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JEE Main & Advanced Chemistry Thermodynamics / रासायनिक उष्मागतिकी Sample Paper Topic Test - Thermo Chemistry (Thermodynamics)

  • question_answer
    Use the bond enthalpies in the table to determine \[\Delta H{}^\circ \] for the formation of hydrazine, \[{{N}_{2}}{{H}_{4}}(g)\] from nitrogen and hydrogen in standard state according to the equation :  \[{{N}_{2}}(g)+2{{H}_{2}}(g)\xrightarrow{{}}{{N}_{2}}{{H}_{4}}(g)\]
    Bond Bond enthalpies
    \[N-N\] \[159\text{ }kJ\text{ }mo{{l}^{-1}}\]
    \[N=N\] \[418\text{ }kJ\text{ }mo{{l}^{-1}}\]
    \[N\equiv N\] \[941\text{ }kJ\text{ }mo{{l}^{-1}}\]
    \[H-H\] \[436\text{ }kJ\text{ }mo{{l}^{-1}}\]
    \[H-N\] \[389\text{ }kJ\text{ }mo{{l}^{-1}}\]

    A) \[\Delta H{}^\circ =-425\text{ }kJ\]

    B) \[\Delta H{}^\circ =-\text{ }98\text{ }kJ\]

    C) \[\Delta H{}^\circ =+\text{ }98\text{ }kJ\]

    D) \[\Delta H{}^\circ =+\text{ }711\,kJ\]

    Correct Answer: C

    Solution :

    [c] \[\Delta H{{{}^\circ }_{rxn}}\,\,=\Delta H{{{}^\circ }_{f}}\,\,[{{N}_{2}}{{H}_{4}}]=[(941+2\times 436)\]\[-(159+4\times 389)]\] \[\Delta H{}^\circ =+98\,kJ\]


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