A) \[+2900 KJ\]
B) \[-2900 kJ\]
C) \[-16.11kJ\]
D) \[+16.11kJ\]
Correct Answer: C
Solution :
The standard enthalpy of the combustion of glucose can be calculated by the eqn. |
\[{{\operatorname{C}}_{6}}{{H}_{12}}{{O}_{6}}\left( s \right)+6{{O}_{2}}\left( g \right)\to 6C{{O}_{2}}\left( g \right)+6{{H}_{2}}O\left( 1 \right)\] |
\[\Delta {{\operatorname{H}}_{C}}=6\times \Delta {{\operatorname{H}}_{t}}\left( C{{O}_{2}} \right)+6\times \Delta {{\operatorname{H}}_{f}}\left( {{H}_{2}}O \right)-\]\[\Delta {{\operatorname{H}}_{f}}\left[ {{\operatorname{C}}_{6}}{{H}_{12}}{{O}_{6}} \right]\] |
\[\Delta \operatorname{H}{}^\circ =6\left( -400 \right)+6-\left( -300 \right)-\left( -1300 \right)\] |
\[\Delta \operatorname{H}{}^\circ =-2900kJ/mol\] |
For one gram of glucose, enthalpy of combustion will be, |
\[\Delta \operatorname{H}{}^\circ =-\frac{2900}{180}=-16.11kJ/g\] |
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