Given the reaction below at \( 25^{\circ} \mathrm{C} \), calculate the enthalpy of formation for sucrose. . \( 1 \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11(\mathrm{~s})}+12 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 12 \mathrm{CO}_{2(\mathrm{~g})}+11 \mathrm{H}_{2} \mathrm{O}_{(l)} \quad \mathrm{H}_{\mathrm{nx}}=-5648 \mathrm{~kJ} / \mathrm{mol} \) Enter the VALUE only (and watch your sign).

To calculate the enthalpy of formation for sucrose (\( \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \)), we can use the given reaction and the enthalpy change (\( \Delta H \)) for the combustion of sucrose. The reaction is: \[ 1 \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11(\mathrm{s})} + 12 \mathrm{O}_{2(\mathrm{g})} \rightarrow 12 \mathrm{CO}_{2(\mathrm{g})} + 11 \mathrm{H}_{2} \mathrm{O}_{(l)} \quad \Delta H = -5648 \mathrm{~kJ} \] The enthalpy of formation (\( \Delta H_f \)) of a compound is defined as the change in enthalpy when one mole of the compound is formed from its elements in their standard states. For the combustion reaction, the enthalpy change is negative because it is an exothermic reaction. To find the enthalpy of formation of sucrose, we can use the following relationship: \[ \Delta H_f (\text{sucrose}) = -\Delta H_{\text{combustion}} = -(-5648 \, \text{kJ}) = 5648 \, \text{kJ/mol} \] Thus, the enthalpy of formation for sucrose is: \[ \boxed{5648} \]