The standard enthalpies of formation of ${\mathrm{CO}}_{2\left(\mathrm{g}\right)},{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{\ell }\right)},\mathrm{glucos}{\mathrm{e}}_{\left(\mathrm{s}\right)}\text{ at }{25}^0\mathrm{C}$ are $-400\mathrm{kJ}/\text{ mole, }-300\mathrm{kJ}/\mathrm{mole}\text{, and }1300\mathrm{kJ}/\mathrm{mole}$respectively. The standard enthalpy of combustion per gram of glucose at 25° C is 

$\begin{array}{l}{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+6{\mathrm{O}}_2\to 6{\mathrm{CO}}_2+6{\mathrm{H}}_2\mathrm{O}\\ \left(\mathrm{s}\right) \left(\mathrm{g}\right) \left(\mathrm{g}\right) \left(\mathrm{\ell }\right)\\ {\mathrm{\Delta H}}_{\text{combustion }}=6\times {\mathrm{\Delta H}}_{\mathrm{f}}\left({\mathrm{CO}}_2\right)+6\times {\mathrm{\Delta H}}_{\mathrm{f}}\left({\mathrm{H}}_2\mathrm{O}\right)-{\mathrm{\Delta H}}_{\mathrm{f}}\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)\\ {\mathrm{\Delta H}}^0=6(-400)+6(-300)-(-1300)=-2900\mathrm{kJ}/\mathrm{mol}\end{array}$
For one gram of glucose enthalpy of
Combustion ${\mathrm{\Delta H}}^0=-\frac{2900}{180}=-16.11\mathrm{kJ}/\mathrm{gr}$