Calculate the standard enthalpy change for reaction, $\begin{aligned} & \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}_{(\ell)}+3 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+3 \mathrm{H}_2 \mathrm{O}_{(\ell)} \\ & \text {If,}\\ & \text {} \begin{aligned} \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right) & =-280 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right) & =-390 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right) & =-285 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} \end{aligned}$

The standard enthalpy change for the reaction

$\mathrm{C}_2\mathrm{H}_5\mathrm{OH}_{(\ell)} + 3\mathrm{O}_{2(\mathrm{g})} \rightarrow 2\mathrm{CO}_{2(\mathrm{g})} + 3\mathrm{H}_2\mathrm{O}_{(\ell)}$

is calculated using the formula
$\Delta H^{\circ}_{\mathrm{rxn}} = \sum n\Delta H^{\circ}_{\mathrm{f}}(\mathrm{products}) - \sum m\Delta H^{\circ}_{\mathrm{f}}(\mathrm{reactants})$

Using the given enthalpies of formation:
$\Delta_{\mathrm{f}}H^{\circ}(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}) = -280\,\mathrm{kJ\,mol^{-1}}$
$\Delta_{\mathrm{f}}H^{\circ}(\mathrm{CO}_2) = -390\,\mathrm{kJ\,mol^{-1}}$
$\Delta_{\mathrm{f}}H^{\circ}(\mathrm{H}_2\mathrm{O}) = -285\,\mathrm{kJ\,mol^{-1}}$
and noting that $\Delta_{\mathrm{f}}H^{\circ}(\mathrm{O}_2) = 0\,\mathrm{kJ\,mol^{-1}}$ for the elemental form.

The sum for products is
$2 \times (-390\,\mathrm{kJ}) + 3 \times (-285\,\mathrm{kJ}) = -780 - 855 = -1635\,\mathrm{kJ}$

The sum for reactants is
$1 \times (-280\,\mathrm{kJ}) + 3 \times (0\,\mathrm{kJ}) = -280\,\mathrm{kJ}$

The enthalpy change is therefore
$\Delta H^{\circ}_{\mathrm{rxn}} = -1635 - (-280) = -1355\,\mathrm{kJ}$

This value corresponds to option C among the given choices.