Calculate the standard enthalpy change of following reaction. $\mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}+3 \mathrm{O}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+2 \mathrm{H}_2 \mathrm{O}_{(\ell)}$ If $\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{C}_2 \mathrm{H}_4\right)=-52 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $\begin{aligned} \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right) & =-390 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right) & =-286 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$

The standard enthalpy change, $\Delta_{\mathrm{r}} H^{\circ}$, is calculated using the formula:

$\Delta_{\mathrm{r}} H^{\circ} = \sum n \Delta_{\mathrm{f}} H^{\circ}(\text{products}) - \sum m \Delta_{\mathrm{f}} H^{\circ}(\text{reactants})$

Given the reaction $\mathrm{C}_2 \mathrm{H}_{4(\mathrm{g})} + 3 \mathrm{O}_{2(\mathrm{g})} \longrightarrow 2 \mathrm{CO}_{2(\mathrm{g})} + 2 \mathrm{H}_2 \mathrm{O}_{(\ell)}$ and the standard enthalpies of formation: $\Delta_{\mathrm{f}} H^{\circ}(\mathrm{C}_2 \mathrm{H}_4) = -52 \ \mathrm{kJ \ mol}^{-1}$, $\Delta_{\mathrm{f}} H^{\circ}(\mathrm{CO}_2) = -390 \ \mathrm{kJ \ mol}^{-1}$, $\Delta_{\mathrm{f}} H^{\circ}(\mathrm{H}_2 \mathrm{O}) = -286 \ \mathrm{kJ \ mol}^{-1}$, with $\Delta_{\mathrm{f}} H^{\circ}(\mathrm{O}_2) = 0$.

Substituting into the formula:

$\Delta_{\mathrm{r}} H^{\circ} = [2 \times (-390) + 2 \times (-286)] - [1 \times (-52) + 3 \times 0] = [-780 - 572] - [-52 + 0] = -1352 - (-52) = -1300 \ \mathrm{kJ}$

Final answer: $\boxed{\text{B}}$