Calculate standard enthalpy change of reaction $\begin{aligned} & \mathrm{C}_2 \mathrm{H}_{2(\mathrm{~g})}+\frac{5}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_2 \mathrm{O}_{(\ell)} \text { if } \\ & \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right)=-393 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{H}_2 \mathrm{O}\right)=-286 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{C}_2 \mathrm{H}_2\right)=227 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$

The standard enthalpy change for the reaction is calculated using the enthalpy of formation formula:

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

For the combustion $\mathrm{C}_2 \mathrm{H}_{2(\mathrm{g})} + \frac{5}{2} \mathrm{O}_{2(\mathrm{g})} \rightarrow 2 \mathrm{CO}_{2(\mathrm{g})} + \mathrm{H}_2 \mathrm{O}_{(\ell)}$, substituting the given values yields:

$\Delta_{\mathrm{r}} \mathrm{H}^{\circ} = [2 \times (-393\ \mathrm{kJ}) + 1 \times (-286\ \mathrm{kJ})] - [1 \times (227\ \mathrm{kJ}) + \frac{5}{2} \times (0\ \mathrm{kJ})]$

Simplifying the expression:

$\Delta_{\mathrm{r}} \mathrm{H}^{\circ} = [-786 - 286] - [227 + 0] = -1072 - 227 = -1299\ \mathrm{kJ}$

The standard enthalpy change is $-1299\ \mathrm{kJ}$, corresponding to answer choice C.