• # question_answer The standard enthalpies of formation of $C{{O}_{2}}(g)$ ${{H}_{2}}O(l)$ and glucose(s) at $25{}^\circ C$ are $-\,400kJ/mol,$$-\,300kJ/mol$ and $-\,1300kJ/mol,$ respectively. The standard enthalpy of combustion per gram of glucose at $25{}^\circ C$is A) $+\,2900\,kJ$ B) $-\,2900\,kJ$ C) $-16.11\,kJ$ D) $+16.11\,kJ$

 Given $H\underset{C{{O}_{2}}}{\mathop{{}^\circ }}\,={{\Delta }_{f}}H{}^\circ \left( C{{O}_{2}} \right)=-400kJmo{{l}^{-1}}$ $H\underset{{{H}_{2}}O}{\mathop{{}^\circ }}\,={{\Delta }_{f}}H{}^\circ \left( {{H}_{2}}O \right)=-300kJmo{{l}^{-1}}$ $H\underset{glu\cos e}{\mathop{{}^\circ }}\,={{\Delta }_{f}}H{}^\circ \left( glu\cos e \right)$ $=-1300\ \,kJ\,\,mo{{l}^{-1}}$ $H\underset{{{O}_{2}}}{\mathop{{}^\circ }}\,={{\Delta }_{f}}H{}^\circ \left( {{O}_{2}} \right)=0.00$
 ${{C}_{6}}{{H}_{12}}{{O}_{6}}\left( s \right)+6{{O}_{2}}\to 6C{{O}_{2}}\left( g \right)+6{{H}_{2}}O\left( l \right)$ ${{\Delta }_{c}}H{}^\circ \left( glu\operatorname{cose} \right)$ $=6\left[ {{\Delta }_{f}}H{}^\circ \left( C{{O}_{2}} \right)+{{\Delta }_{f}}H{}^\circ \left( {{H}_{2}}O \right) \right]$ $-\left[ {{\Delta }_{f}}H{}^\circ \left( {{C}_{6}}{{H}_{12}}{{O}_{6}} \right)+6{{\Delta }_{f}}H{}^\circ \left( {{O}_{2}} \right) \right]$ $=6\left[ -400-300 \right]-\left[ -1300+6\times 0 \right]$ $=-2900kJmo{{l}^{-1}}$ Molar mass of ${{C}_{6}}{{H}_{12}}{{O}_{6}}=180gmo{{l}^{-1}}$ Thus, standard heat of combustion of glucose per gram $=\frac{-2900}{180}=-16.11kJ{{g}^{-1}}$