}} = -\left[1 \times 611 + 4 \times 414\right]$$

$$\Delta H_{\text{form,bonds}} = -\left[611 + 1656\right] = -2267\ \text{kJ}$$ $$-(3)$$

Step 3 - Combine the two steps to obtain $$\Delta H_f^{\circ}$$.

Using Hess’s law:

$$\Delta H_f^{\ominus} = \Delta H_{\text{atom,reactants}} + \Delta H_{\text{form,bonds}}$$

Substituting from $$(2)$$ and $$(3)$$:

$$\Delta H_f^{\ominus} = 2292 - 2267 = 25\ \text{kJ mol}^{-1}$$

Therefore, the standard enthalpy of formation of ethene is $$\mathbf{+25\ kJ\ mol^{-1}}$$ (nearest integer value).