The different types of enthalpy changes are: (i) Enthalpy of combustion (ii) Enthalpy of vaporization (iii) Enthalpy of ionization (iv) Enthalpy of atomization Given the enthalpies of combustion: $$ \begin{aligned} \text{C(s)} + \text{O}_2\text{(g)} &\rightarrow \text{CO}_2\text{(g)}; \Delta H_1 = -406 \text{ kJ mol}^{-1} \\ \text{H}_2\text{(g)} + \frac{1}{2} \text{O}_2\text{(g)} &\rightarrow \text{H}_2\text{O(l)}; \Delta H_2 = -284 \text{ kJ mol}^{-1} \\ \text{C}_{12}\text{H}_{22}\text{O}_{11}\text{(s)} + 12 \text{O}_2\text{(g)} &\rightarrow 12 \text{CO}_2\text{(g)} + 11 \text{H}_2\text{O(l)}; \Delta H_3 = -5638.82 \text{ kJ mol}^{-1} \end{aligned} $$ Using the relation for enthalpy of combustion: $$ \Delta H_c = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants}) $$ Let $\Delta H_f$ be the enthalpy of formation of sugar ($\text{C}_{12}\text{H}_{22}\text{O}_{11}$). Since the enthalpy of formation of elements in their standard state is zero: $$ \begin{aligned} \Delta H_3 &= [12 \times \Delta H_f(\text{CO}_2) + 11 \times \Delta H_f(\text{H}_2\text{O})] - [\Delta H_f(\text{sugar}) + 12 \times \Delta H_f(\text{O}_2)] \\ -5638.82 &= [12 \times (-406) + 11 \times (-284)] - [\Delta H_f(\text{sugar}) + 0] \\ -5638.82 &= [-4872 - 3124] - \Delta H_f(\text{sugar}) \\ -5638.82 &= -7996 - \Delta H_f(\text{sugar}) \\ \Delta H_f(\text{sugar}) &= -7996 + 5638.82 \\ \Delta H_f(\text{sugar}) &= -2357.18 \text{ kJ mol}^{-1} \end{aligned} $$ The enthalpy of formation of sugar is $-2357.18 \text{ kJ mol}^{-1}$.