Chemistry - Thermochemistry Question with Solution | TestHub

Heat of combustion : It is the amount of heat evolved or absorbed (i.e. change in enthalpy) when one mole of the substance is completely burnt in air or oxygen.

Standard enthalpy of combustion of glucose reaction is given as
​${\text{C}}_6{\text{H}}_{12}{\text{O}}_{6\left(\text{s}\right)}+6{\text{O}}_{2\left(\text{g}\right)}\to 6{\text{CO}}_{2\left(\text{g}\right)}+6{\text{H}}_2\text{O}\left(\mathit{\text{l}}\right)$
$\Delta {\text{H}}_{\text{C}}^{\text{o}}={\Sigma }_{\text{Product}}\Delta {\text{H}}_{\text{f}}^{\text{o}}-{\Sigma }_{\text{Reactant}}\Delta {\text{H}}_{\text{f}}^{\text{o}}$
           $={\left\{6\left(-400\right)+6\left(-300\right)\right\}}_{\text{Product}}-{\left\{-1300\times 1+6\times 0\right\}}_{\text{Reactant}}$
           $=-2400-1800+1300=-{\text{2900 kJ mole}}^{-1}$
Molecular mass of glucose $=72+12+96=\text{180 g}/\text{mole}$
$\Delta {\text{H}}_{\text{C}}^{\text{o}}/\text{gram of glucose}=\frac{-2900}{180}=-\text{16.11 kJ}/\text{g}$.