A) \[+\,2900\,kJ\]
B) \[-\,2900\,kJ\]
C) \[-16.11\,kJ\]
D) \[+16.11\,kJ\]
Correct Answer: C
Solution :
| 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 |
| \[=\frac{-2900}{180}=-16.11kJ{{g}^{-1}}\] |
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