The standard enthalpies of formation of  ${\mathrm{CO}}_2\left(\mathrm{g}\right),{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\text{ and }$${\text{ glucose(s) at 25}}^{\circ }\mathrm{C} are -400\mathrm{kJ}/\mathrm{mol},-300\mathrm{kJ}/\mathrm{mol} and -1300$ kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25° is

The standard enthalpy of the combustion of glucose can be calculated by the eqn. 
$$\begin{gathered} {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\left(\mathrm{s}\right)+6{\mathrm{O}}_2\left(\mathrm{g}\right)\to 6{\mathrm{CO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ {\mathrm{\Delta H}}_{\mathrm{C}}^{\mathrm{o}}=6\times {\mathrm{\Delta H}}_{\mathrm{f}}\left({\mathrm{CO}}_2\right)+6\times {\mathrm{\Delta H}}_{\mathrm{f}}\left({\mathrm{H}}_2\mathrm{O}\right)-{\mathrm{\Delta H}}_{\mathrm{f}}\left[{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right] \\ {\mathrm{\Delta H}}_{\mathrm{C}}^{\mathrm{o}}=6(-400)+6(-300)-(-1300) \\ {\mathrm{\Delta H}}_{\mathrm{C}}^{\mathrm{o}}=-2900\mathrm{kJ}/\mathrm{mol} \end{gathered}$$
For one gram of glucose, enthalpy of combustion 
$\Delta H^{\circ }=-\frac{2900}{180}=-16.11\mathrm{kJ}/\mathrm{gm}$