The standard enthalpies of formation of CO₂(g), H₂O(l) and glucose (s) at 25°C are –400 kJ/mole, –300 kJ/mole, –1300 kJ/mole respectively. The standard enthalpy of combustion per gram of glucose at 25°C is:

C₆H₁₂O₆(s) + 6O₂(g) → $6C{O}_2$ (g) + 6H₂O(l)
${\Delta }_r^{0}H=\left(6{\Delta }_f^0{H}_{\left(C{O}_2\right)}+6{\Delta }_f^0{H}_{\left(H_{2}O\right)}\right)-\left({\Delta }_f^0{H}_{\left(C_6{H}_{12}{O}_6\right)}\right)$ 
= 6 x (–400) + 6(–300) – (–1300)
= –2400 – 1800 + 1300 = –2900 kJ/mol
Hence, standard enthalpy of combustion per gram of glucose $=\frac{-2900}{180}kJ/g=-16.11kJ/g$