The enthalpy of combustion ($\Delta_{c}H^{\circ}$) is calculated using the standard molar enthalpies of formation ($\Delta_{f}H^{\circ}$) of the products and reactants . First, we write the balanced thermochemical equation for the complete combustion of ethene ($C_{2}H_{4}$): $C_{2}H_{4}(g) + 3O_{2}(g) \rightarrow 2CO_{2}(g) + 2H_{2}O(l)$ .

According to thermodynamics, the reaction enthalpy is the sum of the enthalpies of the products minus the sum of the enthalpies of the reactants: $\Delta_{r}H^{\circ} = [2 \times \Delta_{f}H^{\circ}(CO_{2}, g) + 2 \times \Delta_{f}H^{\circ}(H_{2}O, l)] - [\Delta_{f}H^{\circ}(C_{2}H_{4}, g) + 3 \times \Delta_{f}H^{\circ}(O_{2}, g)]$ .

By convention, the standard enthalpy of formation for an element in its reference state, such as $O_{2}(g)$, is taken as zero .

Substituting the values provided in the problem: $\Delta_{c}H^{\circ} = [2(-394 \text{ kJ mol}^{-1}) + 2(-286 \text{ kJ mol}^{-1})] - [52 \text{ kJ mol}^{-1} + 3(0)]$ $\Delta_{c}H^{\circ} = [-788 - 572] - 52$ $\Delta_{c}H^{\circ} = -1360 - 52 = -1412 \text{ kJ mol}^{-1}$.

The negative sign confirms that combustion is an exothermic process . This result matches Option B.