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KVPY Sample Paper KVPY Stream-SX Model Paper-20

  • question_answer
    The standard enthalpies of formation of \[{{\operatorname{CO}}_{2}}\](g), \[{{\operatorname{H}}_{2}}O\left( l \right)\] and glucose(s) at \[25{}^\circ C\] are \[-400 kJ/mol,\] \[-300 kJ/mol\] and\[-1300 kJ/mol,\]respectively. The standard enthalpy of combustion per gram of glucose at \[25{}^\circ C\] is

    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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